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

Solution: A little understanding on the theoretical side of logarithms is necessary for solving such questions. If you do not have much theoretical understanding of log, you should know the following at least :
log_10(100) = 2 (You have to know this)
log_10(1000) = 3 (It is easy to remember such things, compared to the formulas)
This means, 10^2 = 100, 10^3 = 1000. That means log_x(y) = a means x^a = y.Also you need to know, log_x(y) = log y/log x (Both to the same base, 10 or e or anything)
Now, moving on to the question,
We know, a * log_z (y) = a * b. So, log_z (y) = b.
We know, b * log_x (z) = a * b. So, log_x (z) = a.
We know, log_y (x) = a * b
Substituting equations 1 and 2 in equation 3, we get,
log_y (x) = log_z(y) * log_x (z)
log x/log y = log y/log z * log z/log x
log x/ log y = log y / log x
log x^2 = log y^2
log x = log y or log x =  log y
x = y or x = 1/y
We know that log y x = a * b
So a * b = log y y or log y 1/y (Substituting for x = 1/y, in the above equation)
So a * b = 1 or a * b = 1.
See from the options, which value of (a, b) does not obey the above.

Q13. (CAT 2008)
If the roots of the equation x^3 – ax2 + bx – c =0 are three consecutive integers, then what is the smallest possible value of b ?
(1) 1/√3
(2) 1
(3) 0
(4) 1
(5) 1/√3

Solution: For an equation with power 3, the coefficient of x is the sum of the individual product of the roots.
Let (n1), n, (n+1) be the roots of the equation. So, n(n1) + n(n+1) + (n1)(n+1) = b
n^2 – n + n^2 + n + n^2 – 1 = b
3n^2 – 1 = b.
We know that n^2 is positive, the minimum value of n^2 will be 0 when n = 0. So, the minimum value of b is 1 when n = 0.

Q14. (CAT 2005)
For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?
x^2 – y^2 = 0
(x – k)^2 + y^2 = 1
(1) 2
(2) 0
(3) √2
(4) √2

Solution: From 1st equation x^2 = y^2.
Substituting for y^2 in the 2nd equation, we get, (x – k)^2 + x^2 = 1
x^2 + k^2 – 2kx + x^2 = 1
2x^2  2kx + (k^2 – 1) = 0
For unique solution, Discriminant D should be 0. (D = b^2 – 4ac)
D = 4k^2 – 8 (k^2 – 1) = 8 – 4k^2
So, for unique solution, 8 – 4k^2 is 0
4k^2 = 8 => k^2 = 2 => k = √2 or √2
We require the positive solution, substituting, we get positive solution for k = √2

Q15. (CAT 2005)
If x ≥ y and y > 1, then the value of the expression logx(x/y) + logy(y/x) can never be
(1) 1
(2) 0.5
(3) 0
(4) 1

Solution: Let us take everything to the same base, so that it is easier.
log_x(x/y) = (log x – log y)/ log x
log_y(y/x) = (log y – log x)/ log y
Adding these 2, we have (log x – log y)/ log x + (log y – log x)/ log y,
1 – (log_y/log x) + 1 – (log x/ log y)
2  (logy/log x)  (log x/ log y)
2 – (log_x y + log_y x) (As, log y/log x = log_x y and log x/ log y = logy x)
Now, we know y > 1 and x ≥ y. So, log_x y < 1, and logy x > 1 (You can try out by taking examples as x = 100 and y = 10 or similar).
So, we have the expression 2  (log x y + log y x) as 2 – (a number greater than 1).
So, our result is lesser than 1 always.

Q16. (CAT 2004)
f1 (x) = x when 0 ≤ x ≤ 1
= 1 when x ≥ 1
= 0 otherwise
f2 (x) = f1(–x) for all x
f3 (x) = –f2(x) for all x
f4 (x) = f3(–x) for all xHow many of the following products are necessarily zero for every x:
f1(x) f2 (x)
f2 (x) f3 (x)
f2(x) f4 (x)
a. 0
b. 1
c. 2
d. 3Which of the following is necessarily true?
a. f4(x) = f1(x) for all x
b. f1(x) = –f3(–x) for all x
c. f2(–x) = f4(x) for all x
d. f1(x) + f3(x) = 0 for all x

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