Quant Boosters  Gaurav Sharma  Set 1

Method 1
We have 17! = 2^15 x 3^6 x 5^3 x 7^2 x 11 x 13 x 17
17! = (2^3 x 5^3)(2^12 x 3^6 x 7^2 X 11 x 13 x 17)
Since (2^3 x 5^3) ends in zero the right most nonzero digit in 17! Will be the unit digit of (2^12 x 3^6 x 7^2 X 11 x 13 x 17) i.e, 6Method 2
If number is of the form 5a + b then it’s rightmost nonzero digit will be 2^a x a! x b!
Here 17 = (5 x 3) + 2
A = 3 and b = 2
So, rightmost nonzero digit is 2^3 x 3! X 2! = 6

Q5) How many natural numbers less than 30 have a composite number of factors?
a) 13
b) 11
c) 14
d) 10

Number of natural numbers less than 30 = 29
Number with only 1 factor = 1
Numbers with 2 factors = all primes below 30 = 10
Numbers with 3 factors = Numbers of the form a^2 where a is prime, so 2^2 = 4, 3^2 = 9, 5^2 = 25 = 3 numbers
Numbers with 5 factors = numbers of the form a^4 where a is prime. Only 2^4 is in this range
So, numbers less than 30 have a composite number of factors are = 29 – 1 – 10 – 3 – 1 = 14

Q6) Aman and eight of his friends took a test of 100 marks. Each of them got a different integer score and the average of their scores was 86. The score of Aman was 91 and it was more than that of exactly three of his friends. What could have been the maximum possible absolute difference between the scores of two of his friends?
a) 83
b) 73
c) 86
d) 44

Aman’s score is 91. In order to maximize the absolute different thake the score of the friend who scored highest as 100 marks. Now, 5 of his friends scored marks greater than 91, one of those is the one who scored 100 marks. Let the marks of other 4 friends be 99, 98, 97 and 96 (Maximum values)
And 3 of his friends scored less than 91 marks
Assume two of them scored 90 and 89 marks (maximum value)
Their average marks are 86
Now the lowest score will be 86 x 9 – ( 100 + 99 + 98 + 97 + 96 + 91 + 80 + 89) = 14
Hence absolute difference = 100 – 14 = 86

Q7) 4x + 2y + z = 28 where x, y and z are real numbers. Find the maximum value of x^1/2 * y^1/4 * z ^1/8
a) 5^3/4
b) 4^7/8
c) 5^3/4
d) 4^8/7

Using AM ≥ GM
(x + x + x +x + y + y + z) / 7 ≥ (x^4 * y^2 * z)^1/7
(4x + 2y + z) / 7 ≥ (x^4 * y^2 * z)^1/7
28/7 ≥ (x^4 * y^2 * z)^1/7
4^7 ≥ (x^4 * y^2 * z)
4^(7/8 ) ≥ (x^4 * y^2 * z)^1/8
Maximum value of (x^4 * y^2 * z)^1/8 is 4^(7/8 )

Q8) F(a,b) = HCF(a,b)/LCM(a,b)
F(1/F(a,b), C) = 1/12
If a, b and c are distinct positive integers such that any tow of them are coprime, then what is the sum of a, b and c ?
a) 6
b) 7
c) 8
d) 9

a, b and c are distinct positive integers such that any two of them are coprime
HCF(a,b) = 1 and LCM(a,b) = ab
F(a,b) = 1/ab
Now F(1/F(a,b),c) = F(ab,c) = 1/abc = 1/12
abc = 12 = 1 x 3 x 4
Hence sum = 1 + 3 + 4 = 8

Q9) The sum of the roots of the quadratic equation ax^2 + bx + c = 0 is equal to the sum of squares of their reciprocals. If a, b and c are real numbers and a # 0 then bc^2, ca^2 and ab^2 are in
a) G.P
b) A.P
c) H.P
d) None of these

Let the roots of the equation be p and q
Then, p + q = b/a and pq = c/a
Given p + q = 1/p^2 + 1/q^2 = (p^2 + q^2)/p^2q^2
P + q = [(p + q)^2 – 2pq]/(pq)^2
b/a = [ (b/a)^2 – 2c/a] / (c/a)^2
2ca^2 = ab^2 + bc^2
bc^2, ca^2 and ab^2 are in A.P

Q10) Two horses are tethered at the mid points of two adjacent sides of a square field. Each of them is tied with a rope that does not allow it to go beyond the centre of the field for grazing. If the length of a side of the field is 8m what is the ratio of areas grazed by both of them together to that of the area grazed by them individually?

Area grazed by the horses together = Area of leaf AEGF
= 2 [ Area of quadrant AEGH – Area of triangle AGH]
= 2 [ Pi/4 (4)^2 – 1/2 * (4)^2] = 8 ( Pi – 2)Area grazed individually = Area (Semicircle AGB ) + Area (Semicircle AGD) – 2 [Area grazed together]
= 1/2 * (4)^2 + 1/2 * (4)^2 – 2( 8(Pi – 2))
= 16Pi – 16Pi + 32 = 32
So required ratio = 8(Pi – 2) : 32 = (pi – 2) : 4

Q11) Each of the four girls A, B, C and D had a few chocolates with her. A first gave 1/2 of the chocolates with her to B, B gave 1/3rd of what she then had to C and C gave 1/4th of what she then had to D. Finally, all the four girls had an equal number of chocolates. If A had 90 chocolates more than B initially, find the difference between number of chocolates that C and D initially had.
a) 10
b) 30
c) 15
d) Cannot be determined

Let number of chocolates with A, B, C and D be a, b, c and d respectively. The transaction tables are as follows
We have, a/2 = 2/3 (b + a/2) = > a = 4b
And we are given a  b = 90 = > a = 120 and b = 30
So everyone had 120 – 120/2 = 60 chocolates at the end.
Putting values of a and b in 3/4 * (c + 1/3(b + a/2)) = 60
We have c = 50
Also d + 1/4 * (c + 1/3(b+a/2)) = 60
We get d = 40
Hence difference between values of c and d is 10

Q12) How many natural numbers when expressed in base 5, base 4 and base 3 form three – digit, four digit and five digit numbers respectively?
a) 44
b) 67
c) 38
d) 72

Three digit numbers in base 5 are from 5^2 to 5^3 – 1
From 25 – 124
Four digit numbers in base 4 are from 4^3 to 4^4 – 1
From 64 to 225
Five digit numbers in base 3 are from 3^4 to 3^5 – 1
From 81 – 242So, numbers common to all the intervals are from 81 – 124
Required answer is 124 – 81 + 1 = 44

Q13) If fresh grapes contain 90% water and 10% pulp by weight and 10 kg of fresh grapes yield 2.5 kg of dry grapes, then find the percentage of pulp by weight in dry grapes.
a) 20%
b) 40%
c) 75%
d) 80%

If the weight of fresh grapes is 10 kg then, it contains 9 kgs of water and 1 kg of pulp (by weight).
If 10 kg of fresh grapes yield 2.5 kg of dry grapes, the quantity of pulp would still remain the same.
So in 2.5 kg of dried grapes there is 1 kg of pulp
Percentage of pulp is 100/2.5 = 40%

Q14) A trader has one weighing stone each of weights 1, 2, 4, 8, 16 … and 1024 kg. If the trader gave another trader some of the stones, weighing a total of 2007 kg, what is the weight of the heaviest weighing stone remaining with him?
a) 32 kg
b) 64 kg
c) 16 kg
d) Cannot be determined