A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

Answer: Option A

Explanation:

Ratio of times taken by A and B = 1:3
Means B will take 3 times which A will do in 1 time

If difference of time is 2 days, B takes 3 days
If difference of time is 10 days, B takes (3/2) * 10 =15 days

Answer: Option D

Explanation:

A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6

We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)

So A 1 day work =
$$\begin{aligned} \frac{1}{6}- \frac{1}{12} \\ = \frac{1}{12} \end{aligned}$$

Similarly C 1 day work =
$$\begin{aligned} \frac{1}{6}- \frac{1}{8} \\ = \frac{4-3}{24} \\ = \frac{1}{24} \end{aligned}$$

So A and C 1 day work =

$$\begin{aligned} \frac{1}{12} + \frac{1}{24} \\ = \frac{3}{24} \\ = \frac{1}{8} \end{aligned}$$

So A and C can together do this work in 8 days

Answer: Option A

Explanation:

To calculate the answer we need to get 1 man per day work and 1 woman per day work.

Let 1 man 1 day work =x
and 1 woman 1 days work = y.
=> 6x+5y = 1/6
and 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90

(9 men + 15 women)'s 1 days work =
(9/54) + (15/90) = 1/3

9 men and 15 women will finish the work in 3 days

Answer: Option D

Explanation:

In this question we having, A's work, B's work and A+B+C work. We need to calculate C's work.
We can do it by,
(A+B+C)'s work - (A's work + B's work).

Let's solve it now:

C's 1 day work =
$$\begin{aligned} \frac{1}{4}- \left(\frac{1}{16} +\frac{1}{12} \right) \\ =\left(\frac{1}{4} - \frac{7}{48} \right) \\ = \frac{5}{48} \end{aligned}$$

So C can alone finish the job in 48/5 days,
Which is =
$$\begin{aligned} 9\frac{3}{5}days \end{aligned}$$

Answer: Option C

Explanation:

Suppose B takes x dáys to do the work.
As per question A will take
$$\begin{aligned} 2* \frac{3}{4} * x = \frac{3x}{2}days \end{aligned}$$

(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30 days