Quantitative Aptitude - Arithmetic Ability Questions

 Investment ratio of A, B & C = (4000 ×6+3000×9):(5500 ×6+4000×9):4

Given length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

Then, the diagonal of the rectangle D = 32 + 42 = 25 = 5

Then, it implies side of square = 5 cm

We know that Area of square = S x S = 5 x 5 = 25 sq.cm.

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second Or

Case II: 4 students in the first car and 4 in the second

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is

8C3+8C4 = 56 + 70 = 126.

F = Rs. 8100

R = 5%

T = 3 months = 1/4 years

BD=FTR100=8100×14×5100=Rs.101.25

TD=FTR100+TR=8100×14×5100+14×5=Rs.100

Therefore BD - TD = 101.25-100 = Rs.1.25

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.

Speed of train relative to man = 67 - 7 = 60 km/hr.
= 60 x 5/18 = 50/3 m/sec.

Time taken to pass the man = 600 x 3/50 = 36 sec.