# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 8

• 00000 to 99999 and 100000 is what is required. As sum of digits has been asked, we do not need place values. So, 100000/10 will be the occurrence of each digit at each place. So, 45 * 10000 * 5 = 2250000. Also, we have to add 1 that comes because of 100000. So, total of 2250001.

• Q26) P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible?
a. 8
b. 9
c. 10
d. 11

• P = 30k + x
x cannot be a multiple of 2, 3 or 5. Also, x has to be less than 30. So, the possible remainders are 1, 7, 11, 13, 17, 19, 23, and 29. Total of 8 values.

• Q27) Raman took classes for 20 consecutive days. On nth day, there were (n + 1) students present in his class, where n is a natural number. Each day Raman distributed Rs.1100 equally among the students present in the class. Student ‘x’, who attended all the 20 classes of Raman, donated 1/(m + 1) th part of the amount received by him (from Raman on that particular day) to a charity at the end of each day, where ‘m’ is the number of students attending the class on that day. What is the total amount donated by student ‘x’ to the charity?
a. Rs. 300
b. Rs. 1000
c. Rs. 666
d. Rs. 500

• On the first day, the student got Rs. 1100/2 and gave away 1/3 of what he got. So, amount donated = 1100/6
On the second day, the student got Rs. 1100/3 and gave away ¼ of what he got. So, amount donated = 1100/12
On the third day, he would have donated 1100/20, on the fourth day, 1100/30 and so on.
So, total amount donated will be
1100 (1/6 + 1/12 + 1/20 + … + 1/462)
1100 (1/2 – 1/3 + 1/3 – ¼ + ¼ - 1/5 + … + 1/21 – 1/22)
1100 * 10/22
Rs. 500.

• Q28) Let M be a three-digit number denoted by ‘ABC’ where A, B and C are numerals from 0 to 9. Let N be a number formed by reversing the digits of M. It is known that M – N + (396 × C) is equal to 990. How many possible values of M are there which are greater than 300?
a. 10
b. 18
c. 30
d. 20

• 99A – 99C + 396C = 990
99A + 297C = 990
A + 3C = 10
C = 1, 2, 3 and A = 7, 4, 1. For each value of A and C, B can take 10 values. Also, A has to be greater than or equal to 3. So, 4B2 and 7B1 will be the numbers. B can take any value from 0 to 9. So, 20 values in total.

• Q29) The product of three positive integers is 6 times their sum. One of these integers is the sum of the other two integers. If the product of these three numbers is denoted by P, then find the sum of all distinct possible values of P.
a. 432
b. 252
c. 144
d. 336

• Let a = b + c
(b + c)bc = 6(2b + 2c)
bc = 12.
The values of a, b and c could be 13, 12, 1 or 8, 6, 2 or 7, 4, 3. So, the distinct products will be 156 + 96 + 84 = 336.

• Q30) P1, P2 and P3 are three consecutive prime numbers and P1 × P2 × P3 = 190747. What is the value of P1 + P2 + P3?

• Can start by understanding that the numbers will be close to the cube root of 190740 which should lie between 50 and 60 (between 125000 and 216000). Can start with the smallest prime numbers and check. It will be satisfied for 53 * 59 * 61. Total of 53 + 59 + 61 = 173.

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