Modern Maths(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

MODERN MATHS MCQs

Total Questions : 95 | Page 9 of 10 pages

Find the value of x such that: $^{7} C_{x - 1} +^{7} C_{x} =^{8} C_{x + 2}$

Discuss Question

Answer: Option B. -> 3
:
B

Soln: We have $^{n} C_{r - 1} + n C_{r} =^{n + 1} C_{r}$
$^{7} C_{x - 1} +^{7} C_{x} =^{8} C_{x}$
$^{8} C_{x} =^{8} C_{x + 2} =^{8} C_{(8 - x - 2)}$

Either x = x + 2 or x + 2 = (8 - x - 2)

But $x \neq x + 2$

2x + 2 = 8

x = 3

Hence option (b)

Alternatively, go from answer options.

Question 82.

At a board meeting, each person shook hands with everyone present. Totally there were 45 handshakes. Mid- way through the meeting, the 3 foreign delegates left as they had a flight to catch. The number of women is now 1 more than the number of men. When the meeting concluded, each person shook hands with only the person of the same gender. What were the total number of handshakes in the meeting?

Discuss Question

Answer: Option B. -> 54
:
B

Answer = b

Initially $^{n} C_{2}$ = 45 => n=10

3 people leave, implies that 7 people are left.

number of women = one more than number of men

no. of women = 4. men =3

Handshakes at the end = $^{4} C_{2} +^{3} C_{2} = 9$
Total handshakes = 45+9 =54.

Question 83.

Find the value of x for which the value of $^{16} C_{x}$ is maximum for any natural number x.

Discuss Question

Answer: Option D. -> 8
:
D

Soln: Maximum value of $^{16} C_{x}$ occurs when x = $\frac{16}{2}$ = 8.

Hence option (d)

Question 84.

If $^{n} C_{r} = 5$ and $^{n} P_{r} = 120$ , then determine the value of n, r.

5, 3
4, 2
7, 4
6, 4
5, 4

Discuss Question

Answer: Option E. -> 5, 4
:
E

Soln: $\frac{^{n} P_{r}}{^{n} C_{r}}$ = $\frac{120}{5}$ = 2

$^{n} C_{r} = \frac{(^{n} P_{r})}{(r!)}$

$\frac{^{n} P_{r}}{[\frac{^{n} P_{r}}{r!}]}$

r! = 24

r = 4

Given $^{n} P_{r}$ = 120

$^{n} P_{4}$ = 120

$\frac{n!}{(n - 4)!} = 120$

n(n-1)(n-2)(n-3) = 120

n = 5

Hence option (e)

Question 85.

Determine the no. of terms in the expansion of $(a + b + c + d)^{5}$ and $(1 + a + a^{2} + a^{3})^{5}$

56, 16
32, 24
165, 32
32, 32
5, 4

Discuss Question

Answer: Option A. -> 56, 16
:
A

No. of terms in expansion of $(a + b + c + d)^{5}$ = Number of solution of (a + b + c + d) = 5, $^{8} C_{3}$ = 56

No. of terms in expansion of $(1 + a + a^{2} + a^{3})^{5} = (5 \times 3) + 1 = 16$

Hence option (a)

Question 86.

How many different three digit numbers can be formed with the digits 1, 2, 3, 4, 5 and 6 so that none of the digits are repeated?

Discuss Question

Answer: Option A. -> 120
:
A

Suppose we start with the unit’s place. Unit place can be filled by any of the 6 digits given, so it can be filled in 6 ways. Now coming the tens place, it can be filled by any of the six digits except the one which has been used at unit’s place. So, it can be filled in 5 ways. Similarly the hundred’s place can be filled by any of the six digits except those two which have been used at unit’s and tens’ place. So, it can be filled in 4 ways.

4 ways 5 ways 6 ways

So, total numbers of numbers possible = $6 \times 5 \times 4 = 120$ . Hence option (a)

Question 87.

Find the value of ‘n’ if $^{n} P_{6} = 12^{n} P_{4}$ .

Discuss Question

Answer: Option A. -> 8
:
A

Soln:
$\frac{n!}{(n - 6)!} = \frac{12 \times n!}{(n - 4)!}$

$(n - 4)! = 12 \times (n - 6)!$
$(n - 4) (n - 5) = 4 \times 3$

n = 8

Hint: go from answer options. Only option(a) satisfies given equation.

Question 88.

The number of natural numbers which lie between $10^{8}$ and $10^{9}$ which have products of their digits as 6 is___.

Discuss Question

Answer: Option A. -> 8
:

$10^{8}$ <n< $10^{9}$

n is a 9 digit number

Product of 6 is possible in the following cases

Exactly one digit is 6 and the remaining digits are 1

Number of possible numbers= 9

One digit is 2 and the other is 3, remaining are 1 = $^{9} P_{2}$ = 72

Total possible numbers = 72+9 = 81

Question 89.

5 Nephrologists decide to hold daily meetings such that (i) At least one Nephrologist attend each day. (ii) A different set of Nephrologists must attend on different days. (iii) On day N for each $1 \leq d < N$ , at least one Nephrologist must attend who was present on day d. How many maximum days can meetings be held?

14
16
20
None of these
10

Discuss Question

Answer: Option B. -> 16
:
B

We need to find the largest possible number of subsets of {1, 2, 3, 4, 5} such that no 2 subsets are disjoint. Fix one element from the set to be presented in each subset and we can have $2^{4}$ such possibilities.

Because one element is common. So, remaining 4 can be selected $^{4} C_{0}$ + $^{4} C_{1}$ + $^{4} C_{2}$ + $^{4} C_{3}$ + $^{4} C_{4}$ = $2^{4}$ = 16 ways. There are total 16 days.