RD Sharma solution class 7 chapter 1 Integers Exercise 1.2

Exercise 1.2

Page-1.8

Question 1:

Divide:
(i) 102 by 17
(ii) −85 by 5
(iii) −161 by −23
(iv) 76 by −19
(v) 17654 by −17654
(vi) (−729) by (−27)
(vii) 21590 by −10
(viii) 0 by −135

Answer 1:

(i) 102  ÷ 17 = $\frac{\left|102\right|}{\left|17\right|} = \frac{102}{17} =$6
(ii) −85  ÷  5 = $-\frac{\left|-85\right|}{\left|5\right|} = -\frac{85}{5} =$ − 17
(iii) −161  ÷ ( − 23)  = $\frac{\left|-161\right|}{\left|-23\right|} =\frac{161}{23}=$ 7
(iv) 76   ÷  − 19 =  $-\frac{\left|76\right|}{\left|-19 \right|} = -\frac{76}{19} =$− 4
(v) 17654   ÷  (− 17654) =  $- \frac{\left|17654\right|}{\left|-17654\right|}= -\frac{17654}{17654}=$− 1
(vi) (−729)  ÷  (− 27) = $\frac{\left|-729\right|}{\left|-27\right|} = \frac{729}{27}=$  27
(vii) 21590   ÷  − 10 = $-\frac{\left|21590\right|}{\left|-10\right|} = -\frac{21590}{10}=$− 2159
(viii) 0 ​ ÷ ​(−135) = $-\frac{\left|0\right|}{\left|-135\right|}= -\frac{0}{135}= 0$

Question 2:

Fill in the blanks:
(i) 296 ÷ ... = −148
(ii) −88 ÷ ... = 11
(iii) 84 ÷ ... = 12
(iv) ....... ÷ −5 = 25
(v) ....... ÷ 156 = −2
(vi) ....... ÷ 567 = −1

Answer 2:

(i) 296  ÷ $-$148 = $-\frac{\left|296\right|}{\left|-148\right|}= -\frac{\left|296\right|}{\left|148\right|}=-\frac{296}{148}= -2$
∴ $296 ÷ (-2) = -148$

(ii) − 88  ÷ 11 = $-\frac{\left|-88\right|}{\left|11\right|}=-\frac{\left|88\right|}{\left|11\right|}=-\frac{88}{11}= -8$
∴ $-88 ÷ -8 = 11$

(iii) 84  ÷ 12 = $\frac{\left|84\right|}{\left|12 \right|}= \frac{84}{12}=7$
∴ $84÷7 = 12$

(iv) $25\times (-5) = -125$

∴ $-125÷-5 = 25$

(v) $156\times (-2) = -312$

∴ $-312÷156 = -2$

(vi) $567\times (-1) = -567$

∴ $-567 ÷ 567 = -1$

Question 3:

Which of the following statements are true?
(i) 0 ÷ 4 = 0
(ii) 0 ÷ (−7) = 0
(iii) −15 ÷ 0 = 0
(iv) 0 ÷ 0 = 0
(v) (−8) ÷ (−1) = −8
(vi) −8 ÷ (−2) = 4

Answer 3:

(i)
$$\begin{gathered} \mathrm{LHS} =\frac{\left|0\right|}{\left|4\right|}=\frac{0}{4}=0 \\ = \mathrm{RHS} \end{gathered}$$

Because LHS is equal to RHS, the equation is true.

(ii)
$$\begin{gathered} \mathrm{LHS} = -\frac{\left|0\right|}{\left|-7\right|}=-\frac{0}{7}=-0=0 \\ = \mathrm{RHS} \end{gathered}$$

Because LHS is equal to RHS, the equation is true.

(iii)
$$\begin{gathered} \mathrm{LHS} = -\frac{\left|-15\right|}{\left|0\right|}=-\frac{15}{0}= \mathrm{Not} \mathrm{defined} \\ \ne \mathrm{RHS} \end{gathered}$$
Because LHS is not equal to RHS, the equation is false.

(iv)
$$\begin{gathered} \mathrm{LHS} = \frac{\left|0\right|}{\left|0\right|}= \frac{0}{0}= \mathrm{Not} \mathrm{Defined} \\ \ne \mathrm{RHS} \end{gathered}$$

Because LHS is not equal to RHS, the equation is false.

(v)
$$\begin{gathered} \mathrm{LHS} =\frac{\left|-8\right|}{\left|-1\right|}= \frac{8}{1}=8 \\ \ne \mathrm{RHS} \end{gathered}$$
Because LHS and RHS are not equal, the equation is false.

(vi)
$$\begin{gathered} \mathrm{LHS} = \frac{\left|-8\right|}{\left|-2\right|} = \frac{8}{2}=4 \\ = \mathrm{RHS} \end{gathered}$$

Because LHS is equal to RHS, the equation is true.