# Quant Marathon by Gaurav Sharma - Set 10

• Compute 1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^4 + 8/10^5 + 13/10^6 + …

1. 100/89
2. 100/81
3. 100/99
4. None of these

S = 1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^4 + 8/10^5 + 13/10^6 + …

S/10 = 1/10^1 + 1/10^2 + 2/10^3 + 3/10^4 + 5/10^5 + 8/10^6 + 13/10^7 + …

S – S/10 = 1/10^0 + 0 + 1/10^2 + 1/10^3 + 2/10^4 + 3/10^5 + 5/10^6

S – S/10 = 1/10^0 + 1/10^2 (1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^6)

S – S/10 = 1 + S/10^2

S (1 – 1/10 – 1/10^2) = 1

S = 100/89

In the given figure what is the radius of the inscribed circle

1. 3/2
2. 5/2
3. 7/5
4. None of these

Ar(ACD) + Ar(CDB ) = Ar(ABC)

5r/2 + 3r/2 = 12/2

8r = 12

r = 3/2

How many numbers from 1 – 2000 are such that at least two of their digits are same?

1. 800
2. 757
3. 758
4. 750

Single digit number with no digit repeated = 9

Two digit numbers with no digit repeated = 9C1 x 9C1 = 81

Three digit numbers with no digit repeated = 9C1 x 9C1 x 8C1 = 9 x 9 x 8 = 648

Four digit numbers less than 2000, with no digit repeated = 9C1 x 8C1 x 7C1 x 1 = 9 x 8 x 7 = 504

Total numbers with no digit repeated = 9 + 81 + 684 + 504 = 1242

Numbers with at least two digits same = 2000 – Numbers with no repetition

2000 – 1242 = 758

A triangle has sides 6, 7 and 8. The line through it’s incenter parallel to the shortest side is drawn to meet other two sides at P and Q. Then find the length of the line segment PQ

1. 4
2. 30/7
3. 15/7
4. Cannot be determined

Area of triangle = r x s

21r/2 = 6h/2 = 3h

r/h = 2/7

APQ and ABC are similar thus,

( h – r)/h = PQ/6

Or 1 – r/h = PQ/6 = > 1 – 2/7 = PQ/6

PQ = 30/7

N = 99^3 – 36^3 – 63^3. How many factors does N have?

1. 48
2. 84
3. 96
4. 134

N = 99^3 – 36^3 – 63^3

N = 99^3 + (– 36) ^3 + (– 63) ^3

N = 3 x 99 x 36 x 63 [if a + b + c = 0, then a^3 + b^3 + c^3 = 3abc]

N = 2^2 x 3^7 x 7 x 11

Number of factors of N = (2 + 1) (7 + 1) (1 + 1) (1 + 1) = 96

A person is travelling on the given grid and has to go from A to B through C – D. in each move he can take a step in either the north direction of the east direction. Following the given instructions, in how many different ways can he travel from A to B

Number of ways to travel from A to C = 6! / (3! X 3!) = 20

Number of ways to travel from C to D = 1

Number of ways to travel from D to B = 6!/(4! X 2!) = 15

Total ways = 20 x 15 = 300

AC is the diameter of a circle with center O. OE and OF are perpendicular to AD and AB respectively such that A – F – B and A – E – D. If perimeter of ABCD is x, then what will be the perimeter of AEOF?

1. x/2
2. 2x/3
3. x/3
4. root(2)x/3

Triangle AEO ~ Triangle ADC ( AA Similarity)

EO/DC = AE/AD = AO/AC = 1/2

Similarly, Triangle AFO ~ Triangle ABC (by AA Similarity)

FO/BC = AF/AB = AO/AC = 1/2

AD + DC + CB + AB = x

1/2 (AD + DC + CB + AB ) = x/2

AE + OE + OF + AF = x/2

How many positive divisors does 7^12 + 7^13 + 7^14 + 7^15 have

1. 200
2. 199
3. 195
4. 197

7^12 + 7^13 + 7^14 + 7^15 = 7^12 (1 + 7 + 49 + 343)

= 7^12 x 400

= 7^12 x 2^4 x 5^2

Number of factors:

(12 + 1) (4 + 1) (2 + 1) = 195

There are 10 numbers and none of them are divisible by 3. What will be the remainder when sum of their squares is divided by 3?

a) 0
b) 1
c) 2
d) 1 or 2

The Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1. In other words, a perfect square leaves remainder 0 or 1 on division by 3.

Here it is said that none of them is divisible by 3. Hence each square term will give a reminder of 1

So for 10 terms, 1 + 1 + 1 … + 1 (10 times) = 10

10 mod 3 = 1

Find the value of [root (1)] x [root (2)] x [root (3)] x … x [root (100)]

1. 2^87 x 3^59 x 5^13 x 7^14
2. 2^88 x 3^58 x 5^12 x 7^15
3. 2^81 x 3^59 x 5^12 x 7^17
4. 2^79 x 3^61 x 5^21 x 7^12

To simplify, find the number of 7’s in the product [root (1)] x [root (2)] x [root (3)] x … x [root (100)]

7s will be from [root (49)] to [root (63)]

i.e 63 – 48 = 15

From the given options only b satisfies the condition.

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