Quantitative Aptitude Prep Tips for MBA

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a month ago

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P
Payal Gupta

Contributor-Level 10

Given quadratic equation x2 + ax + b = 0

Then product of roots αβ = b

Sum of roots α + β = − a

Next quadratic equation bx2 + ax + 1 = 0

Then product of roots = 1 b = 1 α β

Hence, clearly by visualising options

New roots we will b1α and 1β

New answer posted

a month ago

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P
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

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

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

Thus 4P = 2R

PR=12 . (1)

= 5Q = 4R = QR=45 . (2)

From (1) & (2),

P/RQ/R=1/24/5

PQ=58

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

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

Hence, the area of? MQR = 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

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

Let n = the number of terms.

Then,  25716=n2 [17+ (1238)]=n2*458

40716=37n16 n=40737=11

Let d be the common difference.

Then 1238  (= the eleventh term) = 17 + 10d

10d=123817=2358

d=2358*10=4716

New answer posted

a month ago

0 Follower 3 Views

P
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

(0.7A+B)11= (0.3A+B)7

1.6A = 2B  A5=B4

Volume of A : B = 5 : 4.

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

Area of Δ ABE = 7 cm2

Area of ABEF = 14 cm2

Area of ABCD = 14 * 4 = 56 cm2

(As CE = 3 * BE) = 56 cm2

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

I

II

III

2 : 5

3 : 4

4 : 5

 

 

 

Hence new ratio

(27+37+49) (57+47+59)

= 73 : 116

New answer posted

a month ago

0 Follower 2 Views

P
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,  50y+ [x50]3y/5=t+3..... (2)

Also,  100y+ [x100]3y/5=t+2..... (3)

On solving the above equations, we get speed is 1003 km/hr time is 6 hrs, and distance is 200 km.

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

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

55x=55y12..... (1)

Also,  55x=51y110..... (2)

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

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