(A) Given: $(A + B) = 1/30$,$(B + C) = 1/24$,$(C + A) = 1/20$ units per day.Adding these,we get $2(A + B + C) = 1/30 + 1/24 + 1/20 = (4 + 5 + 6)/120 =

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(A) Given: $(A + B) = 1/30$,$(B + C) = 1/24$,$(C + A) = 1/20$ units per day.Adding these,we get $2(A + B + C) = 1/30 + 1/24 + 1/20 = (4 + 5 + 6)/120 = 15/120 = 1/8$.So,$(A + B + C) = 1/16$ units per day.Work done by $A, B, C$ in $10$ days $= 10 \times (1/16) = 10/16 = 5/8$.Remaining work $= 1 - 5/8 = 3/8$.Efficiency of $A = (A + B + C) - (B + C) = 1/16 - 1/24 = (3 - 2)/48 = 1/48$.Time taken by $A$ to finish the remaining work $= (3/8) / (1/48) = (3/8) \times 48 = 18$ days.

Competitive Quantitative Aptitude Time and Work and Wages Time and Work

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$A$ and $B$ can do a piece of work in $30$ days,while $B$ and $C$ can do the same work in $24$ days and $C$ and $A$ in $20$ days. They all work together for $10$ days when $B$ and $C$ leave. How many days more will $A$ take to finish the work?

A
$18$
B
$24$
C
$30$
D
$36$

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Quantitative Aptitude Questions

Chapter

Time and Work and Wages

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Time and Work

$A$ team of $30$ men is supposed to do a work in $38$ days. After $25$ days,$5$ more men were employed and the work was finished one day earlier. How many days would it have been delayed if $5$ more men were not employed? (in days)

Easy

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$A$ does $\frac{2}{5}$ of a work in $9$ days. Then $B$ joined him and they together completed the remaining work in $6$ days. $B$ alone can finish the whole work in (in days):

Medium

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$A$ and $B$ working separately can do a piece of work in $9$ and $12$ days,respectively. If they work for a day alternately with $A$ beginning,the work would be completed in (in days):

Medium

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$A$ and $B$ can complete a piece of work in $45$ and $40$ days,respectively. They began the work together,but $A$ leaves after some days and $B$ completed the remaining work in $23$ days. After how many days did $A$ leave?

Medium

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$A, B$ and $C$ can complete a work in $2 \, h$. If $A$ does the job alone in $6 \, h$ and $B$ in $5 \, h$,how long will it take for $C$ to finish the job alone? (in $h$)

Easy

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$A$ and $B$ can do a piece of work in $30$ days,while $B$ and $C$ can do the same work in $24$ days and $C$ and $A$ in $20$ days. They all work together for $10$ days when $B$ and $C$ leave. How many days more will $A$ take to finish the work?

Similar Questions

$A$ team of $30$ men is supposed to do a work in $38$ days. After $25$ days,$5$ more men were employed and the work was finished one day earlier. How many days would it have been delayed if $5$ more men were not employed? (in days)

$A$ does $\frac{2}{5}$ of a work in $9$ days. Then $B$ joined him and they together completed the remaining work in $6$ days. $B$ alone can finish the whole work in (in days):

$A$ and $B$ working separately can do a piece of work in $9$ and $12$ days,respectively. If they work for a day alternately with $A$ beginning,the work would be completed in (in days):

$A$ and $B$ can complete a piece of work in $45$ and $40$ days,respectively. They began the work together,but $A$ leaves after some days and $B$ completed the remaining work in $23$ days. After how many days did $A$ leave?

$A, B$ and $C$ can complete a work in $2 \, h$. If $A$ does the job alone in $6 \, h$ and $B$ in $5 \, h$,how long will it take for $C$ to finish the job alone? (in $h$)

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