# Free DI and LR - 15 Practice Test - CAT

Find out the sum of all the boxes of the set E.

A. 61
B. 60
C. 55
D. 49
E. 31

#### SOLUTION

Solution : E

Sum of the top most boxes (119) is 10 more than the sum of bottom most boxes (109); it means that the total number of the boxes between the bottom most box  and top most box in all the five sets  is 119 – 109 = 10. As each set has different number of boxes so this sum 10 should be a sum of five

different natural numbers i.e. 0+1+2+3+4 = 10. So, the number of boxes between the bottom most boxes and the top most boxes for the 5 sets should be 0, 1, 2, 3 & 4. Hence there are 1, 2, 3, 4 & 5

boxes in the sets A, B, C, D, and E.

Sum of the boxes of set B is 81 which is a sum of 1 or 2 or 3 or 4 or 5 consequent natural

numbers. If B is having 1 or 2 boxes than it would be impossible to make a sum of 109 from all

bottom most boxes. So they can be 26, 27 & 28 only i.e. B has only 3 boxes.

Now as no two consecutive sets have consecutive number of boxes, so A and C cannot have

2 or 4 boxes. Let us assume that A has only one box so its number would be 21. Also one of the

sets will have a bottom most box numbered 29 (as B’s top most box is numbered 28 and counting

cannot stop there as in that case sum of the bottom most boxes will not be 109) and one will have 22 as the bottom most box number. So, the bottom most box number of the 5th set should be 109 - (21+ 22 + 26 + 29) = 11 which is impossible because in that case the sum of top most boxes will not be 119, so set A has 5 boxes and hence C, D, E have 1, 4 and 2 boxes respectively.

As the top most number of A is 21 and bottom most number of B is 26 that can be reached only if

the counting from A goes to the set with 4 boxes. So, the 3 bottom most box numbers obtained

are 17, 22 and 26. Hence the other two bottom most box numbers are 29 and 15.

So, the table is like:-

Answer option E

Find out the numbers of boxes in set C.

A. 1
B. 2
C. 3
D. 4
E. 5

#### SOLUTION

Solution : A

Option A is the correct answer.

What is the highest number assigned to a box by Appu?

A.  28
B. 29
C. 30
D. 31
E. 32

#### SOLUTION

Solution : B

Sum of the top most boxes (119) is 10 more than the sum of bottom most boxes (109); it means that the total number of the boxes between the bottom most box  and top most box in all the five sets  is 119 – 109 = 10. As each set has different number of boxes so this sum 10 should be a sum of five

different natural numbers i.e. 0+1+2+3+4 = 10. So, the number of boxes between the bottom most boxes and the top most boxes for the 5 sets should be 0, 1, 2, 3 & 4. Hence there are 1, 2, 3, 4 & 5

boxes in the sets A, B, C, D, and E.

Sum of the boxes of set B is 81 which is a sum of 1 or 2 or 3 or 4 or 5 consequent natural

numbers. If B is having 1 or 2 boxes than it would be impossible to make a sum of 109 from all

bottom most boxes. So they can be 26, 27 & 28 only i.e. B has only 3 boxes.

Now as no two consecutive sets have consecutive number of boxes, so A and C cannot have

2 or 4 boxes. Let us assume that A has only one box so its number would be 21. Also one of the

sets will have a bottom most box numbered 29 (as B’s top most box is numbered 28 and counting

cannot stop there as in that case sum of the bottom most boxes will not be 109) and one will have 22 as the bottom most box number. So, the bottom most box number of the 5th set should be 109 - (21+ 22 + 26 + 29) = 11 which is impossible because in that case the sum of top most boxes will not be 119, so set A has 5 boxes and hence C, D, E have 1, 4 and 2 boxes respectively.

As the top most number of A is 21 and bottom most number of B is 26 that can be reached only if

the counting from A goes to the set with 4 boxes. So, the 3 bottom most box numbers obtained

are 17, 22 and 26. Hence the other two bottom most box numbers are 29 and 15.

So, the table is like:-

Answer option B

If sum of all top most boxes and the bottom most boxes are 134 and 114, respectively, what is the sum of all the boxes of all the sets?

A. 20
B. 24
C. 25
D. 30
E. 5

#### SOLUTION

Solution : C

Option C is the correct answer.

What is Sally’s house number?

A. 64
B. 81
C. 25
D. 24

#### SOLUTION

Solution : B

Since Sally thinks she has enough information, we deduce that Sam answered his house number was a perfect square greater than 50 (answering Yes to both). There are two of these {64, 81} and Sally must live in one of them in order to have decided she knew where Sam lives. Sam answered only the second question truthfully, so his house number is greater than 50, but not a perfect square.
Since Sam answered Sue’s second question truthfully, he had to have answered yes to “Is it greater than 25?” Sue was able to deduce Sam’s number, so he also must have said it was a perfect cube. Cubes greater than 25: {27, 64}. Sue must live in one of these houses to deduce Sam’s number.
Since Sam’s number is greater than 50 and is less than Sue’s number, she must live in 64. Since Sue and Sally are not roommates (we’re told there are three numbers), Sally must live in 81.

Given fact: the sum of their numbers is a perfect square multiplied by two.
Sue + Sally + Sam = 2p (for p an integer)
Or, 64+81+Sam = 2p2
Applying the constraint that Sam’s number is greater than 50 and less than 64, it looks like Sam = 55 (p=10).
In summary,
Sam = 55, Sue = 64, Sally = 81

Hence Answer option B

What is Sue’s house number?

A. 64
B. 81
C. 55
D. 54

#### SOLUTION

Solution : A

Option A is the correct answer.

What is Sam’s house number?

A. 45
B. 55
C. 65
D. 72

#### SOLUTION

Solution : B

Option B is the correct answer.

Who likes to play on the terrace?

A. Ram
B. Ghanshyam
C. Benjo
D. Cannot be Determined.

#### SOLUTION

Solution : D Cannot be Determined. Option d

What colored marbles does Shyam like?

A. Magenta
B. Yellow
C. Either magenta or Yellow
D. None of these

#### SOLUTION

Solution : A

Option a i.e. Magenta

If Ram doesn’t like to play in the laboratory then where does he like to play?

A. Canteen
B. Terrace
C. Corridor
D. Cannot be Determined

#### SOLUTION

Solution : B

If Ram doesn’t like to play in laboratory then he must like to play on the terrace.Option b

If basic colored marbles cost Rs.10 each and others cost Rs. 12 then by what percentage is Shyam’s investment more than Ram?

A. 20%
B. 0%
C. 25%
D. Cannot be Determined

#### SOLUTION

Solution : D

Since number of marbles is not known thus we cannot fins the answer. Option d

If Ghanshyam likes to play on the terrace then where does Benjo play?

A. Laboratory
B. Canteen
C. Either a or b
D. Cannot be Determined

#### SOLUTION

Solution : B

If Ghanshyam plays on terrace then Benjo must play in the canteen. Option b