Quantitative Aptitude Prep Tips for MBA

Five years ago the ratio of their ages was 5:3 with the total age being 64 years. Their individual ages will be 40 and 24 years.

Today their ages will be 45 years and 29 years and after 9 years their ages will be 54 and 38 years.

Required ratio will be 27 : 19.

Let X be the units digit of the number and Y be the tens digit.

The number is 10Y + X

The reversed number is 10X + Y

(10Y + X) – (10X + Y) = 18 or 9Y – 9X = 18

Digits = X + Y = 10

on solving X = 4, Y = 6

612 = 212 * 312. For perfect cubes, power must be multiple of 3.

Therefore, power of 2 equal 20, 23, 26, 29, 212 and power of 3 equals 30, 33, 36, 39, 312 So, number of divisors equal 5* 5 = 25.

New rent = 2730 per month.  Rs. 2720 * 12 = Rs. 32640 per year

Net for X = 70 * 4 + 50 * 3 + 66 * 5 = 760

For Y = 50 * 4 + 70 * 8 = 760

For Z = 40 * 7 + 24 * 5 = 400

The shares will 12920, 12920 and 6800.

Let the required time be t hours after 9 am. The clock would have lost time = 4t min.

Now according to the problem

After 7 hours, clock will lose = 4 * 7 = 28 min.

At 4 pm clock will show 3:32 pm and it will lose 2 minutes more till it reaches 4 : 00 pm.

So, actual time will be 4 : 30 pm

When T2 is at 45 points, T3 is at 40 points, that is, T2 can beat T3 by 5 points in a 45 point game. In a 765 point game, T2 can beat T3 by $\frac{765}{45}*5$ $\frac{}{}$ = 85 points.

Selling price = Rs. 16.56

Cost price for 38/% profit = 1.38 C.P. = 16.56

C.P. = $\frac{15}{1.25}$ $\frac{}{}$ = Rs. 12

$\begin{array}{l}\frac{Chemical one }{Chemical second}=\frac{\left(13.5-12\right)}{\left(12-9\right)}\\ =\frac{1.5}{3}=\frac{1}{2}\end{array}$

Let the price of P2 per kg be Rs. X. Then the price of P1 per kg = Rs. 3X.

1 kg of P3 P3contains

$\left(\frac{2}{7}\right)*1,i.e.,\frac{2}{7}kgofAand\frac{5}{7}kgofB$

Price of 1 kg of P3 = $\left(\frac{2}{7}\right)*3X+\left(\frac{5}{7}\right)X=\left(\frac{11}{7}\right)X$ $\frac{}{}\frac{}{}\frac{}{}$

P2y the given condition,

$\begin{array}{l}\frac{11X}{7}=10.4-1.6=Rs.8.8\\ \Rightarrow X=8.8*\left(\frac{7}{11}\right)=5.6\end{array}$

Hence, the price of P1 per kg = 3X = 3 * 5.6 = Rs. 16.8

Price of P2 per KG = Rs5.6

Let X = working time

$\frac{74}{100}*5574300$

Solve to get X = 6 hours.

It is given that 40 men left after $\frac{1}{4}th$ $$ of the time, that is, after 60 days. After 60 days, the camp had food for 160 men for 180 days.

This amount is to be consumed by 120 men since 40 men left in the camp.

Therefore, 160 * 180 = 120 * y or y = 240 days

But in the question, we have to find the total number of days that the food will last.

Therefore, number of days = 60 + 240 = 300 days.