Real numbers class 10 questions with solutions

Q 1 :
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

$a b$
$a^{2} b^{2}$
$a^{3} b^{2}$
$a^{3} b^{3}$

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4

(3) $a^{3} b^{2}$

Q 2 :
Assertion (A): $6^{n}$ never ends with the digit zero, where n is natural number.

Reason (R): Any number ends with digit zero, if its prime factor is of the form $2^{m} \times 5^{n}$ , where m, n are natural numbers.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true but Reason (R) is false.
Assertion (A) is false but Reason (R) is true.

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(1)

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

$6^{n} = {(2 \times 3)}^{n} = 2^{n} \times 3^{n}$ , Its prime factors do not contain 5 i.e., of the form $2^{m} \times 5^{n}$ ,

where m, n are natural numbers. Hence, $6^{n}$ never ends with the digit zero.

Q 3 :
Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

15
30
35
72

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(3)

$a = p^{3} q^{4} and b = p^{2} q^{3}$

$HCF (a, b) = p^{2} q^{3}$ ...(i)

and $LCM (a, b) = p^{3} q^{4}$ ...(ii)

But gives: $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s}$

From eq. (i), $p^{m} q^{n} = p^{2} q^{3}$

So, $m = 2$ and $n = 3$

From eq. (ii), $p^{r} q^{s} = p^{3} q^{4}$

So, $r = 3$ and $s = 4$

$∴$ $(m + n) (r + s) = (2 + 3) (3 + 4) = 35 .$

Q 4 :
If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

$2 a^{2} b^{2}$
$180 a^{2} b^{2}$
$12 a^{2} b^{2}$
$180 a^{3} b^{4}$

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(4)

Given, $p = 18 a^{2} b^{4}$ and $q = 20 a^{3} b^{2}$

$LCM (p, q) = LCM (18 a^{2} b^{4}, 20 a^{3} b^{2}) = 180 a^{3} b^{4}$

Q 5 :
Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340

Reason (R) : HCF is always a factor of LCM

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
Assertion (A) is true but reason (R) is false.
Assertion (A) is false but reason (R) is true.

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(2)

Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

Q 6 :
Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

$p^{8} q^{6}$
$p^{3} q^{2}$
$p^{5} q^{4}$
$p^{5} q^{2} + p^{3} q^{4}$

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(3) $p^{5} q^{4}$

Q 7 :
If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

1650
1600
165
1625

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(1)

Given, HCF = 40 and LCM = 252 × k

We know that, LCM × HCF = Product of two number

$\Rightarrow 40 \times 252 \times k = 2520 \times 6600$

$\Rightarrow k = \frac{2520 \times 6600}{40 \times 252}$

$\Rightarrow k = 1650$

Topic Question Set

Q 1 :
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

Q 2 :
Assertion (A): $6^{n}$ never ends with the digit zero, where n is natural number.

Reason (R): Any number ends with digit zero, if its prime factor is of the form $2^{m} \times 5^{n}$ , where m, n are natural numbers.

Q 3 :
Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

Q 4 :
If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

Q 5 :
Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340

Reason (R) : HCF is always a factor of LCM

Q 6 :
Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

Q 7 :
If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

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Topic Question Set

Q 1 : If two positive integers p and q can be expressed as p=ab2 and q=a3b; a,b being prime numbers, then LCM (p, q) is

Q 2 : Assertion (A): 6n never ends with the digit zero, where n is natural number. Reason (R): Any number ends with digit zero, if its prime factor is of the form 2m×5n, where m, n are natural numbers.

Q 3 : Let a and b be two positive integers such that a=p3q4 and b=p2q3, where p and q are prime numbers. If HCF(a,b)=pmqn and LCM(a,b)=prqs, then (m+n)(r+s)=

Q 4 : If two positive integers p and q can be expressed as p=18 a2b4 and q=20 a3b2, where a and b are prime numbers, then LCM (p, q) is :

Q 5 : Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340 Reason (R) : HCF is always a factor of LCM

Q 6 : Two positive integers m and n are expressed as m=p5q2 and n=p3q4 , where p and q are prime numbers. The LCM of m and n is :

Q 7 : If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is

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Q 1 :
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

Q 2 :
Assertion (A): $6^{n}$ never ends with the digit zero, where n is natural number.

Reason (R): Any number ends with digit zero, if its prime factor is of the form $2^{m} \times 5^{n}$ , where m, n are natural numbers.

Q 3 :
Let a and b be two positive integers such that $a = p^{3} q^{4} and b = p^{2} q^{3}$ , where p and q are prime numbers. If $HCF (a, b) = p^{m} q^{n}$ and $LCM (a, b) = p^{r} q^{s},$ then $(m + n) (r + s) =$

Q 4 :
If two positive integers p and q can be expressed as $p = 18 a^{2} b^{4} and q = 20 a^{3} b^{2},$ where a and b are prime numbers, then LCM (p, q) is :

Q 5 :
Assertion (A): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340

Reason (R) : HCF is always a factor of LCM

Q 6 :
Two positive integers m and n are expressed as $m = p^{5} q^{2}$ and $n = p^{3} q^{4}$ , where p and q are prime numbers. The LCM of m and n is :

Q 7 :
If the HCF(2520, 6600) = 40 and LCM(2520, 6600) = 252 × k, then the value of k is