S.chand publication New Learning Composite mathematics solution of class 8 Chapter 1 Rational numbers Exercise 1F

 Exercise 1F

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Q1 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 1

A group of friends hike $5\frac{3}{4}$ km, stops for lunch, and then hike another $3\frac{1}{5}$ km. How far did they hike?

Sol :

Before lunch hiking distance $=5\frac{3}{4}$km $=\frac{23}{4}$ km

After lunch hiking distance $=3\frac{1}{5}=\frac{16}{5}$ km 

Total distance$=\frac{23}{4}+\frac{16}{5}=\frac{115+64}{20}$

$=8\frac{19}{20}$km

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Q2 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 2

Ms Joshi walks her dog $\frac{4}{5}$ km each day. What is the total distance that Ms Joshi walks her dog in 6 days?

Sol :

Ms Joshi walks her dog $\frac{4}{5}$ km each day

Total distance that Ms Joshi walks her dog in 6 days$=6\times \frac{4}{5}=\frac{24}{5}$ $=4\frac{4}{5}$km

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Q3 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 3

Priya completes $\frac{1}{20}$ of her painting each day. How much of her painting does she complete 5 days?

Sol :

Priya completes $\frac{1}{20}$ of her painting each day

Priya completes a portion of painting in 5 day $=5\times \frac{1}{20}=\frac{1}{4}$

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Q4 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 4

An oxygen tank contained $219\frac{2}{3}$ L of oxygen before $32\frac{1}{3}$ L were used. If the tank can hold $245\frac{3}{8}$ L, how much space in the tank is unused?

Sol :

An oxygen tank contained $219\frac{2}{3}=\frac{659}{3}$ L 

Oxygen used$=32\frac{1}{3}=\frac{97}{3}$ L

Capacity of tank$245\frac{3}{8}=\frac{1963}{8}$ L

Space unused in the tank $=\frac{1963}{8}-\left[\frac{659}{3}-\frac{97}{3}\right]$

$=\frac{1963}{8}-\frac{562}{3}=\frac{5883-4432}{24}$

$=\frac{1393}{24}=58\frac{1}{24}$ L

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Q5 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 5

A water pipe has an outside diameter of 3.5 cm and a wall thickness of $\frac{25}{32}$ cm. What is the inside diameter of the pipe?

Sol :

Let ,

Inner diameter(d)=x

Outer diameter(D)=3.5cm

Thickness(t)$=\frac{257}{32}$cm

$t=\frac{3.5-d}{2}$

or $\frac{25}{32}=\frac{3.5-d}{2}$

or 3.5-d$=\frac{25\times 2}{32}$

or -d$=\frac{25}{16}-\frac{35}{10}$

or $d=\frac{35}{10}-\frac{25}{16}=\frac{560-250}{160}$

$=\frac{310}{160}=1\frac{15}{16}$cm

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Q6 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 6

 Find the perimeter and area of a rectangular piece of land measuring $12\frac{1}{2}$m by $5\frac{1}{5}$ m. How much area of the land is left unused if 4 small square flower beds of side length 1 m are made at the center?

Sol :

Length of rectangle(L)$=12\frac{1}{2}$cm  $=\frac{25}{2}$cm

Breadth of rectangle (B)$=5\frac{1}{5}$cm $=\frac{26}{5}$ cm

Perimeter of rectangle=2(L+B)

$=2\left(\frac{25}{2}+\frac{26}{5}\right)$

$=2\left(\frac{125+52}{10}\right)=2\times \frac{177}{10}$

$=\frac{177}{5}=35\frac{2}{5}$cm

∴Area of rectangle(L×B)$=\left(\frac{25}{2}\times \frac{26}{5}\right)$

=65cm²

∴The flowered area=1²=1cm²

∴4 flowered bed area=4×1=4cm²

∴Area of the land is left unused=(65-4)=61cm²

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Q7 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 7

Out of a certain sum of money a boy spends $\frac{3}{5}$, and then $\frac{1}{4}$ of the remainder. He has Rs 15 left. How much amount had he at first?

Sol :

Total money=x

Money spend $=\frac{3x}{5}$

Remainder$=\frac{1}{4}\left(x-\frac{3x}{5}\right)$

ATQ,

$x-\left[\frac{3x}{5}+\frac{1}{4}\left(x-\frac{3x}{5}\right)\right]=15$

or $x-\left[\frac{3x}{x}+\frac{1}{4}\left(\frac{5x-3x}{5}\right)\right]=15$

or $x-\left[\frac{3x}{5}+\frac{2x}{20}\right]=15$

or $x-\left[\frac{12x+2x}{20}\right]=15$

or $x-\frac{14x}{20}=15$

or $\frac{20x-14x}{20}=15$

or 6x=15×20

or $x=\frac{515\times 20}{6}=50$

Ans 50

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Q8 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 8

$10\frac{1}{2}$ tonnes of sand are to be shared between a number of builders. One of them receives $\frac{4}{7}$ of the total and the remaining sand is shared by 3 builders. How much does the fourth builder receive and how much sand does each of the other 3 builders receive?

Sol :

Total sand $=10\frac{1}{2}=\frac{21}{2}$ tonnes

Fourth builder get$=\frac{21}{2}\times \frac{4}{7}$=6 tonnes

Rest sand$=\frac{21}{2}-6=\frac{21-12}{2}=\frac{9}{2}$

Distributing into 3$=\frac{9}{2\times 3}=\frac{3}{2}$ tonnes 

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Q9 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 9

A man aged 27 marries a woman aged 24, he dies at the age of 81 and she dies at the age of 91. For what fraction of his life is the man married? For what fraction of her life is the woman a widow?

Sol :

Man married life duration$=\frac{81-27}{81}=\frac{2}{3}$

Woman widow life duration$=\frac{91-(54+24)}{91}=\frac{91-48}{91}$ $=\frac{13}{91}=\frac{1}{7}$

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Q10 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper

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Question 10

Shalini has to cut out circles a diameter $1\frac{1}{4}$ cm from an aluminium strip of dimensions $8\frac{3}{4}$ cm by $1\frac{1}{4}$ cm. How many full circles can Shalini cut? Also calculate the wastage of the aluminium strip.

Sol :

Circle diameter(d)$=1\frac{1}{4}$cm

Strip length$=8\frac{3}{4}=\frac{35}{4}$cm

Strip breadth$=1\frac{1}{4}=\frac{5}{4}$

∴Number of full circles which can be cut from the strip$=\frac{35}{4}\div \frac{5}{4}$

$=\frac{35}{4}\times \frac{4}{5}$

=7

$r=\dfrac{\frac{5}{4}}{2}=\frac{5}{8}$

Area of circle=πr²=$\pi\times \frac{5}{8}\times \frac{5}{8}$

$=\frac{22}{7} \times \frac{5}{8}\times \frac{5}{8}$cm²

Area of 7 circles $=7\times\frac{22}{7} \times \frac{5}{8}\times \frac{5}{8}$

$=\frac{275}{32}$cm²

Area of strip$=\frac{35}{4}\times \frac{5}{4}=\frac{175}{16}$cm²

∴Wastage=(Area of strip-Area of 7circles)

$=\frac{175}{16}-\frac{275}{32}$

$=\frac{175\times 2-275}{32}=\frac{350-275}{32}$

$=\frac{75}{32}=2\frac{11}{32}$ cm²