M.C.Q (1 Marks) - Maths STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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15 questions · auto-graded multiple-choice test.

MCQ 11 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
$72$

Answer

Correct option: C.

$7200$

Given that total number of vowels $= 4$
and number of consonants $= 5$
The total of words formed by $2$ vowels and $3$ consonents
$=\ ^4\text{C}_2\times\ ^5\text{C}_3$
$=\frac{4!}{2!\ 2!}\times\frac{5!}{3!\ 2!}$
$=\frac{4\times3\times2!}{2\times1\times2!}\times\frac{5\times4\times3!}{3!\times2}$
Now permutation of $2$ vowels and $3$ consonants $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
So, the total number of words $= 60 \times 120 = 7200.$

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MCQ 21 Mark

The number of $5-$digit telephone numbers having atleast one of their digits repeated is.

A
$90,000$
B
$10,000$
C
$30,240$
✓
$69,760$

Answer

Correct option: D.

$69,760$

Total number of telephone numbers when there is no restriction $= 10^{5 }$
Also number of telephone numbers having all digits different $=\ ^{l0}P_{5 }$
Required number of ways $= 10^5 – ^{10}P_5 $
$= 1000000 – 10 \times 9 \times 8 \times 7 \times 6 $
$= 1000000 - 30240 $
$= 69760$

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MCQ 31 Mark

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is.

✓
$94$
B
$126$
C
$128$
D
None

Answer

Correct option: A.

$94$

Number of men $= 4$
Number of women $= 6$
We are given that the committee includes $2$ men and exactly twice as many women as men.
Thus$,$ the possible selection can be $2$ men and $4$ women and $3$ men and $6$ women
So$,$ the number of committee $= \ ^4C_2 \times \ ^6C_4 + \ ^4C_3 \times \ ^6C_6 = 6 \times 15 + 4 1 = 90 + 4 = 94$

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MCQ 41 Mark

The number of triangles that are formed by choosing the vertices from a set of $12$ points$,$ seven of which lie on the same line is.

A
$105$
B
$15$
C
$175$
✓
$185$

Answer

Correct option: D.

$185$

Total number of triangle formed from $12$ points taking $3$ at a time $= \ ^{12}C_3$
But given that out of $12$ points $7$ are collinear. So$,$ these seven points will form no triangle.
$\therefore$ The required number of triangles $= \ ^{12}C_3 – \ ^7C_3$
$=\frac{12!}{3!\ 9!}-\frac{7!}{3!\ 4!}$
$=\frac{12\times11\times10\times9}{3\times2\times1\times9!}-\frac{7\times6\times5\times4!}{3\times2\times1\times4!}$
$=\frac{12\times11\times10}{3\times2}-\frac{7\times6\times5}{3\times2}$
$=220-35$
$=185$

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MCQ 51 Mark

The number of words which can be formed out of the letters of the word $\text{ARTICLE,}$ so that vowels occupy the even place is.

A
$1440$
✓
$144$
C
$7!$
D
$\ ^4C_4 \times \ ^3C_3$

Answer

Correct option: B.

$144$

Total number of letters in the $‘\text{ARTICLE ’}$ is $7$ out which $\text{A, E, I}$ are vowels and $\text{R, T, C, L}$ are consonants
Given that vowels occupy even place
$\therefore$ Possible arrangement can be shown as below
$\text{C, V, C, V, C, V, C}$ i.e. on $2^{nd}, 4^{th}$ and $6^{th}$ places
Therefore, number of arrangement $= ^3P_3 = 3! = 6$ ways
Now consonants can be placed at $1, 3, 5$ and $7^{th}$ place
$\therefore$ Number of arrangement $= \ ^4P_4 = 4! = 24$
So$,$ the total number of arrangements $= 6 \times 24 = 144$

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MCQ 61 Mark

If $^nC_{12}=^nC_8 ,$ then n is equal to.

✓
$20$
B
$12$
C
$6$
D
$30$

Answer

Correct option: A.

$20$

Give that $^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_8[\because\ ^\text{n}\text{C}_\text{r}=\ ^\text{n}\text{C}_\text{n-r}]$
$^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_\text{n-8}$
$\therefore\text{n}-8=12$
$\Rightarrow\text{n}=12+8$
$=20$

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MCQ 71 Mark

Given $5$ different green dyes, four different blue dyes and three different red dyes$,$ the number of combinations of dyes which can be chosen taking at least one green and one blue dye is.
$[$Hint: Possible numbers of choosing or not choosing $5$ green dyes$, 4$ blue dyes and $3$ red dyes are $2^5, 2^4$ and $2^3 ,$ respectively$.]$

A
$3600$
✓
$3720$
C
$3800$
D
$3600$

Answer

Correct option: B.

$3720$

At least one green dye can be chosen in $^5\text{C}_1+\ ^5\text{C}_2\ +\ ^5\text{C}_3+\ ^5\text{C}_4+\ ^5\text{C}_5=22^5-1$ ways.
At least one blue dye can be chosen in $^4\text{C}_1+\ ^4\text{C}_2\ +\ ^4\text{C}_3+\ ^4\text{C}_4=2^4-1$ ways.
Any number of red dyes can be chosen in $^3\text{C}_0+\ ^3\text{C}_1\ +\ ^3\text{C}_2+\ ^3\text{C}_3=2^3$ ways.
so$,$ total nnumber of required selection $=(2^5-1)\times(2^4-1)\times2^3=3720$

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MCQ 81 Mark

A five digit number divisible by $3$ is to be formed using the numbers $0, 1, 2, 3, 4$ and $5$ without repetitions. The total number of ways this can be done is.
$[$Hint: $5$ digit numbers can be formed using digits $0, 1, 2, 4, 5$ or by using digits $1, 2, 3, 4, 5$ since sum of digits in these cases is divisible by $3.]$

✓
$216$
B
$600$
C
$240$
D
$3125$

Answer

Correct option: A.

$216$

We know that a number is divisible by $3$ if the sum of its digits is divisible by $3.$
Now sum of the given six digits is $15$ which is divisible by $3.$
So to form a number of five$-$digit which is divisible by $3$ we can remove either $'O\ '$ or $'3\ '$. If digits $1, 2, 3, 4, 5$ are used then number of required numbers $= 5!$
If digits $0, 1, 2, 4, 5$ are used then first place from left can be filled in $4$ ways and remaining $4$ places can be filled in $4!$ ways.
So in this case required numbers are $4 \times 4!$ ways.
So, total number of numbers $= 120 + 96 = 216$

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MCQ 91 Mark

The number of possible outcomes when a coin is tossed $6$ times is.

A
$36$
✓
$64$
C
$12$
D
$32$

Answer

Correct option: B.

$64$

We know that a coin has Head and Tail $\text{(H, T)}$
$\therefore$ When a coin is tossed $6$ times$,$ then the
Possible outcome $= 2^6 $
$= 64$

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MCQ 101 Mark

The number of different four digit numbers that can be formed with the digits $2, 3, 4, 7$ and using each digit only once is

A
$120$
B
$96$
✓
$24$
D
$100$

Answer

Correct option: C.

$24$

Given digits $2, 3, 4$ and $7,$ we have to form four$-$digit numbers using these digits.
$\therefore$ Required number of ways $= 4P4 = 4! = 4 × 3 × 2 × 1 = 24$

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MCQ 111 Mark

The number of ways in which a team of eleven players can be selected from $22$ players always including $2$ of them and excluding $4$ of them is.

A
$^{16}C_{11}$
B
$^{16}C_5$
✓
$^{16}C_9$
D
$^{20}C_9$

Answer

Correct option: C.

$^{16}C_9$

Total number of players $= 22$
$2$ players are always included and $4$ are always excluding or never included $= 22 – 2 – 4 = 16$
$\therefore$ Required number of selection $= \ ^{16}C_9$

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MCQ 121 Mark

The total number of $9$ digit numbers which have all different digits is.

A
$10!$
B
$9 !$
✓
$9 \times 9!$
D
$10\times 10!$

Answer

Correct option: C.

$9 \times 9!$

We have to form $9-$ digit number which has all different digit.
First digit from the left can be filled in $9$ ways $($excluding $'\ 0\ ’).$
Now nine digits are left including $'\ O\ ’.$
So remaining eight places can be filled with these nine digits in $\ ^9P_S$ ways.
So$,$ total number of numbers $= 9 \times \ ^9P_8 = 9 \times 9!$

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MCQ 131 Mark

The sum of the digits in unit place of all the numbers formed with the help of $3, 4, 5$ and $6$ taken all at a time is.

A
$432$
✓
$108$
C
$36$
D
$18$

Answer

Correct option: B.

$108$

If the unit place is $'3\ '$ then remaining three places can be filled in $3!$ ways.
Thus $'3\ '$ appears in unit place in $3!$ times.
Similarly each digit appear in unit place $3!$ times.
So, sum of digits in unit place $= 3! (3 + 4 + 5 + 6) = 18 \times 6 = 108$

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MCQ 141 Mark

Every body in a room shakes hands with everybody else. The total number of hand shakes is $66.$ The total number of persons in the room is.

A
$11$
✓
$12$
C
$13$
D
$14$

Answer

Correct option: B.

$12$

Between any two person there is one hand shake.
So, number of hand shakes $=\ ^\text{n}\text{C}_2=\frac{\text{n!}}{2!(\text{n}-2)!}=\frac{\text{n}(\text{n}-1)}{2}=66($given$)$
$\Rightarrow\text{n}(\text{n}-1)=132$
$\Rightarrow\text{n}^2-\text{n}-132=0$
$\Rightarrow(\text{n}-12)(\text{n}=11)=0$
$\therefore\text{n}=12$

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MCQ 151 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
$6$
✓
$18$
C
$12$
D
$9$

Answer

Correct option: B.

$18$

To form parallelogram we required a pair of line from a set of $4$ lines and another pair of line from another set of $3$ lines.
Required number of parallelograms $= \ ^4C_2 \times \ ^3C_2 = 6 \times 3 = 18$

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