# Question Bank - Algebra - Shashank Prabhu, CAT 100 Percentiler

• 1.5 * 3 * 4 * x = 1.5 + 3 + 4 + x
x = 0.5
Total cost was Rs. 9. So he got Rs. 41 back. The way in which it could have been given is 20 + 20 + 1 to make sure that the minimum notes /coins were used.

• Q33) Diophantus's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Diophantus died four years after his son. How many years did Diophantus live?
[OA: 84]

• a + b + c = 104
a + 2b + 3c = 204
b + 3c = 134
You will get b and you already know M = 74.

• The number needs to end in a 3 or a 5 or a 7. If the last digit is 3, there are 2 * 3 * 2 = 12 ways of generating the number. If the last digit is 5, there are 2 * 3 * 2 = 12 ways and if the last digit is 7, there would again be 2 * 3 * 2 = 12 ways. So total of 36 ways.

• Line joining the mid points of two diagonals of a trapezium is parallel to the parallel sides and is half of the difference between them. 1/2(x-25) = 5, x = 35 or 1/2(25 - x) = 5, x = 15.

• Q38) If f(x) + f(1 – x) + f(1 + x) + f(2 + x) = 2x for all real values of x, and f(0) = 1, then find the value of f(4)?
[OA: 5]

• Q41) Atul has exactly six sealed bags containing 15, 31, 19, 20, 16 and 18 candies. Out of the six bags with Atul, there are exactly five bags that contained chocolate candies whereas one box contained orange candies. He distributed all the six bags among his three sons in such a manner that his eldest son got the only box with orange candies and the other bags were distributed in such a manner so that other two brothers received the chocolate candies in the ratio of 2 : 1. How many orange candies were there with Atul? (Assume no candies were taken out of the bags)
[OA: 20]

• Q42) DP assigned the first 11 natural numbers as a, b, c, d, e, … , k not necessarily in that order. After some days, she forgot which number corresponds to which letter, but she knew that
a + b + c + d + e + g + h + i + k = 3f + 5j. Which of the following number cannot be the value of j?
a) 9
b) 5
c) 7
d) 3
[OA: Option d]

• Q46) Ria has three types of boxes viz. red, blue and green. She plays a game in which she placed 9 red boxes on the table. She puts 5 blue boxes each, in a few of the red boxes then she puts 5 green boxes each, in few of the blue boxes. If the number of boxes that have been left empty in the game is 41, then how many boxes were used in the game by Ria?
[OA: 49]

• Q47) What is the number of integral values P can take, where 2 ≤ P ≤ 30, such that the product (P – 1) × (P – 2) ….. (3) × (2) × (1) is not divisible by P?
[OA: 11]

• It basically says that to cover 3 gaps (the clock starts striking 4 the first time the bell rings, then one gap, then the second ring, then the second gap, then the third ring, the third gap and the fourth ring which will mark the end of the acitivity), the clock needs 7 seconds. So, to cover 10 gaps, it should take 70/3 seconds.

• Basically we need to check if one or more of the expressions are 0 or if all of them are either 1 or -1 . If a - 1 =0, we have 36 cases. Using the concept of set theory, we get 36+36+36-6-6-6+1=91 cases. Add to that cases when a = 2, b = 2 or 4 and c = 3 or 5 i.e. 1 * 2 * 2 = 4 cases. Total 95 favourable cases out of 216.

• Q60) In an A.P., the pth term is q and (p + q)th term is 0.Then qth term is
a. -p
b. p+q
c. p
d. p-q
[OA: Option c]

• Q63) Find the number of ordered pairs (x, y) where both x and y are non-negative integers such that, x – (1/y) = (x/y) + 1?
[OA: 2]

• The maximum distance covered would be 68 * 11 = 748.
So, the maximum reading on the panel would be 30751.
The largest palindrome less than this is 30703.
So, maximum of 700 km in 11 hours.

• Q68) If |x-7|=|x|+|x-6|, then how many distinct values of x are possible?
[OA: 2]

• Once you plot the diagram, you will see that when the train travels 2 km, the athlete would travel 6 km. So, when the train is at the end of the tunnel, the train would have covered 2 km after the athlete came out of the tunnel. In that time period, the athlete would have moved ahead by 6 km. So, the distance between the train and the athlete will be 6 km which will be covered in the ratio of their speeds. So, the train will cover 1/4th of the total distance or 1.5 km. If you are good at visualizing it, you need not even draw the diagram.

• Q71) Find the constant term in the expansion of [x^2 - 1/x]^12
[OA: 495]

• The term will be constant when x^2 and x cancel each other out. That will be possible when it is in the form of (x^2)^4*(1/x)^8. This is the term that will have the coefficient 12c4 = 495

• Q74) Let a, b and c be non zero real numbers such that a + b + c = 0. Let q = a^2 + b^2 + c^2 and r = a^4 + b^4 + c^4. Then
a) q^2 > 2r always
b) q^2 = 2r always
c) q^2 < 2r always
d) q^2 - 2r can make both positive and negative values
[OA: Option b]

107

92

32

47

33

89

45

34