Real Numbers — Important Questions

Answer: (c) $4$.

Factorising $196=2\times98=2^2\times7^2$. The exponents are $2$ and $2$, so their sum is $2+2=4$.

Answer: (b) $150$.

Using $\text{HCF}\times\text{LCM}=a\times b$:
$\text{LCM}=\frac{a\times b}{\text{HCF}}=\frac{1800}{12}=150.$

Answer: (b) $3$.

The denominator is $2^3\times5^2$, of the form $2^n5^m$ with $n=3,\ m=2$. The expansion terminates after $\max(n,m)=3$ places, since $\dfrac{23}{200}=0.115$.

Answer: (b) $500$.

The HCF of two numbers must always be a factor of their LCM. Here $600,\ 400,\ 200$ all divide $1200$, but $1200\div500=2.4$ is not an integer, so $500$ cannot be the HCF.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Since $2$ is prime and the square root of every prime is irrational, $\sqrt2$ is irrational — so R correctly explains A.

$96=2^5\times3$ and $404=2^2\times101$.

$\text{HCF}=2^2=4,\qquad \text{LCM}=2^5\times3\times101=9696.$

Check: $\text{HCF}\times\text{LCM}=4\times9696=38784=96\times404.$ ✓

$26=2\times13$ and $91=7\times13$.

$\text{HCF}=13,\qquad \text{LCM}=2\times7\times13=182.$

Verify: $\text{HCF}\times\text{LCM}=13\times182=2366$ and $26\times91=2366.$ ✓

Assume, to the contrary, that $\sqrt5$ is rational. Then $\sqrt5=\dfrac{a}{b}$ where $a,b$ are integers with no common factor other than $1$ and $b\neq0$.

Squaring: $5=\dfrac{a^2}{b^2}\Rightarrow a^2=5b^2$. So $5$ divides $a^2$, hence $5$ divides $a$. Write $a=5c$.

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

Now $5$ divides both $a$ and $b$, contradicting that they have no common factor. Therefore $\sqrt5$ is irrational.

Subtract the remainder first: $245-5=240$ and $1029-5=1024$.

The required number is $\text{HCF}(240,1024)$.

$240=2^4\times3\times5$ and $1024=2^{10}$, so $\text{HCF}=2^4=16$.

The largest such number is $\boxed{16}$.

They toll together again after $\text{LCM}(9,12,15)$ minutes.

$9=3^2,\ 12=2^2\times3,\ 15=3\times5$, so
$\text{LCM}=2^2\times3^2\times5=180\text{ minutes}=3\text{ hours}.$

So they next toll together at $8{:}00+3=\mathbf{11{:}00}$ a.m.

Assume, to the contrary, that $3+2\sqrt5$ is rational. Then $3+2\sqrt5=\dfrac{p}{q}$ for integers $p,q\ (q\neq0)$.

Rearranging:
$2\sqrt5=\frac{p}{q}-3=\frac{p-3q}{q}\Rightarrow \sqrt5=\frac{p-3q}{2q}.$

The right-hand side is a ratio of integers, hence rational, so $\sqrt5$ would be rational.

This contradicts the given fact that $\sqrt5$ is irrational. Therefore $3+2\sqrt5$ is irrational.

For the least area, each stack must hold the greatest possible equal number, which is $\text{HCF}(420,130)$.

$420=2^2\times3\times5\times7$ and $130=2\times5\times13$, so
$\text{HCF}=2\times5=10.$

So $10$ barfis go in each stack.

Kaju stacks $=\dfrac{420}{10}=42$, and badam stacks $=\dfrac{130}{10}=13$.

The maximum number of identical kits is $\text{HCF}(144,90)$.

$144=2^4\times3^2$ and $90=2\times3^2\times5$, so $\text{HCF}=2\times3^2=18$.

(i) Maximum kits $=18$.
(ii) Pens per kit $=\dfrac{144}{18}=8$.
(iii) Pencils per kit $=\dfrac{90}{18}=5$.