Quantitative Aptitude - Arithmetic Ability Questions

In case B, odd times even will give even

Perimeter = Distance covered in 8 min. = (12000/60) * 8 = 1600m 

Let length = 3x metres and breadth = 2x metres.  

Then, 2(3x + 2x) = 1600 or x = 160.  

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) = 153600 

n (A) = (60/100 * 96) = 288/5, n (B) = (30/100 * 96) = 144/5, n (A ∩ B) = (15/100 * 96) = 72/5.

n(A ∪ B) = n (A) + n (B) - n (A ∩ B) = 72.

So, people who had either or both types of lunch = 72.

Hence, people who had neither type of lunch = (96 - 72)= 24.

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, 50C3

(Men 14 : 16) : (Hrs/Day 12 : 7) :: 48 : x
14 * 12 * x = 16 * 7 * 48 or x = 32

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.

Manju borrows Rs. 5000 for 2 years at 4% p.a. simple interest 

She also lends it at 6 1⁄4% p.a for 2 years 

=> Total Gain = 6 1/4% − 4% = 2 1/4%

So her gain in the transaction for 1 year

= The simple interest she gets for Rs.5000 for 1 year at 2 1⁄4% per annum

= PTR100=5000×94×1100 = Rs. 112.5/ year.

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360