1. A can do a work in 20 days and B in 30 days. In how many days they can complete the work if they work together.

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par A's one day's work =  \frac{1}{20}

\par B's one day's work =  \frac{1}{30}

\par (A + B)'s one day work = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}

\par A and B can complete the work in 12 days

 

\par \underline{Shortcut Method}

 

\par If A can do a work in t_{1} \par days and B can do it in t_{2}$ days

\par then A and B Together can do same work in \frac{t_{1 \times } t _ {2}}{t_{1}+ t_{2}} days.

\frac{20 \times 30}{20+50} = \frac{600}{50} = 12

2. A, B and C complete a work in 6, 8 and 12 days respectively. In how many days they can complete the work together.

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par A's one day's work =  \frac{1}{6}

\par B's one day's work =  \frac{1}{8}

\par C's one day's work =  \frac{1}{12}

\par (A + B + C)'s one day work = \frac{1}{6} + \frac{1}{8}+\frac{1}{12} = \frac{4+3+2}{24} = \frac{9}{24} = \frac{3}{8}

\par (A+B+C) will complete the work =  1\div \frac{3}{8} = \frac{8}{3} = 2\frac{2}{3}

\par \underline{Shortcut Method}

\frac{6\times 8\times 12}{6\times 8+6\times 12+8\times 12} = \frac{8}{3} = 2\frac{2}{3}                                          \lbrack \frac{t_{1 \times }t_{2 \times }t_{3}}{t_{1} t_{2}+t_{1} t_{3}+t_{2}t_{3}} \rbrack

3. A and B together complete a work in 12 days. B alone can complete the work in 28 days. A alone can do the work in how many days.

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par (A+B)'s one day's work = \frac{1}{12}

\par B's one day work = \frac{1}{28}                                                                       \lbrack (A+B)- (B) = A\rbrack

\par A's one day work = \frac{1}{12} - \frac{1}{28} = \frac{7-3}{84} = \frac{4}{84} = \frac{1}{21}

\par So, A can do it in 21 days

\par \underline{Shortcut Method}

\frac{12\times 28}{28-12} = \frac{336}{16} = 21                                                                                 t_{2} = \frac{T \times t_{1}}{t_{1}-T}

4. A, B complete a work in 25, 20 days respectively. after working together for 5 days A left. B complete the work in

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par (A+B)'s 5 days work =  5\left [\frac{1}{25}+\frac{1}{20}\right] =  5 \left [\frac{4+5}{100}\right] =  5 \left [ \frac{9}{100}\right] =  \frac{9}{20}

\par Remaining work =  1 - \frac{9}{20} = \frac{11}{20}

\par B can complete \frac{11}{20}  \par work =  \frac{11}{20} \times 20 = 11s

5. A, B can dig a well in 10 days and 12 days respectively. If Rs.11 is for total work what is A share.

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par A's one day's work =  \frac{1}{10}

\par B's one day's work =  \frac{1}{12}

\par Ratio of (A+B)'s one day work =  \frac{1}{10} : \frac{1}{12} = 6 : 5

\par Amount should be shared in 6 : 5 ratio

\par A's share =  \frac{6}{11} \times 11 \par = Rs. 6

6. A does a work in 10 days and B does the same work in 15 Days. In how many days they together will do the same work.

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par A's one day's work =  \frac{1}{10}

\par B's one day's work = \frac{1}{15}

\par (A + B)'s one days work = \left [ \frac{1}{10}+\frac{1}{15} \right ]  = \frac{1}{6}

\par So, both together will finish the work in 6 days.

\par \underline{Shortcut method}

\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6

7. A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day ?

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par A's one day's work =  \frac{1}{18}

\par B's one day's work =  \frac{1}{9}

\par (A+B)'s one day's work = \frac{1}{18}+\frac{1}{9} = \frac{1}{6}

8. A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par One minute's work of both the punctures =  \frac{1}{9}+\frac{1}{6} = \frac{5}{18}

\par So, both punctures will make the tyre flat in \frac{18}{5} = 3\frac{3}{5} minutes

9. A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par (A + B + C)'s one day's work =  \frac{1}{24}+\frac{1}{6}+\frac{1}{12} = \frac{7}{24}

\par So, A, B and C together will complete the job in \frac{24}{7} = 3\frac{3}{7} \par days

10. A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together ?

A. 10 days

B. 11 days

C. 12 days

D. 15 days

Answer: Option B

Explanation:

			

\par One day's work of the three persons = \frac{1}{15}+\frac{1}{20}+\frac{1}{25} = \frac{47}{300}

\par So, all the three together will complete the work in \frac{300}{47} = 6.4  \par days