S.chand books class 8 maths solution chapter 1 Rational Numbers exercise 1 D

Exercise 1 D

---

Q1 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 1

Find a rational number between a and b, if

(i) $a=\dfrac{1}{6} , b=\dfrac{1}{5}$

Sol :

$=\frac{1}{2}(a+b)$

$=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{5}\right)$

$\begin{array}{l|l}2&5,6\\ \hline 3& 5,3\\ \hline 5 & 5,1 \\ \hline & 1,1\end{array}$

L.C.M of 5 and 6 is 30

$=\frac{1}{2}\left( \frac{1 \times 5+1 \times 6}{30}\right)$

$=\frac{1}{2} \times\left(\frac{5+6}{30}\right)$

$\frac{1}{2} \times \frac{11}{30}=\frac{11}{60}$

---

(ii) $a=\dfrac{1}{8} , b=\dfrac{7}{12}$

Sol :

$a=\frac{1}{8}, b=\frac{7}{12}$

$=\frac{1}{2}\left(\frac{1}{8}+\frac{7}{12}\right)$

$\begin{array}{l|l}2&8,12\\ \hline 2& 4,6\\ \hline 2& 2,3\\ \hline 3& 1,3\\ \hline &1,1\end{array}$

L.C.M of 8,12 is 

=2×2×2×3

=24

$=\frac{1}{2} \times\left(\frac{1 \times 3+7 \times 2}{24}\right)$

$=\frac{1}{2} \times\left(\frac{3+14}{24}\right)$

$=\frac{1}{2} \times \frac{17}{24}$

$=\frac{17}{48}$

---

(iii) $a=\dfrac{-3}{4} , b=\dfrac{-2}{3}$

Sol :

$a=\frac{-3}{4}, b=-\frac{2}{3}$

$=\frac{1}{2} \times\left(\frac{-3}{4}+\frac{-2}{3}\right)$

$=\frac{1}{2} \times\left(\frac{-3 \times 3+(-2) \times(4)}{12}\right)$

$=\frac{1}{2} \times\left(\frac{-9-8}{12}\right)$

$=\frac{1}{2} \times-\frac{17}{12}$

$=\frac{-17}{24}$

---

(iv) $a=-\dfrac{4}{9} , b=\dfrac{11}{6}$

Sol :

$a=-\frac{4}{9}, b=\frac{11}{6}$

$=\frac{1}{2}\left(-\frac{4}{9}+\frac{11}{6}\right)$

$\begin{array}{l|l}2&9,6\\ \hline 3& 9,3 \\ \hline 3& 3,1 \\ \hline & 1,1\end{array}$

L.C.M of 9 and 6 is 

=2×3×3

=18

$=\frac{1}{2} \times\left(\frac{-4 \times 2 + 11 \times 3}{18}\right)$

$=\frac{1}{2} \times\left(\frac{-8+33}{18}\right)$

$=\frac{1}{2} \times \frac{25}{18}=\frac{25}{36}$

---

Q2 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 2

Find the two rational numbers between -2 and 2 .

Sol :

Two rational numbers are 0 , 1

---

Q3 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 3

Find the three rational numbers between $-\dfrac{7}{2}\text{ and }-2$

Sol :

Let q₁ , q₂ and q₃ are three required rational numbers

q₁=$\dfrac{1}{2}\left(\dfrac{-7}{2}+\dfrac{-2}{1}\right)$

=$\dfrac{1}{2}\left(\dfrac{-7\times 1 -2\times 2}{2}\right)$

=$\dfrac{1}{2}\left(\dfrac{-7-4}{2}\right)$

=$\dfrac{1}{2}\times \dfrac{-11}{2}=\dfrac{-11}{4}$

 

q₂=$\dfrac{1}{2}\left(\dfrac{-11}{4}+\dfrac{-2}{1}\right)$

=$\dfrac{1}{2}\left(\dfrac{-11\times 1-2\times 4}{4}\right)$

=$\dfrac{1}{2}\left(\dfrac{-11-8}{4}\right)$

=$\dfrac{1}{2}\times \dfrac{-19}{4}=\dfrac{-19}{8}$

 

q₃=$\dfrac{1}{2}\left(\dfrac{-19}{8}+\dfrac{-2}{1}\right)$

=$\dfrac{1}{2}\left(\dfrac{-19\times 1-2\times 8}{8}\right)$

=$\dfrac{1}{2}\left(\dfrac{-19-16}{8}\right)$

=$\dfrac{1}{2}\times \dfrac{-35}{8}=\dfrac{-35}{16}$

 

---

Q4 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 4

Find four rational numbers between -1 and $-\dfrac{1}{2}$ .

Sol :

Let q₁ , q₂ , q₃ and q₄ are four required rational numbers

q₁=$\dfrac{1}{2}\left(\dfrac{-1}{1}+\dfrac{-1}{2}\right)$

=$\dfrac{1}{2}\left(\dfrac{-1\times 2-1\times 1}{2}\right)$

=$\dfrac{1}{2}\left(\dfrac{-2-1}{2}\right)$

=$\dfrac{1}{2}\times \dfrac{-3}{2}=\dfrac{-3}{4}$

 

q₂=$\dfrac{1}{2}\left(\dfrac{-3}{4}+\dfrac{-1}{2}\right)$

L.C.M of 4  and 2 is 4

=$\dfrac{1}{2}\left(\dfrac{-3\times 1-1\times 2}{4}\right)$

=$\dfrac{1}{2}\left(\dfrac{-3-2}{4}\right)$

=$\dfrac{1}{2}\times \dfrac{-5}{4}=\dfrac{-5}{8}$

 

q₃=$\dfrac{1}{2}\left(\dfrac{-5}{8}+\dfrac{-1}{2}\right)$

L.C.M of 8  and 2 is 8

=$\dfrac{1}{2}\left(\dfrac{-5\times 1-1\times 4}{8}\right)$

=$\dfrac{1}{2}\left(\dfrac{-5-4}{8}\right)$

=$\dfrac{1}{2}\times \dfrac{-9}{8}=\dfrac{-9}{16}$

 

q₄=$\dfrac{1}{2}\left(\dfrac{-9}{16}+\dfrac{-1}{2}\right)$

L.C.M of 16 and 2 is 16

=$\dfrac{1}{2}\left(\dfrac{-9\times 1-1\times 8}{16}\right)$

=$\dfrac{1}{2}\left(\dfrac{-9-8}{16}\right)$

=$\dfrac{1}{2}\times \dfrac{-17}{16}=\dfrac{-17}{32}$

 

---

Q5 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 5

Find six rational numbers between -1 and 0

Sol :

Let q₁ , q₂ , q₃ , q₄ , q₅ and q₆ are six required rational numbers

q₁=$\dfrac{1}{2}\left(-1+0\right)$

=$\dfrac{-1}{2}$

 

q₂=$\dfrac{1}{2}\left(\dfrac{-1}{2}+0\right)$

=$\dfrac{1}{2}\left(\dfrac{-1+0}{2}\right)$

=$\dfrac{-1}{4}$

 

q₃=$\dfrac{1}{2}\left(\dfrac{-1}{4}+0\right)$

=$\dfrac{1}{2}\left(\dfrac{-1+0}{4}\right)$

=$\dfrac{-1}{8}$

 

q₄=$\dfrac{1}{2}\left(\dfrac{-1}{8}+0\right)$

=$\dfrac{1}{2}\left(\dfrac{-1+0}{8}\right)$

=$\dfrac{-1}{16}$

 

q₅=$\dfrac{1}{2}\left(\dfrac{-1}{16}+0\right)$

=$\dfrac{1}{2}\left(\dfrac{-1+0}{16}\right)$

=$\dfrac{-1}{32}$

 

q₆=$\dfrac{1}{2}\left(\dfrac{-1}{32}+0\right)$

=$\dfrac{1}{2}\left(\dfrac{-1+0}{32}\right)$

=$\dfrac{-1}{64}$

 

---

Q6 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 6

Insert 10 rational numbers between $\dfrac{-1}{2}\text{ and }\dfrac{7}{8}$ .

Sol :

Taking equivalent rationals $\dfrac{-1}{2}\times \dfrac{4}{4}=\dfrac{-4}{8}$ for $\dfrac{-1}{2}$

So , 10 rational numbers are $\dfrac{-3}{8},\dfrac{-2}{8},\dfrac{-1}{8},\dfrac{-0}{8},\dfrac{1}{8}$ , $\dfrac{2}{8},\dfrac{3}{8},\dfrac{4}{8},\dfrac{5}{8},\dfrac{6}{8}$

 

---

Q7 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 7

Insert 100 rational numbers between $\dfrac{-3}{13}\text{ and }\dfrac{9}{13}$ .

Sol :

Taking equivalent rationals $\dfrac{-3}{13}\times \dfrac{10}{10}=\dfrac{-30}{130}$ and $\dfrac{9}{13}\times \dfrac{10}{10}=\dfrac{90}{130}$

So , ten rational numbers are $\dfrac{-29}{130},\dfrac{-28}{130}...\dfrac{-1}{130},0,\dfrac{1}{130},...\dfrac{70}{130}$

 

---

Q8 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper

OPEN IN YOUTUBE

Question 8

Find (i) Six (ii) Sixty

(iii) Six hundred rational numbers between $\dfrac{-5}{8}\text{ and }\dfrac{3}{8}$

Sol :

(i)

six rational numbers are $\dfrac{-4}{8},\dfrac{-3}{8},\dfrac{-2}{8},\dfrac{-1}{8},0,\dfrac{1}{8}$

(ii)

Taking equivalent rationals $\dfrac{-5}{8}\times \dfrac{10}{10}=\dfrac{-50}{80}$ and  $\dfrac{3}{8}\times \dfrac{10}{10}=\dfrac{30}{80}$

Sixty rational numbers are

$\dfrac{-49}{80},\dfrac{-48}{80},..0..\dfrac{10}{80}$

 

(iii)

Taking equivalent rationals $\dfrac{-5}{8}\times \dfrac{100}{100}=\dfrac{-500}{800}$ and  $\dfrac{3}{8}\times \dfrac{100}{100}=\dfrac{300}{800}$

Six hundred rational numbers are

$\dfrac{-499}{800},\dfrac{-498}{80},..0..\dfrac{99}{800},\dfrac{100}{80}$

---