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#### Question 1:

Write each of the following in exponential form:
(i) ${\left(\frac{3}{2}\right)}^{-1}×{\left(\frac{3}{2}\right)}^{-1}×{\left(\frac{3}{2}\right)}^{-1}×{\left(\frac{3}{2}\right)}^{-1}$
(ii) ${\left(\frac{2}{5}\right)}^{-2}×{\left(\frac{2}{5}\right)}^{-2}×{\left(\frac{2}{5}\right)}^{-2}$

#### Question 2:

Evaluate:
(i) 5−2
(ii) (−3)−2
(iii) ${\left(\frac{1}{3}\right)}^{-4}$
(iv) ${\left(\frac{-1}{2}\right)}^{-1}$

---> (an = 1/(an))
$=\frac{1}{25}$

---> (an = 1/(an))
$=\frac{1}{9}$

(iii) ${\left(\frac{1}{3}\right)}^{-4}=\frac{1}{{\left(1/3\right)}^{4}}$              ---> (an = 1/(an))
$=\frac{1}{1/81}$
= 81

---> (a−1 = 1/(a))
$=-2$

#### Question 3:

Express each of the following as a rational number in the form $\frac{p}{q}:$
(i) 6−1
(ii) (−7)−1
(iii) ${\left(\frac{1}{4}\right)}^{-1}$
(iv) $\left(-4{\right)}^{-1}×{\left(\frac{-3}{2}\right)}^{-1}$
(v) ${\left(\frac{3}{5}\right)}^{-1}×{\left(\frac{5}{2}\right)}^{-1}$

---> (a−1 = 1/a)

---> (a−1 = 1/a)
=$\frac{-1}{7}$

---> (a−1 = 1/a)
$=4$

(iv)  $\left(-4{\right)}^{-1}×{\left(\frac{-3}{2}\right)}^{-1}=\frac{1}{-4}×\frac{1}{-3/2}$           ---> (a−1 = 1/a)
$=\frac{1}{-4}×\frac{2}{-3}$
$=\frac{1}{6}$

---> (a−1 = 1/a)
$=\frac{5}{3}×\frac{2}{5}$
$=\frac{2}{3}$

#### Question 4:

Simplify:
(i) ${\left\{{4}^{-1}×{3}^{-1}\right\}}^{2}$
(ii) ${\left\{{5}^{-1}÷{6}^{-1}\right\}}^{3}$
(iii) ${\left({2}^{-1}+{3}^{-1}\right)}^{-1}$
(iv) ${\left\{{3}^{-1}×{4}^{-1}\right\}}^{-1}×{5}^{-1}$
(v) $\left({4}^{-1}-{5}^{-1}\right)÷{3}^{-1}$

---> (a−1 = 1/a)
$={\left(\frac{1}{12}\right)}^{2}$
$=\frac{\left(1{\right)}^{2}}{\left(12{\right)}^{2}}$                         --->((a/b)n = (an)/(bn))
$=\frac{1}{144}$

---> (a−1 = 1/a)
$={\left(\frac{6}{5}\right)}^{3}$
$=\frac{216}{125}$                         --->((a/b)n = (an)/(bn))

---> (a−1 = 1/a)
$={\left(\frac{1}{12}\right)}^{-1}×\frac{1}{5}$
$=12×\frac{1}{5}$                              ---> (a−1 = 1/a)
$=\frac{12}{5}$

---> (a−1 = 1/a)

$=\frac{1}{20}×3$

$=\frac{3}{20}$

#### Question 5:

Express each of the following rational numbers with a negative exponent:
(i) ${\left(\frac{1}{4}\right)}^{3}$
(ii) ${3}^{5}$
(iii) ${\left(\frac{3}{5}\right)}^{4}$
(iv) ${\left\{{\left(\frac{3}{2}\right)}^{4}\right\}}^{-3}$
(v) ${\left\{{\left(\frac{7}{3}\right)}^{4}\right\}}^{-3}$

#### Question 6:

Express each of the following rational numbers with a positive exponent:
(i) ${\left(\frac{3}{4}\right)}^{-2}$
(ii) ${\left(\frac{5}{4}\right)}^{-3}$
(iii) ${4}^{3}×{4}^{-9}$
(iv) ${\left\{{\left(\frac{4}{3}\right)}^{-3}\right\}}^{-4}$
(v) ${\left\{{\left(\frac{3}{2}\right)}^{4}\right\}}^{-2}$

---> (a−1 = 1/a)

---> (a−1 = 1/a)

---> (am x an = am+n)

---> ((am)n = amn)

---> ((am)n = amn)

#### Question 7:

Simplify:
(i) $\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$
(ii) $\left({3}^{2}-{2}^{2}\right)×{\left(\frac{2}{3}\right)}^{-3}$
(iii) ${\left\{{\left(\frac{1}{2}\right)}^{-1}×\left(-4{\right)}^{-1}\right\}}^{-1}$
(iv) ${\left[{\left\{{\left(\frac{-1}{4}\right)}^{2}\right\}}^{-2}\right]}^{-1}$
(v) ${\left\{{\left(\frac{2}{3}\right)}^{2}\right\}}^{3}×{\left(\frac{1}{3}\right)}^{-4}×{3}^{-1}×{6}^{-1}$

---> (an = 1/(an))

---> (an = 1/(an))

---> (a−1 = 1/a)

---> (a−1 = 1/a)
=-2

--> ((a/b)n = an/(bn))
---> (an = 1/(an))

---> (a−1 = 1/a)

---> ((a/b)n = an/(bn)) and (an = 1/(an))

---> ((a/b)n = an/(bn))

#### Question 8:

By what number should 5−1 be multiplied so that the product may be equal to (−7)−1?

Expressing in fraction form, we get:
5−1 = 1/5 (using the property a−1 = 1/a)
and
(−7)−1 = −1/7 (using the property a−1 = 1/a).
We have to find a number x such that

Multiplying both sides by 5, we get:

Hence, 5−1 should be multiplied by −5/7 to obtain (−7)−1.

#### Question 9:

By what number should ${\left(\frac{1}{2}\right)}^{-1}$ be multiplied so that the product may be equal to ${\left(\frac{-4}{7}\right)}^{-1}?$

Expressing in fractional form, we get:
(1/2)−1 = 2,       ---> (a−1 = 1/a)
and
(−4/7)−1 = −7/4     ---> (a−1 = 1/a)
We have to find a number x such that

Dividing both sides by 2, we get:

Hence, (1/2)−1 should be multiplied by −7/8 to obtain (−4/7)−1.

#### Question 10:

By what number should (−15)−1 be divided so that the quotient may be equal to (−5)−1?

Expressing in fractional form, we get:
(−15)−1 = −1/15,      ---> (a−1 = 1/a)
and
(−5)−1 = −1/5           ---> (a−1 = 1/a)
We have to find a number x such that

Solving this equation, we get:

Hence, (−15)−1 should be divided by 1/3 to obtain (−5)−1.

#### Question 11:

By what number should ${\left(\frac{5}{3}\right)}^{-2}$ be multiplied so that the product may be ${\left(\frac{7}{3}\right)}^{-1}?$

Expressing as a positive exponent, we have:
---> (a−1 = 1/a)
---> ((a/b)n = (an)/(bn))

and
(7/3) 1 = 3/7.                ---> (a−1 = 1/a)
We have to find a number x such that

Multiplying both sides by 25/9, we get:

Hence, (5/3)−2 should be multiplied by 25/21 to obtain (7/3)−1.

#### Question 12:

Find x, if
(i) ${\left(\frac{1}{4}\right)}^{-4}×{\left(\frac{1}{4}\right)}^{-8}={\left(\frac{1}{4}\right)}^{-4x}$
(ii) ${\left(\frac{-1}{2}\right)}^{-19}×{\left(\frac{-1}{2}\right)}^{8}={\left(\frac{-1}{2}\right)}^{-2x+1}$
(iii) ${\left(\frac{3}{2}\right)}^{-3}×{\left(\frac{3}{2}\right)}^{5}={\left(\frac{3}{2}\right)}^{2x+1}$
(iv) ${\left(\frac{2}{5}\right)}^{-3}×{\left(\frac{2}{5}\right)}^{15}={\left(\frac{2}{5}\right)}^{2+3x}$
(v) ${\left(\frac{5}{4}\right)}^{-x}÷{\left(\frac{5}{4}\right)}^{-4}={\left(\frac{5}{4}\right)}^{5}$
(vi) ${\left(\frac{8}{3}\right)}^{2x+1}×{\left(\frac{8}{3}\right)}^{5}={\left(\frac{8}{3}\right)}^{x+2}$

(i) We have:

x = 3

(ii) We have:

x = 6

(iii) We have:

x = 1/2

(iv) We have:

x = 10/3

(v) We have:

x = −1

(vi) We have:

x = −4

#### Question 13:

(i) If $x={\left(\frac{3}{2}\right)}^{2}×{\left(\frac{2}{3}\right)}^{-4}$, find the value of x−2.
(ii) If $x={\left(\frac{4}{5}\right)}^{-2}÷{\left(\frac{1}{4}\right)}^{2}$, find the value of x−1.

(i) First, we have to find x.

--->(a−1 = 1/a)

Hence, x−2 is:

--->(a−1 = 1/a)

(ii) First, we have to find x.
---> ((a/b)n = (an)/(bn))

---> (a0 = 1)
Hence, the value of x−1 is:

--->(a−1 = 1/a)
--->(a−1 = 1/a)

#### Question 14:

Find the value of x for which 52x ÷ 5−3 = 55.

We have:

--->

Hence, x is 1.

#### Question 1:

Express the following numbers in standard form:
(i) 6020000000000000
(ii) 0.00000000000943
(iii) 0.00000000085
(iv) 846 × 107
(v) 3759 × 10−4
(vi) 0.00072984
(vii) 0.000437 × 104
(viii) 4 ÷ 100000

To express a number in the standard form, move the decimal point such that there is only one digit to the left of the decimal point.
(i) 6020000000000000 = 6.02 x 1015      (The decimal point is moved 15 places to the left.)
(ii) 0.0000000000943 = 9.43 x 10−12     (The decimal point is moved 12 places to the right.)
(iii) 0.00000000085 = 8.5 x 10−10     (The decimal point is moved 10 places to the right.)
(iv) 846 x 107 = 8.46 x 102 x 107 = 8.46 x 109     (The decimal point is moved two places to the left.)
(v) 3759 x 10−4 = 3.759 x 103 x 10−4 = 3.759 x 10−1     (The decimal point is moved three places to the left.)
(vi) 0.00072984 = 7.984 x 10−4     (The decimal point is moved four places to the right.)
(vii) 0.000437 x 104 = 4.37 x 10−4 x 104 = 4.37 x 100 = 4.37     (The decimal point is moved four places to the right.)
(viii) 4/100000 = 4 x 100000−1 = 4 x 10−5     (Just count the number of zeros in 1,00,000 to determine the exponent of 10.)

#### Question 2:

Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.02 × 10−6
(iii) 4.5 × 104
(iv) 3 × 10−8
(v) 1.0001 × 109
(vi) 5.8 × 102
(vii) 3.61492 × 106
(viii) 3.25 × 10−7

(i) 4.83 x 107 = 4.83 x 1,00,00,000 = 4,83,00,000
(ii) 3.02 x 10−6 = 3.02/106 = 3.02/10,00,000 = 0.00000302
(iii) 4.5 x 104 = 4.5 x 10,000 = 45,000
(iv) 3 x 10−8 = 3/108 = 3/10,00,00,000 = 0.00000003
(v) 1.0001 x 109 = 1.0001 x 1,00,00,00,000 = 1,00,01,00,000
(vi) 5.8 x 102 = 5.8 x 100 = 580
(vii)  3.61492 x 106 = 3.61492 x 10,00,000 = 3614920
(viii) 3.25 x 10−7 = 3.25/107 = 3.25/1,00,00,000 = 0.000000325

#### Question 1:

Square of $\left(\frac{-2}{3}\right)$ is
(a) $-\frac{2}{3}$
(b) $\frac{2}{3}$
(c) $-\frac{4}{9}$
(d) $\frac{4}{9}$

(d) 4/9
To square a number is to raise it to the power of 2. Hence, the square of (−2/3) is
---> ( (a/b)n =  (an)/(bn) )

#### Question 2:

Cube of $\frac{-1}{2}$ is
(a) $\frac{1}{8}$
(b) $\frac{1}{16}$
(c) $-\frac{1}{8}$
(d) $\frac{-1}{16}$

(c) -1/8
The cube of a number is the number raised to the power of 3. Hence the cube of −1/2 is

$\frac{\left(-1{\right)}^{3}}{{2}^{3}}$        ---> ( (a/b)n = (an)/(bn
$=\frac{-1}{8}$

#### Question 3:

Which of the following is not equal to ${\left(\frac{-3}{5}\right)}^{4}?$
(a) $\frac{\left(-3{\right)}^{4}}{{5}^{4}}$
(b) $\frac{{3}^{4}}{\left(-5{\right)}^{4}}$
(c) $-\frac{{3}^{4}}{{5}^{4}}$
(d) $\frac{-3}{5}×\frac{-3}{5}×\frac{-3}{5}×\frac{-3}{5}$

(c)  −(34/54)

.

#### Question 4:

Which  of the following is not reciprocal of ${\left(\frac{2}{3}\right)}^{4}?$
(a) ${\left(\frac{3}{2}\right)}^{4}$
(b) ${\left(\frac{2}{3}\right)}^{-4}$
(c) ${\left(\frac{3}{2}\right)}^{-4}$
(d) $\frac{{3}^{4}}{{2}^{4}}$

(c) (3/2)−4
The reciprocal of is .
Therefore, option (a) is the correct answer.
Option (b) is just re-expressing the number with a negative exponent.
Option (d) is obtained by working out the exponent.
Hence,option (c) is not the reciprocal of  .

#### Question 5:

Which of the following numbers is not equal to $\frac{-8}{27}?$
(a) ${\left(\frac{2}{3}\right)}^{-3}$
(b) $-{\left(\frac{2}{3}\right)}^{3}$
(c) ${\left(-\frac{2}{3}\right)}^{3}$
(d) $\left(\frac{-2}{3}\right)×\left(\frac{-2}{3}\right)×\left(\frac{-2}{3}\right)$

(a) (2/3)-3

We can write as . It can be written in the forms given below.

---> work out the minuses

Hence, option (b) is equal to .

We can also write:

Hence, option (c) is also equal to .

We can also write:

Hence, option (d) is also equal to $-\frac{8}{27}$.

This leaves out option (a) as the one not equal to $-\frac{8}{27}$.

#### Question 6:

${\left(\frac{2}{3}\right)}^{-5}$ is equal to
(a) ${\left(\frac{-2}{3}\right)}^{5}$
(b) ${\left(\frac{3}{2}\right)}^{5}$
(c) $\frac{2x-5}{3}$
(d) $\frac{2}{3×5}$

(b)${\left(\frac{3}{2}\right)}^{5}$

Rearrange (2/3)−5 to get a positive exponent.

#### Question 7:

${\left(\frac{-1}{2}\right)}^{5}×{\left(\frac{-1}{2}\right)}^{3}$ is equal to
(a) ${\left(\frac{-1}{2}\right)}^{8}$
(b) $-{\left(\frac{1}{2}\right)}^{8}$
(c) ${\left(\frac{1}{4}\right)}^{8}$
(d) ${\left(-\frac{1}{2}\right)}^{15}$

(a) (−1/2)8

We have:

#### Question 8:

${\left(\frac{-1}{5}\right)}^{3}÷{\left(\frac{-1}{5}\right)}^{8}$ is equal to
(a) ${\left(-\frac{1}{5}\right)}^{5}$
(b) ${\left(-\frac{1}{5}\right)}^{11}$
(c) $\left(-5{\right)}^{5}$
(d) ${\left(\frac{1}{5}\right)}^{5}$

(c)  (−5)5

We have:

#### Question 9:

${\left(\frac{-2}{5}\right)}^{7}÷{\left(\frac{-2}{5}\right)}^{5}$ is equal to
(a) $\frac{4}{25}$
(b) $\frac{-4}{25}$
(c) ${\left(\frac{-2}{5}\right)}^{12}$
(d) $\frac{25}{4}$

(a) 4/25

We have:

#### Question 10:

${\left\{{\left(\frac{1}{3}\right)}^{2}\right\}}^{4}$ is equal to
(a) ${\left(\frac{1}{3}\right)}^{6}$
(b) ${\left(\frac{1}{3}\right)}^{8}$
(c) ${\left(\frac{1}{3}\right)}^{24}$
(d) ${\left(\frac{1}{3}\right)}^{16}$

(b) (1/3)8

We have:

---> ( (am)n = amxn)

#### Question 11:

${\left(\frac{1}{5}\right)}^{0}$ is equal to
(a) 0
(b) $\frac{1}{5}$
(c) 1
(d) 5

(c) 1

We have:

---> (a0 = 1, for every non-zero rational number a.)

#### Question 12:

${\left(\frac{-3}{2}\right)}^{-1}$ is equal to
(a) $\frac{2}{3}$
(b) $-\frac{2}{3}$
(c) $\frac{3}{2}$
(d) none of these

We have:

--> (a−1 = 1/a)

#### Question 13:

${\left(\frac{2}{3}\right)}^{-5}×{\left(\frac{5}{7}\right)}^{-5}$ is equal to
(a) ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{-10}$
(b) ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{-5}$
(c) ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{25}$
(d) ${\left(\frac{2}{3}×\frac{5}{7}\right)}^{-25}$

We have:

---> ((a x b)n = an x bn)

#### Question 14:

${\left(\frac{3}{4}\right)}^{5}÷{\left(\frac{5}{3}\right)}^{5}$ is equal to

(a)  ${\left(\frac{3}{4}÷\frac{5}{3}\right)}^{5}$

We have:

--->

#### Question 15:

For any two non-zero rational numbers a and b, a4 ÷ b4 is equal to
(a) (a ÷ b)1
(b) (a ÷ b)0
(c) (a ÷ b)4
(d) (a ÷ b)8

This is one of the basic exponential formulae, i.e. .

#### Question 16:

For any two rational numbers a and b, a5 × b5 is equal to
(a) (a × b)0
(b) (a × b)10
(c) (a × b)5
(d) (a × b)25

(c) (a x b)5
an x bn = (a x b)n
Hence,
a5 x b5 = (a x b)5

#### Question 17:

For a non-zero rational number a, a7 ÷ a12 is equal to
(a) a5
(b) a−19
(c) a−5
(d) a19

(c) a−5

Hence,

#### Question 18:

For a non zero rational number a, (a3)−2 is equal to
(a) a9
(b) a−6
(c) a−9
(d) a1

(b) a−6

We have:

---> ((am)n = am x n)

#### Question 1:

Express each of the following as a rational number of the form $\frac{p}{q},$ where p and q are integers and q ≠ 0.
(i) 2−3
(ii) (−4)−2
(iii) $\frac{1}{{3}^{-2}}$
(iv) ${\left(\frac{1}{2}\right)}^{-5}$
(v) ${\left(\frac{2}{3}\right)}^{-2}$

We know that . Therefore,

(i)

(ii)

(iii)

(iv)

(v)

#### Question 2:

Fiind the value of each of the following:
(i) 3−1 + 4−1
(ii) (30 + 4−1) × 22
(iii) (3−1 + 4−1 + 5−1)0
(iv) ${\left\{{\left(\frac{1}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$

(i) We know from the property of powers that for every natural number a, a−1 = 1/a. Then:
---> (a−1 = 1/a)

(ii) We know from the property of powers that for every natural number a, a−1 = 1/a.
Moreover, a0 is 1 for every natural number a not equal to 0. Then:

(iii) We know from the property of powers that for every natural number a, a−1 = 1/a.
Moreover, a0 is 1 for every natural number a not equal to 0. Then:
$\left({3}^{-1}+{4}^{-1}+{5}^{-1}\right)=1$          ---> (Ignore the expression inside the bracket and use a0 = 1 immediately.)

(iv) We know from the property of powers that for every natural number a, a−1 = 1/a. Then:
---> (a−1 = 1/a)

=$-1$                       ---> (a−1 = 1/a)

#### Question 3:

Find the value of each of the following:
(i) ${\left(\frac{1}{2}\right)}^{-1}+{\left(\frac{1}{3}\right)}^{-1}+{\left(\frac{1}{4}\right)}^{-1}$
(ii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$
(iii) (2−1 × 4−1) ÷ 2−2
(iv) (5−1 × 2−1) ÷ 6−1

(i)
${\left(\frac{1}{2}\right)}^{-1}+{\left(\frac{1}{3}\right)}^{-1}+{\left(\frac{1}{4}\right)}^{-1}=\frac{1}{1/2}+\frac{1}{1/3}+\frac{1}{1/4}$          --> (a−1 = 1/a)
=2+3+4
=12

(ii)
${\left(\frac{1}{2}\right)}^{-2}+\left(\frac{1}{3}\right)$-2+14-2=11/22+11/32+11/42        --> (an = 1/(an))
= $\frac{1}{1/4}+\frac{1}{1/9}+\frac{1}{1/16}$              --> ((a/b)n = (an/bn))
= 4+9+16
=29

(iii)
$\left({2}^{-1}×{4}^{-1}\right)÷{2}^{-2}=\left(\frac{1}{2}×\frac{1}{4}\right)÷\frac{1}{{2}^{2}}$   --> (an = 1/(an))

=$\frac{1}{8}×4$
= 2

(iv)
$\left({5}^{-1}×{2}^{-1}\right)÷{6}^{-1}=\left(\frac{1}{5}×\frac{1}{2}\right)÷\frac{1}{6}$             --> (an = 1/(an))
$=\frac{1}{10}×6$

$=\frac{3}{5}$

#### Question 4:

Simplify:
(i) ${\left({4}^{-1}×{3}^{-1}\right)}^{2}$
(ii) ${\left({5}^{-1}÷{6}^{-1}\right)}^{3}$
(iii) ${\left({2}^{-1}+{3}^{-1}\right)}^{-1}$
(iv) ${\left({3}^{-1}×{4}^{-1}\right)}^{-1}×{5}^{-1}$

(i)
---> (a−1 = 1/a)

---> ((a/b)n = (an)/(bn) )
$=\frac{1}{24}$

(ii)
---> (a−1 = 1/a)

=  ${\left(\frac{6}{5}\right)}^{3}$
=  $\frac{{\left(6\right)}^{3}}{{\left(5\right)}^{3}}$                         ---> ((a/b)n = (an)/(bn) )

= $\frac{216}{125}$

(iii)
---> (a−1 = 1/a)
=  ${\left(\frac{5}{6}\right)}^{-1}$
$=\frac{1}{5/6}$                          ---> (a−1 = 1/a)
$=\frac{6}{5}$

(iv)
---> (a−1 = 1/a)

$=\frac{12}{5}$                                      ---> (a−1 = 1/a)

#### Question 5:

Simplify:
(i) $\left({3}^{2}+{2}^{2}\right)×{\left(\frac{1}{2}\right)}^{3}$
(ii) $\left({3}^{2}-{2}^{2}\right)×{\left(\frac{2}{3}\right)}^{-3}$
(iii) $\left[{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right]÷{\left(\frac{1}{4}\right)}^{-3}$
(iv) $\left({2}^{2}+{3}^{2}-{4}^{2}\right)÷{\left(\frac{3}{2}\right)}^{2}$

(i)
$\left({3}^{2}+{2}^{2}\right)×{\left(\frac{1}{2}\right)}^{3}=\left(9+4\right)×\frac{1}{8}=\frac{13}{8}$

(ii)
---> (a−1=1/(an))
---> ((a/b)n = (an)/(bn))
$=5×\frac{27}{8}$
$=\frac{135}{8}$

(iii)
--->(a-n = 1/(an))
= $\left(27-8\right)÷64$
$=19×\frac{1}{64}$
$=\frac{19}{64}$

(iv)
$\left({2}^{2}+{3}^{2}-{4}^{2}\right)÷{\left(\frac{3}{2}\right)}^{2}=\left(4+9-16\right)×\frac{9}{4}$                    ---> ((a/b)n = (an)/(bn))

$=-3×\frac{9}{4}$
$=\frac{-27}{4}$

#### Question 6:

By what number should 5−1 be multiplied so that the product may be equal to (−7)−1?

Using the property a−1 = 1/a for every natural number a, we have 5−1 = 1/5 and (−7)−1 = −1/7. We have to find a number x such that
$\frac{1}{5}×x=\frac{-1}{7}$
Multiplying both sides by 5, we get:
$x=\frac{-5}{7}$
Hence, the required number is −5/7.

#### Question 7:

By what number should ${\left(\frac{1}{2}\right)}^{-1}$ be multiplied so that the product may be equal to ${\left(-\frac{4}{7}\right)}^{-1}?$

Using the property a−1 = 1/a for every natural number a, we have (1/2)−1 = 2 and (−4/7)−1 = −7/4. We have to find a number x such that
$2x=\frac{-7}{4}$
Dividing both sides by 2, we get:
$x=\frac{-7}{8}$
Hence, the required number is −7/8.

#### Question 8:

By what number should (−15)−1 be divided so that the quotient may be equal to (−5)−1?

Hence, (−15)−1 should be divided by $\frac{1}{3}$ to obtain (−5)−1.