Solved ? - Yes

Topic - Quant Mixed Bag

Source - Genius Tutorial Preparation Forum ]]>

Solved ? - Yes

Topic - Quant Mixed Bag

Source - Genius Tutorial Preparation Forum ]]>

a) 125

b) 150

c) 100

d) 126 ]]>

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 ]]>

a) 14

b) 15

c) 16

d) 17 ]]>

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

]]>a) a + b + c = 0

b) a^2 + b^2 + c^2 = ab + bc + ca

c) a = b = c = 1

d) Either (a) or (b) ]]>

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 ]]>

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

a) 13

b) 11

c) 14

d) 10 ]]>

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 ]]>

a) 83

b) 73

c) 86

d) 44 ]]>

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

]]>a) 5^3/4

b) 4^7/8

c) 5^3/4

d) 4^8/7 ]]>

(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(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 ]]>

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 ]]>

a) G.P

b) A.P

c) H.P

d) None of these ]]>

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 ]]>

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

a) 10

b) 30

c) 15

d) Cannot be determined ]]>

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

a) 44

b) 67

c) 38

d) 72 ]]>

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 – 242

So, numbers common to all the intervals are from 81 – 124

Required answer is 124 – 81 + 1 = 44

a) 20%

b) 40%

c) 75%

d) 80% ]]>