Topic - Quant Mixed Bag

Solved ? : Yes

Source : Collection of solved questions from Hemant Yadav, Math faculty - FIITJEE ]]>

Topic - Quant Mixed Bag

Solved ? : Yes

Source : Collection of solved questions from Hemant Yadav, Math faculty - FIITJEE ]]>

a) boxes and balls are all distinct and only 1 ball in one box

b) boxes are same and balls are distinct and only 1 ball per box

c) balls are same and boxes are different and only 1 ball per box

d) balls and boxes are same and only 1 ball per box

e) repeat all above questions with any number of balls in any box ]]>

(a) 6018

(b) 671676

(c) 1007514

(d) 1008016

(e) 2015028

[OA : 1008016]

]]>[OA : 0]

]]>That means if a number leaves 3, 5 or 6 as remainder with 7, it can never be a perfect square.

Here, all the 6 six numbers when divided by 7 will leave a remainder of 1, 2, 3, 4, 5, 6 in some order. So product of the 6 numbers = 7k + 720 = 7n - 1

But no perfect sq = -1(mod 7)

Hence, no solution

For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18

N = 25, S(N) = 2 + 5 = 7

How many natural numbers, N, exist such that S(N) + S(N^2) = 2011

[OA: 0]

]]>N can be of the form 3k, 3k + 1 or 3k + 2

for 3k the expression is clearly divisible by 3 but 2011 is 3k + 1.

For 3k + 1 square is also in the form 3k + 1. So sum is 3k + 2.

For N of the form 3k + 2 the square is 3k + 4 or 3k + 1.Hence sum is 3n

So no solutions for S(N) + S(N^2) = 2011 ]]>

[OA : 0]

]]>So x^2 – 1 = (x - 1)(x + 1) = P(1)P(2)P(3)…P(n) = 2k

But (x – 1) and (x + 1) are either both odd or both even. So RHS must be odd or multiple of 4. As RHS is neither odd nor multiple of 4, hence no positive integral solution. ]]>

[OA : 2258]

]]>There are total 10 distinct no’s, out of which 2 will not be used. Alice chooses the no’s and Bob places them

Now, if Alice chooses any no. that is not 0 or 9 on her first call, Bob will manage to keep the difference within 2000. (suppose Alice chooses 3, then bob replace one of the stars in the 4th place with 3 like this ---- - 3---. Now whenever Alice chooses 2 or 4, Bob will replace the corresponding place with it like this 2--- - 3--- or 4--- - 3---.

So she will choose neither, then she must choose any of 5 or 6 and bob will pick it because he knows Alice will not choose 2 or 4 as long as the star in the corresponding place is not replaced)

Now, if Alice chooses any of 0 and 9, say 0, and bob places it in the 4th place, she will not choose 1 and 2 and will be able to make the difference cross 2000. so bob will never put 0 in the 4th place. He will place it in x3 in order to make the difference between X and Y minimum. now if the next choice is neither 1 nor 9 and greater than or equal to 5, he will place it in x4 and otherwise in y4 and thus will be able to keep the difference down below 2000. so the next choice must be 1 or 9. Bob will place it in y3 to keep the difference between x3 and y3 minimum, proceeding like this, the largest difference that can be obtained is 6012-3754=2258

If the first chosen no. is 9, then the largest difference is 3987-6245=-2258

So their best play would give 2258 as the difference.

[OA : 32]

]]>Also, f(1) = 1, f(2) = 2

So, f(n) = 2^(n – 1)

Hence, f(6) = 2^5 = 32 ]]>

[OA : 2N - 4]

]]>(2N - 4) telephone calls, for N > 3

Divide the N secret agents into two groups. If N is odd, one group will contain one extra agent.

Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. Similarly in second group, agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. After (N - 2) calls, two agents in each the group will know anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and Z1 & Z2 from group 2.

Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4 agents will know everything.

Now (N - 4) telephone calls are reqiured for remaining (N - 4) secret agents.

Total telephone calls require are

= (N - 2) + 2 + (N - 4)

= 2N - 4

[OA : 4/9]

]]>After joining PQ we will get a trapezium. So to fulfil the required condition sum of the parallel sides of the trapezium should be less than 2a/3 or greater than 4a/3, where a is the side of the square.

For AP = 0 to a/3, probability of choosing Q will constantly decrease from 2/3 to 1/3

For AP = a/3 to 2a/3, it will be 1/3

For AP = 2a/3 to a, it will constantly increase from 1/3 to 2/3

Now, we can get the area as 4/9

Another way can be let AB = 1.

The requirement is the event (AP + DQ) > 4/3 or (AP + DQ) < 2/3. That is the mean of the two sides is either > 2/3 or < 1/3.

Let X be the random variable for the continuous uniform distribution of AP. The distribution of DQ is independent but also belongs to X.

Let Y = X + X (the sum of the two opposite sides).

P(Y > 4/3) is the area of the triangle from Y = 4/3 to Y = 2.

P(Y > 4/3) = (1/2) * (2/3) * (2/3)

= 2/9

Similarly P(Y < 2/3) is the area from Y = 0 to Y = 2/3 and

P(Y < 2/3) = 2/9

The events Y > 4/3 and Y < 2/3 are mutually exclusive so the probability of the union is the sum of the probabilities of both events.

P(Y > 4/3 or Y < 2/3) = 2/9 + 2/9 = 4/9

]]>i) n has exactly k digits (in decimal representation)

ii) all the digits of n are odd.

iii) n is divisible by 5

iv) the number m = n/5 has k odd digits.

[OA : 3^k - 1]

]]>As 'n' has 'k' digits, 2n will have either 'k' digits or 'k + 1' digits. Now 'm' has got to have 'k' odd digits, hence 2n must have 'k + 1' digits with the last digit obviously 0 (as 'n' is divisible by 5) and all the remaining digits odd. So, 'm' is also a 'k' digit number and combining that with the 4th point we can say that 'm' is a 'k' digit number with all of its digits odd.

Hence, 'n' must have its digits such that each digit of 'n' when multiplied by 2 gives a carry of 1, if this doesn't hold then we will get an even digit in 'm', which is not acceptable.

As each digit of 'n' when multiplied by 2, gives a carry forward of 1, each digit of 'n' should be greater than equal of 5 with the last digit always 5 (as 'n' is a multiple of 5)

For k=1, possible cases =1

For k=2, last digit can be selected in 1 way and 2nd last digit can be selected in 3 ways ( 5 or 7 or 9)

For k=3, last digit in 1 way, other 2 digits in 3*3 ways

Generalizing, 'n' can be formed in 3^(k - 1) ways

]]>[OA : 1]

]]>[OA : 27/32]

]]>P(0) = (1/2)² + 2(1/3)(1/2)² + 3(1/3)²(1/2)² + ....

= (1/2)²{1 + 2/3 + 3/9 + 4/27 + ....)

S = 1 + 2/3 + 3/9 + 4/27 + ...

S/3 = 1/3 + 2/9 + 3/27 + ....

=> 2S/3 = 1 + 1/3 + 1/9 + 1/27 + ....

=> S = 9/4

P(0) = (1/4)(9/4) = 9/16

P(1) = (2!/(1!*1!))(1/6)(1/2)² + (3!/1!*1!*1!))(1/3)(1/6)(1/2)² + (4!/(2!*1!*1!)(1/3)²(1/6)(1/2)² + ....

= (1/6)(1/2)²{2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... )

S = 2 + 6(1/3) + 12(1/3)² + 20(1/3)³ + 30(1/3)^4 + ... .....(1)

=> S/3 = 2/3 + 6(1/3)² + 12(1/3)³ + .......... (2)

(1) - (2) => 2S/3 = 2 + 4/3 + 6/3² + 8/3³ + .... ............(3)

=> 2S/9 = 2/3 + 4/3² + 6/3³ + ..... ...........(4)

(3) - (4) => 4S/9 = 2 + 2/3 + 2/3² + 2/3³ + .... = 3

P(1) = (1/6)(1/2)²(27/4)

= 9/32

=> Probability of A winning = 9/16 + 9/32 = 27/32

]]>It is clear that p(n) is a product of 8 consecutive numbers and divisible by atleast 8!. That means minimum power of 2 in p(n) is 7, i.e, p(n) is divisible by 2^7.

But last non-zero digit has to be an odd number, so p(n) should be divisible by 5^7.

So 15625 (5^6) or its multiple has to be one of the consective numbers, and we have to make sure that there is no multiple of 16 among the 8 numbers.

Multiple of 16 nearest to 15625 are 15616 and 15632.

So, least value of n = 15620.

p(n) = 15618 * 15619 * 15620 * 15621 * 15622 * 15623 * 15624 * 15625

After removing (2^7)*(5^7), we will get

7809 * 15619 * 781 * 15621 * 7811 * 15623 * 1953 * 1

Unit digit = unit digit of 9 * 9 * 1 * 1 * 1 * 3 * 3 * 1 = 9

]]>[OA : Only one solution]

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