Download App

Maths 4 Mark Question

Linear Equations in one variable · Maharashtra Board STD 6 (MSBSHSE - Semi English Medium). 14 questions.

Questions

4 Mark Question

Take a timed test

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

Question 14 Marks
Solve the following equation and verify the answer:
$\frac{\text{3x}}{10}-4=14$
Answer
$\frac{\text{3x}}{10}-4=14$ $\Rightarrow\frac{\text{3x}}{10}=14+4$(Transposing -4 to R.H.S.)
$\Rightarrow\frac{\text{3x}}{10}=18$ $\Rightarrow\frac{\text{3x}}{10}\times=18\times10$(Multiplying both sides by 10)
$\Rightarrow3x=180\Rightarrow\frac{\text{3x}}{3}=\frac{180}{3}$(Dividing both sides by 3)
$\Rightarrow \text{x}=60$ So, $\text{x}=60$ is a solution of the given equation. Check: Substituting $\text{x}=60$ in the given equation, we get $\text{L.H.S.}=\frac{3\times60}{10}-4=(3\times6)-4$ = 18 - 4 = 14and R.H.S. = 14
$\therefore$ When x = 60, we have L.H.S. = R.H.S.
View full question & answer→
Question 24 Marks
Solve the following equation and verify the answer:
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$
Answer
$\frac{\text{2x}}{5}-\frac{3}{2}=\frac{\text{x}}{2}+1$ Multiplying each term by 10, the L.C.M. of 5 and 2, we get $\frac{\text{2x}}{5}\times10-\frac{3}{2}\times10$ $=\frac{\text{x}}{2}\times10+1\times10$⇒ 4x - 15 = 5x + 10
⇒ 4x - 5x = 10 + 15
(transposing 5x to L.H.S.)
⇒ -x = 25
⇒ x = -25
(multiplying both sides by -1)
So, x = -25 is a solution of the given equation.
Check: Substituting x = -25 in the given equation, we get
$\text{L.H.S.}=\frac{2\times(-25)}{5}-\frac{3}{2}=-10-\frac{3}{2}$
$=\frac{-20-3}{2}=\frac{-23}{2}$
$\text{R.H.S.}=\frac{-25}{2}+1=\frac{-25+2}{2}=\frac{-23}{2}$
$\therefore$ When x = -25, we have
L.H.S. = R.H.S.
View full question & answer→
Question 34 Marks
Solve the following equation and verify the answer:
6(1 - 4x) + 7(2 + 5x) = 53
Answer
6(1 - 4 x) + 7(2 + 5x) - 53 ⇒ 6 - 24x + 14 + 35x = 53 (Removing brackets) ⇒ -24x + 35x + 14 + 6 = 53 ⇒ 11x + 20 = 53 ⇒ 11x = 53 - 20 ⇒ 11x = 33 (Transposing 20 to R.H.S.) $\Rightarrow\frac{\text{11x}}{11}=\frac{33}{11}$(Dividing both sides by 11)
⇒ x = 3 So, x = 3 is a solution of the given equation. Check: Substituting x = 3 in the given equation, we get, L.H.S. = 6 (1 - 4 × 3) + 7(2 + 5 × 3) = 6(1 - 12) + 7(2 + 15) = 6 × (-11) + 7 × 17 = -66 + 119 = 53 = R.H.S. $\therefore$ When x = 3, we have L.H.S. = R.H.S.
View full question & answer→
Question 44 Marks
Solve the following equation and verify the answer:
5(x - 1) + 2(x + 3) + 6 = 0
Answer
5(x - 1) + 2(x + 3) + 6 = 0 ⇒ 5(x - 1) + 2(x + 3) = -6 (Transposing 6 to R.H.S.) ⇒ 5x - 5 + 2x + 6 = -6 (Removing brackets) ⇒ 5x + 2x - 5 + 6 = -6 ⇒ 7x + 1 = -6 ⇒ 7x = -6 - 1 (Transposing 1 to R.H.S.) ⇒ 7x = -7 $\Rightarrow\frac{\text{7x}}{7}=\frac{-7}{7}$(Dividing both sides by 7)
⇒ x = -1 So, x = -1 is a solution of the given equation. Check: Substituting x = -1 in the given equation, we get L.H.S. = 5(-1 - 1) + 2(-1 + 3) + 6 = 5 × (-2) + 2 × 2 + 6 = 10+ 4 +6 = -10 + 10 = 0 = R.H.S.$\therefore$ When x = -1, we have
L.H.S. = R.H.S.
View full question & answer→
Question 54 Marks
Solve the following equation and verify the answer:
$\frac{\text{2m}}{3}+8=\frac{\text{m}}{2}-1$
Answer
$\frac{\text{2m}}{3}+8=\frac{\text{m}}{2}-1$ Multiplying each term by 6, the L.C.M. of 2 and 3, we get $\frac{\text{2m}}{3}\times6+8\times6=\frac{\text{m}}{2}\times6-1\times6$ ⇒ 4m + 48 = 3m - 6 ⇒ 4m - 3m = -6 - 48(Transposing 3m to L.H.S. and 48 to R.H.S.)
⇒ m = -54 So, m = -54 is a solution of the given equation. Check: Substituting m = -54 in the given equation, we get$\text{L.H.S.}=\frac{-54}{2}-1=-27-1=-28$
$=-36+8=-28$
$\text{R.H.S}=\frac{-54}{2}-1=-27-1=-28$
$\therefore$ When m = -54, we have L.H.S. = R.H.S.
View full question & answer→
Question 64 Marks
Solve the following equation and verify the answer:
3(x + 2) - 2(x - 1) = 7
Answer
3(x + 2) - 2(x - 1) = 7
⇒ 3x + 6 - 2x + 2 = 7
(Removing brackets)
3x - 2x + 6 + 2 = 7
x + 8 = 7
x = 7 - 8
(Transposing 8 to R.H.S.)
x = -1 is a solution of the given equation.
Check: Substituting x = -1 in the given equation, we get
L.H.S. = 3 ( -1 + 2) - 2( -1 - 1)
= 3 × 1 + ( -2 × - 2)
= 3 + 4 = 7 and
R.H.S. = 7
When x = -1, we have
L.H.S. = R.H.S.
View full question & answer→
Question 74 Marks
Solve the following equation and verify the answer:$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$
Answer
$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$Multiplying each term by 12, the L.C.M. of 4, 6, 2, we get
$\frac{\text{n}}{4}\times12-5\times12$
$=\frac{\text{n}}{6}\times12+\frac{1}{2}\times12$
⇒ 3n - 60 = 2n + 6
⇒ 3n - 2n = 6 + 60
(Transposing 2n to L.H.S. and -60 to R.H.S.)
⇒ n = 66
So, n = 66 is a solution of the given equation.
Check: Substituting n = 66 in the given equation, we get
$\text{L.H.S.}=\frac{66}{4}-5=\frac{33}{2}-5$
$=\frac{33-10}{2}=\frac{23}{2}$
$\text{R.H.S.}=\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}$
$=\frac{22+1}{2}=\frac{23}{2}$
$\therefore$ When n = 66, we have L.H.S. = R.H.S.
View full question & answer→
Question 84 Marks
Solve the following equation and verify the answer:
$\frac{3}{4}(\text{x}-1)=\text{x}-3$
Answer
$\frac{3}{4}(\text{x}-1)=\text{x}-3$ $\Rightarrow\frac{3}{4}(\text{x}-1)\times4=(\text{x}-3)\times4$ (Multiplying both sides by 4) ⇒ 3(x - 1) = 4(x - 3) ⇒ 3x - 3 = 4x - 12 (Removing brackets) ⇒ 3x - 4x = -12 + 3 (Transposing 4x to L.H.S. and -3 to R.H.S.) ⇒ -x = -9 ⇒ x = 9 So, x = 9 is a solution of the given equation. Check: Substituting x = 9 in the given equation, we get $\text{L.H.S}=\frac{3}{4}(9-1)=\frac{3}{4}\times8=6$$\text{R.H.S.}= 9 - 3 = 6$
$\therefore$ When x = 9, we have L.H.S. = R.H.S.
View full question & answer→
Question 94 Marks
Solve the following equation and verify the answer:
16(3x - 5) - 10(4x - 8) = 40
Answer
16(3x - 5) - 10(4x - 8) = 40 ⇒ 48x - 80 - 40x + 80 = 40 (Removing brackets) ⇒ 48x - 40x - 80 + 80 = 40 ⇒ 8x = 40$\Rightarrow\frac{\text{8x}}{8}=\frac{40}{8}$
(Dividing both sides by 8)
⇒ x = 5 So, x = 5 is a solution of the given equation. Check: Substituting x = 5 in the given equation, we get L.H.S. = 16(3 × 5 - 5) - 10(4 × 5 - 8) = 16(15 - 5) - 10(20 - 8) = 16 × 10 - 10 × 12 = 160 - 120 = 40 = R.H.S. $\therefore$ When x = 5, we have L.H.S. = R.H.S.
View full question & answer→
Question 104 Marks
Solve the following equation and verify the answer:
3(2 - 5x) - 2(1 - 6x) = 1
Answer
3(2 - 5x) - 2(1 - 6x) = 1 ⇒ 6 - 15x - 2 + 12x = 1 (Removing brackets) ⇒ 6 - 2 - 15x + 12x = 1 ⇒ 4 - 3x = 1 -3x = 1 - 4 (Transposing 4 to R.H.S.) ⇒ -3x = -3$\Rightarrow\frac{\text{-3x}}{-3}=\frac{-3}{-3}$
(Dividing both sides by -3) ⇒ x = 1 So, x = 1 is a solution of the given equation. Check: Substituting x = 1 in the given equation, we get L.H.S. = 3(2 - 5 × 1) - 2(1- 6 × 1) = 3(2 - 5) - 2(1 - 6) = [3 × (-3)] + [-2 × (-5)] = -9 + 10 = 1 = R.H.S. $\therefore$ When x = 1, we have L.H.S. = R.H.S.
View full question & answer→
Question 114 Marks
Solve the following equation and verify the answer:
$\text{2x}-\frac{1}{2}=3$
Answer
$\text{2x}-\frac{1}{2}=3$
$\Rightarrow\text{2x}=3+\frac{1}{2}$
(Transporting $-\frac{1}{2}$ to R.H.S.)
$\Rightarrow\text{2x}=\frac{7}{2}$
$\Rightarrow\frac{\text{2x}}{2}=\frac{7}{2}\times\frac{1}{2}$
(Dividing both sides by 2)
$\Rightarrow\text{x}=\frac{7}{4}$
$\therefore\text{x}=\frac{7}{4}$ is a solution of the given equation.
Check: Substituting $\text{x}=\frac{7}{4}$ in the given equation, we get
$\text{L.H.S.}=2\times\frac{7}{4}-\frac{1}{2}$
$=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3$
and R.H.S. = 3
$\therefore$ When $\text{x}=\frac{7}{4},$ we have
L.H.S. = R.H.S.
View full question & answer→
Question 124 Marks
Solve the following equation and verify the answer:
$\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
Answer
$\frac{\text{x}}{2}=\frac{\text{x}}{3}+1$
$\Rightarrow\frac{\text{x}}{2}-\frac{\text{x}}{3}=1$
(Transpoting $\frac{\text{x}}{3}$ L.H.S.)
Multiplying each term by 6, the L.C.M. of 2 and 3, we get
$\frac{\text{x}}{2}\times6-\frac{\text{x}}{3}\times6=1\times6$
$\Rightarrow3\text{x}-2\text{x}=6$
$\Rightarrow\text{x}=6$
$\therefore\text{x}=6$ is a solution of the given equation.
Check: Substituting x = 6 in the given equaation, we get
$\text{L.H.S.}=\frac{6}{2}=3$ and
$\text{R.H.S.}=\frac{6}{3}+1=2+1=3$
$\therefore$ When x = 6, we have L.H.S. = R.H.S.
View full question & answer→
Question 144 Marks
Solve the following equation and verify the answer:
$\frac{\text{x-3}}{5}=-2=\frac{\text{2x}}{5}$
Answer
$\frac{\text{x-3}}{5}=-2=\frac{\text{2x}}{5}$ multiplying each term by 5, we get $\frac{(\text{x}-3)\times5}{5}-(2\times5)=\frac{\text{2x}}{5}\times5$ ⇒ x - 3 - 10 = 2x ⇒ x - 13 = 2x ⇒ x - 2x = 13 (Transposing 2x to L.H.S. and -13 to R.H.S.) ⇒ -x = 13 ⇒ x = -13 (Multiplying both sides by -1) So, x = -13 is a solution of the given equation. Check: Substituting x = -13 in the given equation, we get $\text{L.H.S.}=\frac{-13-3}{5}-2=\frac{-16}{5}-2$$=\frac{-16-10}{5}=\frac{-26}{5}$
$\text{R.H.S.}=\frac{2\times(-13)}{5}=\frac{-26}{5}$ $\therefore$ When x = -13, we have L.H.S. = R.H.S.
View full question & answer→
Generate Paper FreeAll Linear Equations in one variable topics