We are back with another set of CAT questions. Try to solve this set in five minutes.
Q.1 Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (CAT 2007)
(1) 3 (2) 2 (3) 4 (4) 0 (5) 1
Q.2 Consider the set S = {1, 2, 3… 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least three elements. (CAT 2006)
(1) 3 (2) 4 (3) 6 (4) 7 (5) 8
Q.3 Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise, and Re. 1.01 to three different persons? (CAT 2003)
(1) 19 (2) 20 (13) 17 (4) 18
Q.4 and Q.5 Let S be the set of all pairs (i, j) where 1 ≤ i ≤ j < n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1,2) and (2, 3) are also friends, but (1,4) and (2, 3) are enemies.
Q.4 For general n, how many enemies will each member of S have? (CAT 2007)
Q.5 For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members? (CAT 2007)
Solutions:
Q.1 Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (CAT 2007)
Solution:
Let the number be xxyy. As the number is a perfect square, it doesn’t end with 2, 3, 7, 8. So combinations left are ending with 11, 55, 66, 99, 00 – Not a single number because of respective properties.
Ending with 44 will be the answer.
Also, 1000x + 100x + 10y + y = 1100x + 11y = 11(100x + y). The number is divisible by 11. Which means the square root has to be a multiple of 11.
Only 1 number satisfies this condition. Answer = 1
Q.2 Consider the set S = {1, 2, 3… 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least three elements. (CAT 2006)
Solution:
Using given conditions, we need to find n if a = 1 and Tn = 1000
Tn = a + (n-1) d
1000 = 1 + (n-1) d
(n-1) d = 999
In other words, find factors of 999 = 3 * 333 = 3*3*111 = 3*3*3*37
Hence, 3, 9, 27, 37, 111, 333, 999.
Answer = 7
Q.3 Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise, and Re. 1.01 to three different persons? (CAT 2003)
Solution:
In all such questions, start with maximum possible denomination. We will do something similar to what we do for HCF
| Denomination | 69 | 78 | 101 | Number |
| 50 | 19 | 28 | 51 | 3 |
| 25 | 19 | 28 | 26 | 1 |
| 10 | 9 | 8 | 6 | 5 |
| 5 | 4 | 8 | 6 | 1 |
| 2 | 2 | 4 | 3 | 9 |
Answer = 19
Q.4 and Q.5 Let S be the set of all pairs (i, j) where 1 ≤ i ≤ j < n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1,2) and (2, 3) are also friends, but (1,4) and (2, 3) are enemies.
Q.4 For general n, how many enemies will each member of S have? (CAT 2007)
Solution:
Use substitution to solve this question. Start with n=5
Pairs are {(1, 2), (1, 3), (1, 4), (1,5), (2, 3), (2, 4), (2,5), (3, 4), (3,5), (4,5)}
For (1,2) – enemies would be (3,4),(3,5) and (4,5)
Options C and D are left. Similarly, use n=6 and find the answer as D
Q.5 For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members? (CAT 2007)
Solution:
Use substitution to solve this question. Start with n=5
Pairs are {(1, 2), (1, 3), (1, 4), (1,5), (2, 3), (2, 4), (2,5), (3, 4), (3,5), (4,5)}
Taking two members as (1,2) and (1,3) their common friends would be (1, 4), (1,5), (2, 3)
Options D and E are left. Similarly, use n=6 and find the answer as D
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