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Topic - Quant Mixed Bag

Source - Genius Tutorial Preparation Forum ]]>

Solved ? - Not yet

Topic - Quant Mixed Bag

Source - Genius Tutorial Preparation Forum ]]>

a) 100/89

b) 100/81

c) 100/99

d) None of these ]]>

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 ]]>

a) 3/2

b) 5/2

c) 7/5

d) None of these

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

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

8r = 12

r = 3/2

a) 800

b) 757

c) 758

d) 750 ]]>

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) 4

b) 30/7

c) 15/7

d) 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

a) 48

b) 84

c) 96

d) 134 ]]>

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 ]]>

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 ]]>

a) x/2

b) 2x/3

c) x/3

d) 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

a) 200

b) 199

c) 195

d) 197 ]]>

= 7^12 x 400

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

Number of factors:

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

a) 0

b) 1

c) 2

d) 1 or 2 ]]>

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 ]]>

a) 2^87 x 3^59 x 5^13 x 7^14

b) 2^88 x 3^58 x 5^12 x 7^15

c) 2^81 x 3^59 x 5^12 x 7^17

d) 2^79 x 3^61 x 5^21 x 7^12 ]]>

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

i.e 63 – 48 = 15

From the given options only b satisfies the condition ]]>

similarly b has to be multiple of 3.

so we can say A + B + C = 100

non negative integer solutions = 102C2 ]]>

So ab = 3776

A + B = 123

(A, B) = (64, 59) ]]>

a) 0

b) 1

c) 2

d) 3

e) More than 3 ]]>