Class 10 Maths: Real Numbers – Extra Questions with Answers

Ans: B

Solution: Find the greatest common divisor (GCD) of 288 and 168.

288 = 2^5 * 3^2
168 = 2^3 * 3 * 7
GCD(288, 168) = 2^3 * 3 = 24

Answer: 24

Ans: C

Solution: The prime factorization of $q$ is $q$. The prime factorization of $2q$ is $2 \cdot q$. The LCM of $q$ and $2q$ is $2 \cdot q$.
Answer: $2q$

Ans: A

Solution:
18 = 2 x 3^2
24 = 2^3 x 3
36 = 2^2 x 3^2

HCF(18, 24, 36) = 2 x 3 = 6
LCM(18, 24, 36) = 2^3 x 3^2 = 72
Answer: HCF = 6, LCM = 72

Ans: B

Solution:
$\frac{33}{2^4 \times 5^3} = \frac{33 \times 5}{2^4 \times 5^3 \times 5} = \frac{33 \times 5}{2^4 \times 5^4} = \frac{165}{10^4} = \frac{165}{10000} = 0.0165$
Answer: 0.0165

Ans: B

Solution:
1. Convert the dimensions to cm: 16 m 16 cm = 1616 cm, 8 m 8 cm = 808 cm.
2. Find the greatest common divisor (GCD) of 1616 and 808, which is the side length of the largest square tile. GCD(1616, 808) = 808.
3. Calculate the number of tiles needed: (1616 / 808) * (808 / 808) = 2 * 1 = 2

Answer: 2

Ans: D

Solution:
The decimal 12.345 can be written as $\frac{12345}{1000}$.
We simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 5.
$\frac{12345}{1000} = \frac{12345 \div 5}{1000 \div 5} = \frac{2469}{200}$.
Now we check if the fraction can be simplified further. The factors of 200 are $2^3 \cdot 5^2$. Since 2469 is not divisible by 2 or 5, we have $\frac{p}{q} = \frac{2469}{200}$, where $p=2469$ and $q=200$.
The prime factors of $q=200$ are 2 and 5.

Answer: 2, 5

Ans: C

Solution:
From the given conditions, we know that for any two numbers $a$ and $b$, we have $LCM(a, b) \times HCF(a, b) = a \times b$.
Let $x$ be the unknown number.
Using the given information, we have $LCM(x, 24) = 72$ and $HCF(x, 24) = 3$.
Therefore, $x \times 24 = LCM(x, 24) \times HCF(x, 24)$.
$x \times 24 = 72 \times 3$
$24x = 216$
$x = \frac{216}{24}$
$x = 9$

Also, since $HCF(x, 24) = 3$, then $x$ must be divisible by $3$.
$x = 9$ since $9 = 3 \times 3$.
$HCF(9, 24) = 3$ and $LCM(9, 24) = 72$.
Answer: 9

Ans: B

Solution:
Let the numbers be 2x, 3x, and 4x. Since their GCD is 8, x = 8. The numbers are 16, 24, and 32. The largest number is 32. The square root of 32 is $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$.
Since we are told the GCD is 8, then the numbers are $2(8)=16, 3(8)=24, 4(8)=32$. The largest number is 32. $\sqrt{32} = 4\sqrt{2}$.

Answer: $4\sqrt{2}$

Ans: B

Solution:
120 = 2³ × 3 × 5
216 = 2³ × 3³
HCF(120, 216) = 2³ × 3 = 24
LCM(120, 216) = 2³ × 3³ × 5 = 1080
HCF × LCM = 24 × 1080 = 25920
120 × 216 = 25920

Answer: HCF = 24, LCM = 1080, Product Verification: True

Ans: C

Solution:
Let the number be $x$.
Then we have:
$x \equiv 3 \pmod{10}$
$x \equiv 5 \pmod{12}$
$x \equiv 8 \pmod{15}$
We can rewrite the congruences as:
$x = 10a + 3$
$x = 12b + 5$
$x = 15c + 8$
From the first congruence, $x = 10a+3$. Substituting into the second congruence,
$10a+3 \equiv 5 \pmod{12}$
$10a \equiv 2 \pmod{12}$
$5a \equiv 1 \pmod{6}$
Since $5 \cdot 5 \equiv 1 \pmod{6}$, multiply by 5,
$25a \equiv 5 \pmod{6}$
$a \equiv 5 \pmod{6}$
So $a = 6k+5$ for some integer $k$.
Then $x = 10(6k+5)+3 = 60k+50+3 = 60k+53$.
Now, substitute into the third congruence,
$60k+53 \equiv 8 \pmod{15}$
$60k \equiv 8-53 \pmod{15}$
$60k \equiv -45 \pmod{15}$
$0 \equiv 0 \pmod{15}$
This doesn’t give any additional constraint.
So we can take $k=0$.
Then $x = 60(0)+53 = 53$.
Check:
$53 \div 10 = 5$ remainder $3$
$53 \div 12 = 4$ remainder $5$
$53 \div 15 = 3$ remainder $8$

Answer: 53