Question

A works twice as fast as B and B works twice as fast as C. All three working together can complete a work in 4 days with the help of D. If D alone can complete this work in 16 days, then in how many days will A complete 75% of the work alone?​

Answer 1

Answer:

Right Answer is: C

A alone can complete the 75% of the total work in = 28/4 = 7 days.

Step-by-step explanation:

Hope it helped you.

Answer 2

A works twice as fast as B and B works twice as fast as C

• Let efficiency of C = 1 m
• Efficiency of B = 2 m
• Efficiency of A = 4 m

Given that D alone can finish the same task in 16 days.

As total work is 100%

So work done by D in 4 days = 25%

It is also given that all three working together can finish total task in 4 days with the help of D

So work done by (A+B+C) in 4 days = 100% - 25% = 75%

Efficiency of (A+B+C) = 1+2+4 = 7 m

So work done by (A+B+C) in 4 days = 7×4 m

${ \red{ \bf{ 7 × 4 m = 75percent \: of\: total \:work . . . . (1) }}}$

Now we will find that in how many days will A alone finish 75% of the same task.

Efficiency of A = 4 units

Let A will finish 75% of the same task in x days.

${ \red{ \bf{ So,\: 4x \:units \:= \:75percent \:of \:total \:work . . . . (2) }}}$

${ \blue{ \bf{ Compare\: (1)\: and\: (2) }}}$

${ \blue{ \bf{ x = 7 \:days\: Ans}}}$