Oxford New Enjoying Mathematics Class 7 Answers

I believe in numbers. The ones you can see and the ones you can't. The real and the imaginary. The rational and the irrational and every point on lines that go on forever. Numbers have never let me down.They don't waffle. They don't lie. They don't pretend to be what they are not. They are timeless.
Amy Harmon

Numbers–

Numbers play an important role in our day-to-day life. Waking up in the morning to going to bed we come across these numbers. when we wake up in the morning, we look at the clock which shows the time in number 7: 00 AM. In breakfast, lunch how many chapatis we take that is also represented in number. Early human beings count their animals with sticks or stones. 1 cow for 1 stick, 2 cows for 2 sticks, and so on.

Numbers are categorised as –

Natural numbers or Counting Numbers
Whole Numbers
Integers
rational numbers
Irrational numbers
Real numbers

In this class we have to know up to rational numbers. In higher classes we need to know about irrational numbers and Real numbers.

Natural Numbers or Counting Numbers– Starting with the number 1,2,3,………….

2. Whole Numbers– Whole Numbers start with zero. Whole numbers consists of natural numbers with zero.

3. Integers- Integers are positive and negative. Positive Integers are 1,2,3,……………… and Negative Integers are -1,-2,-3,………………………..

4. Rational numbers- Rational numbers are of the form

\frac{p}{q},\; \; p,q\; \epsilon\; Z, and\; q\neq 0

Note- 0 is also a rational number.

\frac{0}{1},\; \;\frac{0}{6}\;\: \frac{0}{100}

Which of the following are Rational Numbers?

a)\; \; \frac{-1}{0} \; is \; not \; a\; rational \; number

b)\; \; \frac{0}{-1} \; is \; \; a\; rational \; number.\;As\; -1\neq 0

c)\; \; \frac{1}{-1} \; is \; \; a\; rational \; number.\;As\; -1\neq 0

Equivalent Rational numbers-

When we multiply the same number in both numerator and denominator, we get the Equivalent Rational numbers.

example-

\; \frac{2}{3}= \; \frac{2\times 2}{3\times 2}=\frac{4}{6}

here, 4/6 is equivalent to 2/3

Standard Form of a Rational Number-

Standard form of a rational number can be obtained by dividing by the HCF with numerator and denominator.

Example- Express in standard form.

a) \; \frac{60}{130}= \; \frac{6}{13}

a) \; \frac{85}{102}= \; \frac{85\div 17}{102\div 17}=\frac{5}{6}

Comparison of Rational number-

Having equal Denominator (Like Rational Numbers)-

Look at the Numerator only. If the Numerator one rational number is greater than the second rational number , then the First rational number is bigger rational number between two.

Having Different unequal Denominators(Unlike Rational numbers)-

Find out their LCM.

2. Convert into Equivalent Rational numbers by multiplying same number both in Numerator and denominator by equating with LCM of denominators.

3. Now both are having equal Denominators . Now compare the like rational numbers. look at the Numerator and decide which is smaller rational number or bigger rational number.

Having Equal Numearor–

Just look at the denominator . Smaller the number in denominator is the bigger rational number.

2. Bigger the number n denominator is the smaller rational number.

Example- Circle the greater rational number

a) \; \frac{8}{10} \; and\; \frac{8}{15}

Here, both have equal numerator i.e 8

Look at the denominator, we know 10 < 15. But here denominator 10 is smaller number. so smaller the number in denominator will be bigger rational number .

So, \; \frac{8}{10} \; >\; \frac{8}{15}

Absolute Value of a Rational number-

Absolute Value of positive rational number will be Positive rational number

2. Absolute value of Negative rational number will be Positive Rational number.

Exercise 1 A

State which of the following are rational numbers. Give reasons.

a) \; \frac{3}{7}

Solution Yes this is a rational number . As both 3 and 7 are integers and 7 is not equal to zero.

b) \; \frac{123}{478}

solution: \; \frac{123}{478} \; is\; a\; rational\; number\; 123,478\; \epsilon \; Z, \; 478\; \neq 0

c) \; \frac{1}{1000}

Solution: \; \frac{1}{1000} \; is\; a\; rational\; number\; 1,1000\; \epsilon \; Z, \; 1000\; \neq 0

d) \; \frac{0}{68}

Solution: \; \frac{0}{68} \; is\; a\; rational\; number\; 0,68\; \epsilon \; Z, \; 68\; \neq 0

e) \; \frac{893}{0}

Solution: \; \frac{893}{0} \; is\; not a\; rational\; number\; 893, 0\; \epsilon \; Z, \; 0\; = 0

f) \; \frac{-8}{9}

Solution: \; \frac{-8}{9} \; is\; not a\; rational\; number\; -8, 9\; \epsilon \; Z, \; 9\; \neq 0

g) \; \frac{-6}{-25}

Solution: \; \frac{-6}{-25} \; is\; not a\; rational\; number\; -6, -25\; \epsilon \; Z, \; -25\; \neq 0

h) \; \frac{18}{-25}

Solution: \; \frac{18}{-25}=\frac{18\times (-1)}{(-25)\times (-1)}=\frac{-18}{25} \; is\; a\; rational\; \; number\;\; \; \; -18,\; 25\; \epsilon \; Z, \; 25\; \neq 0

2) Which of the following are in standard form? If they are not, convert them to standard form.

a)\; \frac{4}{6}

HCF of 4 and 6 is 2. Dividing numerator and denominator by 2 we get

Solution: \; \frac{4}{6}=\frac{4\div 2}{6\div 2}=\frac{2}{3}

b) \; \frac{4}{-9}

Solution: \; \frac{4}{-9}=\frac{4\times (-1)}{\left ( -9 \right )\times (-1)}=\frac{-4}{9}

c) \; \frac{-11}{-13}

Solution: \; \frac{-11}{-13}=\frac{(-11)\times (-1)}{(-13)\times (-1)}=\frac{11}{13}

HCF of 11 and 13 is 1.

Standard\; form \; of \; \frac{-11}{-13}\; is\; \frac{11}{13}

d) \; \frac{-21}{-28}\;

HCF of 21 and 28 is 7. Dividing by numerator and denominator by 7 we get

solution: \; \frac{-21}{-28}\; =\frac{21}{28}=\frac{21\div 7}{28\div 7}=\frac{3}{4}

Standard\; form \; of \; \; \frac{-21}{-28}\; is\; \frac{3}{4}

e) \; \; \frac{-42}{48}\;

Solution- HCF of 42 and 48 is 6. Dividing Numerator and Denominator by HCF we get

\; \; \frac{-42}{48}\;=\frac{-42\div 6}{48\div 6}=\frac{-7}{8}

Standard\; form \; of\; \; \frac{-42}{48}\;is\; \frac{-7}{8}

f)\; \; \frac{18}{15}\;

Solution- HCF of 18 and 15 is 3. Dividing both numerator and denominator by 3 we get

\; \; \frac{18}{15}\;=\frac{18\div 3}{15\div 3}=\frac{6}{5}

\;Standard \; form \; of\; \frac{18}{15}\; is\; \frac{6}{5}

g)\; \frac{-346}{692}\;

Solution:-HCF of 346 and 692 is 346. Dividing numerator and denominator by 346 we get

\; \frac{-346}{692}\; =\frac{-346\div 346}{692\div 346}=\frac{-1}{2}

\; standard \; form \; of \; \frac{-346}{692}\; is\; \frac{-1}{2}

h) \; \frac{12}{6}\;

Solution- HCF of 12 and 6 is 6. Dividing both numerator and denominator by 6 we get

\; \frac{12}{6}\;=\frac{12\div 6}{6\div 6}=\frac{2}{1}

standard\; form \; of\; \frac{12}{6}\; is\; \frac{2}{1}

3) fill in the blanks:

a)\; \frac{4}{5}\;=\; \frac{-12}{}=\frac{20}{}

Solution- convert into equivalent rational numbers by multiplying same number with numerator and denominator we get

Multiplying numerator 4 with -3 we get -12 . so we have to multiply same number -3 with denominator .

\; \frac{4}{5}\;=\;\frac{4\times (-3)}{5\times (-3)}= \frac{-12}{-15}

Multiplying numerator 4 with 5 we get 20. so we have to multiply same number 5 with denominator.

\; \frac{4}{5}\;=\;\frac{4\times 5}{5\times5}= \frac{20}{25}

b)\; \frac{-5}{7}\;=\;\frac{-15}{}= \frac{-35}{}

solution- Multiplying numerator -5 with 3 we get -15. we have to multiply same number 3 with denominator .

\; \frac{-5}{7}\;=\;\frac{(-5)\times 3}{7\times 3}= \frac{-15}{21}

Multiplying numerator -5 with number 7 we get -35. so we have to multiply same number 7 with denominator.

\; \frac{-5}{7}\;=\;\frac{(-5)\times 7}{7\times 7}= \frac{-35}{49}

c)\; \frac{-3}{}\;=\;\frac{6}{-16}= \frac{}{24}

Solution- Dividing numerator 6 with -2 we get -3. So we have to divide denominator -16 with -2 we get 8

\; \;\frac{6}{-16}= \frac{6\div (-2)}{(-16)\div (-2)}=\frac{-3}{8} \; \;\frac{6}{-16}= \frac{6\div (-2)}{(-16)\div (-2)}=\frac{-3}{8}

Now multiplying Denominator 8 with number 3 we get 24. so we have to multiply numerator -3 with same number 3

\; \;\frac{-3}{8}= \frac{(-3)\times 3}{8\times 3}=\frac{-9}{24}

d)\; \;\frac{16}{24}= \frac{8}{}=\frac{-32}{}

Solution:- Dividing numerator 16 with 2 we get 8. So we have to divide the denominator by 2 .

\; \;\frac{16}{24}= \frac{16\div 2}{24\div 2}=\frac{8}{12}

Multiplying numerator 16 with -2 we get -32. so we have to multiply denominator with -2 .

\; \;\frac{16}{24}= \frac{16\times (-2)}{24\times (-2)}=\frac{-32}{-48}

4) Express the following as rational numbers with positive denominators.

a)\; \;\frac{12}{(-29)}

Multiplying both numerator and denominator with -1 we get

\; \;\frac{12}{(-29)}=\frac{12\times \left ( -1 \right )}{(-29)\times (-1)}=\frac{-12}{29}

b)\; \;\frac{3}{(-8)}

solution:-\; \;\frac{3}{(-8)}=\frac{3\times \left ( -1 \right )}{(-8)\times (-1)}=\frac{-3}{8}

c)\; \;\frac{(-16)}{(-24)}

Solution – Multiplying both numerator and denominator by -1 we get

\; \;\frac{(-16)}{(-24)}=\frac{(-16)\times (-1)}{\left ( -24 \right )\times (-1)}=\frac{16}{24}

d)\; \;\frac{(-6)}{(-19)}

Solution:\; \;\frac{(-6)}{(-19)}=\frac{(-6)\times (-1)}{\left ( -19 \right )\times (-1)}=\frac{6}{19}

5) Convert the rational number 4/9 into equivalent rational numbers with the following numbers as the numerators.

a) 20

Solution- Multiplying numerator 4 with 5 we get 20. so we have to multiply same number 5 with denominator.

\; \;\frac{4}{9}=\frac{4\times 5}{9\times 5}=\frac{20}{45}

b) (-16)

Solution- Multiplying numerator 4 with -4 we get -16. So we have to multiply same number -4 with denominator.

\; \;\frac{4}{9}=\frac{4\times (-4)}{9\times (-4)}=\frac{-16}{-36}

c) 12

Solution- Multiplying both numerator and denominator by 3 we get 12. So we have to multiply same number 3 with denominator.

\; \;\frac{4}{9}=\frac{4\times 3}{9\times 3}=\frac{12}{27}

d) -24

Solution- Multiplying both numerator and denominator by -6 we get -24. so we have to multiply same number -6 with denominator

\; \;\frac{4}{9}=\frac{4\times (-6)}{9\times (-6)}=\frac{-24}{-54}

7) Fill in the blanks =. <, >

a)\; \;\frac{-11}{3} \; , \; \frac{-2}{3}

Solution:- We know that 11 > 2, but -11 < -2

so,\; \;\frac{-11}{3} \; < \; \frac{-2}{3}

b)\; \;\frac{-7}{3} \; < \; \frac{-12}{3}

Solution : – We know 7 < 12 but -7 > -12

so,\; \;\frac{-7}{3} \; > \; \frac{-12}{3}

c)\; \;\frac{3}{3} \; , \; \frac{-5}{3}

Solution:- We know 3/3 =1

We\; know\; \;1 \; > \; \frac{-5}{3}

\; so,\; \;\frac{3}{3} \; > \; \frac{-5}{3}

d)\; \;\frac{-9}{3} \; , \; \frac{0}{3}

Solution:- We know, every negative integer is always less than zero.

So we have -9 < 0

so,\; \;\frac{-9}{3} \; < \; \frac{0}{3}

e)\; \;\frac{-10}{3} \; , \; \frac{-5}{3}

Solution- We know, 10 > 5, but -10 < -5

so,\; \;\frac{-10}{3} \; < \; \frac{-5}{3}

f)\; \;\frac{8}{3} \; , \; \frac{2}{3}

Solution:- We know 8 > 2

so,\; \;\frac{8}{3} \; > \; \frac{2}{3}

g)\; \;\frac{6}{3} \; , \; \frac{-6}{3}

Solution- We know every positive integer is always greater than negative integer

We have 6 > -6

so,\; \;\frac{6}{3} \; > \; \frac{-6}{3}

h)\; \;\frac{12}{3} \; , \; \frac{-1}{3}

Solution:- We know every positive integer is always greater than negative integer

We have 12 > (-1)

so,\; \;\frac{12}{3} \; > \; \frac{-1}{3}

8) Determine which of the two rational numbers is greater in each case.

a)\; \;\frac{-3}{7} \; , \; \frac{2}{7}

Solution- We know, every positive integer is always greater than negative integer.

We have, 2 > (-3)

so,\; \;\frac{2}{7} \; > \; \frac{-3}{7}

so greater rational number is 2 / 7

b)\; \;\frac{-4}{9} \; , \; \frac{-5}{6}

Solution:- LCM of 9 and 6 is 3 x 3 x 2=18

Converting the rational numbers to equivalent rational numbers by equating to LCM by multiplying same number both in numerator and denominator we get

\; \;\frac{-4}{9}=\frac{(-4)\times 2}{9\times 2} =\frac{-8}{18}

\; \;\frac{-5}{6}=\frac{(-5)\times 3}{6\times 3} =\frac{-15}{18}

Now comparing the rational numbers with equal denominator, look at the numerator.

We know, 8 < 15, But -8 > -15

so, \; \frac{-8}{18} \: > \; \frac{-15}{18}

so, \; \frac{-4}{9} \: > \; \frac{-5}{6}

c) \; \frac{1}{2} \: , \; \frac{4}{7}

Solution:- LCM of 2 and 7 is 14.

Converting the rational numbers into equivalent rational numbers by equating to LCM by multiplying same number with numerator and denominator.

\frac{1}{2} = \: \frac{1\times 7}{2\times 7}= \; \frac{7}{14}

\frac{4}{7} = \: \frac{4\times 2}{7\times 2}= \; \frac{8}{14}

Now comparing equivalent rational numbers, looking at the numerator .

We know, 8 > 7

so,\; \frac{8}{14} > \: \frac{7}{14}

so,\; \frac{4}{7} > \: \frac{1}{2}

d)\; \frac{-7}{11} , \: \frac{5}{8}

Solution:- LCM of 11 and 8 is 88

Converting into equivalent rational numbers, equating to LCM by multiplying same number in both numerator and denominator

\; \frac{-7}{11}= \: \frac{(-7)\times 8}{11\times 8}=\frac{-56}{88}

\; \frac{5}{8}= \: \frac{5\times 11}{8\times 11}=\frac{55}{88}

We know, 55 > -56

so,\; \frac{55}{88} > \: \frac{-56}{88}

so,\; \frac{5}{8} > \: \frac{-7}{11}

e)\; \frac{-3}{-13} , \: \frac{-5}{-21}

Solution:- LCM of 13 and 21 is 273

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number in both numerator and denominator

\; \frac{-3}{-13}= \: \frac{3}{13}=\frac{3\times 21}{13\times 21}=\frac{63}{273}

\; \frac{-5}{-21}= \: \frac{5}{21}=\frac{5\times 13}{21\times 13}=\frac{65}{273}

Now comparing Equivalent rational number having equal denominator, look at the numerator

We know, 65 > 63

so,\; \frac{65}{273}\; >\; \frac{63}{273}

so,\; \frac{-5}{-21}\; >\; \frac{-3}{-13}

9) fill in the boxes using symbols <, =, >

a)\; \frac{-7}{12}\; ,\; \frac{-5}{-8}

Solution:-

We \; know\; \frac{-5}{-8}\;= \; \frac{5}{8}

We know, negative rational number is always less than positive rational number.

so,\; \frac{-7}{12}\;< \; \frac{5}{8}

so,\; \frac{-7}{12}\;< \; \frac{-5}{-8}

b)\; \frac{-4}{9}\; , \; \frac{-3}{7}

Solution- LCM of 9 and 7 is 63

Converting into Equivalent rational numbers equating to LCM by multiplying same number with numerator and denominator.

\; \frac{-4}{9}\; = \; \frac{(-4)\times 7}{9\times 7}=\frac{-28}{63}

\; \frac{-3}{7}\; = \; \frac{(-3)\times 9}{7\times 9}=\frac{-27}{63}

we know , -28 < -27

\; \frac{-28}{63}\; < \; \frac{-27}{63}

so,\; \frac{-4}{9}\; < \; \frac{-3}{7}

c)\; \frac{-7}{8}\; , \; \frac{14}{17}

Solution- We know, negative rational number is always less than positive rational number.

so,\; \frac{-7}{8}\; < \; \frac{14}{17}

d)\; \frac{-2}{9}\; , \; \frac{8}{-36}

Solution:- LCM of 9 and 36 is 36

Now converting into equivalent rational numbers, equating to LCM by multiplying same number with both numerator and denominator

\; \frac{-2}{9}\;= \; \frac{(-2)\times 4}{9\times 4}=\frac{-8}{36}

\; \frac{8}{-36}\;= \; \frac{8\times (-1)}{(-36)\times (-1)}=\frac{-8}{36}

so,\; \frac{-8}{36}\;= \; \frac{8}{(-36)}

so,\; \frac{-2}{9}\;= \; \frac{8}{(-36)}

e)\; \frac{5}{8}\;, \; \frac{25}{45}

Solution- LCM of 8 and 45 is 360

Converting into equivalent rational number, equating to LCM by multiplying same number with both numerator and denominator.

\; \frac{5}{8}\;= \; \frac{5\times 45}{8\times 45}=\frac{225}{360}

\; \frac{25}{45}\;= \; \frac{25\times 8}{45\times 8}=\frac{200}{360}

Now comparing equivalent rational numbers , look at the numerator

we \; have\; \frac{225}{360}\;> \frac{200}{360}

so,\; \frac{5}{8}\;> \frac{25}{45}

f)\; \frac{4}{6}\;> \frac{1}{12}

Solution- LCM of 6 and 12 is 12.

Converting into equivalent rational number equating to LCM by multiplying same number with numerator and denominator.

\; \frac{4}{6}\;= \frac{4\times 2}{6\times 2}=\frac{8}{12}

\; \frac{1}{12}\;= \frac{1\times 1}{12\times 1}=\frac{1}{12}

Now comparing the equivalent rational number, look at the numerator.

\; \frac{8}{12}\;> \frac{1}{12}

Now,\; \frac{4}{6}\;> \frac{1}{12}

g)\; \frac{8}{9}\; , \frac{8}{19}

Solution:- LCM of 9 and 19 is 171

Converting rational numbers into equivalent rational numbers by equating to LCM by multiplying same number with both numerator and denominator we get

\; \frac{8}{9}\;= \frac{8\times 19}{19\times 9}=\frac{151}{171}

\; \frac{8}{19}\;= \frac{8\times 9}{19\times 9}=\frac{72}{171}

Now comparing equivalent rational numbers having equal denominator, look at the numerator

We know 151 > 72

\; \frac{151}{171}\; > \frac{72}{171}

so,\; \frac{8}{9}\; > \frac{8}{19}

h)\; \frac{19}{35}\; , \frac{19}{20}

Solution- LCM of 35 and 20 is 5 x 7 x 4=140

Converting into equivalent rational numbers by equating to LCM by multiplying same number with both numerator and denominator

\; \frac{19}{35}\; = \frac{19\times 4}{35\times 4}=\frac{76}{140}

\; \frac{19}{20}\; = \frac{19\times 7}{20\times 7}=\frac{133}{140}

Now comparing equivalent rational numbers having equal denominator, look at the denominator

We know 133 > 76

\; \frac{133}{140}\; > \frac{76}{140}

so,\; \frac{19}{20}\; > \frac{19}{35}

10. Arrange the following in ascending order.

\; \frac{2}{5}\; , \frac{-1}{2} , \frac{8}{-15}, \frac{-3}{-10}

Solution:- LCM of denominators 5, 2, 15, 10 is 5 x 3 x 2=30

Converting into equivalent rational numbers equating to LCM by multiplying same number with both numerator and denominator

\; \frac{2}{5}\; = \frac{2\times 6}{5\times 6}= \frac{12}{30}

\; \frac{-1}{2}\; = \frac{-1\times 15}{2\times 15}= \frac{-15}{30}

\; \frac{-8}{15}\; = \frac{-8\times 2}{15\times 2}= \frac{-16}{30}

\; \frac{3}{10}\; = \frac{3\times 3}{10\times 3}= \frac{9}{30}

Now comparing equivalent rational numbers having equal denominator, look at the numerator

Comparing equivalent rational numbers in ascending orders are

\; \frac{-16}{30}\; < \frac{-15}{30} < \frac{9}{30} < \frac{12}{30}

so,\; \frac{-8}{15}\; < \frac{-1}{2} < \frac{-3}{-10} < \frac{2}{5}

11) Arrange the following in descending order

\; \frac{-7}{10}\; , \frac{8}{-15} , \frac{19}{30} , \frac{-2}{-5}

Solution- LCM of 10, 15, 30 and 5 is 3 x 5 x 2 = 30

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number with both same numerator and denominator

\; \frac{-7}{10}\;= \frac{(-7)\times 3}{10\times 3}= \frac{-21}{30}

\; \frac{8}{-15}\;= \frac{(-8)\times 2}{15\times 2}= \frac{-16}{30}

\; \frac{19}{30}\;= \frac{19\times 1}{30\times 1}= \frac{19}{30}

\; \frac{-2}{-5}\;= \frac{2}{5}= \frac{2\times 6}{5\times 6}=\frac{12}{30}

Now comparing equivalent rational numbers having equal denominator, look at the numerator.

Comparing the rational numbers in descending orders are

\; \frac{19}{30}\;> \frac{12}{30} > \frac{-16}{30} > \frac{-21}{30}

so,\; \frac{19}{30}\;> \frac{-2}{-5} > \frac{-8}{15} > \frac{-7}{10}

New Enjoying Mathematics Class 8 Solutions

Addition Of Rational numbers

Want to Learn Vedic maths tricks

Having equal denominator- Add the numerator only keeping the denominator same.

2. Having different denominators– Find the LCM of denominators. Converting into equivalent rational numbers, then add the rational numbers.

Properties of Rational Numbers

Closure property-

if\; \frac{a}{b}\;, \frac{c}{d}\; rational\; numbers\frac{a}{b}+\frac{c}{d}\; also\; rational \; number

2. Commutative Property–

if\; \frac{a}{b}\;, \frac{c}{d}\; rational\; numbers\frac{a}{b}+\frac{c}{d}\;=\frac{c}{d}+\frac{a}{b}

3. Associative Property-

if\; \frac{a}{b}\;, \frac{c}{d}\;, \frac{e}{f} \; are\; rational\; numbers(\frac{a}{b}+\frac{c}{d})+\frac{e}{f}\;=\frac{a}{b}+(\frac{c}{d}+\frac{e}{f})

4. Additive Identity–

if\; \frac{a}{b}\;, is\;a\: rational\; numbers\; \frac{a}{b}+0\;=\frac{a}{b}

5. Additive Inverse-

if\; \frac{a}{b}\;, is\;a\: rational\; numbers\; \frac{a}{b}+\left ( -\frac{a}{b} \right ) =0\;=\frac{a}{b}

\left ( -\frac{a}{b} \right ) is\; additive\; Inverse \; of \;\frac{a}{b}\; and\; vice \; versa

Exercise 1 B

Add the following.

a) \; \frac{2}{9}+\frac{5}{9}

\mathbf{Solution-} \; \frac{2}{9}+\frac{5}{9}=\frac{2+5}{9}=\frac{7}{9}

b) \; \frac{11}{15}+\frac{-7}{15}

\mathbf{Solution-} \; \frac{11}{15}+\frac{-7}{15}=\frac{11-7}{15}=\frac{4}{15}

c) \; \frac{(-5)}{12}+\frac{(-8)}{12}

solution- \; \frac{(-5)}{12}+\frac{(-8)}{12}=\frac{-5-8}{12}=\frac{-13}{12}

d) \; \frac{7}{40}+\frac{(-23)}{40}

solution- \; \frac{7}{40}+\frac{(-23)}{40}=\frac{7-23}{40}=\frac{-16}{40}

e) \; \frac{-8}{25}+\frac{4}{25}

solution- \; \frac{-8}{25}+\frac{4}{25}=\frac{-8+4}{25}=\frac{-16}{40}

f) \; \frac{22}{35}+\frac{-15}{35}

solution:-\; \frac{22}{35}+\frac{-15}{35}=\frac{22-15}{35}=\frac{7}{35}=\frac{1}{5}

2) Find the sum of :

a)\; \frac{3}{5}\; and \; \frac{-2}{5}

solution-\; \frac{3}{5}\; + \; \frac{-2}{5}=\frac{3-2}{5}=\frac{1}{5}

b)\; \frac{-5}{8}\; and \; \frac{1}{4}

Solution- LCM of 8 and 4 is 8

Converting into equivalent rational numbers equating to LCM by multiplying same number both in numerator and denominator

\frac{-5}{8}\;= \; \frac{-5\times 1}{8\times 1}=\frac{-5}{8}

\frac{1}{4}\;= \; \frac{1\times 2}{4\times 2}=\frac{2}{8}

\frac{-5}{8}+\frac{1}{4}\;= \; \frac{-5}{8}+\frac{2}{8}=\frac{-5+2}{8}=\frac{-3}{8}

c)\; \frac{3}{7}\; and\: \frac{-1}{14}\;

Solution- LCM of 7 and 14 is 14

Converting into equivalent rational numbers equating to LCM by multiplying same number in both numerator and denominator

\; \frac{3}{7}\; =\frac{3\times 2}{7\times 2} =\: \frac{6}{14}\;

\; \frac{-1}{14}\; =\frac{-1\times 1}{14\times 1} =\: \frac{-1}{14}\;

\;\frac{3}{7} +\frac{-1}{14}\; =\frac{6}{14}- \frac{1}{14}\;=\frac{6-1}{14}=\frac{5}{14}

d)\;\frac{-3}{11}\; and\; \frac{15}{22}\;

Solution- LCM of 11 and 22 is 22

Converting into equivalent rational numbers equating to LCM by multiplying same number with both numerator and denominator

\;\frac{-3}{11}\; =\frac{-3\times 2}{11\times 2} =\; \frac{-6}{22}\;

\;\frac{15}{22}\; =\frac{15\times 1}{22\times 1} =\; \frac{15}{22}\;

now\;\frac{-3}{11}+\frac{15}{22}\; =\frac{-6}{22} +\; \frac{15}{22}\;=\frac{-6+15}{22}=\frac{9}{22}

e)\;\frac{8}{25} \; and\; \frac{8}{50}\;

Solution- LC M of 25 and 50 is 50

Converting into equivalent rational number

Converting into equivalent rational numbers equating to LCM by multiplying same number both in numerator and denominator.

\;\frac{8}{25} \; =\; \frac{8\times 2}{25\times 2}\;=\frac{16}{50}

\;\frac{8}{50} \; =\; \frac{8\times 1}{50\times 1}\;=\frac{8}{50}

Now,\;\frac{8}{25} \; +\; \frac{8}{50}=\frac{16}{50}+\frac{8}{50}=\frac{16+8}{50}=\frac{24}{50}

f) \;\frac{-3}{18} \; ,\; \frac{5}{6}

Solution- LCM of 18 and 6 is 18

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number both in numerator and denominator.

\;\frac{-3}{18} \;=\; \frac{-3\times 1}{18\times 1}=\frac{-3}{18}

\;\frac{5}{6} \;=\; \frac{5\times 3}{6\times 3}=\frac{15}{18}

Now,\;\frac{-3}{18} \; +\; \frac{5}{6}=\frac{-3}{18}+\frac{15}{18}=\frac{-3+15}{18}=\frac{12}{18}

3) Simplify:

a)\;\frac{-7}{11} \; +\; \frac{1}{6}

Solution- LCM of 11 and 6 is 66.

Now converting the rational numbers into equivalent rational numbers equating to LCM by multiplying same number with numerator and denominator.

\;\frac{-7}{11} \;=\; \frac{-7\times 6}{11\times 6}=\frac{-42}{66}

\;\frac{1}{6} \;=\; \frac{1\times 11}{6\times 11}=\frac{11}{66}

Now,\;\frac{-7}{11} \;+\; \frac{1}{6}=\frac{-42}{66}+\frac{11}{66}=\frac{-42+11}{66}=\frac{-31}{66}

b)\;\frac{-3}{7} \;+\; \frac{2}{5}

solution- LCM of 7 and 5 is 35

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number in both numerator and denominator

\;\frac{-3}{7} \;=\; \frac{-3\times 5}{7\times 5}=\frac{-15}{35}

\;\frac{2}{5} \;=\; \frac{2\times 7}{5\times 7}=\frac{14}{35}

now,\;\frac{-3}{7} \;+\; \frac{2}{5}=\frac{-15}{35}+\frac{14}{35}=\frac{-15+14}{35}=\frac{-1}{35}

c)\;\frac{-7}{9} \;+\; \frac{2}{7}

solution- LCM of 9 and 7 is 63

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number in both numerator and denominator

\;\frac{-7}{9} \;=\; \frac{(-7)\times 7}{9\times 7}=\frac{-49}{63}

\;\frac{2}{7} \;=\; \frac{2\times 9}{7\times 9}=\frac{18}{63}

now,\;\frac{-7}{9} \;+\frac{2}{7}=\; \frac{-49}{63}+\frac{18}{63}=\frac{-49+18}{63}=\frac{-31}{63}

d)\;\frac{3}{4} \;+\frac{-2}{5}

solution- LCM of 4 and 5 is 20

Converting rational numbers into equivalent rational numbers equating to LCM by multiplying same number both in numerator and denominator

\;\frac{3}{4} \;=\frac{3\times 5}{4\times 5}=\frac{15}{20}

\;\frac{-2}{5} \;=\frac{(-2)\times 4}{5\times 4}=\frac{-8}{20}

now\;\frac{3}{4} \;+\; \frac{-2}{5}=\frac{15}{20}\frac{-8}{20}=\frac{15-8}{20}=\frac{7}{20}

e)\;\frac{3}{5} \;+\; \frac{1}{6}

Solution- LCM of 5 and 6 is 30

Converting into equivalent rational numbers equating LCM by multiplying same number both in numerator and denominator

\;\frac{3}{5} \;=\; \frac{3\times 6}{5\times 6}=\frac{18}{30}

\;\frac{1}{6} \;=\; \frac{1\times 5}{6\times 5}=\frac{5}{30}

now\;\frac{3}{5}+\frac{1}{6} \;=\;\frac{18}{30}+ \frac{ 5}{30}=\frac{18+5}{30}=\frac{23}{30}

f)\;\frac{-18}{20}+\frac{6}{11}

Solution- LCM of 20 and 11 is 220

converting into equivalent rational numbers equating to LCM by multiplying same number both in numerator and denominator we get

\;\frac{-18}{20}=\frac{(-18)\times 11}{20\times 11}=\frac{-198}{220}

\;\frac{6}{11}=\frac{6\times 20}{11\times 20}=\frac{120}{220}

now\;\frac{-18}{20}+\frac{6}{11}=\frac{-198}{220}+\frac{120}{220}=\frac{-198+120}{220}=\frac{-78}{220}

g)\;\frac{13}{60}+\frac{-3}{36}

solution- LCM of 60 and 36 is 36 x 10 = 360

Converting into equivalent rational numbers equating to LCM by multiplying same number n numeraor and denominator

\;\frac{13}{60}=\frac{13\times 6}{60\times 6}=\frac{78}{360}

\;\frac{-3}{36}=\frac{-3\times 10}{36\times 10}=\frac{-30}{360}

now\;\frac{13}{60}+\frac{-3}{36}=\frac{78}{360}+\frac{-30}{360}=\frac{78-30}{360}=\frac{48}{360}=\frac{8}{60}

h)\;\frac{6}{40}+\frac{1}{25}

solution- LCM of 40 and 25 is 1000.

Converting into equivalent rational numbers, equating to LCM by multiplying same number with both numerators and denominators

\;\frac{6}{40}=\frac{6\times 25}{40\times 25}=\frac{150}{1000}

\;\frac{1}{25}=\frac{1\times 40}{40\times 25}=\frac{40}{1000}

\;\frac{6}{40}+\frac{1}{25}=\frac{150}{1000}+\frac{40}{1000}=\frac{150+40}{1000}=\frac{190}{1000}

i)\;\frac{16}{18}+\frac{-16}{27}

Solution- LCM of 18 and 27 is 9 x 2 x 3=54

Converting into equivalent rational numbers, equating LCM by multiplying same number in both numerator and denominator

\;\frac{16}{18}=\frac{16\times 3}{18\times 3}=\frac{48}{54}

\;\frac{16}{27}=\frac{16\times 2}{27\times 2}=\frac{32}{54}

Now,\;\frac{16}{18}+\frac{-16}{27}=\frac{48}{54}+\frac{-32}{54}=\frac{48-32}{54}=\frac{16}{54}=\frac{8}{27}

j)\;\frac{-21}{81}+\frac{5}{18}

Solution- LCM of 81 and 18 is 2 x 3 x 3 x 3 x 3=162

Converting into equivalent rational numbers, equating to LCM by multiplying same number both in numerator and denominator

\;\frac{-21}{81}=\frac{-21\times 2}{81\times 2}=\frac{-42}{162}

\;\frac{5}{18}=\frac{5\times 9}{18\times 9}=\frac{45}{162}

now,\;\frac{-21}{81}+\frac{5}{18}=\frac{-42}{162}+\frac{45}{162}=\frac{-42+45}{162}=\frac{3}{162}

4) Name the properties of addition implied in the following statements.

a)\;\frac{3}{4}+\frac{1}{3}=\frac{1}{3}+\frac{3}{4}

Solution- Addition of rational number is Commutative

b)\;\left ( \frac{7}{12}+\frac{3}{7} \right )+\frac{1}{4}=\frac{7}{12}+\left ( \frac{3}{7}+\frac{1}{4} \right )

Solution- Addition if rational numbers is associative.

c)\;\frac{-3}{8}+\frac{1}{9}=\frac{1}{9}+\frac{-3}{8}

Solution- Addition of rational numbers is commutative

d)\;\frac{-7}{16}+0=\frac{-7}{16}=0+\frac{-7}{16}

Solution- Additive Identity of rational number

e)\;\frac{-8}{11}+\frac{8}{11}=0

Solution- Additive Inverse property

f)\;\left ( \frac{-4}{7}+\frac{-3}{8} \right )+\frac{4}{9}=\frac{-4}{7}+\left ( \frac{-3}{8}+\frac{4}{9} \right )

Solution- Associative Property of rational number

5) Fill in the boxes. Also mention the property used.

a)\;\frac{3}{7}+\frac{4}{9}=\frac{4}{9}+……..

\textup{Solution}\;\frac{3}{7}+\frac{4}{9}=\frac{4}{9}+\frac{3}{7} \textup{\; \; Commutative Property}

b)\;\frac{6}{13}+\left ( \frac{4}{9}+\frac{7}{8} \right )=\left ( \frac{6}{13}+\frac{4}{9} \right )+……

\;\frac{6}{13}+\left ( \frac{4}{9}+\frac{7}{8} \right )=\left ( \frac{6}{13}+\frac{4}{9} \right )+\frac{7}{8} \textup{\; \; Associative Property}

c)\;\frac{-7}{13}+\frac{-2}{9}=\frac{-2}{9}+……….

\textup{solution-}\;\frac{-7}{13}+\frac{-2}{9}=\frac{-2}{9}+\frac{-7}{13} \; \; \textup{Commutative Property}

\textup{solution-}\;\frac{-4}{11}+\left ( \frac{-2}{9}+\frac{-8}{13} \right )=\left ( \frac{-4}{11}+…… \right )+\frac{-8}{13} \; \; \; \textup{Associative Property}

e)\;\frac{-7}{18}+………=\frac{-7}{18}

\textup{solution-}\;\frac{-7}{18}+0=\frac{-7}{18}\; \; \textup{Additive Identity}

f)\;\frac{-7}{26}+…….=0\; \;

solution-)\;\frac{-7}{26}+\frac{7}{26}=0\; \; \textup{Additive Inverse}

g)\;…….+0=\frac{3}{4}

\textup{solution-}\;\frac{3}{4}+0=\frac{3}{4} \; \; \textup{Additive Identity}

h)\;……….+\frac{2}{3}=0 \; \;

\textup{solution-}\;\frac{-2}{3}+\frac{2}{3}=0 \; \; \textup{Additive Inverse}

6) Show that

\left ( \frac{-2}{5}+\frac{4}{9} \right )+\frac{-3}{4}=-\frac{2}{5}+\left ( \frac{4}{9}+\frac{-3}{4} \right )

Solution- LCM of 5 and 9 is 45

Now converting into equivalent rational number by multiplying same nuber in both numerator and denominator.

\left ( \frac{-2}{5}+\frac{4}{9} \right )=\frac{(-2)\times 9+4\times 5}{45}=\frac{-18+20}{45}=\frac{2}{45}

Now LHS=

\left ( \frac{-2}{5}+\frac{4}{9} \right )+\frac{-3}{4}=\frac{2}{45}+\frac{-3}{4}=\frac{8-135}{180}=\frac{-127}{180}

Now

\frac{4}{9} +\frac{-3}{4}=\frac{4\times 4-3\times 9}{36}=\frac{-11}{36}

\textup{RHS-}\frac{-2}{5}+\left ( \frac{4}{9}+\frac{-3}{4} \right )=-\frac{2}{5}+\frac{-11}{36}=\frac{-72-55}{180}=\frac{-127}{180}

Now we have LHS= RHS

7) Verify that

a+(b+c)=(a+b)+c\; \; by \; taking\; a=-\frac{4}{5},\; b=\frac{7}{8},\; and\; c=\frac{3}{4}

solution- LHS-

a+(b+c)=\; \; \; -\frac{4}{5}+(\; \frac{7}{8}\;+\frac{3}{4})

=-\frac{4}{5}+\frac{(7+6)}{8}=-\frac{4}{5}+\frac{13}{8}=\frac{(-4)\times 8+13\times 5}{5\times 8}

=\frac{-32+65}{40}=\frac{33}{40}

RHS-

\; (a+b)+c=(\frac{-4}{5}+\frac{7}{8})+\frac{3}{4}=\frac{(-4)\times 8+7\times 5)}{40}+\frac{3}{4}

=\frac{-32+35}{40}+\frac{3}{4}=\frac{3}{40}+\frac{3}{4}=\frac{3+30}{40}=\frac{33}{40}

Now we have LHS=RHS

8) Add as fast as possible using associative and commutative property

a) 76+598+224

Solution- 76+598+224

= (76+224)+598 (Associative property)

= 300+598

=898

b)\; \frac{7}{38}+\frac{17}{20}+\frac{31}{38}

\textup{solution:-}\; \frac{7}{38}+\frac{17}{20}+\frac{31}{38}=\left ( \frac{7}{38}+\frac{31}{38} \right )+\frac{17}{20}=\frac{38}{38}+\frac{17}{20}=1+\frac{17}{20}=1\frac{17}{20}

c)\; 782+890+218

\textup{solution-}\; 782+890+218=(782+218)+890

=1000+890=1890

Summary

Oxford New Enjoying Mathematics Class 8 Solutions Chapter 1

Article Name

Oxford New Enjoying Mathematics Class 8 Solutions Chapter 1

Description

We solve Exercise 1A, 1B, of Chapter 1 Rational Numbers of Oxford New Enjoying Mathematics Class 8 Solutions. Discuss properties and addition

Author

Mamta Mund

Publisher Name

https://www.mathsgrade.com/

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