M.C.Q (1 Marks)

MCQ 511 Mark

If two positive integers $a$ and $b$ are expressible in the form $a = pq^2$ and $b = p^2q; p, q$ being prime numbers, then $\text{LCM (a, b)}$ is:

A
$pq$
B
$p^3q^3$
C
$p^3q^2$
✓
$p^2q^2$

Answer

Correct option: D.

$p^2q^2$

$A$ and $b$ are two positive integers and $a = pq^2$ and $b = p^2q$, where $p$ and $q$ are prime numbers, then $\text{LCM} = p^2q^2.$

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MCQ 521 Mark

If two positive integers $tn$ and $n$ arc expressible in the form $m = pq^3$ and $n = p^3q^2,$ where $p, q$ are prime numbers, then $\text{HCF} (m, n)$ =

A
$pq$
✓
$pq^2$
C
$p^3q^3$
D
$p^2q^3$

Answer

Correct option: B.

$pq^2$

$m$ and $n$ are two positive integers and $m = pq^3$ and $n = pq^2$, where $p$ and $q$ are prime numbers, then $\text{HCF} = pq^2.$

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MCQ 531 Mark

If $n$ is any natural number, then $6^n - 5^n$ always ends with:

✓
$1$
B
$3$
C
$5$
D
$7$

Answer

Correct option: A.

$1$

$n$ is any natural number and $6^n - 5^n$
We know that $6^n$ ends with $6$ and $5^n$ ends with $5$
$6^n - 5^n$ will end with $6 - 5 = 1$

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MCQ 541 Mark

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
$1 = 1$
$2 = 1 \times 2$
$3 = 1 \times 3$
$4 = 1 \times 2 \times 2$
$5 = 1 \times 5$
$6 = 1 \times 2 \times 3$
$7 = 1 \times 7$
$8 = 1 \times 2 \times 2 \times 2$
$9 = 1 \times 3 \times 3$
$10 = 1 \times 2 \times 5$
$\text{LCM}$ of number $1$ to $10 = \text{LCM} (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$
$= 1 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520$

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MCQ 551 Mark

The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:

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

Answer

Correct option: D.

Four decimal place.

Rational number $=\frac{14587}{1250}=\frac{14587}{2^1\times5^4}$
$\begin{array}{c|c}2 &1250\\\hline 5 & 625\\\hline 5 & 125\\\hline5 & 25\\\hline5&5\\\hline&1 \end{array}$
$=\frac{14587}{10\times5^3}\times\frac{(2)^3}{(2)^3}$
$=\frac{14587\times8}{10\times1000}$
$=\frac{116696}{10000}=11.6696$
Hence, given rational number will terminate after four decimal places.

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MCQ 561 Mark

The sum of the exponents of the prime factors in the prime factorisation of $196$, is:

A
$1$
B
$2$
✓
$4$
D
$6$

Answer

Correct option: C.

$4$

$\begin{array}{c|c}2 &196\\\hline 2 & 98\\\hline 7 & 49\\\hline7 & 7\\\hline&1 \end{array}$
$= 2 \times 2 \times 7 \times 7$
$= 2^2 \times 7^2$
Sum of exponents $= 2 + 2 = 4$

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MCQ 571 Mark

If $a = 2^3 \times 3, b = 2 \times 3 \times 5, c = 3^n \times 5$ and $\text{LCM} (a, b, c) = 2^3 \times 3^2 \times 5$, then $n =$

A
$1$
✓
$2$
C
$3$
D
$4$

Answer

Correct option: B.

$2$

$a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5 \text { and } \operatorname{LCM}(a, b, c)=2^3 \times 3^2 \times 5 \\ \therefore 3^n=3^2 \Rightarrow n=2$

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MCQ 581 Mark

The $\text{LCM}$ of two numbers is $1200.$ Which of the following cannot be their $\text{HCF}?$

A
$600$
✓
$500$
C
$400$
D
$200$

Answer

Correct option: B.

$500$

$\text{LCM}$ of two number $= 1200$
Their $\text{HCF}$ of these two numbers will be the factor of $1200$
$500$ cannot be its $\text{HCF}$.

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MCQ 591 Mark

$3.\overline{27}$ is:

A
An integer.
✓
A rational number.
C
A natural number.
D
An irrational number.

Answer

Correct option: B.

A rational number.

$3.\overline{27}$ is a rational number.

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MCQ 601 Mark

The decimal expansion of the rational number $\frac{33}{2^2\times5}$ will terminate after:

A
One decimal place.
✓
Two decimal places.
C
Three decimal places.
D
More than $3$ decimal places.

Answer

Correct option: B.

Two decimal places.

$\frac{33}{2^2\times5}$
Multiply and divide the expansion by $5$
$\frac{33\times5}{2^2\times5^2}=\frac{165}{10^2}=1.65$
Hence, the decimal expansion of the rational number $\frac{33}{2^3\times5}$ will terminate after two decimal places.

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MCQ 611 Mark

If the $\text{HCF}$ 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.

$2$

Use Euclid's algorithm to find the $\text{HCF}$ of $65$ and $117.$
By Euclid's algorithm,
$b = aq + r, 0 \leq r < a$
$\Rightarrow 117 = 65 \times 1 + 32$
$\Rightarrow 65 = 52 \times 1 + 13$
$\Rightarrow 52 = 13 \times 4 + 0$
$\therefore \text{HCF} (65, 117) = 13$
It is given that $\text{HCF} (65, 117) = 65m - 117.$
$\Rightarrow 65m - 117 = 13$
$\Rightarrow 65m = 130$
$\Rightarrow m = 2$
Hence, the correct option is option $B.$

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MCQ 621 Mark

The $\text{LCM}$ and $\text{HCF}$ of two rational numbers are equal, then the numbers must be:

A
Prime.
B
Co$-$prime.
C
Composite.
✓
Equal.

Answer

Correct option: D.

Equal.

$\text{LCM}$ and $\text{HCF}$ of two rational numbers are equal. Then those must be equal.

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MCQ 631 Mark

If the sum of $\text{LCM}$ and $\text{HCF}$ of two numbers is $1260$ and their $\text{LCM}$ is $900$ more than their $\text{HCF,}$ then the product of two numbers is:

A
$203400$
✓
$194400$
C
$198400$
D
$205400$

Answer

Correct option: B.

$194400$

Given that sum of $\text{LCM}$ and $\text{HCF} = 1260$
$\text{LCM + HCF} = 1260 .....(1)$
Let two numbers be $a$ and $b$ and $\text{HCF} (a, b) = x$
According to question:
Put value of $\text{HCF}$ and $\text{LCM}$ in equation $(1)$
$\Rightarrow 900 + x + x = 1260$
$\Rightarrow 2x = 1260 - 900$
$\Rightarrow 2x = 360$
$\Rightarrow\ \text{x}=\frac{360}{2}$
$\Rightarrow x = 180 ......(2)$
Now, $\text{LCM} \times \text{HCF} =$ Product of two numbers
Product of two number $= (x + 900)(x)$
$= (180 + 900)(180)$
$= 1080 \times 180$
$= 194400$

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MCQ 641 Mark

If $p_1$_ and $p_2$_ are two odd prime numbers such that $p_1 > p_2$, then $\text{p}^2_1-\text{p}^2_2$ is:

✓
An even number.
B
An odd number.
C
An odd prime number.
D
A prime number.

Answer

Correct option: A.

An even number.

Let the two odd prime numbers $p_1$ and $p_2$ be $5$ and $3$.
Then,
$\text{p}^2_1=5^2$
$=25$
And
$\text{p}^2_2=3^2$
$=9$
Thus,
$\text{p}^2_1-\text{p}^2_2=25-9$
$=16$
16 is even number.
Take another example, with $p_1$_ and $p_2$_ be $11$ and $7$.
Then,
$\text{p}^2_1=11^2$
$=121$
And
$\text{p}^2_2=7^2$
$=49$
Thus,
$\text{p}^2_1-\text{p}^2_2=121-49$
$=72$
72 is even number.
Thus, we can say that $\text{p}^2_1-\text{p}^2_2$ is even number
In general the square of odd prime number is odd.
Hence the difference of square of two prime numbers is odd
Hence the correct choice is $(a).$

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MCQ 651 Mark

$n^2 - 1$ 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 even 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: $n =$ 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 661 Mark

The smallest number by which $\sqrt{27}$ should be multiplied so as to get a rational number is

A
$\sqrt{27}$
B
$3\sqrt{3}$
✓
$\sqrt{3}$
D
$3$

Answer

Correct option: C.

$\sqrt{3}$

$\sqrt{27}=\sqrt{3\times3\times3}$
$=3\sqrt{3}$
Out of the given choices $\sqrt{3}$ is the only smallest number by which if we multiply $\sqrt{27}$ we get a rational number.
Hence, the correct choice is $(c).$

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Maths M.C.Q (1 Marks)