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

Solution: We have to consider positive and negative numbers for all the cases.
f1(x) is positive for positive numbers, and 0 for negative numbers. (0 for x = 0)
f2(x) is 0 for positive numbers, and positive for negative numbers. (0 for x = 0)
f3(x) is 0 for positive numbers, and negative for negative numbers. (0 for x = 0)
f4(x) is negative for positive numbers, and 0 for negative numbers. (0 for x = 0)
So, we see that, out of the 3 products in the question, f1(x) * f2(x) and f2(x) * f4(x) are always zero, for any x.Second sub question, f4(x) = f3(x) = f2(x) = f1(x).
Hence, 1st option is false.
–f3(x) = f2(x) = f1(x). Hence this is true.

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

Solution: 1st let us find the value of log_0.0008 (5).
Let log_0.0008 (5) = y.
So 5 = (0.008)^y.
5 = (8/1000)^y
5 = (2/10)^3y
5 = (1/5)^3y
5 = (5)^(3y)
3y = 1. y = 1/3.
So, 1 + log_0.0008 (5) = 1 – 1/3 = 2/3.Now, the options need us to find a relation between M & N, and hence it should be true for all values of M & N.
We have the equation :1/3 log_3 (M) + 3 log_3 (N) = 2/3.
For simplicity, let us split and give, 1/3 log_3 (M) = 1/3 and 3 log_3 (N) = 1/3.
That means log_3 (M) = 1, and M = 3.
Also, it means, log_3 (N) = 1/9. N = 3^(1/9).
N^9 = 3 and M = 3.
So M * N^9 = 3 * 3 = 9.
This is correct as per the 2nd option.

Q18. (CAT 2003)
If both a and b belong to the set {1, 2, 3, 4}, then the number of equations of the form ax^2 + bx + 1 = 0 having real roots is
a. 10
b. 7
c. 6
d. 12

Solution: For real roots, discriminant should be greater than or equal to 0.
D = b^2 – 4ac. Here, c = 1.
So D = b^2 – 4a ≥ 0.
If b = 1, a cannot be any number among 1, 2, 3, 4 as D < 0.
If b = 2, a can be 1 from the set. For other numbers D < 0.
If b = 3, a can be 1 or 2 from the set. For other numbers D < 0.
If b = 4, a can be 1, 2, 3 or 4.
totally there are 0 + 1 + 2 + 4 = 7 numbers possible for which D ≥ 0, and hence real roots.

Q19. (CAT 2003)
If log_10 (x) – log_10 (√x) = 2 log_x (10), then a possible value of x is given by:
a. 10
b. 1/100
c. 1/1 000
d. None of these

Solution: If you are not sure of the basics of logarithms, follow this method :
Let us suppose log10x = y.
Then x = 10y.
x^(1/y) = 10
logx10 = 1/yIf you are pretty thorough with logarithms, then you know that if log_10 (x) = y, then log_x (10) = 1/y. Also, we know that log_10 (√x) is same as log_10 (x)^1/2, which is same as 1/2 log_10 (x).
So, we have log_10 (x) – log_10 (√x) = 2 log_x (10)
y  y/2 = 2/y
y/2 = 2/y
y^2 = 4
y = 2 or y = 2.
If y = 2, then log_10 (x) = 2, and x = 10^2 = 100.
If y = 2, then log_10 (x) = 2, and x = 10^(2) = 1/100

Q20. (CAT 2003)
What is the sum of ‘n’ terms in the series: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + …
(1) log (n^(n1)/ m^(n + 1))^n/2
(2) log (m^m/ n^n)^n/2
(3) log (m^(1  n)/ n^(1  m))^n/2
(4) log (m^(n+1)/ n^(n  1))^n/2

Solution: log m + log(m^2/n) + log(m^3/n^2) + log (m^4/n^3) + ….
= log m + 2 log m – log n + 3 log m – 2 log n + 4 log m – 3 log n + …
= log m ( 1 + 2 + …n ) – log n ( 1 + 2 + ….(n1)) (This is an AP)
= (n)(n+1)/2 * log m  (n1)(n)/2 * log n
= log [m^(n)(n+1)/2  log n^((n1)n/2)]
=log (m^[(n+1)/2]/n^[(n1)/2])^n
= log (m^(n+1)/n^(n1))^(n/2)

Q21. (CAT 2003)
If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?
a. 2^1/3
b. 2^2/3
c. 2^1/4
d. 2^3/4

Solution: Since all the options of the answer are in the power of 2, let us take x = 2^a and z = 2^b.
So, we get y = 2^[(a + b)/2] (As y is AM of x and z)
We know xyz = 4.
So, 2^a * 2^b * 2^[(a + b)/2] = 2^2.
i.e. 2^[(3a +3b)/2] = 2^2.
So, (3a + 3b)/2 = 2.
3a + 3b = 4.
a + b = 4/3.
(a + b)/2 = 2/3.
We know that y = 2(a + b)/2 , and (a + b)/2 is 2/3.

Q22. (CAT 2002)
For three numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?
(1) 3 √13
(2) √19
(3) 13/3
(4) √15

Solution: (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2xz
25 = x^2 + y^2 + z^2 + 2(xy +yz + xz)
25 = x^2 + y^2 + z^2 + 6
x^2 + y^2 + z^2 = 19.
For x to be maximum, y^2, z^2 should be minimum. Minimum value of a “square” integer is 0.
Hence y^2, z^2 = 0. x^2 will be 19, and x will be √19.

Q23. (CAT 2002)
If x^2 + 5y^2 + z^2 = 2y (2x + z), then which of the following statements are necessarily true?
I. x = 2y II. x = 2z III. 2x = z
(1) Only I
(2) Only II
(3) Only III
(4) Only I and II

Solution: x^2 + 5y^2 + z^2 = 2y (2x + z)
x^2 + 5y^2 + z^2  4xy – 2yz = 0. This can be rewritten as:
x^2 + 4y^2 – 4xy + y^2 + z^2 2yz = 0
(x  2y)^2 + (y – z)^2 = 0
Sum of two “square numbers” can be 0, only if both of them are 0.
Hence x – 2y = 0 and y – z = 0.
Hence x = 2y and y = z.
x = 2y = 2z.

Q24. (CAT 2002)
The number of roots of (a^2/x) + (b^2/(x1)) = 1
(1) 1
(2) 2
(3) 3
(4) None of these

Solution: We essentially have a quadratic equation.
a^2 (x1) + b^2 x = x(x1)
x^2 – x(1 + a^2+ b^2) + a^2 = 0
This is a quadratic equation, and discriminant D = b^2 – 4ax = (1 + a^2+ b^2)^2 – 4a^2
D is not equal to 0, and hence they will not have equal roots.
So, they have 2 roots as any quadratic equation.

Q25. (CAT 2002)
Mayank, Mirza, Little and Jagbir bought a motorbike for 60 Dollar. Mayank contributed half of the total amount contributed by others, Mirza contributed onethird of total amount contributed by others, and Little contributed onefourth of the total amount contributed by others. What was the money paid by Jagbir?
(1) 12 Dollar
(2) 13 Dollar
(3) 18 Dollar
(4) 20 Dollar

Solution: Let Mayank’s contribution be x. So, the remaining amount is 60 – x.
Mayank’s contribution is half of other’s contribution.
So, x = (60 – x)/2
60 = 3x => x = 20.
So Mayank’s contribution = 20.
Similarly Mirza’s contribution y = (60 – y)/3 => y = 60/4 = 15
And Little’s contribution z = (60z)/4. => z = 60/ 5 =12
Hence Jagbir’s contribution is 60 – 20 – 15 – 12 = 13.

Q26. (CAT 2002)
If f(x) = log((1+x)/(1x)), then f(x) + f(y) =
(1) f(x + y)
(2) f(1 + xy)
(3) (x + y) f(1 + xy)
(4) f ((x +y) / (1+xy))