Aptitude & reasoning

 

This module will teach you the basics of direct and indirect proportions. These concepts will further help you in time and work questions.

Important Formulas – chain rule

• Direct Proportion

  Two quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent.
  Examples

  1. Cost of the goods is directly proportional to the number of goods. (More goods, More cost)
  2. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)

• Indirect Proportion (inverse proportion)

  Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa.

Examples

• 1. Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)
  2. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)

 

Solved Examples

Level 1

1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?
A. Rs. (xd/y)	B. Rs. x/d
C. Rs. (yd/x)	D. Rs. y/d

 

Answer : Option C

Explanation :

cost of x metres of wire = Rs. d

cost of 1 metre of wire = Rs.(d/x)

cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

2. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
A. 50	B. 30
C. 40	D. 10

 

Answer : Option B

Explanation :

Meal for 200 children = Meal for 120 men

Meal for 1 child = Meal for 120/200 men

Meal for 150 children = Meal for (120×150)/200 men=Meal for 90 men

Total mean available = Meal for 120 men

Renaming meal = Meal for 120 men – Meal for 90 men = Meal for 30 men

 

3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?
A. 26	B. 22
C. 12	D. 24

 

Answer : Option D

Explanation :
Let the required number of days be x

More men, less days (indirect proportion)

Hence we can write as

Men36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x

⇒12×2=x

⇒x=24

4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel?
A. 15	B. 12
C. 21	D. 9

 

Answer : Option D

Explanation :

Let the number of revolutions made by the larger wheel be x

More cogs, less revolutions (Indirect proportion)

Hence we can write as

Cogs 6:14}: x: 21⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9

5. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day?
A. 10	B. 12
C. 8	D. 15

 

Answer : Option B

Explanation :

Let the required hours needed be x

More pumps, less hours (Indirect proportion)
More Days, less hours (Indirect proportion)

Hence we can write as

Pumps  3:4

::x:8

Days                      2:1

⇒3×2×8=4×1×x

⇒3×2×2=x

⇒x=12

6. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?
A. 9	B. 12
C. 10	D. 13

 

Answer : Option D

Explanation :
Let the required number of days be x

More persons, less days (indirect proportion)
More hours, less days (indirect proportion)

Hence we can write as

Persons                39:30

::x:12

Hours    5:6
⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13

7. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?
A. 205	B. 200
C. 180	D. 195

 

Answer : Option D

Explanation :

Let the required number of seconds be x

More cloth, More time, (direct proportion)

Hence we can write as

Cloth         0.128:25} :: 1:x

⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195

 

8. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows?
A. 49	B. 32
C. 36	D. 41

 

Answer : Option A

Explanation :

15 cows ≡ 21 goats

1 cow ≡21/15 goats

35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats

 

Level 2

 

1. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?
A. 1	B. 40
C. 20	D. 26

 

Answer : Option B

Explanation :

Assume that in x days, one cow will eat one bag of husk.

More cows, less days (Indirect proportion)
More bags, more days (direct proportion)
Hence we can write as

Cows    40:1         ::x:40

Bags     1:40

⇒40×1×40=1×40×x ⇒x=40

2. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?
A. 65 paise	B. 70 paise
C. 52 paise	D. 48 paise

 

Answer : Option D

Explanation :
Let 200 gm potato costs x paise

Cost of ¼ Kg potato = 60 Paise
=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = ¹⁰⁰⁰/₄ gm = 250 gm)

More quantity, More Paise (direct proportion)

Hence we can write as

Quantity  200:250} :: x:60

⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48

3. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time, each person’s now working 9 hours a day.
A. 160	B. 150
C. 24	D. 56

 

Answer : Option D

Explanation :

Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed = ²/₅

Remaining days = 56 – 30 = 26
Remaining Work to be completed = 1 – ²/₅ = ³/₅
Let the total number of persons who do the remaining work = x
Number of hours each person needs to be work per day = 9

More days, less persons(indirect proportion) More hours, less persons(indirect proportion)
More work, more persons(direct proportion)

Hence we can write as

Days     30:26

Hours    8:9                                   ::x:104

Work     35:25
⇒30×8×3/5×104=26×9×2/5×x

⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)

=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160

Number of additional persons required = 160 – 104 = 56

 

4. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?
A. x2/y2 units	B. y3/x2 units
C. x3/y2 units	D. y2/x2 units

 

Answer : Option B

Explanation :
Let amount of work completed by y men working y hours per in y days = w units

More men, more work(direct proportion)
More hours, more work(direct proportion)
More days, more work(direct proportion)

Hence we can write as

Men                      x:y

Hours    x:y          ::x:w

Days                      x:y
⇒x³w=y³x ⇒w=y³x/x³=y³/x²

5. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions?
A. 12.5 m	B. 10.5 m
C. 14	D. 12

 

Answer : Option A

Explanation :
Let the required height of the building be x meter

More shadow length, More height (direct proportion)

Hence we can write as

Shadow length 40.25:28.75}:: 17.5:x

⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250

= (2875×7)/1610=2875/230=575/46=12.5

 

6. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples?
A. Rs. 2500	B. Rs. 2300
C. Rs. 2200	D. Rs. 1400

 

Answer : Option A

Explanation :

Let the required price be x

More apples, More price (direct proportion)

Hence we can write as

Apples 357:(49×12)} :: 1517.25:x

⇒357x = (49×12)×1517.25⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51

= (7×4×1517.25)/17

=7×4×89.25≈2500

7. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?
A. 20 metric tonnes	B. 22 metric tonnes
C. 24 metric tonnes	D. 26 metric tonnes

 

Answer : Option D

Explanation :

Let required amount of coal be x metric tonnes

More engines, more amount of coal (direct proportion)

If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)

More hours, more amount of coal(direct proportion)

Hence we can write as

Engines                                                                9:8

rate of consumption                       13:14                     ::24:x

hours                                                                    8:13
⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13⇒x=2×13=26

8. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?
A. 1900	B. 1800
C. 1940	D. 2000

 

Answer : Option A

Explanation :

Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 – 15 =39)
Let x number of people came after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days

More men, Less days (Indirect Proportion)⇒Men        2000:(2000+x)}  ::  20:39

⇒2000×39=(2000+x)20⇒100×39=(2000+x)⇒3900=2000+x⇒x=3900−2000=1900

Surds

A surd is a square root which cannot be reduced to a rational number.

For example,  is not a surd.

However  is a surd.

If you use a calculator, you will see that  and we will need to round the answer correct to a few decimal places. This makes it less accurate.

If it is left as , then the answer has not been rounded, which keeps it exact.

Here are some general rules when simplifying expressions involving surds.

 

 

 

1. a^mx aⁿ = a^m + n

am	= am – n
an

• (a^m)ⁿ= a^mn

 

1. (ab)ⁿ= aⁿbⁿ

 

	a		n	=	an
	b				bn

1. a⁰= 1

 

 

Questions

Level-I

 

1.	(17)3.5 x (17)? = 178
	A. 2.29 B. 2.75 C. 4.25 D. 4.5

 

2.	If a x – 1 = b x – 3 , then the value of x is: b a
	A. 1 2 B. 1 C. 2 D. 7 2

 

3.	Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
	A. 1.45 B. 1.88 C. 2.9 D. 3.7

 

4.	If 5a = 3125, then the value of 5(a – 3) is:
	A. 25 B. 125 C. 625 D. 1625

 

5.	If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:
	A. 0 B. 2 C. 4 D. 6

 

.6.	(256)0.16 x (256)0.09 = ?
	A. 4 B. 16 C. 64 D. 256.25

 

7.	The value of [(10)150 ÷ (10)146]
	A. 1000 B. 10000 C. 100000 D. 106

 

8.	1  + 1 + 1 = ? 1 + x(b – a) + x(c – a) 1 + x(a – b) + x(c – b) 1 + x(b – c) + x(a – c)
	A. 0 B. 1 C. xa – b – c D. None of these

 

9.	(25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
	A. 8.5 B. 13 C. 16 D. 17.5 E. None of these

 

10.	(0.04)-1.5 = ?
	A. 25 B. 125 C. 250 D. 625

 

 

Level-II

 

11.	(243)n/5 x 32n + 1 = ? 9n x 3n – 1
	A. 1 B. 2 C. 9 D. 3n

 

12.	1 + 1 = ? 1 + a(n – m) 1 + a(m – n)
	A. 0 B. 1 2 C. 1 D. am + n

 

13.	If m and n are whole numbers such that mn = 121, the value of (m – 1)n + 1 is:
	A. 1 B. 10 C. 121 D. 1000

 

14.	xb (b + c – a) . xc (c + a – b) . xa (a + b – c) = ? xc xa xb
	A. xabc B. 1 C. xab + bc + ca D. xa + b + c

 

15. If 5√5 * 5³÷ 5^-3/2= 5^a+2 , the value of a is:
    A. 4
    B. 5
    C. 6
    D. 8

16.(132)7 ×(132)? =(132)11.5. A. 3 B. 3.5 C. 4 D. 4.5     17. (ab)x−2=(ba)x−7. What is the value   of x ?   A. 3 B. 4 C. 3.5 D. 4.5       18. (0.04)-2.5 = ?   A. 125 B. 25 C. 3125 D. 625

Answers

Level-I

Answer:1 Option D

 

Explanation:

Let (17)^3.5 x (17)^x = 17⁸.

Then, (17)^{3.5 + x} = 17⁸.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

 

Answer:2 Option C

 

Explanation:

Given	a		x – 1	=		b		x – 3
	b					a

 

a

x – 1

=

a

-(x – 3)

=

a

(3 – x)

b

b

b

x – 1 = 3 – x

2x = 4

x = 2.

 

 

Answer:3 Option C

 

Explanation:

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

0.48z = 1.40

z =	140	=	35	= 2.9 (approx.)
	48		12

 

Answer:4 Option A

 

Explanation:

5^a = 3125        5^a = 5⁵

a = 5.

5^{(a – 3)} = 5^{(5 – 3)} = 5² = 25.

 

 

Answer:5 Option C

 

Explanation:

3^x – y = 27 = 3³        x – y = 3 ….(i)

3^x + y = 243 = 3⁵        x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4

 

 

Answer:6 Option A

 

Explanation:

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

= (256)^0.25

= (256)^(25/100)

= (256)^(1/4)

= (4⁴)^(1/4)

= 4^4(1/4)

= 4¹

= 4

Answer:7 Option B

 

Explanation:

(10)150 ÷ (10)146 =	10150
	10146

= 10^{150 – 146}

= 10⁴

= 10000.

 

Answer:8 Option B

 

Explanation:

Given Exp. =	1  + 1  + 1 1 + xb + xc xa xa 1 + xa + xc xb xb 1 + xb + xa xc xc

 

=	xa	+	xb	+	xc
	(xa + xb + xc)		(xa + xb + xc)		(xa + xb + xc)

 

=	(xa + xb + xc)
	(xa + xb + xc)

= 1.

 

Answer:9 Option B

 

Explanation:

Let (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^x.

Then,	(52)7.5 x (5)2.5	= 5x
	(53)1.5

 

	5(2 x 7.5) x 52.5	= 5x
	5(3 x 1.5)

 

	515 x 52.5	= 5x
	54.5

5^x = 5^{(15 + 2.5 – 4.5)}

5^x = 5¹³

x = 13.

 

Answer:10 Option B

 

Explanation:

(0.04)-1.5 =		4		-1.5
		100

 

=		1		-(3/2)
		25

= (25)^(3/2)

= (5²)^(3/2)

= (5)^{2 x (3/2)}

= 5³

= 125.

 

Level-II

 

Answer:11 Option C

 

Explanation:

Given Expression	= (243)(n/5) x 32n + 1 9n x 3n – 1
	= (35)(n/5) x 32n + 1 (32)n x 3n – 1
	= (35 x (n/5) x 32n + 1) (32n x 3n – 1)
	= 3n x 32n + 1 32n x 3n – 1
	= 3(n + 2n + 1) 3(2n + n – 1)
	= 33n + 1 33n – 1
	= 3(3n + 1 – 3n + 1)   = 32   = 9.

Answer:12 Option C

 

Explanation:

1	+	1	=	1  + 1 1 + an am 1 + am an
1 + a(n – m)		1 + a(m – n)

 

=	am	+	an
	(am + an)		(am + an)

 

=	(am + an)
	(am + an)

= 1.

 

Answer:13 Option D

 

Explanation:

We know that 11² = 121.

Putting m = 11 and n = 2, we get:

(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10³ = 1000.

 

Answer:14 Option B

 

Explanation:

Given Exp.	= x(b – c)(b + c – a) . x(c – a)(c +a – b) . x(a – b)(a + b – c)
	= x(b – c)(b + c) – a(b – c)  .  x(c – a)(c + a) – b(c – a) .  x(a – b)(a + b) – c(a – b)
	= x(b2 – c2 + c2 – a2 + a2 – b2)  .   x–a(b – c) – b(c – a) – c(a – b)
	= (x0 x x0)
	= (1 x 1) = 1.

 

Answer:15 option C

 

Answer:16

Explanation

am.an=am+n

(132)⁷ × (132)^x = (132)^11.5

=> 7 + x = 11.5

=> x = 11.5 – 7 = 4.5

 

 

Answer:17

Explanation:

an=1a−n

(ab)x−2=(ba)x−7⇒(ab)x−2=(ab)−(x−7)⇒x−2=−(x−7)⇒x−2=−x+7⇒x−2=−x+7⇒2x=9⇒x=92=4.5

 

Answer:18

Explanation:

a−n=1/an

(0.04)−2.5=(1/.04)2.5=(100/4)2.5=(25)2.5=(52)2.5=(52)(5/2)=55=3125

 

 

1. Alligation

   It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

2. Mean Price

   Mean price is the cost price of a unit quantity of the mixture

 

3. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

4. Rule of Alligation

   If two ingredients are mixed, then

   (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity of cheaper (c)	Cost Price(CP) of a unit quantity of dearer (d)
Mean Price
(m)
(d – m)	(m – c)

5. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres	B. 29.16 litres
C. 28 litres	D. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43	B. 34
C. 32	D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar	Concentration of alcohol in 2nd Jar
40%	19%
Mean
26%
7	14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8	B. 8 : 7
C. 6 : 7	D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kind	Cost of 1 kg rice of 2nd kind
9.3	10.80
Mean Price
10
10.8-10 = .8	10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3	B. 2 : 2
C. 1 : 2	D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water	Cost Price of 1 litre of milk
0	12
Mean Price
8
12-8=4	8-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg	B. 2 Kg
C. .5 Kg	D. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of manganese in the mixture : 20	Percentage concentration of manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 10	20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg	B. 1400 Kg
C. 1600 Kg	D. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1	% Profit by selling part2
8	12
Net % Profit
11
12 – 11 = 1	11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre	B. 2 litre
C. 1 litre	D. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water in pure water : 100	% Concentration of water in the given mixture : 10
Mean % concentration
20
20 – 10 = 10	100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300	B. 400
C. 600	D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part	Profit% by selling 2nd part
8	18
Net % profit
14
18-14=4	14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50	B. Rs.170.5
C. Rs.175.50	D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea	Cost of 1 kg of 2nd kind of tea
130.50	x
Mean Price
153
(x – 153)	22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres	B. 7litres, 4 litres
C. 6litres, 6 litres	D. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can	CP of 1 litre mix in 1st can
1/2	3/4
Mean Price
5/8
3/4 – 5/8 = 1/8	5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4	B. 4 : 3
C. 9 : 7	D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel A	CP of 1 litre mixture from vessel B
5/7	7/13
Mean Price
8/13
8/13 – 7/13 = 1/13	5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg	B. 63 kg
C. 58 Kg	D. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind	CP of 1 kg sugar of 2nd kind
Rs. 9	Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.4	9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23	B. 21
C. 19	D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%	B. 20%
C. 22%	D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk	CP of 1 litre water
1	0
CP of 1 litre mixture
4/5
4/5 – 0 = 4/5	1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%

 

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

• Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c² = a² + b² where c is the length of the hypotenuse and a and b are the lengths of the other two sides

• Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by π

π≈3.14≈227

• Geometric Shapes and solids and Important Formulas

Geometric Shapes	Description	Formulas
	Rectangle l = Length b = Breadth d= Length of diagonal	Area = lb Perimeter = 2(l + b) d = √l2+b2
	Square a = Length of a side d= Length of diagonal	Area= a*a=1/2*d*d Perimeter = 4a d = 2√a
	Parallelogram b and c are sides b = base h = height	Area = bh Perimeter = 2(b + c)
	Rhombus a = length of each side b = base h = height d1, d2 are the diagonal	Area = bh(Formula 1) Area = ½*d1*d2 (Formula 2 ) Perimeter = 4a
	Triangle a , b and c are sides b = base h = height	Area = ½*b*h (Formula 1) Area(Formula 2)                         = √S(S−a)(S−b)(S−c              where S is the semiperimeter S  =(a+b+c)/2 (Formula 2 for area          -Heron’s formula) Perimeter = a + b + c Radius of incircle of a triangle of area A =AS where S is the semiperimeter =(a+b+c)/2
	Equilateral Triangle a = side	Area = (√3/4)*a*a               Perimeter = 3a Radius of incircle of an equilateral                                                                  triangle of side a = a/2*√3 Radius of circumcircle of an equilateral triangle of side a = a/√3
Base a is parallel to base b	Trapezium(Trapezoid in American English) h = height	Area = 12(a+b)h
	Circle r = radius d = diameter	d = 2r Area = πr2 = 14πd2 Circumference = 2πr = πd
	Sector of Circle r = radius θ = central angle	Area  = (θ/360) *π*r*r Arc Length, s = (θ/180)* π*r In the radian system for angular measurement, 2π radians = 360° => 1 radian = 180°π => 1° = π180 radians Hence, Angle in Degrees = Angle in Radians × 180°π Angle in Radians = Angle in Degrees × π180°
	Ellipse Major axis length = 2a Minor axis length = 2b	Area = πab Perimeter ≈
	Rectangular Solid l = length w = width h = height	Total Surface Area = 2lw + 2wh + 2hl = 2(lw + wh + hl) Volume = lwh
	Cube s = edge	Total Surface Area = 6s2 Volume = s3
	Right Circular Cylinder h = height r = radius of base	Lateral Surface Area = (2 π r)h Total Surface Area = (2 π r)h + 2 (π r2) Volume = (π r2)h
	Pyramid h = height B = area of the base	Total Surface Area = B +                 Sum of  the areas of the triangular sides Volume = 1/3*B*h
	Right Circular Cone h = height r = radius of base	Lateral Surface Area=πrs where s is the slant height =√r*r+h*h Total Surface Area                                 =πrs+πr2
	Sphere r = radius d = diameter	d = 2r Surface Area =4πr*r=πd*d Volume =4/3πr*r*r=16πd*d*d

 

• Important properties of Geometric Shapes
  1. Properties of Triangle
     1. Sum of the angles of a triangle = 180°
     2. Sum of any two sides of a triangle is greater than the third side.

• The line joining the midpoint of a side of a triangle to the positive vertex is called the median

1. The median of a triangle divides the triangle into two triangles with equal areas
2. Centroid is the point where the three medians of a triangle meet.
3. Centroid divides each median into segments with a 2:1 ratio

• Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
• An equilateral triangle is a triangle in which all three sides are equal

1. In an equilateral triangle, all three internal angles are congruent to each other
2. In an equilateral triangle, all three internal angles are each 60°
3. An isosceles triangle is a triangle with (at least) two equal sides

• In isosceles triangle, altitude from vertex bisects the base.

 

1. Properties of Quadrilaterals
2. Rectangle
   1. The diagonals of a rectangle are equal and bisect each other
   2. opposite sides of a rectangle are parallel

• opposite sides of a rectangle are congruent

1. opposite angles of a rectangle are congruent
2. All four angles of a rectangle are right angles
3. The diagonals of a rectangle are congruent
4. Square

• All four sides of a square are congruent
• Opposite sides of a square are parallel

1. The diagonals of a square are equal
2. The diagonals of a square bisect each other at right angles
3. All angles of a square are 90 degrees.

• A square is a special kind of rectangle where all the sides have equal length

1. Parallelogram

• The opposite sides of a parallelogram are equal in length.
• The opposite angles of a parallelogram are congruent (equal measure).

1. The diagonals of a parallelogram bisect each other.

• Each diagonal of a parallelogram divides it into two triangles of the same area

1. Rhombus

• All the sides of a rhombus are congruent
• Opposite sides of a rhombus are parallel.
• The diagonals of a rhombus bisect each other at right angles

1. Opposite internal angles of a rhombus are congruent (equal in size)

• Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
• If each angle of a rhombus is 90°, it is a square

Other properties of quadrilaterals

• The sum of the interior angles of a quadrilateral is 360 degrees
• If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
• A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
• Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
• Each diagonal of a parallelogram divides it into two triangles of the same area
• A square is a rhombus and a rectangle.

1. Sum of Interior Angles of a polygon
   3. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

 

 

Solved Examples

Level 1

1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
2. 4.04 %
3. 2.02 %
4. 4 %
5. 2 %

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

 

2. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
3. 30 %
4. 28 %
5. 32 %
6. 26 %

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100−20))/100
=(100×80)/100=80

Lost 10% of breadth
=> New breadth= (Original breadth × (100−10))/100
=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
5. 25 % Increase
6. 25 % Decrease
7. 50 % Decrease
8. 50 % Increase

Answer : Option D

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

5. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
6. 14 metres
7. 20 metres
8. 18 metres
9. 12 metres

Answer : Option B

Explanation:

lb = 460 m² ——(Equation 1)

Let the breadth = b
Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400⇒b=√400=20 m

 

6. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
7. equal to ½
8. equal to ¾
9. greater than 1
10. equal to 1

Answer : Option C

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

 

Hence greater than 1 is the more suitable choice from the given list

================================================================
Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a² …(1)

From the diagram, sin θ = h/a
=> h = a sin θ

Area of the rhombus = ah = a × a sin θ = a² sin θ …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a² /a²sinθ=1/sinθ

Since 0° < θ < 90°, 0 < sin θ < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

 

7. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
8. 37500 m²
9. 30500 m²
10. 32500 m²
11. 40000 m²

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length
⇒b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
⇒2(l+(3/5)* l)=800⇒l+(3/5)* l =400⇒(8/5)* l =400⇒l/5=50⇒l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

 

8. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
9. 45%
10. 44%
11. 40%
12. 42%

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

Increase in 20% of breadth
=> New breadth= (Original breadth × (100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

9. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
10. 814
11. 802
12. 836
13. 900

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm²

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902)  1517  (1

 

-902

—————–

615)  902  (1

 

• 615

————–

 

287)  615 (2

 

-574

—————–

 

41)  287  (7

 

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm²

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

 

Level 2

1. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
2. 126 sq. ft.
3. 64 sq. ft.
4. 100 sq. ft.
5. 102 sq. ft.

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

 

2. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
3. 400
4. 365
5. 385
6. 315

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700−x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
⇒x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

 

3. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
4. 18 cm
5. 16 cm
6. 40 cm
7. 20 cm

Answer : Option C

Explanation :

Let breadth = x cm
Then length = 2x cm
Area = lb = x × 2x = 2x²

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x² + 75
=> 2x² + 10x – 5x – 25 = 2x² + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

 

4. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
5. 142000
6. 112800
7. 142500
8. 153600

Answer : Option D

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860     (∵ 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m²

 

5. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
6. Rs.3500
7. Rs. 4200
8. Insufficient Data
9. Rs. 4400

Answer : Option C

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

 

6. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
7. 144√3−48π cm²
8. 121√3−36π cm²
9. 144√3−36π cm²
10. 121√3−48π cm²

Answer : Option A

Explanation :
Area of an equilateral triangle = (3/√4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral Δ ABC = (3/√4)*a *a = (3/√4)*24*24=144√3 cm2⋯ (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm² —-(2)

From (1) and (2),
144√3=36r⇒r=144√3/36=4√3−−−−(3)

Area of a circle = πr2 where = radius of the circle
From (3), the area of the inscribed circle = πr2=π(4√3)* (4√3)=48π⋯(4)

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
144√3−48π cm2

 

7. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
8. √11600 cm
9. √14400 cm
10. √10000 cm
11. √12040 cm

Answer : Option A

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC² = AB² + BC² = 40² + 80² = 1600 + 6400 = 8000
⇒AC = √8000 cm

Consider the right angled triangle ACG

AG² = AC² + CG²
(√8000) ²+60²=8000+3600=11600
=> AG = √11600 cm
=> Length of the longest rod = √11600cm

 

8. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
9. 30
10. 44
11. 56
12. 60

Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280/5 = 56

 

• Percentage

  Percent means for every 100

  So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

  percent is denoted by the symbol %. For example, x percent is denoted by x%

• x%=x/100

  Example : 25%=25/100=1/4

• To express x/y as a percent,we have x/y=(x/y×100)%

  Example : 1/4=(1/4×100)%=25%

• If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
• If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these	B. A is smaller than B.
C. Relationship between A and B cannot be determined.	D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these	B. 10% of a
C. 4% of a	D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1	B. 1 : 2
C. 1 : 1	D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A. 4%	B. 6%
C. 5%	D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57%	B. 50%
C. 52%	D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420	B. 700
C. 220	D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 %	B. 45 %
C. 500/11 %	D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023%	B. 20%
C. 21%	D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800	B. 2700
C. 2100	D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1^st candidate got 55% of the total valid votes.

Hence the 2^nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200	B. 7500
C. 7000	D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100	B. 102
C. 110	D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000	B. 30000
C. 32000	D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64%	B. 32%
C. 34%	D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375	B. Rs. 34,000
C. Rs. 82,150	D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
A. 8%	B. 7%
C. 10%	D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60%	B. 70%
C. 80%	D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

Fractions
Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc.

The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms.

• A fraction is unity, when its numerator and denominator are equal.
• A fraction is equal to zero if its numerator is zero.
• The denominator of a fraction can never be zero.
• The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4)
• When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5
• When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other.
• When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term.

Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc.

 

Improper Fraction:  A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc.

 

Like fraction: Fractions in which denominators are same is called like fractions.

e.g. 1/12, 5/12, 7/12, 13/12 etc.

 

Unlike fraction: Fractions in which denominators are not same is called, unlike fractions.

e.g. 1/12, 5/7, 7/9 13/11 etc.

 

Compound Fraction: Fraction of a fraction is called a compound fraction.

e.g. 1/2 of 3/4 is a compound fraction.

 

Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions.

 

Continued fraction: Fraction that contain additional fraction is called continued fraction.

e.g.

 

 

 

Rule: To simplify a continued fraction, begin from the bottom and move upwards.

 

Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc.

 

Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g.

 

 

Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal.

 

Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal.

 

Conversion of recurring decimal into proper fraction: 

CASE I: Pure recurring decimal

 

Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g.

 

CASE II: Mixed recurring decimal

In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g.

 
Questions

Level-I

1.	Evaluate : (2.39)2 – (1.61)2 2.39 – 1.61
	A. 2 B. 4 C. 6 D. 8

 

2.	What decimal of an hour is a second ?
	A. .0025 B. .0256 C. .00027 D. .000126

 

3.	The value of (0.96)3 – (0.1)3 is: (0.96)2 + 0.096 + (0.1)2
	A. 0.86 B. 0.95 C. 0.97 D. 1.06

 

4.	The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is: 0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04
	A. 0.0125 B. 0.125 C. 0.25 D. 0.5

 

5.	If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
	A. 0.172 B. 1.72 C. 17.2 D. 172
6.	When 0.232323….. is converted into a fraction, then the result is:
	A. 1 5 B. 2 9 C. 23 99 D. 23 100

 

7.	.009 = .01 ?
	A. .0009 B. .09 C. .9 D. 9

 

8.	The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
	A. 0.02 B. 0.2 C. 0.04 D. 0.4

 

9.	(0.1667)(0.8333)(0.3333) is approximately equal to: (0.2222)(0.6667)(0.1250)
	A. 2 B. 2.40 C. 2.43 D. 2.50

 

10.	3889 + 12.952 – ? = 3854.002
	A. 47.095 B. 47.752 C. 47.932 D. 47.95
11.	Level-II     0.04 x 0.0162 is equal to:
	A. 6.48 x 10-3 B. 6.48 x 10-4 C. 6.48 x 10-5 D. 6.48 x 10-6

 

12.	4.2 x 4.2 – 1.9 x 1.9 is equal to: 2.3 x 6.1
	A. 0.5 B. 1.0 C. 20 D. 22

 

13.	If 144 = 14.4 , then the value of x is: 0.144 x
	A. 0.0144 B. 1.44 C. 14.4 D. 144

 

 

 

14.	The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
	A. 2010 B. 2011 C. 2012 D. 2013

 

15.	Which of the following are in descending order of their value ?
	A. 1 , 2 , 3 , 4 , 5 , 6 3 5 7 5 6 7 B. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 7 6 7 C. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 6 7 7 D. 6 , 5 , 4 , 3 , 2 , 1 7 6 5 7 5 3
16.	Which of the following fractions is greater than 3 and less than 5 ? 4 6
	A. 1 2 B. 2 3 C. 4 5 D. 9 10

 

17.	The rational number for recurring decimal 0.125125…. is:
	A. 63 487 B. 119 993 C. 125 999 D. None of these

 

18.	617 + 6.017 + 0.617 + 6.0017 = ?
	A. 6.2963 B. 62.965 C. 629.6357 D. None of these

 

19.	The value of 489.1375 x 0.0483 x 1.956 is closest to: 0.0873 x 92.581 x 99.749
	A. 0.006 B. 0.06 C. 0.6 D. 6

 

20.	0.002 x 0.5 = ?
	A. 0.0001 B. 0.001 C. 0.01 D. 0.1

 

 

 

 

 

Answers

Level-I

Answer:1 Option B

 

Explanation:

Given Expression =	a2 – b2	=	(a + b)(a – b)	= (a + b) = (2.39 + 1.61) = 4.
	a – b		(a – b)

 

Answer:2 Option C

 

Explanation:

Required decimal =	1	=	1	= .00027
	60 x 60		3600

 

 

Answer:3 Option A

 

Explanation:

Given expression	= (0.96)3 – (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= a3 – b3 a2 + ab + b2
	= (a – b)   = (0.96 – 0.1)   = 0.86

Answer:4 Option B

 

Explanation:

Given expression =	(0.1)3 + (0.02)3	=	1	= 0.125
	23 [(0.1)3 + (0.02)3]		8

 

 

 

 

Answer:5 Option C

 

Explanation:

29.94	=	299.4
1.45		14.5

 

=		2994	x	1		[ Here, Substitute 172 in the place of 2994/14.5 ]
		14.5		10

 

=	172
	10

= 17.2

 

 

Answer:6 Option C

 

Explanation:

0.232323… = 0.23 =	23
	99

 

Answer:7 Option C

 

Explanation:

Let	.009	= .01;     Then x =	.009	=	.9	= .9
	x		.01		1

 

 

Answer:8 Option C

 

Explanation:

Given expression = (11.98)² + (0.02)² + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x   = 0.04

 

Answer:9 Option D

 

Explanation:

Given expression	= (0.3333) x (0.1667)(0.8333) (0.2222) (0.6667)(0.1250)
	= 3333 x 1 x 5 6 6 2222 2 x 125 3 1000
	= 3 x 1 x 5 x 3 x 8 2 6 6 2
	= 5 2
	= 2.50

 

Answer:10 Option D

 

Explanation:

Let 3889 + 12.952 – x = 3854.002.

Then x = (3889 + 12.952) – 3854.002

= 3901.952 – 3854.002

= 47.95.

 

Level-II

Answer:11 Option B

 

Explanation:

4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10^-4

 

Answer:12 Option B

 

Explanation:

Given Expression =	(a2 – b2)	=	(a2 – b2)	= 1.
	(a + b)(a – b)		(a2 – b2)

 

 

Answer:13 Option A

 

Explanation:

144	=	14.4
0.144		x

 

	144 x 1000	=	14.4
	144		x

 

x =	14.4	= 0.0144
	1000

 

 

Answer:14 Option B

 

Explanation:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

z =	2.50	=	250	= 10.
	0.25		25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

 

 

 

Answer:15 Option D

Answer:16 Option C

 

Explanation:

3	= 0.75,	5	= 0.833,	1	= 0.5,	2	= 0.66,	4	= 0.8,	9	= 0.9.
4		6		2		3		5		10

Clearly, 0.8 lies between 0.75 and 0.833.

	4	lies between	3	and	5	.
	5		4		6

 

 

 

Answer:17 Option C

 

Explanation:

0.125125… = 0.125 =	125
	999

 

 

Answer:18 Option C

 

Explanation:

617.00

6.017

0.617

+  6.0017

——–

629.6357

———

 

Answer:19 Option B

 

Explanation:

489.1375 x 0.0483 x 1.956		489 x 0.05 x 2
0.0873 x 92.581 x 99.749		0.09 x 93 x 100

 

=	489
	9 x 93 x 10

 

=	163	x	1
	279		10

 

=	0.58
	10

= 0.058  0.06.

 

Answer:20 Option B

 

Explanation:

2 x 5 = 10.

Sum of decimal places = 4

0.002 x 0.5 = 0.001

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

1. x km/hr = (x×5)/18 m/s

 

2. y m/s = (y×18)/5 km/hr

 

3. Speed = distance/time, that is, s = d/t

 

4. velocity = displacement/time, that is, v = d/t

 

5. Time taken by a train x meters long to pass a pole or standing man or a post
   = Time taken by the train to travel x meters.

 

6. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

7. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

8. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

9. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

10. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

11. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres	B. 160 metres
C. 200 metres Answer : Option C	D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec	B. 99 s
C. 89 s	D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s	B. 12 s
C. 9.8 s	D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr	B. 69 km/hr
C. 74 km/hr	D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3	B. 2 :1
C. 4 : 3	D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m	B. 190 m
C. 210 m	D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m	B. 10 a.m.
C. 9.10 a.m.	D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42	B. 36
C. 28	D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m	B. 245 m
C. 235 m	D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr	B. 180 km/hr
C. 152 km/hr	D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data	B. 3 : 1
C. 1 : 3	D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46	B. 36
C. 18	D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these	B. 280 meter
C. 240 meter	D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m	B. 54 m
C. 50 m	D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m	B. 360 m
C. 480 m	D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m	B. 500 m
C. 260 m	D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m	B. 2779 m
C. 1329 m	D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

 

Ratio

Introduction:

Ratio is the relation which one quantity bears to another of the same kind. The ratio of two quantities a and b is the fraction a/b and we write it as a: b.

In the ratio a: b, we call a as the first term or antecedent and b, the second term or consequent.

 

Note: The multiplication or division of each term of a ratio by the same non- zero number does not affect the ratio.

 

Compound Ratio: – It is obtained by multiplying together the numerators for new numerator and denominators for new denominator.

 

 

Example 1. If the ratios are 4:3, 15:20, 2:6 and 3:5 find the compound ratio?

 

 

 

 

 

 

 

 

Example2. If we divide 4185 into two parts such that they are in ratio 7:2, then find the values of both the parts?

Sol 2. Let the actual variable be 7x and 2x.

So, the 1^st part = 7 ×465=3255

The 2^nd part = 2 ×465=930

 

 

Note:

The ratio of first , second and third quantities is given by

ac : bc : bd

 

If the ratio between first and second quantity is a:b and third and fourth is c:d .

Similarly, the ratio of first, second, third and fourth quantities is given by
ace : bce : bde : bdf
If the ratio between first and second quantity is a: b and third and fourth is c:d.

 

                                                 Proportion

 
Introduction:-
Four quantities are said to be proportional if the two ratios are equal i.e.  the A, B, C and D are proportion. It is denoted by “::” it is written as A : B : C : D where A and D are extremes and B and C are called means .
                             Product of the extreme = Product of the means

 

 

Direct proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also increases (or decreases).

Example 1. If 5 pens cost Rs 10 then 15 pen cost?

Sol 1. It is seen that if number of pens increases then cost also increases. So,

5 pens: 15 pens:: Rs 10 : required cost

 

 

Inverse proportion: – The two given quantities are so related that if one quantity increases (or decreases) then the other quantity also decreases (or increases).

Example 2.If 10 men can do a work in 20 days then in how many days 20 men can do that work?

Sol 2. Here if men increase then days should decrease, so this is a case of inverse proportion, so

10 men: 20 men :: required days : 20 days

 

 

Rule of three: It Is the method of finding 4th term of a proportion if all the other three are given, if ratio is a:b :: c:d then ,

 

 

 

                                             ALLIGATION

Introduction:-

The word allegation means linking. It is used to find:

1. The proportion in which the ingredients of given price are mixed to produce a new mixture at a given price.
2. The mean or average value of mixture when the price of the two or more ingredients and the proportion in which they are mixed are given.

Mathematical Formula:

 

For two ingredient:-

 

 

Example 1: If the rice at Rs 3.20 per kg and the rice at Rs 3.50 per kg be mixed then what should be their proportion so that the new mixture be worth Rs 3.35 per kg ?

Sol 1: CP of 1 kg of cheaper rice                          CP of 1 kg of dearer rice

Hence they must be mixed in equal proportion i.e. 1:1

 

 

Example 2: Find out the ratio of new mixture so that it will cost Rs 1.40 per kg from the given three kinds of rice costing Rs 1.20, Rs 1.45 and Rs 1.74?

 

Sol 2: 1^st rice cost = 120, 2^nd rice cost = 145 and 3^rd rice cost = 174 paisa.

From the above rule: we have,

Therefore, three rice must be mixed in 39: 20: 20 ratios to have a new mixture of rice.

 

 

Questions

Level-I

..	1.   A  and B together have Rs. 1210. If  of A’s amount is equal to  of B’s amount, how much amount does B have?
	A. Rs. 460 B. Rs. 484 C. Rs. 550 D. Rs. 664

 

2.	Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
	A. 2 : 5 B. 3 : 5 C. 4 : 5 D. 6 : 7

 

 

3.	A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?
	A. Rs. 500 B. Rs. 1500 C. Rs. 2000 D. None of these

 

 

 

4.	Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
	A. 2 : 3 : 4 B. 6 : 7 : 8 C. 6 : 8 : 9 D. None of these

 

 

5.	In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:
	A. 20 litres B. 30 litres C. 40 litres D. 60 litres
6.	The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
	A. 8 : 9 B. 17 : 18 C. 21 : 22 D. Cannot be determined

 

7.	Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s salary?
	A. Rs. 17,000 B. Rs. 20,000 C. Rs. 25,500 D. Rs. 38,000

 

8.	If 0.75 : x :: 5 : 8, then x is equal to:
	A. 1.12 B. 1.2 C. 1.25 D. 1.30

 

 

9.	The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:
	A. 20 B. 30 C. 48 D. 58

 

 

	10 .If Rs. 782 be divided into three parts, proportional to  :  : , then the first part is:
	A. Rs. 182 B. Rs. 190 C. Rs. 196 D. Rs. 204

 

 

 

 

Answers

1. Answer:Option B

 

Explanation:

	4	A	=	2	B
	15			5

 

A =		2	x	15	B
		5		4

 

A =	3	B
	2

 

	A	=	3
	B		2

A : B = 3 : 2.

B’s share = Rs.		1210 x	2		= Rs. 484.
			5

 

 

 

 

2 .Answer: Option C

 

Explanation:

Let the third number be x.

Then, first number = 120% of x =	120x	=	6x
	100		5

 

Second number = 150% of x =	150x	=	3x
	100		2

 

Ratio of first two numbers =		6x	:	3x		= 12x : 15x = 4 : 5.

 

 

3 .Answer: Option C

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x – 3x = 1000

x = 1000.

B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

 

 

4 .Answer: Option A

 

Explanation:

 

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

 

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

 

		140	x 5x		,		150	x 7x		and		175	x 8x
		100					100					100

 

7x,	21x	and 14x.
	2

 

The required ratio = 7x :	21x	: 14x
	2

 

14x : 21x : 28x

 

2 : 3 : 4.

 

 

 

 

 

 

 

 

 

 

 

5 .Answer: Option D

 

Explanation:

Quantity of milk =		60 x	2	litres = 40 litres.
			3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

 

 

Then, milk : water =		40		.
		20 + x

 

Now,		40		=	1
		20 + x			2

 

20 + x = 80

 

x = 60.

Quantity of water to be added = 60 litres.

 

6 .Answer: Option C

 

Explanation:

 

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

 

Their increased number is (120% of 7x) and (110% of 8x).

 

		120	x 7x		and		110	x 8x
		100					100

 

	42x	and	44x
	5		5

 

The required ratio =		42x	:	44x		= 21 : 22

 

7 .Answer: Option D

 

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

 

Then,	2x + 4000	=	40
	3x + 4000		57

57(2x + 4000) = 40(3x + 4000)

 

6x = 68,000

 

3x = 34,000

 

Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

 

 

 

8 .Answer: Option B

 

Explanation:

(x x 5) = (0.75 x 8)    x =		6		= 1.20
		5

 

 

 

 

 

 

 

 

9 .Answer: Option B

 

Explanation:

Let the three parts be A, B, C. Then,

 

A : B = 2 : 3 and B : C = 5 : 8 =		5 x	3		:		8 x	3		= 3 :	24
			5					5			5

 

A : B : C = 2 : 3 :	24	= 10 : 15 : 24
	5

 

B =		98 x	15		= 30.
			49

 

 

 

10 .Answer: Option D

 

 

Explanation:

 

Given ratio =  :  :  = 6 : 8 : 9.

 

1st part = Rs.		782 x	6		= Rs. 204

 

 

 

 

Level-II

11.	The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
	A. 3 : 3 : 10 B. 10 : 11 : 20 C. 23 : 33 : 60 D. Cannot be determined     Answer: Option C   Explanation: Let A = 2k, B = 3k and C = 5k. A’s new salary = 115 of 2k = 115 x 2k = 23k 100 100 10   B’s new salary = 110 of 3k = 110 x 3k = 33k 100 100 10   C’s new salary = 120 of 5k = 120 x 5k = 6k 100 100    New ratio 23k : 33k : 6k = 23 : 33 : 60 10 10
12.	If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
	A. 2 : 5 B. 3 : 7 C. 5 : 3 D. 7 : 3 Answer: Option C   Explanation: Let 40% of A = 2 B 3   Then, 40A = 2B 100 3   2A = 2B 5 3   A = 2 x 5 = 5 B 3 2 3 A : B = 5 : 3.

 

13.	The fourth proportional to 5, 8, 15 is:
	A. 18 B. 24 C. 19 D. 20     Answer: Option B   Explanation: Let the fourth proportional to 5, 8, 15 be x. Then, 5 : 8 : 15 : x 5x = (8 x 15)   x = (8 x 15) = 24. 5

 

14.	Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:
	A. 27 B. 33 C. 49 D. 55 Answer: Option B   Explanation: Let the numbers be 3x and 5x. Then, 3x – 9 = 12 5x – 9 23 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.

 

15.	In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
	A. 50 B. 100 C. 150 D. 200 Answer: Option C   Explanation: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively. Then, sum of their values = Rs. 25x + 10 x 2x + 5 x 3x = Rs. 60x 100 100 100 100   60x = 30     x = 30 x 100 = 50. 100 60 Hence, the number of 5 p coins = (3 x 50) = 150.