Let the work by (x-1) men in (x+1) days = 9z
Then, work done by 1 man in 1 days = [9z/(x-1) * 1/(x+1)]
Let the work done by (x+2) men in (x-1) days = 10z
Therefore Work done by 1 man in 1 days = 10z/(x+2)(x-1)
Therefore
9z/(x-1)(x+1) = 10z/(x+2)(x-1)
Solve
x=8
Option 5)
Three tap together drain a complete tank in 5 hours. If first drains this tank in 10 hours alone, then minimum hours taken by second tap will be (in any condition)
Let time taken by first, second and third tap are t1, t2, t3
Then according to given condition
1/t1 + 1/t2 + 1/t3 = 1/5 (i)
and
1/t1 = 1/10
so
1/t2 + 1/t3 = 1/5 - 1/10 = 1/10
for t3 = Infinite
t2(min) = 10
Option 1)
A plot of land is in the shape of a right angled isosceles triangle. The length of the hypotenuse is 50*sqrt(2). The cost of fencing it at Rs.3 per metre will be:
OPtion
1) Less than Rs.300
2) Less than Rs.100
3) Less than Rs.400
4) Less than Rs.200
5) More than Rs.700
6) More than Rs.500
7) More than Rs.600
8) Less than Rs.500
9) More than Rs.800
10)None of these
Solution
Let each one of equal sides be x metres.
Then x^2+x^2 = (50*sqrt(2))^2
x = 50 m.
Perimeter = x + x + 50*sqrt(2) = 100 + 50*sqrt(2)
Therefore cost of fencing = Rs.3 * (100 + 50*sqrt(2))
= 511.5
More than Rs.500
Let cost of each fan = Rs.100 & sale = 100 fans.
Money receipt = Rs.(100*100) = Rs.10000
New cost per fan = Rs.70 and New sale = 120 fans
New Money Receipt = Rs.(70*120) = Rs.8400
Decrease in money receipt = (1600/10000)*100 = 16%