Solve the Following Question.(4 Marks)

Question 14 Marks

How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer

Let the quantity of water to be added to solution = x liters.

$\therefore$ 25% (1125 + x) < 45% of 1125

$\Rightarrow\frac{25}{100}(1125+\text{x})<\frac{45}{100}\times1125$

$\Rightarrow1125+\text{x}<\frac{45}{25}\times1125$

⇒ 1125 + x < 25 × 45

⇒ 1125 + x < 2025

⇒ x < 2025 - 1125

⇒ x < 900 ...(i)

and 45% of 1125 < 30% (1125 + x)

$\Rightarrow\frac{45}{100}\times1125<\frac{30}{100}(1125)+\text{x}$

$\Rightarrow\frac{45}{30}\times1125<1125+\text{x}$

$\Rightarrow\frac{3}{2}\times1125<1125+\text{x}$

⇒ 1.5 × 1125 < 1125 + x

⇒ 1687.5 < 1125 + x

⇒ 1687.5 - 1125 < x

⇒ 562.5 < x ...(ii)

using (i) and (ii), we get 562.5 < x < 900

hence, quantity of water lies between 562.5 liters and 900 Liters.

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Question 24 Marks

Solve the following system of equations in R.
$\frac{|\text{x}-2|-1}{|\text{x}-2|-2}\leq0$

Answer

$\frac{|\text{x}-2|-1}{|\text{x}-2|-2}\leq0$
Let $\text{y}=|\text{x}-2|$
$\Rightarrow\frac{\text{y}-1}{\text{y}-2}\leq0$
$\Rightarrow1\leq\text{y}<2 $
$\Rightarrow1\leq|\text{x}-2|<2$
$\Rightarrow\text{x}\in[-2+2,-1+2]\cup[1+2,2+2]$
$\Rightarrow\text{x}\in[0,1]\cup[3,4]$
The solution set is $[0,1]\cup[3,4].$

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Question 34 Marks

To receive grade 'A' in a course, one must obtain an average of 90 marks or more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94 marks in first four paper, find the minimum marks that she must score in the last paper to get grade 'A' in the course.

Answer

Suppose shikha scores x marks in the fifth paper. Then,

$90\leq\frac{87+95+92+94+\text{x}}{5}$

$\Rightarrow90\times5\leq182+186+\text{x}$

$\Rightarrow450\leq368+\text{x}$

$\Rightarrow450-368\leq\text{x}$

$\Rightarrow82\leq\text{x}$

Hence, the maximum marks is required in the last paper is 82.

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Question 44 Marks

Solve the following system of equations in R.
$|\text{x}-1|+|\text{x}-2|+|\text{x}-3|\geq6$

Answer

We have,
$|\text{x}-1|+|\text{x}-2|+|\text{x}-3|\geq6$
Case 1: $|\text{x}-1|\geq0$
$\text{x}\geq1$
$\Rightarrow\text{x}-1-(\text{x}-2)-(\text{x}-3)-6\geq0$
$\Rightarrow-\text{x}+4-6\geq0$
$\Rightarrow-\text{x}\geq2$
$\Rightarrow\text{x}\leq-2$
$\Rightarrow(-\infty,-2]\ ...(\text{ii})$
Case 2: $|\text{x}-2|\geq0$
$\Rightarrow\text{x}-1+\text{x}-2-(\text{x}-3)-6\geq0$
$\text{x}\geq6$
$\Rightarrow[6,\infty)\ ...(\text{ii})$
Case 3: When $|\text{x}-3|\geq0$
$\text{x}\geq3$
$\Rightarrow\text{x}-1+\text{x}-2+\text{x}-3-6\geq0$
$\Rightarrow3\text{x}-12\geq0$
$\Rightarrow3\text{x}\geq12$
$\Rightarrow\text{x}\geq4$
$\therefore\text{x}\in[4,\infty)$
Also
$\Rightarrow|\text{x}-1|<0$
$\Rightarrow\text{x}<1$
$\Rightarrow-(\text{x}-1)-(\text{x}-2)-(\text{x}-3)-6\geq0$
$\Rightarrow-3\text{x}\geq0$
$\Rightarrow\text{x}\leq0$
$\Rightarrow|\text{x}-2|<0$
$\text{x}<2$
$\Rightarrow(\text{x}-1)-(\text{x}-2)-(\text{x}-3)-6\geq0$
$\Rightarrow\text{x}-1-\text{x}+2-\text{x}+3-6\geq0$
$\Rightarrow-\text{x}-2\geq0$
$\Rightarrow-\text{x}\geq2$
$\Rightarrow\text{x}\leq-2$
$\Rightarrow|\text{x}-3|<0$
$\Rightarrow\text{x}<3$
$\Rightarrow(\text{x}-1)+(\text{x}-2)-(\text{x}-3)-6\geq0$
$\Rightarrow\text{x}-6\geq0$
$\Rightarrow\text{x}\geq6 $
Combining all cases we get $(-\infty,0]\cup,[4,\infty)$ as the solution set.

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Question 54 Marks

Solve the following system of equations in R.
$2\text{x}+6\geq0,4\text{x}-7<0$

Answer

Consider the first inequation,
$2\text{x}+6\geq0$
$2\text{x}\geq-6$
$\text{x}\geq\frac{-6}{2}$
$\text{x}\geq-3\ ...(\text{i})$
Consider the secound inequation,
4x - 7 < 0
4x < 7
$\text{x}<\frac{7}{4}\ ...(\text{ii})$
From (i) and (ii), $\Big[-3,\frac{7}{4}\Big]$ is the solution of the simultaneous equations.

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Question 64 Marks

Solve the following system of equations in R.
$|\text{x}+1|+|\text{x}|>3$

Answer

$|\text{x}+1|+|\text{x}|>3$
Case 1: when $-\infty<\text{x}<-1$
|x + 1| = -(x + 1) and |x| = -x
$\therefore$ |x + 1| + |x| > 3
⇒ -(x + 1) - x > 3
⇒ -2x > 4
⇒ x < -2
But, $-\infty<\text{x}<-1$
$\therefore$ The solution set of the given inequation is $(-\infty,-2)$
Case 2: When $-1\leq\text{x}<0$
|x + 1| = (x + 1) and |x| = -x
$\therefore$ |x +1| + |x| > 3
⇒ (x + 1) - x > 3
⇒ 1 > 3
Which is not true
Case 3: when $0<\text{x}<\infty$
|x + 1| = (x + 1) and |x| = x
$\therefore$ |x + 1| + |x| > 3
⇒ (x + 1) + x > 3
⇒ 2x > 2
⇒ x > 1
But, $0<\text{x}<\infty$
$\therefore$ The solution set of the given inequation is $(1, \infty)$
Combining case 1, case 2 and case ,
we obtain that the solution set of given in equality is $(-\infty,-2)\cup(1,\infty)$

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Question 74 Marks

Solve the following system of equations in R.
$\Big|\frac{3\text{x}-4}{2}\Big|\leq\frac{5}{12}$

Answer

We have,
$\Big|\frac{3\text{x}-4}{2}\Big|\leq\frac{5}{12}\leq0$
Case 1: when $|3\text{x}-4|\geq0$
$\frac{|3\text{x}-4|}{}-\frac{5}{12}\leq0$
$\Rightarrow\frac{|3\text{x}-4|}{2}-\frac{5}{12}\leq0$
$\Rightarrow\frac{6(3\text{x}-4)-5}{12}\leq0$
$\Rightarrow18\text{x}-24-5\leq0$
$\Rightarrow18\text{x}-29\leq0$
$\Rightarrow\text{x}\leq\frac{29}{18}\ ...(\text{ii})$
Case 2: when |3x - 4| < 0
$\frac{|3\text{x}-4|}{2}-\frac{5}{12}\leq0$
$\Rightarrow-6(3\text{x}-4)-5\leq0$
$\Rightarrow-18\text{x}+24-5\leq0$
$\Rightarrow-18\text{x}+19\leq0$
$\Rightarrow-18\text{x}\geq19 $
$\Rightarrow\text{x}\geq\frac{19}{18}\ ...(\text{iii})$
Combining (ii) and (iii) to get solution set as $\Big(\frac{19}{18},\frac{29}{18}\Big)$

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Question 84 Marks

Solve the following system of equations in R.
$\frac{1}{|\text{x}|-3}<\frac{1}{2}$

Answer

We have,
$\frac{1}{|\text{x}|-3}-\frac{1}{2}<0\ ...(\text{i})$
Case 1: when $|\text{x}|\geq0\Rightarrow\text{x}\geq0$
$\Rightarrow\frac{1}{\text{x}-3}-\frac{1}{2}<0$
$\Rightarrow\frac{2-(\text{x}-3)}{2(\text{x}-3)}<0$
$\Rightarrow\frac{2-\text{x}+3}{2\text{x}-6}<0$
$\Rightarrow\frac{-\text{x}+5}{2\text{x}-6}<0$
$\Rightarrow\text{x}+5<0$
$\Rightarrow-\text{x}<-5$
$\Rightarrow\text{x}>5\ ...(\text{ii})$
Case2: when |x| < 0, x < 0
$\Rightarrow\frac{1}{-\text{x}-3}-\frac{1}{2}<0$
$\Rightarrow\frac{2-(-\text{x}-3)}{2(-\text{x}-3)}<0$
$\Rightarrow2+\text{x}+3<0$
$\Rightarrow\text{x}+5<0$
$\Rightarrow\text{x}<-5\ ...(\text{iii})$
Combining (ii) and (iii) we get $(-\infty,-5)\cup(-3,3)\cup(5,\infty)$ as the solution set.

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Maths Solve the Following Question.(4 Marks)

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