RS Aggarwal 2019,2020 solution class 7 chapter 5 Exponents Test Paper 5

Test Paper 5

Page-96

Question 1:

Write the reciprocal of:

(i) ${\left(\frac{2}{3}\right)}^4$
(ii) ${\left(\frac{-3}{5}\right)}^{61}$
(iii) 2⁵
(iv) (−5)⁶

Answer 1:

We know that the reciprocal of ${\left(\frac{a}{b}\right)}^m$ is ${\left(\frac{b}{a}\right)}^m$.

(i) Reciprocal of ${\left(\frac{2}{3}\right)}^4$= ${\left(\frac{3}{2}\right)}^4$

(ii) Reciprocal of ${\left(\frac{-3}{5}\right)}^{61}={\left(\frac{-5}{3}\right)}^{61}$

(iii) Reciprocal of 2⁵ = Reciprocal of ${\left(\frac{2}{1}\right)}^5={\left(\frac{1}{2}\right)}^5$

(iv) Reciprocal of (−5)⁶ = Reciprocal of ${\left(\frac{-5}{1}\right)}^6={\left(\frac{-1}{5}\right)}^6$

Question 2:

By what number should we multiply (−6)⁻¹ to obtain a product equal to 9⁻¹?

Answer 2:

Let the required number be x.
(−6)^-1 × x = (9)^-1
⇒  $\frac{1}{-6}\times x=\frac{1}{9}$
∴ x = $\frac{1}{9}\times \left(-6\right)=\frac{\left(-2\right)}{3}$
Hence, the required number is $\frac{-2}{3}$.

Question 3:

By what number should (−20)⁻¹ be divided to obtain (−10)⁻¹?

Answer 3:

Let the required number be x.
(−20)^-1 ÷ x = (−10)^-1
⇒  $\frac{1}{\left(-20\right)}\times \frac{1}{x}=\frac{1}{\left(-10\right)}$
⇒ $\frac{1}{\left(-20x\right)}=\frac{1}{\left(-10\right)}$
∴ x = $\frac{\left(-10\right)}{\left(-20\right)}=\frac{1}{2}=2^{-1}$
Hence, the required number is 2^-1.

Question 4:

(i) Express 2000000 in standard form.
(ii) Express 6.4 × 10⁵ in usual form.

Answer 4:

(i) 2000000 = 2.000000 × 10⁶         [since the decimal point is moved 6 places to the left]
                    = 2 × 10⁶     

(ii) 6.4 × 10⁵ = 6.4 × 100000
                      = 640000

Question 5:

Simplify: $\frac{16\times 2^{n+1}-8\times 2^n}{16\times 2^{n+2}-4\times 2^{n+1}}$

Answer 5:

We have:
    $\frac{16\times 2^{n+1}-8\times 2^n}{16\times 2^{n+2}-4\times 2^{n+1}}$

⇒ $\frac{2^4\times 2^{n+1}-2^3\times 2^n}{2^4\times 2^{n+2}-2^2\times 2^{n+1}}$

⇒ $\frac{2^3\times \left(2^{n+2}-2^n\right)}{2^3\times \left(2^{n+3}-2^n\right)}$

⇒ $\frac{2^n\times 2^2-2^n}{2^n\times 2^3-2^n}$

⇒ $\frac{2^n\left(2^2-1\right)}{2^n\left(2^3-1\right)}=\frac{4-1}{8-1}=\frac{3}{7}$

Question 6:

If 2^n-7 × 5^n-4 = 1250, find the value of n.

Answer 6:

We have:
    $2^{n-7}\times 5^{n-4}=1250$
⇒ $\frac{2^n}{2^7}\times \frac{5^n}{5^4}=2\times 5^4$                      [since 1250 = 2 × 5⁴]
⇒ $\frac{2^{\mathrm{n}}\times 5^{\mathrm{n}}}{2^7\times 5^4}=2\times 5^4$
⇒  $2^n\times 5^n=2\times 5^4\times 2^7\times 5^4$           [using cross multiplication]
⇒  $2^n\times 5^n=2^{1+7}\times 5^{4+4}$               [since a^m × aⁿ = a^m+n ]
⇒ $2^n\times 5^n=2^8\times 5^8$
⇒ ${\left(2\times 5\right)}^n={\left(2\times 5\right)}^8$                      [since aⁿ × bⁿ = (a × b)ⁿ ]
⇒ ${10}^n={10}^8$
⇒ n = 8

Question 7:

Mark (✓) against the correct answer
${\left(\frac{3}{4}\right)}^0=?$

(a) 0
(b) $\frac{4}{3}$
(c) 1
(d) none of these

Answer 7:

(c)  1
We know:
(a)⁰ = 1
∴ ${\left(\frac{3}{4}\right)}^0=1$

Question 8:

Mark (✓) against the correct answer
${\left(\frac{-3}{4}\right)}^{-3}=?$

(a) $\frac{27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{-27}{64}$
(d) $\frac{-64}{27}$

Answer 8:

(d) $\frac{-64}{27}$

${\left(\frac{-3}{4}\right)}^{-3}={\left(\frac{4}{-3}\right)}^3$                           $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n\right]$

           = $\frac{4^3}{{\left(-3\right)}^3}$

           = $\frac{4\times 4\times 4}{\left(-3\right)\times \left(-3\right)\times \left(-3\right)}=\frac{64}{\left(-27\right)}$

           = $\frac{64\times -1}{-27\times -1}=\frac{-64}{27}$
             
                                      

Question 9:

Mark (✓) against the correct answer
${\left(\frac{-5}{3}\right)}^{-1}=?$

(a) $\frac{3}{5}$
(b) $\frac{-3}{5}$
(c) $\frac{5}{3}$
(d) $\frac{-5}{3}$

Answer 9:

(b) $\frac{-3}{5}$

${\left(\frac{-5}{3}\right)}^{-1}={\left(\frac{3}{-5}\right)}^1$                                       $\left[s\mathrm{ince} {\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n\right]$
           = $\left(\frac{3}{-5}\times \frac{-1}{-1}\right)=\frac{-3}{5}$

Question 10:

Mark (✓) against the correct answer
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}=?$

(a) 19
(b) $\frac{1}{19}$
(c) −19
(d) $\frac{-1}{19}$

Answer 10:

(a) 19

$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}=\left\{{\left(\frac{3}{1}\right)}^3-{\left(\frac{2}{1}\right)}^3\right\}$                                    $\left[s\mathrm{ince} {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
                        = $\left\{{\left(3\right)}^3-{\left(2\right)}^3\right\}$
                        = (27 − 8) = 19

Question 11:

Mark (✓) against the correct answer
$\left({\frac{-2}{3}}^{10}\right)÷{\left(\frac{-2}{3}\right)}^8=?$

(a) $\frac{4}{9}$
(b) $\frac{-4}{9}$
(c) ${\left(\frac{-2}{3}\right)}^{18}$
(d) none of these

Answer 11:

(a) $\frac{4}{9}$


${\left(\frac{-2}{3}\right)}^{10}÷{\left(\frac{-2}{3}\right)}^8$ = ${\left(\frac{-2}{3}\right)}^{10-8}$            $\left[s\mathrm{ince} a^m÷a^n=a^{m-n}\right]$
                              = ${\left(\frac{-2}{3}\right)}^2$
                              = $\frac{{\left(-2\right)}^2}{3^2}=\frac{4}{9}$

Question 12:

Which of the following numbers is in standard form?

(a) 32.63 × 10⁴
(b) 326.3 × 10³
(c) 3.263 × 10⁵
(d) none of these

Answer 12:

(c) 3.263 x 10⁵

A given number is said to be in standard form if it can be expressed as k x 10ⁿ, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 3.263 x 10⁵

Question 13:

Fill in the blanks.

(i) If 9 × 3ⁿ = 3⁶, then n = ...... .
(ii) (8)⁰ = ?
(iii) ${\left(\frac{a}{b}\right)}^{-16}=......$
(iv) (−2)⁻⁵ = ......

Answer 13:

(i) If 9 × 3ⁿ = 3⁶, then n = 4.
Explanation:
If 9 × 3ⁿ = 3⁶
⇒ 3² × 3ⁿ = 3⁶
⇒ 3^{(2 + n)} = 3⁶
Equating the powers:
⇒ ( 2 + n) = 6
⇒ n = (6 - 2) = 4

(ii) (8)⁰ = 1
Explanation:
By definition, we have a⁰ = 1 for every integer a.
∴ (8)⁰ = 1

(iii) ${\left(\frac{a}{b}\right)}^{-16}$ = ${\left(\frac{b}{a}\right)}^{16}$
Explanation:
We know:
  ${\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1$

(iv) (−2)⁻⁵ = $\frac{-1}{32}$
Explanation:
(−2)⁻⁵ = ${\left(\frac{-2}{1}\right)}^{-5}={\left(\frac{1}{-2}\right)}^5$         $\left[\mathrm{Since} {\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$
= $\frac{{\left(1\right)}^5}{{\left(-2\right)}^5}=\frac{1\times 1\times 1\times 1\times 1}{\left(-2\right)\times \left(-2\right)\times \left(-2\right)\times \left(-2\right)\times \left(-2\right)}=\frac{1}{-32}$

Question 14:

Write 'T' for true and 'F' for false for each of the following.

(i) 645 in standard form is 6.45 × 10².
(ii) 27000 in standard form is 27 × 10^3.
(iii) (3⁰ + 4⁰ + 5⁰) = 12.
(iv) Reciprocal of 5⁶ is 6⁵.
(v) If 5⁻¹ × x = 8⁻¹, then x = $\frac{8}{5}$.

Answer 14:

(i) True
645 = 6.45 x 10²                  [since the decimal point is moved 2 places to the left]

(ii) False
 27000 = 2.7 x 10⁴               [since the decimal point is moved 4 places to the left]

(iii) False
(3⁰ + 4⁰ + 5⁰) = 1               [since a⁰ = 1 for every integer a]

(iv) False
Reciprocal of 5⁶ = Reciprocal of ${\left(\frac{5}{1}\right)}^6={\left(\frac{1}{5}\right)}^6$

(v) False
5⁻¹ × x = 8⁻¹ 
⇒ $\frac{1}{5}\times x=\frac{1}{8}$
⇒ x = $\left(\frac{1}{8}\times 5\right)=\frac{5}{8}$