Exercise 2D
Q1 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper
Q1 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper
Question 1
Find the reciprocal of :
(a) $\frac{1}{3}$
(b) $\frac{7}{8}$
(c) $2 \frac{2}{3}$
(d) 5
Sol :
(a) Reciprocal of $\frac{1}{3}=3$
(b) Reciprocal of $\frac{7}{8}=\frac{8}{7}=1\frac{1}{7}$
(c) Reciprocal of 2$\frac{2}{3}=\frac{3}{8}$
(d) Reciprocal of $5=\frac{1}{5}$
(e) Reciprocal of $6\frac{1}{6}=\frac{6}{37}$
Q2 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper
Question 2
By taking following examples, show that the product of a number and its reciprocal
(a) $\frac{25}{37}$
(b) $5 \frac{1}{9}$
(c) $7 \frac{1}{7}$
(d) $1 \frac{11}{35}$
Sol :
(a) Reciprocal of $\frac{25}{37}=\frac{37}{25}$
∴$\frac{25}{37}\times \frac{37}{25}$
=1
(b) Reciprocal of $5\frac{1}{9}=\frac{46}{9}$
∴$\frac{46}{9}\times \frac{9}{46}$
=1
(c) $7\frac{1}{7}=\frac{50}{7}$
∴$\frac{50}{7}\times \frac{7}{50}$
=1
(d) $1\frac{11}{35}=\frac{46}{35}$
Reciprocal of $\frac{46}{35}=\frac{35}{46}$
=1
Q3 | Ex-2D | Class 7 | Schand New Learning Composite | Fractions and Decimals | Chapter 2| myhelper
Question 3
Think and answer- in defining reciprocal of we have imposed the condition $a\not =0,~ b\not=0$ why ? Give reason.
Sol :
Because division by zero is not defined
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