Quant Boosters  Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta  Set 3

The numbers of students in Batch A and Batch B were in the ratio 2 : 3 in January and 5 : 8 in February. The numbers of students in Batch A and Batch B increased from February to March at rates that were twice and thrice respectively, of rates at which they increased from January to February. If the ratio of the total number of students in these two batches in February and January was 26 : 5, then find the ratio of number of students in Batch A and Batch B in March?
(1) 25 : 64
(2) 25 : 84
(3) 45 : 112
(4) 35 : 92Let the number of students in Batch A in January be 2x . Let number of students in Batch A in February be 5y.
Total numbers of students in Batch A and Batch B combined in January and February are 5x and 13y respectively. Therefore 13y : 5x = 26:5 or y : x = 2:1 or y = 2x.
From here, we can compile the following table:If 5x + 2y + z = 81, where x, y and z are distinct positive integers, then find the difference between the maximum and the minimum possible value of x + y + z.
(1) 44
(2) 48
(3) 55
(4) 60
(5) 64Minimum possible value of x + y + z will be when we maximize the value of ‘x’. Maximum possible value of x will be 15 and since x, y and z are distinct positive integers, y = 1 and z = 4. So,minimum possible value of x+y+z = 15+1+ 4 = 20.
Maximum possible value of x + y + z will be when the value of z is maximized. Maximum possible value of z will be when y = 2 and x = 1, i.e. z = 72.
Required difference is 75 – 20 = 55.How many twodigit numbers have exactly four factors?
(1) 30
(2) 31
(3) 32
(4) 28Since, the twodigit number has exactly four factors, therefore the number has to be a product of two prime numbers or a perfect cube.
Case I:
When one of the numbers in the product is 2.
The other number in the product can be 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
There are 13 such possible products.Case II:
When one of the numbers in the product is 3.
The other number in the product can be 5, 7, 11, 13, 17, 19, 23, 29 and 31.
There are 9 such possible products.Case III:
When one of the numbers in the product is 5.
The other number in the product can be 7, 11, 13, 17 and 19. There are 5 such possible products.Case IV:
When one of the numbers in the product is 7. The other number in the product can be 11 and 13. There are 2 such possible products.Case V:
When the twodigit number is a perfect cube. There is only one such number, i.e. 27.
Therefore, in all there are 30 twodigit numbers that have exactly 4 factors.There were twelve teams participating in a football tournament where each team played exactly one match with each of the other eleven teams. Each match was played between two teams. The winner of each match was awarded 3 points while no point was awarded to the team that lost the match. In case of a tie, both teams got 1 point each. If the aggregate number of points awarded to all the teams at the end of the tournament was 189, then how many matches ended in a tie?
(1) 7
(2) 8
(3) 9
(4) 6
(5) 4Total number of matches played in the football tournament = 11 + 10 ...+ 2 + 1 = 66.
Maximum possible number of points awarded to all the teams is 66 × 3 = 198. This is possible if no match ended in a tie. But the total points awarded to all the teams is 189.
So the number of matches that ended in a tie = 198 – 189 = 9. As a tied match generates only 2 points, which is one less than number of points generated by a win/loss match.There are two numbers A and B. A can be expressed as a product of 13 and a twodigit prime number and B can be expressed as a product of 17 and a twodigit prime number. If the unit’s digit of the product of A and B is 7, then how many distinct products of A and B are possible?
(1) 48
(2) 55
(3) 80
(4) 110
(5) 120Assume the numbers to be 13×N and 17×M, where N and M are twodigit prime numbers.
There are two possible ways in which 7 can be the unit’s digit of the product i.e. 1 × 7 or 3 × 9.
Case I: If units digit of N is 3, then units digit of M will be 9. Then, N = 13, 23, 43, 53, 73 or 83 and M = 19, 29, 59, 79, 89
So, number of distinct products = 6 × 5 = 30
Case II: If units digit of N is 1, then units digit of M will be 7 Then N = 11, 31, 41, 61 or 71 and M = 17, 37, 47, 67 or 97
So, number of distinct products = 5 × 5 = 25
Case III: If unit digits of N is 7, then units digit of M will be 1. Then N = 17, 37, 47, 67 or 97 and M = 11, 31, 41, 61 or 71
So number of distinct products = 5 × 5 = 25
Case IV: If unit digit of N is 9, then unit digit of M will be 3. Then N=19,29,59,79 or 89 and M =13,23,43,53,73 or 83
So number of distinct products = 5 × 6 = 30
Here case III and case IV will give the same products as case II and case I respectively.
∴ Total number of distinct products = 55.A and B play a dice game using two dice viz. ‘X’ and ‘Y’. Die ‘X’ has 1,2,3,4,5 and 7 printed on its six faces whereas die ‘Y’ has 2, 3, 4, 5, 6 and 8 printed on its six faces. There is only one number printed on every face of the two dice. In turns each of A and B rolls both the dice simultaneously and records the product of the two numbers appearing on the top of the two dice, as their respective scores. If the sum of the scores of players A and B is an even number in a round then how many distinct scores A could have in that round?
(1) 32
(2) 18
(3) 22
(4) 24Let X1 and Y1 be the sets having all the numbers printed on
die X and die Y respectively. X1 ={1,2,3,4,5,7}
Y1 = {2, 3, 4, 5, 6, 8}
For the sum of scores of players A and B in a particular round to be even, the individual scores of both A and B should either be odd or even.
Case I: A and B both have scores that are odd numbers. Each of A and B could have any of the following 7 scores: 3, 5, 9, 15, 21, 25 and 35.
Case II: A and B both have scores that are even numbers. Each of A and B could have any of the following 17 scores 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 30, 32, 40, 42 and 56.
Hence, each of A and B could have 17 + 7 = 24 distinct score in that round.A person receives payment P(n) on day ‘n’. If P(n) = 2 × P(n – 1) and the person receives Rs.12 on day two, then what is the total payment received by that person for 10 days starting from the day one?
This is nothing but a geometric progression whose common ratio is 2 and P(2) = 12.
∴P(1) = 6
⇒ P(1) + P(2) + ... + P(10) = 6 + 12 + ... till 10 terms
Ans  6138There are three natural numbers X, Y, and Z such that the LCM of (X, 120) is 1320, LCM of (Y, 120) is 1680 and LCM of (Z, 120) is 1800. Which of the following statements is true?
(1) X, Y, and Z all three can be perfect squares.
(2) Only Y and Z can be perfect squares.
(3) Only Y can be both a perfect square and a perfect cube.
(4) Z is definitely a perfect square.
(5) Z can be a perfect square.Factorizing the numbers:
120=23 ×31×51 1320=120×11= 2^3×3^1× 5^1×11
⇒ X definitely has to be a multiple of 11 but it cannot be a perfect square as X cannot contain (112)
1680 = 120 × 2 × 7 = 2^4 × 3^1 × 5^1 × 7^1
⇒ Y should be multiple of 2^4 × 7^1 but it cannot be a perfect square as Y cannot contain 72.
1800=120×3×5= 2^3×3^2×5^2.
Since the LCM of 120 and Z is 1800, therefore Z can be a perfect square. One of the possible values of Z can be 3^2 × 5^2. But it is not necessary that Z has to be perfect square as it can be 2×3^2 ×5^2 also.
Hence, option (5) is the correct choice.A trader used to make 5% profit on an item selling at usual marked price. One day, he trebled the marked price of the item and finally offered a discount of 30%. Find the percentage profit he made on the item that day.
(1) 120.5%
(2) 100%
(3) 99.5%
(4) 94.5%
(5) None of theseLet the earlier cost price of the item = Rs.100 ⇒ Earlier marked price = Rs.105.
On that day, 30% discount is offered on Rs. 3 × 105 = Rs.315 Thus, new selling price = Rs.220.50
New Profit percentage = 120.50.If ‘a’ is an odd number and ‘b’ is an even number, then what is the total number of solutions of the equation ab = 2a + b + 598?
a) 16
b) 12
c) 5
d) 6
e) 8The given equation is ab + 2 = 2a + b + 600.
⇒ ab – 2a – b + 2 = 600 ⇒ (a – 1)(b – 2) = 600
Now, it is given that ‘a’ is an odd number and ‘b’ is an even number which implies that both (a – 1) and (b – 2) are even numbers.
Therefore, the possible pairs of values of (a – 1) and (b – 2) that satisfy the given equation are (2, 300); (4, 150); (6, 100); (10, 60) (12, 50). (20, 30); (30, 20); (50, 12); (60, 10); (100, 6); (150, 4); and (300, 2).
Therefore, there are 12 solutions for the given equation. Hence, option (2) is the correct choice.The sum of the coefficients of the polynomial (x – 1)^7 (x – 2)^2 (x – 4) is
(a) 0
(b) 16
(c) –16
(d) None of thesePutting x = 1 in the expansion (1+x)^n = nC0 + nC1 x + nC2 x^2 + ... + nCx x^n, we get,
2^n = nC0 + nC1 x + nC2 + ... + nCn.
We kept x = 1, and got the desired result i.e. ∑nr = 0 Cr = 2^n