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Class 12th Quantitative Aptitude Prep Tips for MBA

Three solid metal spheres of radii 3 cm, 4 cm, 5 cm respectively, are melted together. The metal is recast as a single solid sphere. Find the percentage reduction in the area of surface resulting from this.

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Payal Gupta

Contributor-Level 10

Let R be the radius of the recast sphere

$\frac{4}{3} π R^{3} = \frac{4}{3} π (3^{3} + 4^{3} + 5^{3})$

R³ = 216 $\Rightarrow$ R = 6

Total surface area of 3 spheres

$= 4 π (3^{2} + 4^{2} + 5^{2}) = 4 π * 50 = 200 π^{}$

Surface area of the recast sphere = 4π (6²)=144

% reduction in area $= \frac{(200 π - 144 π)}{200 π} * 100 % = 28 %$

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Class 12th Quantitative Aptitude Prep Tips for MBA

In a geometric progression, product of 11th term, 18th term and 20th term is equal to the product of 28th term and 5th term. Which of the following is necessarily true about that geometric progression?

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Payal Gupta

Contributor-Level 10

ar¹⁰ * ar¹⁷ * ar¹⁹ = ar²⁷ * ar⁴

a³r⁴⁶ = a²r³¹

ar¹⁵ = 1

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Class 12th Quantitative Aptitude Prep Tips for MBA

What kind of triangle is formed by points (4,3), (6,-2) and (-11, -3)?

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Payal Gupta

Contributor-Level 10

Let A (4, 3), B (6, −2), C (−11, −3) be the vertices of triangle ABC.

AB² = (6 – 4)² + (−2 −3)² = 29

BC² = (−11 −6)² + [−3 – (−2)]² = 290

AC² = (−11 −4)² + [ (−3 −3)]² = 261

As AB²+ AC²= BC²

ABC is a right-angled triangle.

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Class 12th Maths

Find the equation of the line having the y-intercept equal to −3 and perpendicular to the line of equation 2x = 8 – 3y.

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Payal Gupta

Contributor-Level 10

Slope of given line = $- \frac{2}{3}$ perpendicular line has slope = $\frac{2}{3}$

Now c = −3 (given)

Putting in general equation y = mx + c

2y = 3x ⇒ 3x-2y=6

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Class 12th Quantitative Aptitude Prep Tips for MBA

The gross annual sales of a firm are Rs. 7500. The cost of materials and manufacture is 35% of the gross sales. Rent, rates and taxes amount to Rs. 772 ; advertising to Rs. 1880 ; office expenses to Rs 272 ; manager's salary Rs. 400 plus 1% of gross sales. Find the % profit.

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Payal Gupta

Contributor-Level 10

Total expenses = (35% of 7500) + 712 + 1880 + 272 + 400 + 75 = 6024

Total receipts = 7500 Total profit = 7500 – 6024 = 1476

% profit = $(\frac{1476}{6024}) * 100 = 24.5 %$

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Class 12th Quantitative Aptitude Prep Tips for MBA

Rs.1720 is divided among 45 persons consisting of men, women and children. The sum of men's, women's and children's shares are as 12 : 15 : 16 but the individual shares of a man, woman & child are as 6 : 5 : 4. Find respective share of 1 man, 1 women and 1 child.

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Payal Gupta

Contributor-Level 10

We can see that men, women & children are in the ratio = $\frac{12}{6}, \frac{15}{5}, \frac{16}{4},$ or 2: 3: 4.Number of men, women & children = 10, 15, 20. Now if each man gets 6a, we have10 * 6a + 15 * 5a + 20 * 4a = 1720 or a = 8. So, the answers are Rs. 48, 40, 32.

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Class 12th Quantitative Aptitude Prep Tips for MBA

Three glasses have capacity of 750 litre, 1950 litre, and 600 litres. 2046 litre of wine is poured into them so that the same fraction of each is filled. The volume filled in the largest glass will be

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Payal Gupta

Contributor-Level 10

Only possibility is if 1950 is divided by a multiple of 13, because the other glasses have a capacity that is not a multiple of 13.

Hence 1209 is the only choice.

Alternative method:

As the glasses filled by same fractions,

$\frac{a}{750} = \frac{b}{1950} = \frac{c}{600} = \frac{a + b + c}{750 + 1950 + 600}$

a + b + c = 2046 $∴ \frac{b}{1950} = \frac{2046}{3300}$

b = 1209 litres

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Class 12th Quantitative Aptitude Prep Tips for MBA

A sum is to be divided among men and women. Total number of men and women is 95. The sums of men's share and women's share are in ratio 8 : 9 but their individual shares are in ratio 4 : 3. Find number of men.

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Raj Pandey

Contributor-Level 9

Let the number of men be x

$\frac{4 x}{3 (95 - x)} = \frac{8}{9}$

9 x = 570 – 6 x

x = 38, so number of men is 38

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10 months ago

Class 12th Quantitative Aptitude Prep Tips for MBA

Two trains T1 and T2 start from Delhi to Kanpur and from Kanpur to Delhi respectively. After passing each other, they take 4 hours 2 minutes and 3 hours 20 minutes to reach Kanpur and Delhi respectively. If T1 is moving at 45 km/hr, the speed of T2 is

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Raj Pandey

Contributor-Level 9

$\frac{T 1 speed}{T 2 speed} = \sqrt[]{\frac{Time taken by T 2 to reach Chennai}{Time taken by T 1 to reach Yeshvantpur}}$

$or \frac{45}{T 2 speed} = \sqrt[]{\frac{200}{242}} = \frac{10}{11}$

T2 speed = $\frac{99}{2}$ $\frac{}{}$ =49.5 km/hr.

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Class 12th Quantitative Aptitude Prep Tips for MBA

Rohit buy toffees at 18 toffees for Rs. 108 and sell it at a loss of as much money as Rohit paid for 3 toffees. What is the SP per toffee?

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Raj Pandey

Contributor-Level 9

As per the problem, we have CP of 18? SP of 18 = CP of 3

CP of 15 = SP of 18

But CP of 15 toffees = Rs. 90

So SP of 18 toffees = Rs. 90

SP per toffee $= \frac{90}{18} = Rs. ? 5$

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