# Question Bank - 100 Algebra Questions From Previous CAT Papers (Solved)

• We will try to solve by taking sample values for n.. now there is sqrt(n) appearing in the equation so to make life easy we will consider a perfect square which lies in the given range (we can take 36 or 49). Another important point to consider is that we need to get the extreme cases for x (highest or lowest). Observing the function, x will increase as n increases.. so min(x) happens when n = 36 and its a perfect square too.

if n = 36, x = ((36)^2 + 2 * 6 * 40 + 16)/ (36 + 4 * 6 + 4)
= 1792/64 = 28 (which is the least value x can take)
Please note that option 4 does not include 28 as it says 28 < x ( not 28 < = x )

• Q67. (CAT 2006)

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 with 1000 and have at least 3 elements?

(1) 3
(2) 4
(3) 6
(4) 7
(5) 8

• We know that it should be an AP, 1st element is 1, and last element 1000. a = 1, a + (n-1) d = 1000 ; (n-1) d = 999. So 999 is the product of (n-1) and d. So, we need to see the factors of 999 to see what all values/how many of those values (n-1) and d can assume. Prime factorize 999. The prime factors are 3, 3, 3, and 37. (999 = 3 * 3 * 3 * 37). (Other than 1 and 999)
So, we can have
(n - 1) = 3, d = (3 * 3 * 37)
(n - 1) = 3 * 3, d = (3 * 37)
(n - 1) = 3 * 3 * 3, d = (37)
(n - 1) = 37, d = (3 * 3 * 3)
(n - 1) = 37 * 3, d = (3 * 3)
(n - 1) = 37 * 3 * 3, d = (3)

Other than this, we can have 1 and 999 also as factors of 999.
If we have n-1 as 1, n is 2. But that means only 2 terms in AP. (Not possible according to question)
If we have n-1 = 999, n = 1000. d = 1. This is possible.
So, we have totally, 7 possible combinations.

2nd approach:

The factors of 999 are 1,3,9,27,37,111,333,999. We can assign each of these values to d or (n-1), and find the other one correspondingly and proceed. Again in this case, (n-1) cannot be 1. All other cases are possible. Both the approaches are essentially the same. Just that you need not spend time multiplying the factors in the 1st approach.

• @master_shifu get your sources right. You are typing integers x y z in question and giving answer as root 19. Then again, typing CAT but the question is from XAT. Also misleading people by giving wrong answers when answer is 13/3.

Dont post if you can't provide right answers even, forget explanations.

• @visheshsahni

No one is perfect here yaar and if you find a mistake in a solution and know the right answer, least thing to do is to share the solution. That's how a forum (should) function right ? :)
Corrected the question.

• @zabeer agreed bro. But go through this thread. There are 5-6 more wrong questions. And I have few friends who actually believed the solution. Really appreciate the collection and greatful for it, but many people would follow solutions as it is. Anger is on the part that a friend who showed this actually did same mistake in mock saying that he learnt the solution from here.

Hope you understand. No offense.

• @master_shifu Assumption of y = 2^[(a + b)/2] looks wrong. Y = 2^(a-1) + 2^(b-1)

• @master_shifu If both y and x are zero, then xy + yz + xz would be equal to zero. The answer stated by you is wrong.

• This post is deleted!

• Q17. (CAT 2003)
If 1/3 log_3M + 3 log_3N = 1 + log_0.0008 (5), then:
(1) M^9 = N/9
(2) N^9 = 9/M
(3) M^3 = 3/N
(4) N^9 = 3/M

Correction:
log_0.008(5)

21

13

22

11

21

21

20

11