Quantitative Aptitude - Arithmetic Ability Questions

P( only one of them will be selected) = p[(E and not F) or (F and not E)] 

 = PE∩F∪F∩E 

= PEPF+PFPE

 =17×45+15×67=27

Let the price of 1 litre of diesel be Rs.x and let Karthik initially buy 'y' litres of diesel.
Therefore, he would have spent Rs. xy on diesel.

When the price of diesel increases by 25%, the new price per litre of diesel is 1.25x.

Karthik intends to increase the amount he spends on diesel by 15%.
i.e., he is willing to spend xy + 15% of xy = 1.15xy

Let the new quantity of diesel that he can get be 'q'.
Then, 1.25x x q = 1.15xy
Or q = (1.15/1.25)y = 0.92y.

As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what he used to get earlier.

Or a reduction of 8%.

Given the salaries are in the ratio of 4 : 5 : 6

Now the salaries are increased by 50%, 60% and 50% respectively

The New ratio of salaries will be

4 x (150/100) : 5 x (160/100) : 6 x (150/100)

= 6 : 8 : 9

The new ratio of salaries of Akhil, Arun and Karthik is 9 : 8 : 6

A+7B=112

it is clear that A = 14, then it becomes A>=B, but A is smallest angle) given
so, range of A is 0.0001 to 13.9999 ( I am taking upto 4 decimal places)
so, range of B becomes 14 to 16 ( after rounding off to 4 decimal places)
so, range of C becomes 152 to 164 ( after rounding off to 4 decimal places)

Given, 52416312=168 ⇔52416168=312 

Now , 52.4160.0168=5241616.8=52416168×10 =312×10=3120

Probability of atleast one failure
= 1 - no failure > 31/32
= 1 - P5 > 31/32
=  P5<1/32
= p < 1/2
Also p > 0
Hence p lies in [0,1/2].

Let no. of boys be 'B' and no. of girls be 'G'

Here from given data,

B + G = 1650 .....(1)

B - G = 400 .....(2)

From (1) & (2), we get

2B = 2050

=> B = 1025

=> G = 1650 - 1025 = 625

Hence, the % of girls = 625 x 100/1650

=> 38% (approx).