Quantitative Aptitude - Arithmetic Ability Questions

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%.

From the given data,

containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.

Biggest measure that can measure all these different quantities exactly can be given by

HCF of (1365, 1560, 1755) 

That can be found as 

HCF of 1365, 1755 = 195

Now, HCF of 195 and 1560 = 195

Hence, biggest measure that can measure all the given different quantities exactly is 195 lit.

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.

X= 2/3Y 

=> Y= 3/2X

Aage of X after 6 years = 46 years

So present age of X = 46-6 = 40 years

Now, present age of Y = 3/2 * 40 = 60 years.

Let the present age of Deepa = 5p

Let the present age of Hyma = 6p

After four years their ratio = 6 : 7

=> 5p + 46p + 4 = 67

=> 35p + 28 = 36p + 24

=> p = 4

Therefore, the difference between their ages = 24 - 20 = 4 years.

 C.I. when interest compounded yearly

Rs.5000*1+4100*1+12*4100

 Rs.5000*2625*5150

=Rs.5034

C.I. when interest iscompounded half-yearly 

 Rs.5000*1+21003

= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04

From given data,

R : M = 5 : 7

M : A = 5 : 7

R : M : A = 25 : 35 : 49

25 + 35 + 49 = 109

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.