# Quant Marathon by Gaurav Sharma - Set 1

• How many natural numbers between 1 and 700 are not divisible by 7 and leave a remainder of 3 when divided by 4 ?

1. 125
2. 150
3. 100
4. 126

Numbers of the form 4k + 3 between 1 to 700 are:

3,7,11,15,19,23,27 … 699 ( 175 terms)

Number of the form 7p are 7, 35, … 679 ( 25 terms)

Hence the total number of terms of the form 4k + 3 and not divisible by 7 are 175 – 25 = 150

How many points with the integer co-ordinates lie inside the triangle whose vertices are (0,3), (-5,-2), (5,-2)?

1. 14
2. 15
3. 16
4. 17

We can draw the triangle on coordinates axes as shown

Now from the graph it is clear that the number of points with integer coordinates inside the triangle on line:

Y = 2 is 1

Y = 1 is 3

Y = 0 is 5

Y = -1 is 7

Hence the number of points is 16

If a, b and c are non-zero real numbers such that

a^2/b*c + b^2/a*c + c^2/b*a = 3, which of the following is necessarily true?

1. a+b+c = 0
2. a^2 + b^2 + c^2 = ab + bc + ca
3. a = b = c = 1
4. Either (a) or (b)

Given a^2/b*c + b^2/a*c + c^2/b*a = 3

a^3/abc + b^3/abc + c^3/abc = 3

a^3 + b^3 + c^3 – 3abc = (a + b + c) (a^2 + b^2 + c^2 – ab – ac – bc)

Hence either a + b + c = 0 or a^2 + b^2 + c^2 = ab + bc + ca

Hence option (d) is correct

Find the right most non-zero digit in 17!

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 non-zero 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, 6

Method 2

If number is of the form 5a + b then it’s rightmost non-zero digit will be 2^a x a! x b!

Here 17 = (5 x 3) + 2

A = 3 and b = 2

So, rightmost non-zero digit is 2^3 x 3! X 2! = 6

How many natural numbers less than 30 have a composite number of factors?

1. 13
2. 11
3. 14
4. 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

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?

1. 83
2. 73
3. 86
4. 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

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

1. 5^3/4
2. 4^7/8
3. 5^3/4
4. 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 )

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, them what is the sum of a, b and c ?

1. 6
2. 7
3. 8
4. 9

a, b and c are distinct positive integers such that any two of them are co-prime

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

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

1. G.P
2. A.P
3. H.P
4. 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

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 – ½ (4)^2] = 8 ( Pi – 2)

Area grazed individually = Area (Semicircle AGB ) + Area (Semicircle AGD) – 2 [ Area grazed together]

= ½(4)^2 + ½(4)^2 – 2( 8(Pi – 2))

= 16Pi – 16Pi + 32 = 32

So required ratio = 8(Pi – 2) :32 = (pi – 2) : 4

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.