Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 1

Let total work be 120 units. A does 15 units per day, B does 10 units per day, C does 4 units per day. In a set of 3 days, they finish off 58 units of work. So, 120/58 = 2 sets of 3 days plus an additional 4 units to go which can be done by A and B together in the quickest time of 4/25 days. So total of 6 4/25.

Q16) Fourteen fruits and twenty two flowers are to be distributed among 10 people in such a way that each person gets something. Anyone who gets more than two flowers cannot get more than one fruit and anyone who gets more than one fruit cannot get more than three flowers. What is the maximum number of flowers that one can get?

One person can get 22 flowers and the rest of the fruits can easily be distributed. A question specifically designed for those who think that CAT does not ask questions that have obvious answers.

Q17) If N is a natural number less than 100, then for how many values of N are the numbers 6N + 1 and 15N + 2 relatively prime?
(a) 16
(b) 10
(c) 33
(d) None of these

Let there be two numbers ax and bx where x is the hcf. So, axbx=(ab)x will be divisible by x as ab will be an integer. Also, (ab)xbx = (a2b)x which will also be divisible by x as a2b is an integer.
So, if two numbers share a common factor, the same factor divides the difference (d) between the two numbers and also the difference between d and the smaller number. Say the two numbers are 111 and 153. The difference is 42 which is divisible by the hcf of 111 and 153 ie. 3. Also, 11142=69 is also divisible by 3. Remember that we are taking absolute values only.
Applying the same concept, If 6N + 1 and 15N + 2 are divisible by x, then their difference i.e. 9N + 1 will always be divisible by x. Similarly (9N + 1) – (6N + 1) i.e. 3N will also be divisible by x. If x divides 3N then it can also divide 6N. So, x divides 6N and 6N + 1 both i.e. two consecutive numbers. Hence x cannot be anything but 1. So for all the values of N, the given two numbers will be coprime.

Q18) A tank of capacity 165 liters, has N taps, numbered from 1 to N fitted to it. At the start of the nth minutes, where 1 < = n < = N, the tap numbered n, which fills the tank at the rate of n litres per minute, is opened. If it takes exactly N minutes to fill the tank in this manner, find N.
a. 8
b. 9
c. 10
d. 11

So essentially, the first pipe keeps on filling the tank throughout the duration of work ie. n minutes, the second pipe fills it for (n1) minutes and so on. So, we will get 1+(1+2)+(1+2+3)+(1+2+3+4)...(1+2+...n)=165
Now there are two ways to go about this, one is you use summation. The general formula here is E[n(n+1)/2)] which will be 1/2[n(n+1)(2n+1)/6+n(n+1)/2] which on simplification gives n(n+1)(n+2)/6 as the generic formula (something you might want to memorize). Substitution or trial and error or simply understanding that 9 * 10 * 11=990 should be enough.

Q19) A shopkeeper sells a radio at 10% discount and yet manages a 10% profit. Find his profit/loss percentage, if he were to instead sell the same radio at a discount of 20%.

Let list price be 100.
10% discount means that the product was sold for 90
10% profit meaning that the cost price was 900/11
20% discount means that the product was sold for 80
Loss is (900/1180)/(900/11) = 20/900 or 2 2/9 % loss

Q20) S is the sum of the factorials of the first eight whole numbers. Find the remainder when 5^S is divided by 13.

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 1520 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 onethird 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 coprime, the number of distinct meeting points is given by ab
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 67639=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?