Quantitative Aptitude Prep Tips for MBA

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

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

Hence, the required number of gold coins:

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

= 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 close to each other as possible.

a = 10, b = 10. c = 11 is one possibility, but a, b, c have to be distinct. So, this can be ruled out.

The close options are,

a, b, c = 9, 10, 12; product = 1080

a, b, c = 8, 11, 12; product = 1056

Maximum product = 1080

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 = $\frac{5!*3!}{2!*2!}$

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

Log_x 8x = $\frac{4}{3}$

Log_x 8 + log_xx = $\frac{4}{3}$

Log_x 8 = $\frac{1}{3}$

x = 8³

x = 512

Option (b) and (c) both satisfies the given condition, but 997918 is the larger of the two.

Since x > 0 then at x = 1, we get y = 2 and for the rest values we get y > 2

Also for 0 < x < 1 ; x+ $\frac{1}{x}$ > 2

and for x > 1 ; x+ $\frac{1}{x}$ > 2

Hence, y ≥ 2

We know that,

$a^2+a^2={\left(4\sqrt[]{2}\right)}^2,a^2=16,a=4$

Now, 2 (area of square) = (square of diagonal)

So, for new area 32 (16 * 2)

2 * 32 = d²

8 = d

Final ratio of wages = 5 * 4 : 7 * 3 : 9 * 2 = 20 : 21 : 18

Total wages received by P = $\frac{20}{59}*5900=2000$

Daily wage of P = $\frac{2000}{5}=400$