MCQ - MATHS STD 11 Science Questions - Vidyadip

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27 questions · auto-graded multiple-choice test.

MCQ 11 Mark

If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r+4}}$ is then ${^\text{r}}\text{C}_{\text{3}}$ equal to:

A
54
✓
56
C
58
D
none of these.

Answer

Correct option: B.

56

$\text{r}+\text{r}+4=20$
$\Rightarrow 2\text{r}+4=20$
$\Rightarrow 2\text{r}=16$
$\Rightarrow \text{r}=8$
Now,
${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$
$=\frac{8\times7\times6}{3\times2\times1}=56$

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MCQ 21 Mark

The value of $({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+.....+({^\text{7}}\text{C}_{\text{6}}+{^\text{7}}\text{C}_{\text{7}})$ is:

A
$2^{7}-1$
✓
$2^{8}-2$
C
$2^{8}-1$
D
$2^{8}$

Answer

Correct option: B.

$2^{8}-2$

$({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{3}}+{^\text{7}}\text{C}_{\text{4}})+......$
$=1+2\times{^\text{7}}\text{C}_{\text{1}}+2\times{^\text{7}}\text{C}_{\text{2}}+2\times{^\text{7}}\text{C}_{\text{3}}+2\times{^\text{7}}\text{C}_{\text{4}}+2\times{^\text{7}}\text{C}_{\text{5}}..$
$=2+2^{2}({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})$
$=2+2^{2}(7+\frac{7}{2}\times6+\frac{7}{3}\times\frac{6}{2}\times5)$
$=2+252$
$=254$
$=2^{8}-2$

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MCQ 31 Mark

Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?

✓
60
B
20
C
15
D
125

Answer

Correct option: A.

60

60

Solution:
Three persons can take 5 seats in $^5C_3$ ways. Moreover 3 persons can sit in 3! ways.
Required number of ways ${^\text{5}}\text{C}_{\text{3}}\times3!$
$=10\times6=60$

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MCQ 41 Mark

If ${^\text{15}}\text{C}_{3\text{r}}={^\text{15}}\text{C}_{\text{r+3}},$ is then equal to:

A
5
B
4
✓
3
D
2

Answer

Correct option: C.

3

$3\text{r}+\text{r}+3=15$
$\Rightarrow 4\text{r}+3=15$
$\Rightarrow 4\text{r}=12$
$\Rightarrow \text{r}=3$

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MCQ 51 Mark

If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r-10}}$ is then ${^\text{18}}\text{C}_{\text{r}}$ equal to:

A
4896
✓
816
C
1632
D
None of these.

Answer

Correct option: B.

816

$\text{r}+\text{r}-10=20$
$\Rightarrow 2\text{r}-10=20$
$\Rightarrow 2\text{r}=30$
$\Rightarrow \text{r}=15$
Now,
${^\text{18}}\text{C}_{\text{r}}={^\text{18}}\text{C}_{\text{15}}$
$\therefore\ {^\text{18}}\text{C}_{\text{15}}={^\text{18}}\text{C}_{\text{3}}$
$\therefore\ {^\text{18}}\text{C}_{\text{3}}=\frac{18}{3}\times\frac{17}{2}\times16$
$=816$

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MCQ 61 Mark

If $^{(\text{a}^2-\text{a})}\text{C}_{\text{2}}=^{(\text{a}^2-\text{a})\text{}}\text{C}_{\text{4}},$ is then x:

A
2
✓
3
C
4
D
None of these

Answer

Correct option: B.

3

$\text{a}^{2}-\text{a}=2+4$
$\Rightarrow \text{a}^{2}-\text{a}-6=0$
$\Rightarrow \text{a}^{2}-3\text{a}+2\text{a}-6=0$
$\Rightarrow \text{a}(\text{a}-3)+2(\text{a}-3)=0$
$\Rightarrow (\text{a}+2)(\text{a}-3)=0$
$\Rightarrow \text{a}=-2,\text{a}=3$
$\text{a}=3$

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MCQ 71 Mark

If $={^\text{43}}\text{C}_{\text{r-6}}={^\text{43}}\text{C}_{\text{3r+1}},$ then the value of r is is:

✓
12
B
8
C
6
D
10

Answer

Correct option: A.

12

$\text{r}-6+3\text{r}+1=43$
$\Rightarrow 4\text{r}-5=43$
$\Rightarrow 4\text{r}=48$
$\Rightarrow \text{r}=12$

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MCQ 81 Mark

A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is:

A
112
✓
140
C
164
D
None of these.

Answer

Correct option: B.

140

Suppose there are two friends, A and B, who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting 6 guests $={^\text{8}}\text{C}_{\text{6}}=28$
If one of them attends the party, then the number of ways of selecting 6 guests $=2.{^\text{8}}\text{C}_{\text{5}}=112$
Total number of ways = 112 + 28 = 140

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MCQ 91 Mark

There are 13 players of cricket, out of which 4 are bowlers. In how many ways a team of eleven be selected from them so as to include at least two bowlers?

A
72
✓
78
C
42
D
None of these.

Answer

Correct option: B.

78

4 out of 13 players are bowlers.
In other words, 9 players are not bowlers.
A team of 11 is to be selected so as to include at least 2 bowlers.
Number of ways $={^\text{4}}\text{C}_{\text{2}}\times{^\text{9}}\text{C}_{\text{9}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{9}}\text{C}_{\text{8}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}$
$=6+36+36$
$=78$

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MCQ 101 Mark

${^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{3}}+{^\text{5}}\text{C}_{\text{4}}+{^\text{5}}\text{C}_{\text{5}}$is equal to:

A
30
✓
31
C
32
D
33

Answer

Correct option: B.

31

${^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{3}}+{^\text{5}}\text{C}_{\text{4}}+{^\text{5}}\text{C}_{\text{5}}$
$={^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{1}}+{^\text{5}}\text{C}_{\text{5}}$
$=2\times{^\text{5}}\text{C}_{\text{1}}+2\times{^\text{5}}\text{C}_{\text{2}}+{^\text{5}}\text{C}_{\text{5}}$
$= 2\times5+2\times\frac{5!}{2!3!}+1$
$=10+20+1$
$=31$

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MCQ 111 Mark

How many different committees of 5 can be formed from 6 men and 4 women on which exact 3 men and 2 women serve?

A
6
B
20
C
60
✓
120

Answer

Correct option: D.

120

Number of committes that can be formed $={^\text{6}}\text{C}_{\text{3}}\times{^\text{4}}\text{C}_{\text{2}}$
$=\frac{6!}{3!3!}\times\frac{4!}{2!2!}$
$=\frac{6\times5\times4}{3\times2}\times\frac{4\times3}{2}$
$=120$

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MCQ 121 Mark

In how many ways can a committee of 5 be made out of 6 men and 4 women containing at least one women?

✓
246
B
222
C
186
D
None of these.

Answer

Correct option: A.

246

Required number of ways $={^\text{4}}\text{C}_{\text{1}}\times{^\text{6}}\text{C}_{\text{4}}+{^\text{4}}\text{C}_{\text{2}}\times{^\text{6}}\text{C}_{\text{3}}+{^\text{4}}\text{C}_{\text{3}}\times{^\text{6}}\text{C}_{\text{2}}+{^\text{4}}\text{C}_{\text{4}}\times{^\text{6}}\text{C}_{\text{1}}$
$=60+120+60+6$
$=246$

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MCQ 131 Mark

There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two of them is:

A
45
✓
40
C
39
D
38

Answer

Correct option: B.

40

Number of straight lines formed by joining the 10 points if we take 2 points at a time $={^\text{10}}\text{C}_{\text{2}}=\frac{10}{2}\times\frac{9}{1}=45$
Number of straight lines formed by joining the 4 points if we take 2 points at a time $={^\text{4}}\text{C}_{\text{2}}=\frac{4}{2}\times\frac{3}{1}=6$
But, 4 collinear points, when joined in pairs, give only one line.
Required number of straight lines = 14 - 6 + 1 = 40

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MCQ 141 Mark

If ${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}},$ is then:

A
2m = n
B
2m = n(n + 1)
✓
2m = n(n - 1)
D
2n = m(m - 1)

Answer

Correct option: C.

2m = n(n - 1)

${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{\text{m!}}{1!(\text{m}-1)!}=\frac{\text{n!}}{2!(\text{n}-2)!}$
$\Rightarrow \frac{\text{m}(\text{m}-1)!}{(\text{m}-1)!}=\frac{\text{n}(\text{n}-1)(\text{n}-2)!}{2!(\text{n}-2)!}$
$\Rightarrow2\text{m}=\text{n}(\text{n}-1)$

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MCQ 151 Mark

If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then n:

A
3
B
4
✓
5
D
6

Answer

Correct option: C.

5

${^\text{n+1}}\text{C}_{\text{3}}=2\times{^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{(\text{n}+1)!}{3!(\text{n-2})!}=2\times\frac{\text{n}!}{2!(\text{n}-1)!}$
$\Rightarrow \text{n+1}=6$
$\Rightarrow \text{n}=5$

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MCQ 161 Mark

There are 12 points in a plane. The number of the straight lines joining any two of them when 3 of them are collinear is:

A
62
B
63
✓
64
D
65

Answer

Correct option: C.

64

Number of straight lines joining 12 points if we take 2 points at a time $={^\text{12}}\text{C}_{\text{2}}$
$=\frac{12!}{2!10!}=66$
Number of straight lines joining 3 points if we take 2 points at a time $={^\text{3}}\text{C}_{\text{2}}=3$
Required number of straight lines = 66 - 3 + 1 = 64

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MCQ 171 Mark

If ${^\text{n}}\text{C}_{\text{r}}+{^\text{n}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}},$ is then x:

A
$\text{r}$
B
$\text{r}-1$
C
$\text{n}$
✓
$\text{r}+1$

Answer

Correct option: D.

$\text{r}+1$

We have,
${^\text{n}}\text{C}_{\text{r}}+{^\text{n}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$
$\Rightarrow {^\text{n+1}}\text{C}_{\text{r+1}}={^\text{n+1}}\text{C}_{\text{x}}$
$\Rightarrow \text{r}+1=\text{x}$
${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$
$\Rightarrow \text{n}=\text{x}+\text{y},\text{x}=\text{y}$

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MCQ 181 Mark

The number of diagonals that can be drawn by joining the vertices of an octagon is:

✓
20
B
28
C
8
D
16

Answer

Correct option: A.

20

An octagon has 8 vertices.
The number of diagonals of a polygon is given by $\frac{\text{n}(\text{n}-3)}{2}$
Number of diagonals of an octagon $=\frac{\text{8}(\text{8}-3)}{2}=20$

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MCQ 191 Mark

If $C_0 + C_1 + C_2 + ... + C_n = 256$, then $^{2n}C_2$ is equal to:

A
56
✓
120
C
28
D
91

Answer

Correct option: B.

120

120

Solution:
If set S has n elements, then C (n, k)C n, k is the number of ways of choosing k elements from S.
Thus, the number of subsets of SS of all possible values is given by,
$\text{C}(\text{n},0)+\text{C}(\text{n},1)+\text{C}(\text{n},3)+.....+\text{C}(\text{n},\text{n})=2^\text{n}$
Comparing the given equation with the above equation:
$2^\text{n}=256$
$\Rightarrow 2^\text{n}=2^{8}$
$\Rightarrow \text{n}=8$
$\therefore {^\text{2n}}\text{C}_{\text{2}}={^\text{16}}\text{C}_{\text{2}}$
$\Rightarrow {^\text{16}}\text{C}_{\text{2}}=\frac{16!}{2!4!}=\frac{16\times15}{2}=120$

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MCQ 201 Mark

The number of ways in which a host lady can invite for a party of 8 out of 12 people of whom two do not want to attend the party together is:

A
$2\times{^\text{11}}\text{C}_{\text{7}}+{^\text{10}}\text{C}_{\text{8}}$
B
${^\text{10}}\text{C}_{\text{8}}+{^\text{11}}\text{C}_{\text{7}}$
✓
${^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}$
D
None of these.

Answer

Correct option: C.

${^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}$

A host lady can invite 8 out of 12 people in ways. Two out of these 12 people do not want to attend the party together.
Number of ways $={^\text{12}}\text{C}_{\text{8}}-{^\text{10}}\text{C}_{\text{6}}.$

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MCQ 211 Mark

If ${^\text{20}}\text{C}_{3\text{r+1}}={^\text{20}}\text{C}_{\text{r-1}},$ is then r equal to:

✓
10
B
11
C
19
D
12

Answer

Correct option: A.

10

$\text{r}+\text{1}+\text{r}-1=20$ $[\therefore \ ^\text{n}\text{C}_{\text{x}} = \ ^\text{n}\text{C}_{\text{y}} \Rightarrow \text{n} = \text{x} + \text{y} \text{ or } \text{x = y}]$
$\Rightarrow 2\text{r}=20$
$\Rightarrow \text{r}=10$

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MCQ 221 Mark

Among 14 players, 5 are bowlers. In how many ways a team of 11 may be formed with at least 4 bowlers?

A
265
B
263
✓
264
D
275

Answer

Correct option: C.

264

Among 14 players, 5 are bowlers.
A team of 11 players has to be selected such that at least 4 bowlers are included in the team.
Required number of ways $={^\text{5}}\text{C}_{\text{4}}\times{^\text{9}}\text{C}_{\text{7}}+{^\text{5}}\text{C}_{\text{5}}\times{^\text{9}}\text{C}_{\text{6}}$
$=180+84$
$=264$

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MCQ 231 Mark

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:

A
6
B
9
C
12
✓
18

Answer

Correct option: D.

18

A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines.
Two parallel lines from the set of four parallel lines can be chosen in ${^\text{4}}\text{C}_{\text{2}}$ ways and two parallel lines from the set of 3 parallel lines can be chosen in ${^\text{3}}\text{C}_{\text{2}}$ ways.
Number of parallelograms that can be formed $={^\text{4}}\text{C}_{\text{2}}\times{^\text{3}}\text{C}_{\text{2}}$
$=\frac{4!}{2!2!}\times\frac{3!}{2!1!}$
$=6\times3=18$

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MCQ 241 Mark

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to:

A
60
B
120
✓
7200
D
None of these.

Answer

Correct option: C.

7200

7200

Solution:
2 out of 4 vowels can be chosen in $^4C_2$ ways and 3 out of 5 consonants can be chosen in $^5C_3$ ways.
Thus, there are $({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})$ groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters that can be arranged in 5! ways.
Required number of words $=({^\text{4}}\text{C}_{\text{2}}\times{^\text{5}}\text{C}_{\text{3}})\times5!$
$=60\times120=7200$

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MCQ 251 Mark

Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the number of circles that can be drawn so that each contains at least 3 of the given points is:

A
216
✓
156
C
172
D
None of these.

Answer

Correct option: B.

156

We need at least three points to draw a circle that passes through them.
Now, number of circles formed out of 11 points by taking three points at a time $={^\text{11}}\text{C}_{\text{3}}=165$
Number of circles formed out of 5 points by taking three points at a time $={^\text{5}}\text{C}_{\text{3}}=10$
It is given that 5 points lie on one circle.
Required number of circles = 165 - 10 + 1 = 156.

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MCQ 261 Mark

If C(n, 12) = C(n, 8) is then r equal to:

✓
231
B
210
C
252
D
303

Answer

Correct option: A.

231

${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}}$
$\Rightarrow \text{n}=12+8=20$
${^\text{n}}\text{C}_{\text{x}}={^\text{n}}\text{C}_{\text{y}}$
$\Rightarrow \text{n}=\text{x}+\text{y}, \text{x}=\text{y}$
Now,
${^\text{22}}\text{C}_{\text{n}}={^\text{22}}\text{C}_{\text{20}}$
$=\frac{22}{2}\times\frac{21}{1}$
$=231$

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MCQ 271 Mark

If ${^\text{n}}\text{C}_{\text{12}}={^\text{n}}\text{C}_{\text{8}},$ is then n:

✓
20
B
12
C
6
D
30

Answer

Correct option: A.

20

$\text{n}=12+8=20$

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