Quantitative Aptitude - Arithmetic Ability Questions

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

                = 25C3  = 2300.

         n(E)= 10C1*15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

I = prt = [15000 × 0.07 × (214/365) ]=615.52

Future value, S = P + I = $15 000 + $615.52 = $15 615.52

Literally any percentage means divide by 100.

So 125% = 125100. That can be reduced to 5/4.

To convert to a decimal, you can do long division, dividing 5 into 4.

7 × 5 = 35

35 – 1 = 34

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000

We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.

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.

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.