Quantitative Aptitude - Arithmetic Ability Questions

Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs

Hence, if bus runs at its own speed, the time saved = 22 - 19.8 = 2.2 hrs.

$$\begin{gathered} 112 × 54 \\ 112 × 1024 \\ = 112 × 1000024 \\ = 1120000/16 \end{gathered}$$

= 70,000.

S = { 1, 2, 3, 4, 5, 6 } 

=> n(S) = 6

E = { 3, 6}

=> n(E) = 2

Therefore, P(E) = 2/6 =1/3

Clearly total persons are increased in  28/35 :: 52/65 = 4:5.

As time is inversely proportional to men, so total time will decrease in the ratio 5:4.

Hence, 22.5 x 4/5 = 18 days.

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

63.5% of 8924.2 +? % of 5324.4 = 6827.5862

=> 63.5% of 8925 + ? % of 5325 = 6830

=> 6830 ~= 5320 + ?(5325)/100

=> ? ~= 22

Given L1 = 140 m

L2 = 160 m

S1 = 60 km/hr

S2 = 80 km/hr

From the question we get,

S1 + S2 = (L1 + L2) / T

=> (60 + 80) 5/18 m/s = 140 + 160/T

=> T = 54/7 = 7.71 sec