Quantitative Aptitude Prep Tips for MBA

Let the total work is 300 units (LCM of 150,100 and 60)

A's per day work = 2 units

B's per day work = 3 units

C's per day work = 5 units

(A+ B + C)'s per day work = 2 + 3 + 5 = 10 units

Time taken = 300/10 = 30 Days

Let y = 2^111x then, $\frac{1}{4}{y}^3+4y=2y^2+1$  

y³ – 8y² +16y – 4 = 0

Let three real roots be a, b and c then abc = and corresponding values of x be x₁,   x₂, and x_3\

$2^{111 (x_1 + x_2 + x_3)}= 4$

111 (x₁ + x₂ + x₃) = 2

x₁ + x₂ + x₃ = 2/111

m + n = 2 + 111 = 113

p and q are the roots of the equation x² – 4ax + 3 = 0

pq = 3

Similarly the constant term of x² – 2bx + c = 0

Product of $\left(\frac{2}{p}-q\right)and\left(\frac{2}{q}-p\right)$

Hence, c = $\frac{4}{pq}-2-2+pq=\frac{1}{3}$

f (x) = $\frac{x}{4x^2+7x+9}$ ………. (1)

Divide by x in numerator and denominator equation (1) can be written as f (x) = $\frac{1}{4x+7x+\frac{9}{x}}$

f (x) is maximum, when the denominator is minimum we have to find the minimum value of $4x+\frac{9}{x}$ , so, denomination will also minimum

we know that, AM ≥ GM

$\frac{4x+\frac{9}{x}}{2}\ge \sqrt[]{4x*\frac{9}{x}}$

$\frac{4x+\frac{9}{x}}{2}=6$

$4x+\frac{9}{x}=12$

The minimum value of $4x+\frac{9}{x}$ is 12

Therefore, the maximum value of f (x) = $\frac{1}{12+7}=\frac{1}{19}$

For Aman,

SP of 1 book = 14Rs.

CP of 1 book = $\frac{100}{(100+x)}*14$

For Baman,

SP of 1 book = 12Rs.

CP of 1 book = $\frac{100}{(100+y)}*12$

Price of all notebooks is same, therefore.

$\frac{100}{(100+x)}*14$ = $\frac{100*12}{100+y}$

100 + 7y = 6a

100 = 12y – 7y; [x=2y]

5y = 100

y = 20

a = 40

CP each notebook = $\frac{100*12}{100+20}$ = 10Rs.

In his sales with Daman, he sold 10 books to her for Rs. 130

CP of 10books = Rs. 100

Profit percent = $\frac{130-100}{30}*100$

=30percent

a + b + c + d = 7

a ³ 0, b³ 0, c ³ 1, d ³ 1

(a, b, c, d are the numbers of balls)

Let c = c' + 1 = 0, 0 ≤ c' ≤ 5

d = d' + 1 = 0, 0 ≤ d' ≤ 5

a + b + c' + d' = 5

Numbers of solution = ^{5 + 4 – 1}C_4–1

= ⁸C₃ = 56

Let f (x) = ax² + bx + c

f (0)= 2

2 = a (0)² + b (0) + c

c = 2

f (– 2) = 3

a (–2)² +b (–2) + 2 = 3

4a – 2b = 1                    ………. (1)

The given equation has a maximum value at x = –2

$\frac{-b}{2a}=-2$

b = 4a

put in equation 1

we get, $a=\frac{-1}{4},b=-1$

f (6)= $\frac{-1}{4}{(6)}^2+(-1)6+2$

= –13

P₁ + P₂ + P₃ = 59

P₁ and P₂ → minimize

Then P₃ → minimum i.e. 47

So, only possible value of P₃ can take → 29, 31, 37, 41, 43, and 47

Total 6 values