Quantitative Aptitude - Arithmetic Ability Questions

Let P's speed = x km/hr. Then, Q's speed = (7 - x) km/ hr.

So, 24/x + 24/(7 - x) = 14

x2 - 98x + 168 = 0

(x - 3)(x - 4) = 0 => x = 3 or 4.

Since, P is faster than Q,

so P's speed = 4 km/hr and Q's speed = 3 km/hr.

In this type of question, we will first calculate the distance covered in given time.

Distance covered will be : Number of revolutions x Circumference 

So we will be having distance and time, from which we can calculate the speed. 

Radius of wheel = 70 / 2 = 35 cm

Distance covered in 40 revolutions will be 40*circumferene=40*2πr=40*2*227*35=8800cm = 88 m

Distance covered in 1 sec  = 8810=8.8m/s=8.8*185km/hr=31.68km/hr

Take a square of side 4cm and a rectangle having l=6cm and b=2cm

then perimeter of square = perimeter of rectangle

area of square = 16 sq.cm

area of rectangle = 12 sq.cm

Hence a >b

Let the numbers be x and y. Then, 2x + 3y = 39...(i)and 3x + 2y = 30 ...(ii)

On solving (i) and (ii), we get :  x = 6 and y = 9.

larger number = 9

Number of runs made by running = 120 - (3 x 4 + 8 x 6) = 120 - (60) = 60 

Required percentage = [(60/120)*100] % = 50%

Let the original price of the cycle be 100. After the first reduction the price will be 75.

This new price is then reduced by 20% = 0.8 x 75 = 60

60 represents a reduction of 40 percent on the original.

when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm

Let the side of the square plot be 'a' ft.
Given area of the plot (a x a) = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 x 58 = Rs. 3944.