Quantitative Aptitude - Arithmetic Ability Questions

20% of a = b 

=> (20/100) * a = b 

 b% of 20 =(b/100) x 20 = [(20a/100)  / 100] x 20= 4a/100 = 4% of a.

Here given series follows a pattern that

0.25, 8, 128, 1024, ?

0.25 x 32 = 8

8 x 16 = 128

128 x 8 = 1024

1024 x 4 = 4096

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Here the given series is 1, 3, 6, 11, 20, 37, ??,

and the logic behind the series is

1 x 2 + 1 = 3

3 x 2 + 0 = 6

6 x 2 + (-1) = 11

11 x 2 + (-2) = 20

20 x 2 + (-3) = 37

37 x 2 + (-4) = 70

Hence, the next number in the given series is 70.

As per the BODMAS rule, the priority in which the operations should be done is:

Note: Addition and subtraction can be treated on same priority (from left to right) when they are in consecutive order. ? = 1234 + 2345 – 3456 + 4567 ? = 3579 - 3456 + 4567 ? = 123 + 4567 ? = 4690

Let the number be x.

Then, ideally he should have multiplied by  x by 5/3. Hence Correct result was x * (5/3)= 5x/3. 

By mistake he multiplied x by 3/5 . Hence the result with error  = 3x/5 

Then, error = (5x/3 - 3x/5) = 16x/15 

Error %  = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

1, 1 + 1² = 2, 2 + 2² = 6, 6 + 3² = 15, 15 + 4² = 31, 31 + 5² = 56, 56 + 6² = 92

Last number of given series must be 92 not 91.