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 + …
 100/89
 100/81
 100/99
 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
 3/2
 5/2
 7/5
 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?
 800
 757
 758
 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
 4
 30/7
 15/7
 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?
 48
 84
 96
 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?
 x/2
 2x/3
 x/3
 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
 200
 199
 195
 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 2The 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)]
 2^87 x 3^59 x 5^13 x 7^14
 2^88 x 3^58 x 5^12 x 7^15
 2^81 x 3^59 x 5^12 x 7^17
 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.