4 Mark Question

Question 14 Marks

Solve the following equations :

$\frac{b+(b+1)+(b+2)}{4}=21$

Answer

$\frac{b+(b+1)+(b+2)}{4}=21$
$
\therefore \frac{b+(b+1)+(b+2)}{4} \times 4=21 \times 4
$
...[Multiplying both the sides by 4 ]
$
\begin{aligned}
& \therefore b+b+1+b+2=84 \\
& \therefore 3 b+3=84 \\
& \therefore 3 b+3-3=84-3
\end{aligned}
$
...[ Subtracting 3 from both the sides]
$\therefore 3 b=81$
$\therefore \backslash \backslash$ frac $\{3$ b\} $\{3\}=\backslash \operatorname{frac}\{81\}\{3\}[/$ latex ...[Dividing both the sides by 3 ] $\therefore b=27$

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Question 24 Marks

Solve the following equations :

$\frac{y-4}{3}+3 y=4$

Answer

$
\begin{aligned}
& \text {} \frac{y-4}{3}+3 y=4 \\
& \therefore \frac{y-4}{3} \times 3+3 y \times 3=4 \times 3
\end{aligned}
$
...[Multiplying both the sides by 3 ]
$
\begin{aligned}
& \therefore y-4+9 y=12 \\
& \therefore 10 y-4=12 \\
& \therefore 10 y-4+4=12+4
\end{aligned}
$
...[Adding 4 on both the sides]
$
\begin{aligned}
& \therefore 10 y=16 \\
& \therefore \frac{10 y}{10}=\frac{16}{10} \ldots[\text { Dividing both the sides by } 10] \\
& \therefore y=\frac{8}{5}
\end{aligned}
$

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Question 34 Marks

Solve the following equations :

$\frac{x-9}{x-5}=\frac{5}{7}$

Answer

$\frac{x-9}{x-5}=\frac{5}{7}$
$\therefore \frac{x-9}{x-5} \times 7(x-5)=\frac{5}{7} \times 7(x-5)$
...[Multiplying both the sides by $7(x-5)$ ]
$\therefore 7(x-9)=5(x-5)$
$\therefore 7 x-63=5 x-25$
$\therefore 7 x-63+63=5 x-25+63$
... [Adding 63 on both the sides]
$\therefore 7 x=5 x+38$
$\therefore 7 x-5 x=5 x+38-5 x$
...[Subtracting $5 x$ from both the sides]
$\therefore 2 x=38$
$\therefore \frac{2 x}{2}=\frac{38}{2} \ldots$ [Dividing both the sides by 2 ]
$\therefore x=19$

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Question 44 Marks

Solve the following equations :

$3(y+8)=10(y-4)+8$

Answer

$
\begin{aligned}
& \text {} 3(y+8)=10(y-4)+8 \\
& \therefore 3 y+24=10 y-40+8 \\
& \therefore 3 y+24=10 y-32 \\
& \therefore 3 y+24-24=10 y-32-24
\end{aligned}
$
...[Subtracting 24 from both the sides]
$
\begin{aligned}
& \therefore 3 y=10 y-56 \\
& \therefore 3 y-10 y=10 y-56
\end{aligned}
$
...[Subtracting 10y from both the sides]
$
\begin{aligned}
& \therefore-7 y=-56 \\
& \therefore \frac{-7 y}{-7}=\frac{-56}{-7} \ldots \text { [Dividing both the sides by }-7 \text { ] } \\
& \therefore y=8
\end{aligned}
$

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Question 54 Marks

Solve the following equations :

$13 x-5=\frac{3}{2}$

Answer

$13 x-5=\frac{3}{2}$
$
\therefore \quad(13 x-5) \times 2=\frac{3}{2} \times 2
$
...[Multiplying both the sides by 2 ]
$
\begin{array}{ll}
\therefore & 26 x-10=3 \\
\therefore & 26 x-10+10=3+10
\end{array}
$
... [Adding 10 on both the sides]
$
\begin{aligned}
& \therefore \quad 26 x=13 \\
& \therefore \quad \frac{26 x}{26}=\frac{13}{26} \\
& \therefore \quad x=\frac{1}{2}
\end{aligned}
$
...[Dividing both the sides by 26]

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Question 64 Marks

Solve the following equations :

$\frac{y}{7}+\frac{y-4}{3}=2$

Answer

$\begin{aligned} & \text {} \frac{y}{7}+\frac{y-4}{3}=2 \\ & \therefore \quad \frac{y \times 3}{7 \times 3}+\frac{(y-4) \times 7}{3 \times 7}=2 \\ & \therefore \quad \frac{3 y}{21}+\frac{7 y-28}{21}=2 \\ & \therefore \quad \frac{3 y+7 y-28}{21}=2 \\ & \therefore \quad \frac{10 y-28}{21}=2 \\ & \therefore \quad \frac{10 y-28}{21} \times 21=2 \times 21 \\ & \text {...[Multiplying both the sides by 21] } \\ & \therefore 10 y-28=42 \\ & \therefore \quad 10 y-28+28=42+28 \\ & \text {...[Adding } 28 \text { on both the sides] } \\ & \therefore \quad 10 y=70 \\ & \therefore \quad \frac{10 y}{10}=\frac{70}{10} \\ & \therefore \quad y=7 \\ & \end{aligned}$

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Question 74 Marks

Solve the following equations :

$\frac{9 x}{8}+1=10$

Answer

$
\begin{aligned}
& \text {} \frac{9 x}{8}+1=10 \\
& \therefore \quad \frac{9 x}{8}+1-1=10-1
\end{aligned}
$
...[Subtracting 1 from both the sides]
$
\begin{aligned}
& \therefore \quad \frac{9 x}{8}=9 \\
& \therefore \quad \frac{9 x}{8} \times 8=9 \times 8
\end{aligned}
$
...[Multiplying both the sides by 8 ]
$
\begin{array}{ll}
\therefore & 9 x=72 \\
\therefore & \frac{9 x}{9}=\frac{72}{9} \\
\therefore & x=8
\end{array}
$
...[Dividing both the sides by 9 ]

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Question 84 Marks

Solve the following equations :

$5(x-3)=3(x+2)$

Answer

$5(x-3)=3(x+2)$
$
\begin{aligned}
& \therefore 5 x-15=3 x+6 \\
& \therefore 5 x-15+15=3 x+6+15
\end{aligned}
$
...[Adding 15 on both the sides]
$
\begin{aligned}
& \therefore 5 x=3 x+21 \\
& \therefore 5 x-3 x=3 x+21-3 x
\end{aligned}
$
...[Subtracting $3 x$ from both the sides]
$
\therefore 2 x=21
$
$\therefore \frac{2 x}{2}=\frac{21}{2} \ldots$ [Dividing both the sides by 2 ]
$
\therefore x=\frac{21}{2}
$

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Question 94 Marks

Solve the following equations :

$3 x+12=2 x-4$

Answer

$3 x+12=2 x-4$
$
\therefore 3 x+12-12=2 x-4-12
$
...[Subtracting 12 from both the sides]
$
\begin{aligned}
& \therefore 3 x=2 x-16 \\
& \therefore 3 x-2 x=2 x-16-2 x
\end{aligned}
$
...[Subtracting $2 x$ from both the sides]
$
\therefore x=-16
$

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Question 104 Marks

Solve the following equations :

$2 m+7=9$

Answer

$2 m+7=9$
$\therefore 2 m+7-7=9-7$
...[Subtracting 7 from both the sides]
$\therefore 2 m=2$
$\therefore \frac{2 m}{2}=\frac{2}{2}$ [Dividing both the sides by 2 ]
$\therefore \mathrm{m}=1$

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Question 114 Marks

Solve the following equations :

$17 p-2=49$

Answer

$
\begin{aligned}
& \text {} 17 p-2=49 \\
& \therefore 17 p-2+2=49+2
\end{aligned}
$
...[Adding 2 on both the sides]
$
\begin{aligned}
& \therefore 17 p=51 \\
& \left.\therefore \frac{17 p}{17}=\frac{51}{17} \ldots \text {..[Dividing both the sides by } 17\right] \\
& p=3
\end{aligned}
$

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Question 124 Marks

Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
i. x – 4 = 3, x = – 1, 7, – 7
ii. 9m = 81, m = 3, 9, -3
iii. 2a + 4 = 0, a = 2, – 2, 1
iv. 3 – y = 4, y = – 1, 1, 2

Answer

i. x – 4 = 3 ….(i)
Substituting x = – 1 in L.H.S. of equation (i),
L.H.S. = (-1) – 4
= – 5
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),
L.H.S. = (7) – 4
= 3
R.H.S. = 3
∴ L.H.S. = R.H.S.
∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i),
L.H.S. = (- 7) – 4
= -11
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 7 is not the solution of the given equation.

ii. 9m = 81 …(i)
Substituting m = 3 in L.H.S. of equation (i),
L.H.S. = 9 × (3)
= 27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),
L.H.S. = 9 × (9)
= 81
R.H.S. = 81
∴L.H.S. = R.H.S.
∴m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i),
L.H.S. = 9 × (- 3)
= -27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = – 3 is not the solution of the given equation.

iii. 2a + 4 = 0 …..(i)
Substituting a = 2 in L.H.S. of equation (i),
L.H.S. = 2 (2) + 4
= 4 + 4
= 8
R.H.S. = 0
∴L.H.S. ≠ R.H.S.
∴a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),
L.H.S. = 2 (-2)+ 4
= -4 + 4
= 0
R.H.S. = 0
∴L.H.S. = R.H.S.
∴a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),
L.H.S. = 2(1)+ 4
= 2 + 4
= 6
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴a = 1 is not the solution of the given equation.

iv. 3 – y = 4 …(i)
Substituting y = -1 in L.H.S. of equation (i),
L.H.S. = 3 – (- 1)
= 3 + 1
= 4
R.H.S. = 4
∴L.H.S. = R.H.S.
∴y = – 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),
L.H.S. = 3-(1)
= 2
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),
L.H.S. = 3-(2)
= 1
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 2 is not the solution of the given equation.

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Maths 4 Mark Question

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