Quantitative Aptitude - Arithmetic Ability Questions

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

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

Converting each of the given fractions into decimal form, we get

2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833 

Since 0.833>0.8>0.75>0.66 

So, 5/6 > 4/5 > 3/4 > 2/3 

Therefore,  Required Difference = 5/6 - 2/3 = 1/6

a^3 + b^3 = (a+ b) * (a^2 + b^2 - ab)

[ (10.3)^3 + (1)^3 ] / [(10.3)^2  + (1)^2 - (1 * 10.3)]

 => a + b = 10.3 + 1 = 11.3 

Let K = Kalyan's monthly salary
N = Nandu's monthly salary
S = Sarnesh's monthly salary.
Then from given data,
12% of K = 16% of N
S = N/2 => N = 2S
K = 16/12 x N = 16/12 x 2S
= 16/6 x 1.08/12 = 16/6 x 0.09 = 0.24 lacs = 24,000

Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.

Let marks obtained by N = x

M's marks = 0.80x

L's marks = 1.25 (0.80x)

K's marks = (0.90) (1.25) (0.80x)

= 0.9x

But 0.9x = 360 ⇒ x = 400

Percentage of Marks obtained by N

= 400 x 100/500 = 80%

Let the cost price of 1 article = Rs. 1

From the given data,

Then, the selling price of 1 article = 22/18 = 11/9

Then, Profit = SP - CP = 11/9 - 1 = 2/9

Required, profit % = Profit/CP x 100

= [(2/9)/1] x 100

= 200/9

= 22.222%.