Quantitative Aptitude - Arithmetic Ability Questions

Let the total applicants be 100

Then, 20% got 85 marks

i.e 20 x 85 = 1700

and 25% got 95 marks

 i.e 25 x 95 = 2375

Now, the remaining applicants are 55 and let the average marks scored by them be x.

Therefore, 2375 + 1700 + 55x  =  60×100 

6000 - 4075 = 55x

55x=1925

x=35.

As the work is same
M1D1 = M2D2

Here Let the number of men employed was M
=> 11M = 7(M+4)
=> 11M - 7M = 28
=> M = 7

Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93.

Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4

According to the question,

Difference between square of the average of first two odd number and the of the average last

two odd numbers is 396

i.e, x+3 and x-3

$$\begin{gathered} ⇒x + 32 - x - 32 = 396 \\ ⇒ x2 + 9 + 6x - x2 + 6x - 9 = 396 \\ ⇒ 12x = 396 \\ ⇒ x = 33 \end{gathered}$$

Hence, the smallest odd number is 33 - 4 = 29.

Let length, breadth and height of the room be 7, 3, 1 unit respectively.

Area of walls = 2(l+b)xh = 2(7+3)x1 = 20 sq. unit.

Now, length, breadth and height of room will become 3.5, 6 and 2 respectively. 

Area of walls = 2(l+b)xh = 2(3.5+6)x2 = 38 sq. unit. 

% Increase in the area of walls = (38-20)x100/20 = 90%.

Let C.P. of each Magazine be Rs. 1

C.P. of P articles = Rs. P. 

S.P. of P articles = Rs. 40.

Profit = Rs. (40 - P)

Now, Gain% = ProfitC.P×100 = 40-PP×100 

Here 40-PP×100= 25

-->P = 32.

Given monday population = 3000
Per day 20% increase in population.

Tuesday population = 3000 + 3000x20/100 = 3000 + 600 = 3600

Wednesday population =3600 + 3600x20/100 = 3600 + 720 = 4320

Thursday population = 4320 + 4320x20/100 = 4320 + 864 = 5184

So that day is Thursday.