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

Exercise 1 E

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


⇒x × (-3) = -26

On transposing


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


L.C.M of 8 and 12 is =2×2×2×3 =24

\begin{tabular}{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{tabular}

\dfrac{3\times 3-5\times 2}{24}


then lets find reciprocal of

\dfrac{-15}{8}\times \dfrac{16}{27}


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}


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


Cross multiply

⇒-25×8=15x ×2




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




On transposing


L.C.M of 20 and 5 is =2×2×5 =20

\begin{tabular}{l|l}  2 & 20,5 \\  \hline 2 & 10,5 \\  \hline 5 & 5,5 \\  \hline& 1,1  \end{tabular}

\dfrac{131\times 1-14\times 4}{20}=x


x=\dfrac{75}{20}=\dfrac{15}{4}=3\dfrac{3}{4} cm


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}



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

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


\dfrac{2\times 2+1}{2}=\dfrac{1\times 4+1}{4}+\dfrac{1}{6}+x



On transposing


L.C.M of 2, 4 and 6 is =2×2×3 =12

\begin{tabular}{l|l}  2 & 2,4,6 \\ \hline  2 & 1,2,3\\ \hline3&1,1,3\\ \hline &1,1,1  \end{tabular}

x=\dfrac{5\times 6-5\times 3-1\times 2}{12}


x=\dfrac{13}{12}=1\dfrac{1}{12} hours

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


On transposing


L.C.M of 7 and 3 is 21

1=\dfrac{x\times 7-2x\times 3}{21}


Cross multiplication

⇒21=x or


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


\dfrac{1\times 8+5}{8}+\dfrac{1}{2}+\dfrac{2\times 4+3}{4}


L.C.M of 8,2 and 4 is =2×2×2 =8

\begin{tabular}{l|l}  2 & 8,2,4 \\ \hline  2 & 4,1,2\\ \hline 3&2,1,1\\ \hline &1,1,1  \end{tabular}
\dfrac{13\times 1+1\times 4+11\times 2}{8}



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}



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