Aptitude & reasoning

 

Introduction

Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.

Lender and Borrower

The person giving the money is called the lender and the person taking the money is the borrower.

Principal (sum)

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

Interest

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).

Amount (A)

The total of the sum borrowed and the interest is called the amount and is denoted by A

• The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.

 

• Let Principal = P, Rate = R% per annum and Time = T years. Then

  Simple Interest, SI = PRT/100

 

• From the above formula , we can derive the followings

  P=100×SI/RT

  R=100×SI/PT

  T=100×SI/PR

 

Some Formulae

1. If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
2. The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
3. If an amount P₁is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
   R=(P1R1+P2R2)/ (P1+P2)
4. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_nrespectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
5. If a certain sum of money P lent out for a certain time T amounts to P₁at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES

LEVEL 1

1.       Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%	B. 6%
C. 4%	D. 7%

Ans. Let rate = R%

Then, Time, T = R years

P = Rs.1400

SI = Rs.686

SI= PRT/100⇒686 = 1400 × R × R/100⇒686=14 Rx R ⇒49=Rx R ⇒R=7

i.e.,Rate of Interest was 7%. (D)

2.       How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years	B. 3 years
C. 1 year	D. 4 years

 

 

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)

3.       A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700	B. Rs. 690
C. Rs. 650	D. Rs. 698

 

 

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 – 117 = Rs.698 (D)

 

4.       A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
A. Rs. 2323	B. Rs. 1223
C. Rs. 2563	D. Rs. 2353

 

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)

5.       What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
A. 3 : 2	B. 1 : 3
C. 2 : 3	D. 3 : 1

 
Solution 1
Let Principal = P

Rate of Interest = R%

Required Ratio = (PR×5/100)/ (PR×15/100) =1:3 (B)
Solution 2

Simple Interest = PRT100

Here Principal(P) and Rate of Interest (R) are constants

Hence, Simple Interest ∝ T

Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)

6.       A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%	B. 12%
C. 8%	D. 5%

 

 

Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205

Simple Interest for 5 years = 22053×5=Rs.3675

Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125

R = 100×SI/PT=100×3675/(6125×5) =12% (B)

7.       A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%	B. 10%
C. 7%	D. 8%

 

 

Ans. Let the rate of interest per annum be R%

Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200

⇒5000×R×2/100+3000×R×4/100=2200

⇒100R + 120R=2200⇒220R=2200⇒R=10

i.e, Rate = 10%. (B)

8.       In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
A. 4 years	B. 6 years
C. 8 years	D. 9 years

 

 

Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years

150×6×n/100=800×4.5×2/100

150×6×n=800×4.5×2

n=8 years (C)

 

LEVEL 2

1.        Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400	B. Rs. 7200
C. Rs. 6500	D. Rs. 7500

 

 

Ans. Let the investment in scheme A be Rs.x

and the investment in scheme B be Rs. (13900 – x)

We know that SI = PRT/100

Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100

Total interest =Rs.3508

Thus, 28x/100+22(13900−x)/100 = 3508

28x+305800−22x=350800

6x = 45000

x=45000/6=7500

Investment in scheme B = 13900 – 7500 = Rs.6400 (A)

2.       A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
A. 45 years	B. 60 years
C. 40 years	D. 50 Years

 
Solution 1
Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

⇒ (400−x) = x×10×y/100

⇒ (400−x) = xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

⇒ (200−x) = x×4×y/100

⇒ (200−x) = xy/25— (equation 2)

(equation 1)/(equation2)

⇒(400−x) / (200−x) = (xy/10)/(xy/25)

⇒ (400−x)/ (200−x) =25/10

⇒ (400−x)/ (200−x) =52

⇒800−2x = 1000−5x

⇒200=3x

⇒x =200/3 Substituting this value of x in Equation 1, we get,

(400−200/3) = (200y/3)/10

⇒ (400−200/3) = 20y/3

⇒1200−200=20y

⇒1000=20y

y=1000/20=50 years (D)

Solution 2
If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 200

T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)

=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)

3.       Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
A. Rs. 25000	B. Rs. 15000
C. Rs. 10000	D. Rs. 20000

 

Ans. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by

R= (P1R1+P2R2)/(P1+P2)

P₁ = Rs. 12000, R₁ = 10%

P₂ =? R₂ = 20%

R = 14%

14 = (12000×10+P2×20)/ (12000+P2)

12000×14+14P2 =120000+20P2

6P2=14×12000−120000=48000

⇒P2=8000

Total amount invested = (P₁ + P₂) = (12000 + 8000) = Rs. 20000 (D)

4.       A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
A. Rs. 200	B. Rs. 600
C. Rs. 400	D. Rs. 500

 

 

Ans. This means, simple interest at 4% for that principal is Rs.120

P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)

5.       The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 1⁄2% rate is
A. Rs. 27.30	B. Rs. 22.50
C. Rs. 28.80	D. Rs. 29

 

 

Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year

Rate, R = 7.5%=15/2%

SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100

= 2730/100 = 27.30 (A)

6.       A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
A. Rs.1200	B. Rs.1400
C. Rs.2200	D. Rs.2800

Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_n respectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R1T1:1/R2T2:⋯1/RnTn

 

T₁ = 1 , T₂ = 2, T₃ = 3

R₁ = 5 , R₂ = 5, R₃ = 5

Share of Vikas : Share of Vijay : Share of Viraj

= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2

Total amount is Rs. 7700

Share of Vikas = 7700×6/11=700×6 = 4200

Share of Viraj = 7700×2/11=700×2=1400

Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)

 

7.       David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A. Rs.5000	B. Rs.2000
C. Rs.6000	D. Rs.3000

 

 

Ans. Let x, y and x be his investments in A, B and C respectively. Then

Then, Interest on x at 10% for 1 year

+ Interest on y at 12% for 1 year

+ Interest on z at 15% for 1 year

= 3200

x×10×1/100+y×12×1/100+z×15×1/100=3200

⇒10x+12y+15z=320000−−−(1)

Amount invested in Scheme C was 240% of the amount invested in Scheme B

=>z=240y/100 = 60y/25=12y/5−−−(2)

Amount invested in Scheme C was 150% of the amount invested in Scheme A

=>z=150x/100=3x/2

=>x=2z/3=2/3×12y/5=8y/5−−−(3)

From(1),(2) and (3),

10x + 12y + 15z = 320000

10(8y/5)+12y+15(12y/5)=320000

16y+12y+36y=320000

64y=320000

y=320000/64=10000/2=5000

i.e.,Amount invested in Scheme B = Rs.5000 (A)

 

 

Boat and stream problems is a sub-set of time, speed and distance type questions where in relative speed takes the foremost role. We always find several questions related to the above concept in SSC common graduate level exam as well as in bank PO exam. Upon listing the brief theory of the issue below we move to the various kinds of problems asked in the competitive examination.

Important Formulas – Boats and Streams

• Downstream
  In running/moving water, the direction along the stream is called downstream.
• Upstream
  In running/moving water, the direction against the stream is called upstream.

 

• Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then

  Speed downstream = (u+v) km/hr
  Speed upstream = (u – v) km/hr

 

• Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

  Speed in still water =1/2*(a+b)km/hr
  Rate of stream = 1/2*(a−b) km/hr

Some more short-cut methods

• Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey

  = (Speed downstream × Speed upstream)/Speed in still water=((x+y)(x−y))/xkm/hr

 

• Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance

  =((x* x –y* y)*t)/2ykm

 

• A man rows a certain distance downstream in t₁ hours and returns the same distance upstream in t₂ If the speed of the stream is y km/hr, then the speed of the man in still water

  =y((t2+t1) / (t2−t1)) km/hr

 

• A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places

  =t((x* x –y* y))/2xkm

 

• A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then

  x=y*((n+1)/(n−1))

 

 

Solved Examples

 

Level 1

 

1. A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:
A. 8.5 km/hr	B. 10 km/hr.
C. 12.5 km/hr	D. 9 km/hr

 

Answer : Option B

Explanation :

Man’s speed with the current = 15 km/hr

=>speed of the man + speed of the current = 15 km/hr

speed of the current is 2.5 km/hr

Hence, speed of the man = 15 – 2.5 = 12.5 km/hr

man’s speed against the current = speed of the man – speed of the current

= 12.5 – 2.5 = 10 km/hr

2. In one hour, a boat goes 14 km/hr along the stream and 8 km/hr against the stream. The speed of the boat in still water (in km/hr) is:
A. 12 km/hr	B. 11 km/hr
C. 10 km/hr	D. 8 km/hr

 

Answer : Option B

Explanation :

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr and Rate of stream =1/2(a−b) km/hr
Speed in still water = 1/2(14+8) kmph = 11 kmph.

3. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water?
A. 2 hr 30 min	B. 2 hr
C. 4 hr	D. 1 hr 15 min

 

Answer : Option A

Explanation :

Speed upstream = 2/2=1 km/hr

Speed downstream = 1/(20/60)=3 km/hr

Speed in still water = 1/2(3+1)=2 km/hr

Time taken to travel 5 km in still water = 5/2= 2 hour 30 minutes

4. Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is:
A. 700 hours	B. 350 hours
C. 1400 hours	D. 1010 hours

 

Answer : Option A Explanation : Speed downstream = (14 + 1.2) = 15.2 kmph Speed upstream = (14 – 1.2) = 12.8 kmph Total time taken = 4864/15.2+4864/12.8 = 320 + 380 = 700 hours

 

5. The speed of a boat in still water in 22 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 24 minutes is:
A. 9.4 km	B. 10.2 km
C. 10.4 km	D. 9.2 km

 

Answer : Option C

Explanation :

Speed downstream = (22 + 4) = 26 kmph

Time = 24 minutes = 24/60 hour = 2/5 hour

distance travelled = Time × speed = (2/5)×26 = 10.4 km

6. A boat covers a certain distance downstream in 1 hour, while it comes back in 11⁄2 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
A. 14 kmph	B. 15 kmph
C. 13 kmph	D. 12 kmph

 

Answer : Option B

Explanation :

Let the speed of the boat in still water = x kmph

Given that speed of the stream = 3 kmph

Speed downstream = (x+3) kmph

Speed upstream = (x-3) kmph

He travels a certain distance downstream in 1 hour and come back in 1¹⁄₂ hour.

ie, distance travelled downstream in 1 hour = distance travelled upstream in 1¹⁄₂ hour

since distance = speed × time, we have

(x+3)×1=(x−3)*3/2

=> 2(x + 3) = 3(x-3)

=> 2x + 6 = 3x – 9

=> x = 6+9 = 15 kmph

7. A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 54 km downstream
A. 5 hours	B. 4 hours
C. 3 hours	D. 2 hours

 

Answer : Option D

Explanation :

Speed of the boat in still water = 22 km/hr

speed of the stream = 5 km/hr

Speed downstream = (22+5) = 27 km/hr

Distance travelled downstream = 54 km

Time taken = distance/speed=54/27 = 2 hours

 

8. A boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water?
A. 5 kmph	B. 4.95 kmph
C. 4.75 kmph	D. 4.65

 

Answer : Option B

Explanation :

Speed downstream = 22/4 = 5.5 kmph

Speed upstream = 22/5 = 4.4 kmph

Speed of the boat in still water = (½) x (5.5+4.42) = 4.95 kmph

9. A man takes twice as long to row a distance against the stream as to row the same distance in favor of the stream. The ratio of the speed of the boat (in still water) and the stream is:
A. 3 : 1	B. 1 : 3
C. 1 : 2	D. 2 : 1

 

Answer : Option A

Explanation :

Let speed upstream = x

Then, speed downstream = 2x

Speed in still water = (2x+x)2=3x/2

Speed of the stream = (2x−x)2=x/2

Speed of boat in still water: Speed of the stream = 3x/2:x/2 = 3 : 1

 

Level  2

1. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
A. 10	B. 6
C. 5	D. 4

 

Answer : Option C

Explanation :

Speed of the motor boat = 15 km/hr

Let speed of the stream = v

Speed downstream = (15+v) km/hr

Speed upstream = (15-v) km/hr

Time taken downstream = 30/(15+v)

Time taken upstream = 30/(15−v)

total time = 30/(15+v)+30/(15−v)

It is given that total time is 4 hours 30 minutes = 4.5 hour = 9/2 hour

i.e., 30/(15+v)+30/(15−v)=9/2

⇒1(15+v)+1(15−v)=(9/2)×30=3/20

⇒(15−v+15+v)/(15+v)(15−v)=3/20

⇒30/(15*15−v*v)=3/20

⇒30/(225−v*v)=3/20

⇒10/(225−v* v)=1/20

⇒225−v* v =200

⇒v* v =225−200=25

⇒v=5 km/hr

2. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
A. 1 km/hr.	B. 2 km/hr.
C. 1.5 km/hr.	D. 2.5 km/hr.

 

Answer : Option A

Explanation :

Assume that he moves 4 km downstream in x hours

Then, speed downstream = distance/time=4/x km/hr

Given that he can row 4 km with the stream in the same time as 3 km against the stream

i.e., speed upstream = 3/4of speed downstream=> speed upstream = 3/x km/hr

He rows to a place 48 km distant and come back in 14 hours

=>48/(4/x)+48/(3/x)=14

==>12x+16x=14

=>6x+8x=7

=>14x=7

=>x=1/2

Hence, speed downstream = 4/x=4/(1/2) = 8 km/hr

speed upstream = 3/x=3/(1/2) = 6 km/hr

Now we can use the below formula to find the rate of the stream

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2*(a+b) km/hr

Rate of stream =12*(a−b) km/hr
Hence, rate of the stream = ½*(8−6)=1 km/hr

3. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?
A. 5 : 6	B. 6 : 5
C. 8 : 3	D. 3 : 8

 

Answer : Option C

Explanation :

Let the rate upstream of the boat = x kmph

and the rate downstream of the boat = y kmph

Distance travelled upstream in 8 hrs 48 min = Distance travelled downstream in 4 hrs.

Since distance = speed × time, we have

x×(8*4/5)=y×4

x×(44/5)=y×4

x×(11/5)=y— (equation 1)

Now consider the formula given below

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr

Rate of stream =1/2(a−b) km/hr
Hence, speed of the boat = (y+x)/2

speed of the water = (y−x)/2

Required Ratio = (y+x)/2:(y−x)/2=(y+x):(y−x)=(11x/5+x):(11x/5−x)

(Substituted the value of y from equation 1)

=(11x+5x):(11x−5x)=16x:6x=8:3

 

4. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?
A. 3.2 km	B. 3 km
C. 2.4 km	D. 3.6 km

 

Answer : Option C

Explanation :

Speed in still water = 5 kmph
Speed of the current = 1 kmph

Speed downstream = (5+1) = 6 kmph
Speed upstream = (5-1) = 4 kmph

Let the requited distance be x km

Total time taken = 1 hour

=>x/6+x/4=1

=> 2x + 3x = 12

=> 5x = 12

=> x = 2.4 km

5. A man can row three-quarters of a kilometer against the stream in 111⁄4 minutes and down the stream in 71⁄2minutes. The speed (in km/hr) of the man in still water is:
A. 4 kmph	B. 5 kmph
C. 6 kmph	D. 8 kmph

 

Answer : Option B

Explanation :

Distance = 3/4 km

Time taken to travel upstream = 11¹⁄₄ minutes

= 45/4 minutes = 45/(4×60) hours = 3/16 hours

Speed upstream = Distance/Time= (3/4)/ (3/16) = 4 km/hr

Time taken to travel downstream = 7¹⁄₂minutes = 15/2 minutes = 15/2×60 hours = 1/8 hours

Speed downstream = Distance/Time= (3/4)/ (1/8) = 6 km/hr

Rate in still water = (6+4)/2=10/2=5 kmph

6. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:
A. 4 mph	B. 2.5 mph
C. 3 mph	D. 2 mph

 

Answer : Option D

Explanation :

Speed of the boat in still water = 10 mph

Let speed of the stream be x mph

Then, speed downstream = (10+x) mph

speed upstream = (10-x) mph

Time taken to travel 36 miles upstream – Time taken to travel 36 miles downstream= 90/60 hours

=>36/(10−x)−36/(10+x)=3/2=>12/(10−x)−12/(10+x)=1/2=>24(10+x)−24(10−x)=(10+x)(10−x)

=>240+24x−240+24x=(100−x* x)=>48x=100− (x* x)=> x* x +48x−100=0

=>(x+50)(x−2)=0=>x = -50 or 2; Since x cannot be negative, x = 2 mph

7. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24-mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
A. 2*1/3 mph	B. 1*1/3 mph
C. 1*2/3 mph	D. 2*2/3 mph

 

Answer : Option D

Explanation :

Let the speed of Rahul in still water be x mph
and the speed of the current be y mph

Then, Speed upstream = (x – y) mph
Speed downstream = (x + y) mph

Distance = 12 miles

Time taken to travel upstream – Time taken to travel downstream = 6 hours

⇒12/(x−y)−12/(x+y)=6

⇒12(x+y)−12(x−y)=6(x*x−y*y)

⇒24y=6(x*x−y*y)

⇒4y= x*x−y*y

⇒x * x =(y* y +4y)⋯(Equation 1)

Now he doubles his speed. i.e., his new speed = 2x

Now, Speed upstream = (2x – y) mph

Speed downstream = (2x + y) mph

In this case, Time taken to travel upstream – Time taken to travel downstream = 1 hour

⇒12/(2x−y)−12/(2x+y)=1

⇒12(2x+y)−12(2x−y)=4*x* x –y* y

⇒24y=4*x* x –y* y

⇒4*x* x = y* y +24y⋯(Equation 2)

(Equation 1 × 4)⇒4x* x =4(y* y +4y)⋯(Equation 3)

(From Equation 2 and 3, we have)

y* y +24y=4(y* y +4y)⇒y* y +24y=4y* y +16y⇒3y* y =8y⇒3y=8

y=8/3 mphi.e., speed of the current = 8/3 mph=2*2/3 mph