RS Aggarwal solution class 8 chapter 1 Rational Numbers Exercise 1G

Exercise 1G

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Q1 | Ex-1G | Class 8 | Rational Numbers | RS AGGARWAL | Chapter 1  | myhelper

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Question 1:

From a rope 11 m long, two pieces of lengths $2\frac{3}{5}\mathrm{m} \mathrm{and} 3\frac{3}{10}\mathrm{m}$ are cut off. What is the length of the remaining rope?

Answer 1:

Length of the rope when two pieces of lengths $2\frac{3}{5} \mathrm{m}$ and $3\frac{3}{10} \mathrm{m}$are cut off = Total length of the rope - Length of the two cut off pieces
$\therefore 11-\left(2\frac{3}{5}+3\frac{3}{10}\right)$
Now,

$$\begin{gathered} 2\frac{3}{5}+3\frac{3}{10}\Rightarrow \left(2+\frac{3}{5}\right)+\left(3+\frac{3}{10}\right) \\ =\frac{13}{5}+\frac{33}{10} \end{gathered}$$
LCM of 5 and 10 is 10, i.e., $\left(5\times 1\times 2\right)$.

 $$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \frac{\left(13\times 2\right)+\left(33\times 1\right)}{10} \\ =\frac{26+33}{10} \\ =\frac{59}{10} \end{gathered}$$

∴​ $2\frac{3}{5}+3\frac{3}{10}=\frac{59}{10}$
Length of the remaining rope $=11-\frac{59}{10}$

                                            $$\begin{gathered} =\frac{110-59}{10} \\ =\frac{51}{10} \\ =5\frac{1}{10} \text{m} \end{gathered}$$

Therefore, the length of the remaining rope is $5\frac{1}{10} \text{m}$.
 

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Question 2:

A drum full of rice weighs $40\frac{1}{6}$kg. If the empty drum weighs $13\frac{3}{4}$kg, find the weight of rice in the drum.

Answer 2:

Weight of rice in the drum = Weight of the drum full of rice - Weight of the empty drum

                                       $$\begin{gathered} =40\frac{1}{6}-13\frac{3}{4} \\ =\left(40+\frac{1}{6}\right)-\left(13+\frac{3}{4}\right) \\ =\frac{241}{6}-\frac{55}{4} \\ =\frac{241}{6}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of} \frac{55}{4}\right) \\ =\frac{482-165}{12} \\ =\frac{317}{12} \\ =26\frac{5}{12} \text{kg} \end{gathered}$$
Therefore, the weight of rice in the drum is $26\frac{5}{12} \text{kg}$.

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Question 3:

A basket contains three types of fruits weighing $19\frac{1}{3}\mathrm{kg}$ in all. If $8\frac{1}{9}\mathrm{kg}$ of these be apples, $3\frac{1}{6}\mathrm{kg}$ be oranges and the rest pears, what is the weight of the pears in the basket?

Answer 3:

Weight of pears in the basket = Weight of the basket containing three types of fruits - (Weight of apples + Weight of oranges)
 =$19\frac{1}{3}-\left(8\frac{1}{9}+3\frac{1}{6}\right)$
Now,

$$\begin{gathered} \left(8\frac{1}{9}+3\frac{1}{6}\right)\Rightarrow \left(8+\frac{1}{9}\right)+\left(3+\frac{1}{6}\right) \\ =\frac{73}{9}+\frac{19}{6} \end{gathered}$$

LCM of 9 and 6 is 18, that is, $\left(3\times 3\times 2\right)$.

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \frac{\left(73\times 2\right)+\left(19\times 3\right)}{18} \\ =\frac{146+57}{18} \\ =\frac{203}{18} \end{gathered}$$

∴​ $8\frac{1}{9}+3\frac{1}{6}=\frac{203}{18}$
Now,
Weight of pears in the basket = $19\frac{1}{3}-\frac{203}{18}$
                                            $$\begin{gathered} =\left(19+\frac{1}{3}\right)-\frac{203}{18} \\ =\frac{58}{3}-\frac{203}{18} \\ =\frac{58}{3}+\left(\mathrm{Additive} \mathrm{inverse} \mathrm{of}\frac{203}{18}\right) \\ =\frac{348-203}{18} \\ =\frac{145}{18} \\ =8\frac{1}{18} \text{kg} \end{gathered}$$
 ​Therefore, the weight of the pears in the basket is $8\frac{1}{18} \text{kg}$.

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Question 4:

On one day a rickshaw puller earned Rs 160. Out of his earnings he spent Rs $26\frac{3}{5}$ on tea and snacks, Rs $50\frac{1}{2}$ on food and Rs $16\frac{2}{5}$ on repairs of the rickshaw. How much did he save on that day?

Answer 4:

Total earning = ₹160
Money spent on tea and snacks = ₹$26\frac{3}{5}$
Money spent on food = ₹$50\frac{1}{2}$
Money spent on repairs = ₹$16\frac{2}{5}$
Let the savings be ₹x.
Money spent on tea and snacks + Money spent on food + Money spent on repairs + Savings = Total earning
So, $26\frac{3}{5}$ + $50\frac{1}{2}$ + $16\frac{2}{5}$ + x = 160
$$\begin{gathered} \Rightarrow 26\frac{3}{5}+50\frac{1}{2}+16\frac{2}{5}+x=160 \\ \Rightarrow \frac{133}{5}+\frac{101}{2}+\frac{82}{5}+x=160 \\ \Rightarrow \frac{266+505+164}{10}+x=160 \\ \Rightarrow \frac{935}{10}+x=160 \\ \Rightarrow x=160-\frac{935}{10} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{1600-935}{10} \\ \Rightarrow x=\frac{665}{10}=₹66\frac{1}{2} \end{gathered}$$
So, the savings are ₹$66\frac{1}{2}$.

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Question 5:

Find the cost of $3\frac{2}{5}$ metres of cloth at Rs $63\frac{3}{4}$ per metre.

Answer 5:

Cost of 1 m of cloth = ₹$63\frac{3}{4}$
So, cost of $3\frac{2}{5}$ m of cloth
= $63\frac{3}{4}$ × $3\frac{2}{5}$
$$\begin{gathered} =\frac{255}{4}\times \frac{17}{5} \\ =₹216\frac{3}{4} \end{gathered}$$
So, the cost of $3\frac{2}{5}$ m of cloth is ₹$216\frac{3}{4}$.

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Question 6:

A car is moving at an average speed of $60\frac{2}{5}$ km/hr. How much distance will it cover in $6\frac{1}{4}$ hours?

Answer 6:

Speed = $60\frac{2}{5} \mathrm{km}/\mathrm{h}$
Time = $6\frac{1}{4}$ h
We know that
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Speed}\times \mathrm{Time}=\mathrm{Distance} \\ \Rightarrow \mathrm{Distance}=60\frac{2}{5}\times 6\frac{1}{4} \\ \Rightarrow \mathrm{Distance}=\frac{302}{5}\times \frac{25}{4} \\ \Rightarrow \mathrm{Distance}=377\frac{1}{2} \mathrm{km} \end{gathered}$$
Hence, the distance covered in $6\frac{1}{4}$ h is $377\frac{1}{2}\mathrm{km}$.

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Question 7:

Find the area of a rectangular park which is $36\frac{3}{5}$ m long and $16\frac{2}{3}$ m broad.

Answer 7:

Area of the rectangular park = Length of the park $\times$ Breadth of the park     (∵ Area of rectangle = Length $\times$ Breadth)

$$\begin{gathered} =36\frac{3}{5}\times 16\frac{2}{3} \\ =\left(36+\frac{3}{5}\right)\times \left(16+\frac{2}{3}\right) \\ =\frac{183}{5}\times \frac{50}{3} \\ =\frac{183\times 50}{5\times 3} \\ =\frac{9150}{15} \\ =610 {\text{m}}^2 \end{gathered}$$

Therefore, the area of the rectangular park is $610 {\text{m}}^2$.

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Question 8:

Find the area of a square plot of land whose each side measures $8\frac{1}{2}$ metres.

Answer 8:

Area of the square plot = Side $\times$ Side = ${\left(\mathrm{Side}\right)}^2$ = a²  (Because the area of the square is $a^2$, where a is the side of the square)
                                                                                                   
 $$\begin{gathered} =8\frac{1}{2}\times 8\frac{1}{2} \\ =\left(8+\frac{1}{2}\right)\times \left(8+\frac{1}{2}\right) \\ =\frac{17}{2}\times \frac{17}{2} \\ =\frac{17\times 17}{2\times 2} \\ =\frac{289}{4} \\ =72\frac{1}{4} {\text{m}}^2 \end{gathered}$$
Therefore, the area of the square plot is $72\frac{1}{4} {\text{m}}^2$.

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Question 9:

One litre of petrol costs ₹ $63\frac{3}{4}$ . What is the cost of 34 litres of petrol?

Answer 9:

Cost of 1 litre of petrol = ₹$63\frac{3}{4}$
Cost of 34 litres of petrol = $63\frac{3}{4}$ × 34 = $\frac{255}{4}\times 34=₹2167\frac{1}{2}$
So, the cost of 34 litres of petrol is ₹$2167\frac{1}{2}$.

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Question 10:

An aeroplane covers 1020 km in an hour. How much distance will it cover in $4\frac{1}{6}$ hours?

Answer 10:

Distance covered by the aeroplane in $4\frac{1}{6}$ hours = $4\frac{1}{6}\times 1020$
                                                                         $$\begin{gathered} =\left(4+\frac{1}{6}\right)\times 1020 \\ =\frac{25}{6}\times 1020 \\ =\frac{25}{6}\times \frac{1020}{1} \\ =\frac{25\times 1020}{6\times 1} \\ =\frac{25500}{6} \\ =4250 \mathrm{km} \end{gathered}$$

Therefore, the distance covered by the aeroplane is $4250 \mathrm{km}$.

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Question 11:

The cost of $3\frac{1}{2}$ metres of cloth is ₹ $166\frac{1}{4}$ . what is the cost of one metre of cloth?

Answer 11:

Cost of $3\frac{1}{2}$ m of cloth = ₹$166\frac{1}{4}$
So, the cost of 1 m of cloth = $\frac{166{\displaystyle \frac{1}{4}}}{3{\displaystyle \frac{1}{2}}}=\frac{{\displaystyle \frac{665}{4}}}{{\displaystyle \frac{7}{2}}}=\frac{{\displaystyle 665}}{{\displaystyle 4}}\times \frac{2}{7}=\frac{95}{2}=₹47\frac{1}{2}$
Hence, the cost of 1 m of cloth is ₹ $47\frac{1}{2}$.

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Question 12:

A cord of length $71\frac{1}{2}$ m has been cut into 26 pieces of equal length. What is the length of each piece?

Answer 12:

Length of each piece of the cord = $71\frac{1}{2}÷26$
                                                 $$\begin{gathered} =\left(71+\frac{1}{2}\right)÷26 \\ =\frac{143}{2}÷26 \\ =\frac{143}{2}÷\frac{26}{1} \\ =\frac{143}{2}\times \frac{1}{26} \\ =\frac{143\times 1}{2\times 26} \\ =\frac{143}{52} \\ =\frac{9}{4} \\ =2\frac{3}{4} \mathrm{m} \end{gathered}$$

Hence, the length of each piece of the cord is $2\frac{3}{4} \mathrm{metres}$.

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Question 13:

The area of a room is $65\frac{1}{4}{\mathrm{m}}^2.$ If its breadth is $5\frac{7}{16}$ metres, what is its length?

Answer 13:

Area of a room = Length $\times$ Breadth
Thus, we have: 
 $$\begin{gathered} 65\frac{1}{4}=\mathrm{L}\text{ength}\times 5\frac{7}{16} \\ \mathrm{L}\text{ength}=65\frac{1}{4}÷5\frac{7}{16} \end{gathered}$$
            $$\begin{gathered} =\left(65+\frac{1}{4}\right)÷\left(5+\frac{7}{16}\right) \\ =\frac{261}{4}÷\frac{87}{16} \\ =\frac{261}{4}\times \frac{16}{87} \\ =\frac{261\times 16}{4\times 87} \\ =\frac{4176}{348} \\ =12 \text{m} \end{gathered}$$

Hence, the length of the room is 12 metres.

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Question 14:

The product of two fractions is $9\frac{3}{5}.$ If one of the fractions is $9\frac{3}{7},$ find the other.

Answer 14:

Let the other fraction be x.

Now, we have:

$$\begin{gathered} 9\frac{3}{7}\times x=9\frac{3}{5} \\ \Rightarrow x=9\frac{3}{5}÷9\frac{3}{7} \\ =\left(9+\frac{3}{5}\right)÷\left(9+\frac{3}{7}\right) \\ =\frac{48}{5}÷\frac{66}{7} \\ =\frac{48}{5}\times \frac{7}{66} \\ =\frac{48\times 7}{5\times 66} \\ =\frac{336}{330} \\ =\frac{56}{55} \\ =1\frac{1}{55} \end{gathered}$$
Hence, the other fraction is $1\frac{1}{55}$.

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Question 15:

In a school, $\frac{5}{8}$ of the students are boys. If there are 240 girls, find the number of boys in the school.

Answer 15:

If $\frac{5}{8}$of the students are boys, then the ratio of girls is $1-\frac{5}{8}$, that is, $\frac{3}{8}$.

Now, let x be the total number of students.

Thus, we have:

$$\begin{gathered} \frac{3}{8}x=240 \\ \Rightarrow x=240÷\frac{3}{8} \end{gathered}$$

         $$\begin{gathered} =240\times \frac{8}{3} \\ =\frac{240}{1}\times \frac{8}{3} \\ =\frac{240\times 8}{1\times 3} \\ =\frac{1920}{3} \\ =640 \end{gathered}$$

Hence, the total number of students is 640.
Now,
Number of boys = Total number of students - Number of girls
                         $$\begin{gathered} =640-240 \\ =400 \end{gathered}$$

Hence, the number of boys is 400.

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Question 16:

After reading $\frac{7}{9}$ of a book, 40 pages are left. How many pages are there in the book?

Answer 16:

Ratio of the read book = $\frac{7}{9}$
Ratio of the unread book = $1-\frac{7}{9}$

                                      $=\frac{2}{9}$
Let x be the total number of pages in the book.

Thus, we have:
                         
$$\begin{gathered} \frac{2}{9}\times x=40 \\ \Rightarrow x=40÷\frac{2}{9} \end{gathered}$$

        $$\begin{gathered} =40\times \frac{9}{2} \\ =\frac{40}{1}\times \frac{9}{2} \\ =\frac{40\times 9}{1\times 2} \\ =\frac{360}{2} \\ =180 \end{gathered}$$

Hence, the total number of pages in the book is 180.

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Question 17:

Rita had Rs 300. She spent $\frac{1}{3}$ of her money on notebooks and $\frac{1}{4}$ of the remainder on stationery items. How much money is left with her?

Answer 17:

Amount of money spent on notebooks = $300\times \frac{1}{3}$

                                                          $$\begin{gathered} =\frac{300}{1}\times \frac{1}{3} \\ =\frac{300}{3} \\ =100 \end{gathered}$$

∴ Money left after spending on notebooks = $300-100$
                                                                $=200$
Amount of money spent on stationery items from the remainder = $200\times \frac{1}{4}$
                                                                                               $$\begin{gathered} =\frac{200}{1}\times \frac{1}{4} \\ =\frac{200}{4} \\ =50 \end{gathered}$$

∴ Amount of money left with Rita = $200-50$
                                                   $=\mathrm{Rs} 150$

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Question 18:

Amit earns ₹ 32000 per month. He spends $\frac{1}{4}$ of his income of food; $\frac{3}{10}$ of the remainder on house rent and $\frac{5}{21}$ of the remainder on the education of children. How much money is still left with him?

Answer 18:

Amit's income per month = ₹32,000
Money spent on food = $\frac{1}{4} \mathrm{of} ₹32,000=\frac{1}{4}\times ₹32,000=₹8,000$
Remaining amount = ₹32,000 − ₹8,000 = ₹24,000
Money spent on house rent = $\frac{3}{10}\times ₹24,000=\mathrm{Rs} 7,200$
Money left = ₹24,000 − ₹7,200 = ₹16,800
Money spent on education of children = $\frac{5}{21}\times ₹16,800=₹4,000$
Amount of money still left with him = ₹16,800 − ₹4,000 = ₹12,800

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Question 19:

If $\frac{3}{5}$ of a number exceeds its $\frac{2}{7}$ by 44, find the number.

Answer 19:

Let x be the required number.
We know that $\frac{3}{5}$ of the number exceeds its $\frac{2}{7}$ by 44.
That is,

$\frac{3}{5}\times x=\frac{2}{7}\times x+44$
$$\begin{gathered} \frac{3}{5}\times x-\frac{2}{7}\times x=44 \\ \left(\frac{3}{5}-\frac{2}{7}\right)\times x=44 \\ \left(\frac{3}{5}+\mathrm{Additive} \mathrm{inverse} \mathrm{of} \frac{2}{7}\right)\times x=44 \\ \left(\frac{21-10}{35}\right)\times x=44 \end{gathered}$$
                                       $\frac{11}{35}\times x=44$
                                                 $x=44÷\frac{11}{35}$

                                                   $$\begin{gathered} =44\times \frac{35}{11} \\ =\frac{44}{1}\times \frac{35}{11} \\ =\frac{44\times 35}{1\times 11} \\ =\frac{1540}{11} \\ =140 \end{gathered}$$

Hence, the number is 140.

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Question 20:

At a cricket test match $\frac{2}{7}$ of the spectators were in a covered place while 15000 were in open. Find the total number of spectators.

Answer 20:

Ratio of spectators in the open $=1-\frac{2}{7}$
                                               $=\frac{5}{7}$
Total number of spectators in the open = x
Then,$\frac{5}{7}\times x=15000$

$\begin{aligned} \Rightarrow x &=15000 \div \frac{5}{7} \\ &=15000 \times \frac{7}{5} \\ &=\frac{15000}{1} \times \frac{7}{5} \\ &=\frac{15000 \times 7}{1 \times 5} \\ &=\frac{10500}{5} \\ &=21000 \end{aligned}$

Hence, the total number of spectators is 21,000