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

Q58. (CAT 2003 Leaked)
Consider the following two curves in the XY plane
y = x^3 + x^2 + 5
y = x^2 + x + 5
Which of the following statements is true for −2 ≤ x ≤ 2?
(1) The two curves intersect once
(2) The two curves intersect twice
(3) The two curves do not intersect
(4) The two curves intersect thrice

Q59.

Q60.

Q61.

Q62.

Q63. (CAT 2006)
A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?
(1) 3
(2) 4
(3) 5
(4) 6
(5) 7

@master_shifu
f(x)=(x)^3 then f(x)=(x)^3
We know x=x then f(x)=(x)^3=(x)^3=f(x).
Then f is even.or (x)^3=x^3=(x)^3 then
f(x)=f(x) then f is even...

@master_shifu
let h=fog then h(x)=x and hoh(x)=h(x)=x
let k=gof then k(x)=x and hok(x)=h(x)=x
Now koh(x)=k(x)=x
kohok(x)=x then
gofofogogof(x)*fogofog(x)=kohok(x)koh(x)=
xx=x^2
The answer x^2

Q64. (CAT 2006)
What are the values of x and y that satisfy both the equations?
2^(0.7x) * 3^(1.25y) = 8 √6 / 27
4^(0.3x) * 9^(0.2y) = 8 * 81^(1/5)
(1) x = 2, y = 5
(2) x = 2.5, y = 6
(3) x = 3, y = 5
(4) x = 3, y = 4
(5) x = 5, y = 2

On a simple inspection of the question, we see that, the 2nd equation is easier to solve, compared to the first one. First one has 2 & 3 on LHS and 6 & 27 on the RHS, in which 6 is a combination of 2 & 3. So, we cannot map easily. Second equation has 4 & 9 on LHS and 8 & 81 on RHS. So, 4 on LHS should map to 8 on RHS and 9 on LHS should map to 81 on RHS. So, we have a 1 to 1 mapping here.
So, taking the 2nd equation, we have, 4^(0.3x) * 9^(0.2y) = 8 * 81^(1/5)
On Solving for x, we have 4^(0.3x) is related to 8 on RHS.
4^(0.3x) can be written as 4^(3x/10) or 2^(6x/10) or 2^(3x/5) (as 4 is 2^2)
So, 2^(3x/5) is related to 8 (2^3)
2^(3x/5) = 2^3, x = 5.
We see that, there is only 1 option with x =5. Hence that is the answer.
You save a lot of time, as you need not find y, and you need not even bother about the 1st equation.

Q65. (CAT 2005)

method 1 :
Simple observation :
x = root(4 + root(4  y))
y > 2
x = root(4 + z)
z < 2
x < 3 but > 2Options :
 is = 3. Ruled Out
 is < 2 Ruled out
 Could be Correct
 is > 3 Ruled out
Answer : Option 3
Want a bit more explanation on this. Here you go!
Here x = root ( 4+ root ( 4x) )
Since the pattern after every 4 + 4  repeats and it tends to infinity ; we consider the series as x itself .
=> x = root ( 4 + root ( 4x) )
squaring on both sides . x^2 = 4 + root(4x)
=> x^24 = root(4x)
Again squaring on both sides .
=> (x^24)^2 = (4x)
==> x^4 + 16  8x^2 = 4x
==> x^4 8x^2 +x + 12 =0 .Now the easiest way to verify from options .
option 1 : 81  72+3+12 = 24 ; doesn't satisy.
option 4 : 169  8 * 13 + root 13 +12 ==> doesn't satisfy.
option 3 : check it !! :)Generally , here on substituting x=3 we got 24 . now when x=2 we get 2 .
So the ans must be some thing above 2 and below 3 .
So it has to be (root 13 + 1)/ 2

Q66. (CAT 2003 Retest)

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 + (n1) d = 1000 ; (n1) d = 999. So 999 is the product of (n1) and d. So, we need to see the factors of 999 to see what all values/how many of those values (n1) 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 n1 as 1, n is 2. But that means only 2 terms in AP. (Not possible according to question)
If we have n1 = 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 (n1), and find the other one correspondingly and proceed. Again in this case, (n1) 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.

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 56 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.