Quantitative Aptitude - Arithmetic Ability Questions

Let X be the total amount with Ramesh

Given 28% of X = Cash remaining

=> Amount spent = 72% of X = 38460 + 24468 = 62928

=> 72% of X = 62928

=>  X = ?

=> X = 87400

A = P(1 + rt/100)

So, A= 3P (it becomes triples)

3P = P(1 + 8r/100)→r = 25%

Now, when t = 20

⇒A = P(1 + 25*20/100)

⇒A = P(1+ 5)⇒A = 6P

So, it becomes 6 times.

Let the quantity of alcohol and water be 4x liters and 3x liters respectively. Then,

(4x)/(3x+5) = 4/5

=> 20x = 4(3x+5) 

=> x = 2.5 

Quantity of alcohol = (4 * 2.5) liters = 10 liters.

Let the number of bricks be 'N'

10 x 4/100 x 6 x 90/100 = 25/100 x 15/100 x 8/100 x N

10 x 4 x 6 x 90 = 15 x 2 x N => N = 720.

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1