Quantitative Aptitude - Arithmetic Ability Questions

Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,

Then   (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or  7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

Given number series follows a pattern that,

2

2 x 1.5 + 1 = 4

4 x 1.5 + 1 = 7 (not equal to 8)

7 x 1.5 + 1 = 11.5

11.5 x 1.5 + 1 = 18.25

18.25 x 1.5 + 1 = 28.375

Hence, the odd number in the given number series is 8.

The given number series follows a pattern that

5 x 5 – 1 = 24

24 x 4 – 2 = 94

94 x 3 – 3 = 279

279 x 2 – 4 = 554

I.P*R*5100=P⇔R=20

 II.P1+R1002-P-P*R*2100=400=>pR2=4000000

 III.P*R*1100=2000=>PR=200000

 PR2PR=4000000200000⇔R=20

Thus I only or (II and III) give answer.

 Correct answer is (D)

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

4x – (60 – x) = 130

=> 5x = 190 

=> x = 38

A : B = 100 : 90.

A : C = 100 : 87.

B/C = (B/A*A/C) = (90/100*100/87) = 30/29

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs (29/30*180)m =174m

 B beats C by (180 - 174) m = 6 m.