Quantitative Aptitude Prep Tips for MBA

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Payal Gupta

Contributor-Level 10

Let f (x) = px2 + qx + k, where p, q and k are integers, and p ≠ 0

 f (0) = k = 1

 f (x) = px2 + qx + 1

= px2 +qx + k (Differentiate both sides with respect to x)

 f' (x) = 2px + q

For maxima or minima f' (x) = 0, x = q2p

f (x) attains maximum at x = 1

 q = − 2p

f (1) = p + q + 1 = 3

 1 − p = 3

 p = − 2

 q = 4

 f (x) = −2x2 + 4x + 1

f (10) = − 200 + 40 + 1 = −159

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Payal Gupta

Contributor-Level 10

f (x) = x – x2 + 1

g (x) = x2 + b + 3

f (2) g (1) < 0

(a – 4 + 1) (1 + 6 + 3) < 0

(a – 3) (b + 4) < 0

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Payal Gupta

Contributor-Level 10

100! has 24 zeroes.

100! + 200! = 100! [1 + 101 * 102 * … * 200]

Which will again give 24 zeroes at the end.

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Payal Gupta

Contributor-Level 10

So, cars which have atleast one options

= 8 + 4 + 10 + 12 + 2 + 4 + 6 = 46

Hence, cars with no option = 50 – 46 = 4

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Payal Gupta

Contributor-Level 10

Let 2P = 3Q = 4R = 5S = K

Then,  P=K2, Q=K3, R=K4, S=K5

K2+K3+=K4+K5=1540

K=1200

Therefore. R = 300, S = 240

Difference=300 - 240 = 60

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Payal Gupta

Contributor-Level 10

Capacity of the tube

π ( 1 ) 2 ( 7 1 ) + 2 3 π ( 1 ) 3 = 2 0 3 π

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Payal Gupta

Contributor-Level 10

If D < 0 roots are imaginary.

i.e. q2 – 4pr < 0 q2 < 4pr 

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Payal Gupta

Contributor-Level 10

Discriminant = (25)24 (1) (3)=2012=8

So, the roots are real and irrational.

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Payal Gupta

Contributor-Level 10

400 students * 80 days = 240 students * d days

d = 4 0 0 3 = 1 3 3 . 3 3 d a y s  

The food was to last 80 more days but now it is lasting 53.33 days more.

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3 months ago

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P
Payal Gupta

Contributor-Level 10

 

A

B

I

5x

3x

II

4y

7y

5x + 4y = 253         …. (1)

3x + 7y = 161         …. (2)

On solving equations (1) & (2), we get

x = 49 and y = 2.

Hence, type I Alloy = 392 kg, and Type II Alloy = 22 kg

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