Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k
As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)
x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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