Aptitude & reasoning – Meghalaya PCS Free Notes

Height and Distance

June 4, 2017October 26, 2024 by PSC Notes

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

Trigonometric Basics

sinθ=oppositeside/hypotenuse=y/r

cosθ=adjacentside/hypotenuse=x/r

tanθ=oppositeside/adjacentside=y/x

cosecθ=hypotenuse/oppositeside=r/y

secθ=hypotenuse/adjacentside=r/x

cotθ=adjacentside/oppositeside=x/y

From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above

Basic Trigonometric Values

θ in degrees	θ in radians	sinθ	cosθ	tanθ
0°	0	0	1	0
30°	π/6	1/2	3/√2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	3/√2	1/2	√3
90°	π/2	1	0	Not defined

Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

Trigonometry – Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

Trigonometry – Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

Trigonometry – Pythagorean Formulas

sin2θ+cos2θ=1

sec2θ−tan2θ=1

cosec2θ−cot2θ=1

Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

i.e., angle of elevation = AOP

Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight

i.e., angle of depression = AOP

Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

Solved Examples

Level 1

1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m	B. 6.2 m
C. 12.4 m	D. 24.8 m

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m

2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m	B. 400 m
C. 312 m	D. 298 m

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m

3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these	B. 60°
C. 45°	D. 30°

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°

4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these	B. 12 m
C. 14 m	D. 10 m

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m

5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m	B. 138.4 m
C. 46.24 m	D. 160 m

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30° (Because DA || BC)

tan 30°=AC/BC=>tan 30°=80/BC=>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
A. 30°	B. 60°
C. 45°	D. None of these

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation, RPQ = θ

From the right PQR,

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/( √3×√3)=3√3/3=√3

θ=tan−1(3√)=60°

Level 2

1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second	B. 10 min 57 second
C. 14 min 34 second	D. 12 min 23 second

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

=>1=h/x=>h=x——(1)

tan 30°=AB/BD=AB/(BC + CD)=h/(x+y)

=>1/√3=h/(x+y)

=>x + y = √3h

=>y = √3h – x

=>y = √3h−h(∵ Substituted the value of x from equation 1 )

=>y = h(√3−1)

Given that distance y is covered in 8 minutes
i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8—(A)

h∝t—(B)

(A)/(B)=>h(√3−1)/h=8/t

=>(√3−1)=8/t

=>t=8/(√3−1)=8/(1.73−1)=8/.73=800/73minutes ≈10 minutes 57 seconds

2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres	B. 8 metres
C. 10 metres	D. 12 metres

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE=>√3=15/CE=>CE = 15√3— (1)

tan 30°=AB/BD=>1/√3=(15−h)/BD

=>1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

=>(15−h)=(1/√3)×(15/√3)=15/3=5

=>h=15−5=10 m

i.e., height of the electric pole = 10 m

3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m	B. 173 m
C. 273 m	D. 200 m

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, BAD = 30° , BCD = 45°

tan 30° = BD/BA⇒1/√3=100/BA

⇒BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒BC=100

Distance between the two ships = AC = BA + BC
=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273 m

4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m	B. 50 m
C. 66.67 m	D. 33.33 m

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m

XAD = ADB = 30° (∵ AX || BD )
XAE = AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

=>√3=100/CE=>CE = 100/√3— (1)

tan 30°=AB/BD=>1/√3=(100−h)/BD

=>1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1 )

=>(100−h)=1/√3×100/√3=100/3=33.33=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
A. 9 m	B. 10.40 m
C. 15.57 m	D. 12 m

Answer : Option A

Explanation :

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, ABC = 60°, DBC = 30°

Let DC be h.

tan 30°=DC/BC

1/√3=h/BC

h=BC√3—— (1)

tan 60°=AC/BC

√3=(18+h)/BC

18+h=BC×√3—— (2)

(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3

=>3h=18+h=>2h=18=>h=9 m

i.e., the height of the tower = 9 m

6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
A. 0.63 meter/sec	B. 2.16 meter/sec
C. 3.87 meter/sec	D. 0.72 meter/sec

Answer : Option B

Explanation :

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

tan 60°=BA/CA

√3=BA/150

BA=150√3

i.e, the distance travelled by the balloon = 150√3meters

time taken = 2 min = 2 × 60 = 120 seconds

Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
A. 22 m	B. 44 m
C. 33 m	D. None of these

Answer : Option B

Explanation :

Let DC be the wall, AB be the tree.

Given that DBC = 30°, DAE = 60°, DC = 11 m

tan 30°=DC/BC

1/√3=11/BC

BC = 11√3 m

AE = BC =11√3 m—— (1)

tan 60°=ED/AE

√3=ED/11√3[∵ Substituted the value of AE from (1)]

ED =11√3×√3=11×3=33

Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
A. 141 m and 282 m	B. 70.5 m and 141 m
C. 65 m and 130 m	D. 130 m and 260 m

Answer : Option B

Explanation :

Let AB and CD be the poles with heights h and 2h respectively

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD.

Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right ABE,
tanθ=AB/BE and tanθ=h/100

h = 100tanθ—— (1)

From the right EDC,

tan(90−θ)=CD/ED

cotθ=2h/100[∵tan(90−θ)=cotθ]

2h =100cotθ—— (2)

(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]

=>√2h=100

=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5

2h=2×70.5=141

i.e., the height of the poles are 70.5 m and 141 m.

9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
A. 8.65 m	B. 2 m
C. 2.5 m	D. 3.65 m

Answer : Option D

Explanation :

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

From the diagram, AF = BE = 5 m

From the right AFD, tan45°=DF/AF

1=DF/5

DF = 5—— (1)From the right AFC, tan60°=CF/AF

√3=CF/5

CF=5√3—— (2)

Length of the window = CD = (CF – DF)

=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m

10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?
A. 1.2 km	B. 0.6 km
C. 1.4 km	D. 2.7 km

Answer : Option D

Explanation :

Let A be the foot and C be the summit of a mountain.

Given that CAB = 45°

From the diagram, CB is the height of the mountain. Let CB = x

Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60°

i.e., CDE = 60°

From the right ABC,

tan45°=CB/AB

=>1=x/AB[∵ CB = x (the height of the mountain)]

=>AB = x—— (eq:1)

From the right AYD,

sin30°=DY/AD

=>1/2=DY/2(∵ Given that AD = 2)

=> DY=1—— (eq:2)

cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)

From the right CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]

=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]

=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]

=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7

i.e., the height of the mountain = 2.7 km

AGE PROBLEMS

June 4, 2017October 25, 2024 by PSC Notes

Age Problems

Important Formulas on “Problems on Ages”:

If the current age is x, then ntimes the age is nx.
If the current age is x, then age nyears later/hence = x+ n.
If the current age is x, then age nyears ago = x– n.
The ages in a ratio a: bwill be ax and bx.

5. If the current age is x, then	1	of the age is	x	.
	n		n

Example:

A problem with one variable: How old is Al?

Many single-variable algebra word problems have to do with the relations between different people’s ages. For example:

Al’s father is 45. He is 15 years older than twice Al’s age. How old is Al?

We can begin by assigning a variable to what we’re asked to find. Here this is Al’s age, so let Al’s age = x.

We also know from the information given in the problem that 45 is 15 more than twice Al’s age. How can we translate this from words into mathematical symbols? What is twice Al’s age?

Well, Al’s age is x, so twice Al’s age is 2x, and 15 more than twice Al’s age is 15 + 2x.That equals 45, right? Now we have an equation in terms of one variable that we can solve for x: 45 = 15 + 2x.

original statement of the problem:	45 = 15 + 2x
subtract 15 from each side:	30 = 2x
divide both sides by 2:	15 = x

Since x is Al’s age and x = 15, this means that Al is 15 years old.

It’s always a good idea to check our answer:

twice Al’s age is 2 x 15:	30
15 more than 30 is 15 + 30:	45

This should be the age of Al’s father, and it is.

Questions:

Level-I:

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?

2 times

2	1	times
	2

2	3	times
	4

3 times

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

A.	4 years
B.	8 years
C.	10 years
D.	None of these

A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:

A.	14 years
B.	19 years
C.	33 years
D.	38 years

A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?

A.	7
B.	8
C.	9
D.	10
E.	11

Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?

A.	24
B.	27
C.	40
D.	Cannot be determined
E.	None of these

A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

A.	14 years
B.	18 years
C.	20 years
D.	22 years

Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?

A.	16 years
B.	18 years
C.	20 years
D.	Cannot be determined
E.	None of these

The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:

A.	12 years
B.	14 years
C.	18 years
D.	20 years

At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?

A.	12 years
B.	15 years
C.	19 and half
D.	21 years

10.

Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?

A.	16 years
B.	18 years
C.	28 years
D.	24.5 years
E.	None of these

11.

Level-II:

The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

A.	8, 20, 28
B.	16, 28, 36
C.	20, 35, 45
D.	None of these

12.

Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?

A.	2 years
B.	4 years
C.	6 years
D.	8 years

13.

A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?

A.	32 years
B.	36 years
C.	40 years
D.	48 years

14.

Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?

A.	1 year
B.	2 years
C.	25 years
D.	Data inadequate
E.	None of these

15.

The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:

A.	5 : 2
B.	7 : 3
C.	9 : 2
D.	13 : 4

16.

What is Sonia’s present age?
I.	Sonia’s present age is five times Deepak’s present age.
II.	Five years ago her age was twenty-five times Deepak’s age at that time.

A.	I alone sufficient while II alone not sufficient to answer
B.	II alone sufficient while I alone not sufficient to answer
C.	Either I or II alone sufficient to answer
D.	Both I and II are not sufficient to answer
E.	Both I and II are necessary to answer

17.

Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year?
I.	Retirement age is 60 years.
II.	There are 50 employees in the department.

A.	I alone sufficient while II alone not sufficient to answer
B.	II alone sufficient while I alone not sufficient to answer
C.	Either I or II alone sufficient to answer
D.	Both I and II are not sufficient to answer
E.	Both I and II are necessary to answer

18.

Divya is twice as old as Shruti. What is the difference in their ages?
I.	Five years hence, the ratio of their ages would be 9 : 5.
II.	Ten years back, the ratio of their ages was 3 : 1.

A.	I alone sufficient while II alone not sufficient to answer
B.	II alone sufficient while I alone not sufficient to answer
C.	Either I or II alone sufficient to answer
D.	Both I and II are not sufficient to answer
E.	Both I and II are necessary to answer

Answers:

Level-I:

Answer:1 Option A

Explanation:

Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24

Answer:2 Option A

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

Answer:3 Option A

Explanation:

Let the son’s present age be x years. Then, (38 – x) = x

2x = 38.

x = 19.

Son’s age 5 years back (19 – 5) = 14 years.

Answer:4 Option D

Explanation:

Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B’s age = 2x = 10 years.

Answer:5 Option A

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

Answer:6 Option D

Explanation:

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

Answer:7 Option A

Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

Answer:8 Option D

Explanation:

Let the present ages of son and father be x and (60 –x) years respectively.

Then, (60 – x) – 6 = 5(x – 6)

54 – x = 5x – 30

6x = 84

x = 14.

Son’s age after 6 years = (x+ 6) = 20 years..

Answer:9 Option B

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26 4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

Answer:10 Option D

Explanation:

Let Rahul’s age be x years.

Then, Sachin’s age = (x – 7) years.

	x – 7	=	7
	x		9

9x – 63 = 7x

2x = 63

x = 31.5

Hence, Sachin’s age =(x – 7) = 24.5 years.

Answer:11 Option B

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

Answer:12 Option C

Explanation:

Mother’s age when Ayesha’s brother was born = 36 years.

Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.

Required difference = (42 – 36) years = 6 years.

Answer:13 Option C

Explanation:

Let the mother’s present age be x years.

Then, the person’s present age =		2	x		years.
		5

		2	x + 8		=	1	(x + 8)
		5				2

2(2x + 40) = 5(x + 8)

x = 40.

Answer:14 Option D

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R – Q = ?.

Explanation:

R – Q = Q – T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R – Q) = ?

Here we know the value(age) of Q (25), but we don’t know the age of R.

Therefore, (R-Q) cannot be determined.

Answer:15 Option B

Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Answer:16 Option E

Explanation:

I. S = 5D D =	S	….(i)
	5

S – 5 = 25 (D – 5) S = 25D – 120 ….(ii)

Using (i) in (ii), we get S =		25 x	S		– 120
			5

4S = 120.

S = 30.

Thus, I and II both together give the answer. So, correct answer is (E).

Answer:17 Option E

Explanation:

Retirement age is 60 years.
There are 50 employees in the department.

Average age of 50 employees = 30 years.

Total age of 50 employees = (50 x 30) years = 1500 years.

Number of employees next year = 40.

Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.

Average age next year =	940	years = 23	1	years.
	40		2

Thus, I and II together give the answer. So, correct answer is (E).

Answer:18 Option C

Explanation:

Let Divya’s present age be D years and Shruti’s present age b S years

Then, D = 2 x S D – 2S = 0 ….(i)

I.	D + 5	=	9	….(ii)
	S + 5		5

II.	D – 10	=	3	….(iii)
	S – 10		1

From (ii), we get : 5D + 25 = 9S + 45 5D – 9S = 20 ….(iv)

From (iii), we get : D – 10 = 3S – 30 D – 3S = -20 ….(v)

Thus, from (i) and (ii), we get the answer.

Also, from (i) and (iii), we get the answer.

I alone as well as II alone give the answer. Hence, the correct answer is (C).

CLASSIFICATION LEVEL 1

June 4, 2017October 24, 2024 by PSC Notes

Classification involves putting things into a class or group according to particular characteristics so it’s easier to make sense of them, whether you’re organizing your shoes, your stock portfolio, or a group of invertebrates. From all competitive examination classification is one of the most important topics, this pattern come with lot of questions minimum they asking the 4 to 5 question from the classification topic. In the SSC CGL or SSC constable GD examination having the same topics from the reasoning section but the standard of the topic will be different, so most of the candidates preference for this topic to get the best score in the written examination.

Directions: Find the odd one out

A. Square B. Circle C. Rectangle D. Triangle

A. Cotton B. Terene C. Silk D. Wool

A. Light B. Wave C. Heat D. Sound

A. 81 : 243 B. 16 :64 C. 64 : 192 D. 25 : 75

A. 64 : 8 B. 80 : 9 C. 7 : 49 D. 36 : 6

A. 26 : 62 B. 36 : 63 C. 46 : 64 D. 56 : 18

A. ABZY B. BCYX C. CDVW D. DEVU

A. ACE B. FHJ C. KLM D. SUW

Find the wrong number in the series

441, 484, 529, 566, 625

484 B. 529 C. 625 D. 566

Find the wrong number in the series

232, 343, 454, 564, 676

676 B. 454 C. 343 D. 564

SOLUTION TO CLASSIFICATION LEVEL 1

B. Except circle, all others are geometrical figures consisting straight lines.

B. Except terene, all others are natural fibres.

B. Except wave, all others are different form of energy.

B. 81*3=243

64*3=192

25*3=75

But 16*4=64

D. Except D, in each pair one number is square root of the other.

D. Except D, in each pair the position of digits has been interchanged.

C. A+1=B & Z-1=Y

B+1=C & Y-1=X

D+1=E & V-1=U

But C+1=D & V+1=W

C. A+2=C & C+2=E

F+2=H & H+2=J

But K+1=L & L+1=M

D. 21^2=441

22^2=484

23^2=529

25^2=625

But (23.79)^2=566

D. 232+111=343

343+111=454

454+111=565 (but given 564)

DISCOUNT

June 4, 2017October 22, 2024 by PSC Notes

Discount

The discount is referred to the reduction in the price of some commodity or service. It may anywhere appear in the distribution channel in the form of modifications in marked price (printed on the item) or in retail price (set by retailer usually by pasting a sticker on the item) or in list price (quoted for the buyer). The discount is provided for the purpose of increasing sales, to clear out old stock, to encourage distributors, to reward potential customer etc. In short, the discount can serve as a way to attract customers for a particular item or service.

In math, discount is one of the easiest way to raise the customers of particular product. Discounts are a significant element of your online merchandising plan. You build discounts so that you can force sales on items or collection of products to your customers who convene particular conditions. In math, the discount problems can be solved by using discount formula.

The “discount rate” means the interest rate. Discount rate is based on the simple interest rate. To calculate simple interest rate, just find out the interest rate for one period (multiply by amount, interest rate, period) but calculate the discount rate, just multiply by the amount and an interest rate. This is called the define discount rate.

To calculate the discount rate, just multiply the amount by an interest rate. By using the Formula Discount rate DR = pr (p = principal amount, r = interest rate).

What is Discount Rate?

Discount rate is one of the simple ways to increase the customers of particular product. Discounts are a important element of your online merchandising strategy. You make discounts so that you can force sales on products or collection of products to your customers who meet certain particular conditions.

The formula used to calculate the discount is discount = marked price – selling price.

Here,

Selling price is what you actually pay for the item.

Marked price is the normal price of the item without a discount.

Discount is either a dollar rate or a percentage of the marked cost.

Discount Rate Definition

Discount Rate is the cost of the total amount generally less than its original value is called . In other words, a total bill will generally sell at a discount, and the discount rate is annualized percentage of this discount, that is percentage is adjusted to give an annual percentage.

Discount Rate Formula

Formula of the Discount Rate is:

Discount rate DR = pr

where,

p = principal amount
r = interest rate

Questions:

Level-I

1: Ricky purchase the dress. That dress rate was Rs1000 at 10% discount . Find discount rate? And then ricky how many dollars give to cashier?

2: Kalvin purchased land for 50000 dollars at 20% in 2000th year. Then 2004th year that land sales 3000 dollars. How many dollars he loss?

The marked price of a ceiling fan is $ 1250 and the shopkeeper allows a discount of 6% on it. Find the selling price of the fan.
A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain percent?
A dealer purchased a washing machine for $ 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.
How much per cent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, he gains 20%?
Find the single discount equivalent to two successive discounts of 20% and 10%.
A merchant who marked his goods up by 50% subsequently offered a discount of 20% on the marked price. What is the percentage profit that the merchant make after offering the discount?

Applied to a bill for Rs. 1,00,000 the difference between a discount of 40% and two successive discounts of 36% and 4% is:
On a 20% discount sale, an article costs Rs. 596. What was the original price of the article?

Level-II:

A discount of 15% on one article is the same as discount of 20% on a second article. The costs of the
A discount of 2 ½% is given to the customer on marked price of an article. A man bought the article for Rs. 39. The marked price of article is:
Printed price of an article is Rs. 900 but the retailer gets a discount of 40%. He sells the article for Rs. 900. Retailer’s gain percent is:
The marked price of a watch was Rs. 720. A man bought the same watch for Rs. 550.80, after getting two successive discounts. If the first discount was 10%, what was the second discount rate?
A shopkeeper marks his goods 20% above cost price, but allows 30% discount for cash. His net loss is:
A retailer buys 40 pens at the marked price of 36 pens from a wholesaler. If he sells these pens giving a discount of 1%, what is the profit percent?
A pizzeria has a coupon that reads, “Getoff a $9.00 cheese pizza.” What is the discount? What is the sale price of the cheese pizza?

18.In a video store, a DVD that sells for $15 is marked, “10% off.” What is the sale price of the DVD?

Answers:

Level-I:

Solution:1
Here,

Principal amount p = 1000 rs

Interest rate r = 10%

Discount rate DR = pr

DR = 1000*

= 100

The discount amount for the dress is 100.

Discount rate DR = 100.

Dress rate = principal amount – discount rate

= 1000 – 100

=900

Ricky gives 900 rs to cashier

Solution:2
Principal amount p = 50000 dollars

Interest rate r = 20%

Discount rate DR = pr

DR = 50000 x 2010020100 in 2000th year

= 10000

Discount rate DR = 1000 dollars in 2000th year.

The discount amount is 10000 dollars.

Discount rate DR = 50000*30/100 in 2004th year

Discount rate =15000 dollars.

The discount amount is 15000 dollars.

Loss Discount rate in 2004th year – Discount rate in 2000th year

=15000 dollars – 10000 dollars

=5000 dollars

Kalvin 5000 dollars losses in that land.

Solution:3

Marked price = $ 1250 and discount = 6%.

Discount = 6% of Marked Price

= (6% of $ 1250)

= $ {1250 × (6/100)}

= $ 75

Selling price = (Marked Price) – (discount)

= $ (1250 – 75)

= $ 1175.

Hence, the selling price of the fan is $ 1175.

Solution:4

Let the cost price be $ 100.

Then, marked price = $ 140.

Discount = 25% of Marked Price

= (25% of $ 140)

= $ {140 × (25/100)

= $ 35.

Selling price = (marked price) – (discount)

= $ (140 – 35)

= $ 105.

Gain% = (105 – 100) % = 5%.

Hence, the trader gains 5%.

Solution:5

Cost price of the machine = $ 7660, Gain% = 10%.

Therefore, selling price = [{(100 + gain%)/100} × CP]

= $ [{(100 + 10)/100} × 7660]

= $ [(110/100) × 7660]

= $ 8426.

Let the marked price be $ x.

Then, the discount = 12% of $x

= $ {x × (12/100)}

= $ 3x/25

Therefore, SP = (Marked Price) – (discount)

= $ (x – 3x/25)

= $ 22x/25.

But, the SP = $ 8426.

Therefore, 22x/25 = 8426

⇒ x = (8426 × 25/22)

⇒ x = 9575.

Hence, the marked price of the washing machine is $ 9575

Solution:6

Let the cost price be $ 100.

Gain required = 20%.

Therefore, selling price = $ 120.

Let the marked price be $x.

Then, discount = 25% of $x

= $ (x × 25/100)

= $ x/4

Therefore, selling price = (Marked Price) – (discount)

= $ {x – (x/4)

= $ 3x/4

Therefore, 3x/4 = 120

⇔ x = {120 × (4/3)} = 160

Therefore, marked price = $ 160.

Hence, the marked price is 60% above cost price.

Solution:7

Let the marked price of an article be $ 100.

Then, first discount on it = $ 20.

Price after first discount = $ (100 – 20) = $ 80.

Second discount on it = 10% of $ 80

= $ {80 × (10/100)} = $ 8.

Price after second discount = $ (80 – 8) = $ 72.
Net selling price = $ 72.

Single discount equivalent to given successive discounts = (100 – 72)% = 28%

Solution:8 The easiest way to solve these kinds of problems is to assume a value for the merchant’s cost price.
To make calculations easy, it is best to assume the cost price to be $100.

The merchant marks his goods up by 50%.
Therefore, his marked price (quoted price) = cost price + mark up.
Marked price = $100 + 50% of $100 = 100 + 50 = $150.

The merchant offers a discount of 20% on his marked price.
Discount offered = 20% of 150 = $30.

Therefore, he finally sold his goods for $150 – $30 = $ 120.
We assumed his cost to be $100 and he sold it finally for $120.

Therefore, his profit = $20 on his cost of $ 100.
Hence, his % profit = profit/cost price * 100 = 20/100*100 = 20%.

Solution:9 40% of Rs. 1,00,000 = Rs. 40,000
36% of 1,00,000 = 36000
4% of 36,000 = Rs. 2,560.
Therefore, two successive discounts on Rs. 1,00,000 = 36,000 + 2560 = Rs. 38,560.
Difference between a discount of 40% and two successive discounts of 36% and 4%
= 40,000 – 38,560
= Rs. 1,440

Solution:10 If the selling price of the article is S, then
S – 20% of S = 596
S – S/5 = 596
4S/5 = 596
⇒ S = 596 x 5/4
⇒ S = 745

Level-II

Solution:11Let the prices of two articles be X and Y
From the question 15X/100 = 20Y/100
X/Y = 20/15
Thus the ratio of prices of two articles is 4 : 3
Any two amounts in the ratio 4 : 3 will satisfy the condition.
In the above instance, Rs. 80 and Rs. 60 is the answer.

Solution:12 Formula for Marked Price = 100 x SP/(100 – d%) = 100 x 39/(100 – 2.5%)
= 3900 / 97.5
= Rs. 40.
Marked Price of Article is Rs. 40.

Solution:13 Retailer gets a discount of 40% means he buys it at 60% of the price
60% x 900 = Rs. 540
Profit on selling it at Rs. 900 = 900 – 540 = Rs. 360.
Profit % = (Profit / C.P) x 100 = (360 / 540) x 100 = 66²/₃
Retailer’s Gain percent is 66²/₃

Solution:1410% discount on 720 = Rs. 72
Cost after 1st discount = 720 – 72 = Rs. 648.
Cost after 2nd discount = Rs. 550.80
Therefore 2nd discount = 648 – 550.80 = Rs. 97.20
Discount % = (97.2 x 100)/648 = 15%
Second discount rate = 15%.

Solution:15 Let the cost price be Rs. 100.
M.P. (which is 20% above C.P.) = Rs. 120.
30% discount on Rs. 120 = Rs. 36.
Selling Price = Rs. 120 – 36 = Rs. 84
Cost Price = Rs 100 and Selling Price = Rs 84 {since CP > SP, it is a loss}
Loss% = (16/100) x 100 = 16%.
His net loss percent is 16%.

Solution:16 Assuming the M.P. of each pen to be Rs. 10, the M.P. of 36 pens = Rs. 360
Cost price of 40 pens = Rs. 360 (from the question)
Cost price of each pen = 360/40 = Rs. 9
Selling Price of each pen at a discount of 1% on a marked price of Rs. 10 = 99% x 10 = Rs. 9.90
Profit = 9.90 – 9.00 = Rs. 0.90
Profit % = (0.90/9.00) x 100 = 10%
Profit % = 10%.

Solution:17 The discount is $3.00 and the sale price is $6.00

Solution:18 The rate is 10%. Thus, the customer is paying 90% for the DVD

The sale price is: 0.90 x $15.00 = $13.50

The sale price is $13.50.

SQUARE ROOT & CUBE ROOTS

June 4, 2017October 21, 2024 by PSC Notes

Square Root & Cube Root

Step 1: First of all group the number in pairs of 2 starting from the right.

Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.

Step 3: To get the unit’s place digit of the square root

Remember the following

If number ends in	Unit’s place digit of the square root
1	1 or 9(10-1)
4	2 or 8(10-2)
9	3 or 7(10-3)
6	4or 6(10-4)
5	5
0	0

Lets see the logic behind this for a better understanding

We know,

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

Now, observe the unit’s place digit of all the squares.

Do you find anything common?

We notice that,

Unit’s place digit of both 1² and 9²is 1.

Unit’s place digit of both 2² and 8² is 4

Unit’s place digit of both 3² and 7² is 9

Unit’s place digit of both 4² and 6² is 6.

Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.

Remember,

If first pair of the original number > Result obtained on multiplication then select the greater number out of the two numbers as the unit’s place digit of the square root.

If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.

Let us consider an example to get a better understanding of the method

Example 1: √784=?

Step 1: We start by grouping the numbers in pairs of two from right as follows

7 84

Step 2: To get the ten’s place digit,

We find that nearest square to first group (7) is 4 and √4=2

Therefore ten’s place digit=2

Step 3: To get the unit’s place digit,

We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).

Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,

2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8

Ans:√784=28

Cube roots of perfect cubes

It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.

We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.

Points to remember for speedy calculation of cube roots

To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.

1³	=	1
2³	=	8
3³	=	27
4³	=	64
5³	=	125
6³	=	216
7³	=	343
8³	=	512
9³	=	729
10³	=	1000

From the above cubes of 1 to 10, we need to remember an interesting property.

1³ = 1	=>	If last digit of the perfect cube = 1, last digit of the cube root = 1
2³ = 8	=>	If last digit of the perfect cube = 8, last digit of the cube root = 2
3³ = 27	=>	If last digit of the perfect cube = 7, last digit of the cube root = 3
4³ = 64	=>	If last digit of the perfect cube = 4, last digit of the cube root = 4
5³ = 125	=>	If last digit of the perfect cube =5, last digit of the cube root = 5
6³ = 216	=>	If last digit of the perfect cube = 6, last digit of the cube root = 6
7³ = 343	=>	If last digit of the perfect cube = 3, last digit of the cube root = 7
8³ = 512	=>	If last digit of the perfect cube = 2, last digit of the cube root = 8
9³ = 729	=>	If last digit of the perfect cube = 9, last digit of the cube root = 9
10³ = 1000	=>	If last digit of the perfect cube = 0, last digit of the cube root = 0

It’s very easy to remember the relations given above because

1	->	1	(Same numbers)
8	->	2	(10’s complement of 8 is 2 and 8+2 = 10)
7	->	3	(10’s complement of 7 is 3 and 7+3 = 10)
4	->	4	(Same numbers)
5	->	5	(Same numbers)
6	->	6	(Same numbers)
3	->	7	(10’s complement of 3 is 7 and 3+7 = 10)
2	->	8	(10’s complement of 2 is 8 and 2+8 = 10)
9	->	9	(Same numbers)
0	->	0	(Same numbers)

Also see
8 -> 2 and 2 -> 8
7 -> 3 and 3-> 7

Questions

Level-I

The cube root of .000216 is:

A.	.6
B.	.06
C.	77
D.	87

What should come in place of both x in the equation	x	=	162	.
	128		x

A.	12
B.	14
C.	144
D.	196

The least perfect square, which is divisible by each of 21, 36 and 66 is:

A.	213444
B.	214344
C.	214434
D.	231444

1.5625 = ?

A.	1.05
B.	1.25
C.	1.45
D.	1.55

If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?

A.	13.41
B.	20.46
C.	21.66
D.	22.35

If a = 0.1039, then the value of 4a² – 4a + 1 + 3a is:

A.	0.1039
B.	0.2078
C.	1.1039
D.	2.1039

If x =	3 + 1	and y =	3 – 1	, then the value of (x² + y²) is:
	3 – 1		3 + 1

A.	10
B.	13
C.	14
D.	15

A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

A.	57
B.	67
C.	77
D.	87

The square root of (7 + 35) (7 – 35) is

A.	5
B.	2
C.	4
D.	35

10.

If 5 = 2.236, then the value of	5	–	10	+ 125 is equal to:
	2		5

A.	5.59
B.	7.826
C.	8.944
D.	10.062

Level-II

11.

	625	x	14	x	11		is equal to:
	11		25		196

A.	5
B.	6
C.	8
D.	11

12.

0.0169 x ? = 1.3

A.	10
B.	100
C.	1000
D.	None of these

13.

	3 –	1		2	simplifies to:
		3

None of these

14.

How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ?

A.	1
B.	2
C.	3
D.	None of these

15.

The square root of 64009 is:

A.	253
B.	347
C.	363
D.	803

16. √29929 = ?

A.	173
B.	163
C.	196
D.	186

17. √106.09 = ?
A.	10.6
B.	10.5
C.	10.3
D.	10.2

18. ?/√196 = 5
A.	76
B.	72
C.	70
D.	75

Answers

Level-I

Answer:1 Option B

Explanation:

(.000216)^1/3	=		216		1/3
			10⁶

=		6 x 6 x 6		1/3
		10² x 10² x 10²

=	6
	10²

=	6
	100

= 0.06

Answer:2 Option A

Explanation:

Let	x	=	162
	128		x

Then x² = 128 x 162

= 64 x 2 x 18 x 9

= 8² x 6² x 3²

= 8 x 6 x 3

= 144.

x = 144 = 12.

Answer:3 Option A

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

Answer:4 Option B

Explanation:

1|1.5625( 1.25

|——-

22| 56

| 44

|——-

245| 1225

| 1225

|——-

| X

|——-

1.5625 = 1.25.

Answer:5 Option D

Explanation:

35 + 125 = 17.88

35 + 25 x 5 = 17.88

35 + 55 = 17.88

85 = 17.88

5 = 2.235

80 + 65 = 16 x 5 + 65

= 45 + 65

= 105 = (10 x 2.235) = 22.35

Answer:6 Option C

Explanation:

4a² – 4a + 1 + 3a = (1)² + (2a)² – 2 x 1 x 2a + 3a

= (1 – 2a)² + 3a

= (1 – 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

Answer:7 Option C

Explanation:

x =	(3 + 1)	x	(3 + 1)	=	(3 + 1)²	=	3 + 1 + 23	= 2 + 3.
	(3 – 1)		(3 + 1)		(3 – 1)		2

y =	(3 – 1)	x	(3 – 1)	=	(3 – 1)²	=	3 + 1 – 23	= 2 – 3.
	(3 + 1)		(3 – 1)		(3 – 1)		2

x² + y² = (2 + 3)² + (2 – 3)²

= 2(4 + 3)

= 14

Answer:8 Option C

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77

Answer:9 Option B

Explanation:

(7 + 35)(7 – 35)

(7)² – (35)²

= 49 – 45 = 4 = 2

Answer:10 Option B

Explanation:

5	–	10	+ 125	=	(5)² – 20 + 25 x 55
2		5			25

=	5 – 20 + 50
	25

=	35	x	5
	25		5

=	355
	10

=	7 x 2.236
	2

= 7 x 1.118

= 7.826

Level-II

Answer:11 Option A

Explanation:

Given Expression =	25	x	14	x	11	= 5.
	11		5		14

Answer:12 Option B

Explanation:

Let 0.0169 x x = 1.3.

Then, 0.0169x = (1.3)² = 1.69

x =	1.69	= 100
	0.0169

Answer:13 Option C

Explanation:

	3 –	1		2	= (3)² +		1		2	– 2 x 3 x	1
		3					3				3

= 3 +	1	– 2
	3

= 1 +	1
	3

=	4
	3

Answer:14 Option D

Explanation:

A number ending in 8 can never be a perfect square.

Answer:15 Option A

Explanation:

2 |64009( 253 |4 |———-45 |240 |225 |———-503| 1509 | 1509 |———- | X |———-

64009 = 253.

Answer:16 Option A

Explanation:
√29929 = So, √29929 = 173

Answer:17 Option C

Answer:18 Option C

TIME & DISTANCE

June 4, 2017October 21, 2024 by PSC Notes

In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems.

Important Formulas

Speed=Distance/Time
Distance=Speed×Time
Time=Distance/Speed
To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
x km/hr=(x×5)/18m/s
To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
x m/s=(x×18)/5 km/hr
If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy/(x+y) kmph
Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝1/Time (when distance is constant)
If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a

Solved Examples

Level 1

1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 8.2	B. 4.2
C. 6.1	D. 7.2

Answer : Option D

Explanation :

Distance = 600 meter

time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance/time=600/300=2m/s=(2×18)/5 km/hr=36/5 km/hr=7.2 km/hr

2.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr	B. 14 hr
C. 12 hr	D. 19 hr

Answer : Option A

Explanation :

Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)

distance = 8.5 km

Time = distance/speed=8.5/.5=17 hr.

3.Walking 6/7^th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min	B. 1 hr
C. 2 hr	D. 1 hr 12 min

Answer : Option D

Explanation :

New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=>1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

4.A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
A. 3 km	B. 4 km
C. 5 km	D. 6 km

Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph

Hence, average speed = (2×3×2)/(2+3)=12/5 km/hr .

Total time taken = 5 hours

⇒Distance travelled=(12/5)×5=12 km

⇒Distance between his house and office =12/2=6 km

5.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
A. 80 km	B. 70 km
C. 60 km	D. 50 km

Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance/Time=x/10….. (Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒Speed = Distance/Time=(x+20)/14….. (Equation2)

From Equations 1 and 2,
X/10=(x+20)/14⇒14x=10x+200⇒4x=200⇒x=200/4=50

6.A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?
A. 210 miles	B. 230 miles
C. 250 miles	D. 260 miles

Answer : Option B

Explanation :

Speed1 = 50 miles/hour

Time1 = 2*(1/2) hour=5/2 hour

⇒Distance1 = Speed1 × Time1 = (50×5)/2=25×5=125 miles

⇒Speed2 = 70 miles/hour

Time2 = 1*1/2 hour=3/2 hour

Distance2 = Speed2 × Time2 = 70×3/2=35×3=105 miles

Total Distance = Distance1 + Distance2 =125+105=230 miles

7.Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, ¹¹/₅ seconds after he sees it strike the tree. How far is the man from the wood chopper?
A. 1800 ft	B. 2810 ft
C. 3020 ft	D. 2420 ft

Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s ⇒Time = ¹¹/₅ second

Distance = Speed × Time = 1100 ×11/5=220×11=2420 ft

8.A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in meters)?
A. 1250	B. 1280
C. 1320	D. 1340

Answer : Option A

Explanation :

Speed = 5 km/hr

Time = 15 minutes = 1/4 hour

Length of the bridge = Distance Travelled by the man

= Speed × Time = 5×1/4 km

=5×1/4×1000 metre=1250 metre

Level 2

1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs	B. 8 hrs 45 min
C. 7 hrs 45 min	D. 9 hts 20 min

Answer : Option C

Explanation :

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. 2.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 121 km	B. 242 km
C. 224 km	D. 112 km

Answer : Option C

Explanation :

distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 – x) hr
Distance covered in the first half = 21x
Distance covered in the second half = 24(10 – x)
But distance covered in the first half = Distance covered in the second half
=> 21x = 24(10 – x) => 21x = 240 – 24x => 45x = 240 => 9x = 48 => 3x = 16⇒x=16/3

Hence Distance covered in the first half = 21x=21×16/3=7×16=112 km. Total distance = 2×112=224 km

3.A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
A. 30 km/hr	B. 35 km/hr
C. 25 km/hr	D. 40 km/hr

Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40/60hr+48/3600hr=1+2/3+1/75=126/75hr

Distance = 42 kmSpeed=distance/time=42(126/75) =42×75/126

⇒5/7 of the actual speed = 42×75/126

⇒actual speed = 42×75/126×7/5=42×15/18=7×15/3=7×5=35 km/hr

4.A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
A. 36	B. 38
C. 40	D. 42

Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph

Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes

⇒x/v−x/(v+3)=40/60⇒x[1/v−1/(v+3)]=2/3⇒x[(v+3−v)/v(v+3)]=2/3

⇒2v(v+3)=9x…………….(Equation1)

Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
⇒x/(v−2)−x/v=40/60⇒x[1/(v−2)−1/v]=2/3

⇒x[(v−v+2)/v(v−2)]=2/3⇒x[2/v(v−2)]=2/3

⇒x[1/v(v−2)]=1/3⇒v(v−2)=3x…………….(Equation2)

Equation1/Equation2

⇒2(v+3)/(v−2)=3⇒2v+6=3v−6⇒v=12

Substituting this value of v inEquation1⇒2×12×15=9x

=>x= (2×12×15)/9= (2×4×15)/3=2×4×5=40. Hence distance = 40 km

5.In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?
A. 8 kmph	B. 5 kmph
C. 4 kmph	D. 7 kmph

Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km

We know that distance/speed=time. Hence, 30/x−30/y=2………..(Equation1)

30/y−30/2x=1………..(Equation2)

Equation1 + Equation2⇒30/x−30/2x=3⇒30/2x=3⇒15/x=3⇒5/x=1⇒x=5. Hence Arun’s speed = 5 kmph

6.A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
A. 70.24 km/hr	B. 74. 24 km/hr
C. 71.11 km/hr	D. 72.21 km/hr

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph.

By using the same formula, we can find out the average speed quickly average speed = (2×64×80)/(64+80)=(2×64×80)/144⇒ (2×32×40)/36= (2×32×10)/9⇒ (64×10)/9=71.11 kmph

7.A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?
A. 11.2 kmph	B. 10 kmph
C. 10.2 kmph	D. 10.8 kmph

Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km

Time taken to travel 10 km at an average speed of 12 km/hr = distance/speed=10/12 hr

Time taken to travel 12 km at an average speed of 10 km/hr = distance/speed=12/10 hr

Total time taken =10/12+12/10 hr

Average speed = distance/time=22/(10/12+12/10)=(22×120)/{(10×10)+(12×12)}

(22×120)/244=(11×120)/122=(11×60)/61=660/61≈10.8 kmph

8.An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 123 hours, it must travel at a speed of:
A. 660 km/hr	B. 680 km/hr
C. 700 km/hr	D. 720 km/hr

Answer : Option D

Explanation :

Speed and time are inversely proportional ⇒Speed ∝ 1/Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional

Speed ∝ 1/Time⇒Speed1/Speed2=Time2/Time1

⇒240/Speed2=(1*2/3)5⇒240/Speed2=(5/3)/5⇒240/Speed2=1/3⇒Speed2=240×3=720 km/hr

9.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?
A. 80 kmph	B. 102 kmph
C. 120 kmph	D. 140 kmph

Answer : Option C

Explanation :

Let speed of the car = x kmph

Then speed of the train = x *(100+50)/100=150 x /100=3 x /2 kmph

Time taken by the car to travel from A to B=75/x hours

Time taken by the train to travel from A to B=75/(3 x /2)+12.5/60 hours

Since both start from A at the same time and reach point B at the same time

75/x=75/(3 x /2)+12.5/60⇒25/x=12.5/60⇒x=(25×60)/12.5=2×60=120

TIME AND WORK

In these problems the number of persons, quantity of work done and time taken are important factors. Also time taken by a person depends on the efficiency of that person which comes into picture when different people do the work such as women, children do the work alongside the men. The problems related to time and work can be solved by two major approaches – ratio & proportions and unitary method. Let us proceed to find some formulae related to these questions.

Important Formulas – Time and Work

If A can do a piece of work in n days, work done by A in 1 day = 1/n

If A does 1/n work in a day, A can finish the work in n days

If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2

If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days

If A is thrice as good as B in work, then

Ratio of work done by A and B = 3:1

Ratio of time taken to finish a work by A and B = 1: 3

SOLVED EXAMPLES

Level 1

1.P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15	B. 7/15
C. 11/15	D. 2/11

Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

2.A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in — hours.
A. 12 hours	B. 6 hours
C. 8 hours	D. 10 hours

Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12

Work done by B in 1 hour = 7/12 – 1/2 = 1/12

=> B alone can complete the work in 12 hours

3.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 ½ days	B. 22 days
C. 31 days	D. 22 days

Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 ½ days

4.P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. How many days does P alone need to finish the remaining work?
A. 8	B. 5
C. 4	D. 6

Answer : Option D

Explanation :

Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

5.Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes	B. 7 hour 30 minutes
C. 8 hour 15 minutes	D. 8 hour 30 minutes

Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3

Pages typed by Suresh in 1 hour = 40/5 = 8

Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3

Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4

= 8 ¼ hours = 8 hour 15 minutes

6.P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in — days
A. 1	B. 2
C. 3	D. 4

Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12

Work done by P in 1 day = 2 × (1/12) = 1/6

Work done by P and Q in 1 day = 1/12 + 1/6 = ¼

=> P and Q can finish the work in 4 days

7.A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3	B. 4:3
C. 2:3	D. 2:1

Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16

Work done by 1 woman in 1 day = 1/(16×20)

Work done by 16 men in 1 day = 1/15

Work done by 1 man in 1 day = 1/(15×16)

8.P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?

A. Rs.40

B. Rs.70

C. Rs.90

D. Rs.100

Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)

Amount Earned by P and R in 1 day = 600/5 = 120 —(2)

Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)

(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day

– Amount Earned by P,Q and R in 1 day = 120+130-180 = 70

=>Amount Earned by R in 1 day = 70
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20

= 1/3 :1/4 = 4:3

Level 2

1.P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days	B. 14 days
C. 15 days	D. 9 days

Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 × 1/5 = 1

Ie, the work will be done in 15 days

2.A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in – days if they work together.
A. 18 days	B. 22 ½ days
C. 24 days	D. 26 days

Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days

Hence, if the difference is 2 days, B can complete the work in 3 days

=> if the difference is 60 days, B can complete the work in 90 days

=> Amount of work B can do in 1 day= 1/90

Amount of work A can do in 1 day = 3 × (1/90) = 1/30

Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45

=> A and B together can do the work in 45/2 days = 22 ½ days

3.P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
A. 30 days	B. 25 days
C. 20 days	D. 15 days

Answer : Option B

Explanation :

Work done by P and Q in 1 day = 1/10

Work done by R in 1 day = 1/50

Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50

But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50

=> Work done by P in 1 day = 3/50

=> Work done by Q and R in 1 day = 3/50

Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25

So Q alone can do the work in 25 days

4.6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in – days.

A. 4 days

B. 6 days

C. 2 days

D. 8 days

Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 — (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b = ½

=> 52m + 96b = 1— (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

5.Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm	B. 2 pm
C. 1:00 pm	D. 11 am

Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8

Work done by Q in 1 hour = 1/10

Work done by R in 1 hour = 1/12

Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60

From 9 am to 11 am, all the machines were operating.

Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60.

6.A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in — days.
A. 5 ⁵⁄₁₁	B. 4 ⁵⁄₁₁
C. 6 ⁴⁄₁₁	D. 6 ⁵⁄₁₁

Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours. A and B works 8 hours a day.

Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 ⁵⁄₁₁ days

Pending work = 1- 37/60 = 23/60

Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11

which is approximately equal to 2. Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

7.If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12	B. 14
C. 16	D. 18

Answer : Option C

Explanation :

Wages of 1 woman for 1 day = 21600/(40×30)

Wages of 1 man for 1 day = (21600×2)/(40×30)

Wages of 1 man for 25 days = (21600×2×25)/(40×30)

Number of men = 14400/(21600×2×25)/(40×30)=144/(216×50)/40×30)=144/9=16

8.There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 ¹⁄₄ days	B. 4 ¹⁄₃ days
C. 5 ¹⁄₆ days	D. 6 ¹⁄₅ days

Answer : Option C

Explanation :

Work completed in 1st day = 1/16

Work completed in 2nd day = (1/16) + (1/16) = 2/16

Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16

An easy way to attack such problems is from the choices. You can see the choices are

very close to each other. So just see one by one.

For instance, The first choice given in 3 ¹⁄₄

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16

The work done in 4 days = (1+2+3+4)/16 = 10/16

The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?

The work done in 6 days = (1+2+3+4+5+6)/16 > 1

Hence the answer is less than 6, but greater than 5. Hence the answer is 5 ¹⁄₆ days.

(Just for your reference, work done in 5 days = 15/16)

Pending work in 6th day = 1 – 15/16 = 1/16.

In 6th day, 6 people are working and work done = 6/16.

To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.

Hence total time required = 5 + 1/6 = 5 ¹⁄₆ days

PIPES & CISTERN

June 4, 2017October 15, 2024 by PSC Notes

Pipes and Cistern

Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

If a pipe can fill a tank in xhours, then:

part filled in 1 hour =	1	.
	x

If a pipe can empty a tank in yhours, then:

part emptied in 1 hour =	1	.
	y

If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =		1	–	1		.
		x		y

If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =		1	–	1		.
		y		x

Questions:

Level-I:

Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?

Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

1	13	hours
	17

2	8	hours
	11

3	9	hours
	17

4	1	hours
	2

A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:

4	1	hours
	3

7 hours

8 hours

14 hours

Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:

A.	5 min.
B.	9 min.
C.	10 min.
D.	15 min.

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

A.	6 hours
B.	10 hours
C.	15 hours
D.	30 hours

Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:

A.	60 gallons
B.	100 gallons
C.	120 gallons
D.	180 gallons

A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

A.	20 hours
B.	25 hours
C.	35 hours
D.	Cannot be determined
E.	None of these

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

A.	1 hour
B.	2 hours
C.	6 hours
D.	8 hours

Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?

A.	12 min
B.	15 min
C.	25 min
D.	50 min

10.

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

A.	10 min. 20 sec.
B.	11 min. 45 sec.
C.	12 min. 30 sec.
D.	14 min. 40 sec.

11.

Level-II:

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

A.	81 min.
B.	108 min.
C.	144 min.
D.	192 min.

12.

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?

A.	15 min
B.	20 min
C.	27.5 min
D.	30 min

13.

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

A.	3 hrs 15 min
B.	3 hrs 45 min
C.	4 hrs
D.	4 hrs 15 min

14.

Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

6 hours

6	2	hours
	3

7 hours

7	1	hours
	2

15.

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

A.	10
B.	12
C.	14
D.	16

16.

How much time will the leak take to empty the full cistern?
I.	The cistern is normally filled in 9 hours.
II.	It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.

A.	I alone sufficient while II alone not sufficient to answer
B.	II alone sufficient while I alone not sufficient to answer
C.	Either I or II alone sufficient to answer
D.	Both I and II are not sufficient to answer
E.	Both I and II are necessary to answer

17.

How long will it take to empty the tank if both the inlet pipe A and the outlet pipe B are opened simultaneously?
I.	A can fill the tank in 16 minutes.
II.	B can empty the full tank in 8 minutes.

A.	I alone sufficient while II alone not sufficient to answer
B.	II alone sufficient while I alone not sufficient to answer
C.	Either I or II alone sufficient to answer
D.	Both I and II are not sufficient to answer
E.	Both I and II are necessary to answer

18.

If both the pipes are opened, how many hours will be taken to fill the tank?
I.	The capacity of the tank is 400 litres.
II.	The pipe A fills the tank in 4 hours.
III.	The pipe B fills the tank in 6 hours.

A.	Only I and II
B.	Only II and III
C.	All I, II and III
D.	Any two of the three
E.	Even with all the three statements, answer cannot be given.

Answers:

Level-I:

Answer:1 Option B

Explanation:

Part filled by (A + B + C) in 3 minutes = 3		1	+	1	+	1		=		3 x	11		=	11	.
		30		20		10					60			20

Part filled by C in 3 minutes =	3	.
	10

Required ratio =		3	x	20		=	6	.
		10		11			11

Answer:2 Option C

Explanation:

Net part filled in 1 hour		1	+	1	–	1		=	17	.
		5		6		12			60

The tank will be full in	60	hours i.e., 3	9	hours.
	17		17

Answer:3 Option D

Explanation:

Work done by the leak in 1 hour =		1	–	3		=	1	.
		2		7			14

Leak will empty the tank in 14 hrs.

Answer:4 Option B

Explanation:

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1.

x		2	+	1		+ (30 – x).	2	= 1
		75		45			75

	11x	+	(60 -2x)	= 1
	225		75

11x + 180 – 6x = 225.

x = 9.

Answer:5 Option C

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.

	1	+	1	=	1
	x		(x – 5)		(x – 9)

	x – 5 + x	=	1
	x(x – 5)		(x – 9)

(2x – 5)(x – 9) = x(x – 5)

x² – 18x + 45 = 0

(x – 15)(x – 3) = 0

x = 15. [neglecting x = 3]

Answer:6 Option C

Explanation:

Work done by the waste pipe in 1 minute =	1	–		1	+	1
	15			20		24

=		1	–	11
		15		120

= –	1	. [-ve sign means emptying]
	40

Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallon

Answer:7 Option C

Explanation:

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take	x	and	x	hours respectively to fill the tank.
	2		4

	1	+	2	+	4	=	1
	x		x		x		5

	7	=	1
	x		5

x = 35 hrs.

Answer:8 Option C

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

	x + 6 + x	=	1
	x(x + 6)		4

x² – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6. [neglecting the negative value of x]

Answer:9 Option A

Explanation:

Part filled by A in 1 min =	1	.
	20

Part filled by B in 1 min =	1	.
	30

Part filled by (A + B) in 1 min =		1	+	1		=	1	.
		20		30			12

Both pipes can fill the tank in 12 minutes.

Answer:10 Option D

Explanation:

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

Remaining part =		1 –	7		=	8	.
			15			15

Part filled by B in 1 minute =	1
	20

	1	:	8	:: 1 : x
	20		15

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Level-II:

Answer:11 Option C

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in	x	minutes.
	3

	1	+	3	=	1
	x		x		36

	4	=	1
	x		36

x = 144 min.

Answer:12 Option D

Explanation:

Part filled by (A + B) in 1 minute =		1	+	1		=	1	.
		60		40			24

Suppose the tank is filled in x minutes.

Then,	x		1	+	1		= 1
	2		24		40

	x	x	1	= 1
	2		15

x = 30 min.

Answer:13 Option B

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

Remaining part =		1 –	1		=	1	.
			2			2

	2	:	1	:: 1 : x
	3		2

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

Answer:14 Option C

Explanation:

(A + B)’s 1 hour’s work =		1	+	1		=	9	=	3	.
		12		15			60		20

(A + C)’s hour’s work =		1	+	1		=	8	=	2	.
		12		20			60		15

Part filled in 2 hrs =		3	+	2		=	17	.
		20		15			60

Part filled in 6 hrs =		3 x	17		=	17	.
			60			20

Remaining part =		1 –	17		=	3	.
			20			20

Now, it is the turn of A and B and	3	part is filled by A and B in 1 hour.
	20

Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

Answer:15 Option C

Explanation:

Part filled in 2 hours =	2	=	1
	6		3

Remaining part =		1 –	1		=	2	.
			3			3

(A + B)’s 7 hour’s work =	2
	3

(A + B)’s 1 hour’s work =	2
	21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

=		1	–	2		=	1
		6		21			14

C alone can fill the tank in 14 hours.

Answer:16 Option E

Explanation:

Time taken to fill the cistern without leak = 9 hours.

Part of cistern filled without leak in 1 hour =	1
	9

Time taken to fill the cistern in presence of leak = 10 hours.

Net filling in 1 hour =	1
	10

Work done by leak in 1 hour =		1	–	1		=	1
		9		10			90

Leak will empty the full cistern in 90 hours.

Clearly, both I and II are necessary to answer the question.

Correct answer is (E).

Answer:17 Option E

Explanation:

I. A’s 1 minute’s filling work =	1
	16

II. B’s 1 minute’s filling work =	1
	8

(A + B)’s 1 minute’s emptying work =		1	–	1		=	1
		8		16			16

Tank will be emptied in 16 minutes.

Thus, both I and II are necessary to answer the question.

Correct answer is (E).

Answer:18 Option B

Explanation:

II. Part of the tank filled by A in 1 hour =	1
	4

III. Part of the tank filled by B in 1 hour =	1
	6

(A + B)’s 1 hour’s work =		1	+	1		=	5
		4		6			12

A and B will fill the tank in	12	hrs = 2 hrs 24 min.
	5

So, II and III are needed.

Correct answer is (B).

PROFIT & LOSS

June 4, 2017October 7, 2024 by PSC Notes

Profit and loss

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

IMPORTANT FORMULAE

Gain = (S.P.) – (C.P.)
Loss = (C.P.) – (S.P.)
Loss or gain is always reckoned on C.P.
Gain Percentage: (Gain %)

Gain % =		Gain x 100
		C.P.

Loss Percentage: (Loss %)

Loss % =		Loss x 100
		C.P.

Selling Price: (S.P.)

SP =	(100 + Gain %)	x C.P
	100

Selling Price: (S.P.)

SP =		(100 – Loss %)	x C.P.
		100

Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 + Gain %)

Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 – Loss %)

If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:

Loss % =		Common Loss and Gain %		2	=		x		2	.
		10					10

If a trader professes to sell his goods at cost price, but uses false weights, then

Gain % =		Error	x 100	%.
		(True Value) – (Error)

Questions:

Level-I:

Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:

4	4	%
	7

5	5	%
	11

10%

12%

The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of xis:

A.	15
B.	16
C.	18
D.	25

If selling price is doubled, the profit triples. Find the profit percent.

66	2
	3

100

105	1
	3

120

In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A.	30%
B.	70%
C.	100%
D.	250%

A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

A.	3
B.	4
C.	5
D.	6

The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?

A.	Rs. 2000
B.	Rs. 2200
C.	Rs. 2400
D.	Data inadequate

A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?

A.	Rs. 18.20
B.	Rs. 70
C.	Rs. 72
D.	Rs. 88.25

A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

A.	Rs. 1090
B.	Rs. 1160
C.	Rs. 1190
D.	Rs. 1202

Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?

A.	3.5
B.	4.5
C.	5.6
D.	6.5

10.

Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:

30%

33	1	%
	3

35%

44%

11.

Level-II:

On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

A.	Rs. 45
B.	Rs. 50
C.	Rs. 55
D.	Rs. 60

12.

When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?

A.	Rs. 21,000
B.	Rs. 22,500
C.	Rs. 25,300
D.	Rs. 25,800

13.

100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:

14	2	% gain
	7

15% gain

14	2	% loss
	7

15 % loss

14.

A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:

5	15	% loss
	17

5	15	% gain
	17

6	2	% gain
	3

None of these

15.

A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A.	No profit, no loss
B.	5%
C.	8%
D.	10%
E.	None of these

A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
1%
2%
3%
4%

A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
10%
20%
25%
28%

In terms of percentage profit, which among following the best transaction.
1. P. 36, Profit 17
2. P. 50, Profit 24
3. P. 40, Profit 19
4. P. 60, Profit 29

Answer:1 Option B

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.

Gain % =		300	x 100	%	= 5	5	%
		5500				11

Answer:2 Option B

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

		20 – x	x 100 = 25
		x

2000 – 100x = 25x

125x = 2000

x = 16.

Answer:3 Option B

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y – x) = (2y – x) y = 2x.

Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.

Profit % =		x	x 100	% = 100%

Answer:4 Option B

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

Required percentage =		295	x 100	%	=	1475	% = 70% (approximately).
		420				21

Answer:5 Option C

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs.	6
	5

For Rs.	6	, toffees sold = 6.
	5

For Re. 1, toffees sold =		6 x	5		= 5.
			6

Answer:6 Option A

Explanation:

Let C.P. be Rs. x.

Then,	1920 – x	x 100 =	x – 1280	x 100
	x		x

1920 – x = x – 1280

2x = 3200

x = 1600

Required S.P. = 125% of Rs. 1600 = Rs.		125	x 1600		= Rs 2000.
		100

Answer:7 Option C

Explanation:

C.P. = Rs.		100	x 392		= Rs.		1000	x 392		= Rs. 320
		122.5					1225

Profit = Rs. (392 – 320) = Rs. 72.

Answer:8 Option C

Explanation:

S.P. = 85% of Rs. 1400 = Rs.		85	x 1400		= Rs. 1190
		100

Answer:9 Option C

Explanation:

Cost Price of 1 toy = Rs.		375		= Rs. 31.25
		12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75

Profit % =		1.75	x 100	%	=	28	% = 5.6%
		31.25				5

Answer:10 Option D

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.		5	x 30		= Rs. 25.
		6

S.P. of 30 articles = Rs.		6	x 30		= Rs. 36.
		5

Gain % =		11	x 100	% = 44%.
		25

Answer:11 Option D

Explanation:

(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)

C.P. of 12 balls = S.P. of 17 balls = Rs.720.

C.P. of 1 ball = Rs.		720		= Rs. 60.
		12

Answer:12 Option C

Explanation:

85 : 18700 = 115 : x

x =		18700 x 115		= 25300.
		85

Hence, S.P. = Rs. 25,300.

Answer:13 Option A

Explanation:

C.P. of 1 orange = Rs.		350		= Rs. 3.50
		100

S.P. of 1 orange = Rs.		48		= Rs. 4
		12

Gain% =		0.50	x 100	%	=	100	% = 14	2	%
		3.50				7		7

Answer:14 Option B

Explanation:

C.P. of 1^st transistor = Rs.		100	x 840		= Rs. 700.
		120

C.P. of 2^nd transistor = Rs.		100	x 960		= Rs. 1000
		96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

Gain % =		100	x 100	%	= 5	15	%
		1700				17

Answer:15 Option B

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =		80	x 100	% = 5%.
		1600

Answer:16 Option D

Explanation:

So we have C.P. = 27.50
S.P. = 28.60

Gain = 28.60 – 27.50 = Rs. 1.10

Gain%=(Gain/Cost∗100)%=(1.10/27.50∗100)%=4%

Answer:17 Option B

Explanation:

We know, C.P. = 5000
S.P. = 4000
Loss = 5000 – 4000 = 1000
Loss%=(Loss/Cost∗100)%=(1000/5000∗100)%=20%

Answer:18 Option D

Explanation:

Hint: Calculate profit percent as

Profit% = (profit/cost) * 100

SIMPLIFICATION

June 4, 2017October 4, 2024 by PSC Notes

Simplification

Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly.

Rules of Simplification

V → Vinculum

B → Remove Brackets – in the order ( ) , { }, [ ]

O → Of
D → Division

M → Multiplication

A → Addition

S → Subtraction

Classification

Types	Description
Natural Numbers:	all counting numbers ( 1,2,3,4,5….∞)
Whole Numbers:	natural number + zero( 0,1,2,3,4,5…∞)
Integers:	All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞)
Even & Odd Numbers :	All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞)
Prime Numbers:	It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞)
Composite Numbers:	Natural numbers which are not prime
Co-Prime:	Two natural number a and b are said to be co-prime if their HCF is 1.

Divisibility

Numbers	IF A Number	Examples
Divisible by 2	End with 0,2,4,6,8 are divisible by 2	254,326,3546,4718 all are divisible by 2
Divisible by 3	Sum of its digits is divisible by 3	375,4251,78123 all are divisible by 3. [549=5+4+9][5+4+9=18]18 is divisible by 3 hence 549 is divisible by 3.
Divisible by 4	Last two digit divisible by 4	5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.
Divisible by 5	Ends with 0 or 5	225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.
Divisible by 6	Divides by Both 2 & 3	4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.
Divisible by 8	Last 3 digit divide by 8	746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.
Divisible by 10	End with 0	220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.
Divisible by 11	[Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11	Consider the number 39798847 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3) (23-12) 23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

hence ,represent it as:

dividend = ( divisor✘quotient ) + remainder

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

Rules

Modulus of a Real Number:

Modulus of a real number a is defined as

\|a\| =		a, if a > 0
		–a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

Sum Rules

(1+2+3+………+n) = ¹/₂n(n+1)

(1²+2²+3²+………+n²) = ¹/₆n (n+1) (2n+1)

(1³+2³+3³+………+n³) = ¹/4n² (n+1)²

Questions:

Level-I:

A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

A.	45
B.	60
C.	75
D.	90

There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

A.	20
B.	80
C.	100
D.	200

3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

A.	Rs. 3500
B.	Rs. 3750
C.	Rs. 3840
D.	Rs. 3900

4. If a – b = 3 and a² + b² = 29, find the value of ab.

A.	10
B.	12
C.	15
D.	18

The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?

A.	Rs. 1200
B.	Rs. 2400
C.	Rs. 4800
D.	Cannot be determined
E.	None of these

A sum of Rs. 1360 has been divided among A, B and C such that A gets of what B gets and B gets of what C gets. B’s share is:

A.	Rs. 120
B.	Rs. 160
C.	Rs. 240
D.	Rs. 300

One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?

A.	Rs. 30,000
B.	Rs. 50,000
C.	Rs. 60,000
D.	Rs. 90,000

A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

A.	30 birds
B.	60 birds
C.	72 birds
D.	90 birds

Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:

10.

To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?

A.	10
B.	35
C.	62.5
D.	Cannot be determined
E.	None of these

Level-II:

In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?

A.	160
B.	175
C.	180
D.	195

12.

Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?

A.	256
B.	432
C.	512
D.	640
E.	None of these

13.

A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:

A.	22
B.	23
C.	24
D.	26

14.

(469 + 174)² – (469 – 174)²	= ?
(469 x 174)

A.	2
B.	4
C.	295
D.	643

15.

David gets on the elevator at the 11^th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?

A.	19
B.	28
C.	30
D.	37

Find the value of 1/(3+1/(3+1/(3-1/3)))

A.) 3/10	B.) 10/3
C.) 27/89	D.) 89/27

Find the value of

A.) 3½ 99;	B.) 34/99
C.) 2.131313	D.) 3.141414

18.Find the value of

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

A.) 14/10	B.) 1.35
C.) 13/10	D.) 0

Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)

A.) 0.56	B.) 0.65
C.) 0.54	D.) 0.45

Find the value of

A.) 1.5	B.) -1.5
C.) 1	D.) 0

Answers:

Level-I

Answer:1 Option D

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

Answer:2 Option C

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10 x – y = 20 …. (i)

and x + 20 = 2(y – 20) x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

Answer:3 Option D

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y or y =	5	x.
	2

15x + 2y = 4000

15x + 2 x	5	x = 4000
	2

20x = 4000

x = 200.

So, y =		5	x 200		= 500.
		2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

Answer:4 Option A

Explanation:

2ab = (a² + b²) – (a – b)²

= 29 – 9 = 20

ab = 10.

Answer:5 Option B

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

Divide equation (i) by 2, we get the below equation.

=> x + 2y = 800. — (iii)

Now subtract (iii) from (ii)

x + 6y = 1600 (-)

x + 2y = 800

—————-

4y = 800

—————-

Therefore, y = 200.

Now apply value of y in (iii)

=> x + 2 x 200 = 800

=> x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

Answer:6 Option C

Explanation:

Let C’s share = Rs. x

Then, B’s share = Rs.	x	, A’s share = Rs.		2	x	x		= Rs.	x
	4			3		4			6

	x	+	x	+ x = 1360
	6		4

	17x	= 1360
	12

x =	1360 x 12	= Rs. 960
	17

Hence, B’s share = Rs.		960		= Rs. 240.

Answer:7 Option C

Explanation:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

1	x =	1	(150000 – x)
3		2

	x	+	x	= 75000
	3		2

	5x	= 75000
	6

x =	75000 x 6	= 90000
	5

Savings in Public Provident Fund = Rs. (150000 – 90000) = Rs. 60000

Answer:8 Option A

Explanation:

Let the total number of shots be x. Then,

Shots fired by A =	5	x
	8

Shots fired by B =	3	x
	8

Killing shots by A =	1	of	5	x	=	5	x
	3		8			24

Shots missed by B =	1	of	3	x	=	3	x
	2		8			16

	3x	= 27 or x =		27 x 16		= 144.
	16			3

Birds killed by A =	5x	=		5	x 144		= 30.
	24			24

Answer:9 Option A

Explanation:

Original share of 1 person =	1
	8

New share of 1 person =	1
	7

Increase =		1	–	1		=	1
		7		8			56

Required fraction =	(1/56)	=		1	x	8		=	1
	(1/8)			56		1			7

Answer:10 Option C

Explanation:

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

New capacity of bucket =	2	x
	5

Required number of buckets =	25x
	(2x/5)

25x

=	125
	2

= 62.5

Level-II:

Answer:11 Option B

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

Answer:12 Option C

Explanation:

Let total number of children be x.

Then, x x	1	x =	x	x 16 x = 64.
	8		2

Number of notebooks =	1	x² =		1	x 64 x 64		= 512

Answer:13 Option D

Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140 x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

Answer:14 Option B

Explanation:

Given exp. =	(a + b)² – (a – b)²
	ab

=	4ab
	ab

= 4 (where a = 469, b = 174.)

Answer:15 Option C

Explanation:

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x 120x = 40

x =	1
	3

Number of floors covered by David in (1/3) min. =		1	x 57		= 19.
		3

So, their paths cross at (11 +19) i.e., 30^th floor.

Answer:16 Option ‘C’

Explanation:

1/[3 + (1/(3+1/(3 – 1/3)))]

=> 1/[3 + 1/(3 + 1/(8/3))]

=> 1/[3 + 1/(3 + 3/8)]

=> 1/[3 + 8/27]

=> 1/(89/27)

=> 27/89

Answer:17 Option ‘D’

Explanation:

6/9 + 7/9 + 9/9 + 69/99

2/3 + 7/9 + 1 + 69/99

(66 + 77 + 99 + 69)/99

311/99 => 3.141414

Answer:18 Option ‘A’

Explanation:

((0.1)³ + (0.6)³ + (0.7)³ − (0.3)(0.6)(0.7))/((0.1)² + (0.6)² + (0.7)² − 0.006 − 0.42 − 0.07)

=> (0.1 + 0.6 + 0.7)³/(0.1 + 0.6 + 0.7)²

=> 0.1 + 0.6 + 0.7 => 1.4 = 14/10

Answer:19 Option ‘A’

Answer:20 Option ‘D’

11/30 − [1/6 + 1/5 + [7/12 − 7/12]]

11/30 − [1/6 + 1/5 + [0]]

11/30 − [(5 + 6)/30]

11/30 − 11/30 = 0.

PARTNERSHIP

June 4, 2017October 3, 2024 by PSC Notes

Partnership :

Partnership is an association of two or more parties, they put money for business.

Simple Partnership:

Simple partnership is one in which the capitals of the partners are invested for the same time. The profit or losses are divided among the partners in the ratio of their investments.

Compound Partnership:

Compound Partnership is one which the capitals of the partners are invested for different periods. In such cases equivalent capitals are calculated for a unit time by multiplying the capital with the number of units of time. The profits or losses are then divided in the ratio of these equivalent capitals. Tus the ratio of profits is directly proportional to both capital invested as time.

Working partner:

A partner who participates in the working and manages the business is called a Working Partner.

Sleeping Partner:

A partner who only invests capital but does not participate in the working of the business is called a Sleeping Partner.

Division of Profit and Loss:

1. Rule :When investment of all partners are for the same time, the loss or profit is distributed among partners in the ratio of investment.
Ex. Let P and Q invested Rs. a and b for one year in a business then share of profit and loss be ,

P’s share of profit : Q’s share profit = a : b

2.Rule : When investments are for different time period, then profit ratio is calculated as capital multiplied by length of investment

Ex. P’s share of profit : Q’s share profit = a* t1 : b* t2

Questions with solutions

Level-I

A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdrawns Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years ?

Solution: A : B : C = (40,000 X 36) : (80,000 X 12 + 40,000 X 24) : (120,000 X 24 + 40,000 X 12) = 3: 4: 16

A, B, C enter into a partnership investing Rs. 35,000, Rs.45,000 and Rs.55,000 respectively. The respective shares of A, B, C in an annual profit of Rs.40,500 are ?

Solution : A : B : C = 35000 : 45000 : 55000 = 7 : 9 : 11.

A’s share = Rs (40500 x 7/27) = Rs. 10500

B’s share = Rs.(40500× 9/27) = Rs. 13500

C’s share = Rs.(40500×11/27)= Rs. 16500

In a business, Lucky invests Rs. 35,000 for 8 months and manju invests Rs 42,000 for 10 months. Out of a profit of Rs. 31,570. Manju’s share is 😕

Solution : lucky: Manju = (35000 X 8) : (42,000 X 10) = 2:3
Manju’s share = Rs.3/5×31570 = Rs. 18,942

Amar started a business investing Rs. 70,000. Ramki joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business ?

Solution: Amar : Ramki : Sagar =

(70000 X 36) : (105000 X 30) : (140000 X24) = 12 : 15 : 16.

5 . A begins a business with Rs 450 and is joined afterwards by B with Rs 300. After how many months does B join if the profits at the end of the year is divided in the ratio 2 : 1?

Solution.-.(B) Suppose B joins for x months.

Then, 450 ´12 = 2
300 ´ x 1

x =450× 6
300

x= 9 months

\B joins after (12 – 9) = 3 months.

Shekhar started a business investing Rs. 25,000 in 1999. In 2000, he invested an additional amount of Rs. 10,000 and Rajeev joined him with an amount of Rs. 35,000. In 2001, Shekhar invested another additional amount of Rs. 10,000 and Jatin joined them with an amount of Rs. 35,000. What will be Rajeev’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 1999?.

Solution : Shekhar : Rajeev : Jatin =

(25000 X 12 + 35000 X 12 + 45000 X 12) : (35000 X 24) :   (35000 X 12)
= 1260000   : 840000 : 420000 =   3 : 2 : 1.
Rajeev’s share   = Rs.(150000×26) =   Rs. 50000

A,B and C started a business with Rs.15000, Rs.25000 and Rs.35000 respectively. A was paid 10% of the total profit as a salary and the balance was divided in the ration of investment. If A’s share is Rs.4,200, then C’s share is: ?

Solution : A, B and C must divide their salaries in the ratio :

15,000 : 25,000:35,000 = 3:5:7
Assume total Profit = 100X.

then A share is 10% of 100X for managing business and 3/15 part of 90X for his investment (as the remaining profit is (100X – 10X = 90X)
So total A’s share = 10X + 315 × 90X = 4,200
⇒X = 150
Substituting X = 150 in 90X we get remaining profit for sharing. That is Rs.13,500
Now C’s share = 715×13,500 = Rs.6,300

Level-II

A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855, the total profit is:

A.	Rs. 1425
B.	Rs. 1500
C.	Rs. 1537.50
D.	Rs. 1576

Answer:1 Option B

Explanation:

Let the total profit be Rs. 100.

After paying to charity, A’s share = Rs.		95 x	3		= Rs. 57.
			5

If A’s share is Rs. 57, total profit = Rs. 100.

If A’s share Rs. 855, total profit =		100	x 855		= 1500

A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.

A.	Rs. 1900
B.	Rs. 2660
C.	Rs. 2800
D.	Rs. 2840

Answer: 2 Option B

Explanation:

For managing, A received = 5% of Rs. 7400 = Rs. 370.

Balance = Rs. (7400 – 370) = Rs. 7030.

Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)

= 39000 : 42000 : 30000

= 13 : 14 : 10

B’s share = Rs.		7030 x	14		= Rs. 2660.
			37

3 .A, B and C enter into a partnership in the ratio : : . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:

A.	Rs. 2100
B.	Rs. 2400
C.	Rs. 3600
D.	Rs. 4000

Answer:3 Option D

Explanation:

Ratio of initial investments =		7	:	4	:	6		= 105 : 40 : 36.
		2		3		5

Let the initial investments be 105x, 40x and 36x.

A : B : C =		105x x 4 +	150	x 105x x 8		: (40x x 12) : (36x x 12)
			100

= 1680x : 480x : 432x = 35 : 10 : 9.

Hence, B’s share = Rs.		21600 x	10		= Rs. 4000.
			54

A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:

A.	Rs. 8400
B.	Rs. 11,900
C.	Rs. 13,600
D.	Rs. 14,700

Answer:4 Option D

Explanation:

Let C = x.

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.

So, x + x + 5000 + x + 9000 = 50000

3x = 36000

x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.

A’s share = Rs.		35000 x	21		= Rs. 14,700.
			50

Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?

A.	5 : 7 : 8
B.	20 : 49 : 64
C.	38 : 28 : 21
D.	None of these

Answer:5 Option B

Explanation:

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

Now,	14x	=	5	98x = 40y y =	49	x
	8y		7		20

And,	14x	=	5	112x = 35z z =	112	x =	16	x.
	7z		8		35		5

x : y : z = x :	49	x	:	16	x	= 20 : 49 : 64.
	20			5

A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?

A.	Rs. 7500
B.	Rs. 8000
C.	Rs. 8500
D.	Rs. 9000

Answer:6 Option D

Explanation:

Let B’s capital be Rs. x.

Then,		3500 x 12	=	2
		7x		3

14x = 126000

x = 9000.

A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew of his capital and B withdrew of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:

A.	Rs. 330
B.	Rs. 360
C.	Rs. 380
D.	Rs. 430

Answer:7 Option A

Explanation:

A : B =

4x x 3 +

4x –

x 4x

x 7

5x x 3 +

5x –

x 5x

x 7

= (12x + 21x) : (15x + 28x)

= 33x :43x

= 33 : 43.

A’s share = Rs.		760 x	33		= Rs. 330.
			76

A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?

A.	3 : 5 : 2
B.	3 : 5 : 5
C.	6 : 10 : 5
D.	Data inadequate

Answer:8 Option C

Explanation:

Let the initial investments of A and B be 3x and 5x.

A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?

A.	Rs. 45
B.	Rs. 50
C.	Rs. 55
D.	Rs. 60

Answer:9 Option A

Explanation:

A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.

C’s rent = Rs.		175 x	9		= Rs. 45.
			35

10.

A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?

A.	Rs. 7500
B.	Rs. 9000
C.	Rs. 9500
D.	Rs. 10,000

Answer: 10 Option A

Explanation:

A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.

B’s share = Rs.		25000 x	3		= Rs. 7,500.
			10