Question

3 men and 2 women can do a piece of work in 15 days, 2 men and 3 women can do the same work in 18 days.
The number of day to be taken
by 1 man and 1 women to do the work is :​

Answer 1

Answer:

$40\frac{10}{11}$ days.

Step-by-step explanation:

Given: 3 men and 2 women can do a piece of work in 15 days, 2 men and 3 women can do the same work in 18 days.

To find: Days taken by 1 man and 1 woman to do the work.

First we find ratio between men & women.

Let $M,W$ denotes men & women respectively.

Then as the work done is same or equal so we have:

$(3M+2W)*15=(2M+3W)*18$

$(3M+2W)*5=(2M+3W)*6\\15M+10W=12M+18W\\3M=8W$

$M:W=8:3$

Now find total work to be done.

So, $Work=(3M+2W)*15$

                $=(3*8+2*3)*15\\=(24+6)*15\\=450$

So, days taken by 1 man and 1 woman are:

$(1W+1M)*x=450\\(3+8)*x=450\\11x=450\\x=40\frac{10}{11} \text{days}$

Therefore, number of day to be taken by 1 man and 1 woman to do the work are $40\frac{10}{11}$.

#SPJ2

Answer 2

Answer:

1 men and 1 women can do the same work in $40 \frac{10}{11}.$

Step-by-step explanation:

Given:

3 men and 2 women can do a piece of work in $15$ days, $2$ men and $3$ women can do the same work in $18$ days.

To find:

Days taken by $1$ man and $1$ woman to do the work.

First we find ratio between men & women

Let $M,W$ denotes men & women respectively.

Then as the work done is same or equal so we have:

$(3M+2W)\times 15=(2M+3W)\times 18\\(3M +2W)\times 5=(2M+3W)\times 6\\15M+10W=12M+18W\\3M=8W\\M:W=8:3$

Now find total work to be done.

So,

$Work=(3M+2W)\times 15\\=(3 \times 8+2 \times 3)\times 15\\=(24+6)\times 15\\=450$

So, days taken by 1 man and 1 woman are:

$(1W+1M)\times x=450\\(3+8)\times x=450\\11x=450\\x=40\frac{10}{11} days$

Therefore, number of day to be taken by 1 man and 1 woman to do the work are $40 \frac{10}{11}.$

#SPJ2