ans: 6 men.
2 men and 5 women require 3 days to complete $\frac{1}{4}$ job
They complete $\frac{1}{4×3}$ = $\frac{1}{12}$ job in1 day.
3 men + 5 women require 2 days to complete $\frac{1}{4}$ job;
They complete $\frac{1}{4×2} = \frac{1}{8}$days.
3 men + 5 women = $\frac{1}{8}$ …………………(i)
2 men + 5 women = $\frac{1}{12}$……………………(ii) [ বিয়োগ করি ]
1 man= $\frac{1}{24}$
$\frac{1}{24}$ job is done in 1 day by 1 men
1 job is done in 1 day by 24 men
1 job is done in 4 day by $\frac{24}{4}$ =6 men.
ans: 3:2
Since the man buys maximum number of shirts, so the following calculation may follows:
If maximum no. of shirts is 12, total cost =12 x 70 = 840 which greater than given total 810
If maximum no. of shirts is 11, total cost =11 x 70 = 770 and cost of ties = 810 — 770 = 40
If maximum no. of shirts is 10, total cost = 10 x 70 = 700 and cost of ties = 810 — 700 = 110
If maximum no. of shirts is 9, total cost = 9 x 70 = 630 and cost of ties = 810 — 630 = 180
Here 40 and 140 is not divisible by 30, so these two is not possible. But, 180 is divisible by 3
so number of ties = 180 + 30 = 6
Ratio of shirts to ties = 9 : 6 = 3 : 2
ans: 200 days
After 30 days 4000 soldiers have the food of (190 – 30) = 160 days.
After posting of 800 soldiers, there are (4000 – 800) = 3200 soldiers.
4000 soldiers can eat = 160 days
1 soldiers can eat = 4000 x 160 days
3200 soldiers can eat =$\frac{400\times 160}{3200}$= 200 days
