Exercise 1F
Q1 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q2 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q3 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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 $=5\times \frac{1}{20}=\frac{1}{4}$
Q4 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q5 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q6 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q7 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q8 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
Q9 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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$=\frac{91-(54+24)}{91}=\frac{91-48}{91}$ $=\frac{13}{91}=\frac{1}{7}$
Q10 | Ex-1F |Class 8 |Rational Numbers | S.Chand | New Learning | Composite maths | myhelper
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
