Free DI and LR - 15 Practice Test - CAT 

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

Option A is the correct answer.

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

Option C is the correct answer.

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

Option A is the correct answer.

Option B is the correct answer.

Cannot be Determined. Option d

Option a i.e. Magenta

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

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

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