# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 2

• Total cases = 8c2 * 5c4 + 8c3 * 5c3 = 140 + 560 = 700
When the two men are selected together for sure: 2c2 * 5c4 + 2c2 * 6c1 * 5c3 = 65 cases
So, two men are not selected together will be true in 700 - 65 = 635 cases.

• Q14) If s and t are positive integers such that s/t=64.12, which of the following could be the remainder when s is divided by t?
a) 2
b) 4
c) 8
d) 45

• 6412/100 -> remainder 12
3206/50 -> remainder 6
1603/25 -> remainder 3
if we go to the other side,
12824/200 -> remainder 24
19236/300 -> remainder 36
So, all the remainders are multiples of 3. 45 is the only option that satisfies

• Q15) What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?
a) 271/900
b) 27/100
c) 7/25
d) 1/9
e) 1/10

• There are 20 repetitions of a number in the first 99 natural numbers. So, ignoring the repetitions - twice and thrice (all your 177, 277, and so on) we get 19*8=152+100 terms from (700-799) which gives us 252. Pretty dense concept but it's handy. There are 900 3 digits numbers.
Required answer = 252/900 = 7/25

• Q16) A box contains 20 bulbs. The probability that the box contains exactly 2 defective bulbs is 0.4 and the probability that box contains exactly 3 defective bulbs is 0.6. Bulbs are drawn at random one by one without replacement and tested till the defective bulbs are found. What is the probability that the testing procedure ends at the twelfth testing?

• 2 defective bulbs scenario:
Favorable cases: One faulty bulb in the first 11 picks in 11c1 ways. 12th one will be picked in 1 way only. Total 11 ways
Total cases: 2 defective bulbs in the first two picks in 1 way. In the first three picks in 2c1=2 ways, in the first four picks in 3c1=3 ways and so on till, in the first twenty picks in 19 ways. Total of 190 ways

3 defective bulbs scenario:
Favorable cases: Two faulty bulbs in the first 11 picks in 11c2 ways. 12th one in 1 way only. Total 55 ways.
Total cases: 3 defective bulbs in the first 3 picks in 1 way. In the first four picks in 3c2=3 ways, in the first five picks in 4c2=6 ways and so on till, in the first twenty picks in 19c2=171 ways. Total of 1140 ways

Multiplying by individual weights and adding will give us:
44/1900 + 330/11400
44/1900 + 55/1900
99/1900

The total cases in both scenarios are nothing but 20c2 and 20c3 respectively.
I have put it in detail so that it is easier to understand.

• Q17) Find the smallest positive integer that can be expressed as the sum of 9 consecutive positive integers, sum of 10 consecutive positive integers and sum of 11 consecutive positive integers.

• Sum of 9 consecutive numbers can be thought of as (a-4)+(a-3)...+a+(a+1)+...(a+4) = 9a
Sum of 11 consecutive numbers can be thought of as (b-5)...+b+(b+1)...+(b+5) = 11b
Sum of 10 consecutive numbers will be c + (c + 1) + (c + 2) + ... + (c + 9) = 10c + 45
But it has to be a multiple of 11 and 9 as well. So, c has to be a multiple of 9
Also, 10c + 45 = 11k
So, c could be 1, 12, 23, 34, 45.
So, c has to be 45.
LCM of the 3 comes to 495.

• Q18) A container has 3L of pure wine. 1L from the container is taken out and 2L water is added. The process is repeated several times. After 19 such operations, quantity of wine in mixture is (Answer in the form of a/b)

• After 1 operation 6/3 L wine left
After 2nd operation =6/4 L wine left
After 3rd operation => 6/5 L wine left and so on.
Series =6/(K+2) ...(K is no of operation)
19th operation =6/(19+2) =6/21=> 2/7

• Q19) Iqbal dealt some cards to mushtaq and himself from a full pack of playing cards and laid the rest aside. Iqbal then said to mushtaq. "If you give me a certain number of your cards. I will have four times as many as cards you will have. If I give the same number of cards. I will have thrice as many as you will have of the given choices, which could represent the number of cards with Iqbal?
a) 27
b) 17
c) 31
d) 35

• 31 is correct. One way to go is to take two variables and solve. The other is to understand that the total number of cards dealt would be in the form of 4a as well as 5b and so, a multiple of 20 and so, either 20 or 40. If there are 20 cards, according to the first distribution, the value of I:M would be 16:4. Also, according to the second statement, I:M would be 15:5. So, after taking certain cards, Iqbal gets to 16 and after giving away certain cards, Iqbal gets to 15. This is not possible and so, there were 40 cards to begin with. So, I:M in first case will be 32:8 and in the second case will be 30:10. So, the original value with Iqbal would have been 31.

After the first distribution, let Mushtaq have M cards with him. So, Iqbal will have 4M cards with him. So, total cards will be 5M. Similarly after the second distribution, Mushtaq will have say N cards with him. So, Iqbal will have 3N cards with him. Total number of cards is 4N. So, x = 5M = 4n and so, has to be a multiple of 20.

Let Iqbal have I cards and Mushtaq have M cards. Let Mushtaq give x cards to Iqbal in the first case and let Iqbal give x cards to Mushtaq in the second case:
4(M-x)=(I+x) -> 4M-I=5x
3(M+x)=(I-x) -> 3M-I=-4x
M=9x and I=31x
Remember that M+I < 52 and so, 40x < 52 and so, x=1. Consequently, M=31.

• Q20) How many 3 digit even numbers can you form such that if one of the digits is 5 then the following digit is 7

• Classic trap: IF there is a 5, then there will be a 7. So, all the numbers that do not have 5 have to be counted as well. Such numbers can have the first digit chosen in 8 ways (except 0 and 5), second digit in 9 ways (except 5) and third digit in 5 ways (as it is an even number). So, 9 * 8 * 5=360 additional cases to the existing 5 (570, 572, 574, 576, 578). Total 365.

Is the number 272 acceptable in this scenario? Is 104 acceptable? The answer to both is yes because the condition simply says that if there is a 5, then there definitely should be a 7 which holds true in these cases. The unwritten part is that even if there is no 5, there could still be a three digit even number formed which is our basic requirement.

• Q21) A, B, and C can finish a work in 6, 8, and 15 hours respectively. They started their work together and got Rs 94.60 in total. What would be each one’s share

• Let's say the total work is 120 units. So, in 1 day, A does 20 units of work, B does 15 units and C does 8 units. So, A's share of the total work will be 20/43, B's share will be 15/43 and C's share will be 8/43. The total amount is 94.60 and so, A will get 20/43 * 94.6=44, B will get 15/43 * 94.6=33 and C will get 8/43 * 94.6=17.6.

The key here is to not get scared by an abstract number like 94.60 and solve the question. Remember, if it is given, it is logical and useful in some manner.

• Q22) A car after traveling 18 km from a point A developed some problem in the engine and the speed became 4/5th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
a. 25
b. 30
c. 20
d. 35
e. None of these

• By traveling 12 km more at original speed, you save 9 minutes. To save 45 minutes that were overshot in the first case, you need to travel 60 km more at original speed. So, total distance is 60+18=78. Let original speed be 5x. The difference in time is because of the 60 km travel at 4x. So, 60/4x-60/5x=3/4 and x=4 and so, 5x=20.

• Q23) Two numbers b and C are chosen at random with replacement from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. The probability that x^2 + bx + c > 0 for all x€ R is

61

61

61

61

61

1

61

63