Quantitative Aptitude Prep Tips for MBA

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.

37.5x + 97.5x + 30x = 10230

X = 62

97.5x = 97.5 * 62 = 6045 ml.

Let the total work be 240 units.

Total Work = 240 Units

Pipe 1: 40 hours = 6 units/h

Pipe 1 + Leak: 48 hours = 5 units/h

Leak can empty 1 unit in 1 hour.

Pipe 2 can fill 3 units per hour (half as efficient as the first one)

Therefore, effective filling is 2 units per hour.

Therefore, 120 units (half the tank) will get filled in 60 hours.

Weighted means give us

$\frac{}{}$ $\left(\frac{4}{25}+\frac{7}{40}+\frac{1}{4}+\frac{9}{200}\right)$  max. = 63

$\frac{}{}$ $\frac{\left(32+35+50+9\right)}{200}$  max. = 63

max = $\frac{63*200}{126}$ $\frac{}{}$

= 100

Time taken by two trains will be 9 hours and 6 hours respectively. So, Let the total distance be LCM of (9 and 6) = 18.

Speed of trains will be 2 km/hr & 3 km/hr.

At 9 am distance between two trains is 14 km

Time taken =   $\frac{14}{\left(2+3\right)}$ $\frac{}{}$ = 2 hours 48 minutes

So, trains will meet at 11:48 am.

If Z earns Rs 100a, X earns 120a & Y earns 160a

160a – 100a = 60a = 7980

a = 133

X= 120 * 133 = 15960

Y = 160 * 133 = 21280

Z = 100 * 133 = 13300

L.C.M. of 5, 6, 7, 8

= 35 * 24 = 840

Therefore, the required number = 840x + 3,

Which is exactly divisible by 11.

For x = 2, it is divisible by 11.

Hence,

Required number = 840x + 3

= 840 * 2 + 3

= 1683

32% of CP = Rs. 8320 or CP = Rs. 26000

Now the Aman wants a profit of 62%.

Therefore, he has to sell it at a price of 26000 * 1.62 = 42120

The solid sphere is melted and recast into cones. The volume of material is the same before and after casting.

Volume of the sphere = $\left(\frac{4}{3}\right)\pi r^3$ . (1)

Volume of the right circular cone = $\frac{1}{3}\pi R^{2}H$ . (2)

Diameter of the base of the cone

= $\sqrt[]{2}$ slant height

= $\sqrt[]{2}$ S (say). (3)

4R = 2S = 2 (H² + R²)

2R² = 2H

R = H. (4)

Volume of the cone = $\frac{1}{3}\pi R^{2}H$

$=\frac{1}{3}\pi r^2.(r)$  (since, R = r)

$=\frac{1}{3}\pi r^3$

From equations (1) and (4), it can be seen that the melted material creates exactly 4 cones of the specified dimensions. No material is left over.

(6x² + 43x – 49) – (5x² + 2x – 7) = 0

x² + 41x – 42 = 0

x² + 42x – x – 42 = 0

x (x + 42) – 1 (x + 42) = 0

(x – 1) (x + 42) = 0

x = 1 satisfy both the equations. So, there is only one common root.