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9 months ago

Maths Quantitative Aptitude Prep Tips for MBA

In a school, temporary staff of 1000 people was recruited in which 600 were employed in cleaning department while 400 were employed in teaching department. The average salary of each temporary employee is Rs. 600. Each employee in cleaning department earns Rs. 100 more than that in teaching department. What is the salary of each employee in cleaning department?

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Payal Gupta

Contributor-Level 10

Average salary of each temporary employee is 600.

1000 temporary employee,

Total salary = 6,00,000

let teaching departement = Rs. x / staff and cleaning department salary = x + 100

Now, 600 (x + 100) + 400 (x) = 6,00,000

x=540

Hence answer, x+100= 540+100=640

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New answer posted

9 months ago

Maths Quantitative Aptitude Prep Tips for MBA

If A gives B Rs. 20 then A has 40% of what B has. Instead if B gives Rs. 40 to A then B will have 40% of what A has. What is the amount originally A has?

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Payal Gupta

Contributor-Level 10

We have (A – 20) = 0.4 (B + 20), i.e. A – 0.4B = 28

And (B – 40) = 0.4 (A + 40), i.e. B – 0.4A = 56

Solving we get, A = Rs. 60

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New answer posted

9 months ago

Maths Quantitative Aptitude Prep Tips for MBA

P, Q & R started travelling from the same point x in the same direction. P & Q started at 10:00 a.m. Then R started at 12 : 00 p.m. R overtook P at 2 : 00 p.m. and then doubled his speed. R overtook Q at 3:00 p.m. The ratio of speed of P & Q is

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Payal Gupta

Contributor-Level 10

Let speeds of P, Q & R be P, Q & R km/hr respectively.

Thus 4P = 2R

= $\frac{P}{R} = \frac{1}{2}$ . (1)

= 5Q = 4R = $\frac{Q}{R} = \frac{4}{5}$ . (2)

From (1) & (2),

= $\frac{P / R}{Q / R} = \frac{1 / 2}{4 / 5}$

= $\frac{P}{Q} = \frac{5}{8}$

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New answer posted

9 months ago

Maths Class 12th

In the given figure, PQRS is a square and M and N are midpoints of their respective sides. Find the area of the quadrilateral PMRN.

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Payal Gupta

Contributor-Level 10

Since M is the midpoint of side PQ, the length of MQ is 2.

Hence, the area of? MQR = $\frac{1}{2}$ * 2 * 4 = 4.

Also area of? NSR = 4. Thus, the unshaded area of the figure = 4 + 4 = 8.

Hence, the area of quadrilateral PMRN

= Area of the square PQRS – The unshaded area of the figure

= 16 – 8 = 8

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New answer posted

9 months ago

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

9 months 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

9 months 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

9 months 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

9 months 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

9 months 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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