Quantitative Aptitude - Arithmetic Ability Questions

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%

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

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

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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 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%.

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

80x900/100 + 256 x 4 x ? = 1331 - 99

720 + 1024 x ? = 1232

? = 1232 - 720/1024

? = 512/1024 = 0.5

? = 0.5