Quantitative Aptitude - Arithmetic Ability Questions

Total running distance in four weeks = (24 x 240) + (4 x 400)

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms

From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

Given share of L is 2/7 th of profit. Then, their profits are divided in to 5:2 ratio.

Ratio of K and L is

16000×8x×4=52

x = Rs. 12,800.

Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.

Therefore, Required sum = 1a+1b=a+bab=55600=11120

Let O's share = Rs. P

=> N's share = 75100 x P = 3P4

M's share = 54 x 3P4 = 15P16

$$\begin{gathered} 15P16 + 3P4 + P = 4300 \\ => P = 4300 x 1643 = 1600 \end{gathered}$$ 

The servant worked for 9 days instead of 12 days, he should receive 9/12 of his total payment

Let the price of 1 shirt be Rs. S 

i.e., 3/4 (400 + S).

However, the question states that the servant receive Rs. 280 + S where S is the price of the shirt.

By equating the two equations we get 3/4 (400 + S) = 280 + S.

Therefore, Price of the shirt S = Rs. 80.

Total age of 10 students = 150 years

Total age of 15 students = 240 years

Total age of 5 new students = 240 - 150 = 90 years

 Average age of 5 new students = 90/5 = 18 years.