# Quant Boosters - Ravi Handa - Set 2

• First, let a = |x|, b = |y|, c = |z|.
Now, we need to find the number of positive integral solutions of a + b + c = 15. The number of solutions are 14C2 = 91. Now for each value of a,b and c we will have two values of x, y and z each. Therefore, the total number of solutions = 91 x 2 x 2 x 2= 728.

Now let one of the variables be equal to 0. For example, let x = 0 and |y| and |z| be at least equal to 1. Therefore, we need the positive integral solution of b + c = 15, where b = |y| and c = |z|. The number of solutions is 14C1 = 14. Each of these solutions will give two values of y and z and there are 3 ways in which we can keep one of the variables equal to 0. Therefore, total number of ways are 14 x 2 x 2 x 3 = 168.

Now let two of the variables be equal to 0. In this case, the total number of solutions is equal to 6.

Therefore, the total number of integral solutions = 728 + 168 + 6 = 902.

• Q9) Let g(x) = max (5 − x, x + 2). The smallest possible value of g(x) is
a) 4.0
b) 4.5
c) 1.5
d) None of these.

• To solve such kind of questions, in most cases, all you need to do is to equate the two values inside the function
=> 5 – x = x + 2
=> x = 3/2 = 1.5
To find out the answer to the question, we need to find out the value of g(1.5)
When we put x = 1.5, we get g(x) = max (5 – 1.5, 1.5 + 2) = max (3.5, 3.5) = 3.5
So, our answer is 3.5 Option D – None of these

I would strongly recommend that you watch the below video for better understanding of the solution of the question

• Q10) What is the sum of all the four-digit numbers with distinct non-zero digits such that the sum of their digits is 23 ?
a) 1111 * 23 * 54
b) 1111 * 23 * 48
c) 1111 * 23 * 42
d) 1111 * 23 * 36

• Let us consider that the 4 digit number is abcd
Sum of all its arrangements = 1111 * 6 * (a + b + c + d) = 1111 * 6 * 23
Now, if there are 'n' ways of selecting the values for abcd, your answer would be 1111 * 23 * 6 * n
So, let's try and figure that part out.
9851, 9842, 9761, 9752, 9743, 9653, 8762,8753,8654
are the only ones which are there. So, n = 9
So, our answer is 1111 * 23 * 54

• Q11) If the sum of the first 'n' terms of an Arithmetic Progression is 100 and the sum of the next 'n' terms of the Arithmetic Progression is 300, then what is the ratio of the first term and the common difference?

• Let's say 'n' is 5
Sum of first 5 terms is 100, so 3rd term is 20
Sum of next 5 terms is 300, so the middle of those 5 terms, which will be the 8th term overall is 60
=> 5d = 60 - 20
=> d = 8
=> a = 20 - 2d = 4
Required ratio is 1:2

• Q12) A is 80% more efficient than B who is 60 % more efficient than C. A takes 40 days less than B to complete a work. A starts the work and works for 25 days and then B takes over. B then works for next 30 days and then stops. In how much more time can C complete the remaining work?

• Let us say efficiency of C is 1
=> Efficiency of B = 1.6 = 8/5
=> Efficiency of A = 1.8 * 1.6 = 9/5 of B = 72/25 of A
A takes 40 days less than B because A takes 5/9th time of B
=> A takes 4/9 days less than B
=> B takes 90 days and A takes 50 days
I guess you can work out the rest from here.

• Q13) In how many ways can 10 boys be selected from 20 if Rajesh and Ramesh are not selected together?

• If only Rajesh is selected, 18C9
If only Ramesh, is selected, 18C9
If none of them is selected, 18C10
Total ways = 18C9 + 18C9 + 18C10

• Q14) A alone would take 8 hours more to complete a work than when A and B would work together. If B worked alone it would take him 9/2 hours more to complete the work than when A and B would work together. What time would they take to finish the work if they worked together?

• You can use this concept for such questions,
If A takes 'a' hours more than A & B combined and B takes 'b' hours more than A & B combined, then A & B together will take sqrt(ab) hours to do the job.
You will get the answer directly and quickly.
If you wish to understand how this formula came, assume that they together take 'x' hours.
So, A will take 'x + a' hours and B will take 'x + b' hours.
The equation will be 1/(x+a) + 1/(x+b) = 1/x
Solving this equation, you would get x = sqrt(ab)
Applying the same in this question, x = sqrt(8 * 9/2) = sqrt(36) = 6

• Q15) A man bought an article and sold it at a gain of 5%. If he had bought it for 5% less and sold it for Rs 1 less, he would have made a profit of 10%. Find CP of the article

• Let us say that the cost price is 100x, he sold it for 105x
If he had purchased it at 5% less or at 95x and sold it at 1 rs less or at 105x - 1, he would have gained 10%
=> 1.1 * 95x = 105x - 1
=> 105x - 104.5x = 1
=> 0.5x = 1
=> x = 2
So, the cost price of the article = 100x = 200 Rs.

• Q16) If a^3 + b^3 + c^3 = 36 then the maximum value of 3abc will be?

• When the sum of numbers is constant, the product is maximum when they are equal.
=> When a^3 + b^3 + c^3 = 36
=> Maximum value of a^3 * b^3 * c^3 = 12 * 12 * 12
=> Maximum value of abc = 12
=> Maximum value of 3abc = 36

• Q17) A person was appointed for a 50 days job on a condition that he will be paid Rs. 12 for every working day but he will be fined Rs. 6 for every day he remains absent. After the completion of the work, he got Rs. 510. For how many days, he worked?
(a) 15 days
(b) 45 days
(c) 30 days
(d) 20 days
(e) None of the above

• 12 * present - 6 * absent = 510
=> 2 * present - absent = 85
Also, present + absent = 50
Adding these two equations 3 * present = 85 + 50 = 135
=> present = 135/3 = 45

• Q18) There was 120 Litres of pure milk in a vessel. Some quantity of milk was taken out and replaced with 23 Litres of water in such a way that the resultant ratio of the quantity of milk to that of water in the mixture was 4:1. Again 23 Litres of the mixture was taken out & replaced with 28 Litres of water. What is the ratio of milk to water in the resultant mixture ?
a. 58 : 37
b. 116 : 69
c. 46 : 29
d. 101 : 37
e. 53 : 23

61

42

61

61

63

61

61