M.C.Q (1 Marks)

MCQ 11 Mark

Choose the correct answer from the given four options in the following questions:
If two positive integers $a$ and $b$ are written as $a=x^3 y^2$ and $b=x y^3 ; x, y$ are prime numbers, then $H C F$ ( $a, b$ ) is:

A
$x y$.
B
$x y^2$.
C
$x^3 y^3$.
✓
$x^2 y^2$.

Answer

Correct option: D.

$x^2 y^2$.

Given that, $a=x^3 y^2=x \times x \times x \times y \times y$
and $b=x y^3=x \times y \times y \times y$
$\therefore$ HCF of $a$ and $b=\operatorname{HCF}\left(x^3 y^2 x y^2\right)=x \times y x y=x y^2$
[since, HCF is the product of the smallest power of each common prime facter involved in the numbers]

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MCQ 21 Mark

Choose the correct answer from the given four options in the following questions:$n^2 – 1$ is divisible by 8, if n is:

A
An integer.
B
A natural number.
✓
An odd integer.
D
An even integer.

Answer

Correct option: C.

An odd integer.

Let $a=n^2-1$
Here n can be ever or odd.
Case I:
$n =$ Even i.e., $n =2 k$. where k is an integer.
$\Rightarrow a=(2 k)^2-1$
$\Rightarrow a=4 k^2-1$
At $k=-1,=4(-1)^2-1=4-1=3$, which is not divisible by 8 .
At $k=0, a=4(0)^2-1=0-1=-1$, which is not divisible by 8 , which is not
Case II: $\square$
$n =$ Odd i.e., $n =2 k +1$, where k is an odd integer.
$\Rightarrow a=2 k+1$
$\Rightarrow a=(2 k+1) 2-1$
$\Rightarrow a=4 k^2+4 k+1-1$
$\Rightarrow a=4 k^2+4 k$
$\Rightarrow a=4 k(k+1)$
At $k=-1, a=4(-1)(-1+1)=0$ which is divisible by 8 .
At $k=0, a=4(0)(0+1)=4$ which is divisible by 8 .
At $k=1, a=4(1)(1+1)=8$ which is divisible by 8.
Hence, we can conclude from above two cases, if $n$ is odd, then $n^2-1$ is divisible by 8 .

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MCQ 31 Mark

Choose the correct answer from the given four options in the following questions:
If two positive integers $p$ and $q$ can be expressed as $p=a b^2$ and $q=a^3 b ; a, b$ being prime numbers, then LCM ( $p, q$ ) is:

A
ab.
B
. $a^2 b^2$.
✓
$a^3 b^2$.
D
$a^3 b^3$.

Answer

Correct option: C.

$a^3 b^2$.

Given that, $p = ab ^2= a \times b \times b$
and $q=a^3 b=a \times a \times a \times b$
$\therefore$ LCM of pandq $=\operatorname{LCM}\left(a b^2, a^3 b\right)=a \times b \times b \times a \times a=a^3 b^2$
(since, LCM is the product of the greatest power of each prime factor Invotved in the numbers)

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MCQ 41 Mark

Choose the correct answer from the given four options in the following questions:
The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:

A
One decimal place.
B
Two decimal places.
C
Three decimal places.
✓
Four decimal places.

Answer

Correct option: D.

Four decimal places.

Rational number $=\frac{14587}{1250}=\frac{14587}{2^{1}\times5^{4}}$
$=\frac{14587}{10\times5^{3}}=\frac{(2)^{3}}{(2)^{3}}$
$=\frac{14587\times8}{10\times1000}$
$=\frac{1166969}{10000}=11.6696$
Hence, given rational number will terminate after four decimal places.
$\begin{array}{c|c} 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

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MCQ 51 Mark

Choose the correct answer from the given four options in the following questions:
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:

A
10.
B
100.
C
504.
✓
2520.

Answer

Correct option: D.

2520.

Factors of 1 to 10 numbers,
1 = 1
2 = 1 × 2
3 = 1 × 3
4 = 1 × 2 × 2
5 = 1 × 5
6 = 1 × 2 × 3
7 = 1 × 7
8 = 1 × 2 × 2 × 2
9 = 1 × 3 × 3
10 = 1 × 2 × 5
$\therefore$ LCM of number 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

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MCQ 61 Mark

Choose the correct answer from the given four options in the following questions:
The product of a non-zero rational and an irrational number is:

✓
Always irrational.
B
Always rational.
C
Rational or irrational.
D
One.

Answer

Correct option: A.

Always irrational.

Product of a non-zero rational and an irrational number is always irrational i.e., $\frac{3}{4}\times\sqrt{2}=\frac{3\sqrt{2}}{4}\text{ (irrational)}.$

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MCQ 71 Mark

Choose the correct answer from the given four options in the following questions:
For some integer q, every odd integer is of the form:

A
q.
B
q + 1.
C
2q.
✓
2q + 1.

Answer

Correct option: D.

2q + 1.

We know that, odd integers are 1, 3, 5, ...
So, it can be written in the form of 2q + 1.
where, q = integer = Z
or q = ···, -1, 0, 1, 2, 3, ...
$\therefore$ 2q + 1 = ... -3, -1, 1, 3, 5,
Alternate Answer
Let 'a' be given positive integer. On dividing 'a' by 2, let q be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have
a = 2q + r, where
$0\leq\text{r}<2$
⇒ a = 2q + r, where r = 0 or r = 1
⇒ a = 2q or 2q + 1
when a= 2q + 1 for some integerq, then clearly a is odd.

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MCQ 81 Mark

Choose the correct answer from the given four options in the following questions:The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is:

✓
13.
B
65.
C
875.
D
1750.

Answer

Correct option: A.

13.

Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70 - 5), 117 = (125 - 8), which is divisible by the required number.

Now, required number = HCF of 65, 117 [for the largest number]

117 = 65 × 1 + 52 [$\because$ dividend = divisor × quotient + remainder]

65 = 52 × 1 + 13

52 = 13 × 4 + 0

$\therefore$ HCF = 13

Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.

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MCQ 91 Mark

If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is:

A
4.
✓
2.
C
1.
D
3.

Answer

Correct option: B.

By Euclid's division algorithm,
$\text{b}=\text{aq}+\text{r, }0\leq\text{r}<\text{a}$ [$\because$ dividend = divisor × quotient + remainder]
⇒ 117 = 65 × 1 + 52
⇒ 65 = 52 × 1 + 13
$\therefore$ HCF (65, 117) = 13 .....(i)
Also, given that, HCF (65, 117) = 65m - 117 .....(ii)
From Eqs. (i) and (ii),
65m - 117=13
65m = 130
m = 2.

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MCQ 101 Mark

Choose the correct answer from the given four options in the following questions:
For some integer m, every even integer is of the form:

A
m.
B
m + 1.
✓
2m.
D
2m + 1.

Answer

Correct option: C.

2m.

We know that, even integers are 2, 4, 6, ...
So, it can be written in the form of 2m.
where, m = Integer = Z [since, integer is represented by Z]
or m = ···, -1, 0, 1, 2, 3, ...
$\therefore$ 2m = ···, -2, 0, 2, 4, 6, ...
Alternate Answer
Let 'a' be a positive integer. On dividing 'a' by 2, let m be the quotient and be the remainder. Then, by Euclid's divlslon algonthm, we have
a = 2m + r, where
$\text{a}\leq\text{r}<2\text{ i.e.,}$
r = 0 and r = 1.
⇒ a = 2m or a = 2m + 1
when, a = 2m for some integer m, then clearly a is even.

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Maths M.C.Q (1 Marks)

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