The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 2 Brain Teasers Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 2 Brain Teasers Solutions
Question 1.
A. Tick (✓) the correct option:
(i) The additive inverse of \(\frac{-3}{4}\) is
(a) \(\frac{-3}{4}\)
(b) \(\frac{-4}{3}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{3}{4}\)
Answer:
(d) \(\frac{3}{4}\)
If \(\frac{-3}{4}\) is a rational number, then -(\(\frac{-3}{4}\)) i.e, \(\frac{3}{4}\) is the additive inverse of \(\frac{-3}{4}\) such that \(\frac{-3}{4}+\left[-\left(\frac{-3}{4}\right)\right]=\frac{-3}{4}+\left(\frac{-3}{4}\right)\) = 0
(ii) If x, y and z are rational numbers, then the property (x + y) + z = x + (y + z) is known as
(a) communicative property
(b) associative property
(c) distributive property
(d) closure property
Answer:
(b) associative property
Sum of three rational numbers remains same even of after changing the grouping of the addends, i.e., if x, y and z are three rational numbers then (x + y) + z = x + (y + z) This is known as associative law of addition.
Hence, (b) is the correct answer.
(iii) \(\frac{7}{12} \div\left(\frac{-7}{12}\right)\) is
(a) 1
(b) 7
(c) -1
(d) does not exist
Answer:
(c) -1
\(\frac{7}{12} \div\left(\frac{-7}{12}\right)\)
= \(\frac{7}{12} \times\left(\frac{-12}{7}\right)\)
= -1
Hence, (c) is the correct answer.
(iv) Identity element for subtraction of rational numbers is
(a) 1
(b) 0
(c) -1
(d) -7
Answer:
(d) -7
For all rational numbers x, we have
x – 0 = x
But, 0 – x = -x
Therefore, identity element for subtraction of rational numbers does not exist.
Hence, (d) is the correct answer.
(v) The multiplicative inverse of 6\(\frac{1}{3}\) is
(a) \(\frac{-19}{3}\)
(b) \(\frac{-3}{19}\)
(c) \(\frac{3}{19}\)
(d) \(\frac{19}{3}\)
Answer:
(c) \(\frac{3}{19}\)
6\(\frac{1}{3}\) = \(\frac{19}{3}\)
So, by simply interchanging the numerator and denominator, we can find the multiplicative inverse.
Therefore, multiplicative inverse of 6\(\frac{1}{3}\) i.e., \(\frac{19}{3}\) is \(\frac{3}{19}\).
Hence, (c) is the correct answer.
B. Answer the following questions:
(i) Write all rational numbers whose absolute value is \(\frac{5}{9}\).
Answer:
The absolute value of a rational number written in the following ways.
Absolute value of \(\frac{5}{9}\) is \(\left|\frac{5}{9}\right|=\frac{5}{9}\)
Absolute value of \(\frac{-5}{9}\) is \(\left|\frac{-5}{9}\right|=\frac{5}{9}\)
Therefore, \(\frac{5}{9}\) and \(\frac{-5}{9}\) are the rational numbers whose absolute value is \(\frac{5}{9}\)
(ii) Find the reciprocal of \(\frac{4}{5} \times\left(\frac{3}{-8}\right)\)
Answer:
\(\frac{4}{5} \times\left(\frac{3}{-8}\right)\)
= \(\frac{4 \times 3}{5 \times(-8)}\) [Divide 4 and (-8) by 4 to get rational number in standard form].
= \(\frac{1 \times 3}{5 \times(-2)}=\frac{3}{(-10)}=\frac{-3}{10}\)
Now, by simply interchanging the numerator and denominator, we can find reciprocal.
∴ Reciprocal of \(\frac{-3}{10}\) is \(\frac{10}{-3}\),i.e., \(\frac{-10}{3}\).
(iii) What should be added to \(\frac{-5}{11}\) to get \(\frac{26}{33}\)
Answer:
Here, the sum of the two numbers is \(\frac{26}{33}\) and one of them is \(\frac{-5}{11}\).
Required number = \(\frac{26}{33}-\left(\frac{-5}{11}\right)\)
= \(\frac{26}{33}+\frac{5}{11}=\frac{26+15}{33}=\frac{41}{33}\)
Hence, the required number is \(\frac{41}{33}\).
(iv) Subtract 6\(\frac{2}{3}\) from the sum of \(\frac{-3}{7}\) and 2.
Answer:
Sum of \(\frac{-3}{7}\) and 2 = \(\frac{-3}{7}\) + 2
= \(\frac{-3}{7}+\frac{2}{1}=\frac{-3+14}{7}=\frac{11}{7}\)
Now, we have to subtract 6\(\frac{2}{3}\) i.e., \(\frac{20}{3}\) from \(\frac{11}{7}\).
⇒ \(\frac{11}{7}-\frac{20}{3}=\frac{33-140}{21}=\frac{-107}{21}\)
(v) Find the value of 1 + \(\frac{1}{1+\frac{1}{6}}\)
Answer:
1 + \(\frac{1}{1+\frac{1}{6}}\)
= 1 + \(\frac{1}{\left(\frac{1}{1}+\frac{1}{6}\right)}\)
= 1 + \(\frac{1}{\left(\frac{6+1}{6}\right)}\)
= 1 + \(\frac{1}{\left(\frac{7}{6}\right)}\)
= 1 + (1 × \(\frac{6}{7}\))
= 1 + \(\frac{6}{7}\)
= \(\frac{7+6}{7}\)
= \(\frac{13}{7}\)
Question 2. State whether the following statements are true or false. If false, justify your answer with an example.
(i) If |x| = 0, then x has no reciprocal.
Answer:
True
(ii) If x < y then |x| < |y|
Answer:
False.
Justification:
Take x = – 4 and y = – 3
– 4 < – 3 but |-4| > |-3| = 4 > 3
So it is false.
(iii) If x-1 < y-1
Answer:
False.
Justification:
Take x = 3 and y = 4
Here 3 < 4 and \(\frac{1}{3}>\frac{1}{4}\)
(iv) The negative of a negative rational number is a positive rational number.
Answer:
True
(v) Product of two rational numbers can never be an integer.
Answer:
False
Justification:
Take x = \(\frac{2}{3}\) and y = \(\frac{-3}{2}\)
∴ xy = \(\frac{2}{3} \times \frac{-3}{2}\) = -1 integer
(vi) Product of two integers is never a fraction.
Answer:
True
(vii) If x and y are two rational numbers such that x > y then x – y is always a positive rational number.
Answer:
True
Question 3.
For x = \(\frac{3}{4}\) and y = \(\frac{-9}{8}\), insert a rational number between:
(i) (x + y)-1 and x-1 + y -1
Answer:
(x + y) -1 = \(\left(\frac{3}{4}+\frac{-9}{8}\right)^{-1}=\left(\frac{3}{4}-\frac{9}{8}\right)^{-1}\)
= \(\left(\frac{3 \times 2-9 \times 1}{8}\right)^{-1}\)
= \(\left(\frac{6-9}{8}\right)^{-1}=\left(\frac{-3}{8}\right)^{-1}\)
= \(\frac{-8}{3}\)
x-1 + y-1 = \(\)
= \(\left(\frac{3}{4}\right)^{-1}+\left(\frac{-9}{8}\right)^{-1}\)
= \(\frac{4}{3}-\frac{8}{9}=\left(\frac{4 \times 3-8 \times 1}{9}\right)\)
= \(\frac{12-8}{9}=\frac{4}{9}\)
Rational number between \(\) and \(\)
= \(\frac{1}{2}\left(\frac{-8}{3}+\frac{4}{9}\right)\)
= \(\frac{1}{2}\left(\frac{-8 \times 3+4 \times 1}{9}\right)\)
= \(\frac{1}{2}\left(\frac{-24+4}{9}\right)\)
= \(\frac{1}{2} \times \frac{-2}{9}=\frac{-10}{9}\)
Hence the required rational number = \(\frac{-10}{9}\)
(ii) (x – y)-1 and x-1 – y-1
Answer:
(x – y)-1 = \(\left(\frac{3}{4}-\frac{-9}{8}\right)^{-1}\)
= \(\left(\frac{3}{4}+\frac{9}{8}\right)^{-1}\left(\frac{3 \times 2+9 \times 1}{8}\right)^{-1}\)
= \(\left(\frac{6+9}{8}\right)^{-1}=\left(\frac{15}{18}\right)^{-1}\)
= \(\frac{8}{15}\)
x-1 – y-1 = \(\left(\frac{3}{4}\right)^{-1}-\left(\frac{-9}{8}\right)^{-1}\)
= \(\frac{4}{3}-\left(\frac{-8}{9}\right)\)
= \(\left(\frac{4}{3}+\frac{8}{9}\right)\)
= \(\left(\frac{4 \times 3+8 \times 1}{9}\right)\)
= \(\frac{12+8}{9}=\frac{20}{9}\)
Rational number between \(\frac{8}{15}\) and \(\frac{20}{9}\)
= \(\frac{1}{2}\left(\frac{8}{15}+\frac{20}{9}\right)\)
= \(\frac{1}{2}\left(\frac{3 \times 8+20 \times 5}{45}\right)\)
= \(\frac{1}{2}\left(\frac{24+100}{45}\right)\)
= \(\frac{1}{2} \times \frac{124}{45}=\frac{62}{45}\)
Hence the required rational number is \(\frac{62}{45}\)
Question 4.
Verify that: (x ÷ y)-1 = x-1 ÷ y-1 by taking x = \(\frac{-5}{11}\), y = \(\frac{7}{3}\).
Answer:
L.H.S = (x ÷ y)-1 = \(\left(\frac{-5}{11} \div \frac{7}{3}\right)^{-1}\)
= \(\left(\frac{-5}{11} \times \frac{3}{7}\right)^{-1}\)
= \(\left(\frac{-5 \times 3}{11 \times 7}\right)^{-1}\)
= \(\left(\frac{-15}{77}\right)^{-1}\)
= \(\frac{77}{-15}\)
R.H.S = x-1 ÷ y-1 = \(\left(\frac{-5}{11}\right)^{-1} \div\left(\frac{7}{3}\right)^{-1}\)
= \(\frac{11}{-5} \div \frac{3}{7}\)
= \(\frac{11}{-5} \times \frac{7}{3}\)
= \(\frac{11 \times 7}{-5 \times 3}\)
= \(\frac{77}{-15}\)
L.H.S = R.H.S
Hence Verified.
Question 5.
Verify that |x + y| ≤ |x| + |y| by taking x = \(\frac{2}{3}\), y = \(\frac{-3}{5}\)
Answer:
|x + y| = \(\left|\frac{2}{3}+\left(\frac{-3}{5}\right)\right|\)
= \(\left|\frac{2}{3}-\frac{3}{5}\right|\)
= \(\left|\frac{10-9}{15}\right|\)
= \(\left|\frac{1}{15}\right|\)
= \(\frac{1}{15}\)
Now, |x| + |y| = \(\left|\frac{2}{3}\right|+\left|\frac{-3}{5}\right|\)
= \(\frac{2}{3}+\frac{3}{5}\)
= \(\frac{10+9}{15}\)
= \(\frac{19}{15}\)
As \(\frac{1}{15}<\frac{19}{15}\), so |x + y| ≤ |x| + |y|
Hence, verified.
Question 6.
Find the reciprocals of:
(i) \(\frac{2}{-5} \times \frac{3}{-7}\)
Answer:
Reciprocal of
\(\frac{2}{-5} \times \frac{3}{-7}\)
= \(\left(\frac{2}{-5} \times \frac{3}{-7}\right)^{-1}\)
= \(\left(\frac{2 \times 3}{-5 \times-7}\right)^{-1}\)
= \(\left(\frac{6}{35}\right)^{-1}=\frac{35}{6}\)
(ii) \(\frac{-4}{3} \times \frac{-5}{-8}\)
Answer:
Reciprocal of
\(\left(\frac{6}{35}\right)^{-1}=\frac{35}{6}\)
= \(\left(\frac{-4 x-5}{3 \times-8}\right)^{-1}\)
= \(\left(\frac{20}{-24}\right)^{-1}\)
= \(\frac{-24}{20}=\frac{-6}{5}\) (Standard form)
Question 6.
Simplify:
(i) \(\left|\frac{5}{7}-\frac{2}{3}\right|+\left|\frac{3}{14}-\frac{5}{7}\right|\)
Answer:
(ii) \(\left(\frac{5}{11}\right)^{-1}-\frac{13}{5}+\frac{3}{15}\)
Answer:
(iii) \(\frac{9}{5} \times \frac{-2}{27}+\frac{7}{30}\)
Answer:
(iv) \(\frac{-7}{15} \div\left(\frac{50}{3}\right)^{-1}\)
Answer:
\(\frac{-7}{15} \div\left(\frac{50}{3}\right)^{-1}\)
= \(\frac{-7}{15} \div\left(\frac{3}{50}\right)\)
= \(\frac{-7}{15} \times \frac{50}{3}\)
= \(\frac{-70}{9}\)
Question 7.
Divide:
(i) The sum of \(\frac{5}{21}\) and \(\frac{4}{7}\) by their difference.
Answer:
Sum = \(\frac{5}{21}+\frac{4}{7}\)
= \(\frac{5 \times 1+4 \times 3}{21}\)
= \(\frac{5+12}{21}\)
= \(\frac{17}{21}\)
Difference = \(\frac{5}{21}-\frac{4}{7}\)
= \(\frac{5 \times 1-4 \times 3}{21}\)
= \(\frac{5-12}{21}=\frac{-7}{21}\)
Now \(\frac{17}{21} \div \frac{-7}{21}\)
= \(\frac{17}{21} \times \frac{21}{-7}\)
= \(\frac{-17}{7}\)
(ii) The difference of \(\frac{12}{5}, \frac{-16}{20}\) by their product.
Answer:
Difference of \(\frac{12}{5}\) and \(\frac{-16}{20}\)
= \(\frac{12}{5}-\left(\frac{-16}{20}\right)\)
= \(\frac{12}{5}+\frac{16}{20}\)
= \(\frac{12 \times 4+16 \times 1}{20}\)
= \(\frac{48+16}{20}\)
= \(\frac{64}{20}\)
= \(\frac{16}{5}\)
Product of \(\frac{12}{5}\) and \(\frac{-16}{20}\)
= \(\frac{12}{5} \times \frac{-16}{20}\)
= \(\frac{-48}{25}\)
∴ \(\frac{16}{5} \div \frac{-48}{25}\)
= \(\frac{16}{5} \times \frac{-25}{48}\)
= \(\frac{-5}{3}\)
Question 8.
Find the reciprocal of \(\frac{-2}{3} \times \frac{5}{7}+\frac{2}{9} \div \frac{1}{3} \times \frac{6}{7}\)
Answer:
\(\frac{-2}{3} \times \frac{5}{7}+\frac{2}{9} \div \frac{1}{3} \times \frac{6}{7}\)
= \(\frac{-2}{3} \times \frac{5}{7}+\frac{2}{9} \div \frac{1}{3} \times \frac{6}{7}\)
= \(\frac{-10}{21}+\frac{4}{7}\)
= \(\frac{-10+12}{21}\)
= \(\frac{2}{21}\)
∴ Its reciprocal = \(\left(\frac{2}{21}\right)^{-1}=\frac{21}{2}\)
DAV Class 7 Maths Chapter 2 HOTS
Question 1.
A drum of kerosene oil is \(\frac{3}{4}\) full. When 15 litres of oil is drawn from it, it is \(\frac{7}{12}\) full the total capacity of the drum.
Answer:
Let the total capacity of the drum be x litres.
According to the question, we have
\(\frac{3}{4}\)x – 15 = \(\frac{7}{12}\)x
⇒ \(\frac{3}{4}\)x – \(\frac{7}{12}\)x = 15
⇒ \(\frac{9 x-7 x}{12}\) = 15
⇒ \(\frac{2x}{12}\) = 15
⇒ x = \(\frac{12 \times 15}{2}\)
⇒ x = 6 × 15 = 90
Therefore, the total capacity of the drum is 90 litres.
Question 2.
Find the product of:
(1 – \(\frac{1}{2}\))(1 – \(\frac{1}{3}\))(1 – \(\frac{1}{4}\)) ……… (1 – \(\frac{1}{10}\))
Answer:
DAV Class 7 Maths Chapter 2 Enrichment Questions
Question 1.
Complete the following magic square of multiplication.
Answer:
Additional Questions
Question 1.
Find equivalent forms of rational numbers having a common denominator in the given group of rational numbers. \(\frac{5}{8}, \frac{6}{9}, \frac{5}{36}, \frac{13}{30}\)
Solution:
L.C.M. of 8, 9, 36, 30 = 360
= \(\frac{5 \times 45}{8 \times 45}, \frac{6 \times 40}{9 \times 40}, \frac{5 \times 10}{36 \times 10}, \frac{13 \times 12}{30 \times 12}\)
= \(\frac{225}{360}, \frac{240}{360}, \frac{50}{360}, \frac{156}{360}\)
Question 2.
Verify commutative property for x = \(\frac{-2}{7}\) and y = \(\frac{3}{5}\).
Answer:
Commutative property for rational numbers is
x + y = y + x
L.H.S. = x + y = \(\frac{-2}{7}+\frac{3}{5}\)
= \(\frac{-2 \times 5+3 \times 7}{35}\)
= \(\frac{-10+21}{35}\)
= \(\frac{11}{35}\)
R.H.S = y + x = \(\frac{3}{5}+\left(\frac{-2}{7}\right)\)
= \(\frac{3}{5}-\frac{2}{7}\)
= \(\frac{3 \times 7-2 \times 5}{35}\)
= \(\frac{21-10}{35}\)
= \(\frac{11}{35}\)
L.H.S. = R.H.S.
Hence verified.
Question 3.
Verify associative property for rational number if x = \(\frac{3}{8}\), y = \(\frac{5}{6}\) and z = \(\frac{-2}{3}\).
Answer:
Associative property for rational numbers is
x + (y + z) = (x + y) + z
L.H.S = x + (y + z) = \(\frac{3}{8}+\left(\frac{5}{6}+\frac{-2}{3}\right)\)
= \(\frac{3}{8}+\left(\frac{5}{6}-\frac{2}{3}\right)\)
= \(\frac{3}{8}+\left(\frac{5 \times 1-2 \times 2}{6}\right)\)
= \(\frac{3}{8}+\left(\frac{5-4}{6}\right)\)
= \(\frac{3}{8}+\frac{1}{6}\)
= \(\frac{3 \times 3+1 \times 4}{24}\)
= \(\frac{9+4}{24}\)
= \(\frac{13}{24}\)
R.H.S = (x + y) + z = \(\left(\frac{3}{8}+\frac{5}{6}\right)+\left(\frac{-2}{3}\right)\)
= \(\left(\frac{3 \times 3+5 \times 4}{24}\right)-\frac{2}{3}\)
= \(\left(\frac{9+20}{24}\right)-\frac{2}{3}\)
= \(\frac{29}{24}-\frac{2}{3}\)
= \(\frac{29 \times 1-2 \times 8}{24}\)
= \(\frac{29-16}{24}\)
= \(\frac{13}{24}\)
L.H.S. = R.H.S. Hence verified.
Question 4.
Find the value of x × (y + z) and x × y + x × z. State the property held there in given expressions, if x = \(\frac{5}{7}\), y = \(\frac{3}{-4}\) and z = \(\frac{2}{5}\)
Answer:
x × (y + z) = \(\frac{5}{7} \times\left(\frac{-3}{4}+\frac{2}{5}\right)\)
= \(\frac{5}{7} \times\left(\frac{-3 \times 5+2 \times 4}{20}\right)\)
= \(\frac{5}{7} \times\left(\frac{-15+8}{20}\right)\)
= \(\frac{5}{7} \times \frac{-7}{20}=-\frac{1}{4}\)
x × y + x × z = \(\frac{5}{7} \times \frac{-3}{4}+\frac{5}{7} \times \frac{2}{5}\)
= \(\frac{5 \times-3}{7 \times 4}+\frac{5 \times 2}{7 \times 5}\)
= \(\frac{-15}{28}+\frac{10}{35}\)
= \(\frac{-15 \times 5+4 \times 10}{140}\)
= \(\frac{-75+40}{140}=\frac{-35}{140}\)
= \(\frac{-1}{4}\)
Here x × (y + z) = x × y + x × z = \(\frac{-1}{4}\)
∴ It is distribution of multiplication over addition property for rational numbers.
Question 5.
Find the value of the following \(\frac{3}{5} \times \frac{3}{4} \div \frac{9}{8}+\frac{3}{5}-\frac{4}{11}\) × 0
Answer:
Hence the required value is 1.
Question 6.
State whether x – y = y – x for x = \(\frac{-3}{7}\) and y = \(\frac{2}{5}\).
Answer:
L.H.S = x – y = \(\frac{-3}{7}-\frac{2}{5}\)
= \(\frac{-3 \times 5-2 \times 7}{35}\)
= \(\frac{-15-14}{35}\)
= \(\frac{-29}{35}\)
R.H.S = y – x = \(\frac{2}{5}-\left(\frac{-3}{7}\right)\)
= \(\frac{2}{5}+\frac{3}{7}\)
= \(\frac{2 \times 7+3 \times 5}{35}\)
= \(\frac{14+15}{35}\)
= \(\frac{29}{35}\)
L.H.S = R.H.S
So they are not equal.
Question 7.
The sum of two rational numbers is – 3. If one of them is \(\frac{-2}{7}\), find the other number.
Solution:
Sum of the two rational numbers = – 3
other number = \(\frac{-2}{7}\)
= -3 + (\(\frac{-2}{7}\))
= -3 + \(\frac{2}{7}\)
= \(\frac{-3}{1}+\frac{2}{7}\)
= \(\frac{-3 \times 7+2 \times 1}{7}\)
= \(\frac{-21+2}{7}\)
= \(\frac{-19}{7}\)
Question 8.
For x = \(\frac{5}{6}\), y = \(\frac{-3}{4}\) and z = \(\frac{1}{2}\) verify x × (y – z) = x × y – x × z
Answer:
L.H.S
x × (y – z) = \(\frac{5}{6} \times\left(\frac{-3}{4}-\frac{1}{2}\right)\)
= \(\frac{5}{6} \times\left(\frac{-3 \times 1-1 \times 2}{4}\right)\)
= \(\frac{5}{6} \times\left(\frac{-3-2}{4}\right)\)
= \(\frac{5}{6} \times \frac{-5}{4}=\frac{5 \times-5}{6 \times 4}\)
= \(\frac{-25}{24}\)
R.H.S
x × y – x × z = \(\frac{5}{6} \times \frac{-3}{4}-\frac{5}{6} \times \frac{1}{2}\)
= \(\frac{5 \times-3}{6 \times 4}-\frac{5 \times 1}{6 \times 2}\)
= \(\frac{-15}{24}-\frac{5}{12}\)
= \(\frac{-15-5 \times 2}{24}\)
= \(\frac{-15-10}{24}\)
= \(\frac{-25}{24}\)
L.H.S = R.H.S
Hence verified.
Question 9.
Verify that (x + y)-1 ≠ x-1 + y-1 if x = \(\frac{2}{5}\), y = \(\frac{-1}{3}\)
Answer:
L.H.S ≠ R.H.S
Hence verified
Question 10.
If x = \(\frac{2}{5}\), y = \(\frac{-1}{3}\) verify that (x – y)-1 ≠ x-1 – y-1
Answer:
L.H.S ≠ R.H.S
Hence verified
Question 11.
Verify that (x × y)-1 = x-1 × y-1 for x = \(\frac{2}{11}\) and y = \(\frac{-3}{7}\)
Answer:
L.H.S = (x × y)-1
= \(\left(\frac{2}{11} \times \frac{-3}{7}\right)^{-1}\)
= \(\left(\frac{2 \times-3}{11 \times 7}\right)^{-1}=\left(\frac{-6}{77}\right)^{-1}\)
= \(\frac{77}{-6}=\frac{-77}{6}\)
R.H.S = x-1 × y-1
= \(\left(\frac{2}{11}\right)^{-1} \times\left(\frac{-3}{7}\right)^{-1}\)
= \(\frac{11}{2} \times \frac{-7}{3}\)
= \(\frac{11 \times-7}{2 \times 3}\)
= \(\frac{-77}{6}\)
Question 12.
For x = \(\frac{4}{5}\), y = \(\frac{2}{9}\) and z = \(\frac{1}{3}\) verify that (x + y) ÷ z = x ÷ z + y ÷ z.
Answer:
Question 13.
State whether (x ÷ y) ÷ z = x ÷ (y ÷ z) for x = \(\frac{2}{3}\), y = \(\frac{1}{4}\) and z = \(\frac{1}{2}\).
Answer:
L.H.S = (x ÷ y) ÷ z
= \(\left(\frac{2}{3} \div \frac{1}{4}\right) \div \frac{1}{2}\)
= \(\left(\frac{2}{3} \times \frac{4}{1}\right) \div \frac{1}{2}\)
= \(\left(\frac{2 \times 4}{3 \times 1}\right) \div \frac{1}{2}\)
= \(\frac{8}{3} \div \frac{1}{2}=\frac{8}{3} \times \frac{2}{1}\)
= \(\frac{16}{3}\)
R.H.S = x ÷ (y ÷ z)
= \(\frac{2}{3} \div\left(\frac{1}{4} \div \frac{1}{2}\right)\)
= \(\frac{2}{3} \div\left(\frac{1}{4} \times \frac{2}{1}\right)\)
= \(\frac{2}{3} \div\left(\frac{1 \times 2}{4 \times 1}\right)\)
= \(\frac{2}{3} \div \frac{2}{4}=\frac{2}{3} \times \frac{4}{2}\)
= \(\frac{4}{3}\)
L.H.S ≠ R.H.S
So, (x ÷ y) ÷ z ≠ x ÷ (y ÷ z)
Question 14.
Find three rational number between \(\frac{-2}{3}\) and \(\frac{1}{3}\)
Answer:
First rational number between \(\frac{-2}{3}\) and \(\frac{1}{3}\)
= \(\frac{1}{2}\left(\frac{-2}{3}+\frac{1}{3}\right)\)
= \(\frac{1}{2}\left(\frac{-2+1}{3}\right)\)
= \(\frac{1}{2} \times \frac{-1}{3}=\frac{-1}{6}\)
Second rational number between \(\frac{-2}{3}\) and \(\frac{-1}{6}\)
= \(\frac{1}{2}\left(\frac{-2}{3}-\frac{1}{6}\right)\)
= \(\frac{1}{2}\left(\frac{-2 \times 2-1 \times 1}{6}\right)\)
= \(\frac{1}{2}\left(\frac{-4-1}{6}\right)=\frac{1}{2} \times \frac{-5}{6}\)
= \(\frac{-5}{12}\)
Third rational number between \(\frac{-1}{6}\) and \(\frac{1}{3}\)
= \(\frac{1}{2}\left(\frac{-1}{6}+\frac{1}{3}\right)\)
= \(\frac{1}{2}\left(\frac{-1+1 \times 2}{6}\right)\)
= \(\frac{1}{2}\left(\frac{-1+2}{6}\right)\)
= \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)
∴ Required rational number between \(\frac{-2}{3}\) and \(\frac{1}{3}\) are \(\frac{-1}{6}, \frac{-5}{12}, \frac{1}{12}\)
Question 15.
Divide the sum by the difference of two rational numbers x = \(\frac{-3}{8}\) and y = \(\frac{5}{9}\).
Answer:
Sum of the number x + y
Question 16.
Find the reciprocal of \(\frac{-2}{3} \times \frac{4}{7}-\frac{3}{5} \div \frac{6}{7} \times \frac{2}{5}\)
Answer:
