# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 1

• The Euler coefficient of 13 is 12. So, we have to reduce the power by multiples of 12 as much as possible. Now, all the powers starting from 4! will be in the form of 12x and so, can be ignored. So, 0!+1!+2!+3!=10 remains. So, 5^10 mod 13. 5^2 mod 13 is -1. So, eventual remainder is -1 and so, 12

• Q21) The average age of a newly married couple is 30 years. Thirteen years later, the average age of their family consisting of the two of them and one child is 32 years. A second child is born at that stage. What is the average age (in years) of the two children after 6 more years?
a. 16
b. 22
c. 12
d. 11

• Average should have been 43 of the couple. However, it is coming down to 32 for 3 people. So, 11*2 = 22 is taken away so as to bring the child to 32 years of age. So, child is 10 years old. 2 ppl grow for 6 years so the total age will increase by 12 and will become 22. Average = 11.

The other way is to take complete values and solve. I will just write the steps.
30 * 2 = 60, 13 * 2=26, total 86, but 32 * 3=96, so child=10, 10+6 * 2=22 divided by 2 is 11.
Good aspirants will be able to do these questions orally within 15-20 seconds. The point is not solving a lot of questions quickly but saving time to allocate to questions that are difficult/time consuming.

• Q22) A and B are moving along the circumference of a circle with speeds that are in the ratio 1 : K. They start simultaneously from a point P in the clockwise direction. They meet for the first time at a point Q which is at a distance of one-third the circumference from P, in the clockwise direction. K cannot be equal to
(a) 1/4
(b) 4/7
(c) 4
(d) None of these

• If two entities are moving in the same direction along a circular track, starting from the same position at speeds ax and bx where a and b are co-prime, the number of distinct meeting points is given by |a-b|

Also, the meeting points are equally spaced and so, if they start from a single point and meet for the first time at a distance equal to one third of the circumference of the circle, the second meeting point will be further one third ahead of the first meeting point and the third meeting point with be a further one third ahead of the second meeting point i.e. the original meeting point. So, 3 distinct meeting points in total.

If the ratio is 1: (1/4), the basic ratio is 4:1 and so, 3 distinct points of meeting
If the ratio is 1: (4/7), the basic ratio is 7:4 and so, 3 distinct points of meeting
If the ratio is 1:4, there would be 3 basic points of meeting.

• Q23) N! is completely divisible by 13^52. What is sum of the digits of the smallest such number N?
(a) 11
(b) 15
(c) 16
(d) 19

• This is more of trial n error or going for 13×52 n then scaling down by 39 .
Trial n error
If you have no starting point, find the number of 13s in 650! Using the normal method. It comes to 50+3=53... go back by one multiple of 13 and you are at 52 thirteens n so 637
The 'real' method
13×52=676 will definitely have 52 thirteens. It will also contribute a few more due to the presene of 13^2 and its multiples... 676 covers three multiples of 169... 169, 338 and 507. So we come back by three multiples of 13 n so 676-39=637
In either case, 6+3+7=16

• Q24) n is a number, such that 2n has 28 factors and 3n has 30 factors. find the number of factors of 6n.
a) 35
b) 32
c) 28
d) None

• In the first case, there is a 2 added to the existing number of twos and in the second case, this 2 is removed and a 3 is added to the existing number of threes. If a number has 28 factors, the powers of the individual primes could be (27), (1,13), (3,6), (1,1,6).

Similarly, if a number has 30 factors, the powers of the individual primes could be (29), (1,14), (2,9), (4,5), (1,1,1). The second case has to have one 2 lesser and one 3 more than the first. Only (3,6) and (4,5) satisfy this case and so, the number is in the form of 2^5 * 3^3. 6n becomes 2^6 * 3^4 and so, number of factors is 7 * 5 = 35

• Q25) Anshul and Nitish run between point A and point B which are 6 km apart. Anshul starts at 10 a.m.from A, reaches B, and returns to A. Nitish starts at 10:30 a.m. from B, reaches A, and comes back to B. Their speeds are constant with Nitish’s speed being twice that of Anshul’s. While returning to their starting points they meet at a point which is exactly midway between A and B. When do they meet for the first time?

• Anshul covers 9 km in say x hours. So, Nitish will also cover 9 km in (x - 0.5) hours. As the ratios of their speeds are known, we can form the following equation:

9/(x-0.5) = 2 * 9/x
x = 2x-1
x=1 hour.

Anshul covers 9 km in 1 hour and so, his speed is 9 km/hr
Nitish covers 9 km in 0.5 hours and so, his speed is 18 km/hr

For the first half hour, Anshul covers 4.5 km. So, 1.5 km need to be covered at the effective speed of 27 km/hr and so, will take 1.5/27 = 1/18 more hours ie. 3 1/3 minutes. So, a total of 33 1/3 minutes post 10.

• Q26) What is the number of non-negative integer solutions for the equation x^2 – xy + y^2 = x + y?
(a) 3
(b) 4
(c) 1
(d) None of these

• 2x^2 - 2xy + y^2 = 2x + 2y
(x-y)^2+x^2+y^2-2x-2y+2=2
(x-y)^2+(x-1)^2+(y-1)^2=2
(1,1,0) gives (2,1) or (0,1)
(1,0,1) gives (1,0) or (1,2)
(0,1,1) gives (0,0) or (2,2)

• Q27) A is the set of the first 100 natural numbers. What is the minimum number of elements that should be picked from A to ensure that at least one pair of numbers whose difference is 10 is picked?
(a) 51
(b) 55
(c) 20
(d) 11

• A lot of students get confused in the minimum number of elements to be picked so as to be sure and might end up listing elements as (1, 21, 23, 34... 100) to get 10+1 = 11 elements. There is no real method for such questions but if you can simply think a bit laterally, you should be able to come up with an answer quickly.

• Q28) A game consisting of 50 rounds is played among P, Q and R as follows: Two players play in each round and the player who loses in that round is replaced by the third player in the next round. If the only rounds in which P played against Q are the 3rd, 14th, 25th and 36th, then what can be the maximum number of games won by R?
(a) 40
(b) 42
(c) 41
(d) 36

• The easier way to do this is to understand that P and Q played each other in those 4 matches. Which means that R would have lost in the previous round at least. Other than that there is no condition. So, R loses 4 matches and does not participate in 4 matches. So, max number of wins is 42.

• Q29) 5f(x) + 4f[ (4x + 5) / (x – 4) ] = 9(2x + 1), where x ∈ R and x # 4. What is the value of f(7) ?
(a) -17
(b) -8
(c) -7
(d) None of these

• Simply substitute x=7 and x=11. You will get 2 linear equations with 2 variables. The fact that f(7) is asked should make you suspicious. As f(x) has to be f(7) for you to get an answer, substitute 7 and then you might realize that the next step would be to plug x=11

• Q30) There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairs of parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg, 101 kg and 104 kg. Which of the following can be the weight of one of the parcels?
(a) 40 kg
(b) 45 kg
(c) 48 kg
(d) 53 kg

41

45

61

61

42

61

63