Quant Boosters  Kamal Lohia  Set 1

kamal_lohia last edited by zabeer
Faculty and Content Developer at Tathagat  Delhi College of Engineering
Number of Questions  30
Topic  Quant Mixed Bag
Solved ? : Yes
Source :

Q1) A drop of 20% in the price of rocher ferrero enabled sweta to buy 3 pieces more for Rs 270. what’s the original price (in Rs) of 1 piece of rocher ferrero?

By dropping of 20% prices, the earlier quantity will cost 20% cheaper i.e. she now has 20% of 270 = 54₹ spare money with which she is purchasing 3 more articles. So per piece price is = 54/3 = 18.
And earlier price was = (5/4) * 18 = 22.5

Q2) How many different integer value of [ x^2 / 2014] are possible where x can be any value from 1 to 2014?

From x = 1 to x = 1007, we will get all positive integers from [1^2/2014] to [1007^2/2014] i.e. 0 to 503 which are 504 in all.
From x= 1008 to x = 2014, all terms will yield a different positive integer as difference between any two consecutive terms is greater than 1. So this will lead to another 1007 different integers in addition to 504 earlier... which makes the answer to be 1007 + 504 = 1511.

Q3) N is product of first 50 prime numbers.
A is a factor of N and B is a factor of A.
How many ordered pairs (A, B ) exist?

For every prime number 'p', you have 3 ways to be included (√ ) or not included (×) in (A, B) as follows:
( √ , √ ), (√ , ×) & (×, ×).
Hence 3^50.

Q4) Let M = 3 * 3! + 4 * 4! +.. + 15 * 15!
What is the remainder when M  15 is divided by 14!  2 ?

N(N!) = [N+11] * N! = (N+1)!  N!
3 x 3! = ( 4 – 1) 3! = 4 x 3! – 3! = 4! – 3!
4 x 4! = 5! – 4! and so on
Every terms will get cancelled except 16! – 3!
M – 15 = 16! – 21 = 240 ( 14! 2) + 459
Dividing by (14! – 2) will yield a remainder of 459

Q5) In a world where we operate in base 9 system, find the remainder when 12345 is divided by 17

First convert the numbers in base10 and then divide
You are to find remainder when 12345 (base 9) = 1×9^4 + 2×9^3 + 3×9^2 + 4×9 + 5 is divided by 17 (base 9) = 16.
Now see that 9^2 = 1 mod 16, so also 9^4 = 1 mod 16.
So clearly final remainder (in base 10) will be (1+3+5) + (2+4) ×9 mod 16 = 15 mod 16 = 16 (base 9).

Q6) Minimum value of (1 + x^2)/(1 + x), for x > 0

kamal_lohia last edited by kamal_lohia
Faculty and Content Developer at Tathagat  Delhi College of Engineering
(1 + x²)/(1 + x) = {(x²  1) + 2}/(x + 1)
= x  1 + 2/(x + 1)
= (x + 1) + 2/(x + 1)  2.Now product of first two terms is constant (i.e. 2), so their sum will be least when both are equal to each other (i.e. √2 each). And third term is a constant one, so won't contribute for the minimization of the expression.
Hence required minimum value is 2√2  2

Q7) There are N natural numbers (1, 2.....N). Now one number is removed and average becomes 602/17. Find out which number is removed.

602/17 = 35(7/17).
So N should be near 70. Also N should be of the form 17k + 1. That means N = 69 and if all numbers would have been used, then average should have been (1 + 69)/2 = 35. Now after removing average increased by (7/17). So extra quantity distributed is = 68 * (7/17) = 28. Thus the MISSED number is 35  28 = 7.

Q8) Cindy wishes to arrange her dresses into X piles, each consisting of the same number of Dresses, Y. Each pile will have more than one Dress and no pile will have all the dresses. If there are 13 possible values for Y given all of the restriction, what is the smallest number of dresses she could have?

Number should have exactly 13 ways in which it can be written as product of two integers i.e. X * Y. Also none of X and Y can be 1. So number should have exactly 15 factors. And smallest number is 144.

Q9) f(x) is a polynomial of degree 2001 with roots 1, 2, 3, 4 ... 2001. How many times does the polynomial F(x) = f(x) + 1 crosses the x axis in the interval (1005, 1010)?

There are assumptions in this question. If in the range [1005, 1010], the graph is starting from positive Y, then it'll end also in positive Y and f(x) + 1 will have 4 solutions; 2 each in (1006, 1007) and (1008, 1009).
BUT if in the range [1005, 1010], the graph is starting from negative Y, then it'll end also in negative Y and f(x) + 1 will have 6 solutions; 2 each in (1005, 1006), (1007, 1008) and (1009, 1010). Also, we are assuming here that crests and troughs formed are at least of one unit vertical height.

Q10) The unit's digit of a fivedigit number (having distinct digits) is 1 and is equal to the number of 5's in the number. The ten's digit is equal to the number of 6's in the number. How many such fivedigit numbers are possible?