Real Numbers — CBSE Class 10 Maths Important Questions

Answer: (c) $4$4.

Factorising $196=2\times98=2^2\times7^2$196=2×98=2²×7². The exponents are $2$2 and $2$2, so their sum is $2+2=4$2+2=4.

Answer: (b) $150$150.

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

Answer: (b) $3$3.

The denominator is $2^3\times5^2$2³×5², of the form $2^n5^m$2^n5^m with $n=3,\ m=2$n=3, m=2. The expansion terminates after $\max(n,m)=3$max (n,m)=3 places, since $\dfrac{23}{200}=0.115$23/200=0.115.

Answer: (b) $500$500.

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

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

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

$96=2^5\times3$96=2⁵×3 and $404=2^2\times101$404=2²×101.

$\text{HCF}=2^2=4,\qquad \text{LCM}=2^5\times3\times101=9696.$HCF=2²=4, LCM=2⁵×3×101=9696.

Check: $\text{HCF}\times\text{LCM}=4\times9696=38784=96\times404.$HCF× LCM=4×9696=38784=96×404. ✓

$26=2\times13$26=2×13 and $91=7\times13$91=7×13.

$\text{HCF}=13,\qquad \text{LCM}=2\times7\times13=182.$HCF=13, LCM=2×7×13=182.

Verify: $\text{HCF}\times\text{LCM}=13\times182=2366$HCF× LCM=13×182=2366 and $26\times91=2366.$26×91=2366. ✓

Assume, to the contrary, that $\sqrt5$√5 is rational. Then $\sqrt5=\dfrac{a}{b}$√5=a/b where $a,b$a,b are integers with no common factor other than $1$1 and $b\neq0$b≠0.

Squaring: $5=\dfrac{a^2}{b^2}\Rightarrow a^2=5b^2$5=a²/b²⇒ a²=5b². So $5$5 divides $a^2$a², hence $5$5 divides $a$a. Write $a=5c$a=5c.

Then $(5c)^2=5b^2\Rightarrow 25c^2=5b^2\Rightarrow b^2=5c^2$(5c)²=5b²⇒ 25c²=5b²⇒ b²=5c², so $5$5 divides $b^2$b², hence $5$5 divides $b$b.

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

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

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

$240=2^4\times3\times5$240=2⁴×3×5 and $1024=2^{10}$1024=2¹⁰, so $\text{HCF}=2^4=16$HCF=2⁴=16.

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

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

$9=3^2,\ 12=2^2\times3,\ 15=3\times5$9=3², 12=2²×3, 15=3×5, so
$\text{LCM}=2^2\times3^2\times5=180\text{ minutes}=3\text{ hours}.$LCM=2²×3²×5=180 minutes=3 hours.

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

Assume, to the contrary, that $3+2\sqrt5$3+2√5 is rational. Then $3+2\sqrt5=\dfrac{p}{q}$3+2√5=p/q for integers $p,q\ (q\neq0)$p,q (q≠0).

Rearranging:
$2\sqrt5=\frac{p}{q}-3=\frac{p-3q}{q}\Rightarrow \sqrt5=\frac{p-3q}{2q}.$2√5=p/q-3=(p-3q)/q⇒ √5=(p-3q)/2q.

The right-hand side is a ratio of integers, hence rational, so $\sqrt5$√5 would be rational.

This contradicts the given fact that $\sqrt5$√5 is irrational. Therefore $3+2\sqrt5$3+2√5 is irrational.

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

$420=2^2\times3\times5\times7$420=2²×3×5×7 and $130=2\times5\times13$130=2×5×13, so
$\text{HCF}=2\times5=10.$HCF=2×5=10.

So $10$10 barfis go in each stack.

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

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

$144=2^4\times3^2$144=2⁴×3² and $90=2\times3^2\times5$90=2×3²×5, so $\text{HCF}=2\times3^2=18$HCF=2×3²=18.

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