Reasoning PYQs

Persons – 8; R, S, T, U, W, X, Y and Z;

Months – April, May, June and July;

Dates – 7^th and 28^th;

1. Only two persons gave after U.
2. Only one person gave between U and S.

From this we get the possible arrangements:

3. S and R gave on different dates.
4. R gave after S.

    From this case II is eliminated as we cannot determine the month in which R gave the exam.

5. Five persons gave between W and T.
6. T gave before R and both gave on different dates.

7. Y gave after Z, who gave in the month which has odd number of days.

8. Y and W did not give in the same month.

From this case I is eliminated, as Y and W attended in the same month which is not possible. From this we get the final arrangement as follows.

Except W, all other persons gave the exam in odd numbered date.

W is the odd one out.

Persons – 8; R, S, T, U, W, X, Y and Z;

Months – April, May, June and July;

Dates – 7^th and 28^th;

1. Only two persons gave after U.
2. Only one person gave between U and S.

From this we get the possible arrangements:

3. S and R gave on different dates.
4. R gave after S.

    From this case II is eliminated as we cannot determine the month in which R gave the exam.

5. Five persons gave between W and T.
6. T gave before R and both gave on different dates.

7. Y gave after Z, who gave in the month which has odd number of days.

8. Y and W did not give in the same month.

From this case I is eliminated, as Y and W attended in the same month which is not possible. From this we get the final arrangement as follows.

Here, the number of persons who gave the exam before R is six similarly the number of persons gave the exam after T is also six.

T is the correct answer.

Persons – 8; R, S, T, U, W, X, Y and Z;

Months – April, May, June and July;

Dates – 7^th and 28^th;

1. Only two persons gave after U.
2. Only one person gave between U and S.

From this we get the possible arrangements:

3. S and R gave on different dates.
4. R gave after S.

    From this case II is eliminated as we cannot determine the month in which R gave the exam.

5. Five persons gave between W and T.
6. T gave before R and both gave on different dates.

7. Y gave after Z, who gave in the month which has odd number of days.

8. Y and W did not give in the same month.

From this case I is eliminated, as Y and W attended in the same month which is not possible. From this we get the final arrangement as follows.

X gave the exam on 7^th in April.

Given number – 78356497

By subtracting 2 from every odd digit, we get 58136475.

Odd Digits in the number: 7, 3, 5, 9

2^nd digit from the left end is 8

4^th digit from the right end is 6

8 × 6 = 48

Correct answer is ‘48’.

Confusion PointsQuestion is asked Odd Digit not Odd Place Digit.

By subtracting 2 from every odd digit, we get 58136475.

The least possible Venn diagram for the given statements is as follows:

Conclusions:

I. All Green Can be Black → True (As there is no direct relation between green and black so All Green can be Black is possible but not definite. So, it is true)

II. Some Green is Red → False (As All Red is Blue and No Blue is green, Some green is red is not possible. So, it is false)

Only I follow.

The best possible diagram is, 

Conclusions:

I. All top is side: False (No direct relation between top and side, it is possible but not definite)

II. Some top is not side: False (No direct relation between top and side, it is possible but not definite)

Conclusion I and II form a complementary pair.

Either I or II follows.

Key Points

Conditions for Either - Or 

1. Subject and predicate should be the same
2. Both the individual conclusions should be false
3. One conclusion should be positive and one conclusion should be negative
4. There should either one of the following pairs present in the conclusions

• All + Some not
• Some + No

The least Possible Venn diagram is shown below:

Conclusions:

I. Some crack is not east - False (As crack and east do not have any direct relation it is possible but not definite)

II. Some east is much.  - True (Only a few much is east so some much is east and some much is not east)

III. All much is east. - False (Only a few much is east so some much is east and some much is not east so this is definitely false)

Only II follows.

1. There are four boxes placed between Box P and Box B. Neither Box P nor Box B is placed at the bottom

 

2. There are two boxes placed between box B and Box D.

 

3. Box R is immediately above the box P. (Here, there is no space for the box R to place above the box P)

 

4. Box C placed at the even place.

 

5. Neither Box Q and Box C are placed at the bottom. (Here, box Q is placed at the position of 5^th so that box A is placed at the bottom).

Hence, the final arrangement will be as follows:

 

Hence, Box A is placed at the bottom.

1. There are four boxes placed between Box P and Box B. Neither Box P nor Box B is placed at the bottom

 

2. There are two boxes placed between box B and Box D.

 

3. Box R is immediately above the box P. (Here, there is no space for the box R to place above the box P)

 

4. Box C placed at the even place.

 

5. Neither Box Q and Box C are placed at the bottom. (Here, box Q is placed at the position of 5^th so that box A is placed at the bottom).

Hence, the final arrangement will be as follows:

 

Hence, box D is immediately below box Q.

1. There are four boxes placed between Box P and Box B. Neither Box P nor Box B is placed at the bottom

 

2. There are two boxes placed between box B and Box D.

 

3. Box R is immediately above the box P. (Here, there is no space for the box R to place above the box P)

 

4. Box C placed at the even place.

 

5. Neither Box Q and Box C are placed at the bottom. (Here, box Q is placed at the position of 5^th so that box A is placed at the bottom).

Hence, the final arrangement will be as follows:

 

Therefore, only one is placed between the box Q and Box R.

1. There are four boxes placed between Box P and Box B. Neither Box P nor Box B is placed at the bottom

 

2. There are two boxes placed between box B and Box D.

 

3. Box R is immediately above the box P. (Here, there is no space for the box R to place above the box P)

 

4. Box C placed at the even place.

 

5. Neither Box Q and Box C are placed at the bottom. (Here, box Q is placed at the position of 5^th so that box A is placed at the bottom).

Hence, the final arrangement will be as follows:

 

Hence, all the statements are true.

1. There are four boxes placed between Box P and Box B. Neither Box P nor Box B is placed at the bottom

 

2. There are two boxes placed between box B and Box D.

 

3. Box R is immediately above the box P. (Here, there is no space for the box R to place above the box P)

 

4. Box C placed at the even place.

 

5. Neither Box Q and Box C are placed at the bottom. (Here, box Q is placed at the position of 5^th so that box A is placed at the bottom).

Hence, the final arrangement will be as follows:

 

Hence, Box R is exactly between box D and Box P.

Given series: 824 625 769 983 726

After arranging their digits in increasing order in numbers,

248 256 267 389 679

The second lowest number is 256

second digit of this number is 5

If five numbers are written with their digits in the increasing order, the second  digit of second lowest number is 5.

Given series: 824 625 769 983 726

After arranging the numbers in lowest to highest term;

625    726    769    824    983

third highest number is = 769

highest number is = 983

Third digit of third highest number is 9, and third digit of highest number is 3.

When 9 is divided by 3 answer is 9 / 3 = 3

If third digit of third highest number is divided by the third digit of highest number then the answer is 3.

Given series:  824 625 769 983 726

On adding last two digit we get;

824 (2 + 4) = 6

625 (2 + 5) = 7

769 (6 + 9) = 15

983 (8 + 3) = 11

726 (2 + 6) = 8

After multipliying these resultant by 5 and first digit is substracted from them;

6 * 5 = 30 (30 - 8) = 22

7 * 5 = 35 (35 - 6) = 29

15 * 5 = 75 (75 - 7) = 68

11 * 5 = 55 (55 - 9) = 46

8 * 5 = 40 (40 - 7) = 33

Here we get 2 odd numbers as a result, 29 and 33

2 resultant numbers are odd numbers.

Given series:  824 625 769 983 726

After arranging the numbers in ascending order from left to right we get:

625 726 769 824 983 The number which is exact in middle is 769

sum of squares of all digits are (7 * 7 + 6 * 6 + 9 * 9) = 49 + 36 + 81 = 166

Sum of squares of all digits of middle no. is 166.

Given series:  824 625 769 983 726

Average of all five numbers are;

824 + 625 + 769 + 983 + 726 = 3927 / 5 = 785.4

Numbers which are greater than that of 785.4 is 983 and 824.

Two numbers are greater than that average of all five numbers.

As per the given information, 

1. Eight members M, O, P, R, S, T, U, and V are in a family. Either both the parents of a person are alive or neither of them is alive. There are only three generations in the family.

2. O is the mother of R and T.

3. U is the sister-in-law of R and S is the sister-in-law T.

4. M is the granddaughter of P.

5. V is the only child of R. V is the grandson of O. 

 M is the niece of R.

As per the given information, 

1. Eight members M, O, P, R, S, T, U, and V are in a family. Either both the parents of a person are alive or neither of them is alive. There are only three generations in the family.

2. O is the mother of R and T.

3. U is the sister-in-law of R and S is the sister-in-law T.

4. M is the granddaughter of P.

5. V is the only child of R. V is the grandson of O. 

 If G is married to M then G is the Son-in-law of U.

As per the given information, 

1. Eight members M, O, P, R, S, T, U, and V are in a family. Either both the parents of a person are alive or neither of them is alive. There are only three generations in the family.

2. O is the mother of R and T.

3. U is the sister-in-law of R and S is the sister-in-law T.

4. M is the granddaughter of P.

5. V is the only child of R. V is the grandson of O. 

 P is grandfather of M is the correct statement.