CBSE Class 11 Maths Revision Notes Chapter 6

Q.1 A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

Ans

Let x be the length of the shortest board,
then (x + 3) and 2x are the lengths of the second and third piece, respectively
Thus x + (x + 3) + 2x ≤ 91
⇒ 4x + 3 ≤ 91
⇒ 4x ≤ 88
⇒ x ≤ 22
According to the problem,
2x ≥ (x +3) + 5
⇒ x ≥ 8
The possible lengths of the shortest board are greater than or equal to 8 but less than or equal to 22.

Q.2 A solution is to be kept between 68° F and 77° F. What is the range of temperature in degree Celsius (C) ? if the Celsius(C) / Fahrenheit (F) conversion formula is given by F = (9/5)C + 32.

Ans

Q.3 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 20% but less than 30% acid content?

Ans

Let x litres water is added to 45% solution of acid, therefore,
20% of (1125 + x) < (1125) × 45/100 or (1125 + x)/5 < (1125) × 45/100
Multiplying by 20, we get
4(1125 + x) < 10125 or
4500 + 4x < 10125
Subtracting 4500 from both sides, we get
4x < 10125 – 4500 = 5625 or
x < 1406.25 …(1)
Now,
30% of (1125 + x) > (1125) × 45/100 or 3(1125 + x)/10 > (1125) × 45/100
Multiplying by 20, we get
6(1125 + x) > 10125 or 6750 + 6x > 10125
Subtracting 6750 from both sides, we get
6x > 10124 – 6750 = – 3375 or
x > 562.5 …(2)
Therefore, from (1) and (2), we get
1406.25 < x < 900.

Q.4 A manufacturer has 600 litres of a 10% solution of acid. How many litres of a 20% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%?

Ans

Let x litres of 20% acid solution is required to be added. Then Total mixture = (x + 600) litres
Therefore,
20% × + 10% of 600 > 15% of (x + 600) and
20% × + 10% of 600 < 18% of (x + 600 or
(20/100)x + (10/100) × 600 > (15/100) (x + 600) and
(20/100)x + (10/100) × 600 < (18/100) (x + 600) or
20x + 6000 > 15x + 9000 and
20x +6000 < 18x + 10800 or
5x > 3000 and 2x < 4800 or
x > 600 and x < 2400,
i.e., 600 < x < 2400.
Thus, the number of litres of the 20% solution of acid will have to be more than 600 litres but less than 2400 litres.

Q.5 Solve the following system of inequalities graphically 3x + 2y ≤ 150, x + 4y ≤ 150, x ≤ 15, x, y ≥ 0.

Ans

We first draw the graph of the lines corresponding to given inequalities.
To draw the graph of line, we need at least two solutions.
Two solution for the line 3x + 2y = 150 are:

Two solutions for the line x + 4y = 80 are:

Given x ≤ 15
x = 15 is a line parallel to y-axis and 15 units apart from the origin in the positive direction of x-axis.
Since, x ≥ 0, y ≥ 0 the solution region lies only in the first quadrant.
Thus graphical representation of the required solution is given in figure by the shaded region.

Q.6 Solve the following system of inequalities graphically 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3 where x, y ≥ 0.

Ans

We first draw the graph of the lines corresponding to given inequalities.
To draw the graph of line, we need at least two solutions
Two solutions for the line 4x + 3y = 60 are:

Two solutions for the line y = 2x are:

Given x ≥ 3, x = 3 is a line parallel to y-axis and 3 units apart from origin in the positive direction of x-axis.
Since, x ≥ 0, y ≥ 0. So the solution region lies only in the first quadrant.
Thus graphical representation of the required solution is given in figure by shaded region.

Q.7 Solve the following system of inequalities graphically:

5x + 4y ≤ 40 …(1)
x ≥ 2 …(2)
y ≥ 3 …(3)

Ans

To draw the graph of line, we need at least two solutions.
Two solutions for the line 5x + 4y = 40 are:

Given x ≥ 2 and y ≥ 3
x = 2 is a line parallel to the y-axis and 2 units apart from the origin in the positive direction of x-axis.
y = 3 is a line parallel to the x-axis and 3 units apart from the origin in the positive direction of y-axis.
Since, x ≥ 0, y ≥ 0. So, the solution region lies only in the first quadrant.
Thus graphical representation of the solutions are given in Fig. by shaded region.

Q.8 Solve the following system of inequalities graphically:

x + 2y ≤ 8 …(1)
2x + y ≤ 8 …(2)
x ≥ 0 …(3)
y ≥ 0 …(4)

Ans

To draw the graph of line, we need at least two solutions.
Two solutions for the line x + 2y = 8 are:

and two solution for the line 2x + y = 8 are:

Since x ≥ 0, y ≥ 0, therefore solutions lie in first quadrant.
Thus graphical representation of the solutions are given in figure and every point in the shaded region represents a solution of the given system of inequalities.

Q.9 Solve the following system of inequalities graphically:
2x + y ≥ 6, 3x + 4y < 12.

Ans

To draw the graph of line, we need at least two solutions.
Two solutions for the line 2x + y = 6 …(1)

and two solutions for the line 3x + 4y = 12 …(2)

Thus graphical representation of the solutions are given in figure by shaded region.

Q.10 Solve the system of inequalities:
3x – 7 < 5 + x …(1)
11 – 5x ≤ 1 …(2)
and represent the solutions on the number line.

Ans

From inequality (1), we have
3x – 7 < 5 + x …(1)
or x < 6 …(3)
Also, from inequality (2), we have
11 – 5x ≤ 1 …(2)
– 5x ≤ – 10, .i.e., x ≥ 2 …(4)
We draw the graph of inequalities (3) and (4) on the number line, the values of x, which are common to both,
are shown by a bold line on the number line in figure.

Thus, solutions of the system are real numbers x lying between 2 and 6 including 2, i.e., 2 ≤ x < 6.

Q.11 The longest side of a triangle is 4 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Ans

Let AB be the shortest side of the triangle = x cm
Then the longest side will be = 4x cm
And the third side will be = 4x – 2
Perimeter of the triangle is at least 61 cm.
x + 4x + 4x – 2 ≥ 61
⇒ 9x – 2 ≥ 61
⇒ x ≥ 7.
Hence, the minimum length of the shortest side is 7 cm.

Q.12 Solve the inequality 3y – 5x < 30 graphically in two-dimensional plane.

Ans

Q.13 Solve the inequality y + 8 ≥ 2x graphically in two-dimensional plane.

Ans

Q.14 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.

Ans

Q.15

$\mathbf{\text{Solve}}\frac{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{\ge }\frac{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{5}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We have,\hspace{0.33em}}\frac{(2\mathrm{x}-1)}{3}\ge \frac{(3\mathrm{x}-2)}{4}-\frac{(2-\mathrm{x})}{5}\text{\hspace{0.33em}or}\\ \frac{(2\mathrm{x}-1)}{3}\ge \frac{\left\{5\right(2\mathrm{x}-2)-4(2-\mathrm{x}\left)\right\}}{20}\\ \Rightarrow 20(2\mathrm{x}-1)\ge 3(19\mathrm{x}-18)\\ \Rightarrow 34\ge 17\mathrm{x}\\ \text{Therefore,}2\ge \mathrm{x}\text{or}\mathrm{x}\le 2\\ \text{Hence, the solution set is}(2,\infty )\text{.}\end{array}$

Q.16 Solve 4x + 3 < 6x + 7.

Ans

Q.17

$\mathbf{\text{Solve\hspace{0.33em}}}\mathbf{|}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{|}\mathbf{\le }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We know that}|\mathrm{x}-\mathrm{a}|\le \mathrm{r}\iff \mathrm{a}-\mathrm{r}\le \mathrm{x}\le \mathrm{a}+\mathrm{r}\\ \text{Therefore,}|3\mathrm{x}-2|\le \frac{1}{2}\iff 2-\frac{1}{2}\le 3\mathrm{x}\le 2+\frac{1}{2}\\ \iff \frac{1}{2}2\le \mathrm{x}\le 5/6\\ \iff \mathrm{x}\in [\frac{1}{2},\frac{5}{6}]\end{array}$

Q.18 Solve (x – 3)/(x – 5) > 0.

Ans

Here, (x – 3)/(x – 5) > 0
x = 3, 5 are critical points

Hence from figure (x – 3)/(x – 5) > 0
⇒ x ∈ (– ∞, 3) ∪ (5, ∞).

Q.19 Solve 7x + 9 ≥ 30.

Ans

Here, 7x + 9 ≥ 30.
⇒ 7x ≥ 30 – 9
⇒ x ≥ 3
⇒ x ∈ [3, ∞).

Q.20 Solve 5x – 3 < 3x + 1, where x∈ N.

Ans

Here, 5x – 3 < 3x + 1
5x – 3x < 3 + 1
2x < 4
x < 2
Therefore if x is less than 2 then the value of x is 1.

Q.21 Draw the graph of |x| ≤ 3.

Ans

Q.22

$\begin{array}{l}\mathbf{\text{Solve the system of inequations graphically:}}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\le }\mathbf{8}\\ \mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{8}\\ \mathbf{x}\mathbf{\ge }\mathbf{0}\\ \mathbf{y}\mathbf{\ge }\mathbf{0}\end{array}$

Ans

$\begin{array}{l}\mathrm{x}+2\mathrm{y}\le 8...\left(1\right)\\ \text{Let}\mathrm{x}+2\mathrm{y}=8\\ \text{Put}\mathrm{x}=0,\text{we get}\mathrm{y}=4,\text{i.e., point}(0,4)\\ \text{Put}\mathrm{y}=0,\text{we get}\mathrm{x}=8,\text{i.e., point}(8,0)\\ \text{We shall draw a line through these two points.}\\ \text{Now, put origin}(\mathrm{x}=0,\mathrm{y}=0)\text{in (1), we get}\\ 0+0\le 8\\ \Rightarrow 0\le 8\\ \text{which is true, so we will shade the region containing origin.}\\ 2\mathrm{x}+\mathrm{y}\le 8...\left(2\right)\\ \text{Let}2\mathrm{x}+\mathrm{y}=8\\ \text{Put}\mathrm{x}=0,\text{we get}\mathrm{y}=8,\text{i.e, point}(0,8)\\ \text{Put}\mathrm{y}=0,\text{we get}\mathrm{x}=4,\text{i.e., point}(4,0)\\ \text{We shall draw a line through these two points.}\\ \text{Now, put origin}(\mathrm{x}=0,\mathrm{y}=0)\text{in}\left(2\right),\text{we get}\\ 0+0\le 8\\ \Rightarrow 0\le 8\\ \text{which is true, so we will shade the region containing origin.}\\ \text{Since}\mathrm{x}\ge 0\text{and}\mathrm{y}\ge 0\text{represent the region of first quadrant}\end{array}$

On combining all the three conditions, we get the above shaded region, which represents solution of system of inequalities.

Q.23

$\mathbf{\text{Solve for}}\mathbf{x}\mathbf{:}\frac{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{\ge }\frac{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{5}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ \frac{(2\mathrm{x}-1)}{3}\ge \frac{(3\mathrm{x}-2)}{4}-\frac{(2-\mathrm{x})}{5}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{5(3\mathrm{x}-2)-4(2-\mathrm{x})}{20}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{15\mathrm{x}-10-8+4\mathrm{x}}{20}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{19\mathrm{x}-18}{20}\\ \Rightarrow 40\mathrm{x}-20\ge 57\mathrm{x}-54\\ \Rightarrow 54-20\ge 57\mathrm{x}-40\mathrm{x}\\ \Rightarrow 34\ge 17\mathrm{x}\\ \Rightarrow 17\mathrm{x}\le 34\\ \Rightarrow \mathrm{x}\le 2\\ \Rightarrow -\infty <\mathrm{x}\le 2\\ \Rightarrow \mathrm{X}\in (-\infty ,2]\end{array}$

Q.24

$\mathbf{Solve}\mathbf{the}\mathbf{inequality}\mathbf{2}\mathbf{<}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{<}\mathbf{5}\mathbf{\forall }\mathbf{x}\mathbf{\in }\mathbf{R}\mathbf{.}$

Ans

$\begin{array}{l}\text{Case}-\mathrm{I}\\ \text{On taking first two terms,}\\ 2<3\mathrm{x}-4\\ \Rightarrow 2+4<3\mathrm{x}\\ \Rightarrow 3\mathrm{x}>6\\ \Rightarrow \mathrm{x}>2...\left(1\right)\\ \text{Case}-\mathrm{II}\\ \text{On taking last two terms,}\\ 3\mathrm{x}-4<5\\ \Rightarrow 3\mathrm{x}<4+5\\ \Rightarrow 3\mathrm{x}<9\\ \Rightarrow \mathrm{x}<3...\left(2\right)\\ \text{From equation (1) and (2), we get}\\ 2<\mathrm{x}<3\\ \Rightarrow \mathrm{x}\in (2,3)\end{array}$

Q.25

$\mathbf{\text{Solve for}}\mathbf{x}\mathbf{:}\frac{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{)}}{\mathbf{2}}\mathbf{>}\frac{\mathbf{(}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ \frac{(3\mathrm{x}+5)}{2}>\frac{(5\mathrm{x}+1)}{3}-\frac{(\mathrm{x}-1)}{2}\\ \Rightarrow \frac{(3\mathrm{x}+5)}{2}>\frac{2(5\mathrm{x}+1)-3(\mathrm{x}-1)}{6}\\ \Rightarrow \frac{(3\mathrm{x}+5)}{2}>\frac{10\mathrm{x}+2-3\mathrm{x}+3}{6}\\ \Rightarrow \frac{(3\mathrm{x}+5)}{2}>\frac{7\mathrm{x}+5}{6}\\ \Rightarrow 6(3\mathrm{x}+5)>2(7\mathrm{x}+5)\\ \Rightarrow 18\mathrm{x}+30>14\mathrm{x}+10\\ \Rightarrow 18\mathrm{x}-14\mathrm{x}>10-30\\ \Rightarrow 4\mathrm{x}>-20\\ \Rightarrow \mathrm{x}>-5\\ \Rightarrow -5<\mathrm{x}<\infty \\ \Rightarrow \mathrm{x}\in (-5,\infty )\end{array}$

Q.26

$\mathbf{Solve}\mathbf{the}\mathbf{inequality}\mathbf{:}\mathbf{8}\mathbf{<}\frac{\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{10}}{\mathbf{2}}\mathbf{\le }\mathbf{11}\mathbf{\forall }\mathbf{x}\mathbf{\in }\mathbf{R}\mathbf{.}$

Ans

$\begin{array}{l}\text{Case}-\mathrm{I}\\ \text{On taking first two terms, we get}\\ 8<\frac{3\mathrm{x}+10}{2}\\ \Rightarrow 16<3\mathrm{x}+10\\ \Rightarrow 3\mathrm{x}+10>16\\ \Rightarrow 3\mathrm{x}>16-10\\ \Rightarrow 3\mathrm{x}>6\\ \Rightarrow \mathrm{x}>2...\left(1\right)\\ \text{Case}-\mathrm{II}\\ \text{On taking last two terms, we get}\\ \frac{3\mathrm{x}+10}{2}\le 11\\ \Rightarrow 3\mathrm{x}+10\le 22\\ \Rightarrow 3\mathrm{x}\le 22-10\\ \Rightarrow 3\mathrm{x}\le 12\\ \Rightarrow \mathrm{x}\le 4...\left(2\right)\\ \text{From (1) and (2), we get}\\ 2<\mathrm{x}\le 4\\ \Rightarrow \mathrm{X}\in (2,4]\end{array}$

Q.27

$\mathbf{\text{Find the integral value of}}\mathbf{x}\mathbf{:}\mathbf{7}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{\le }\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{7}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ 7\mathrm{x}-5\le 3\mathrm{x}+7\\ \Rightarrow 7\mathrm{x}-3\mathrm{x}\le 7+5\\ \Rightarrow 4\mathrm{x}\le 12\\ \Rightarrow \mathrm{x}\le 3\\ \Rightarrow \mathrm{x}=-\infty \dots -3,-2,-1,0,1,2,3\end{array}$

Q.28 The marks obtained by Beeru of class XI in first and second terminal examination are 70 and 45 respectively. Find the number of minimum marks he should get in the annual examination to have an average of atleast 60 marks.

Ans

$\begin{array}{l}\text{Let}\mathrm{x}\text{be the marks obtained by Beeru in the\hspace{0.33em}annual examination.}\\ \because \text{\hspace{0.33em}\hspace{0.33em}Average of}3\text{exam marks}\ge 60\\ \Rightarrow \frac{70+45+\mathrm{x}}{3}\ge 60\\ \Rightarrow 115+\mathrm{x}\ge 180\\ \Rightarrow \mathrm{x}\ge 180-115\\ \Rightarrow \mathrm{x}\ge 65\\ \text{Thus, Beeru must obtain a minimum of\hspace{0.33em}65 marks to get an average of at least 60 marks.}\end{array}$

Q.29

$\mathbf{Solve}\mathbf{for}\mathbf{x}\mathbf{:}\frac{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{\ge }\frac{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{5}}\mathbf{\text{\hspace{0.33em}Also show the graph of solutions on number line.}}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ \frac{(2\mathrm{x}-1)}{3}\ge \frac{(3\mathrm{x}-2)}{4}-\frac{(2-\mathrm{x})}{5}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{5(3\mathrm{x}-2)-4(2-\mathrm{x})}{20}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{15\mathrm{x}-10-8+4\mathrm{x}}{20}\\ \Rightarrow \frac{(2\mathrm{x}-1)}{3}\ge \frac{19\mathrm{x}-18}{20}\\ \Rightarrow 40\mathrm{x}-20\ge 57\mathrm{x}-54\\ \Rightarrow 54-20\ge 57\mathrm{x}-40\mathrm{x}\\ \Rightarrow 34\ge 17\mathrm{x}\\ \Rightarrow 17\mathrm{x}\le 34\\ \Rightarrow \mathrm{x}\le 2\\ \Rightarrow -\infty <\mathrm{x}\le 2\\ \Rightarrow \mathrm{x}\in (-\infty ,2]\end{array}$

Thus, here the dark and thick line represent solution of linear inequality.

Q.30

$\mathbf{\text{Solve for}}\mathbf{x}\mathbf{:}\mathbf{37}\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{<}\mathbf{17}\mathbf{\forall }\mathbf{\in }\mathbf{R}$

Ans

We have,
37 – 5x < 17
⇒ –5x < 17 – 37
⇒ –5x < –20
Divide both sides by –5,
⇒ x > 4
Since, on dividing by a – ve number, the sign of inequality changes,
⇒ 4 < × > ∞
⇒ x ∈ (4, ∞)

Q.31 Find the integral value of x: 5x – 5 < 3x + 1.

Ans

We have, 5x – 5 < 3x + 1
⇒ 5x – 3x < 1 + 5
⇒ 2x < 6
⇒ x < 3
⇒ x = –∞ … –3, –2, –1, 0, 1, 2 = (–∞, 2]

Q.32

$\mathbf{\text{Solve for}}\mathbf{x}\mathbf{:}\mathbf{13}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{\le }\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{17}\mathbf{\forall }\mathbf{x}\mathbf{\in }\mathbf{R}$

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ 13\mathrm{x}-5\le 2\mathrm{x}+17\\ \Rightarrow 13\mathrm{x}-2\mathrm{x}\le 17+5\\ \Rightarrow 11\mathrm{x}\le 22\\ \Rightarrow \mathrm{x}\le 2\\ \Rightarrow -\infty <\mathrm{x}\le 2\\ \Rightarrow \mathrm{x}\in (-\infty ,2]\end{array}$