Exercise 1 D
Q1 | Ex-1D | Class 8 | Schand composite mathematics solution | myhelper
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
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
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
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
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
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
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
$\dfrac{-499}{800},\dfrac{-498}{80},..0..\dfrac{99}{800},\dfrac{100}{80}$
