Algebra

1.If a * b = 2a - 3b + ab, then 3 * 5 + 5 * 3 is equal to?

1. 22
2. 24
3. 26
4. 28

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 3 * 5 + 5 * 3 \\ \Rightarrow 3 * 5 = 2 \times 3 - 3 \times 5 + 3 \times 5 \\ \Rightarrow 3 * 5 = 6 - 15 + 15 \\ \Rightarrow 3 * 5 = 6 \\ \Rightarrow 5 * 3 = 2 \times 5 - 3 \times 3 + 3 \times 5 \\ \Rightarrow 5 * 3 = 10 - 9 + 15 \\ \Rightarrow 5 * 3 = 16 \\ ∴ 3 * 5 + 5 * 3 \\ \Rightarrow 6 + 16 \\ \Rightarrow 22 \end{aligned}

$\eqalign{ & 3*5 + 5*3 \cr & \Rightarrow 3*5 = 2 \times 3 - 3 \times 5 + 3 \times 5 \cr & \Rightarrow 3*5 = 6 - 15 + 15 \cr & \Rightarrow 3*5 = 6 \cr & \Rightarrow 5*3 = 2 \times 5 - 3 \times 3 + 3 \times 5 \cr & \Rightarrow 5*3 = 10 - 9 + 15 \cr & \Rightarrow 5*3 = 16 \cr & \therefore 3*5 + 5*3 \cr & \Rightarrow 6 + 16 \cr & \Rightarrow 22 \cr}$

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2.If

p \times q = p + q + \frac{p}{q},

$p \times q = p + q + \frac{p}{q}{\text{,}}$ then the value of 8 × 2 is?

1. 6
2. 10
3. 14
4. 16

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 8 \times 2 \\ = 8 + 2 + \frac{8}{2} \\ = 10 + 4 \\ = 14 \end{aligned}

$\eqalign{ & {\text{8}} \times {\text{2}} \cr & = 8 + 2 + \frac{8}{2} \cr & = 10 + 4 \cr & = 14 \cr}$

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3.The value of

(1 + \frac{1}{x}) (1 + \frac{1}{x + 1})

$\left( {{\text{1 + }}\frac{1}{x}} \right)\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$

(1 + \frac{1}{x + 2}) (1 + \frac{1}{x + 3})

$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$ is?

1. $1 + \frac{1}{x + 4}$ $1 + \frac{1}{{x + 4}}$
2. x + 4
3. $\frac{1}{x}$ $\frac{1}{x}$
4. $\frac{x + 4}{x}$ $\frac{{x + 4}}{x}$

Answer:Option 1

Explanation:

Solution:

\begin{aligned} (1 + \frac{1}{x}) (1 + \frac{1}{x + 1}) (1 + \frac{1}{x + 2}) (1 + \frac{1}{x + 3}) \\ Taking L .C .M of each term . \\ \Rightarrow (\frac{x + 1}{x}) (\frac{x + 1 + 1}{x + 1}) (\frac{x + 2 + 1}{x + 2}) (\frac{x + 3 + 1}{x + 3}) \\ \Rightarrow \frac{1}{x} \times (x + 4) \\ \Rightarrow \frac{x + 4}{x} \end{aligned}

$\eqalign{ & \left( {{\text{1 + }}\frac{1}{x}} \right)\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right) \cr & {\text{Taking L}}{\text{.C}}{\text{.M of each term}}{\text{.}} \cr & \Rightarrow \left( {\frac{{x + 1}}{x}} \right)\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right) \cr & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr}$

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4.If

\frac{2 a + b}{a + 4 b} = 3,

$\frac{{2a + b}}{{a + 4b}} = 3{\text{,}}$ then find the value of

\frac{a + b}{a + 2 b} = ?

$\frac{{a + b}}{{a + 2b}} = ?$

1. $\frac{5}{9}$ $\frac{5}{9}$
2. $\frac{2}{7}$ $\frac{2}{7}$
3. $\frac{10}{9}$ $\frac{{10}}{9}$
4. $\frac{10}{7}$ $\frac{{10}}{7}$

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{2 a + b}{a + 4 b} = 3 (Given) \\ \Rightarrow 2 a + b = 3 (a + 4 b) \\ \Rightarrow 2 a + b = 3 a + 12 b \\ \Rightarrow - a = 11 b \\ \Rightarrow a = - 11 b \\ ∴ \frac{a + b}{a + 2 b} \\ \Rightarrow \frac{- 11 b + b}{- 11 b + 2 b} \\ \Rightarrow \frac{- 10 b}{- 9 b} \\ \Rightarrow \frac{10}{9} \end{aligned}

$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr}$

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5.If a * b = a + b + ab, then 3 * 4 - 2 * 3 is equal to?

1. 6
2. 8
3. 10
4. 12

Answer:Option 1

Explanation:

Solution:

\begin{aligned} a * b = a + b + a b \\ 3 * 4 \\ = 3 + 4 + 3 \times 4 \\ = 19 \\ 2 * 3 \\ = 2 + 3 + 2 \times 3 \\ = 11 \\ ∴ 3 * 4 - 2 * 3 \\ = 19 - 11 \\ = 8 \end{aligned}

$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr}$

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6.If a : b = 2 : 3 and b : c = 4 : 5, find a² : b² : bc = ?

1. 4 : 9 : 45
2. 16 : 36 : 45
3. 16 : 36 : 20
4. 4 : 36 : 40

Answer:Option 1

Explanation:

Solution:

\begin{aligned} a : b = 2 : 3 and b : c = 4 : 5 \\ a b c \\ 2_{\times (4)} 3_{\times (4)} \\ 4_{\times (3)} 5_{\times (3)} \\ \bar{\underline{8 12 15}} \\ ∴ a^{2} : b^{2} : b c \\ \Rightarrow {(8)}^{2} : {(12)}^{2} : (12 \times 15) \\ \Rightarrow 64 : 144 : 180 \end{aligned}

$\eqalign{ & a:b = 2:3{\text{ and }}b:c = 4:5 \cr & a{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}\,\,\,\,\,\,\,\,{\text{ }}{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {\,\,8{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}15{\text{ }}\,\,{\text{ }}} } \cr & \therefore {a^2}:{b^2}:bc \cr & \Rightarrow {\left( 8 \right)^2}:{\left( {12} \right)^2}:\left( {12 \times 15} \right) \cr & \Rightarrow 64:144:180 \cr}$

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7.If

A : B = \frac{1}{2} : \frac{3}{8},

${\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8}{\text{,}}$

B : C = \frac{1}{3} : \frac{5}{9}

${\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9}$ and

C : D = \frac{5}{6} : \frac{3}{4},

${\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4}{\text{,}}$ then find the ratio of A : B : C : D = ?

1. 6 : 4 : 8 : 10
2. 6 : 8 : 9 : 10
3. 8 : 6 : 10 : 9
4. 4 : 6 : 8 : 10

Answer:Option 1

Explanation:

Solution:

\begin{aligned} A : B = \frac{1}{2} : \frac{3}{8} \\ \Rightarrow A : B = 8 : 6 \\ \Rightarrow A : B = 4 : 3 \\ B : C = \frac{1}{3} : \frac{5}{9} \\ \Rightarrow B : C = 9 : 15 \\ \Rightarrow B : C = 3 : 5 \\ C : D = \frac{5}{6} : \frac{3}{4} \\ \Rightarrow C : D = 20 : 18 \\ \Rightarrow C : D = 10 : 9 \\ A : B : C : D \\ 4 : 3 \\ 3 : 5 \\ 10 : 9 \\ \bar{\underline{8 : 6 : 10 : 9}} \end{aligned}

$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} \cr & \Rightarrow {\text{A}}:{\text{B}} = 8:6 \cr & \Rightarrow {\text{A}}:{\text{B}} = 4:3 \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} \cr & \Rightarrow {\text{B}}:{\text{C}} = 9:15 \cr & \Rightarrow {\text{B}}:{\text{C}} = 3:5 \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} \cr & \Rightarrow {\text{C}}:{\text{D}} = 20:18 \cr & \Rightarrow {\text{C}}:{\text{D}} = 10:9 \cr & {\text{A}}:{\text{ B }}:{\text{C }}:{\text{D}} \cr & 4{\text{ }}:{\text{ }}3 \cr & \,\,\,\,\,\,\,\,\,\,3{\text{ }}:{\text{ }}5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{\text{ }}:{\text{ }}9 \cr & \overline {\underline {{\text{ }}8{\text{ }}:6{\text{ }}:10{\text{ }}:{\text{ }}9{\text{ }}} } \cr}$

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8.If

\frac{144}{0.144} = \frac{14.4}{x},

$\frac{{144}}{{0.144}} = \frac{{14.4}}{x}{\text{,}}$ then the value of x is?

1. 144
2. 14.4
3. 1.44
4. 0.0144

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{144}{0.144} = \frac{14.4}{x} \\ \Rightarrow \frac{144 \times 1000}{144} = \frac{144}{x \times 10} \\ \Rightarrow 1000 = \frac{144}{10 x} \\ \Rightarrow x = \frac{144}{1000 \times 10} \\ \Rightarrow x = \frac{144}{10000} \\ \Rightarrow x = 0.0144 \end{aligned}

$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr}$

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9.If 1 < x < 2, then the value of

\sqrt{{(x - 1)}^{2}} + \sqrt{{(x - 3)}^{2}} is?

$\sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {x - 3} \right)}^2}} {\text{ is?}}$

1. 1
2. 2
3. 3
4. 2x - 4

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 1 < x < 2 \\ \sqrt{{(x - 1)}^{2}} + \sqrt{{(x - 3)}^{2}} \\ (Square root cancel with square) \\ ∴ x - 1 + 3 - x \\ = 2 \end{aligned}

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

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10.If a ⊕ b = (a × b) + b, then 5 ⊕ 7 equals to?

1. 12
2. 35
3. 42
4. 50

Answer:Option 1

Explanation:

Solution:

\begin{aligned} a \oplus b = (a \times b) + b \\ \Rightarrow 5 \oplus 7 \\ = (5 \times 7) + 7 \\ = 35 + 7 \\ = 42 \end{aligned}

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

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