Question

A certain work was started by 4 men and 10 women who completed 50% of the work in 6 days . Then another 2 men and 2 women joined them and they could complete two-third of the remaining work in three days. How many men along with 6 women are required to complete the remaining work in two more days ?​

Answer 1

Answer:

8 women are needed to finish the work.

Step-by-step explanation:

Remaining work time-and-work-q-65267.png 4 men + 10 women do 1 work in 12 days.

6 men + 12 women do 1 work in 9 days.

48 men + 120 women = 54 men + 108 women ⇒ 6 men = 12 women ⇒ 1 men = 2 women

∴ In 12 days 1 work requires 9 men

∴ In 1 day 1 work requires 9 × 12 men

∴ In 3 days 1 work requires time-and-work-q-65261.png

∴ In 3 days ⁴⁄₉ work requires time-and-work-q-65255.png men

There are 6 men and 12 women or (12 men equivalent)

So, 4 men equivalent is required additionally

∴ 8 women are needed to finish the work.

Answer 2

Time and Work

Given:

4 men and 10 women finishes 50% of the work in 6 days.

2 Men and 2 women joined them and covers  two-third of remaining work in 3 days.

To find:

No. of men along with 6 women required to finish the remaining work in 2 extra days.

Explanation:

Let the work done by 1 men in 1 day be x.

Let the work done by 1 women in 1 day be y.

As per question,

$6(4x+10y)=\frac{1}{2} ------------------eq(i)\\\\3(4x+10y+2x+2y)=\frac{2}{3}rd \times(1-\frac{1}{2} ) } \\\\=>3(6x+12y)=\frac{1}{3} ----------------eq(ii)$

we can write equation (i) as

$2x+5y=\frac{1}{24}$

On multiplying this with 3 , to use elimination method, it will,

$6x+12y=\frac{1}{24} -----------equation(iii)$

On solving both (ii) and (iii),

$y=\frac{1}{216}\\so,\\x=\frac{1}{108}$

Let the number of men required be p

$2(6x+py)=(1-\frac{1}{2}-\frac{1}{3} )\\\\2(6x+py)=\frac{1}{6}$

Substituting the value of x and y, we get

$p=6$

Hence, the number of men required will be 6.