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

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))

Solution: f(x) + f(y) = log((1+x)/(1x)) + log((1+y)/(1y))
= log (1+x) – log (1x) + log(1+y) – log(1y)
= log (1+x) + log(1+y) – log (1x) – log(1y)
= log ((1+x)(1+y) / (1x)(1y))
= log (1 + x + y + xy / 1 – x – y + xy)Now, look at the options :
1st option :f(x + y) = log (1 + x + y / 1 – x – y)
2nd option : f(1+xy) = log (1 + xy/ 1 – xy)
3rd option : Not true as it involves power of (x + y)
4th option: f (x + y/1+xy) = log (1 + (x + y/1+xy) / 1 – (x + y / 1 + xy)
= log (1 + x + y + xy / 1 – x – y + xy)
Hence 4th option is right.

Q27. (CAT 2002)
Three travelers are sitting around a fire, and are about to eat a meal. One of them has five small loaves of bread, the second has three small loaves of bread. The third has no food, but has eight coins. He offers to pay for some bread. They agree to share the eight loaves equally among the three travelers, and the third traveler will pay eight coins for his share of the eight loaves. All loaves were the same size. The second traveler (who had three loaves) suggests that he be paid three coins and that the first traveler be paid five coins. The first traveler says that he should get more than five coins. How much the first traveler should get?
a. 5
b. 7
c. 1
d. None of these

Solution: There are 8 loaves of bread which are equally shared between 3.
So, each will have 8/3 loaves of bread.
So A which had 5, will now have only 8/3. Hence A lost 5 – 8/3 = 7/3 bread
B had 3 and will now have 8/3. So, B lost 3 – 8/3 = 1/3 bread.
So 8 Rs has to be divided in the ratio of what bread they lost/gave away to the the 3rd person.
Since A gave away 7 times that of what B gave, A should get 7 times the amount that B gets.
Hence, A gets 7 Rs and B gets 1Re.

Q28. (CAT 2002)
A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked. "How many gold coins do we have?" After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 48 times the difference of the numbers is equal to the difference of their squares. The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins the merchant has?
(1) 48
(2) 96
(3) 32
(4) 36

Solution: Let a and b be the 2 unequal numbers he divide the gold coins into, and let a > b.
From the question,
48 * (a – b) = a^2 – b^2.
48 * (a – b) = (a + b) (a – b)
48 * (ab)/(ab) = (a + b)
48 = a + b
As (a – b) is not 0, a – b can be cancelled. [ (a – b ) is not zero, as he divides into unequal numbers]
Hence a + b = 48. The total gold coin he has is 48

Q29.(CAT 2002)
Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs.2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?
(1) 240
(2) 300
(3) 400
(4) None of these