Quantitative Aptitude - Arithmetic Ability Questions

Area of a square SxS = 1024 => S = 32 cm
Length of the rectangle = 32x2 cm = 64 cm
Breadth of the rectangle = 32 - 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5

Let the price of each note book be Rs.x.

Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.

Hence xy = 300

=> y = 300/x

(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy

=>5(300/x) - 10x - 50 = 0 => -150+x2+5x

multiplying both sides by -1/10x

=> x2+15x-10x-150=0

=> x(x + 15) - 10(x + 15) = 0

=> x = 10 or -15

As x>0, x = 10.

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres.

=> the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres

However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity.

i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.

Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

Compound ratio of A:B:C
A:B = 3:4
B:C = 5:6
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A:B:C = 15:20:24

We can divide Rs.118000 in this ratio.

Share of A = 15/59 x 118000 = 30000

Share of B = 20/59 x 118000 = 40000

Share of C = 24/59 x 118000 = 48000

Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

Let the total number of valid votes be 'V'.

70% of V = 70/100 x V = 7V/10

Number of votes secured by the other candidate = V - 7V/100 = 3V/10

Given, 7V/10 - 3V/10 = 172 => 4V/10 = 172

=> 4V = 1720 => V = 430.