Quantitative Aptitude - Arithmetic Ability Questions

End of first year     :  10000 x (3/5) = Rs. 6000
End of second year :  6000 x (3/5) = Rs. 3600
End of third year    :  3600 x (3/5) = Rs. 2160

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. 

Solving the two equations, we get: l = 63 and b = 40. 

Area = (l x b) = (63 x 40) sq.m

= 2520 sq.m

Given A & B in partnership

A Invests 116000 for 12 months 

=> A's share = 116000 x 12 = 13,92,000

B Invests for 6 months

=> B's share = 144000 x 6 = 8,64,000

Their Ratio = 1392 : 864 = 29 : 18

Let the Annual profit = P

Given B's share = Rs. 9000

=> 18/47 x P = 9000

=> P = 9000 x 47/18

=> P = 23,500

Hence, Overall profit = P = Rs. 23,500

Sum of 4 digit numbers = (2+4+6+8) x P33 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x P23 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x P13 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x P03 x (1) = 20 x 1 x 1 = 20 

Adding All , Sum = 147320

Let the age of Manisha and Sudeshna is 5x and 6x years respectively.

According to the given data,

5x+8/6x+8 = 7/8

42x + 56 = 40x + 64

x = 4

Required Difference of their ages = (6 × 4)  –  (5 × 4)

= 4 years.

Given a jeep travels a certain distance taking 6 hrs in forward journey
During the return journey, it takes 4 hrs with an increased speed 12 km/hr
Let 'x' be the distance.
Now, speed is given by distance/time.
Here the difference between both speeds = 12

 x4 - x6 = 12

 3x - 2x12 = 12

=> x = 144 km

Therefore, the distance travelled by the jeep in forward or return journey is x = 144 km.

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800x1860x5= 48 sec.