MCQ

MCQ 11 Mark

A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?

A
320
B
750
C
40
✓
11340

Answer

Correct option: D.

11340

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MCQ 21 Mark

Find the number of triangles which can be formed by joining the angular points of a polygon of $8$ sides as vertices.

A
$16$
✓
$56$
C
$24$
D
$8$

Answer

Correct option: B.

$56$

$56$
A triangle is obtained by joining three vertices.
Number of ways of selecting 3 vertices out of 8 vertices $={ }^8 C _3$
$=\frac{8 \times 7 \times 6}{1 \times 2 \times 3} $
$ =56$

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MCQ 31 Mark

In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

A
12
B
288
✓
144
D
256

Answer

Correct option: C.

144

B G B G B G B 4 boys take their seats in 4! ways. 3 girls take their seats in 3! ways. Required number = 4! × 3! = 24 × 6 = 144

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MCQ 41 Mark

In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

A
9 × 8!
B
8 × 8!
C
9 × 9!
✓
8 × 9!

Answer

Correct option: D.

8 × 9!

Arrange 8 papers in $8 !$ ways and two papers in 9 gaps are arranged in ${ }^9 P _2$ ways.

Required number $=8 !{ }^9 P _2$

= 8! × 9 × 8 = 9! × 8

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MCQ 51 Mark

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together?

A
3! 8!
✓
3! 4! 8! 4!
C
4! 4!
D
8! 4! 4!

Answer

Correct option: B.

3! 4! 8! 4!

3! 4! 8! 4! 8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways. Three groups of Indians, Americans and Englishmen can be permuted in 3! ways. Required number = 3! × 8! × 4! × 4!

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MCQ 61 Mark

A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is

✓
21
B
4
C
42
D
10

Answer

Correct option: A.

Number of ways to select one morning and one evening course $={ }^7 C _1 \times{ }^3 C _1=21$

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MCQ 71 Mark

A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is

A
5
B
3
✓
8
D
15

Answer

Correct option: C.

8 Hint: Number of ways to select one course from available 8 courses (i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

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MCQ 81 Mark

The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is

A
840
B
600
C
720
✓
480

Answer

Correct option: D.

480

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MCQ 91 Mark

The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is

A
80
B
60
✓
40
D
100

Answer

Correct option: C.

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MCQ 101 Mark

There are $10$ persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers is equal to:

A
$2! \times 7!$
✓
$2! \times 8!$
C
$3! \times 7!$
D
$3! \times 8!$

Answer

Correct option: B.

$2! \times 8!$

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MCQ 112 Marks

If at the end of certain meeting, everyone had shaken hands with every one else, it was found that 45 handshakes were exchanged, then the number of members present at the meeting are

✓
10
B
15
C
20
D
21

Answer

Correct option: A.

(a) : Let number of members present $=n$
Since, every person handshake with everyone else
$
\therefore \quad{ }^n C_2=45
$ [Given]
$
\Rightarrow \frac{n !}{(n-2) ! 2 !}=45
$
$
\begin{aligned}
\Rightarrow & n(n-1)=90 \\
\Rightarrow & n^2-n-90=0 \\
\Rightarrow & n^2-10 n+9 n-90=0 \\
\Rightarrow & (n-10)(n+9)=0
\end{aligned}
$
$\Rightarrow n=10$
$\therefore \quad$ There are 10 people in the meeting.

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MCQ 122 Marks

There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is negative number, is

✓
$\frac{496}{1001}$
B
$\frac{505}{1001}$
C
$\frac{490}{1001}$
D
$\frac{504}{1001}$

Answer

Correct option: A.

$\frac{496}{1001}$

(a) : As we know that, product of 4 numbers will be negative if either 1 of them is negative or 3 of them are negative.$\therefore \quad$ Number of ways of choosing 4 numbers so that the product is negative $={ }^6 C _1 \times{ }^8 C _3+{ }^8 C _1 \times{ }^6 C _3$
Total number of ways of choosing 4 numbers out of 14 numbers $={ }^{14} C _4$
$\therefore \quad$ Required probability
$
\begin{aligned}
& =\frac{\text { Possible outcomes }}{\text { Total outcomes }}=\frac{{ }^6 C _1 \times{ }^8 C _3+{ }^8 C _1 \times{ }^6 C _3}{{ }^{14} C _4} \\
& =\frac{496}{1001}
\end{aligned}
$

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MCQ 132 Marks

If in a regular polygon, the number of diagonals are 54 , then the number of sides of the polygon are

A
10
✓
12
C
9
D
6

Answer

Correct option: B.

(b) : Number of diagonal in a polygon having $n$ sides $=\frac{n(n-3)}{2}=54$ (given)
$
\Rightarrow n(n-3)=108 \Rightarrow n^2-3 n-108=0
$
$
\begin{aligned}
& \Rightarrow n^2-12 n+9 n-108=0 \\
& \Rightarrow(n-12)(n+9)=0
\end{aligned}
$
$\Rightarrow n=12 \quad[\because n$ is number of sides $]$
$\therefore \quad$ Polygon has 12 sides.

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MCQ 142 Marks

A linguistic club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this group including the selection of a leader (from among these 4 members) for the team. If the team has to include at most one boy, the number of ways of selecting the team is

A
140
B
320
C
76
✓
380

Answer

Correct option: D.

380

(d) : Since at most 1 boy can be selected, so team can be selected in two ways like as follows :
1: 3 girls and 1 boy are selected
2: 4 girls are selected
Permutations and Combinations
Number of ways of selecting 3 girls and 1 boy $={ }^6 C _3 \times{ }^4 C _1$
$
=\frac{6 \times 5 \times 4}{1 \times 2 \times 3} \times 4=80 \text { ways }
$
Now, number of ways of selecting 4 girls $={ }^6 C_4=15$
$\therefore$ Total number of ways of selecting 4 members
$
=80+15=95 \text { ways }
$
Now, out of 4 member, number of ways selecting a leader $={ }^4 C_1=4$ ways
$\therefore$ Total number of ways of selecting a team
$
=95 \times 4=380
$

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MCQ 152 Marks

It is required to seat 5 men and 4 women in a row so that the men occupy odd places. Then the number of arrangements that are possible is

✓
2880
B
1440
C
144
D
362880

Answer

Correct option: A.

2880

(a) : Total number of people $=9$
Men occupy odd places, so women can occupy four even places.
MWMWMWMWM
4 women can sit in 4 position in 4! ways 5 men can sit in 5 position in 5 ! ways
$\therefore$ Total number of arrangement $=4 ! \times 5 !$

= $24 ! \times 120 !$=2880

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MCQ 162 Marks

In a certain examination, a candidate has to pass in each of the 5 subjects. Hence, the number of ways he can fail is

A
5
B
$2^5$
C
$5 !$
✓
$2^5-1$

Answer

Correct option: D.

$2^5-1$

(d) : To pass the exam a candidate has to pass in each of the 5 subjects which can be done in only 1 way. Hence, number of ways he can fail is $2^5-1$.

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MCQ 172 Marks

The number of ways in which the letters of the word MACHINE can be arranged such that the vowels may occupy only odd positions is

A
288
B
1152
C
625
✓
576

Answer

Correct option: D.

576

(d) : Total letters in the word MACHINE $=7$
No, of vowels $=3( A , I , E )$
Number of ways of arranging vowels $={ }^4 P_3=4 \times 3 \times 2=$ 24
Number of ways of arranging consonants $={ }^4 P_4=4 !=24$
$\therefore$ Required number of ways $=24 \times 24=576$

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