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Average speed = 2t_{1} x t_{2}/( t_{1} + t_{2})
Average speed = 2t_{1} x t_{2}/( t_{1} + t_{2})
= (2x64x80)/(64+80)km
= (2 x 64 x 80 ) / 144 km/hr.
=71.11km/hr.
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At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time
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
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Suppose the man covers first distance in x hrs. and second distance in y hrs.
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
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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
∴ W_{1} x D _{1} = W_{2} x
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
∴ W_{1} x D _{1} = W_{2} x D_{2} W = women, D=days
10 X 15 = 25 x D_{2}
D_{2} = 6
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6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 day
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.