As per the given question ,
We are concerned with solid part of the fruit (pure portion). Assume Q kg of dry fruit is obtained.
? Solid part in fresh fruit = Solid part in dry fruit
? 0·28 × 100 = 0·8 × Q
? Q = 35 kg
? 35 kg of dry fruit can be obtained from 100 kg fresh fruit.
Average speed = 2t1 x t2/( t1 + t2)
= (2x64x80)/(64+80)km
= (2 x 64 x 80 ) / 144 km/hr.
=71.11km/hr.
At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time
= (4/3 of usual time ) ? (usual time ) = 20min.
? 4x/3 -x = 20
? x/3 =20
? x=60 min
Suppose the man covers first distance in x hrs. and second distance in y hrs.
Then , 4x + 5y = 35 and 5x +4y= 37
Solving these equation , we get
X=5 and y=3
Total time taken = (5+3) hrs= 8hrs
Work done by 6 men = work done by 10 women.
Work done by 1 man = work done by 10 / 6 = 5/3 women
? 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women
? W1 x D 1 = W2 x D2 W = women, D=days
10 X 15 = 25 x D2
D2 = 6
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
? 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
? 10 children can do the work in 6 days.
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