Quantitative Aptitude - Arithmetic Ability Questions

Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4

According to the question,

Difference between square of the average of first two odd number and the of the average last

two odd numbers is 396

i.e, x+3 and x-3

$$\begin{gathered} ⇒x + 32 - x - 32 = 396 \\ ⇒ x2 + 9 + 6x - x2 + 6x - 9 = 396 \\ ⇒ 12x = 396 \\ ⇒ x = 33 \end{gathered}$$

Hence, the smallest odd number is 33 - 4 = 29.

K's one day's work = 1/30
L's one day's work = 1/45
(K + L)'s one day's work = 1/30 + 1/45 = 1/18

The part of the work completed in 3 days = 3 (1/18) = 1/6.

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.

S.I=PNR/100

$$\begin{gathered} 2211 x 75 x 112 \\ => 222 x 75 x 112 \end{gathered}$$

 Total number of prime factors is (22+5+2) = 29

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

P1+20100n > 2P 

Now, (6/5 x 6/5 x 6/5 x 6/5) > 2.

So, n = 4 years.