Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & 1 < x < 2 \cr & \sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {3 - x} \right)}^2}} \cr & \left( {{\text{Square root cancel with square}}} \right) \cr & \therefore x - 1 + 3 - x \cr & = 2 \cr} $$

$$\eqalign{ & a \oplus b = \left( {a \times b} \right) + b \cr & \Rightarrow 5 \oplus 7 \cr & = \left( {5 \times 7} \right) + 7 \cr & = 35 + 7 \cr & = 42 \cr} $$

$$\eqalign{ & {10^{0.48}} = x{\text{ }} \cr & {10^{0.70}} = y{\text{ }} \cr & {\text{and }}{x^z} = {y^2} \cr & \therefore {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Rightarrow {10^{0.48z}} = {10^{1.40}} \cr} $$
If a^x = a^y, if base equal power are equal : (x = y)
$$\eqalign{ & \therefore 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} \cr & \Rightarrow z = \frac{{35}}{{12}} \cr & \Rightarrow z = 2.9 \cr} $$

Given 47.2506 = 4A + 7B + 2C + $$\frac{5}{{\text{D}}}$$ + 6E . . . . . . . . (1)
Now, we can write 47.2506 = 40 + 7 + 0.2 + 0.05 + 0.0006 . . . . . . . . (2)
Comparing (1) and (2) we get
4A = 40 ⇒ A = 10
$7B = 7 ⇒ B = 1
2C = 0.2 ⇒ C = 0.1
$$\frac{5}{{\text{D}}}$$ = 0.05 ⇒ D = $$\frac{5}{{0.05}}$$ = 100
and, 6E = 0.0006 ⇒ E = 0.0001
So velue of 5A + 3B + 6C + D + 3E
= (5 × 10) + (3 × 1) + (6 × 0.1) + 100 + (3 × 0.0001)
= 50 + 3 + 0.6 + 100 + 0.0003
= 153.6003

$$\eqalign{ & x*y = {x^2} + {y^2} - xy \cr & {\text{ }}9*11 = {\left( 9 \right)^2} + {\left( {11} \right)^2} - 11 \times 9 \cr & \Rightarrow 9*11 = 81 + 121 - 99 \cr & \Rightarrow 9*11 = 103 \cr} $$

$$\eqalign{ & {3^{x + 3}} + 7 = 250 \cr & \Rightarrow {3^{x + 3}} = 250 - 7 \cr & \Rightarrow {3^{x + 3}} = 243 \cr & \Rightarrow {3^{x + 3}} = {3^5} \cr & \Rightarrow x + 3 = 5 \cr & \Rightarrow x = 2 \cr} $$

$$\frac{1}{4} \times $$ $$\frac{2}{6} \times $$ $$\frac{3}{8} \times $$ $$\frac{4}{{10}} \times $$ $$\frac{5}{{12}} \times $$ . . . . . $$ \times \frac{{31}}{{64}}$$ $$ = \frac{1}{{{2^x}}}$$
Denominator of first term is cut by the numerator of next terms, second term to the next one and so on.
$$\eqalign{ & \Rightarrow {\left( {\frac{1}{2}} \right)^{30}} \times {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{{2^x}}} \cr & \Rightarrow {\left( {\frac{1}{2}} \right)^{30 + 6}} = \frac{1}{{{2^x}}} \cr & \Rightarrow \frac{1}{{{2^{36}}}} = \frac{1}{{{2^x}}} \cr & \Rightarrow x = 36 \cr} $$

$$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \Rightarrow x = 2 + 1 + 2\sqrt 2 \cr & \Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 2 + 1 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 2 - 1 \cr & \therefore \sqrt x - \frac{1}{{\sqrt x }} \cr & = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right) \cr & = \sqrt 2 + 1 - \sqrt 2 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & \because p = 999 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {1000} \right)}^3}} \cr & = 1000 \cr} $$

$$\eqalign{ & {\text{ }}\frac{a}{b} = \frac{7}{9},{\text{ }}\frac{b}{c} = \frac{3}{5} \cr & \therefore a\,\,\,:\,\,\,b\,\,\,:\,\,\,c \cr & \,\,\,\,\,7\,\,\,:\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{3_{ \times 3}}:\,\,\,{5_{ \times 3}} \cr & \overline {{\text{ }}\,\,{\text{ }}7\,\,\,:\,\,\,\,9\,\,:\,\,15{\text{ }}} \cr} $$