Quantitative Aptitude - Arithmetic Ability Questions

In one day work done by K = 1/20

Work done by K in 2 days = 1/10

Work done by K,L,M on 3rd day =1/20 + 1/30 + 1/60 = 1/10

Therefore, workdone in 3 days is = 1/10 + 1/10 = 1/5

1/5th of the work will be completed at the end of 3 days.

The remaining work = 1 - 1/5 = 4/5

1/5 of work is completed in 3 days

Total work wiil be done in 3 x 5 = 15 days.

P=F1+in

Given   1.21x0.91.1x0.11

= 1.21x100x0.9x101.1x10x0.11x100

= 9

= 3.

Sol : In this kind of questions we find the work force required to complete the work in 1 day (or given unit of time) then we equate the work force to find the relationship between the efficiencies (or work rate) between the different workers.

Therefore, 6B+8G = 6 days

 => 6(6B+8G)= 1 day  (inversely proportional)

 => 36B+48G =1         ( by unitary method)

Again  14B+10G = 4days

 => 56B+40G =1

so, here it is clear that either we employ 36B and 48G to finish the work in 1 day or 56B and 40G to finish the same job in 1 day. thus , we can say

 => 36B+48G = 56B+40G

 => G= 2.5B

Thus a Girl is 2.5 times as efficient as a boy.

Now, since  36B+48G = 1

 => 36B+48(2.5 B)=1

 =>156B=1

i.e., to finish the job in 1 day 156 boys are required or the amount of work is 156 boys-days

Again    1G+1B=2.5B+1B=3.5B

Now, since 156 boys can finish the job in 1 day

so 1 boy can finish the job in 156 days

Therefore, 3.5 boys can finish the job in 1×1563.5= 47 days.

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

17C2 = 136 (combination)