Quantitative Aptitude - Arithmetic Ability Questions

Let the required number days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

 Spiders1:7Webs7:1⋮⋮ 7:x 

=> 1 * 7 * x = 7 * 1 * 7 

=> x= 7

Given number series is 9 17 11 19 13 

Here the pattern is

• 9

• 9 + 8 = 17

• 17 - 6 = 11

• 11 + 8 = 19

• 19 - 6 = 13

• 13 + 8 = 21

• 21 - 6 = 15

• 15 + 8 = 23

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.

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

Given series is 

12        16         32          68         ?

Here it follows a pattern that,

12

12 + (2x2) = 16

16 + (4x4) = 32

32 + (6x6) = 68

68 + (8x8) = 132

(150 x 460)/100 + (24 x 650)/100

= 690+156

= 846

One way to deal with fractions is to convert them all to decimals. 

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7.