1. $${\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?$$
a) $$\frac{1}{{256}}$$
b) 16
c) 80
d) 256
Explanation:
$$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$
2. $$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$ 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 :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$ $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$
3. 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:
$$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$
4. 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:
$$\eqalign{ & {\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & 2x - 1 = - 3 \cr & 2x = - 2 \cr & x = - 1 \cr} $$
5.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:
$$\eqalign{ & {\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \frac{{5 \times {5^{\frac{1}{2}}} \times {5^3}}}{{{5^{ - \frac{3}{2}}}}} = {5^{a + 2}} \cr & {5^{\left( {1 + \frac{1}{2} + 3 + \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & {5^6} = {5^{a + 2}} \cr & a + 2 = 6 \cr & a = 4 \cr} $$
6. (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
7.$${\text{If}}\,{\kern 1pt} {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}},$$ then the value of x is
a) $$\frac{1}{2}$$
b) 1
c) 2
d) $$\frac{7}{2}$$
Explanation:
$$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & 2x = 4 \cr & x = 2 \cr} $$
8. 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:
$$\eqalign{ & {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$
9. 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
10. 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.