The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 4 Worksheet 1 Solutions of Application of Percentage offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 4 WS 1 Solutions
Question 1.
Write the base and exponent in each of the following:
(i) \(\left(\frac{-1}{3}\right)^3\)
Answer:
\(\left(\frac{-1}{3}\right)^3\), base = \(\frac{-1}{3}\) and exponent = 3
(ii) \(\left(\frac{-4}{7}\right)^6\)
Answer:
\(\left(\frac{-4}{7}\right)^6\), base = \(\frac{-4}{7}\) and exponent = 6
(iii) \(\left(\frac{2}{9}\right)^5\)
Answer:
\(\left(\frac{2}{9}\right)^5\), base = \(\frac{2}{9}\) and exponent = 5
(iv) \(\left(\frac{15}{19}\right)^3\)
Answer:
\(\left(\frac{15}{19}\right)^3\), base = \(\frac{15}{19}\) and exponent = 3
(v) (-15)4
Answer:
(-15)4, base = -15 and exponent = 4
(vi) \(\frac{-2}{3}\)
Answer:
\(\frac{-2}{3}=\left(\frac{-2}{3}\right)^1\), Here base = \(\frac{-2}{3}\) and exponent = 1
Question 2.
Express the following in exponential form:
(i) \(\frac{5}{6} \times \frac{5}{6}\)
Answer:
\(\frac{5}{6} \times \frac{5}{6}=\left(\frac{5}{6}\right)^2\)
(ii) \(\frac{9}{2} \times \frac{9}{2} \times \frac{9}{2} \times \frac{9}{2}\)
Answer:
\(\frac{9}{2} \times \frac{9}{2} \times \frac{9}{2} \times \frac{9}{2}=\left(\frac{9}{2}\right)^4\)
(iii) \(\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right)\)
Answer:
\(\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right)=\left(\frac{-7}{8}\right)^3\)
(iv) \(\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right)\)
Answer:
\(\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right) \times\left(\frac{-1}{2}\right)=\left(\frac{-1}{2}\right)^5\)
(v) 1.8 × 1.8 × 1.8 × 1.8 × 1.8 × 1.8 × 1.8
Answer:
1.8 × 1.8 × 1.8 × 1.8 × 1.8 × 1.8 × 1.8 = (1.8)7
(vi) \(\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12}\)
Answer:
\(\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12}=\left(\frac{11}{12}\right)^6\)
Question 3.
Express the following as rational numbers in the form
(i) \(\left(\frac{5}{6}\right)^3\)
Answer:
\(\left(\frac{5}{6}\right)^3=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}\)
= \(\frac{125}{216}\)
(ii) \(\left(\frac{-12}{13}\right)^2\)
Answer:
\(\left(\frac{4}{9}\right)^4=\frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9}\)
= \(\frac{256}{6561}\)
(iii) \(\left(\frac{4}{9}\right)^4\)
Answer:
\(\left(-\frac{1}{2}\right)^5=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)\)
= \(-\frac{1}{32}\)
(iv) \(\left(-\frac{1}{2}\right)^5\)
Answer:
\(\left(-\frac{1}{2}\right)^5=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)\)
= \(\left(-\frac{1}{2}\right)^5=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)\)
(v) \(\left(\frac{1}{4}\right)^4\)
Answer:
\(\left(\frac{1}{4}\right)^4=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{256}\)
(vi) \(\left(\frac{3}{5}\right)^3\)
Answer:
\(\left(\frac{3}{5}\right)^3=\frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}\)
= \(\frac{27}{125}\)
Question 4.
Express the following as powers of rational numbers:
(i) \(\frac{81}{625}\)
Answer:
\(\frac{81}{625}=\frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5}\)
= \(\frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}=\left(\frac{3}{5}\right)^4\)
(ii) \(\frac{-8}{125}\)
Answer:
\(\frac{-8}{125}=\frac{-2 \times 2 \times 2}{5 \times 5 \times 5}\)
= \(\left(\frac{-2}{5}\right) \times\left(\frac{-2}{5}\right) \times\left(\frac{-2}{5}\right)=\left(\frac{-2}{5}\right)^3\)
(iii) \(\frac{-343}{512}\)
Answer:
\(\frac{-343}{512}=\frac{-7 \times 7 \times 7}{8 \times 8 \times 8}\)
= \(\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right) \times\left(\frac{-7}{8}\right)=\left(\frac{-7}{8}\right)^3\)
(iv) \(\frac{32}{243}\)
Answer:
\(\frac{32}{243}=\frac{2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3}\)
= \(\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)=\left(\frac{2}{3}\right)^5\)
(v) \(\frac{-1}{216}\)
Answer:
\(\frac{-1}{216}=\frac{-1 \times 1 \times 1}{6 \times 6 \times 6}\)
= \(\left(\frac{-1}{6}\right) \times\left(\frac{-1}{6}\right) \times\left(\frac{-1}{6}\right)=\left(\frac{-1}{6}\right)^3\)
(vi) \(\frac{729}{1000}\)
Answer:
\(\frac{729}{1000}=\frac{9 \times 9 \times 9}{10 \times 10 \times 10}\)
= \(\frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}=\left(\frac{9}{10}\right)^3\).
DAV Class 8 Maths Chapter 4 Worksheet 1 Notes
Exponent is the power raised to any number.
For example:
(i) ab, here, b is the exponent and a is called the base.
(ii) (- 5)1/3, Here \(\frac{1}{3}\) is the power of exponent and – 5 is the base.
It is read as – 5 raised to the power 4
\(\left(\frac{3}{4}\right)^4=\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\)
Laws of exponents:
(i) If m is the Power of any rational number \(\frac{a}{b}\), b ≠ 0 then
\(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\) and \(\left(\frac{a}{b}\right)^0\) = 1
(ii) Reciprocal of \(\left(\frac{a}{b}\right)^m\) is \(\left(\frac{b}{a}\right)^m\) if m is a positive integer.
(iii) xm xn = xm + n, x ≠ 0
(iv) xm ÷ xn = xm – n, x ≠ 0
(v) xm ÷ xn = \(\frac{1}{x^{n-m}}\) if m < n and x ≠ 0
If x is non-zero rational number, then x0 = 1.
(vi) x– m = \(\frac{1}{x^m}\) i.e., x is the reciprocal of x
(vii) \(\left(\frac{p}{q}\right)^{-m}=\left(\frac{q}{p}\right)^m\) if \(\frac{p}{q}\) ≠ 0
(viii) am x bm = (ab)m
(ix) am ÷ bm = \(\left(\frac{a}{b}\right)^m\)
(x) [ab]c = abc
Every number, large or small can be expressed in the form k x l0 where k is a terminating decimal satisfying 1 ≤ k < 10 and n an integer (positive for large numbers and negative for small numbers).
Example 1:
Express the following as rational numbers.
(i) \(\left(\frac{3}{4}\right)^3\)
Answer:
\(\left(\frac{3}{4}\right)^3\)
= \(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\)
= \(\frac{27}{64}\)
(ii) \(\left(\frac{-5}{7}\right)^2\)
Answer:
\(\left(\frac{-5}{7}\right)^2\)
= \(\frac{-5}{7} \times \frac{-5}{7}\)
= \(\frac{25}{49}\)
(iii) \(\left(\frac{-4}{5}\right)^{-2}\)
Answer:
\(\left(\frac{-4}{5}\right)^{-2}\)
= \(\left(\frac{-5}{4}\right)^2\)
= \(\frac{-5}{4} \times \frac{-5}{4}\)
= \(\frac{25}{16}\)
(iv) \(\left(\frac{2}{3}\right)^0\)
Answer:
\(\left(\frac{2}{3}\right)^0\) = 1
Example 2:
Express the following rational numbers in exponential form.
(i) \(\frac{125}{64}\)
Answer:
\(\frac{125}{64}\) = \(\frac{5 \times 5 \times 5}{4 \times 4 \times 4}\)
= \(\frac{5^3}{4^3}\)
= \(\left(\frac{5}{4}\right)^3\)
(ii) \(\frac{625}{1296}\)
Answer:
\(\frac{625}{1296}\)
= \(\frac{5 \times 5 \times 5 \times 5}{6 \times 6 \times 6 \times 6}\)
= \(\frac{5^4}{6^4}\)
= \(\left(\frac{5}{4}\right)^4\)
(iii) \(\frac{343}{512}\)
Answer:
\(\frac{343}{512}\)
= \(\frac{7 \times 7 \times 7}{8 \times 8 \times 8}\)
= \(\frac{7^3}{8^3}\)
= \(\left(\frac{7}{8}\right)^3\)
(iv) \(\frac{64}{729}\)
Answer:
\(\frac{64}{729}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3 \times 3}\)
= \(\frac{2^6}{3^6}\)
= \(\left(\frac{2}{3}\right)^6\)
Example 3:
Find the reciprocal of the following.
(i) \(\left(\frac{3}{5}\right)^{-3}\)
Answer:
Reciprocal of \(\left(\frac{3}{5}\right)^{-3}\) = \(\frac{1}{\left(\frac{3}{5}\right)^{-3}}\)
= \(\left(\frac{3}{5}\right)^3\)
= \(\left(\frac{5}{3}\right)^{-3}\)
(ii) \(\left(\frac{-5}{7}\right)^2\)
Answer:
Reciprocal of \(\left(\frac{-5}{7}\right)^2\) = \(\frac{1}{\left(\frac{-5}{7}\right)^2}\)
= \(\left(\frac{-7}{5}\right)^2\)
(iii) (0)3
Answer:
Reciprocal of (0)3 is not defined.
(iv) \(\left(\frac{1}{5}\right)^{-4}\)
Answer:
Reciprocal of \(\left(\frac{1}{5}\right)^{-4}\) = \(\frac{1}{\left(\frac{1}{5}\right)^4}\)
= (5)– 4
Example 4:
Simplify the following:
(i) \(\left(\frac{4}{5}\right)^6 \div\left(\frac{4}{5}\right)^4\)
Answer:
\(\left(\frac{4}{5}\right)^6 \div\left(\frac{4}{5}\right)^4\) = \(\left(\frac{4}{5}\right)^{6-4}\)
= \(\left(\frac{4}{5}\right)^2\)
= \(\frac{4 \times 4}{5 \times 5}\)
= \(\frac{16}{25}\)
(ii) \(\left(\frac{2}{3}\right)^2 \times\left(\frac{2}{3}\right)^3\)
Answer:
\(\left(\frac{2}{3}\right)^2 \times\left(\frac{2}{3}\right)^3\) = \(\left(\frac{2}{3}\right)^{2+3}\)
= \(\left(\frac{2}{3}\right)^5\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3}\)
= \(\frac{32}{243}\)
Example 5:
Simplify the following:
(i) \(\left[\left(-\frac{1}{5}\right)^2\right]^3\)
Answer:
\(\left[\left(-\frac{1}{5}\right)^2\right]^3\) = \(\left(\frac{-1}{5}\right)^{2 \times 3}\)
= \(\left(\frac{-1}{5}\right)^6\)
= \(\frac{1}{15625}\)
(ii) \(\left[\left(\frac{2}{3}\right)^8\right]^{\frac{1}{4}}\)
Answer:
\(\left[\left(\frac{2}{3}\right)^8\right]^{\frac{1}{4}}\) = \(\left(\frac{2}{3}\right)^{8 \times \frac{1}{4}}\)
= \(\left(\frac{2}{3}\right)^2\)
= \(\frac{4}{9}\)
Example 6:
Find the value of x, if \(\left[\left(\frac{3}{4}\right)^3\right]^2=\left(\frac{3}{4}\right)^{3 x}\)
Answer:
\(\left[\left(\frac{3}{4}\right)^3\right]^2=\left(\frac{3}{4}\right)^{3 x}\)
= \(\left(\frac{3}{4}\right)^{3 \times 2}=\left(\frac{3}{4}\right)^{3 x}\)
= \(\left(\frac{3}{4}\right)^6=\left(\frac{3}{4}\right)^{3 x}\)
Since bases are same, their exponents must be same.
∴ 3x = 6
⇒ x = 2.
Example 7:
By what number should (- 3)3 be multiplied so that the product may be \(\frac{1}{27}\).
Answer:
Let the required number be x
∴ x × (- 3)– 3 = \(\frac{1}{27}\)
x = \(\frac{1}{27}\) ÷ (- 3)– 3
x = \(\frac{1}{27} \div \frac{1}{(-3)^3}\)
x = \(\frac{1}{27}\) × (- 3)– 3
x = \(\frac{1}{27}\) × (- 27)
∴ x = – 1
Example 8:
Find, if the following are true.
(i) 23 × 33 = (2 × 3)3
(ii) (- 2)4 ÷ (- 3)4 = \(\left(\frac{-2}{3}\right)^4\)
(iii) (- 3)2 + (- 2)2 = (- 3 – 2)2
Answer:
(i) 23 × 33 = 8 × 27 = 216
(2 × 3)3 = (6)3
= 6 × 6 × 6
= 216
(ii) (- 2)4 ÷ (- 3)4 = \(\left(\frac{-2}{-3}\right)^4\)
= \(\frac{2^4}{3^4}\)
= \(\frac{16}{81}\)
\(\left(\frac{-2}{3}\right)^4\) = \(\left(\frac{-2}{3}\right) \times\left(\frac{-2}{3}\right) \times\left(\frac{-2}{3}\right) \times\left(\frac{-2}{3}\right)\)
= \(\frac{16}{81}\)
Hence (- 2)4 ÷ (- 3)4 = \(\left(\frac{-2}{-3}\right)^4\)
(iii) (- 3)2 + (- 2)2 = (- 3 – 2)2
(- 3)2 + (- 2)2 = 9 + 4 = 13
(- 3 – 2)2 = (- 5)2 = 25
Hence (- 3)2 + (- 2)2 ≠ (- 3 – 2)2