Time and Work Quiz Set 010

Question 1

A, B and C can independently complete a work in 5, 15 and 13 days respectively. B and C start the work together, but A joins them after 2 days. In how many days will the work be completed?

A

\$4{5/67}\$ days.

B

\$5{5/67}\$ days.

C

\$6{5/67}\$ days.

D

\$7{5/67}\$ days.

Soln.
Ans: a

Use the shortcut formula. If A, B, C can independently complete the job in x, y and z days, and A joins after n days, the work is completed in \${xyz}/{xy + yz + zx}\$ × \$(1 + n/x)\$ days. Putting the various values x = 5, y = 15, z = 13, n = 2, and simplifying, we get \${273/67}\$, which is same as: \$4{5/67}\$.

Question 2

A and B can complete a job in 16 and 64 days. They start together but A leaves after working for 6 days. How long would B take to finish the job counting from the day both A and B started together?

A

40 days.

B

41 days.

C

39 days.

D

43 days.

Soln.
Ans: a

If B takes x days. The total job is A's work in 4 days + B's work in x days = \$6/16\$ + \$x/64\$ = 1. Solving, we get x = 40 days.

Question 3

6 men and 4 women finish a job in 16 days. In how many days will 8 women and 12 men finish that job?

A

8.

B

7.

C

9.

D

10.

Soln.
Ans: a

Since the work force is being doubled proportionately, the time is halved = 8 days.

Question 4

A can do a piece of work in 18 days. B is 20% more efficient than A. In how many days will they complete the work if they work together?

A

\$8{2/11}\$ days.

B

\$8{3/11}\$ days.

C

\$8{4/11}\$ days.

D

\$8{5/11}\$ days.

Soln.
Ans: a

Let us first calculate the one day work of B. One day work of A is given as \$1/18\$. If B is 20% efficient, then one day work of B is \$1/18\$ × \$120/100\$ = \$1/15\$. Putting x = 18 and y = 15 in the shortcut method, we get \${xy}/{x + y}\$ = \${90/11}\$, which is same as: \$8{2/11}\$.

Question 5

If 80 men can do a task in 16 days, how many men are required to complete the task in 10 days?

A

128.

B

11.

C

9.

D

13.

Soln.
Ans: a

If m1 men can do a task in d1 days, and m2 in d2, then we must have m1 × d1 = m2 × d2. Putting m1 = 80, d1 = 16 and d2 = 10, we get m2 = 128. 