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21 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Tick the correct answer in the following: A can do a piece of work in $25$ days, which $B$ alone can do in $20$ days. A started the work and was joined by $B$ after $10$ days. The work lasted for:
  • A
    $12\frac{1}{2}\ \text{days.}$
  • B
    $15\ \text{days.}$
  • $16\frac{2}{3}\ \text{days}.$
  • D
    $14\ \text{days.}$
Answer
Correct option: C.
$16\frac{2}{3}\ \text{days}.$
$A's \ 1$ days work $=\frac{1}{25}$
$B's \ 1$ days work $=\frac{1}{20}$
$A$ and $B's$ days work $=\frac{1}{25}+\frac{1}{20}$
$=\frac{4+5}{100}=\frac{9}{100}$
$A's \ 10$ days work $=\frac{1}{25}\times10=\frac{2}{5}$
Remaining work $=1-\frac{2}{5}=\frac{3}{5}$
$\therefore\frac{3}{5}$ work will be finished by $A$ and $B$ in,
$=\frac{3}{5}\times\frac{100}{9}=\frac{20}{3}\ \text{days}=6\frac{2}{3}\ \text{days,}$
$\therefore$ Whole work was finished in $=10+6\frac{2}{3}$
$=16\frac{2}{3}\ \text{days}.$
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MCQ 21 Mark
Tick the correct answer in the following: $3$ men or $5$ women can do a work in $12$ days. How long will $6$ men and $5$ women take to do it?
  • A
    $6$ days.
  • B
    $5$ days.
  • $4$ days.
  • D
    $3$ days.
Answer
Correct option: C.
$4$ days.
$3$ men $= 5$ women
$1$ men $\frac{5}{3}$ women $5$
$6$ men $=\frac{5}{3}\times6=10$ women,
$\therefore$ Total women in second case,
$= 10 + 5 = 15$ women
Now,
$5$ women $: 15$ women $: 12$ days $: x$
$\therefore$ By inverse proportion,
$5 : 15 : x : 12$
$\text{x}=\frac{5\times12}{15}=4\ \text{days}$
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MCQ 31 Mark
Tick the correct answer in the following: $A $ and $B$ together can do a piece of work in $12$ days; $B$ and $C$ can do it in $20$ days while $C$ and $A$ can do it in $15$ days. $A, B$ and $C$ all working together can do it in
  • A
    $6$ days.
  • B
    $9$ days.
  • $10$ days.
  • D
    $10\frac{1}{2}$ days.
Answer
Correct option: C.
$10$ days.
$A$ and $B's \ 1$ day's work $=\frac{1}{12}$
$B$ and $C's \ 1$ day's work $=\frac{1}{20}$
$C$ and $A's \ 1$ day's work $=\frac{1}{15}$
Adding we get,
$2(A, B$ and $C)'s \ 1$ day's work,
$=\frac{1}{12}+\frac{1}{20}+\frac{1}{15}$
$=\frac{5+3+4}{60}$
$=\frac{12}{60}$
$=\frac{1}{5}$
$\therefore A, B$ and $C's \ 1$ days work,
$=\frac{1}{5}\times\frac{1}{2}$
$=\frac{1}{10}$
$\therefore$ That will finish the work in $= 10$ days.
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MCQ 41 Mark
Mark againts the correct answer in the following: The rates of working of two tapes $A$ and $B$ are in the ratio $2 : 3.$ The ratio of the time taken by $A$ and $B$ respectively to fill a given cistern is:
  • A
    $2 : 3$
  • $3 : 2$
  • C
    $4 : 9$
  • D
    $9 : 4$
Answer
Correct option: B.
$3 : 2$
Rates at which taps $A$ and $B$ work $= 2 : 3$
The ratio of time taken by taps $A$ and $B$ to fill the cistern,
$=\frac{1}{\text{Rate at which taps A and B work}}$
$=\frac{1}{\frac{2}{3}}$
$=\frac{3}{2}$
$=3:2$
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MCQ 51 Mark
Tick the correct answer in the following:$A$ can do a piece of work in $15$ days. $B$ is $50\%$ more efficient than $A. B$ can finish it in:
  • $10\ \text{days}.$
  • B
    $7\frac{1}{2}\ \text{days}.$
  • C
    $12\ \text{days.}$
  • D
    $10\frac{1}{2}\ \text{days.}$
Answer
Correct option: A.
$10\ \text{days}.$
$A's \ 1$ day's work $=\frac{1}{15}$
Then $B's$ day's work $=\frac{1}{10}\times\frac{100+50}{100}$
$=\frac{1}{15}\times\frac{150}{100}$
$=\frac{1}{10}$
$B$ will finish the work in $10$ days.
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MCQ 61 Mark
Tick the correct answer in the following: $A$ can do a job in $16$ days and $B$ can do the same job in $12$ days. With the help of $C,$ they can finish the job in $6$ days only. Then, $C$ alone can finish it in
  • A
    $34$ days.
  • B
    $22$ days.
  • C
    $36 $ dyas.
  • $48$  days.
Answer
Correct option: D.
$48$  days.
$A's \ 1$ day's work $=\frac{1}{16}$
$B's \ 1$ day's work $=\frac{1}{12}$
$\therefore C's \ 1$ days work $=\frac{1}{6}-\Big(\frac{1}{16}+\frac{1}{12}\Big)$
$=\frac{1}{6}-\frac{1}{16}-\frac{1}{12}$
$=\frac{8-3-4}{48}$
$=\frac{1}{48}$
$\therefore C$ can finish the work in $= 48$ days.
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MCQ 71 Mark
Tick the correct answer in the following: A can do a piece of work in $20$ days which $B$ alone can do in $12$ days. $B$ worked at it for $9$ days. A can finish the remaining work in:
  • A
    $3$ days.
  • $5$ days.
  • C
    $7$ days.
  • D
    $11$ days.
Answer
Correct option: B.
$5$ days.
$A's \ 1$ day's work $=\frac{1}{20}$
$B's \ 1$ day's work $=\frac{1}{12}$
Both's $1$ day's work $=\frac{1}{20}+\frac{1}{12}$
$=\frac{3+5}{60}=\frac{8}{60}=\frac{2}{15}$
$B's \ 9$ day's work $=\frac{1}{12}\times9\frac{3}{4}$
Remaining work $=1-\frac{3}{4}=\frac{1}{4}$
$\therefore A$ will finish $\frac{1}{4}$ work in $=20\times\frac{4}{1}$
$= 5$ days.
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MCQ 81 Mark
Mark againts the correct answer in the following: $A$ works twice as fast as $B.$ If both of them can together finish a peice of work in $12$ hours, then $B$ alone can do it in:
  • A
    $24$ hours.
  • B
    $27$ hours.
  • $36$ hours.
  • D
    $18$ hours.
Answer
Correct option: C.
$36$ hours.
Suppose $B$ takes $x$ hours to complete the work.
$\therefore B's \ 1$ hour work $=\frac{1}{\text{x}}$
$A$ works twice as fast as $B.$
$\therefore A's \ 1$ hour work $=\frac{2}{\text{x}}$
$\frac{1}{12}=\frac{1}{}\text{x}+\frac{2}{\text{x}}=\frac{3 }{\text{x}}$
$\Rightarrow\text{x}=36$ hours.
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MCQ 91 Mark
Tick the correct answer in the following: Two pipes can fill a tank in $10$ hours and $12$ hours respectively, while a third pipe empties the full tank in $20$ hours. If all the three pipes operate simultaneously, in how much time will the tank be full?
  • A
    $7$ hours $15$ minutes.
  • B
    $7$ hours $30$ minutes.
  • C
    $7$ hours $45 $ minutes.
  • $8$ hours.
Answer
Correct option: D.
$8$ hours.
Frist inlet pipe's $1$ hour work $=\frac{1}{10}$
Second inlet pipe's $1$ hour work $=\frac{1}{12}$
Third outlet pipe's one hour work $=\frac{1}{20}$
$\therefore$ Three outlet pipe's $1$ hour work,
$=\frac{1}{10}+\frac{1}{12}-\frac{1}{20}$
$=\frac{6+5-3}{60}$
$=\frac{8}{60}$
$=\frac{2}{15}$
$\therefore$ The tank will be full in $\frac{15}{2}$ hours $= 7$
$7$ hours $30$ minutes.
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MCQ 101 Mark
Mark againts the correct answer in the following: $A$ can finish a piece of work in $12$ hours while $B$ can finish it in $15$ hours. How long will both take to finish it, working together?
  • A
    $9\text{ hours}.$
  • $6\frac{2}{3}\text{ hours}.$
  • C
    $6\frac{3}{4}\ \text{hours}.$
  • D
    $8\frac{1}{3}\ \text{hours}.$
Answer
Correct option: B.
$6\frac{2}{3}\text{ hours}.$
$A's \ 1$ hours work $=\frac{1}{12}$
$B's \ 1$ hours work $=\frac{1}{15}$
$(A + B)'s \ 1$ hours work $=\frac{1}{12}+\frac{1}{15}=\frac{9}{60}=\frac{3}{20}$
Time taken by $A$ and $B$ to complete the work together $=\frac{20}{3}=6\frac{2}{3}$ hours.
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MCQ 111 Mark
Mark againts the correct answer in the following: $A$ can do a piece of work in $14$ days and $B$ is $40\%$ more efficient than $A.$ In how many days can $B$ finish it?
  • $10\ \text{days}.$
  • B
    $7\frac{1}{2}\ \text{days}.$
  • C
    $5\frac{1}{4}\ \text{days}.$
  • D
    $5\frac{3}{5}\ \text{days}.$
Answer
Correct option: A.
$10\ \text{days}.$
$A's \ 1$ day work $=\frac{1}{14}$
$B$ is $40\%$ more efficient than $A.$
$\therefore B's \ 1$ day work $=\frac{140}{100}\times\frac{1}{14}=\frac{1}{10}$
$B$ takes $10$ days to complete the work.
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MCQ 121 Mark
Tick the correct answer in the following: The rates of working of $A$ and $B$ are in the ratio $3 : 4.$ The number of days taken by them to finish the work are in the ratio:
  • A
    $3 : 4$
  • B
    $9 : 16$
  • $4 : 3$
  • D
    $16 : 9$
Answer
Correct option: C.
$4 : 3$
Ratio in the rates of working of $A$ and $B = 3 : 4$
Ratio in time $=\frac{1}{3}:\frac{1}{4}$
$=\frac{4:3}{12}$
$=4:3$
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MCQ 131 Mark
Tick the correct answer in the following: $A$ man can do a piece of work in $5$ days. He and his son working together can finish it in $3$ days. In how many days can the son do it alone?
  • A
    $6\frac{1}{2}\ \text{days}$
  • B
    $7\ \text{days}$
  • $7\frac{1}{2}\ \text{days}$
  • D
    $8\ \text{days}$
Answer
Correct option: C.
$7\frac{1}{2}\ \text{days}$
$A$ man's $1$ day work $=\frac{1}{5}$
Man and his son's $1$ days work $=\frac{1}{3}$
$\therefore$ Son's $1$ days work $=\frac{1}{3}-\frac{1}{5}$
$=\frac{5-3}{15}$
$=\frac{2}{15}$
$\therefore$ His son will finish the work,
$=\frac{15}{2}\ \text{day}$
$=7\frac{1}{2}\ \text{days}$
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MCQ 141 Mark
Tick the correct answer in the following: $A$ tap can fill a cistern in $8$ hours and another tap can empty the full cistern in $16$ hours. If both the taps are open, the time taken to fill the cistern is:
  • A
    $5\frac{1}{3}$ hours.
  • B
    $10$ hours.
  • $16$ hours.
  • D
    $20$ hours.
Answer
Correct option: C.
$16$ hours.
Frist tap's $1$ hours work to fill $=\frac{1}{8}$
Second tap's $1$ hours work to empty $=\frac{1}{16}$
Both $1$ hour can fill the cistern
$=\frac{1}{8}-\frac{1}{16}=1$
$=\frac{2-1}{16}$
$=\frac{1}{16}$
$\therefore$ The cistern will fill up in $16$ hours.
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MCQ 151 Mark
Tick the correct answer in the following$: A$ pump can fill a tank in $2$ hours. Due to a leak in the tank it takes $2\frac{1}{3}$ hours to fill the tank. The leak can empty the full tank in:
  • A
    $2\frac{1}{3}$ hours.
  • B
    $7$ hours.
  • C
    $8$ hours.
  • $14 $ hours.
Answer
Correct option: D.
$14 $ hours.
Frist pump's $1$ hours work to fill $=\frac{1}{2}$
Due to leakage, tank is filled in $2\frac{1}{3}$ hours,
$=\frac{7}{3}$ hours,
Both its one hour work $=\frac{3}{7}$
$\therefore$ Leakage's $1$ hour work to empty the tank,
$=\frac{1}{2}-\frac{3}{7}$
$=\frac{7-6}{14}$
$=\frac{1}{14}$
$\therefore$ Leak will empty the tank in $14$ hours.
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MCQ 161 Mark
Tick the correct answer in the following: To complete a work, $A$ takes $50\%$ more time than $B.$ If together they take $18$ days to complete the work, how much time shall $B$ take to do it?
  • $30$ days.
  • B
    $35$ days.
  • C
    $40$ days.
  • D
    $45$ days.
Answer
Correct option: A.
$30$ days.
Let $B$ can do a work in $= x$ days,
Then $A$ can do the work,
$=\text{x}+\frac{\text{x}}{2}=\frac{3}{2}\text{x}\ \text{days}$
$A$ and $B \ 1$ days work $=\frac{1}{18}$
$A's \ 1$ days's work $=\frac{1}{\text{x}}$
And $B's \ 1$ days work $=\frac{2}{3\text{x}}$
$\therefore\frac{1}{\text{x}}+\frac{2}{3\text{x}}=\frac{1}{18}$
$\frac{5}{3\text{x}}=\frac{1}{18}$
$\Rightarrow3\text{x}=5\times18$
$\Rightarrow\text{x}=\frac{5\times18}{3}=30$
$\therefore B$ can do the work in $= 30$ days.
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MCQ 171 Mark
Tick the correct answer in the following: $A$ alone can do a piece of work in $10$ days and $B$ alone can do it in $15$ days. In how many days will $A$ and $B$ together do the same work?
  • A
    $5$ days.
  • $6$ days.
  • C
    $8$ days.
  • D
    $9$ days.
Answer
Correct option: B.
$6$ days.
$A's \ 1$ day's work $=\frac{1}{10}$
$B's \ 1$ day's work $=\frac{1}{15}$
$\therefore$ Both $A$ and $B's \ 1$ day's work,
$=\frac{1}{10}+\frac{1}{15}$
$=\frac{3+2}{30}$
$=\frac{5}{30}$
$=\frac{1}{6}$
$\therefore A$ and $B$ can do the work in $6$ days.
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MCQ 181 Mark
Tick the correct answer in the following: $A$ works twice as fast as $B.$ If both of them can together finish a piece of work in $12$ days, then $B$ alone can do it in:
  • A
    $24$ days.
  • B
    $27$ days.
  • $36$ days.
  • D
    $48$ days.
Answer
Correct option: C.
$36$ days.
Let $B's \ 1$ day's work $= x$
Then $A's \ 1$ day's work $= 2x$
$A$ and $B's \ 1$ day's work $=\frac{1}{12}$
$\therefore\text{x}+2\text{x}=\frac{1}{12}$
$\Rightarrow3\text{x}=\frac{1}{12}$
$\Rightarrow\text{x}=\frac{1}{12\times3}=\frac{1}{36}$
$\therefore B$ can do the work in $36$ days.
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MCQ 191 Mark
Tick the correct answer in the following: Two pipes can fill a tank in $20$ minutes and $30$ minutes respectively. If both the pipes are opened simultaneously, then hte tank will be filled in:
  • A
    $10$ minutes.
  • $12$ minutes.
  • C
    $15$ minutes.
  • D
    $25$ minutes.
Answer
Correct option: B.
$12$ minutes.
Frist pipe $1$ minutes work $=\frac{1}{20}$
Second pipe $1$ minutes work $=\frac{1}{30}$
$\therefore$ Both's $1$ minutes work $=\frac{1}{20}+\frac{1}{30}$
$=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12}$
$\therefore$ Both's $1$ will do the work in $= 12$ minutes.
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MCQ 201 Mark
Tick the correct answer in the following: $A$ does $20\%$ less work than $B.$ If $A$ can finish a piece of work in $7\frac{1}{2}$ hours, then $B$ can finish it in:
  • A
    $5\ \text{hours}.$
  • B
    $5\frac{1}{2}\ \text{hours}.$
  • $6\ \text{hours}.$
  • D
    $6\frac{1}{2}\ \text{hours}.$
Answer
Correct option: C.
$6\ \text{hours}.$
$A's$ hour's work $=\frac{2}{15}$
$A$ and $B's$ ratio in work $=\frac{100-20}{100}:1$
$=\frac{80}{100}:1$
$=\frac{4}{5}:1$
Let ratio be, $\frac{4}{5} x$ and $x,$ then,
$\frac{4}{5}:1=\frac{2}{15}:\frac{1}{\text{x}}$
$1\times\frac{2}{15}=\frac{4}{5}\times\frac{1}{\text{x}}=\frac{2}{15}=\frac{4}{5\text{x}}$
$2\times5\text{x}=4\times15$
$\Rightarrow\text{x}=\frac{4\times15}{2\times5}=6$
$\therefore B$ can finish the work in $6$ hours.
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MCQ 211 Mark
Mark againts the correct answer in the following: $A$ pump can fill a cistern in $2$ hours. Due to a leak in the tank it takes $2\frac{1}{3}$ hours to fill it. The leak can empty the full tank in:
  • A
    $7$ hours.
  • $14$ hours.
  • C
    $8$ hours.
  • D
    $3$ hours.
Answer
Correct option: B.
$14$ hours.
$A$ pump fills a tank in $2$ hours.
Part of tank filled by the pump in one hour $=\frac{1}{2}$
Let $x$ hours be the time required for water to leak from the cistern.
Part of tank drained by the leak in one hour $=-\frac{1}{\text{x}}$
ime required to fill the leaking cistern $=2\frac{1}{3}=\frac{7}{3}$ hours.
Part of tank stored with water in one hour $=\frac{3}{7}$
$\Rightarrow\frac{3}{7}=\frac{1}{2}-\frac{1}{\text{x}}$
$\frac{1}{\text{x}}=\frac{1}{2}-\frac{3}{7}=\frac{7-6}{14}=\frac{1}{14}$
$\text{x}=14$ hours.
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