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7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

In how many way can 7 identical balls be placed in 4 boxes, A, B, C and D such that boxes C and D have at least one ball each?

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Payal Gupta

Contributor-Level 10

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

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

f(x) is a quadratic polynomial with maximum value of 3 at x = 2. If f (0) = 2. Find f (6)

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Payal Gupta

Contributor-Level 10

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

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

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

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

ABCD is a square, having side 12cm, in which ABC and ABD are two quadrant of a circle. If a circle is drawn touching both the quadrants and the square as shown in figure as show in figure given below, then find the area of circle.

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Payal Gupta

Contributor-Level 10

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

If P₁ + P₂ + P₃ = 59, where P₁, P₂ and P₃ are prime numbers and P₁ < P₂ < P₃, then how many values can P₃ take?

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Payal Gupta

Contributor-Level 10

P₁ + P₂ + P₃ = 59

P₁ and P₂ → minimize

Then P₃ → minimum i.e. 47

P₁	P₂	P₃
5	7	47
5	11	43
3	13	43
5	13	41
5	17	37
3	19	37
5	23	31
7	23	29

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

Total 6 values

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

How many pairs of positive integers x, y satisfy $\frac{1}{x} + \frac{6}{y} = \frac{1}{15}$ , where y is a two digit odd integer?

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Payal Gupta

Contributor-Level 10

$\frac{1}{x} = \frac{1}{15} - \frac{6}{y}$

$\frac{1}{x} = \frac{y - 90}{15 y}$

$x = \frac{15 y}{y - 90}$ ______equation 1

i.e. y = 90+y,

1≤ y₁ ≤9

y₁ is odd

Equation 1 becomes

$x = \frac{15 (90 + y_{1})}{y_{1}}$

$= 15 + \frac{90 * 15}{y_{1}}$

y₁ can take 1,3,5,9

y=91, 93, 95, 99

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

An alcohol solution is formed by mixing two solutions. Solution 1 had a concentration of 20 percent and a volume of 3L while solution 2 had a concentration of 40 percent with a volume of 1L. The resultant solution was named solution 3. Three litres of solution 4 which had an unknown concentration was mixed with solution 3. What must be concentration of solution 4 if that the new resultant solution has a concentration of 50 percent?

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Payal Gupta

Contributor-Level 10

Sol1	20percent (0.6)	3L
Sol2	40percent (0.4)	1L
Sol3	1L	4L
Sol4	XL	3L
Sol5	3.5L	7L

1 + x = 3.5

x= 2.5

Required percent y = $\frac{2.5}{3} * 100$ = 83.33 percent

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

Two geometric means a, b are inserted between x and y such that a³+ b³ = n (ab) A,where A is the arithmetic mean of x and y. What is value of n?

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Payal Gupta

Contributor-Level 10

x, a, b, y are in GP

a = xr

b = xr²

y = xr³

r = ${(\frac{y}{x})}^{1 / 3}$

rewrite,

a = x^2/3.y^1/3

b = x^1/3.y^2/3

a * b = x * y

a³+ b³= x²y + y² x

= xy (x+y)

= xy (2A) …………… $[\frac{x + y}{2} = A]$

Hence, n = 2

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of equilateral of the equilateral triangle to the radius of circle.

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Payal Gupta

Contributor-Level 10

Let the side of = a

CD = a/2

Let EC = x

In GEC

GE/EC = tan60 = $\sqrt[]{3}$

GE = x $\sqrt[]{3}$

DE = a/2 – x = FG = HF

GE = HG

$x \sqrt[]{3} = \frac{a}{2} - x + \frac{a}{2} - x$

x $\sqrt[]{3}$ = a – 2x

x = $\frac{a}{\sqrt[]{3} + 2}$

GE = x $\sqrt[]{3}$ = $\frac{a \sqrt[]{3}}{\sqrt[]{3} + 2}$

r = side of square ÷ 2

r= $\frac{a \sqrt[]{3}}{4 + 2 \sqrt[]{3}}$

ratio = a/r

= $\frac{a (4 + 2 \sqrt[]{3})}{a \sqrt[]{3}}$

= $\frac{4 + 2 \sqrt[]{3})}{\sqrt[]{3}} = \frac{4}{\sqrt[]{3}} + 2$

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

A certain sum of money invested at CI becomes Rs 3380 in 2 years & Rs 3515.20 in 3 years. What is the sum invested & what is rate % per annum?

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Vishal Baghel

Contributor-Level 10

(d):3515.2 – 3380 = 135.2 = SI on Rs. 3380 for one year.

By SI formula, 135.2 = $3380 * R * \frac{1}{100}$ $\frac{}{}$ or R = 4 %

Now we have 3380 = P (1.04)2 or P = Rs. 3125

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New answer posted

7 months ago

Maths Quantitative Aptitude Prep Tips for MBA

Duronto Express in motion, 66 metre long overtakes a passenger train 88 metres long travelling at 30 kmph in the opposite direction in 0.168 minute. Find speed of the Duronto Express.

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Vishal Baghel

Contributor-Level 10

(c):Let speed of passenger train be x m/s.

$[(\frac{66 + 88}{8.33 + x})] = 0.168 * 60$

$[30 * \frac{5}{18} = 8.33]$

x = 6.94 m/sec = 25 kmph.

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