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
Explanation:
$$\eqalign{ & {\text{Let}}\,{\left( {25} \right)^{7.5}} \times {\left( 5 \right)^{2.5}} \div {\left( {125} \right)^{1.5}} = {5^x} \cr & {\text{Then}},\,\frac{{{{\left( {{5^2}} \right)}^{7.5}} \times {{\left( 5 \right)}^{2.5}}}}{{{{\left( {{5^3}} \right)}^{1.5}}}} = {5^x} \cr & \frac{{{5^{\left( {2 \times 7.5} \right)}} \times {5^{2.5}}}}{{{5^{\left( {3 \times 1.5} \right)}}}} = {5^x} \cr & \frac{{{5^{15}} \times {5^{2.5}}}}{{{5^{4.5}}}} = {5^x} \cr & {5^x} = {5^{\left( {15 + 2.5 - 4.5} \right)}} \cr & {5^x} = {5^{13}} \cr & x = 13 \cr} $$
10. (0.04)-1.5 = ?
a) 25
b) 125
c) 250
d) 625
Explanation:
$$\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr} $$
11. $${\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?$$
a) $$\frac{1}{{256}}$$
b) 16
c) 80
d) 256
Explanation:
$$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$
12. $$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$ is equal to = ?
a) $$\frac{1}{3}$$
b) $$\frac{5}{{16}}$$
c) $$\frac{3}{8}$$
d) $$\frac{{41}}{{7280}}$$
Explanation:
$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$
Formula :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$ $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$
13. Given that $$\sqrt 5 $$ = 2.236 and $$\sqrt 3 $$ = 1.732, then the value of
a) 0.564
b) 0.504
c) 0.252
d) 0.202
Explanation:
$$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$
14. If $${\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}}$$ then x is equal to ?
a) -2
b) -1
c) 1
d) 2
Explanation:
$$\eqalign{ & {\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & 2x - 1 = - 3 \cr & 2x = - 2 \cr & x = - 1 \cr} $$
15.If $${\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}}$$ then the value of a is = ?
a) 4
b) 5
c) 6
d) 8
Explanation:
$$\eqalign{ & {\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \frac{{5 \times {5^{\frac{1}{2}}} \times {5^3}}}{{{5^{ - \frac{3}{2}}}}} = {5^{a + 2}} \cr & {5^{\left( {1 + \frac{1}{2} + 3 + \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & {5^6} = {5^{a + 2}} \cr & a + 2 = 6 \cr & a = 4 \cr} $$
16. (17)3.5 × (17)? = 178?
a) 2.29
b) 2.75
c) 4.25
d) 4.5
Explanation: Let (17)3.5 × (17)x = 178
Then, (17)3.5 + x = 178
3.5 + x = 8
x = (8 - 3.5)
x = 4.5
17.$${\text{If}}\,{\kern 1pt} {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}},$$ then the value of x is
a) $$\frac{1}{2}$$
b) 1
c) 2
d) $$\frac{7}{2}$$
Explanation:
$$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & 2x = 4 \cr & x = 2 \cr} $$
18. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
a) 1.45
b) 1.88
c) 2.9
d) 3.7
Explanation:
$$\eqalign{ & {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$
19. If 5a = 3125, then the value of 5(a - 3) is:
a) 25
b) 125
c) 625
d) 1625
Explanation: 5a = 3125 ⇔ 5a = 55
⇒ a = 5.
5(a - 3) = 5(5 - 3) = 52 = 25
20. If 3(x - y) = 27 and 3(x + y) = 243, then x is equal to:
a) 0
b) 2
c) 4
d) 6
Explanation: 3x - y = 27 = 33 ⇔ x - y = 3 ....(i)
3x + y = 243 = 35 ⇔ x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.
21. The value of $${\left( {\sqrt 8 } \right)^{\frac{1}{3}}}$$ is = ?
a) 2
b) 4
c) $$\sqrt 2 $$
d) 8
Explanation:
$$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$
22. The value of $${\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}}$$ is = ?
a) $$\frac{4}{9}$$
b) $$\frac{9}{4}$$
c) $$\frac{{16}}{{81}}$$
d) $$\frac{{81}}{{16}}$$
Explanation:
$$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$
23. The value of $${\text{2}}{{\text{7}}^{ - \frac{2}{3}}}$$ lies between = ?
a) 0 and 1
b) 0 and 2
c) 2 and 3
d) 3 and 4
Explanation:
$$\eqalign{ & {\text{ 2}}{{\text{7}}^{ - \frac{2}{3}}} \cr & = {\left( {{3^3}} \right)^{ - \frac{2}{3}}} \cr & = {3^{\left[ {3 \times \left( { - \frac{2}{3}} \right)} \right]}} \cr & = {3^{ - 2}} \cr & = \frac{1}{{{3^2}}} \cr & = \frac{1}{9} \cr & {\text{Clearly}},\,\,0 < \frac{1}{9} < 1 \cr} $$
24. The value of $$\frac{1}{{\sqrt {3.25} + \sqrt {2.25} }}$$ $$ +\, \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }}$$ $$ +\, \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }}$$ $$ +\, \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }}$$ is = ?
a) 1.00
b) 1.25
c) 1.50
d) 2.25
Explanation:
$$\eqalign{ & \frac{1}{{\sqrt {3.25} + \sqrt {2.25} }} \times \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{\sqrt {3.25} - \sqrt {2.25} }} \cr & = \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{3.25 - 2.25}} \cr & = \sqrt {3.25} - \sqrt {2.25} \,......(i) \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }} \cr & = \sqrt {4.25} - \sqrt {3.25} \,.......(ii) \cr & \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }} \cr & = \sqrt {5.25} - \sqrt {4.25} \,.......(iii) \cr & \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }} \cr & = \sqrt {6.25} - \sqrt {5.25} \,.......(iv) \cr & {\text{Now}}\,{\text{add}}\,{\text{all}}\,{\text{them}} \cr & \sqrt {3.25} - \sqrt {2.25} + \sqrt {4.25} - \sqrt {3.25} + \sqrt {5.25} - \sqrt {4.25} + \sqrt {6.25} - \sqrt {5.25} \cr & = \sqrt {6.25} - \sqrt {2.25} \cr & = 2.5 - 1.5 \cr & = 1 \cr} $$
25. $$\frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}}$$ is simplified to = ?
a) -2
b) -1
c) 1
d) 2
Explanation:
$$\eqalign{ & \frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}} \cr & = \frac{{1 + \frac{1}{3}}}{{\frac{1}{3} - 1}} \cr & = \frac{{\frac{4}{3}}}{{ - \frac{2}{3}}} \cr & = - 2 \cr} $$
26. Simplify : $$\left( {\frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }}} \right) = ?$$
a) $$\frac{1}{2} - 5\sqrt 3 $$
b) 2 - $$5\sqrt 3 $$
c) 1
d) 0
Explanation:
$$\eqalign{ & \frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{\frac{{3\left( {2 - \sqrt 3 } \right) - 2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 \,} \right)\left( {2 - \sqrt 3 } \right)}}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{6 - 3\sqrt 3 - 4 - 2\sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\left( {2 - 5\sqrt 3 } \right)}} \cr & = \frac{{2 - 5\sqrt 3 }}{{2 - 5\sqrt 3 }} \cr & = 1 \cr} $$
27. ($$\sqrt 8$$ - $$\sqrt 4 $$ - $$\sqrt 2 $$) Equals to = ?
a) 2 - $$\sqrt 2 $$
b) $$\sqrt 2 $$ - 2
c) 2
d) -2
Explanation:
$$\eqalign{ & \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr & = 2\sqrt 2 - 2 - \sqrt 2 \cr & = 2\sqrt 2 - \sqrt 2 - 2 \cr & = \sqrt 2 - 2 \cr} $$
28. $${\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}}$$ is equal to ?
a) 1
b) 2
c) $$\frac{1}{2}$$
d) $$\frac{1}{{16}}$$
Explanation:
$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$
29. The value of $$\left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right)$$ is = ?
a) $$\frac{3}{2}$$
b) $$\frac{4}{9}$$
c) $$\frac{{16}}{{81}}$$
d) $$\frac{{32}}{{243}}$$
Explanation:
$$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$
30.$$\left[ {{4^3} \times {5^4}} \right] \div {4^5} = ?$$
a) 29.0825
b) 30.0925
c) 35.6015
d) 39.0625
Explanation:
$$\eqalign{ & \frac{{{4^3} \times {5^4}}}{{{4^5}}} \cr & = \frac{{{5^4}}}{{{4^{\left( {5 - 3} \right)}}}} \cr & = \frac{{{5^4}}}{{{4^2}}} \cr & = \frac{{625}}{{16}} \cr & = 39.0625 \cr} $$
31. $$\frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }}$$ is equal to = ?
a) $${\text{1}} - \sqrt 5 + \sqrt 2 + \sqrt {16} $$
b) $${\text{1}} + \sqrt 5 + \sqrt 2 - \sqrt {10} $$
c) $${\text{1}} + \sqrt 5 + \sqrt 2 + \sqrt {10} $$
d) $${\text{1}} - \sqrt 5 - \sqrt 2 + \sqrt {10} $$
Explanation:
$$\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr & = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr & = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr & = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr & {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr} $$
32. $$\left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right]$$ is equal to = ?
a) 32
b) 8
c) 1
d) 0
Explanation:
$$\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr & = \left[ {8 - \left( {{2^3}} \right)} \right] \cr & = \left[ {8 - 8} \right] \cr & = 0 \cr} $$
33. $$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - $$ $$\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + $$ $$\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$ is equal to = ?
a) 3
b) 2
c) 0
d) $$\sqrt 3 $$
Explanation:
$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$
$$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$ $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$ $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$ $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$ $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
$$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$
34. The value of $$\frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}$$ $$\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}$$ $$\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}}$$ is = ?
a) 102
b) 105
c) 107
d) 109
Explanation:
$$\eqalign{ & \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr & = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr & = \left( {{6^2} + {4^3} + {2^1}} \right) \cr & = \left( {36 + 64 + 2} \right) \cr & = 102 \cr} $$
35. $${\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?$$
a) $$\frac{1}{3}$$
b) $$\frac{1}{{48}}$$
c) 1
d) 48
Explanation:
$$\eqalign{ & {\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16 \times 3} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{ - \frac{2}{7}}} \times {\left( 3 \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \cr & = {\left( {16} \right)^{\left( { - \frac{7}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{7}{7}} \right)}} \cr & = {\left( {16} \right)^{ - 1}} \times {\left( 3 \right)^{ - 1}} \cr & = \frac{1}{{16}} \times \frac{1}{3} \cr & = \frac{1}{{48}} \cr} $$.
36. 93 × 62 ÷ 33 = ?
a) 948
b) 972
c) 984
d) 1012
Explanation:
$$\eqalign{ & \frac{{{9^3} \times {6^2}}}{{{3^3}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {3 \times 2} \right)}^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {3 \times 2} \right)}} \times {3^2} \times {2^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {6 + 2} \right)}} \times {2^2}}}{{{3^3}}} \cr & = {3^{\left( {8 - 3} \right)}} \times {2^2} \cr & = {3^5} \times {2^2} \cr & = 243 \times 4 \cr & = 972 \cr} $$
37. $$\left({\frac{{2+\sqrt 3}}{{2-\sqrt3}}+ \frac{{2 - \sqrt 3}}{{2 + \sqrt 3}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$ Simplifies to :
a) 2 - $$\sqrt 3 $$
b) 2 + $$\sqrt 3 $$
c) 16 - $$\sqrt 3 $$
d) 40 - $$\sqrt 3 $$
Explanation:
$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$
$$ = \left\{ {\frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}} \right\}$$
$$ = \left\{ {\frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 }}{{4 - 3}} + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}} \right\}$$
$$\eqalign{ & = \left\{ {14 + \frac{{3 + 1 - 2\sqrt 3 }}{2}} \right\} \cr & = 14 + \frac{{2\left( {2 - \sqrt 3 } \right)}}{2} \cr & = 14 + 2 - \sqrt 3 \cr & = 16 - \sqrt 3 \cr} $$
38. The value of $$\sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} $$ is = ?
a) $$\sqrt 6 - \sqrt 2 $$
b) $$\sqrt 6 + \sqrt 2 $$
c) $$\sqrt 6 - 2$$
d) $${\text{2}} - \sqrt 6 $$
Explanation:
$$\eqalign{ & \sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{\sqrt {36} + \sqrt {24} - \sqrt {24} - \sqrt {16} }}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{6 - 4}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{2}{{5 + \sqrt {24} }} \times \frac{{5 - \sqrt {24} }}{{5 - \sqrt {24} }}} \cr & = \sqrt {\frac{{2\left( {5 - \sqrt {24} } \right)}}{{25 - 24 }}} \cr & = \sqrt {2\left( {5 - 2\sqrt 6 } \right)} \cr & = \sqrt {2\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 3 \times \sqrt 2 } \right\}} \cr & = \sqrt {2{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \cr & = \sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right) \cr & = \sqrt 6 - 2 \cr} $$
39. (19)12 × (19)8 ÷ (19)4 = (19)?
a) 6
b) 8
c) 12
d) None of these
Explanation:
$$\eqalign{ & \frac{{{{\left( {19} \right)}^{12}} \times {{\left( {19} \right)}^8}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{19}^{\left( {12 + 8} \right)}}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{\left( {19} \right)}^{20}}}}{{{{\left( {19} \right)}^4}}} \cr & = {\left( {19} \right)^{\left( {20 - 4} \right)}} \cr & = {\left( {19} \right)^{16}} \cr} $$
Hence the missing number = 16
40. (64)4 ÷ (8)5 = ?
a) (8)8
b) (8)2
c) (8)12
d) None of these
Explanation:
$$\eqalign{ & {\left( {64} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( {{8^2}} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( 8 \right)^{\left( {2 \times 4} \right)}} \div {8^5} \cr & = \frac{{{8^8}}}{{{8^5}}} \cr & = {8^{\left( {8 - 5} \right)}} \cr & = {8^3} \cr} $$
41. $$\left[ {\frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - \left( {0.73} \right) \times \left( {0.27} \right)}}} \right]$$ simplifies to ?
a) 1
b) 0.4087
c) 0.73
d) 0.27
Explanation:
$$\eqalign{ & {\text{let}} \cr & a = 0.73 \cr & b = 0.27 \cr & = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {0.73 + 0.27} \right) \cr & = 1 \cr} $$
42. If $$x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$ and $$y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}$$ then $$\left( {x + y} \right)$$ equals ?
a) 8
b) 16
c) $${\text{2}}\sqrt {15} $$
d) $${\text{2}}\left( {\sqrt 5 + \sqrt 3 } \right)$$
Explanation:
$$\eqalign{ & x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Similarly}} \cr & y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Now, }}x + y \cr & = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr & = \frac{{16}}{2} \cr & = 8 \cr} $$
43. Simplified from of $${\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5}$$ is = ?
a) $$\frac{1}{x}$$
b) x
c) x-5
d) x5
Explanation:
$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$
44. What will come in place of both the question marks in the following question :
$$\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}$$
a) 10
b) $${\text{10}}\sqrt 2 $$
c) $$\sqrt {20} $$
d) 210
Explanation:
$$\eqalign{ & {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr & {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr & {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr & x = 210 \cr} $$
45. The value of $${{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}$$ is = ?
a) $$\sqrt 5 $$
b) 5
c) $${\text{5}}\sqrt 5 $$
d) 25
Explanation:
$$\eqalign{ & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}} \cr & = {5^{0.25}} \times {\left( {{5^3}} \right)^{0.25}} \cr & = {5^{0.25}} \times {5^{\left( {3 \times 0.25} \right)}} \cr & = {5^{0.25}} \times {5^{0.75}} \cr & = {5^{\left( {0.25 + 0.75} \right)}} \cr & = {5^1} \cr & = 5 \cr} $$
46. $${\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {? - 5} \right)}}$$
a) 1
b) 4
c) 6
d) 7
Explanation:
$$\eqalign{ & {\text{Let ,}} \cr & {\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {x - 5} \right)}} \cr & {6^{\left( {x - 5} \right)}} = {\left( 6 \right)^4} \div {\left( {{6^2}} \right)^3} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^4} \div {6^{\left( {2 \times 3} \right)}} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^4} \div {6^6} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^{\left( {4 - 6 + 3} \right)}} \cr & {6^{\left( {x - 5} \right)}} = 6 \cr & x - 5 = 1 \cr & x = 6 \cr} $$
47. The value of $${\left( {256} \right)^{\frac{5}{4}}}$$ is = ?
a) 512
b) 984
c) 1024
d) 1032
Explanation:
$$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$
48. $$\sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } $$ is equal to ?
a) $$\sqrt 2 $$
b) $${\text{2}}\sqrt 2 $$
c) 2
d) 3
Explanation:
$$\eqalign{ & x = \sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } \cr & {x^2} = 2 + \sqrt {2 + \sqrt {2 + .......} } \cr & {x^2} = 2 + x \cr & {x^2} - x - 2 = 0 \cr & x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \cr & \left( {x + 1} \right)\left( {x - 2} \right) = 0 \cr & x = 2 \cr} $$
49.The value of $${\text{2}} + \sqrt {0.09} \, - \,\root 3 \of {0.008} \, - \,75\% $$ of 2.80 is = ?
a) 0
b) 0.01
c) -1
d) 0.001
Explanation:
$$\eqalign{ & {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr & = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr & = 2 + 0.3 - 0.2 - 2.10 \cr & = 2.3 - 2.3 \cr & = 0 \cr} $$
50. The value of $${\left( {3 + 2\sqrt 2 } \right)^{ - 3}}$$ $$ + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} = ?$$
a) 189
b) 180
c) 108
d) 198
Explanation:
$$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$
$$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$ $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$
$$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ { {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$
51. The value of $$\frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}}{\text{is = ?}}$$
a) $$ - \frac{1}{{36}}$$
b) $$\frac{2}{3}$$
c) $$\frac{1}{{13}}$$
d) $$\frac{5}{{13}}$$
Explanation:
$$\eqalign{ & \frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}} \cr & = \frac{{{2^{n - 1}}\left( {1 - 2} \right)}}{{{2^{n + 1}}\left( {{2^3} + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{\left( {n - 1} \right) - \left( {n + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{ - 2}} \cr & = \left( { - \frac{1}{9}} \right).\frac{1}{{{2^2}}} \cr & = \left( { - \frac{1}{9}} \right) \times \frac{1}{4} \cr & = - \frac{1}{{36}} \cr} $$
52. If $$x = 5 + 2\sqrt 6 {\text{,}}$$ then $$\sqrt x - \frac{1}{{\sqrt x }}$$ = is?
a) $${\text{2}}\sqrt 2 $$
b) $${\text{2}}\sqrt 3 $$
c) $$\sqrt 3 + \sqrt 2 $$
d) $$\sqrt 3 - \sqrt 2 $$
Explanation:
$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$
53. $$\left( {4 + \sqrt 7 } \right),$$ expressed as a perfect square, is equal to = ?
a) $${\left( {2 + \sqrt 7 } \right)^2}$$
b) $${\left( {\frac{{\sqrt 7 }}{2} + \frac{1}{2}} \right)^2}$$
c) $$\left\{ {\frac{1}{2}{{\left( {\sqrt 7 + 1} \right)}^2}} \right\}$$
d) $$\left( {\sqrt 3 + \sqrt 4 } \right)$$
Explanation:
$$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$
54. If 3x+y = 81 and 81x-y = 3, then the value of $$\frac{x}{y}$$ is = ?
a) $$\frac{{15}}{{17}}$$
b) $$\frac{{17}}{{30}}$$
c) $$\frac{{15}}{{34}}$$
d) $$\frac{{17}}{{15}}$$
Explanation:
$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$
55. $$\left( {\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}} \right)$$ simplifies to = ?
a) $$\sqrt 5 + \sqrt 6 $$
b) $${\text{2}}\sqrt 5 + \sqrt 6 $$
c) $$\sqrt 5 - \sqrt 6 $$
d) $${\text{2}}\sqrt 5 - 3\sqrt 6 $$
Explanation:
$$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$
$$ \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$
$$\eqalign{ & \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr & \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr & \sqrt 5 - \sqrt 6 \cr} $$
56. $${6^{1.2}} \times {36^?} \times {30^{2.4}} \times {25^{1.3}} = {30^5}$$
a) 0.1
b) 0.7
c) 1.4
d) 2.6
Explanation:
$$\eqalign{ & {\text{Let }}\,{6^{1.2}} \times {36^x} \times {30^{2.4}} \times {25^{1.3}} = {30^5} \cr & {\text{Then,}}\,{6^{1.2}} \times {({6^2})^x} \times {(6 \times 5)^{2.4}} \times {({5^2})^{1.3}} = {30^5} \cr & {6^{1.2}} \times {6^{2x}} \times {6^{2.4}} \times {5^{2.4}} \times {5^{2.6}} = {(6 \times 5)^5} \cr & {6^{\left( {1.2 + 2x + 2.4} \right)}} \times {5^{\left( {2.4 + 2.6} \right)}} = {6^5} \times {5^5} \cr & {6^{\left( {3.6 + 2x} \right)}} \times {5^5} = {6^5} \times {5^5} \cr & 3.6 + 2x = 5 \cr & 2x = 1.4 \cr & x = 0.7 \cr} $$
57.$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ simplifies to = ?
a) $${\text{2}}\sqrt 6 $$
b) $${\text{4}}\sqrt 6 $$
c) $${\text{2}}\sqrt 3 $$
d) $${\text{3}}\sqrt 2 $$
Explanation:
$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$
58. If $$\sqrt 3 $$ = 1.732 is given, then the value of $$\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}$$ is = ?
a) 11.732
b) 13.928
c) 12.928
d) 13.925
Explanation:
$$\eqalign{ & \sqrt 3 = 1.732 \cr & \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}} \cr & 4 + 3 + 4\sqrt 3 \cr & 7 + 4 \times 1.732 \cr & 7 + 6.928 \cr & 13.928 \cr} $$
59. Evaluate: $$16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} $$ if $$\sqrt {12} $$ = 3.46
a) 3.46
b) 10.38
c) 13.84
d) 24.22
Explanation:
$$\eqalign{ & 16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} \cr & 16\sqrt {\frac{{3 \times 4}}{{4 \times 4}}} - 9\sqrt {\frac{{3 \times 4}}{{3 \times 3}}} \cr & 16 \times \frac{{\sqrt {12} }}{4} - \frac{{9\sqrt {12} }}{3} \cr & 4\sqrt {12} - 3\sqrt {12} \cr & \sqrt {12} \cr & 3.46 \cr} $$
60. $${2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = $$ $${\left( {32} \right)^?}$$
a) 5.9
b) 7.7
c) 9.5
d) 13.1
Explanation:
$$\eqalign{ & {\text{Let }}{2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = {\left( {32} \right)^x} \cr & {\text{Then,}}{2^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{3.6}} \times {2^{\left( {2 \times 3.6} \right)}} \times {2^{\left( {2 \times 3.6} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{\left( {3.6 + 7.2 + 7.2} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{18}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{\left( {3.6 + 2.3} \right)}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{5.9}} = {\left( {{2^5}} \right)^x} \cr & x = 5.9 \cr} $$
61. If 32x-y = 3x+y = $$\sqrt {27} {\text{,}}$$ the value of y is = ?
a) $$\frac{1}{2}$$
b) $$\frac{1}{4}$$
c) $$\frac{3}{2}$$
d) $$\frac{3}{4}$$
Explanation:
$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & x = 1 \cr & y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$
62. $$\frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} $$ = ?
a) 8
b) 4
c) $$\frac{1}{2}$$
d) 2
Explanation:
$$\eqalign{ & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} \cr & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {169} } } } }}{2} \cr & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {121} } } }}{2} \cr & \frac{{\sqrt {10 + \sqrt {36} } }}{2} \cr & \frac{{\sqrt {16} }}{2} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$
63. $$\frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }}$$ is equal to = ?
a) 330
b) 355
c) 305
d) 366
Explanation:
$$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & 36 + 49 + 64 + 81 + 100 \cr & 330 \cr} $$
64.Given 2x = 8y+1 and 9y = 3x-9 , then value of x + y is = ?
a) 18
b) 21
c) 24
d) 27
Explanation:
$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & x = 3y + 3 \cr & x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & 2y = x - 9 \cr & x - 2y = 9......({\text{ii}}) \cr & {\text{Subtracting (i) from (ii),}} \cr & {\text{we}}\,{\text{get}}\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & x + y = 21 + 6 = 27 \cr} $$
65. What are the values of x and y that satisfy the equation, $${{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}}{\text{ ?}}$$
a) x = 2.5, y = 6
b) x = 3, y = 5
c) x = 3, y = 4
d) x = 5, y = 2
Explanation:
$$\eqalign{ & {{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}} \cr & \Leftrightarrow \frac{{{{\text{2}}^{0.7x}}}}{{{{\text{3}}^{ 1.25y}}}}{\text{ = }}\frac{{{2^3}{{.2}^{\frac{1}{2}}}{{.3}^{\frac{1}{2}}}}}{{{3^3}}} \cr & \Leftrightarrow \frac{{{2^{\left( {3 + \frac{1}{2}} \right)}}}}{{{3^{\left( {3 - \frac{1}{2}} \right)}}}} = \frac{{{2^{\frac{7}{2}}}}}{{{2^{\frac{5}{2}}}}} = \frac{{{2^{3.5}}}}{{{3^{2.5}}}} \cr & 0.7x = 3.5 \Rightarrow x = \frac{{3.5}}{{0.7}}{\text{ = 5}} \cr & {\text{and }}1.25y = 2.5 \cr & \Rightarrow y = \frac{{2.5}}{{1.25}} = 2 \cr} $$
66. Evaluate : $$\sqrt {20} + \sqrt {12} + \root 3 \of {729} \,\, - $$ $$\frac{4}{{\sqrt 5 - \sqrt 3 }} \,- $$ $$\sqrt {81} = ?$$
a) $$\sqrt 2 $$
b) $$\sqrt 3 $$
c) 0
d) $$2\sqrt 2 $$
Explanation:
$$\sqrt {20} + \sqrt {12} + \root 3 \of {729} - \frac{4}{{\sqrt 5 - \sqrt 3 }} - \sqrt {81} $$
$$ = 2\sqrt 5 + 2\sqrt 3 + 9\, - $$ $$\left( {\frac{4}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right)$$ $$\, - \,9$$
$$\eqalign{ & = 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{{4\left( {\sqrt 5 + \sqrt 3 } \right)}}{2}} \right) - 9 \cr & = 2\sqrt 5 + 2\sqrt 3 + 9 - 2\sqrt 5 - 2\sqrt 3 - 9 \cr & = 0 \cr} $$
67. If $$\frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }} = A + \sqrt B {\text{,}}$$ then B - A is = ?
a) -13
b) $${\text{2}}\sqrt {13} $$
c) 13
d) $${\text{3}}\sqrt 3 - \sqrt 7 $$
Explanation:
$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{2^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ = }}A + \sqrt B \cr & \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = A + \sqrt B \cr & \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = A + \sqrt B \cr & 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = A + \sqrt B \cr & 2\sqrt 3 - 1 = A + \sqrt B \cr & A = - 1\,\,\& \,\,\sqrt B = 2\sqrt 3 \cr & B = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{So}},B - A = 12 - \left( { - 1} \right) = 13 \cr} $$
68. Given that 100.48 = x, 100.70 = y and xz = y2 then the value of z is close to = ?
a) 1.45
b) 1.88
c) 2.9
d) 3.7
Explanation:
$$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & 0.48z = 1.40 \cr & z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & z = 2.9\left( {{\text{approx}}} \right) \cr} $$
69. If m and n are whole numbers such that mn = 121, then the value of (m - 1)n+1 is = ?
a) 1
b) 10
c) 121
d) 1000
Explanation:
$$\eqalign{ & {\text{We know that}} \cr & {\text{1}}{{\text{1}}^2} = 121 \cr & {\text{Putting}} \cr & m = 11\& n = 2 \cr & {\text{we get}} \cr & {\left( {m - 1} \right)^{n + 1}} \cr & = {\left( {11 - 1} \right)^{\left( {2 + 1} \right)}} \cr & = {10^3} \cr & = 1000 \cr} $$
70. 1 + (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) is equal to =
a) $$\frac{{{3^{64}} - 1}}{2}$$
b) $$\frac{{{3^{64}} + 1}}{2}$$
c) 364 - 1
d) 364 + 1
Explanation:
$$1 + \left( {3 + 1} \right)$$ $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$
$$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$
71. $${\left( {\frac{{{x^b}}}{{{x^c}}}} \right)^{\left( {b + c - a} \right)}}.$$ $${\left( {\frac{{{x^c}}}{{{x^a}}}} \right)^{\left( {c + a - b} \right)}}.$$ $${\left( {\frac{{{x^a}}}{{{x^b}}}} \right)^{\left( {a + b - c} \right)}} = ?$$
a) xabc
b) 1
c) xab+bc+ca
d) xa+b+c
Explanation:
$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
$$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$ $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$ $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$
72.If 2n-1 + 2n+1 = 320, then the value of n is = ?
a) 6
b) 8
c) 5
d) 7
Explanation:
$$\eqalign{ & {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr & {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr & {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr & {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr & {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr & n - 1 = 6 \cr & n = 7 \cr} $$
73. 461 + 462 + 463 + 464 is divided by = ?
a) 17
b) 3
c) 11
d) 13
Explanation:
$$\eqalign{ & {4^{61}} + {4^{62}} + {4^{63}} + {4^{64}} \cr & = {4^{61}}\left( {{4^0} + {4^1} + {4^2} + {4^3}} \right) \cr & = {4^{61}} \times 85 \cr} $$
Now check with option 85 is divisible by 17.
74. The value of $${\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}$$ $${\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.$$ $${\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}}$$
a) 1
b) a
c) b
d) c
Explanation:
$$\eqalign{ & {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^0} \cr & = 1 \cr} $$
75. If ax = b, by = c and cz = a, then the value of xyz is =
a) 0
b) 1
c) $$\frac{1}{{{\text{abc}}}}$$
d) abc
Explanation:
$$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$
76.(0.04)2 ÷ (0.008) × (0.2)6 = (0.2)?
a) 5
b) 6
c) 8
d) None of these
Explanation:
$$\eqalign{ & {\text{Let }}{\left( {0.04} \right)^2} \div \left( {0.008} \right) \times {\left( {0.2} \right)^6} = {\left( {0.2} \right)^x} \cr & {\text{Then,}}{\left( {0.2} \right)^x} = {\left[ {{{\left( {0.2} \right)}^2}} \right]^2} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {2 \times 2} \right)}} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & {\left( {0.2} \right)^x} = {\left( {0.2} \right)^4} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {4 - 3 + 6} \right)}} \cr & {\left( {0.2} \right)^x} = {\left( {0.2} \right)^7} \cr & x = 7 \cr} $$
77. The value of $$\frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}}$$ is = ?
a) $$\frac{3}{7}$$
b) $$\frac{7}{3}$$
c) $${\text{1}}\frac{3}{7}$$
d) $${\text{2}}\frac{2}{7}$$
Explanation:
$$\eqalign{ & \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr & = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr & = \frac{{{3^1}}}{{{7^1}}} \cr & = \frac{3}{7} \cr} $$
78. $$\frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}}$$ is equal to = ?
a) 2(n+1)
b) $$\left( {\frac{9}{8} - {2^n}} \right)$$
c) $$\left( { - {2^{n + 1}} + \frac{1}{8}} \right)$$
d) 1
Explanation:
$$\eqalign{ & \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr & = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr & = \frac{8}{8} \cr & = 1 \cr} $$
79. $$\frac{{256 \times 256 - 144 \times 144}}{{112}}$$ is equal to = ?
a) 420
b) 400
c) 360
d) 320
Explanation:
$$\eqalign{ & \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr & = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr & = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr & = 400 \cr} $$
80. $$\sqrt {3\sqrt {3\sqrt {3........} } } $$ is equal to = ?
a) $$\sqrt 3 $$
b) 3
c) $${\text{2}}\sqrt 3 $$
d) $${\text{3}}\sqrt 3 $$
Explanation: When the question is in form
$$\eqalign{ & \Rightarrow \sqrt {n\sqrt {n\sqrt n } } ........\infty \cr & \Rightarrow {\text{So }}n{\text{ is answer }} \cr & \Rightarrow {\text{3}} \cr} $$
81. The value of $$\sqrt {40 + \sqrt {9\sqrt {81} } }$$ is = ?
a) $$\sqrt {111} $$
b) 9
c) 7
d) 11
Explanation:
$$\eqalign{ & \sqrt {40 + \sqrt {9\sqrt {81} } } \cr & \sqrt {40 + \sqrt {9 \times 9} } \cr & \sqrt {40 + 9} \cr & \sqrt {49} \cr & 7 \cr} $$
82. $$\sqrt {8 - 2\sqrt {15} } $$ is equal to = ?
a) $${\text{3}} - \sqrt 5 $$
b) $$\sqrt 5 - \sqrt 3 $$
c) $${\text{5}} - \sqrt 3 $$
d) $$\sqrt 5 + \sqrt 3 $$
Explanation:
$$\eqalign{ & \sqrt {8 - 2\sqrt {15} } \cr & = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr & = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$
83. $$\sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } $$ is equal to = ?
a) $${\text{1}} - \sqrt 3 $$
b) $$\sqrt 3 - 1$$
c) $${\text{2}}\left( {2 - \sqrt 3 } \right)$$
d) $${\text{2}}\left( {2 + \sqrt 3 } \right)$$
Explanation:
$$\eqalign{ & \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr & = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr & = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr & = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr & = \sqrt 3 - 1 \cr} $$
84. If N = $$\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - $$ $$\sqrt {3 - 2\sqrt 2 } {\text{,}}$$ then the value of N is = ?
a) $${\text{2}}\sqrt 2 - 1$$
b) 3
c) 1
d) 2
Explanation:
$$\eqalign{ & {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr & {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
$${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{ & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & {{\text{X}}^2} = 2 \cr & {\text{X = }}\sqrt 2 \cr & {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr & {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr & {\text{N}} = 1 \cr} $$
85. If $$\frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}$$ = 1 then the value of x is = ?
a) $$\sqrt 5 $$
b) 5
c) $${\text{2}}\sqrt 5 $$
d) $${\text{3}}\sqrt 5 $$
Explanation:
$$\eqalign{ & \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr & \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr & \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr & x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr & x = 5{\text{ }} \cr} $$
86.Find the simplest value of $${\text{2}}\sqrt {50} $$ + $$\sqrt {18} $$ - $$\sqrt {72} $$ = ?(given $$\sqrt 2 $$ = 1.414)
a) 4.242
b) 9.898
c) 10.6312
d) 8.484
Explanation:
$$\eqalign{ & {\text{2}}\sqrt {50} {\text{ + }}\sqrt {18} - \sqrt {72} \cr & {\text{2}} \times {\text{5}}\sqrt 2 {\text{ + 3}}\sqrt 2 - 6\sqrt 2 \cr & 13\sqrt 2 - 6\sqrt 2 \cr & 7\sqrt 2 \cr & 7 \times 1.414 \cr & 9.898 \cr} $$
87. 553 + 173 - 723 + 201960 is equal to = ?
a) -1
b) 0
c) 1
d) 17
Explanation:
$$\eqalign{ & {\text{Let }}a = 55, \cr & \,\,\,\,\,\,\,\,\,\,\,b = 17, \cr & \,\,\,\,\,\,\,\,\,\,\,c = - 72 \cr & a + b + c \cr & = 55 + 17 - 72 \cr & = 0 \cr & {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \left( {a + b + c} \right) = 0 \cr & {\text{Answer is }}0. \cr} $$
88. The simplification value of $$\left( {\sqrt 3 + 1} \right)$$ $$\left( {10 + \sqrt {12} } \right)$$ $$\left( {\sqrt {12} - 2} \right)$$ $$\left( {5 - \sqrt 3 } \right)$$ is = ?
a) 16
b) 88
c) 176
d) 132
Explanation:
$$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr} $$
$$ \Rightarrow \left( {\sqrt 3 + 1} \right) \times $$ $$2\left( {5 + \sqrt 3 } \right) \times $$ $$2\left( {\sqrt 3 - 1} \right)$$ $$\left( {5 - \sqrt 3 } \right)$$
$$\eqalign{ & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 + \sqrt 3 } \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left[ {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right]\left[ {{{\left( 5 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr} $$
89. $$\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$$
a) 0
b) $$\frac{1}{2}$$
c) 1
d) am+n
Explanation:
$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{1 + \frac{{{a^n}}}{{{a^m}}}}} + \frac{1}{{1 + \frac{{{a^m}}}{{{a^n}}}}} \cr & = \frac{{{a^m}}}{{{a^m} + {a^n}}} + \frac{{{a^n}}}{{{a^m} + {a^n}}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$
90. $$\frac{1}{{1 + {x^{\left( {b - a} \right)}} + {x^{\left( {c - a} \right)}}}} \,+ $$ $$\frac{1}{{1 + {x^{\left( {a - b} \right)}} + {x^{\left( {c - b} \right)}}}} \,+ $$ $$\frac{1}{{1 + {x^{\left( {b - c} \right)}} + {x^{\left( {a - c} \right)}}}} = ?$$
a) 0
b) 1
c) xa-b-c
d) None of these
Explanation:
$$\eqalign{ & \frac{1}{{1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}}} + \frac{1}{{1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}}} + \frac{1}{{1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}}} \cr} $$
$$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$ $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$ $$\frac{{{x^c}}}{{\left( {{x^a} + {x^{b}} + {x^c}} \right)}}$$
$$\eqalign{ & = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = 1 \cr} $$
91. (3x - 2y) : (2x + 3y) = 5 : 6, then one of the value of $${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}{\text{ is = ?}}$$
a) $$\frac{1}{{25}}$$
b) 5
c) $$\frac{1}{5}$$
d) 25
Explanation:
$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & {\left( 5 \right)^2} \cr & 25 \cr} $$
92. The exponential form of $$\sqrt {\sqrt 2 \times \sqrt 3 } {\text{ is = ?}}$$
a) $${{\text{6}}^{ - \frac{1}{2}}}$$
b) $${{\text{6}}^{\frac{1}{2}}}$$
c) $${{\text{6}}^{\frac{1}{4}}}$$
d) 6
Explanation:
$$\eqalign{ & {\text{The exponential form of }} \cr & \sqrt {\sqrt 2 \times \sqrt 3 } \cr & = \sqrt {{6^{\frac{1}{2}}}} \cr & = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr & = {6^{\frac{1}{4}}} \cr} $$
93. The quotient when 10100 is divided by 575 is
a) 225 × 1075
b) 1025
c) 275
d) 275 × 1025
Explanation:
$$\eqalign{ & {\text{Expression,}} \cr & {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr} $$
94. The value of $$\frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + $$ $$\frac{1}{{1 - \sqrt 2 + \sqrt 3 }}$$ is = ?
a) $$\sqrt 2 $$
b) $$\sqrt 3 $$
c) 1
d) $$4\left( {\sqrt 3 + \sqrt 2 } \right)$$
Explanation:
$$\eqalign{ & \frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + \frac{1}{{1 - \sqrt 2 + \sqrt 3 }} \cr & = \frac{1}{{1 + \sqrt 3 + \sqrt 2 }} + \frac{1}{{1 + \sqrt 3 - \sqrt 2 }} \cr & = \frac{{1 + \sqrt 3 - \sqrt 2 + 1 + \sqrt 3 + \sqrt 2 }}{{{{\left( {1 + \sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{2 + 2\sqrt 3 }}{{4 + 2\sqrt 3 - 2}} \cr & = \frac{{2 + 2\sqrt 3 }}{{2 + 2\sqrt 3 }} \cr & = 1 \cr} $$
95. 21? × 216.5 = 2112.4
a) 18.9
b) 4.4
c) 5.9
d) 13.4
Explanation:
$$\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr} $$
96. $${25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^?}$$
a) 1.6
b) 1.7
c) 3.2
d) None of these
Explanation:
$$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & x = 2.1 \cr} $$
97. $${8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^?}$$
a) 4.8
b) 5.7
c) 5.8
d) 7.1
Explanation:
$$\eqalign{ & {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr & {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr & {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr & {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr & {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr & {2^x} = {2^{5.7}} \cr & x = 5.7 \cr} $$
98. If 3(x+y) = 81 and 81(x-y) = 3, then the value of x is = ?
a) 42
b) $$\frac{{15}}{8}$$
c) $$\frac{{17}}{8}$$
d) 39
Explanation:
$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}} \cr & {{\text{3}}^{x + y}}{\text{ = }}{{\text{3}}^4} \cr & x + y = 4.....(i) \cr & {\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {3^{4x - 4y}}{\text{ = }}{{\text{3}}^1} \cr & 4x - 4y{\text{ = 1}}....{\text{(ii)}} \cr & {\text{From equation (i) and (ii)}} \cr & 4x - 4y = 1 \cr & 4x + 4y = 16 \cr & 8x = 17 \cr & x = \frac{{17}}{8} \cr} $$
99.Simplified from of $${\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} = ?$$
a) x5
b) x-5
c) x
d) $$\frac{1}{x}$$
Explanation:
$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {\left( {{x^{\frac{1}{5}}}} \right)} \right]^{ - 5}} \cr & = {x^{ - \frac{1}{5} \times 5}} \cr & = {x^{ - 1}} \cr & = \frac{1}{x} \cr} $$
100. Find the value of x in the expression : $$\root 4 \of {3x + 1} = 2$$
a) 3
b) 6
c) 4
d) 5
Explanation:
$$\eqalign{ & \root 4 \of {3x + 1} = 2 \cr & {\left( {\root 4 \of {3x + 1} } \right)^4} = {2^4} \cr & {\left( {3x + 1} \right)^{4 \times \frac{1}{4}}} = 16 \cr & 3x = 15 \cr & x = 5 \cr} $$