A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be:

4 | x y = (5 x 1 + 4) = 9

5 | y -1 x = (4 x y + 1) = (4 x 9 + 1) = 37

| 1 -4

Now, 37 when divided successively by 5 and 4, we get

5 | 37

4 | 7 - 2

Respective remainders are 2 and 3

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A family consists of 6 members P, Q, R, S, T, and U . There are two married couples . Q is a doctor and the father of T ,U is grandfather of R and is a contractor . s is grandmother of T and is a housewife. There is one doctor, one contractor -one house wife and two students in the family.

Which of the following is definitely a group of male members?

Q the doctor, is the father of T. S, the house wife, is the grandmother of T and hence the mother of Q. Since there are only two married couples one being that of Q, the grandfather of R i. e, U must be married to S. Thus, R and T will be both children of Q and these must be the student. So P, who remains, Shell be the wife of Q and she alone can be the nurse. Thus, U must be the contractor. Clearly, for definite the males are Q, the father and U, the grandfather..

In following questions, one term in number series is incorrect.

Find out the incorrect number 10, 14, 28, 32, 64, 68, 132

Find out the incorrect number 10, 14, 28, 32, 64, 68, 132

Alternately, the numbers are increased by four and doubled to get the next number>

Thus, 10 + 4 = 14;

14 * 2 = 28;

28 + 4 = 32;

32 * 2 =64 and so on.

So, 132 is wrong and must be replaced by (68 * 2) i.e. 136..

A is the brother of B. B is the brother of C. D is the father of A. Based on these three statements, which of the following statement cannot be definitely true?

Option (D)

A is the brother of B and B is the brother of C. So, C may be the brother or sister of A.

On dividing 2272 as well as 875 by 3-digit number N, we get the same remainder. The sum of the digits of N is:

Clearly, (2272 - 875) = 1397, is exactly divisible by N.

Now, 1397 = 11 x 127

The required 3-digit number is 127, the sum of whose digits is 10.

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The age of three children are in proportion 5:7:8. After six years the sum of there ages is 50. Find their ages six years ago. ( in years).

Let there present ages be 5x,7x,8x years respectively.

Then, (5x+6)+(7x+6)+(8x+6)=50

16x=32

x=2

So six years ago their ages was 4,8,10 years respectively.

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P is the Brother of R;

Q is the son of P ;

T is the brother of S;

S is the daughter of R.

Who are the cousins of Q?

Option (D)

T is the brother of S, who is the daughter of R. So, T and S are the children of R. Now, P is the brother of R and Q is the son of P. So, T and S are the cousins of Q.

The average of 5 quantities is 6. The average of 3 of them is 8. What is the average of the remaining two numbers?

The average of 5 quantities is 6.

Therefore, the sum of the 5 quantities is 5 *6 = 30.

The average of the 3 of these 5 quantities is 8.

Therefore, the sum of these 3 quantities = 3*8 = 24

The sum of the remaining two quantities = 30 - 24 = 6.

Average of these two quantities = 6 / 2 = 3.

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