Quantitative Aptitude - Arithmetic Ability Questions

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

Here the series follows,

194

194 - (1x1 - 1) = 194

194 - (2x2 - 1) = 191

191 - (3x3 - 1) = 183

183 - (4x4 - 1) = 168

168 - (5x5 - 1) = 144

Here the series follows the pattern that, product of succesive prime numbers i.e,

 2 × 3 = 6

 3 × 5 = 15

 5 × 7 = 35

 7 × 11 = 77

 11 × 13 = 143

 13 × 17 = 221

 17 × 19 = 323

So here in place of 77 the number 63 is written.

So the wrong number is 63.

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Saving  = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864.

51 + 52 + 53 + ...........+ 100

 = (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)

 = It is in the form of n(n+1)2 series summation

= n1 = 100 , n2 = 50

=100100+12 - 5050+12

 =  (5050 - 1275) = 3775

2l+bb=51      => 2l + 2b = 5b     => 3b = 2l 

b=(2/3)l

Then, Area = 216 cm² 

=> l x b = 216     => l x (2/3)l =216 

l = 18 cm.

Middle numbers  =  [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.