Explore topic-wise InterviewSolutions in .

Right answer is (B) finitely many or zero

The best explanation: Here the number of equations and the number of unknowns are equal and the system is HOMOGENEOUS, so it may have the zero SOLUTION or INFINITELY many solutions.

The correct ANSWER is (c) one or infinitely many

Best explanation: SINCE the system is homogeneous and there are more EQUATIONS than the number of unknowns, so the possibilities are either a UNIQUE solution or infinitely many SOLUTIONS. However, if the rank r of the system is 5, then it can be a unique solution as well as if r<5, then there are infinitely many solutions.

Right answer is (a) 24

To elaborate: SUPPOSE all of them plucked the different number of ORANGES. A girl can pluck at least 0 oranges and the number of oranges plucks by each student is distinct. So, total number of plucked oranges should be LESS than 100. But 0+1+2…..+19+20 = 210>200a contradiction.

Thus there exist TWO girls who plucked the same number of oranges. If thus there exist two girls who plucked the same number of oranges. It means each girl of remaining 18 students plucked different number of oranges. Number of oranges Plucked by 18 students = 0+1+2+3…+17 = 153 oranges. Number of oranges plucked by remaining 2 student = 200 – 153 = 47.Both students plucked same number of oranges. So, Number of oranges plucked by one of them = 47/2=24.

Right choice is (b) an = 3 * 7^n – 5*3^n

For EXPLANATION: The characteristic polynomial is x^2−6x+8. By solving the characteristic equation, x^2−6x+8=0 we GET x=2 and x=4, these are the characteristic roots. Therefore we know that the SOLUTION to the recurrence RELATION has the form an=a*2^n+b*4^n, for some constants a and b. Now, by using the initial conditions a0 and a1 we have: a=7/2 and b=-1/2. Therefore the solution to the recurrence relation is: an = 4 * 2^n – 1*3^n = 7/2 * 2^n – 1/2*3^n.

Correct choice is (d) 1437

The best explanation: When n=1, a1=17a0+30, Now a2=17a1+30*2. By SUBSTITUTION, we get a2=17(17a0+30)+60. Then REGROUPING the TERMS, we get a2=1437, where a0=3.

The correct choice is (b) 6N

The best explanation: CHECK for the left side of the equation with all the options into the RECURRENCE RELATION. Then, we GET that 6n is the required solution to the recurrence relation an=5an-1 + 6an-2.

The correct answer is (a) 7/2*2^n−1/2*6^n

The best explanation: Rewrite the recurrence relation bn-8bn-1+12bn-2=0. Now from the characteristic equation: x^2−8x+12=0 we have x: (x−2)(x−6)=0, so x=2 and x=6 are the characteristic roots. Therefore the SOLUTION to the recurrence relation will have the FORM: bn=b2^n+c6^n. To find b and c, set n=0 and n=1 to get a system of TWO equations with two UNKNOWNS: 3=b2^0+c6^0=b+c, and 4=b2^1+c6^1=2b+6c. Solving this system gives c=-1/2 and b=7/2. So the solution to the recurrence relation is, bn=7/2*2^n−1/2*6^n.

The CORRECT ANSWER is (c) 141

Best explanation: When n=1, a1=2a0+3, Now a2=2a1+3. By SUBSTITUTION, we get a2=2(2a0+3)+3.

Regrouping the terms, we get a4=141, where a0=6.

The correct OPTION is (b) an=6(6^n)+6/7n6^n

Explanation: The characteristic equation of the recurrence relation is → x^2−4x-12=0

So, (x-6)(x+2)=0. Only the characteristic root is 6. THEREFORE the solution to the recurrence relation will have the form: an=a.6^n+b.n.6^n. To find a and b, set n=0 and n=1 to get a system of TWO equations with two unknowns: 6=a6^0+b.0.6^0=a and 7=a6^1+b.1.6^1=2a+6b. Solving this system gives a=6 and b=6/7. So the solution to the recurrence relation is, an=6(6^n)−6/7n6^n.

Right answer is (b) an = 7/2*2^n−1/2*6^n

For EXPLANATION: The characteristic equation of the recurrence relation is → x^2−20x+36=0

So, (x-2)(x-18)=0. Hence, there are two REAL roots x1=2 and x2=18. Therefore the solution to the recurrence relation will have the FORM: an=a2^n+b18^n. To find a and b, SET n=0 and n=1 to get a system of two equations with two UNKNOWNS: 4=a2^0+b18^0=a+b and 3=a2^1+b6^1=2a+6b. Solving this system gives b=-1/2 and a=7/2. So the solution to the recurrence relation is,

an = 7/2*2^n−1/2*6^n.

The correct option is (a) 10399

The EXPLANATION is: an=5n+an-1

= 5n + 5(n-1) + … + an-2

= 5n + 5(n-1) + 5(n − 2) +…+ a1

= 5n + 5(n-1) + 5(n − 2) +…+ 4 [SINCE, a1=4]

= 5n + 5(n-1) + 5(n − 2) +…+ 5.1 – 1

= 5(n + (n − 1)+…+2 + 1) – 1

= 5 * n(n+1)/ 2 – 1

 an = 5 * n(n+1)/ 2 – 1

Now, n=64 so the answer is a64 = 10399.

Correct CHOICE is (d) 5.2

Explanation: By using the BINOMIAL Theorem, the terms are of the form ^6CN * (4x)^6-n * (B/x)^n.

For the term to be independent of x, we need x^6-n(1/x)^n = x^0⇒ x^6-n(x-1)^n = x^0⇒ x^6-nx^-n = x^0 ⇒ 6 – n = n ⇒ 2n = 6 and n = 3. Thus, we have a CONSTANT term of ^6C3 * 3^3 * b^3 = 8000

20 * 27 * b^3 = 80000

540 * b^3 = 80000

b^3 = 148.14 ⇒ b= 5.2.

Correct option is (b) 8

To elaborate: Consider the binomial expansion of (m+1)^71 and (m-1)^71 which gives these two

expressions below respectively: 1) m^51 + ^51C1m^50 + ^51C2m^49 + ^51C3m^48 + … + ^51C50m^1 + ^51C51m^0

2) m^51 – ^51C1m^50 + ^51C2m^49 – ^51C3m^48 + … + ^51C50m^1 – ^51C51m^0 .

By TAKING the difference we have, 2(^51C1m^50 + ^51C3m^48 – ^51C5m^46 + … + ^51C50m^2 – ^51C51m^0 ).

In this case, m = \(\sqrt{51}\) and 2(^51C1m^50 + ^51C3m^48 – ^51C5m^46 + … + ^51C50m^2 – ^51C51m^0 ).

Consider, module 10 on the POWERS(for any NATURAL number n): (51)^n ≡ (51 mod 10^n) ≡ 1 gives 2(^51C1 + ^51C3 + ^51C5 + … + ^51C50 + ^51C51). Now, by ADDING the odd terms of the 51^st row of the Pascal Triangle 2.(\(\frac{1}{2}\) * 2^51) = 2^51 = 2^(51 mod 4) = 2^3 = 8.

Correct ANSWER is (c) 1320

Best EXPLANATION: The coefficient of the 8^th TERM is ^11C8 = 165. HENCE, the 8^th term of the expansion is 165 * 2^3 * x^8 = 1320x^8, where the coefficient is 1320.

Right answer is (a) 792

Best explanation: NOTE that, the “x” in the BINOMIAL has to be CHOSEN 5 times out of 12. Thus, the COEFFICIENT of the term x^5y^7 must be equal to the number of combinations of 5 objects out of 12: ^12C5 = 792.

The correct option is (a) \(\frac{5}{16}\)

Explanation: By using the binomial distribution, to calculate how LIKELY to win Rs. 45 (or equivalently, the likelihood the coin comes up tails 3 TIMES). The POSSIBLE outcomes of this game are to win Rs. 45. Therefore, the required probability is \(\frac{^6C_3}{26} = \frac{5}{16}\).

The correct OPTION is (a) 15

To explain: Here making a 4-digit NUMBER is equivalent to filling 4 PLACES with 6 numbers. So, the number of ways of filling all the four places is ^6C4 = 15. Hence, the total possible 4-digit numbers from the above 6 numbers are 15.

Correct option is (c) 24310

For EXPLANATION I would SAY: The number of ways to choose 9 CUPCAKES out of a SET of 17 is ^17C9 = \(\frac{17!}{9!(17-9)!}\) = 24,310.

The correct CHOICE is (a) 45360x^4

Explanation: We know that in the expansion of (x+y)^n, if n is even then the MIDDLE term is (n/2 + 1)^th term. HENCE, the middle term in the expansion of (x/2 + 6y)^8 is(8/2+1)^th = 5^th term.

Now, assuming that x^5 occurs in the (r+1)^th term of the expansion (x/2+6y)^8, we obtain Tr+1 =^nCrx^n-ry^r = ^8C4(x/2)^4(6y)^4 = 45360x^4.

The correct choice is (d) 56

The explanation is: We can use the formula ^nCk = \(\FRAC{n!}{K!(n-k)!}\) to CALCULATE the value of ^8C5 = \(\frac{8!}{5!(8-5)!}\) = 56.

The correct answer is (B) 59136y^6

To explain: By assuming that x^7 occurs in the (r+1)^TH TERM of the expansion (x-2Y)^12, we obtain Tr+1 = ^nCra^n-rb^r = ^12C6 x^6 (2y)^6 = 59136y^6.

Correct CHOICE is (c) 10*(\(\FRAC{x}{3}\))

To elaborate: By using Binomial THEOREM,the expression \(\left(\frac{x}{3} + \frac{1}{x}\right)^5\) can be expanded as \(\left(\frac{x}{3} + \frac{1}{x}\right)^5 = ^5C_0(\frac{x}{3})^5 + ^5C_1(\frac{x}{3})^4(\frac{1}{x})^1 + ^5C_2(\frac{x}{3})^3(\frac{1}{x})^2\)

\(+ ^5C_3(\frac{x}{3})^2(\frac{1}{x})^3 + ^5C_4(\frac{x}{3})^1(\frac{1}{x})^4 \)

= \((\frac{x}{3})^5 + 5.(\frac{x}{3}) + 10.(\frac{x}{3}) + 10.(\frac{1}{3X}) + 5(\frac{1}{3x^3})\). Hence, the middle term is 10*(\(\frac{x}{3}\)).

Right ANSWER is (d) 129024

To EXPLAIN: It is known that (r+1)^TH TERM, in the binomial expansion of (a+b)^n is given by, Tr+1 = ^nCra^n-rb^r. Assuming that x^7 occurs in the (r+1)^th term of the expansion (x+4)^9, we obtain Tr+1 = 129024x^4.

Right OPTION is (c) 8(y^3 + y^1)

The BEST explanation: By using Binomial theorem,the expression (y+1)^4 – (y-1)^4 can be expanded as = (y+1)^4 = ^4C0y^4 + ^4C1y^3 + ^4C2y^2 + ^4C3y^1 + ^4C4y^0 and (y-1)^4 = ^4C0y^4 – ^4C1y^3 + ^4C2y^2 – ^4C3y^1 + ^4C4y^0. Now, (y+1)^4 – (y-1)^4 = (^4C0y^4 + ^4C1y^3 + ^4C2y^2 + ^4C3y^1 + ^4C4y^0)– (^4C0y^4 – ^4C1y^3 + ^4C2y^2 – ^4C3y^1 + ^4C4y^0) = 2(^4C1y^3 + ^4C3y^1) = 8(y^3 + y^1).

Right choice is (c) 0.1597

The best I can EXPLAIN: Probability of success p=0.6 and q=0.4. X=success in making a telephone CALL. HENCE, the probability of 11 successes in 20 attempts = P(X=11) = ^20C11(0.6)^11(0.4)^20 – 11 = 0.1597.

Right OPTION is (a) \(\frac{4}{9}\)

To explain: In this experiment, throwing a die ANYTHING other than a 4 is a success and ROLLING a 4 is failure. SINCE there are two trials, the required probability is

b(2; 2, \(\frac{5}{6}\)) = ^2C2 * (\(\frac{4}{6}\))^2 * (\(\frac{2}{6}\))^0 = \(\frac{4}{9}\).

Correct answer is (c) 0.2

For EXPLANATION I would say: A success is a bug-free COMPILATION, and a failure is the finding out of a bug. The PROGRAMMER has 0, 1, 2, or 3 failures and so her probability of FINISHING the program is :Pr(X=0) + Pr(X=1) + Pr(X=2) + Pr(X=3) = (0.95)^0(0.05) + (0.95)^0(0.05) + (0.95)^0(0.05) + (0.95)^0(0.05) = 0.2.

The correct answer is (c) 3.051 * 10^-5

Easiest explanation: ACCORDING to the NULL hypothesis it is a FAIR coin and so in that case the probability of FLIPPING at least 59% tails is = ^15C0(0.5)^15 = 3.051 * 10^-5.

Right choice is (b) 1555

Explanation: First we NEED to FIND the sum of all possible numbers and then DIVIDE it by the total such numbers possible to GAIN an average of all the numbers. So, we have (n-1)!(sum of digits)(1111…n times)/n!. Here n = 4. THEREFORE, (5-1)!(2+3+5+7+11)(1111)/5! = 1555.

Correct answer is (B) 44

The best I can explain: This is a case of derangement of 5 boys and 5 girls. The REQUIRED number of ways can be described as D = 5! (1 – \(\FRAC{1}{1!} + \frac{1}{2!} – \frac{1}{3!} + \frac{1}{4!} – \frac{1}{5!}) = 120(\frac{11}{30})\) = 44 ways.

Correct choice is (b) 1610496

Explanation: The NUMBER of WAYS in which the 5 correct envelopes can be selected = ^12C5 = 864

Derangement of the REMAINING 7 envelopes & letters = 1864 (derangement value for 7 is 1864)

Total No of ways of arrangement = 1864 * 864 = 1610496.

The correct CHOICE is (d) 24

To elaborate: Let the box covers be A, B, C, D and E. The possible ways for the covers to not be in the exact order of A, B, C, D, E are: 4! = 24 ways. (Since correct order i.e., A, B, C, D and E must be ELIMINATED from such arrangements).

The correct ANSWER is (a) NP-complete

For EXPLANATION: Computational COMPLEXITY of derangements isNP-complete in order to determine whether a given permutation GROUP consists of any derangements or not.

Correct choice is (c) 1854

For explanation: This can be SOLVED by the derangement formula:

!N = n!(1 – \(\frac{1}{1!} + \frac{1}{2!} – \frac{1}{3!} + … + \frac{(-1)^N1}{n!}\)) ⇒ 7! = 1854.

Correct choice is (a) 133496

The BEST explanation: By the derangements formula, the NUMBER of possible derangements should be 9!(\(\frac{1}{0!} – \frac{1}{1!} + … – \frac{1}{9!}\)) = 133496. Hence, there are a total of ways to GIVE the gifts to him such that no one distributes the real gifts.

Correct option is (d) 36

Explanation: The PLACE of 2, 4, 6 is SPECIFIED i.e. each of them will GET their place out of the last 3 places only. So 1, 3, 5 will automatically get one of the places in the first 3 places. This MUST ensure that 2, 4 and 6 occupies one of the last 3 places each and 1, 3 and 5 one of 1st 3 places each. Hence, 1, 3 and 5 can be arranged in 3! ways and 2, 4 and 6 also in 3! Ways. So, no of such derangements = 3! * 3! = 6 * 6 = 36.

Correct choice is (B) infinite

For explanation I WOULD say: Since m1 = 4, x2 = −7, x3 = 0 is a solution of the system, the system is not inconsistent. Thus the only POSSIBILITY is infinitely many SOLUTIONS.

Right option is (d) infinite

Best explanation: A homogeneous system is consistent. The rank is R = 3 and the number of variables is n = 4. Hence there is n – r = 1 FREE variable. THUS there are infinitely many solutions.

The CORRECT option is (c) infinitely MANY

Explanation: The possibilities for the solution SET for any homogeneous system is either a unique solution or infinitely many solutions. Since the homogeneous system has the zero solution and y1 = 5, y2 = −3, Y3 = 2 is another solution, it has at least two distinct solution. Thus the only possibility is infinitely many solutions.

Right answer is (c) zero

The best I can EXPLAIN: Since the rank of this homogeneous system(which is ALWAYS consistent) and the number of UNKNOWNS are equal, the only possible solution is zero and it is a UNIQUE solution.

The CORRECT OPTION is (B) INFINITELY many

Explanation: SINCE m