This is the first post in the CAT 2016 Sprint Preparation Series – PnC, Series, Probability. We have posted 20 questions from previous year CAT papers, forums, mock tests, and other entrances that are on par with the level of difficulty you can expect in CAT 2016. We will be posting the solutions and traps/things to look at while solving similar questions so that you are avoid making silly mistakes during the test.
You can go through the entire series by clicking on this link: CAT 2016 Sprint Preparation Series by Learningroots.
CAT 2016 Sprint Preparation Series – PnC, Series, Probability
1. How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?
2. Ten points are marked on a straight-line and 11 points are marked on another straight-line. How many triangles can be constructed with vertices from among the above points?
3. If there are 10 positive real numbers n1 < n2 < n3… < n10 , how many triplets of these numbers (n1, n2, n3), (n2, n3, n4), … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?
4. How many numbers from 1 to 100 are present such that they are divisible by 3 or 5 but not by both 3 and 5?
5. Find the sum of the infinite series:
a. 679/1296
b. 679/216
c. 339/1296
d. None of these
6. Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?
a. 228
b. 216
c. 294
d. 192
7. Ramesh hits the target 3 out of 4 times and Suresh hits the target 2 out of 3 times. What is the probability that the target would be hit when they both try?
8. In how many ways can 34300 be written as a product of 8 different integers?
9. 10 points lie in a plane, of which 4 points are collinear. Barring these 4 points no three of the 10 points are collinear. How many quadrilaterals can be drawn?
10. How many divisors of 9000 are even but not divisible by 4? Also what is the sum of all such divisors?
11. A bag contains 100 red balls and a certain number of green balls. Two persons A and B draw a ball one by one with replacement. The first one to draw a green ball is the winner. A picks up the ball first and we also know that because of this A is three times more likely to win. Find the number of green balls.
12. The probability that India wins a match against South Africa is 2/5. If there is a five match series and there is no concept of a draw in any match, what is the probability that India would win at least 3 matches?
13. How many three-digit numbers are there such that no two adjacent digits of the number are consecutive?
14. In how many ways can 4 distinct books be put into 3 identical bags?
15. Fifty white marbles are placed in a row. What is the minimum number of black marbles that need to be placed between the white marbles such that each marble (white or black) has at least one neighbour of the other colour?
16. Find the summation of the infinite series:
17. In Cricket, a batsman can score 0, 1, 2, 3, 4 or 6 runs of a ball. What is the number of distinct sequences in which exactly 30 runs can be scored in an over of six balls? Assume that all the runs are scored by batsmen only and there are no extra balls/runs.
18. Six people are in a queue at the ticket counter of The Comedy Store. All of them want to buy a ticket which costs Rs. 50. Each of the six people has exactly one note with him/her. Three of them have a 50 rupee-note and the other three have a 100 rupee-note. There are no notes of any kind at the ticket counter initially. In how many different ways can these six people can be arranged in the queue such that there is no problem of extending change at the ticket counter?
a. 144
b. 252
c. 216
d. 180
19. Find the sum of all the digits in the integers from 1 to 100000.
a. 5000050000
b. 2250001
c. 225001
d. None of these
20. All three-digit numbers, in which the ten’s digit is a natural number and is a perfect square, are formed using the digits 1 to 9. What is the sum of all such numbers?
Solutions
1. For a number to be divisible by 125, the last three digits have to be divisible by 125. So, the last 3 digits have to be either 375 or 875. Total number of 5 digit numbers will be 2! + 2! = 4.
2. You can either take 2 points from the first line and 1 from the second line or 2 points from the second line and 1 from the first line:
10c2 * 11c1 + 11c2 * 10c1
1045 triangles in total.
3. Three different numbers can be chosen in 10c3 = 120 ways. A triplet can be arranged in 6 different ways out of which, exactly one combination would be correct. So, we have 120 possible solutions in total.
4. Classic trap in this question. Numbers divisible by 3 will be 33, numbers divisible by 5 will be 20. Overlap will be those divisible by 15 and so 6 numbers. Total of 33 + 20 – 6 = 47 numbers will be divisible by either 3 or by 5. However, the question has another criterion that needs to be satisfied. The numbers should not be divisible by both 3 and 5 and so, we will have to eliminate those cases that are divisible by 15. So, we get 47 – 6 = 41 as our final answer.
5. The best way to think about this type is by taking successive differences of the numerators.
1—-13—-30—-53—-83—-121
—12—-17—-23—-30—-38—
——-5——6——7——8—–
———–1——1—–1——-
Not sure about the alignment, but I guess you get the point.
As the third level of difference comes to be 1, you have to do the traditional ‘divide by the denominator’ step thrice.
S = 1/7 + 13/7^2 + 30/7^3 + 53/7^4 + 83/7^5 + 121/7^6 …
7S = 1 + 13/7 + 30/7^2 + 53/7^3 + 83/7^4 …
Subtracting the first expression from the second expression
6S = 1 + 12/7 + 17/7^2 + 23/7^3 + 30/7^4 …
Again doing the same thing we did above
42S = 7 + 12 + 17/7 + 23/7^2 + 30/7^3 …
Subtracting again,
36S = 18 + 5/7 + 6/7^2 + 7/7^3 …
Again multiplying by 7
252S = 126 + 5 + 6/7 + 7/7^2 …
Subtracting again,
216S = 113 + 1/7 + 1/7^2 + 1/7^3 …
216S = 113 + (1/7)/(6/7)
S = 679/1296
6. If the first and the third digits are odd, the number can be formed in 3 * 2 * 2 * 2 * 1 = 24 ways.
If the first and fifth digits are odd, the number can be formed in 3 * 2 * 2 * 2 * 1 = 24 ways.
If the third and fifth digits are odd, the number can be formed in 3 * 2 * 2 * 2 * 1 = 24 ways.
Now, in the first case, the last digit could be either 2 or 4 and so, the total will be 2 * 12 + 4 * 12 = 72.
In the second and third cases, the last digit could be either 1, 3 or 5 and so, the total will be 16 * ( 1 + 3 + 5) = 144.
Overall, it will be 216.
7. The probability is 1 – P(when no one hits the target)
1 – (1/4 * 1/3) = 11/12.
8. 34300 = 2^2 * 5^2 * 7^3
The keyword here is integers and so, the integers can be positive or negative (anyway we cannot have 8 distinct positive integers whose product will be 34300).
The way in which this can be done is
(-1)*(1)*(-2)*(2)*(-5)*(5)*(-7)*(7)… and you will be left with another 7.
This 7 can go with any of the numbers except with -1 and 1 as then it would be another copy of -7 or 7.
So, a total of 6 ways.
9. 10c4 – 4c4 – 4c3*6c1 = 185 quadrilaterals.
10. 9000 = 2^3 * 3^2 * 5^3
The power of 2 in each factor has to be one. The rest can vary from 0-2 and 0-3 respectively. So, total of 1 * 3 * 4 = 12 factors.
Sum of these factors will be (2^1)(3^0+3^1+3^2)(5^0+5^1+5^2+5^3)=2*13*156=4056.
11. Let there be x green balls.
The chances of A winning are: x/(100+x) or 100/(100+x) * 100/(100+x) * x/(100+x) and so on…
The chances of B winning are: 100/(100+x) * x/(100+x) or 100/(100+x) * 100/(100+x) * 100/(100+x) * x/(100+x) and so on…
So, using the concept of infinite GP, we get:
{x/(100+x)}/(1-(100/(100+x)^2)) = 3*(100x/(100+x)^2)/(1-(100/(100+x)^2))
So, x/(100+x) = 300x/(100+x)^2
100x + x^2 = 300x
Or, x = 200.
12. India wins 5 matches: (2/5)^5 = 32/3125
India wins 4 matches: (2/5)^4*(3/5)*5c4 = 240/3125
India wins 3 matches: (2/5)^3*(3/5)^2*5c3 = 720/3125
Total probability of: 992/3125.
13. If the middle digit is anything from 2 to 8, we will get something like _ 2 _ as the number. Now, the first digit can be anything except 0, 1 and 3 so, it can be chosen in 7 ways. The units digit can be anything except 1 and 3 and so, can be chosen in 8 ways. Total of 56 ways. Because there are 7 cases, total of 392 ways.
If the middle digit is 1, the number will be _ 1 _ and so, can be done in 8 * 8 = 64 ways.
If the middle digit 0 or 9, the number is _ 0 _ and so, can be done in 8 * 9 * 2 = 144 ways
Total of 600 ways.
14. The books can be split as either
4-0-0: There is only 1 way in which this can be done
3-1-0: 3 books out of 4 can be chosen in 4c3 ways. The last book can go into either of the remaining bags so it won’t matter. Total of 4 ways.
2-1-1: 2 books out of 4 can be chosen in 4c2 ways. The last two books can go to either bags individually and so it won’t matter. Total of 6 ways.
2-2-0: 2 books out of 4 can be chosen in 4c2 ways. However, as soon as we choose 2 books, we are automatically forming another set of 2 books. So total cases will be 4c2/2! = 3 ways.
If you are unable to understand the reason why this was divided by 2, you can take 4 variables and try it out on your own writing down cases.
Total of 14 ways.
15. The arrangement would be WBWWBWW… So, 25 black marbles in total.
16. Again this is a typical series question:
1/(1*4) + 1/(4*7) + 1/(7*10) + 1/(10*13)…
1/3 * {(4 – 1)/(1*4) + (7 – 4)/(4*7) + (10 – 7)/(7*10) + (13 – 10)/(10*13)…}
1/3 * {1 – ¼ + ¼ – 1/7 + 1/7 – 1/10}
For n terms, the general form will be:
1/3 * (1 – 1/n)
(n – 1)/3n
17. 30 can be formed by these cases:
666660: Total of 6 ways.
666642: Total of 6!/4! = 30 ways.
666633: Total of 6!/4!2! = 15 ways.
666444: Total of 6!/3!3! = 20 ways.
Total of 71 ways.
18. The cases will be:
50-50-50-100-100-100
50-50-100-50-100-100
50-50-100-100-50-100
50-100-50-50-100-100
50-100-50-100-50-100
As the 6 people have to be arranged among themselves, it will be 3! * 3! * 5 = 180 ways.
19. There will be 10^5 + 1 numbers in total.
In the first part, there will be an equal number of 0s, 1s, 2s… 9s in each place. This number is 10000. So, there will be (0 + 1 + 2 + … 9) * 10000 in total for the first place. Similarly we get the same number for the other places as well.
Total of 45 * 5 * 10000 = 2250000
Adding the last 1 to it, we get 2250001.
20. Tens digit could be 1 or 4 or 9.
Total numbers will be 9 * 3 * 9 = 243.
1, 4 and 9 will be equally split among these.
So, 27*(1 + 2 + 3 … 9)*100 + 81*(1 + 4 + 9)*10 + 27*(1 + 2 + 3 … 9)*1
134055.
You can go through the entire series by clicking on this link: CAT 2016 Sprint Preparation Series by Learningroots.
