Quantitative Aptitude - Arithmetic Ability Questions

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.

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 

Clearly total persons are increased in  28/35 :: 52/65 = 4:5.

As time is inversely proportional to men, so total time will decrease in the ratio 5:4.

Hence, 22.5 x 4/5 = 18 days.

Let original length = x and original breadth = y.

Original area = xy.

New length = x/2 and New breadth = 3y.

New area = (x/2 * 3y) = (3/2 )xy

Therefore,  Increase % = [(1/2)xy * (1/xy) * 100] % = 50%

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

Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs

Hence, if bus runs at its own speed, the time saved = 22 - 19.8 = 2.2 hrs.

Given L1 = 140 m

L2 = 160 m

S1 = 60 km/hr

S2 = 80 km/hr

From the question we get,

S1 + S2 = (L1 + L2) / T

=> (60 + 80) 5/18 m/s = 140 + 160/T

=> T = 54/7 = 7.71 sec