Quantitative Aptitude - Arithmetic Ability Questions

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

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.

FV=P(1+r/n)^nt

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.

A = P + Prt

Let Principal = Rs. P and Rate = R% p.a. Then,

Amount= Rs.P1+R1004

 C.I = P1+R1004-1

 P1+R1004-1=1491

Clearly, it does not give the answer

Let the sum be Rs. x. Then

C.I=x[1+4/100)^2-x]=[626/675x-x]

S.I=x^2/25

C.I-S.I=1