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Maths 2 Marks Questions

Linear Equations in One Variable · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 12 questions.

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2 Marks Questions

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12 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Verify by substitution that:
$x = 4$ is the root of $3x - 5 = 7$
Answer
$x = 4$ is the root of $3x - 5 = 7$
Now, substituting $x = 4$ in place of $‘x’$ in the given equation:
$3x - 5 = 7$
$3(4) - 5 = 7$
$12 - 5 = 7$
$7 = 7$
Since, $L.H.S. = R.H.S.$
Hence, $x = 4$ is the root of $3x - 5 = 7$
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Question 22 Marks
Verify by substitution that: $x = 8$ is the root of $\frac12\text{x} + 7 = 11$
Answer
$x = 8$ is the root of $12x + 7 = 11$
Now, substituting $x = 8$ in place of $‘x’$ in the given equation:
$12x + 7 = 11, 12(8) + 7 =11 4 + 7 = 11 11 = 11$
Since, $L.H.S. = R.H.S.$
Hence, $x = 8$ is the root of $12x + 7 = 11$
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Question 32 Marks
Verify by substitution that:
$x = 3$ is the root of $5 + 3x = 14$
Answer
$x = 3$ is the root of $5 + 3x = 14$
Now, substituting $x = 3$ in place of $‘x’$ in the given equation:
$5 + 3x = 14,$
$5 + 3(3) = 14$
$5 + 9 = 14$
$14 = 14$
Since, $L.H.S. = R.H.S.$
Hence, $x = 3$ is the root of $5 + 3x = 14$
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Question 42 Marks
Verify by substitution that: $x = 4$ is the root of $\frac{3\text{x}}{2}=6$
Answer
$x = 4$ is the root of $\frac{\text{3x}}{2}=6$
 Now, substituting $x = 4$ in place of $‘x’$ in the given equation:
$\frac{\text{3x}{}}{2}=6,$ $\frac{3\times4}{2}=6$ $\frac{12}{2}=6$ $6=6$
Since, $L.H.S. = R.H.S.$
Hence, $x = 4$ is the root of $\frac{3\text{x}}{2}=6$
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Question 52 Marks
Verify by substitution that: $y = 2$ is the root of $y - 3 = 2y - 5$
Answer
$y = 2$ is the root of $y - 3 = 2y - 5.$
Now, substituting $y = 2$ in place of $‘y’$ in the given equation:
$y - 3 = 2y - 5, 2 – 3 = 2(2) - 5 -1 = 4 - 5 -1 = -1$
Since, $L.H.S. = R.H.S.$
Hence, $y = 2$ is the root of $y - 3 = 2y - 5.$
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Question 62 Marks
Verify by substitution that: $x = 2$ is the root of $3x - 2 = 8x - 12$
Answer
$x = 2$ is the root of $3x - 2 = 8x - 12$
Now, substituting $x = 2$ in place of $‘x’$ in the given equation:
$3x - 2 = 8x - 12, 3(2) - 2 $
$= 8(2) - 12 6 - 2 $
$= 16 - 12 4 = 4$
Since, $L.H.S. = R.H.S.$
 Hence, $x = 2$ is the root of $3x - 2 = 8x - 12$
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