Quantitative Aptitude Prep Tips for MBA

(a):Let G be (X, Y) then $X=\frac{\left\{3+5+(-3)\right\}}{3}=\frac{5}{3}$ $\frac{}{}\frac{}{}$

$Y=\frac{(7+5+2)}{3}=\frac{14}{3}$

$\Rightarrow Gis \left(\frac{5}{3},\frac{14}{3}\right)$

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

17 – 12 = 5

(c) :As per the problem:

Average score of 20 candidate = 25 marks

Total score of 20 candidate = 500 marks

Let the score of the topper be x. Then,

$\frac{(500-x)}{19}=23$

500 – x = 437

x = 500 – 437 = 63 marks

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

12 * 9 – 17 * 5 = 23

f (3n) – f (3n– 3) = n

n = 1

f (3) – f (0) = 1

f (0) = 0

n = 2

$\begin{array}{l}f\left(6\right)–f\left(3\right)=2\\ ?\text{?}??\end{array}$

f (3n) – f (3) = 2 + 3 + 4+…………….+ n

f (3n) = 1 + 2 + 3 +…………….+ n

= $n\frac{\left(n+1\right)}{2}$

$?$ f (1312) – f (3 * 311)

$\frac{3^{11}}{2}\left(3^{11}+1\right)=\frac{3^{22}+3^{11}}{2}$

Hence, remainder $\left[\frac{\frac{3^{22}+3^{11}}{2}}{10}\right]$ = 8

$1-log10_{10}5=\frac{1}{3}\left[log10_{10}\frac{x*5^{1/3}}{2}\right]$

3 [log101010 – log10105] = $log10_{10}\frac{x*5^{1/3}}{2}$

log101023 = $log10_{10}\frac{x*5^{1/3}}{2}$

23 = $\frac{x*5^{1/3}}{2}$

x = $\frac{2^4}{5^{1/3}}$

Hence, m + 3n = $4+3*\frac{1}{3}$ = 5

Let the interest rate be r percent interest earned from 1.2 Lacs at the end of year

= 1,200,000 * r /100

= 1200r

From 1.8 lakh, = 8/12 * 1, 80,000 * 2r/100

= 2400r

Total interest = 3600r = 54000

r = 15 percent

Let the quotient obtained when the numbers is divided by 3, 5 and 6 be x₁, x₂ and x₃

Respectively

N = 3 x₁ + 1

x₁ = 5 x₂ + 3

x₂ = 6 x₃ + 2

N = 3 [5 (6 x₃ + 2) + 3] + 1

=90 x₃ + 40 < 1000

x₃ can take 11 values

Assuming that

b – c = p b = c + p

c – a = q c = a + q

Hence b = c + p

= a + q + p

$?$ $y+\frac{1}{(b-c)a(c-a)}$ can be written as a + q + p + $\frac{1}{a.p.q}$

Apply AM GM in equally concept in above equation

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 4\sqrt[]{a*p*q*\frac{1}{a.p.q}}$

$\frac{a+q+p\frac{1}{a.p.q}}{4}\ge 1$

$a+q+p\frac{1}{a.p.q}\ge 4$

Hence, minimum value is 4