2 Marks Questions

Question 12 Marks

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation,
$\text{C}=\frac{5\text{F}-160}{9}$
What is the numerical value of the temperature which is same in both the scales?

Answer

$\text{C}=\frac{5\text{F}-160}{9}$ putting C = F, in the given relation, we get $\text{F}=\frac{5\text{F}-160}{9}\Rightarrow9\text{F}=5\text{F}-160$ $\Rightarrow4\text{F}=160$ $\therefore\text{F}=\frac{-160}{4}=-40^\circ$Hence, the numarical value of the temperature which is same in both the scales is -40.

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Question 22 Marks

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation,
$\text{c}=\frac{5\text{F}-160}{9}$
If the temperature is 86°F, what is the temperature in Celsius?

Answer

$\text{c}=\frac{5\text{F}-160}{9}$
putting F = 86°, we get $\text{C}=\frac{5(86)-160}{9}=\frac{430-160}{9}=\frac{270}{9}=30^\circ$
Hence, the temperature in Celsius is 30°C.

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Question 32 Marks

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation,
$\text{c}=\frac{5\text{F}-160}{9}$
If the temperature is 35°C, what is the temperature in Fahrenheit?

Answer

$\text{c}=\frac{5\text{F}-160}{9}$
putting C = 35°, we get $35^\circ=\frac{5\text{(F)}-160}{9}\Rightarrow315^\circ=5\text{F}-160$
$\Rightarrow5\text{F}=315+160=475$
$\therefore\text{F}=\frac{475}{5}=95^\circ$
Hence, the temperature in Fahrenheit is 95°F.

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Question 42 Marks

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation,
$\text{c}=\frac{5\text{F}-160}{9}$
If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

Answer

$\text{c}=\frac{5\text{F}-160}{9}$
putting C = 0°, we get
$0=\frac{5\text{F}-160}{9}\Rightarrow0=5\text{F}-160$
$\Rightarrow5\text{F}=160$
$\therefore\text{F}=\frac{160}{5}=32^\circ$
Now, putting F = 0°, we get
$\text{C}=\frac{5\text{F}-160}{9}\Rightarrow\text{C}=\frac{5(0)-160}{9}=\Big(-\frac{160}{9}\Big)^\circ$
If the temperature in Fahrenheit is 32° and if the temperature is 0°F, then the temperature in Celsius is $\Big(-\frac{160}{9}\Big)^\circ\text{C.}$

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Question 52 Marks

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?

Answer

y varies of directly as x.
$\Rightarrow\text{y}\propto\text{x},$
$\therefore\text{y}=\text{kx}$
Substituting y = 12 when x = 4, we get,
$12=\text{k}\times4\Rightarrow\text{k}=12\div4=3$
Hence, the required eqution is y = 3x.
The value of y when x = 5 is y = 3 × 5 = 15.

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Question 62 Marks

Find the solution of the linear equation x+2y = 8 which represents a point on:

x-axis
y-axis

Answer

We have, x + 2y = 8 ...(i)

When the point is on the X-axis, then put y = 0 in Eq. (i), we get

x + 2(0) = 8
⇒ x = 8 Hence, the required point is (8, 0).

When the point is on the Y-axis, then put x = 0 in Eq. (i), we get

0 + 2y = 8
$⇒\text{y}=\frac{8}{2}= 4$
Hence, the required point is (0, 4).

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Question 72 Marks

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution.

Answer

The value of c for which the liner eqution 2x + cy = 8 has equal values of x and y i.e., x = y for solution is,
$2\text{x}+\text{cy}=8\Rightarrow2\text{x}+\text{cx}=8\ [\because\text{y}=\text{x}]$
$\Rightarrow\text{cx}=8-2\text{x}$
$\therefore\text{c}=\frac{8-2\text{x}}{\text{x}},\text{x}\neq0$

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Question 82 Marks

Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and K-axes?

Answer

The given eqution is 3x + 4y = 6.
To draw the graph of this eqution. we need atleast two points lying on the graph of $4\text{y}=6-3\text{x}$
$\Rightarrow\text{y}=\frac{6-3\text{x}}{4}$
When x = 2, then $\text{y}=\frac{6-3\times2}{4}=\frac{6-6}{4}=0$
When x = 0, then $\text{y}=\frac{6-3\times0}{4}=\frac{6}{4}=\frac{3}{2}$

x	2	0
y	0	$\frac{3}{2}$

Here, we find two points $\text{A}\Big(0,\frac{3}{2}\Big)$ and B(2, 0) and join then, to get the line AB.
Line AB is the required graph.

You can see that the graph (line AB) cuts the X-axis at the point (2, 0) and the Y-axis at the point $\Big(0,\frac{3}{2}\Big).$

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