Quantitative Aptitude Prep Tips for MBA

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P
Payal Gupta

Contributor-Level 10

Volume of smaller cone

=13π (3)2*9=27π

=13π (5)2*15=125π

 Volume of the solid = 125π – 27π = 98 π

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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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Payal Gupta

Contributor-Level 10

Total S.P. will be

8* [2.36x]+16100 [x4+1]120=18.88x+4.8x+19.2

= [x8]216=27x

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

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Payal Gupta

Contributor-Level 10

Let A = abc and B = cba

Therefore, B – A = 100c + 10b + a – (100a + 10b + c) = 99 (c – a). B – A is a multiple of 7.

Therefore, c – a = 7 (a, c) (1, 8) or (2, 9).

Hence, number is between 108 to 198 or 209 to 299.

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Payal Gupta

Contributor-Level 10

Let there be x litres of wine in the beginning

(x8x)2=81100

 x = 80 litres

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Payal Gupta

Contributor-Level 10

Suppose the hound catches the rabbit in t minutes

Number of jumps by the rabbit = 35t & distance covered = 20 * 35t = 700 t cm.

Similarly Distance covered by hound =  25t * 60 = 1500t cm.

Now, 1500 t – 700t = 7000 cm or t = 3 5 4 min.

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Payal Gupta

Contributor-Level 10

When S1 runs 1900 metres, S3 would run 1862 metres. Hence, the start given is  38/1900 * 2000 = 40 metres

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Raj Pandey

Contributor-Level 9

In this problem, we are looking for the LCM of the divisors, that is, 3 and 4 or any of their multiple to which the desired remainder 2 can be added. The smallest such number is 2. The others will be 14, 26, 38, 50, 62, 74, 86 and 98, that is there are 9 such numbers

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Raj Pandey

Contributor-Level 9

( A B ) * ( B C ) * ( C D ) = A D

5 6 * 1 3 1 4 * 9 4 = 1 9 5 1 1 2

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Raj Pandey

Contributor-Level 9

The new time is 1 3  of the old time. If time becomes 1 3 , then new speed should have been   3 1 of the initial speed. Therefore

3 S S S * 1 0 0 = 2 0 0 %

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