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

Exercise 1 E

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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

⇒$x=\dfrac{-26}{-3}=\dfrac{26}{3}=8\dfrac{2}{3}$

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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)

⇒$\dfrac{-1}{24}\div \dfrac{-9}{10}$

⇒$\dfrac{-1}{24}\times  \dfrac{-10}{9}$

⇒$\dfrac{-10\times -1}{24\times 9}=\dfrac{10}{216}=\dfrac{5}{108}$

 

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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

⇒$x=\dfrac{-200}{30}=-6\dfrac{2}{3}$

 

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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

 

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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)

⇒$8\times x=\dfrac{25}{2}$

⇒$x=\dfrac{25}{2}\div 8$

⇒$x=\dfrac{25}{2}\times \dfrac{1}{8}$

⇒$x=\dfrac{25}{16}=1\dfrac{9}{16}$

 

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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

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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

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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

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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

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

⇒$\dfrac{13\times 1+1\times 4+11\times 2}{8}$

⇒$\dfrac{13+4+22}{8}=\dfrac{39}{8}$

⇒$=\dfrac{39}{8}=4\dfrac{7}{8}$

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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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