KRISHNA MOHAN: Quantitative Aptitude

Q1. A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?

(a) A and B

(b) A and C

(d) A and D

(e) C and D

Ans.(d)

Sol. Work done in one day by A, B and C are 1/4,1/8,1/16 and 1/32 respectively.

Using answer choices, we note that the pair of B and C does 3/16 of work in one day; the pair of A and D does 1/4+1/32=9/32 of work in one day

Hence, A and D take 32/9 days

B and C take 16/3=32/6 days

Hence, the first pair must comprise of A and D

Directions (2-6): Read the passage below solve the questions based on it.

There are infinite pipes attached to a very big tank. Pipes are numbered like P₁,P₂,P₃………… and so on. Its also known that efficiency of every subsequent pipe is half the efficiency of earlier pipe, i.e., efficiency of P₂ pipe is half the efficiency of P₁ pipe and so on. However, in case of P₁ pipe, this rule is not true since there is no earlier pipe. When all the pipes are working together, the tank gets filled in 2 hours.

Q2. How much time will P₄ take of fill the tank working alone?

(a) 8 h

(b) 16 h

(d) 4 h

(e) Cannot be determined

Ans.(c)

Sol. Pipe P_4 will take 32 hours. Hence, option (c) is the answer.

When all the pipes are used together, tank gets filled in 2 hours. Hence in one hour, tank get ½ = 50% filled.

Assume that pipe P_1 fills x% of the tank in one hour.

So, pipe P_2 fills x/2% of the tank in one hour; pipe P_3 fills x/4% of the tank in one hour and so on.

According to the question,

x+x/2+x/4+ ………………………….. till infinity = 50%

This is an infinite GP.

Sum of infinite GP = a/(1 - r)=x/(1 - 1/2) = 2x

Hence 2x = 50% ⇒ x = 25%

So, 1st tank fills 25% of the tank in One hour. Hence 1st tank fills the tank in 100/25 = 4 hrs.

So, 2nd pipe fills the tank in 8 hours; 3rd pipe fills the tank in 16 hours and so on.

Q3. What is difference in the time taken by P₅ and P₆?

(a) 32 h

(b) 64 h

(d) 16 h

(e) Cannot be determined

Ans.(b)

Sol. Pipe P₅ will take 64 hours and Pipe P₆ will take 128 hours.

Hence difference in the time taken = 64 hours. Hence, option (b) is the answer.

When all the pipes are used together, tank gets filled in 2 hours. Hence in one hour, tank get ½ = 50% filled.
Assume that pipe P_1 fills x% of the tank in one hour.
So, pipe P₂ fills x/2% of the tank in one hour; pipe P₃ fills x/4% of the tank in one hour and so on.
According to the question,
x+x/2+x/4+ ………………………….. till infinity = 50%
This is an infinite GP.
Sum of infinite GP = a/(1 - r)=x/(1 - 1/2) = 2x
Hence 2x = 50% ⇒ x = 25%
So, 1st tank fills 25% of the tank in One hour. Hence 1st tank fills the tank in 100/25 = 4 hrs.

So, 2nd pipe fills the tank in 8 hours; 3rd pipe fills the tank in 16 hours and so on.

Q4. How many pipes can fill the tank within 100 hours working alone?

(a) 4

(b) 5

(d) 7

(e) Infinite

Ans.(b)

Sol. From pipe P1 to pipe P5, each of the pipes can fill the tank independently within 100 hours. Hence 5 pipes can till the tank independently within 100 hours. Hence, option (b) is the answer.

When all the pipes are used together, tank gets filled in 2 hours. Hence in one hour, tank get ½ = 50% filled.

Assume that pipe P_1 fills x% of the tank in one hour.

So, pipe P2 fills x/2% of the tank in one hour; pipe P3 fills x/4% of the tank in one hour and so on.

According to the question,

x+x/2+x/4+ ………………………….. till infinity = 50%

This is an infinite GP.

Sum of infinite GP = a/(1 - r)=x/(1 - 1/2) = 2x

Hence 2x = 50% ⇒ x = 25%

So, 1st tank fills 25% of the tank in One hour. Hence 1st tank fills the tank in 100/25 = 4 hrs.

So, 2nd pipe fills the tank in 8 hours; 3rd pipe fills the tank in 16 hours and so on.

Q5. Due to some technical problem, only four pipes P₁,P₂,P₃and P₄ are in working condition. These pipes are now paired up to obtain two pairs of pipes. Now it is found that one pair of pipe is taking 2/3rd of the time taken by the other pair to fill the tank independently. Which of the following is one of the two pairs?

(a) P₁,P₂

(b) P₁,P₃

(d) P₂,P₄

(e) Cannot be determined

Ans.(c)

Sol. Efficiency of pipe P1 = 25%, Efficiency of pipe P2= 12.5%, Efficiency of pipe P3 = 6.25%, Efficiency of pipe P4 = 3.12%. If we club (pipe P1 and pipe P4) in one group and (pipe P2 and pipe P3) in other group, it satisfies the condition. Hence, option (c) is the answer.

When all the pipes are used together, tank gets filled in 2 hours. Hence in one hour, tank get ½ = 50% filled.
Assume that pipe P1 fills x% of the tank in one hour.
So, pipe P2 fills x/2% of the tank in one hour; pipe P3 fills x/4% of the tank in one hour and so on.
According to the question,
x+x/2+x/4+ ………………………….. till infinity = 50%
This is an infinite GP.
Sum of infinite GP = a/(1 - r)=x/(1 - 1/2) = 2x
Hence 2x = 50% ⇒ x = 25%
So, 1st tank fills 25% of the tank in One hour. Hence 1st tank fills the tank in 100/25 = 4 hrs.

So, 2nd pipe fills the tank in 8 hours; 3rd pipe fills the tank in 16 hours and so on.

Q6. How much time will P₂ take to fill the tank working alone?

(a) 4

(b) 8

(d) 32

(e) 12

Ans.(b)

Sol. Pipe P2 will take 8 hours in filling the tank. Hence, option (b) is the answer.

Q7. Ram finishes a work in 7 days. Rahim finishes the same job in 8 days and Robert in 6 days. They take turns to finish the work. Ram worked on the first day, Rahim on the second day and Robert on the third and then again Ram and so on. Who was working on the last day when work got finished?

(a) Ram

(b) Rahim

(d) Rahim and Robert

(e) Cannot be determined

Ans.(a)

Sol. Three day’s work = 1/7+1/8+1/6=73/168

Six day’s work = 73/84

Seventh day work = 1/7, done by Ram

Since 73/84+1/7=85/84 > 1, therefore, Ram was working on the last day.

Q8. Construction of a road was entrusted to a civil engineer. He has to finish the work in 124 days for which he employed 120 workers. Two-third of the work was completed in 64 days. How many workers can be reduced now without affecting the completion of the work on time?

(a) 56

(b) 64

(d) 24

(e) None of these

Ans.(a)

Sol. 2/3rd of the work was completed in 64 days by 120 workers.

1/3rd of the work was completed in 32 days by 120 workers.

Also 1/3rd of the work is to be completed in 60 days by (120 – x) workers, where x is the number of men reduced in order to finish the work on schedule.

So, (120 – x) × 60 = 120 ⇒ x = 56

Q9. Two workers earned Rs. 225 first worked for 10 days and the second for 9 days. How much did each of them get daily if the first worker got Rs. 15 more for working 5 days than the second worker got for working 3 days?

(a) Rs. 11.70; Rs. 12.00

(b) Rs. 10.80; Rs. 13.00

(d) Rs. 12.60; Rs. 11.00

(e) None of these

Ans.(b) Sol. Let A got Rs. x per day and B got Rs. y per day. So, 10x + 9y = 225 and 5x = 3y + 15 ⇒ x = 10.80, y = 13.

Q10. Two pipes A and B can fill a tank in 20 and 30 h respectively. Both the pipes are opened to fill the tank but when the tank is 1/3 full, a leak develops in the tank. Due to this leakage one-third of the water supplied by pipes A and B goes waste. What is the total time to fill the tank if the leak if not closed?

(a) 12 h

(b) 16 h

(d) 20 h

(e) None of these

Ans.(b)

Sol. Let us assume total work = 180 (we are not assuming it to be LCM of 20 and 30 = 60 because in that case 1/3rd of A + B will be fractional)

Time taken to fill 1/3rd of the tank = 180/ (9 + 6) = 4 h

Due to leakage, net inflow = 2/3 (9 + 6) = 10 units

Time taken to fill remaining 120 units = 12 h

So total time taken = 12 + 4 = 16 h

Q11. In how many days will 10 men finish the job? To answer the question which of the following information(s) is/are sufficient (assume equal efficiency of work)?

A. 20 men can finish the same job in 20 days.

B. 40 men can finish the job in 1/4 time of what 10 men take.

C. 50 men can finish double the job in 16 days.

(a) Only (A) alone is sufficient

(b) Only (C) alone is sufficient

(d) (A) and (B) together or (C) alone is sufficient

(e) (A) or (C) alone is sufficient

Ans.(e)

Sol. (A) alone: (Use M1 D1=M2 D2)

10 × N = 20 × 20 ⇒ N = 40 days.

(C) alone: (Use M1 D1 W2=M2 D2 W1)

10 × N × 2 = 50 × 16 × 1 ⇒ N = (50 × 16)/20 = 40 days

Statement B provides the same information as given in the question.

Q12. In how many days will 22 men finish the work?

A. 6 women and 8 men can finish a work in 10 days.

B. 6 women work as much as 3 men in the same time.

C. Each women takes twice more time taken by a man to finish the work.

(a) A alone is sufficient

(b) A and B together are only sufficient

(d) All A, B and C together are necessary

(e) All even together are not sufficient

Ans.(c)

Sol. 2 women = 1 man

(6 women + 8 men) = 3 + 8) men = 11 men

⇒ (11 × 10)/12 = 12 days

Q13. Amit can do a job in 9 days. To find the number of days in which Gagan can do the same job, which of the following information(s) is/are necessary / sufficient?

P. Gagan is 50% more efficient than Amit.

Q. Amit and Gagan together can do the job in 3(3/5) days.

R. Amit is 33(1/3)% less efficient than Gagan.

(a) Only P alone is sufficient

(b) Only q alone is sufficient

(d) Any one of the three statements is sufficient

(e) All the three statements are necessary

Ans.(d)

Sol. Since Ganga is 50% more efficient than Amit, he will do the job in 9

(100/(100+50))=9 2/3 = 6 days

From statement Q:

Ganga will do the job in (18/5×9)/(9-18/5)=(18×9)/(45-18)

=(18×9)/27=6 days

From statement R:

Ganga will do the job in ((100-100/3)/10)=9(200/300)

= 6 days

Directions (14-15): Read the passage below and solve the questions based on it.

Tank at a water supply station is filled with water by several pumps. At first, three pumps of the same capacity are turned on; 2.5 h later, two more pumps (both the same) of a different capacity are set into operation. After 1 h, the additional pumps were set into operation; the tank was almost filled to its capacity (15 m^3 were still lacking); in another hour the tank was full. One of the two additional pumps could have filled the tank in 40 h.

Q14. What is the volume of the tank?

(a) 60 m^3

(b) 80 m^3

(d) 90 m^3

(e) None of these

Ans.(a)

Sol. Given that, in the last hour, tank lacks 15 m^3. Hence 25% of the tank = 15m^3. Hence 100% of the tank = 60m^3.

Hence, option (a) is the answer.

Assume that the three pumps initially put are named pump A and two pumps added later on are named pump B.

Assume the percentage of tank filled in by one pump A = y% per hour.

Hence total percentage of tank filled in three pumps in one hour = 3y% per hour.

Hence total percentage of tank filled in three pumps in 2.5 hours = 2.5 × 3y% = 7.5y%

After 2.5 hours, 2 more pumps (both same) of a different capacity are set into operation. Given that “one of the two additional pumps could have filled the tank in 40 hours”. Hence percentage of tank fill in by one pump B = (100%)/40 = 2.5% per hour.

Hence total percentage of tank filled by two pumps B in one hour = 2 × 2.5 % per hour = 5% per hour.

Hence total percentage of tank filled by two pumps B in two hours = 2 × 5% per hour = 10 % per hour.

So, 90% of the tank is filled in by pipes A in (2.5 + 1 + 1) hrs. = 4.5 hrs.

So, in One hour, percentage of tank filled by 3pipes A = (90%)/4.5 = 20% per hour

Hence in one hour, total percentage of tank filled by (3 pipes A and 2 pipes B) together = 20% + 5% = 25%

Q15. How much water does one of the first three pumps emit in an hour?

(a) 5 m^3

(b) 4 m^3

(d) 2 m^3

(e) None of these

Ans.(b)

Sol. In One hour, percentage of tank filled by 3 pipes

A = (90%)/4.5 = 20% per hour

Hence, in One hour, percentage of tank filled by 1 pipe

A = (20%)/3 per hour

Since 100% of the tank = 60m^3 ⇒ 1% of the tank = 0.6 m^3

⇒ (20%)/3 of the tank = 20/3 × 0.6 = 4 m^3.

Hence, option (b) is the answer.