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Maths 4 Mark Question

Linear Equations in One Variable · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 21 questions.

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Question 14 Marks
$\text{x}+\frac12=\frac72$
Answer
$\text{x}+\frac12=\frac72$
Subtracting $\frac12$ from both sides, we get
$\text{x}+\frac12-\frac12=\frac72-\frac12$
$\text{x}=\frac72-\frac12=\frac62$
$\text{x}=3$
Verification:
Substituting x = 3 is L.H.S., we get L.H.S. $=3+\frac{6+1}{2}=72$ and R.H.S. = 72
L.H.S. = R.H.S.
Hence, verified.
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Question 24 Marks
6x + 5 = 2x + 17
Answer
We have
6x + 5 = 2x + 17
Transposing 2x to L.H.S. and 5 to R.H.S., we get
6x - 2x = 17 - 5
4x = 12
Dividing both sides by 4, we get
$\frac{4\text{x}}{4}=\frac{12}{4}$
$\text{x}=3$
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
L.H.S. = R.H.S.
Hence, verified.
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Question 34 Marks
$\text{m}-\frac{\text{m}-1}{2}=1-\frac{\text{m}-2}{3}$
Answer
$\text{m}-\frac{\text{m}-1}{2}=1-\frac{\text{m}-2}{3}$
$=\frac{2\text{m}-\text{m}+1}{2}=\frac{3-\text{m}+2}{3}$
$=\frac{\text{m}+1}{2}=\frac{5-\text{m}}{3}$
$=\frac{\text{m}+1}{2}=\frac{5}{3}-\frac{\text{m}}{3}$
$=\frac{\text{m}}{2}+\frac12=\frac53-\frac{\text{m}}{3}$
Transposing $\frac{\text{m}}{3}$ to L.H.S. and $\frac12$ to R.H.S., we get
$=\frac{\text{m}}{2}+-\frac{\text{m}}{3}=\frac{5}{3}-\frac12$
$=\frac{3\text{m}+2\text{m}}{6}=\frac{10-3}{6}$
Multiplying both sides by 6, we get
$=\frac{5\text{m}}{6}\times6=\frac{7}{6}\times6$
$=5\text{m}=7$
Dividing both sides by 5, we get
$=\frac{5\text{m}}{5}=\frac75$
$=\text{m}=\frac75$
Verification:
Substituting $\text{m}=\frac75$ on both sides, we get
$\frac75-\frac{7-5}{10}=1-\frac{7-10}{15}$
$\frac75-\frac{2}{10}=\frac{15+3}{15}$
$\frac{14-2}{10}=\frac{15+3}{15}$
$\frac{12}{10}=\frac{18}{15}$
$\frac{6}{5}=\frac{6}{5}$
$\text{L.H.S.} =\text{R.H.S.}$
Hence, verified.
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Question 44 Marks
5(x - 2) +3(x + 1) = 25
Answer
5(x - 2) + 3(x + 1) = 25
On expanding the brackets, we get
(5 × x) - (5 × 2) + 3 × x + 3 × 1 = 25
5x - 10 + 3x + 3 = 25
5x + 3x - 10 + 3 = 25
8x - 7 = 25
Adding 7 to both sides, we get
8x - 7 + 7 = 25 + 7
8x = 32
Dividing both sides by 8, we get
$\frac{8\text{x}}{8}=\frac{32}{8}$
$\text{x}=4$
Verification:
Substituting x = 4 in L.H.S., we get
= 5(4 - 2) + 3(4 + 1) = 5(2) + 3(5) = 10 + 15 = 25 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.
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Question 54 Marks
$\frac45-\text{x}=\frac35$
Answer
$\frac45-\text{x}=\frac35$
Subtracting $\frac45$ from both sides, we get
$\frac{4}{5}-\text{x}-\frac45=\frac35-\frac45$
$-\text{x}=\frac35-\frac45$
$-\text{x}=-\frac15$
Dividing both sides by -1, we get
$-\text{x}\times(-1)=-\frac15\times(-1)$
$\text{x}=\frac15$
Verification:
Substituting $\text{x}=\frac15$ in L.H.S., we get
$\text{L.H.S.}=\frac{4}{5}-\frac15=\frac{4-1}{5}=\frac35,$ and $\text{R.H.S.} = \frac35$
L.H.S. = R.H.S.
Hence, verified.
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Question 64 Marks
$\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
Answer
$\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
Transposing $\frac{\text{x}}{2}$ to L.H.S., we get
$\frac{\text{x}}{2}-\frac{\text{x}}{3}=1$
$\frac{\text{3x}-2\text{x}}{6}=1$
$\frac{\text{x}}{6}=1$
Multiplying both sides by 6, we get
$\frac{\text{x}}{6}\times6=1\times6$
$\text{x}=6$
Verification:
Substituting x = 6 in the given equation, we get
$\frac{6}{6}=\frac63+1$
$3=2+1$
$3=3$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.
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Question 74 Marks
3(x - 3) = 5(2x + 1)
Answer
3(x - 3) = 5(2x + 1)
On expanding the brackets on both sides, we get
= 3 × x - 3 × 3
= 5 × 2x + 5 × 1
= 3x - 9 = 10x + 5
Transposing 10x to L.H.S. and 9 to R.H.S., we get
= 3x - 10x = 9 + 5
= -7x = 14
Dividing both sides by 7, we get
$=\frac{-7\text{x}}{7}$
$=\frac{-14}{7}$
$=\text{x}=-2$
Verification:
Substituting x = -2 on both sides, we get
3(-2 - 3) = 5{2(-2) +1}
3(-5) = 5(-3)
-15 = -15
L.H.S. = R.H.S.
Hence, verified.
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Question 84 Marks
$\text{x}-\frac{\text{x}}{4}-\frac{1}{2}=3+\frac{\text{x}}{4}$
Answer
$\text{x}-\frac{\text{x}}{4}-\frac{1}{2}=3+\frac{\text{x}}{4}$
Transposing $\frac{\text{x}}{4}$ to L.H.S. and $-\frac12$ to R.H.S., we get
$=\text{x}-\frac{\text{x}}{4}-\frac{\text{x}}{4}=3+\frac12$
$=\frac{4\text{x}-\text{x}-\text{x}}{4}=\frac{6+1}{2}$
$=\frac{2\text{x}}{4}=\frac72$
Multiplying both sides by 4, we get
$=\frac{\text{2x}}{4}\times4=\frac72\times4$
$=2\text{x}=14$
Dividing both sides by 2, we get
$=\frac{\text{2x}}{2}=\frac{14}2{}$
$=\text{x}=7$
Verification:
Substituting x = 7 on both sides, we get
$7-\frac{7}{4}-\frac12=3+\frac74$
$\frac{2-7-2}{4}=\frac{12+7}{4}$
$=\frac{19}{4}=\frac{19}{4}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.
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Question 94 Marks
7 + 4y = - 5
Answer
7 + 4y = -5
Subtracting 7 from both sides, we get
7 + 4y - 7 = -5 -7
4y = -12
Dividing both sides by 4, we get
$\text{y}=\frac{-12}{4}$
$\text{y}=-3$
Verification:
Substituting y = -3 in L.H.S., we get
L.H.S. = 7 + 4y = 7 + 4(-3) = 7 - 12 = -5, and R.H.S. = -5
L.H.S. = R.H.S.
Hence, verified.
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Question 104 Marks
$\frac{3}{4}(\text{x}-1)=\text{x}-3$
Answer
$\frac{3}{4}(\text{x}-1)=\text{x}-3$
On expanding the brackets on both sides, we get
$=\frac{\text{3x}}{4}-\frac34=\text{x}-3$
Transposing $\frac34\text{x}$ to R.H.S. and 3 to L.H.S., we get
$\Rightarrow3-\frac34=\text{x}-\frac34\text{x}$
$\Rightarrow\frac{12-3}{4}=\frac{4\text{x}-3\text{x}}{4}$
$\Rightarrow\frac{9}{4}-=\frac{\text{x}}{4}$
Multiplying both sides by 4, we get
$\Rightarrow\text{x}=9$
Verification:
Substituting x = 9 on both sides, we get
$\frac{3}{4}(9-1)=9-3$
$\frac34\times8=6$
L.H.S. = R.H.S.
Hence, verified.
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Question 114 Marks
3(x + 2) - 2(x - 1) = 7
Answer
3(x + 2) - 2(x - 1) = 7
On expanding the brackets, we get
3x + 3 × 2 - 2 × x + 2 × 1 = 7
3x + 6 - 2x + 2 = 7
3x - 2x + 6 + 2 = 7
x + 8 = 7
Subtracting 8 from both sides, we get
x + 8 - 8 = 7 - 8
x = -1
Verification:
Substituting x = -1 in L.H.S., we get
L.H.S. = 3(x + 2) - 2(x - 1) = 3(-1 + 2) - 2(-1 -1) = (3 × 1) - (2 × -2) = 3 + 4 = 7, and R.H.S. = 7
L.H.S. = R.H.S.
Hence, verified.
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Question 124 Marks
$\frac{\text{5x}-1}{3}-\frac{2\text{x}-2}{3}=1$
Answer
$\frac{\text{5x}-1}{3}-\frac{2\text{x}-2}{3}=1$
$\frac{5\text{x}-1-2\text{x}+2}{3}=1$
$\frac{3\text{x}+1}{3}=1$
Multiplying both sides by 3, we get 3
$\frac{3\text{x}+1}{3}\times3=1\times3$
$=\text{3x}+1=3$
Subtracting 1 from both sides, we get
= 3x + 1 - 1 = 3 - 1
= 3x = 2
Dividing both sides by 3, we get
= 3x + 1 - 1 = 3 -1
= 3x = 2
Dividing both sides by 3, we get
$=\frac{\text{3x}}{3}=\frac{2}{3}$
$=\text{x}=\frac23$
Verification:
Substituting $\text{x}=\frac23$ in L.H.S., we get
$=\frac{5\big(\frac23\big)-1}{3}-\frac{2\big(\frac23\big)-2}{3}$
$=\frac{\frac{1}3-1}{3}-\frac{\frac43-3}{3}$
$=\frac{\frac{10-3}{3}}{3}-\frac{\frac{4-6}{3}}{3}$
$=\frac{7}{3\times3}-\Big(\frac{-2}{3\times3}\Big)$
$=\frac79+\frac29$
$=\frac99=1=\text{R.H.S.}$
L.H.S. = R.H.S.
Hence, verified.
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Question 134 Marks
10 - y = 6
Answer
10 - y = 6
Subtracting 10 from both sides, we get
10 - y - 10 = 6 - 10
-y = -4
Multiplying both sides by -1, we get
-y × -1 = - 4 × - 1
y = 4
Verification:
Substituting y = 4 in L.H.S., we get
L.H.S. = 10 - y = 10 - 4 = 6 and R.H.S. = 6
L.H.S. = R.H.S.
Hence, verified.
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Question 144 Marks
$0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
Answer
$0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
$\frac{5}{10}\text{x}+\frac{\text{x}}{3}=\frac{25\text{x}}{100}+7$
$\frac{\text{x}}{2}+\frac{\text{x}}{3}=\frac{\text{x}}{4}+7$
Transposing $\frac{\text{x}}{4}$ to L.H.S., we get
$\frac{\text{x}}{2}+\frac{\text{x}}{3}-\frac{\text{x}}{4}=7$
$\frac{65\text{x}+4\text{x}-\text{3x}}{12}=7$
Multiplying both sides by 12, we get
$\frac{7\text{x}}{12}\times12=7\times12$
$=\text{7x}=84$
Dividing both sides by 7, we get
$=\frac{7\text{x}}{7}=\frac{84}{7}$
$=\text{x}=12$
Verification:
Substituting x = 12 on both sides, we get
$0.5(12) + \frac{12}3 = 0.25(12) + 7$
6 + 4 = 3 + 7
10 = 10
L.H.S. = R.H.S.
Hence, verified.
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Question 154 Marks
3x - 2(2x - 5) = 2(x + 3) - 8
Answer
3x - 2(2x - 5) = 2(x + 3) - 8
On expanding the brackets on both sides, we get
= 3x - 2 × 2x + 2 × 5 = 2 × x + 2 × 3 - 8
= 3x - 4x + 10 = 2x + 6 - 8
= -x + 10 = 2x - 2
Transposing x to R.H.S. and 2 to L.H.S., we get
= 10 + 2 = 2x + x
= 3x = 12
Dividing both sides by 3, we get
$=\frac{3\text{x}}{3}=\frac{12}{3}$
$=\text{x}=4$
Verification:
Substituting x = 4 on both sides, we get
3(4) - 2{2(4) - 5} = 2(4 + 3) - 8
12 - 2(8 - 5) = 14 - 8
12 - 6 = 6
6 = 6
L.H.S. = R.H.S.
Hence, verified.
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Question 164 Marks
$\frac{6\text{x}-2}{9}+\frac{3\text{x}+5}{18}=\frac13$
Answer
$\frac{6\text{x}-2}{9}+\frac{3\text{x+5}}{18}=\frac13$
$=\frac{6\text{x}(2)-2(2)+\text{3x}+5}{18}=\frac13$
$=\frac{12\text{x}-4+3\text{x}+5}{18}=\frac13$
$=\frac{15\text{x}+1}{18}=\frac13$
Multiplying both sides by 18, we get
$=\frac{15\text{x}+1}{18}\times18=\frac12\times18$
$=15\text{x}+1=6$
Transposing 1 to R.H.S., we get
= 15x = 6 - 1
= 15x = 5
Dividing both sides by 15, we get
$=\frac{15\text{x}}{15}=\frac{5}{15}$
$=\text{x}=\frac13$
Verification:
Substituting $\text{x}=\frac13$ both sides, we get
$\frac{6\big(\frac13\big)-2}{9}+\frac{3\big(\frac13\big)+5}{18}=\frac13$
$\frac{2-2}{9}+\frac{1+5}{18}=\frac13$
$0+\frac{6}{18}=\frac13$
$\frac13=\frac13$
L.H.S. = R.H.S.
Hence, verified.
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Question 174 Marks
$2\text{y}-\frac12=-\frac13$
Answer
$2\text{y}-\frac12=-\frac{1}{3}$
Adding $\frac12$ to both sides, we get
$\text{2y}-\frac{1}{2}+\frac12=-\frac13+\frac12$
$\text{y}=\frac{-2+3}{6}$
$\text{2y}=\frac{1}{6}$
Dividing both sides by 2, we get
$\frac{2\text{y}}{2}=\frac{16}{2}$
$\text{y}=\frac{1}{12}$
Verification:
Substituting $\text{y}=\frac{1}{12}$ in L.H.S. we get
$\text{L.H.S.}=2\frac{1}{12}-\frac12$
$=\frac16-\frac12$
$=\frac{1-3}{6}$
$=\frac{-2}{6}$
$=-\frac{1}{3},$ and R.H.S. $=-\frac13$
L.H.S. = R.H.S.
Hence, verified.
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Question 184 Marks
$0.6\text{x}+\frac45=0.28\text{x}+1.16$
Answer
$0.6\text{x} + \frac45 = 0.28\text{x} + 1.16$
Transposing 0.28x to L.H.S. and 45 to R.H.S., we get
= 0.6x - 0.28x = 1.16 - 45
= 0.32x = 1.16 - 0.8
= 0.32x = 0.36
Dividing both sides by 0.32, we get
= 0.32 × 0.32 = 0.360.32
= x = 98
Verification:
Substituting x = 98 on both sides, we get
$0.6\Big(\frac98\Big) + 45 = 0.28\Big(\frac98\Big) + 1.16$
$\frac{5.4}{8} + \frac45 = \frac{2.52}8 + 1.16$
$0.675 + 0.8 = 0.315 + 1.16$
$1.475 = 1.475$
L.H.S. = R.H.S.
Hence, verified.
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Question 194 Marks
2(5x - 3) - 3(2x - 1) = 9
Answer
We have
2(5x - 3) - 3(2x - 1) = 9
Expanding the brackets, we get
2 × 5x - 2 × 3 - 3 × 2x + 3 × 1 = 9
10x - 6 - 6x + 3 = 9
10x - 6x - 6 + 3 = 9
4x - 3 = 9
Adding 3 to both sides, we get
4x - 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
$\frac{4\text{x}}{4}=\frac{12}{4}$
Thus, x = 3.
Verification:
Substituting x = 3 in L.H.S., we get
= 2(5 × 3 - 3) - 3(2 × 3 - 1)
= 2 × 12 - 3 × 5
= 24 - 15
= 9
L.H.S. = R.H.S.
Hence, verified.
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Question 204 Marks
$\text{x}-\frac{1}{3}=\frac{2}{3}$
Answer
$\text{x}-\frac13=\frac23$
Adding $\frac13$ to both sides, we get
$\text{x}-\frac13+\frac13=\frac23+\frac13$
$\Rightarrow\text{x}=\frac23+\frac13$
$\Rightarrow\text{x}=\frac33$
$\Rightarrow\text{x}=1$
Verification:
Substituting x = 1 in L.H.S., we get
$\text{L.H.S.}=1-\frac{1}{3}=\frac{3-1}{2}=\frac{2}{3},$ and $\text{R.H.S.}=\frac23$
$\text{L.H.S.}= \text{R.H.S.}$
Hence, verified.
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Question 214 Marks
$\frac{\text{x}}{2}+\frac32=\frac{2\text{x}}{5}-1$
Answer
$\frac{\text{x}}{2}+\frac{3}{2}=\frac{2\text{x}}5-1{}$
Transposing $\frac{2\text{x}}{5}$ to L.H.S. and $\frac32$ to R.H.S., we get
$=\frac{\text{x}}{2}-\frac{2\text{x}}{5}=-1-\frac32$
$=\frac{5\text{x}-4\text{x}}{10}=\frac{-2-3}{2}$
$=\frac{\text{x}}{10}=\frac{-5}{2}$
Multiplying both sides by 10, we get
$=\frac{\text{x}}{10}\times10=\frac{-5}{2}\times10$
$=\text{x}=-25$
Verification:
Substituting x = -25 in the given equation, we get
$\frac{-25}{2}+\frac32=\frac{2(-25)}{5}-1$
$\frac{-22}{2}=-10-1$
$-11=-11$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.
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