Real Numbers: Class 10 PYQs 2018–2026 with solutions

Answer: (C)

A is true ($\sqrt{3}+\sqrt{5}$ is irrational). R is false: $\sqrt{2}$ and $-\sqrt{2}$ are irrational but their sum is 0, which is rational.

Answer: (B)

$36m^2=18m\times 2m$, so $18m$ divides $36m^2$, and HCF $=18m$: A is true. R is a true general fact, but it does not explain why the HCF here equals $18m$.

Answer: (C) 98

$98=2\times 7^2$, $28=2^2\times 7$. HCF $m=2\times 7=14$, LCM $n=2^2\times 7^2=196$.

$n-7m=196-98=98$.

Answer: (C) any odd number

$(-1{)}^8=1$, so $(-1{)}^n=-1$, which happens exactly when $n$ is odd.

Answer: (D) 1.857142

$\sqrt{3}\approx 1.732$ and $\sqrt{5}\approx 2.236$. 1.857142 is a terminating decimal (rational) and lies between them. (A) is below $\sqrt{3}$ and non-repeating, (B) is above $\sqrt{5}$, and $\pi$ is irrational and too large.

Answer: 400 (none of the options). The board gave full marks to every candidate for this question.

LCM$(16,20,50)=2^4\times 5^2=400=20^2$, which is already a perfect square, so the least such number is 400.

(Among the options, only 3600 $=60^2$ is a perfect square divisible by all three, but it is not the least.)

Answer: (A) 5

$4004=2^2\times 7\times 11\times 13$. Sum of exponents $=2+1+1+1=5$.

Answer: (D) 10

$40=2^3\times 5$, $110=2\times 5\times 11$, $360=2^3\times 3^2\times 5$. HCF $=2\times 5=10$.

Answer: (A) 1650

HCF × LCM = product of the numbers: $40\times 252k=2520\times 6600$

$$k=\frac{2520\times 6600}{40\times 252}=10\times 165=1650$$

Answer: (B) 13

LCM $=\frac{65\times 104}{13}=520=40x\Rightarrow x=13$.

Suppose $2+3\sqrt{5}$ is rational. Then $2+3\sqrt{5}=\frac{a}{b}$, where $a,b$ are integers and $b\ne 0$.

$$3\sqrt{5}=\frac{a}{b}-2=\frac{a-2b}{b}\Rightarrow \sqrt{5}=\frac{a-2b}{3b}$$

Since $a$, $b$ are integers, $\frac{a-2b}{3b}$ is rational, so $\sqrt{5}$ would be rational.

This contradicts the fact that $\sqrt{5}$ is irrational. Hence $2+3\sqrt{5}$ is irrational.

$210=2\times 3\times 5\times 7$ and $55=5\times 11$, so HCF $=5$.

$$5=210\times 5+55m\Rightarrow 55m=5-1050=-1045\Rightarrow m=-19$$

Suppose $2-5\sqrt{3}$ is rational, say $2-5\sqrt{3}=\frac{a}{b}$, where $a,b$ are integers, $b\ne 0$.

$$5\sqrt{3}=2-\frac{a}{b}=\frac{2b-a}{b}\Rightarrow \sqrt{3}=\frac{2b-a}{5b}$$

The right side is rational (a ratio of integers), so $\sqrt{3}$ would be rational, which contradicts the given fact.

Hence $2-5\sqrt{3}$ is irrational.

A number ends with 0 only if it is divisible by 10, i.e. its prime factorisation contains both 2 and 5.

$(15{)}^n=(3\times 5{)}^n=3^n\times 5^n$. It has no factor 2.

By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), $15^n$ can never have 2 as a factor. So $15^n$ cannot end with the digit 0.

They toll together again after LCM(9, 12, 15) minutes.

$9=3^2$, $12=2^2\times 3$, $15=3\times 5$. LCM $=2^2\times 3^2\times 5=180$ minutes.

They will toll together after 180 minutes = 3 hours.

$48=2^4\times 3$, $60=2^2\times 3\times 5$, $65=5\times 13$.

LCM $=2^4\times 3\times 5\times 13=3120$.

The number must be a multiple of 3120: $3120\times 3=9360$ and $3120\times 4=12480>10000$.

Greatest such number $=9360$.

The side of the largest square is HCF(156, 216).

$156=2^2\times 3\times 13$, $216=2^3\times 3^3$. HCF $=2^2\times 3=12$.

Side of each square = 12 cm.

Number of squares $=\frac{156}{12}\times \frac{216}{12}=13\times 18=$ 234.

First, $\sqrt{5}$ is irrational. Suppose $\sqrt{5}=\frac{a}{b}$ with $a,b$ coprime integers, $b\ne 0$. Then $a^2=5b^2$, so 5 divides $a^2$ and hence $a$. Write $a=5c$: $25c^2=5b^2\Rightarrow b^2=5c^2$, so 5 divides $b$ too. This contradicts $a,b$ being coprime, so $\sqrt{5}$ is irrational.

Now suppose $\frac{1}{\sqrt{5}}$ is rational, $\frac{1}{\sqrt{5}}=\frac{p}{q}$ with integers $p,q$, $p\ne 0$, $q\ne 0$. Then $\sqrt{5}=\frac{q}{p}$, which is rational. This contradicts the irrationality of $\sqrt{5}$.

Hence $\frac{1}{\sqrt{5}}$ is irrational.

Suppose $\sqrt{3}$ is rational: $\sqrt{3}=\frac{a}{b}$, where $a,b$ are coprime integers and $b\ne 0$.

Squaring: $3b^2=a^2$. So 3 divides $a^2$, and therefore 3 divides $a$ (3 is prime).

Let $a=3c$. Then $3b^2=9c^2\Rightarrow b^2=3c^2$, so 3 divides $b^2$ and hence $b$.

Now 3 divides both $a$ and $b$, which contradicts that they are coprime.

Hence $\sqrt{3}$ is irrational.

Suppose $\sqrt{5}$ is rational: $\sqrt{5}=\frac{a}{b}$, with $a,b$ coprime integers, $b\ne 0$.

Squaring: $a^2=5b^2$. So 5 divides $a^2$, hence 5 divides $a$ (5 is prime). Let $a=5c$.

Then $25c^2=5b^2\Rightarrow b^2=5c^2$, so 5 divides $b^2$ and hence $b$.

So 5 is a common factor of $a$ and $b$, contradicting that they are coprime.

Therefore $\sqrt{5}$ is irrational.

$12^n=(2^2\times 3{)}^n=2^{2n}\times 3^n$.

A number ending in 0 or 5 is divisible by 5, so its prime factorisation must contain 5.

By the Fundamental Theorem of Arithmetic the factorisation of $12^n$ is unique and contains only 2 and 3. So $12^n$ cannot end with 0 or 5.

Suppose $\sqrt{2}+\sqrt{5}=m$, where $m$ is rational. Squaring: $7+2\sqrt{10}=m^2\Rightarrow \sqrt{10}=\frac{m^2-7}{2}$.

The right side is rational, so $\sqrt{10}$ would be rational.

But $\sqrt{10}$ is irrational: if $\sqrt{10}=\frac{a}{b}$ in lowest terms, $a^2=10b^2$, so 2 and 5 divide $a$; writing $a=10c$ gives $b^2=10c^2$, so 2 and 5 divide $b$ too, a contradiction.

Hence $\sqrt{2}+\sqrt{5}$ is irrational.