Question

1 A can do a certain work in the same time in
which B and C together can do it. If A and B
together could do it in 10 days and C alone in 50
days, then B alone could do the work in
1. 15 days 2. 20 days 3. 25 days 4. 30 days​

Answer 1

Answer:

If A completes the work in n days as such by anology B & C together complete the work in same time

Now efficiency of A = 1/n th part of work per day

Combined efficiency of B & C = efficiency of A

E(A) + E(B) + E(C) = 1/n + 1/n = 2/n

Efficiency of A & B = 1/10

Efficiency of C = 1/50

We can write as

E(A) + E(B) +E(C) = 1/10 + 1/50

6/50 = 2/n

n = 100/6

Since E(A) + E(B) = 1/10

E(A) = 6/100

E(B) = 1/10 -6/100 = 4/100 = 1/25

Therefore B will complete the work alone in 25 days

Step-by-step explanation:

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Answer 2

Given :

• ( A + B ) = 10 days
• C = 50 days

According to the question :

Let,

( A + B )'s 1 day work = 1 / 10

C's 1 day work = 1 / 50

☞ ( A + B + C )'s 1 day work ,

=> 1 / 10 + 1 / 50 [ LCM = 50 ]

=> 6 / 50

=> 3 / 25

• A's 1 day work = ( B + C )'s 1 day work [ given ]

Let the work be ' x '

=> 2x ( A's 1 day work ) = 3 / 25

=> A's 1 day work = 3 / 25 × 2

=> A's 1 day work = 3 / 50

To find :

B's 1 day work = ( A + B )'s day - A's 1 day work

=> 1 / 10 - 3 / 50 [ LCM = 50 ]

=> 2 / 50

=> 1 / 25

$\huge\underline{\underline{\texttt{\pink{B=25 days}}}}$

$\huge\underline{\underline{\texttt{\pink{Option 3}}}}$