The morgridge college of education at the university of denver Interview Questions, Process, and Tips

First nos series is 7,9,11,?
ie odd number siries ie 7,9,11,13
Second number series is 16,15,14
ie 1 less the previous number 16,15,14,13
Ans —-series is 7,16,9,15,11,14,13,13

55967

find out what is the problem and taking the bad out always works.

20

5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000

1

Let\: the\: total\: price \: be\: Rs. X\: then,

B= 2x/7 & A= \left ( x-2 \right )/7

So, A:B = 5x/7 : 2x/7 = 5 : 2

Let \: B’s \: capital\; be\: Rs.Y\: then\: \left ( 16000\ast 8 \right )/4y= 5/2

=> (16000\ast 8\ast 2)/(4\ast 5) = y

=> y = Rs. 12800

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240

Hockey

answer 10

Ramu’s Mother-in-law means Ramu’s wife’s mother
Only daughter means Ramu’s wife
And daughter’s son means Ramu’s Son

Answer : Son

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3

time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20

As it is a right angled triangle so we know the hypotenuse must be 13 and 5 & 13 are base and height (in any order).
Area of triangle = 0.5* base * height = 0.5 * 5 * 12 = 30

If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.

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