Question Bank - 100 Algebra Questions From Previous CAT Papers (Solved)
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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.
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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
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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.
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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
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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.
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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
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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
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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^(n-1)/ 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
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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 + ….(n-1)) (This is an AP)
= (n)(n+1)/2 * log m - (n-1)(n)/2 * log n
= log [m^(n)(n+1)/2 - log n^((n-1)n/2)]
=log (m^[(n+1)/2]/n^[(n-1)/2])^n
= log (m^(n+1)/n^(n-1))^(n/2)
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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
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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.
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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
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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.
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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
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Solution: x^2 + 5y^2 + z^2 = 2y (2x + z)
x^2 + 5y^2 + z^2 - 4xy – 2yz = 0. This can be re-written 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.
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Q24. (CAT 2002)
The number of roots of (a^2/x) + (b^2/(x-1)) = 1
(1) 1
(2) 2
(3) 3
(4) None of these
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Solution: We essentially have a quadratic equation.
a^2 (x-1) + b^2 x = x(x-1)
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.
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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 one-third of total amount contributed by others, and Little contributed one-fourth 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
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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 = (60-z)/4. => z = 60/ 5 =12
Hence Jagbir’s contribution is 60 – 20 – 15 – 12 = 13.
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Q26. (CAT 2002)
If f(x) = log((1+x)/(1-x)), 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))
