Surds and Indices Questions and Answers Part-1

1. Simplify : (3)8 × (3)4 = ?
a) (27)3
b) (27)5
c) (729)2
d) (729)3

Explanation: 38 × 34
= 3(8 + 4)
= 312
= (36)2
= (729)2

2. Simplify : $$\frac{{343 \times 49}}{{216 \times 16 \times 81}} = ?$$
a) $$\frac{{{7^5}}}{{{6^7}}}$$
b) $$\frac{{{7^5}}}{{{6^8}}}$$
c) $$\frac{{{7^6}}}{{{6^7}}}$$
d) $$\frac{{{7^4}}}{{{6^8}}}$$

Explanation:
\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr}

3.Simplify : $$\frac{{16 \times 32}}{{9 \times 27 \times 81}} = ?$$
a) $${\left( {\frac{2}{3}} \right)^9}$$
b) $${\left( {\frac{2}{3}} \right)^{11}}$$
c) $${\left( {\frac{2}{3}} \right)^{12}}$$
d) $${\left( {\frac{2}{3}} \right)^{13}}$$

Explanation:
\eqalign{ & \frac{{16 \times 32}}{{9 \times 27 \times 81}} \cr & = \frac{{{2^4} \times {2^5}}}{{{3^2} \times {3^3} \times {3^4}}} \cr & = \frac{{{2^{\left( {4 + 5} \right)}}}}{{{3^{\left( {2 + 3 + 4} \right)}}}} \cr & = \frac{{{2^9}}}{{{3^9}}} \cr & = {\left( {\frac{2}{3}} \right)^9} \cr}

4. Given $$\sqrt 2$$ = 1.414, the value of $$\sqrt 8$$ $$\, +$$ $${\text{2}}\sqrt {32}$$ $$\, -$$ $$3\sqrt {128}$$ $$\,\, +$$ $${\text{4}}\sqrt {50}$$  is = ?
a) 8.484
b) 8.526
c) 8.426
d) 8.876

Explanation:
\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr}

5. $${9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^?}$$
a) 3
b) 4
c) 5
d) 6

Explanation:
\eqalign{ & {\text{Let}}\,{9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & {\left( 3 \right)^x} = {3^5} \cr & {\left( 3 \right)^x} = 5 \cr}

6. (256)0.16 × (256)0.09 = ?
a) 4
b) 16
c) 64
d) 256.25

Explanation:
\eqalign{ & = {\left( {256} \right)^{0.25}} \cr & = {\left( {256} \right)^{\frac{{25}}{{100}}}} \cr & = {\left( {256} \right)^{\frac{1}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{1}{4}}} \cr & = {4^1} \cr & = 4 \cr}

7. The value of [(10)150 ÷ (10)146]
a) 1000
b) 10000
c) 100000
d) 10

Explanation:
\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr}

8. $$\frac{1}{{1 + {x^{\left( {b - a} \right)}} + {x^{\left( {c - a} \right)}}}}$$    $$+ \frac{1}{{1 + {x^{\left( {a - b} \right)}} + {x^{\left( {c - b} \right)}}}}$$    $$+ \frac{1}{{1 + {x^{\left( {b - c} \right)}} + {x^{\left( {a - c} \right)}}}} = ?$$
a) 0
b) 1
c) xa - b - c
d) None of these

Explanation: Given exp. =
$$= \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} +$$   $$\frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} +$$   $$\frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}}$$
$$= \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} +$$   $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} +$$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}}$$
\eqalign{ & = \frac{{ {{x^a} + {x^b} + {x^c}} }}{{ {{x^a} + {x^b} + {x^c}} }} \cr & = 1 \cr}

9. (25)7.5 × (5)2.5 ÷ (125)1.5 = 5?
a) 8.5
b) 13
c) 16
d) 17.5

\eqalign{ & {\text{Let}}\,{\left( {25} \right)^{7.5}} \times {\left( 5 \right)^{2.5}} \div {\left( {125} \right)^{1.5}} = {5^x} \cr & {\text{Then}},\,\frac{{{{\left( {{5^2}} \right)}^{7.5}} \times {{\left( 5 \right)}^{2.5}}}}{{{{\left( {{5^3}} \right)}^{1.5}}}} = {5^x} \cr & \frac{{{5^{\left( {2 \times 7.5} \right)}} \times {5^{2.5}}}}{{{5^{\left( {3 \times 1.5} \right)}}}} = {5^x} \cr & \frac{{{5^{15}} \times {5^{2.5}}}}{{{5^{4.5}}}} = {5^x} \cr & {5^x} = {5^{\left( {15 + 2.5 - 4.5} \right)}} \cr & {5^x} = {5^{13}} \cr & x = 13 \cr}
\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr}