Exercise 1 E
Q1 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 1
The product of two rational numbers is -26 . If one of the numbers is -3, find the other.
Sol :
Let x be the other rational number
A.T.Q
⇒x × (-3) = -26
On transposing
⇒$x=\dfrac{-26}{-3}=\dfrac{26}{3}=8\dfrac{2}{3}$
Q2 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 2
Divide the sum of $\dfrac{3}{8}\text{ and }\dfrac{-5}{12}$ by the reciprocal of $\dfrac{-15}{8}\times \dfrac{16}{27}$
Sol :
First lets find sum
⇒ $ \dfrac{3}{8}+\dfrac{-5}{12}$
L.C.M of 8 and 12 is =2×2×2×3 =24
$ \begin{array}{c|c}
2 & 8,12 \\
\hline 2 & 4,6 \\
\hline 2 & 2,3 \\
\hline 3 & 1,3 \\
\hline 2 & 4,6 \\
\hline& 1,1
\end{array} $
⇒$\dfrac{3\times 3-5\times 2}{24}$
⇒$\dfrac{9-10}{24}=\dfrac{-1}{24}$..(i)
then lets find reciprocal of
⇒$\dfrac{-15}{8}\times \dfrac{16}{27}$
⇒$\dfrac{-240}{216}=\dfrac{-10}{9}$
then reciprocate it ⇒$\dfrac{-9}{10}$..(ii)
According to question lets divide
From (i) and (ii)
⇒$\dfrac{-1}{24}\div \dfrac{-9}{10}$
⇒$\dfrac{-1}{24}\times \dfrac{-10}{9}$
⇒$\dfrac{-10\times -1}{24\times 9}=\dfrac{10}{216}=\dfrac{5}{108}$
Q3 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 3
By what rational number should $-12\dfrac{1}{2}$ be divided to get $1\dfrac{7}{8}$ ?
Sol :
According to question
Let x be the number which can divide
and $-12\dfrac{1}{2}=-\dfrac{12\times 2+1}{2}=-\dfrac{25}{2}$
$1\dfrac{7}{8}=\dfrac{1\times 8+7}{8}=\dfrac{15}{8}$
⇒$-\dfrac{25}{2}\div x=\dfrac{15}{8}$
On transposing
⇒$-\dfrac{25}{2}=\dfrac{15}{8}\times x$
⇒$-\dfrac{25}{2}=\dfrac{15x}{8}$
Cross multiply
⇒-25×8=15x ×2
⇒-200=30x
⇒$x=\dfrac{-200}{30}=-6\dfrac{2}{3}$
Q4 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 4
The perimeter of isosceles triangle is $6\dfrac{11}{20}$ cm . If one of its equal sides is $1\dfrac{2}{5}$ cm, find the third side.
Sol :
Note: An isosceles triangle have 2 equal sides and one unequal side
⇒Equal sides are given $1\dfrac{2}{5}=\dfrac{5\times 1+2}{5}=\dfrac{7}{5}$
⇒and let unequal (third) side is x
⇒Also , perimeter of isosceles triangle is $6\dfrac{11}{20}=\dfrac{6\times 20+11}{20}=\dfrac{131}{20}$
⇒perimeter of isosceles triangle= sum of all sides of triangle
⇒$\dfrac{131}{20}=\dfrac{7}{5}+\dfrac{7}{5}+x$
⇒$\dfrac{131}{20}=\dfrac{7+7}{5}+x$
⇒$\dfrac{131}{20}=\dfrac{14}{5}+x$
On transposing
⇒$\dfrac{131}{20}-\dfrac{14}{5}=x$
L.C.M of 20 and 5 is =2×2×5 =20
$ \begin{array}{l|l}
2 & 20,5 \\
\hline 2 & 10,5 \\
\hline 5 & 5,5 \\
\hline& 1,1
\end{array} $
⇒$\dfrac{131\times 1-14\times 4}{20}=x$
⇒$\dfrac{131-56}{20}=x$
⇒$x=\dfrac{75}{20}=\dfrac{15}{4}=3\dfrac{3}{4}$ cm
Q5 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 5
You participate in a $12\dfrac{1}{2}$ km relay run for charity . Your team has 8 people and each person runs the same distance . How many km does each person run ?
Sol :
Total distance$=\dfrac{12\times 2+1}{2}=\dfrac{25}{2}$..(i)
Let x distance travel by each person then total distance traveled is 8x..(ii)
From (i) and (ii)
⇒$8\times x=\dfrac{25}{2}$
⇒$x=\dfrac{25}{2}\div 8$
⇒$x=\dfrac{25}{2}\times \dfrac{1}{8}$
⇒$x=\dfrac{25}{16}=1\dfrac{9}{16}$
Q6 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 6
A postcard is $10\dfrac{2}{5}$ cm long and $7\dfrac{4}{5}$ cm wide . Find the area of the postcard .
Sol :
⇒Post card is in the shape of rectangle
⇒Length$=\dfrac{10\times 5+2}{5}=\dfrac{52}{5}$
⇒Breadth$=\dfrac{7\times 5+4}{5}=\dfrac{39}{5}$
⇒Area of rectangle=Length×Breadth
⇒$\dfrac{52}{5}\times \dfrac{39}{5}$
⇒$\dfrac{52\times 39}{5\times 5}=\dfrac{2028}{25}$
⇒$\dfrac{2028}{25}=81\dfrac{3}{25}$ cm2
Q7 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 7
Rhea went to see two one-act plays. With an intermission of $\dfrac{1}{6}$ hour , the evening lasted $2\dfrac{1}{2}$ hours . The first play was $1\dfrac{1}{4}$ hours long . How long did the second play last ?
Sol :
According to question
Total time = First play + intermission + Second play
$2\dfrac{1}{2}=1\dfrac{1}{4}+\dfrac{1}{6}+x$
$\dfrac{2\times 2+1}{2}=\dfrac{1\times 4+1}{4}+\dfrac{1}{6}+x$
$\dfrac{4+1}{2}=\dfrac{4+1}{4}+\dfrac{1}{6}+x$
$\dfrac{5}{2}=\dfrac{5}{4}+\dfrac{1}{6}+x$
On transposing
$x=\dfrac{5}{2}-\dfrac{5}{4}-\dfrac{1}{6}$
L.C.M of 2, 4 and 6 is =2×2×3 =12
$
\begin{array}{l|l}
2 & 2,4,6 \\ \hline
2 & 1,2,3\\ \hline3&1,1,3\\ \hline &1,1,1
\end{array}
$
$x=\dfrac{5\times 6-5\times 3-1\times 2}{12}$
$x=\dfrac{30-15-2}{12}=\dfrac{13}{12}$
$x=\dfrac{13}{12}=1\dfrac{1}{12}$ hours
Q8 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 8
If $\dfrac{1}{3}$ of a number exceeds its $\dfrac{2}{7}$ by 1 , find the number .
Sol :
According to question
Let the number be x , then
⇒$\dfrac{1}{3}\times x=\dfrac{2}{7}\times x+1$
⇒$\dfrac{x}{3}=\dfrac{2x}{7}+1$
On transposing
⇒$1=\dfrac{x}{3}-\dfrac{2x}{7}$
L.C.M of 7 and 3 is 21
⇒$1=\dfrac{x\times 7-2x\times 3}{21}$
⇒$1=\dfrac{7x-6x}{21}$
Cross multiplication
⇒21=x or
⇒x=21
Q9 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 9
Vaibhav cycles $1\dfrac{5}{8}$ km from school to a book store , $\dfrac{1}{2}$ km more to the grocery store and then $2\dfrac{3}{4}$ km home . Find the total distance cycled by Vaibhav . Is it more or less than 5 km . If more or less , find the difference .
Sol :
Total Distance cycled = distance of school to bookstore+ distance of bookstore to grocery store+distance of grocery store to home
⇒$1\dfrac{5}{8}+\dfrac{1}{2}+2\dfrac{3}{4}$
⇒$\dfrac{1\times 8+5}{8}+\dfrac{1}{2}+\dfrac{2\times 4+3}{4}$
⇒$\dfrac{13}{8}+\dfrac{1}{2}+\dfrac{11}{4}$
L.C.M of 8,2 and 4 is =2×2×2 =8
$\begin{array}{l|l}
2 & 8,2,4 \\ \hline
2 & 4,1,2\\ \hline 3&2,1,1\\ \hline &1,1,1
\end{array}
$
⇒$\dfrac{13\times 1+1\times 4+11\times 2}{8}$
⇒$\dfrac{13+4+22}{8}=\dfrac{39}{8}$
⇒$=\dfrac{39}{8}=4\dfrac{7}{8}$
Q10 | Ex-1E | Class 8 | Schand composite mathematics solution | myhelper
Question 10
Concert ticket usually cost $120\dfrac{4}{5}$ per person . For students they are priced at $\dfrac{1}{4}$ of the normal cost . How much will 6 tickets cost for students ?
Sol :
Ticket price for student$=\dfrac{1}{4}\times 120\dfrac{4}{5}$
$=\dfrac{1}{4}\times \dfrac{120\times 5+4}{5}$
$=\dfrac{1}{4}\times \dfrac{604}{5}$
$=\dfrac{1\times 604}{4\times 5}=\dfrac{604}{20}=\dfrac{151}{5}$
For 6 students = 6×Cost of single ticket
⇒$6\times \dfrac{151}{5}=\dfrac{906}{5}$
⇒$\dfrac{906}{5}=181\dfrac{1}{5}$
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