Time and Work – Quantitative Aptitude Question

Time and Work – Quantitative Aptitude Question

Time and Work – Quantitative Aptitude Question

Question 1:

A can do a work in 10 days and B can do the same work in 15 days. How long will they take to complete the work if both work together?

Options: A) 6 days
B) 5 days
C) 8 days
D) 12 days

Answer: A) 6 days

Explanation:

  • A’s work rate = 1/10 (work done per day)
  • B’s work rate = 1/15 (work done per day)

Together, they can complete (1/10 + 1/15) of the work per day.

LCM of 10 and 15 = 30, so:

  • A’s rate = 3/30
  • B’s rate = 2/30

Together, they complete (3 + 2) = 5/30 of the work per day. Thus, they will take 30/5 = 6 days to complete the work.


Question 2:

If 8 men can do a piece of work in 12 days, how long will it take for 6 men to do the same work?

Options: A) 14 days
B) 16 days
C) 18 days
D) 20 days

Answer: B) 16 days

Explanation: The work done is inversely proportional to the number of men. So, the relationship can be represented as:Men1×Days1=Men2×Days2

Substituting the given values:8×12=6×Days2 96=6×Days2​ Days2=96/6=16 days


Question 3:

A pipe can fill a tank in 8 hours, and another pipe can empty the tank in 12 hours. How long will it take to fill the tank if both pipes are open?

Options: A) 24 hours
B) 20 hours
C) 48 hours
D) 48/5 hours

Answer: D) 48/5 hours

Explanation:

  • First pipe’s rate = 1/8 (fills the tank in 8 hours)
  • Second pipe’s rate = -1/12 (empties the tank in 12 hours)

When both pipes are open, the combined rate of filling the tank is:18−1/12

LCM of 8 and 12 = 24:1/8=3/24,1/12=2/24

So, combined rate = 3/24−2/24=1/24

Thus, the time taken to fill the tank = 24 hours.


Question 4:

A and B together can complete a work in 20 days. If A works alone for 10 days and then B completes the remaining work in 15 days, how long would A take to complete the work alone?

Options: A) 30 days
B) 40 days
C) 50 days
D) 60 days

Answer: B) 40 days

Explanation: Let A’s rate be 1/x and B’s rate be 1/y, where x is the number of days A takes and y is the number of days B takes to complete the work alone.

From the given information:

  • A and B together take 20 days, so 1x+1y=1/20
  • A works for 10 days, so the work completed by A is 10/x
  • B works for 15 days, so the work completed by B is 15/y

The total work completed is 1, so:10/x+15/y=1

From 1x+1y=1/20 we substitute 1/y=1/20−1/x into the second equation and solve for x, finding that x=40

Thus, A alone will take 40 days to complete the work.


Question 5:

A and B can complete a work in 18 days, B and C in 24 days, and A and C in 30 days. How many days will it take for A, B, and C to complete the work together?

Options: A) 12 days
B) 15 days
C) 10 days
D) 20 days

Answer: B) 15 days

Explanation: Let the rates of A, B, and C be 1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c}a1​,b1​,c1​, where a, b, and c represent the number of days each person takes to complete the work alone.

The given equations are:1a+1b=118,1b+1c=124,1a+1c=130\frac{1}{a} + \frac{1}{b} = \frac{1}{18}, \quad \frac{1}{b} + \frac{1}{c} = \frac{1}{24}, \quad \frac{1}{a} + \frac{1}{c} = \frac{1}{30}a1​+b1​=181​,b1​+c1​=241​,a1​+c1​=301​

Adding all three equations:(1a+1b+1b+1c+1a+1c)=118+124+130\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{c} \right) = \frac{1}{18} + \frac{1}{24} + \frac{1}{30}(a1​+b1​+b1​+c1​+a1​+c1​)=181​+241​+301​

Simplifying the left side:2(1a+1b+1c)=118+124+1302 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{1}{18} + \frac{1}{24} + \frac{1}{30}2(a1​+b1​+c1​)=181​+241​+301​

The right-hand side is simplified by finding the LCM:118+124+130=560+2.560+260=9.560=19120\frac{1}{18} + \frac{1}{24} + \frac{1}{30} = \frac{5}{60} + \frac{2.5}{60} + \frac{2}{60} = \frac{9.5}{60} = \frac{19}{120}181​+241​+301​=605​+602.5​+602​=609.5​=12019​

Thus:1a+1b+1c=19240\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{19}{240}a1​+b1​+c1​=24019​

The time taken by A, B, and C together is the reciprocal of 19240\frac{19}{240}24019​, which is 15 days.

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