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Maths Quantitative Aptitude Prep Tips for MBA

The first term of a series in A.P. is 17, the last term is $- 12 \frac{3}{8}$ and the sum of the series is $25 \frac{7}{16}$ Find the common difference.

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Payal Gupta

Contributor-Level 10

Let n = the number of terms.

Then, $25 \frac{7}{16} = \frac{n}{2} [17 + (- 12 \frac{3}{8})] = \frac{n}{2} * 4 \frac{5}{8}$

$\Rightarrow \frac{407}{16} = 37 \frac{n}{16}$ $\Rightarrow n = \frac{407}{37} = 11$

Let d be the common difference.

Then $- 12 \frac{3}{8}$ (= the eleventh term) = 17 + 10d

$\Rightarrow 10 d = - 12 \frac{3}{8} - 17 = - \frac{235}{8}$

$\Rightarrow d = \frac{- 235}{8 * 10} = - \frac{47}{16}$

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

Two container A & B contain oil. 60% of contents of container A was transferred to container B. Then 50% contents of container B was transferred to A. The quantity of oil in container A to that of container B is now 11 : 7. What was ratio of initial volumes of oil in containers of A and B?

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Payal Gupta

Contributor-Level 10

Let oil in containers be A & B.

After 1st operation

Container A = 0.4 A

Container B = 0.6 A + B

After 2nd operation

Container A = 0.4 A + 0.3 A + 0.5 B

Container B = 0.3 A + 0.5 B

= $\frac{(0.7 A + B)}{11} = \frac{(0.3 A + B)}{7}$

1.6A = 2B $\Rightarrow$ $\frac{A}{5} = \frac{B}{4}$

Volume of A : B = 5 : 4.

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 cm²; EC = 3(BE). The area of ABCD (in cm²) is

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Payal Gupta

Contributor-Level 10

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

There are three glasses of equal capacity. The ratio of milk & water in the three is 2 : 5, 3 : 4, 4 : 5. The mixture of these three is put into a bucket. Find ratio of water to milk in the bucket.

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Payal Gupta

Contributor-Level 10

I	II	III
2 : 5	3 : 4	4 : 5

Hence new ratio

$\frac{(\frac{2}{7} + \frac{3}{7} + \frac{4}{9})}{(\frac{5}{7} + \frac{4}{7} + \frac{5}{9})}$

= 73 : 116

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

A car running from point P₁ to point P₂ meets with an accident 50 km from point P₁, after which it travels with three- fifths of its original velocity and arrives 3 hours late at point P₂; if the accident had occurred 50 km further on, it would have been only 2 hours late. Find the distance from point P₁ to point P₂.

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Payal Gupta

Contributor-Level 10

Let the distance be x km and speed be y km/hr and t be the time in hours. Then the equation will be x = yt …. (1),

Therefore, $\frac{50}{y} + \frac{[x - 50]}{3 y / 5} = t + 3 . . . . . (2)$

Also, $\frac{100}{y} + \frac{[x - 100]}{3 y / 5} = t + 2 . . . . . (3)$

On solving the above equations, we get speed is $\frac{100}{3}$ km/hr time is 6 hrs, and distance is 200 km.

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New answer posted

a year ago

Maths Class 12th

Two runners ride from A to B a distance of 55 km, and the first arrives 30 minutes before the second. They then ride from B to A, the first giving the second a start of 4 km and yet arrive 6 minutes before him. Find the speed of A.

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Payal Gupta

Contributor-Level 10

Let the speed of A be x km/hr and speed of B be y km/hr, then:

$\frac{55}{x} = \frac{55}{y} - \frac{1}{2} . . . . . (1)$

Also, $\frac{55}{x} = \frac{51}{y} - \frac{1}{10} . . . . . (2)$

On solving both equation (1) & (2) we get x = 11 km/hr and y = 10 km/hr.

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

There is a right circular cone of base radius 5 cm. This cone is cut into two portions by a plane parallel to its base such that the portion above the plane is a cone of base radius 3 cm and the portion below the plane has a height of 6 cm. Find the volume of the bottom solid.

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Payal Gupta

Contributor-Level 10

Volume of smaller cone

$= \frac{1}{3} π {(3)}^{2} * 9 = 27 π$

$= \frac{1}{3} π {(5)}^{2} * 15 = 125 π$

$\Rightarrow$ Volume of the solid = 125π – 27π = 98 π

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

A Banana seller sells x dozen Bananas. If he had sold 100 dozen Bananas more at 20 paise a dozen less or if he had sold 120 dozen less at 30 paise a dozen more, he would have received the same amount for his Bananas. How many dozens does he sell?

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Payal Gupta

Contributor-Level 10

Px= (P-0.2) (x+100)= (P+0.3) (x-120)

Solving x=1000

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

A shopkeeper buys 216 packets of biscuits at the rate of Rs. $\frac{x}{8}$ per pkt. He sells 8 dozen at the rate of Rs. 2.36 x per dozen and the rest at 16 $(\frac{x}{4} + 1)$ paise each. What is the value of profit?

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Payal Gupta

Contributor-Level 10

Total S.P. will be

$8 * [2.36 x] + \frac{16}{100} [\frac{x}{4} + 1] 120 = 18.88 x + 4.8 x + 19.2$

$= [\frac{x}{8}] 216 = 27 x$

Profit = 23.68x + 19.2 − 27x = 19.2 - 3.32x

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New answer posted

a year ago

Maths Quantitative Aptitude Prep Tips for MBA

8 litres of wine are drawn from a cask and replaced by 8 litres of water. 8 litres of the mixture are next drawn and again replaced by 8 litres of water. If the ratio of wine to water in the cask is now 81 : 19, how much wine was in the cask at first? (There was no water in the cask at first.)

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Payal Gupta

Contributor-Level 10

Let there be x litres of wine in the beginning

${(\frac{x - 8}{x})}^{2} = \frac{81}{100}$

$\Rightarrow$ x = 80 litres

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