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Maths 2 Mark Question

Linear Equations · Maharashtra Board STD 8 (MSBSHSE - English Medium). 15 questions.

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2 Mark Question

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15 questions · timed · auto-graded

Question 12 Marks
Solve:
$\frac{3\text{x}}{5\text{x}+2}=-4$
Answer
$\frac{3\text{x}}{5\text{x}+2}=-4$
$\Rightarrow\frac{3\text{x}​​}{5\text{x}+2}=\frac{-4}{1}$
$\Rightarrow1(3\text{x})=-4(5\text{x}+2)$ (by cross multiplication)
$\Rightarrow3\text{x}=-20\text{x}-8$
$\Rightarrow3\text{x}+20\text{x}=-8$
$\Rightarrow23\text{x}=-8$
$\Rightarrow\text{x}=\frac{-8}{23}$
$\therefore\text{x}=\frac{-8}{23}$
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Question 22 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$ find the value of:
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$
Answer
From the first part:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=18$
Squaring both the sides:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=18^2$
$\Rightarrow(​​\text{x}^2)^2+2\times(\text{x}^2)\times\Big(\frac{1}{\text{x}^2}\Big)+\Big(\frac{1}{\text{x}^2}\Big)^2=324$
$\Rightarrow\text{x}^4+2+\frac{1}{\text{x}^4}=324$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=324-2$
$\therefore\text{x}^4+\frac{1}{\text{x}^4}=322$
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Question 32 Marks
Factorise:
$1-6 x+9 x^2$
Answer
$1-6 x+9 x^2$
$=9 x^2-6 x+1$
$=9 x^2-3 x-3 x+1$
$=3 x(3 x-1)-1(3 x-1)$
$=(3 x-1)(3 x-1)$
$=(3 x-1)^2$
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Question 42 Marks
Solve: $\frac{2\text{x}+7}{3\text{x}+5}=\frac{15}{17}$
Answer
$\frac{2\text{x}+7}{3\text{x}+5}=\frac{15}{17}$ $\Rightarrow17 ( 2\text{x} + 7 ) = 15 ( 3\text{x} + 5 ) $ $\Rightarrow34\text{x} + 119 = 45\text{x} + 75$ $\Rightarrow119 - 75 = 45\text{x} - 34\text{x}$ $\Rightarrow 44 = 11\text{x}$ $\Rightarrow \text{x} = \frac{44}{11}=4$ $\therefore\text{x}=4$
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Question 52 Marks
Factorise:
$63 x^2 y^2-7$
Answer
$=7\left(9 x^2 y^2-1\right)$
$=7\left((3 x y)^2-(1)^2\right)$
[according to the formula $a ^2- b ^2=( a + b )( a - b )$ ]
$=7(3 x y+1)(3 x y-1)$
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Question 72 Marks
Solve:
5x + 7 = 2x - 8
Answer
$5\text{x} + 7 = 2\text{x}-8$
$\Rightarrow5\text{x}-\text{2x}=-8-7$
$\Rightarrow3\text{x}=-15$
$\Rightarrow\text{x}=\frac{-15}{3}=-5$
$\therefore\text{x}=-5$
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Question 82 Marks
Find the following products:
$(4 a+5 b) \times(5 a-6 b)$
Answer
$(4 a+5 b) \times(5 a-6 b)$
$=20 a^2-24 a b+25 a b-30 b^2$
$=20 a^2+a b-30 b^2$
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Question 92 Marks
Solve:
2z - 1 = 14 - z
Answer
$2\text{z} - 1 = 14 - \text{z}$
$\Rightarrow2\text{z}+\text{z}=14+1$
$\Rightarrow3\text{z}=15$
$\Rightarrow\text{z}=\frac{15}{3}=5$
$\therefore\text{z}=5$
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Question 112 Marks
Solve:
$\frac{\text{x}-5}{2}-\frac{\text{x}-3}{5}=\frac{1}{2}$
Answer
$\frac{\text{x}-5}{2}-\frac{\text{x}-3}{5}=\frac{1}{2}$
$\Rightarrow10\Big(\frac{\text{x}-5}{2}\Big)-10\Big(\frac{\text{x}-3}{5}\Big)=10\Big(\frac{1}{2}\Big)$
(multiplying throughout by 10, which is the L.C.M. of 2, 2 and 5):
$\Rightarrow5(\text{x}-5)-2(\text{x}-3)=5$
$\Rightarrow5\text{x}-2$
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Question 122 Marks
Two numbers are such that the ratio between them is 3 : 5. If each is increased by 10, the ratio between the new numbers so formed is 5 : 7. Find the original numbers.
Answer
Let us consider x as the common multiple of both the number.
Then, first number = 3x
Second number = 5x
$\therefore\frac{3\text{x}+10}{5\text{x}+10}=\frac{5}{7}$
$\Rightarrow7(3\text{x}+10)=5(5\text{x}+10)$ (by cross multiplication)
$\Rightarrow 21\text{x} + 70 = 25\text{x} + 50$
$\Rightarrow 21\text{x} - 25\text{x} = 50 - 70$
$\Rightarrow -4\text{x} = -20$
$\Rightarrow \text{x} =\frac{ -20}{-4} = 5$
Therefore, the common multiple of both the numbers is 5.
First number = 3x = 3 × 5 = 15
Second number = 5x = 5 × 5 = 25
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Question 132 Marks
Find the following products:
$\left(6 x^2-x+8\right) \times\left(x^2-3\right)$
Answer
$\left(6 x^2-x+8\right) \times\left(x^2-3\right)$
$=6 x^4-18 x^2-x^3+3 x+8 x^2-24$
$=6 x^4-x^3-18 x^2+8 x^2+3 x-24$
$=6 x^4-x^3-10 x^2+3 x-24$
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Question 152 Marks
Solve:
8x + 3 = 27 + 2x
Answer
$8\text{x} + 3 = 27 + 2\text{x}$
$\Rightarrow8\text{x}-\text{2x}=27-3$
$\Rightarrow6\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{6}=4$
$\therefore\text{x}=4$
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