Question

‘A’is 40% less efficient than ‘B'who can do the same work in 20% less time than 'C'. IfA and B togethe
can complete 80% of work in 12 days, then in how many days 60% of work can be completed by B and C together.
1.2 days
2. 4 days
3. 6 days
4.8 days
5.10 days​

Answer 1

Answer:

70 days

Step-by-step explanation:

Let C's efficiency be 100, B's will be 80 and A's 112. Reducing we get the ratios as 28:20:25

No.of days required for work ( inversely proportional) LCM =700

Days: A:B:C = 25 : 35: 28

When difference of days between A and C is 3 , C takes 28 days. Hence when actual difference is 6 days C will take 6*28/3 = 56 days. Therefore A will take 50 days.

B is 40% less efficient than A and hence takes 50*35/25 = 70 days.

Answer 2

Concept :

Efficiency is the proportion of user-inputted work (input work) that is converted into output work by a machine (output work). Because some of the attempt to evaluate is needed to remove friction, the return work is always lesser than the input work. As a result, efficiency is never $100$ percent.

Given:

A’is $40$% less efficient than ‘B'who can do the same work in $20$% less time than 'C'. If A and B together can complete 80% of work in $12$ days

Find:

how many days 60% of work can be completed by B and C together.

Solution:

According to the problem,

Let C's efficiency be $100$, B's will be $80$ and A's $112$. Reducing we get the ratios as $28:20:25$

No.of days required for work ( inversely proportional) LCM =$700$

Days: A:B:C = $25 : 35: 28$

When difference of days between A and C is $3$ , C takes $28$ days. Hence when actual difference is $6$ days C will take $6*28/3 = 56$ days. Therefore A will take $50$ days.

B is 40% less efficient than A and hence takes $50*35/25 = 70$ days.

Hence the answer is $70$ days

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