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Practice each question, then open the answer or worked solution.
Pluto is 59,1,30,00, 000 m from the sun. Express this in the standard form.
Ans: Distance between Pluto and Sun = 59,1,30,00, 000
Standard form of 59,1,30,00, 000 = 5913 x 106
= 5.913 x 106 x 103
Use, am × an = am+n
Therefore, 5.913 x 106 x 103 = 5.913 x 109
Special balances can weigh something as 0.00000001 gram. Express this number in the standard form.
Ans: Weight = 0.00000001 g
Standard form= 0.00000001 = 0.1 x 10-7 g
= 1 x 10-7 x 10-1 g [Use, am × an = am+n]
= 1.0 x 10-8 g
A sugar factory has annual sales of 3 billion 720 million kilograms of sugar. Express this number in the standard form.
Ans: Annual sales of a sugar factory = 3 billion 720
million kilograms = 3720000 kg
Standard form of-
3720000 = 372 x 10 x 10 x 10 x 10 = 372 x 104 kg
Use am × an = am+n
= 372 x 104 x 102 = 3.72 x 106 kg.
The number of red blood cells per cubic millimetre of blood is approximately 5.5 million. If the average body contains 5 litres of blood, what is the total number of red cells in the body? Write the standard form. (1 litre = 1,00,000 mm3)
Ans: The average body contains 5 L of blood.
Also, the number of red blood cells per cubic millimetre of blood is approximately 5.5 million.
Blood contained by body = 5 L = 5 x 100000 mm3
Red blood cells = 5 x 100000 mm3
Blood = 5.5 x 1000000 x 5 x 100000= 55 x 5 x 105 + 5
= 275 x 1010 = 2.75 x 1010 x 102 = 2.75 x 1012
Express each of the following in standard form:
(a) The mass of a proton in gram is $\dfrac{{1673}}{{{\mathbf{1000000000000000000000000000}}}}$
(b) A Helium atom has a diameter of 0.000000022 cm.
(c) Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.
(d) Human body has 1 trillion of cells which vary in shape and size.
Ans: Standard form = $\dfrac{{1673}}{{{{10}^{27}}}} = 1673 \times {10^{ - 27}}g$
= $1.673 \times {10^{ - 27}} \times {10^3}g$
= $1.673 \times {10^{ - 27 + 3}}g$
= $1.673 \times {10^{ - 24}}g$
Ans: Standard form= 0.22 x 10-7 cm = 2.2 x 10-1 x 10-7
Using the law of exponent am × an = am+n,
= 2.2 x 10-1-7 = 2.2 x 10-8 cm
Therefore, the standard form is 2.2 x 10-8 cm.
Ans: Standard form = 0.334 x 10-20 = 3.34 x 10-20 x 10-1
[Use am × an = am+n]
3.34 x 10-20 x 10-1 = 3.34 x 10-21
Ans: Cells in human body = 1 trillion
As, 1 trillion = 1000000000000
Standard form of 1000000000000 = 10 x 10 x 10 x 10 x 10 x
10 x 10 x 10 x 10 x 10 x 10 x 10 = 1012
(e) Express 56 km in m.
Ans: Given, 56 km = 56 x 1000 m= 56000 m
Standard form of 56000 m = 56 x 103 = 5.6 x 103 x 101 = 5.6 x 104 m
(f) Express 5 tons in g.
Ans: Given, 5 tons = 5 x 100 kg = 5 x 100 x 1000 g = 500000g
Standard form of 500000
= 5 x 10 x 10 x 10 x 10 x 10 = 5 x 105
(g) Express 2 years in seconds.
Ans: 2 yr = 2 x 365 days
= 2 x 365 x 24h
= 2 x 365 x 24 x 60min
= 2 x 365 x 24 x 60 x 60s
= 63072000 s
Standard form of
63072000 = 63072 x 10 x 10 x 10 = 63072 x 103
63072 x 103 = 6.3072 x 104 x 103
Since, am × an = am+n
6.3072 x 104 x 103 = 6.3072 x 107 s
(h) Express 5 hectares in cm2 (1 hectare = 10000 m2)
Ans: 5 hec = 5 x 10000 m2
= 5 x 10000 x 100 x 100 cm2
= 5 x 104+2+2 [Use am × an = am+n]
= 5 x 108 cm2
Find x so that ${\left( {\dfrac{2}{9}} \right)^3} \times {\left( {\dfrac{2}{9}} \right)^{ - 6}} = {\left( {\dfrac{2}{9}} \right)^{2x - 1}}$
Ans: Use am × an = am+n
$\Rightarrow {\left( {\dfrac{2}{9}} \right)^3} \times {\left( {\dfrac{2}{9}} \right)^{ - 6}} = {\left( {\dfrac{2}{9}} \right)^{2x - 1}}$
$\Rightarrow {\left( {\dfrac{2}{9}} \right)^{3 - 6}} = {\left( {\dfrac{2}{9}} \right)^{2x - 1}}$
$\Rightarrow {\left( {\dfrac{2}{9}} \right)^{ - 3}} = {\left( {\dfrac{2}{9}} \right)^{2x - 1}}$
$\Rightarrow - 3 = 2x - 1$
$\Rightarrow 2x = - 3 + 1$
$\Rightarrow x = \dfrac{{ - 2}}{2} = - 1$
By what number should ${\left( {\dfrac{{ - 3}}{2}} \right)^{ - 3}}$ be divided so that the quotient may be
${\left( {\dfrac{4}{{27}}} \right)^{ - 2}}$?
Ans: Let ${\left( {\dfrac{{ - 3}}{2}} \right)^{ - 3}}$ is divided by x to get quotient ${\left( {\dfrac{4}{{27}}} \right)^{ - 2}}.$
Then,
$\Rightarrow {\left( {\dfrac{{ - 3}}{2}} \right)^{ - 3}} \div x = {\left( {\dfrac{4}{{27}}} \right)^{ - 2}}$
$\Rightarrow {\left( {\dfrac{2}{{ - 3}}} \right)^3} \times \dfrac{1}{x} = {\left( {\dfrac{{27}}{4}} \right)^2}$
$\Rightarrow x = \dfrac{{{2^3} \times {4^2}}}{{{{\left( { - 3} \right)}^3} \times {{\left( {27} \right)}^2}}} = \dfrac{{{2^3} \times {{\left( {{2^2}} \right)}^2}}}{{ - {{\left( 3 \right)}^3} \times {{\left( {{3^3}} \right)}^2}}} = \dfrac{{{2^{3 + 4}}}}{{ - {3^{3 + 6}}}} = \dfrac{{{2^7}}}{{ - {3^9}}}$
In questions 118 and 119, find the value of n.
$\dfrac{{{6^n}}}{{{6^{ - 2}}}} = {6^3}$
Ans: Use ${a^m} \div {a^n} = {\left( a \right)^{m - n}}$
$\dfrac{{{6^n}}}{{{6^{ - 2}}}} = {6^3}$
$\Rightarrow {6^{n - \left( { - 2} \right)}} = {6^3}$
$\Rightarrow {6^{n + 2}} = {6^3}$
On comparing both sides
n + 2= 3
n = 1
Therefore, the value of n is 1.
$\dfrac{{{2^n} \times {2^6}}}{{{2^{ - 3}}}} = {2^{18}}$
Ans: Use ${a^m} \times {a^n} = {\left( a \right)^{m + n}}$, ${a^m} \div {a^n} = {\left( a \right)^{m - n}}$ and ${a^{ - m}} = \dfrac{1}{{{a^m}}}$.
$\Rightarrow \dfrac{{{2^n} \times {2^6}}}{{{2^{ - 3}}}} = {2^{18}}$
$\Rightarrow {2^{n + 6 - \left( { - 3} \right)}} = {2^{18}}$
$\Rightarrow {2^{n + 9}} = {2^{18}}$
On comparing both sides,
⇒n + 9 = 18
⇒n = 9
$\dfrac{{125 \times {x^{ - 3}}}}{{{5^{ - 3}} \times 25 \times {x^{ - 6}}}}$
Ans: Use ${a^m} \times {a^n} = {\left( a \right)^{m + n}}$, ${a^m} \div {a^n} = {\left( a \right)^{m - n}}$ and ${a^{ - m}} = \dfrac{1}{{{a^m}}}$.
$\dfrac{{125 \times {x^{ - 3}}}}{{{5^{ - 3}} \times 25 \times {x^{ - 6}}}} = \dfrac{{{5^3} \times {x^{ - 3}}}}{{{5^{ - 3}} \times {5^2} \times {x^{ - 6}}}}$
= ${5^{3 + 3 - 2}} \times {x^{ - 3 + 6}}$
= ${5^4} \times {x^3} = 625{x^3}$