Exercise 1D RS Aggarwal Class 9 Number Systems Best Solution

Exercise 1D RS Aggarwal Class 9 contains a total of eight questions. The questions are based on the following topics:

Real Numbers
Completeness Property
Density property
Addition Properties of Real Numbers
Multiplication Properties of Real Numbers
Existence of Multiplicative inverse
More Results on Real Numbers

Exercise 1D RS Aggarwal Class 9 Solutions

The first question of Exercise 1D RS Aggarwal Class 9 is based on addition of irrational numbers.

1. Add.

(i) $(2 \sqrt{3} - 5 \sqrt{2})$ and $(\sqrt{3} + 2 \sqrt{2})$

Answer

We have
$(2 \sqrt{3} - 5 \sqrt{2})$ + $(\sqrt{3} + 2 \sqrt{2})$
= $2 \sqrt{3} - 5 \sqrt{2}$ + $\sqrt{3} + 2 \sqrt{2}$
= $2 \sqrt{3} + \sqrt{3} - 5 \sqrt{2} + 2 \sqrt{2}$
= $\sqrt{3} (2 + 1) - \sqrt{2} (5 - 2) = 3 \sqrt{3} - 3 \sqrt{2}$
= $3 (\sqrt{3} - \sqrt{2})$ .

(ii) $(2 \sqrt{2} + 5 \sqrt{3} - 7 \sqrt{5})$ and $(3 \sqrt{3} - \sqrt{2} + \sqrt{5})$

Answer

We have
$(2 \sqrt{2} + 5 \sqrt{3} - 7 \sqrt{5})$ + $(3 \sqrt{3} - \sqrt{2} + \sqrt{5})$
= $2 \sqrt{2} + 5 \sqrt{3} - 7 \sqrt{5}$ + $3 \sqrt{3} - \sqrt{2} + \sqrt{5}$
= $2 \sqrt{2} - \sqrt{2} + 5 \sqrt{3} + 3 \sqrt{3} - 7 \sqrt{5} + \sqrt{5}$
= $\sqrt{2} (2 - 1) + \sqrt{3} (5 + 3) - \sqrt{5} (7 - 1)$
= $\sqrt{2} + 8 \sqrt{3} - 6 \sqrt{5}$ .

(iii) $(\frac{2}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + 6 \sqrt{11})$ and $(\frac{1}{3} \sqrt{7} + \frac{3}{2} \sqrt{2} - \sqrt{11})$

Answer

We have
$(\frac{2}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + 6 \sqrt{11})$ + $(\frac{1}{3} \sqrt{7} + \frac{3}{2} \sqrt{2} - \sqrt{11})$
= $\frac{2}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + 6 \sqrt{11}$ + $\frac{1}{3} \sqrt{7} + \frac{3}{2} \sqrt{2} - \sqrt{11}$
= $\frac{2}{3} \sqrt{7} + \frac{1}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + \frac{3}{2} \sqrt{2} + 6 \sqrt{11} - \sqrt{11}$
= $\sqrt{7} (\frac{2}{3} + \frac{1}{3}) - \sqrt{2} (\frac{1}{2} - \frac{3}{2}) + \sqrt{11} (6 - 1))$
= $\sqrt{7} (\frac{2 + 1}{3}) - \sqrt{2} (\frac{1 - 3}{2}) + \sqrt{11} (6 - 1))$
= $\sqrt{7} (\frac{3}{3}) - \sqrt{2} (\frac{- 2}{2}) + 5 \sqrt{11}$
= $\sqrt{7} + \sqrt{2} + 5 \sqrt{11}$

The second question of Exercise 1D RS Aggarwal Class 9 is based on operations of multiplication on irrational numbers.

2. Multiply.

(i) $3 \sqrt{5}$ by $2 \sqrt{5}$

Answer

We have
$3 \sqrt{5} \times 2 \sqrt{5}$
= $3 \times 2 \times \sqrt{5} \times \sqrt{5}$
= $6 \times {(\sqrt{5})}^{2}$
= $6 \times 5$
= $30$

(ii) $6 \sqrt{15}$ by $4 \sqrt{3}$

Answer

We have
$6 \sqrt{15} \times 4 \sqrt{3}$
= $6 \times 4 \times \sqrt{15 \times 3}$
= $24 \sqrt{5 \times 3 \times 3}$
= $24 \times 3 \sqrt{5}$
= $72 \sqrt{5}$

(iii) $2 \sqrt{6}$ by $3 \sqrt{3}$

Answer

We have
$2 \sqrt{6} \times 3 \sqrt{3}$
= $2 \times 3 \times \sqrt{6 \times 3}$
= $6 \sqrt{2 \times 3 \times 3}$
= $6 \times 3 \sqrt{2}$
= $18 \sqrt{2}$

(iv) $3 \sqrt{8}$ by $3 \sqrt{2}$

Answer

We have
$3 \sqrt{8} \times 3 \sqrt{2}$
= $3 \times 3 \times \sqrt{8 \times 2}$
= $9 \times \sqrt{2 \times 2 \times 2 \times 2}$
= $9 \times < 2 \times 2$
= $36$

(v) $\sqrt{10}$ by $\sqrt{40}$

Answer

We have
$\sqrt{10} \times \sqrt{40}$
= $\sqrt{10 \times 40}$
= $\sqrt{2 \times 5 \times 2 \times 2 \times 2 \times 5}$
= $2 \times 2 \times 5$
= $20$

(vi) $3 \sqrt{28}$ by $2 \sqrt{7}$

Answer

We have
$3 \sqrt{28} \times 2 \sqrt{7}$
= $3 \times 2 \sqrt{28 \times 7}$
= $6 \sqrt{2 \times 2 \times 7 \times 7}$
= $6 \times 2 \times 7$
= $84$

The third question of Exercise 1D RS Aggarwal Class 9 is based on operations of division on irrational numbers.

3. Divide

(i) $16 \sqrt{6}$ by $4 \sqrt{2}$

Answer

We have
$16 \sqrt{6} \div 4 \sqrt{2}$ = $\frac{16 \sqrt{6}}{4 \sqrt{2}}$ = $4 \sqrt{\frac{6}{2}}$ = $4 \sqrt{3}$

(ii) $12 \sqrt{15}$ by $4 \sqrt{3}$

Answer

We have
$12 \sqrt{15} \div 4 \sqrt{3}$ = $\frac{12 \sqrt{15}}{4 \sqrt{3}}$ = $3 \sqrt{\frac{15}{3}}$ = $3 \sqrt{5}$

(iii) $18 \sqrt{21}$ by $6 \sqrt{7}$

Answer

We have
$18 \sqrt{21} \div 6 \sqrt{7}$ = $\frac{18 \sqrt{21}}{6 \sqrt{7}}$ = $3 \sqrt{\frac{21}{7}}$ = $3 \sqrt{3}$

The fourth question of Exercise 1D RS Aggarwal Class 9 is based on simplification of irrational numbers.

4. Simplify

(i) $(3 - \sqrt{11}) (3 + \sqrt{11})$

Answer

We have
$(3 - \sqrt{11}) (3 + \sqrt{11})$
= $3^{2} - {(\sqrt{11})}^{2}$ . . . [ $∵ (a - b) (a + b) = a^{2} - b^{2}]$
= $9 - 11$
= −4.

(ii) $(- 3 + \sqrt{5}) (- 3 - \sqrt{5})$

Answer

We have
$(- 3 + \sqrt{5}) (- 3 - \sqrt{5})$
= ${(−3)}^{2} - {(\sqrt{5})}^{2}$ . . . [ $∵ (a - b) (a + b) = a^{2} - b^{2}]$
= $9 - 5$
= 4.

(iii) ${(3 - \sqrt{3})}^{2}$

Answer

We have
${(3 - \sqrt{3})}^{2}$
= $3^{2} - 2 \times 3 \times \sqrt{3} + {(\sqrt{3})}^{2}$ ... $[∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}]$
= $9 - 6 \sqrt{3} + 3$
= $12 - 6 \sqrt{3}$ .

(iv) ${(\sqrt{5} - \sqrt{3})}^{2}$

Answer

We have
${(\sqrt{5} - \sqrt{3})}^{2}$
= ${(\sqrt{5})}^{2} - 2 \times \sqrt{5} \times \sqrt{3} + {(\sqrt{3})}^{2}$ ... $[∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}]$
= $25 - 2 \sqrt{15} + 3$
= $28 - 2 \sqrt{15}$ .

(v) $(5 + \sqrt{7}) (2 + \sqrt{5})$

Answer

We have
$(5 + \sqrt{7}) (2 + \sqrt{5})$
= $5 (2 + \sqrt{5}) + \sqrt{7} (2 + \sqrt{5})$
= $10 + 5 \sqrt{5} + 2 \sqrt{7} + \sqrt{7 \times 5}$
= $10 + 5 \sqrt{5} + 2 \sqrt{7} + \sqrt{35}$ .

(vi) $(\sqrt{5} - \sqrt{2}) (\sqrt{2} - \sqrt{3})$

Answer

We have
$(\sqrt{5} - \sqrt{2}) (\sqrt{2} - \sqrt{3})$
= $\sqrt{5} (\sqrt{2} - \sqrt{3}) - \sqrt{2} (\sqrt{2} - \sqrt{3})$
= $\sqrt{5 \times 2} - \sqrt{5 \times 3} - {(\sqrt{2})}^{2} + < \sqrt{2 \times 3}$
= $\sqrt{10} - \sqrt{15} - 2 + \sqrt{6}$ .

The fifth question of Exercise 1D RS Aggarwal Class 9 is based on simplification of irrational numbers.

5. Simplify $(3 + \sqrt{3}) (2 + \sqrt{2})$ .

Answer

We have
$(3 + \sqrt{3}) (2 + \sqrt{2})$
= $3 (2 + \sqrt{2}) + \sqrt{3} (2 + \sqrt{2})$
= $6 + 3 \sqrt{2} + 2 \sqrt{2} + \sqrt{3 \times 2}$
= $6 + 5 \sqrt{2} + \sqrt{6}$ .

The sixth question in Exercise 1D RS Aggarwal Class 9 pertains to the determination of whether a given number is rational or irrational.

6. Examine whether the following numbers are rational or irrational:

(i) $(5 - \sqrt{5}) (5 + \sqrt{5})$

Answer

$∵ (5 - \sqrt{5}) (5 + \sqrt{5})$ = $5^{2} - {(\sqrt{5})}^{2}$ = $25 - 5$ = $20$
$∴ (5 - \sqrt{5}) (5 + \sqrt{5})$ is a rational number.

(ii) ${(\sqrt{3} + 2)}^{2}$

Answer

We have
${(\sqrt{3} + 2)}^{2}$
= ${(\sqrt{3})}^{2} + 2 \times \sqrt{3} \times 2 + 2^{2}$
= $3 + 4 \sqrt{3} + 4$
= $7 + 4 \sqrt{3}$ .
Since $7 + 4 \sqrt{3}$ is an irrational number.
So, ${(\sqrt{3} + 2)}^{2}$ is an irrational number.

(iii) $\frac{2 \sqrt{13}}{3 \sqrt{52} - 4 \sqrt{117}}$

Answer

We have
$\frac{2 \sqrt{13}}{3 \sqrt{52} - 4 \sqrt{117}}$
= $\frac{2 \sqrt{13}}{3 \sqrt{2 \times 2 \times 13} - 4 \sqrt{3 \times 3 \times 13}}$
= $\frac{2 \sqrt{13}}{3 \times 2 \sqrt{13} - 4 \times 3 \sqrt{13}}$
= $\frac{2 \sqrt{13}}{6 \sqrt{13} - 12 \sqrt{13}}$
= $\frac{2 \sqrt{13}}{\sqrt{13} (6 - 12)}$
= $\frac{2}{- 6}$
= $\frac{1}{- 3}$ .
Hence, $\frac{2 \sqrt{13}}{3 \sqrt{52} - 4 \sqrt{117}}$ is a rational number.

(iv) $\sqrt{8} + 4 \sqrt{32} - 6 \sqrt{2}$

Answer

We have
$\sqrt{8} + 4 \sqrt{32} - 6 \sqrt{2}$
= $\sqrt{2 \times 2 \times 2} + 4 \sqrt{2 \times 2 \times 2 \times 2 \times 2} - 6 \sqrt{2}$
= $2 \sqrt{2} + 4 \times 2 \times 2 \times \sqrt{2} - 6 \sqrt{2}$
= $2 \sqrt{2} + 16 \sqrt{2} - 6 \sqrt{2}$
= $12 \sqrt{2}$ .
Hence, $\sqrt{8} + 4 \sqrt{32} - 6 \sqrt{2}$ is an irrational number.

The seventh question in Exercise 1D RS Aggarwal Class 9 addresses a word problem that involves irrational numbers.

7. On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by $(5 + \sqrt{11}) (5 - \sqrt{11})$ .

(i) Find the number of chocolates distributed by her.

Answer

The number of chocolates distributed by Reema is
$(5 + \sqrt{11}) (5 - \sqrt{11})$ .
$∵ (5 + \sqrt{11}) (5 - \sqrt{11})$ = $5^{2} - {(\sqrt{11})}^{2}$ = $25 - 11$ = $14$ .
Hence, Reema distributed 14 chocolates.

(ii) Write the moral values depicted here by Reema.

Answer

The moral values depicted here by Reema is

To help the poor and needy and to make the deprived children happy.

The eighth question in Exercise 1D RS Aggarwal Class 9 pertains to the simplification of irrational numbers.

8. Simplify.

(i) $3 \sqrt{45} - \sqrt{125} + \sqrt{200} - \sqrt{50}$

Answer

We have
$3 \sqrt{45} - \sqrt{125} + \sqrt{200} - \sqrt{50}$
= $3 \sqrt{3 \times 3 \times 5} - \sqrt{5 \times 5 \times 5} + \sqrt{2 \times 2 \times 2 \times 5 \times 5} - \sqrt{2 \times 5 \times 5}$
= $3 \times 3 \sqrt{5} - 5 \sqrt{5} + 2 \times 5 \sqrt{2} - 5 \sqrt{2}$
= $9 \sqrt{5} - 5 \sqrt{5} + 10 \sqrt{2} - 5 \sqrt{2}$
= $4 \sqrt{5} + 5 \sqrt{2}$ .

(ii) $\frac{2 \sqrt{30}}{\sqrt{6}} - \frac{3 \sqrt{140}}{\sqrt{28}} + \frac{\sqrt{55}}{\sqrt{99}}$

Answer

We have
$\frac{2 \sqrt{30}}{\sqrt{6}} - \frac{3 \sqrt{140}}{\sqrt{28}} + \frac{\sqrt{55}}{\sqrt{99}}$
= $2 \sqrt{\frac{30}{6}} - 3 \sqrt{\frac{140}{28}} + \sqrt{\frac{55}{99}}$
= $2 \sqrt{5} - 3 \sqrt{5} + \sqrt{\frac{5}{9}}$
= $- \sqrt{5} + \frac{\sqrt{5}}{\sqrt{9}}$
= $- \sqrt{5} + \frac{\sqrt{5}}{3}$
= $- \sqrt{5} (1 - \frac{1}{3})$
= $- \frac{2}{3} \sqrt{5}$ .

(iii) $\sqrt{72} + \sqrt{800} - \sqrt{18}$

Answer

We have
$\sqrt{72} + \sqrt{800} - \sqrt{18}$
= $\sqrt{2 \times 2 \times 2 \times 3 \times 3} + \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5} - \sqrt{3 \times 3 \times 2}$
= $6 \sqrt{2} + 20 \sqrt{2} - 3 \sqrt{2}$
= $\sqrt{2} (6 + 20 - 3)$
= $23 \sqrt{2}$

Exercise 1D RS Aggarwal Class 9 Solutions

1. Add.

(i) (23−52) and (3+22)

Answer

(ii) (22+53−75) and (33−2+5)

Answer

(iii) (237−122+611) and (137+322−11)

Answer

2. Multiply.

(i) 35 by 25

Answer

(ii) 615 by 43

Answer

(iii) 26 by 33

Answer

(iv) 38 by 32

Answer

(v) 10 by 40

Answer

(vi) 328 by 27

Answer

3. Divide

(i) 166 by 42

Answer

(ii) 1215 by 43

Answer

(iii) 1821 by 67

Answer

4. Simplify

(i) (3−11) (3+11)

Answer

(ii) (−3+5) (−3−5)

Answer

(iii) (3−3) 2

Answer

(iv) (5−3)2

Answer

(v) (5+7) (2+5)

Answer

(vi) (5−2) (2−3)

Answer

5. Simplify (3+3) (2+2).

Answer

6. Examine whether the following numbers are rational or irrational:

(i) (5−5) (5+5)

Answer

(ii) (3+2)2

Answer

(iii) 213352−4117

Answer

(iv) 8+432−62

Answer

7. On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by (5+11) (5−11).

(i) Find the number of chocolates distributed by her.

Answer

(ii) Write the moral values depicted here by Reema.

Answer

8. Simplify.

(i) 345−125+200−50

Answer

(ii) 2306−314028+5599

Answer

(iii) 72+800−18

Answer

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(i) $(2 \sqrt{3} - 5 \sqrt{2})$ and $(\sqrt{3} + 2 \sqrt{2})$

(ii) $(2 \sqrt{2} + 5 \sqrt{3} - 7 \sqrt{5})$ and $(3 \sqrt{3} - \sqrt{2} + \sqrt{5})$

(iii) $(\frac{2}{3} \sqrt{7} - \frac{1}{2} \sqrt{2} + 6 \sqrt{11})$ and $(\frac{1}{3} \sqrt{7} + \frac{3}{2} \sqrt{2} - \sqrt{11})$

(i) $3 \sqrt{5}$ by $2 \sqrt{5}$

(ii) $6 \sqrt{15}$ by $4 \sqrt{3}$

(iii) $2 \sqrt{6}$ by $3 \sqrt{3}$

(iv) $3 \sqrt{8}$ by $3 \sqrt{2}$

(v) $\sqrt{10}$ by $\sqrt{40}$

(vi) $3 \sqrt{28}$ by $2 \sqrt{7}$

(i) $16 \sqrt{6}$ by $4 \sqrt{2}$

(ii) $12 \sqrt{15}$ by $4 \sqrt{3}$

(iii) $18 \sqrt{21}$ by $6 \sqrt{7}$

(i) $(3 - \sqrt{11}) (3 + \sqrt{11})$

(ii) $(- 3 + \sqrt{5}) (- 3 - \sqrt{5})$

(iii) ${(3 - \sqrt{3})}^{2}$

(iv) ${(\sqrt{5} - \sqrt{3})}^{2}$

(v) $(5 + \sqrt{7}) (2 + \sqrt{5})$

(vi) $(\sqrt{5} - \sqrt{2}) (\sqrt{2} - \sqrt{3})$

5. Simplify $(3 + \sqrt{3}) (2 + \sqrt{2})$ .

(i) $(5 - \sqrt{5}) (5 + \sqrt{5})$

(ii) ${(\sqrt{3} + 2)}^{2}$

(iii) $\frac{2 \sqrt{13}}{3 \sqrt{52} - 4 \sqrt{117}}$

(iv) $\sqrt{8} + 4 \sqrt{32} - 6 \sqrt{2}$

7. On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by $(5 + \sqrt{11}) (5 - \sqrt{11})$ .

(i) $3 \sqrt{45} - \sqrt{125} + \sqrt{200} - \sqrt{50}$

(ii) $\frac{2 \sqrt{30}}{\sqrt{6}} - \frac{3 \sqrt{140}}{\sqrt{28}} + \frac{\sqrt{55}}{\sqrt{99}}$

(iii) $\sqrt{72} + \sqrt{800} - \sqrt{18}$