Quantitative Aptitude Prep Tips for MBA

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 +

(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.

Q can be 0 or 5, but 65 is not divisible by 4.

So, Q must be 0.

Now, sum of the digit should be divisible by 9.

So, P is 8.

Let y gold coins were given to 1st son.

Number of gold coins given to 2nd son = $\frac{5}{7}y$

Number of gold coins given to 3rd son = $\frac{3}{7}y$

So,  $y-\frac{3}{7}y=60\Rightarrow \frac{4y}{7}=60$

$\Rightarrow y=\frac{60\times 7}{4}=15\times 7=105$

Hence, the required number of gold coins:

$=105+\frac{5}{7}\times 105+\frac{3}{7}\times 105$

= 105 + 75 + 45 = 225

The sum of three numbers is given; the product will be maximum if the numbers are equal.

So, if a + b + c is defined, abc will be maximum when all three terms are equal. In this instance, however, with a, b, c being distinct integers, they cannot all be equal. So, we need to look at a, b, c to be as

Coefficient of x49 = − Sum of the roots of f (x)

= − (1 + 3 + 7 +9+ . + 99)

− (50)²= −2500

(15, 3)! = Product of 3 consecutive natural numbers starting from 15, which is at least divisible by 3!

Hence, H = 3!

Consider the vowels A, E as one unit. Besides this, we have 2R, 1N, 1G that is a total of 5 things out of which the 2 R are identical.

Required number of ways $=\frac{5!}{2!}$

Now A, A and E can be arranged within themselves in ways. $\frac{3!}{2!}$

Required answer = $\frac{5!\times 3!}{2!\times 2!}$

 $\frac{lo{g}_{y}8x}{lo{g}_{y}x}=\frac{4}{3}$

Log_x 8x = $\frac{4}{3}$

Log_x 8 + log_xx = $\frac{4}{3}$

Log_x 8 = $\frac{1}{3}$

x = 8³

x = 512