Linear Inequalities — NCERT Solutions

$3x-2<2x+1$

Subtract $2x$:

$x-2<1$

Add $2$:

$x<3$.

So the solution is $x<3$.

Solve the inequality:

$5x - 3 \geq 3x - 5$

Subtract $3x$ from both sides:

$2x - 3 \geq -5$

Add 3 to both sides:

$2x \geq -2$

Divide by 2:

$x \geq -1$

So the solution is all real numbers greater than or equal to $-1$.

Solve:

$3(1-x) < 2(x+4)$

Expand both sides:

$3 - 3x < 2x + 8$

Bring like terms together:

$-3x - 2x < 8 - 3$

$-5x < 5$

Divide by $-5$ and reverse the inequality sign:

$x > -1$

So the solution is $x > -1$.

Solve:

$\dfrac{x}{2} \geq \dfrac{5x-2}{3} - \dfrac{7x-3}{5}$

Take LCM $30$ and multiply throughout by $30$:

$15x \geq 10(5x-2) - 6(7x-3)$

$15x \geq 50x - 20 - 42x + 18$

$15x \geq 8x - 2$

$7x \geq -2$

$x \geq -\dfrac{2}{7}$

So the solution is $x \geq -\dfrac{2}{7}$.

Let the marks in the third test be $x$.

Average of at least 60 means:

$\dfrac{70+75+x}{3} \geq 60$

$145+x \geq 180$

$x \geq 35$

So the minimum marks required in the third test is 35.

Let the fifth exam marks be $x$.

To get Grade A, average must be at least 90:

$\dfrac{87+92+94+95+x}{5} \geq 90$

$368+x \geq 450$

$x \geq 82$

So the minimum marks required is 82.

Let the smaller odd number be $x$, so the next consecutive odd number is $x+2$.

Since both are smaller than 10 and their sum is more than 11:

$x < 10$ and $x+(x+2) > 11$

$2x+2 > 11$

$2x > 9$

$x > 4.5$

The odd positive integers less than 10 and greater than 4.5 are $5$ and $7$.

So the required pairs are:

• $(5,7)$
• $(7,9)$

Let the smaller even number be $x$, so the next consecutive even number is $x+2$.

Given both are larger than 5 and their sum is less than 23:

$x > 5$ and $x+(x+2) < 23$

$2x+2 < 23$

$2x < 21$

$x < 10.5$

The even positive integers greater than 5 and less than 10.5 are $6, 8, 10$.

So the required pairs are:

• $(6,8)$
• $(8,10)$

Let the shortest side be $x$ cm.

Then the longest side is $3x$ cm and the third side is $3x-2$ cm.

Perimeter is at least 61 cm, so:

$x + 3x + (3x-2) \geq 61$

$7x - 2 \geq 61$

$7x \geq 63$

$x \geq 9$

So the minimum possible length of the shortest side is 9 cm.

Let the shortest board be $x$ cm.

Then the other two lengths are $x+3$ and $2x$.

Total length cannot exceed 91 cm:

$x+(x+3)+2x \leq 91$

$4x+3 \leq 91$

$4x \leq 88$

$x \leq 22$

Also, the third piece is at least 5 cm longer than the second:

$2x \geq (x+3)+5$

$2x \geq x+8$

$x \geq 8$

So the possible lengths of the shortest board are:

$8 \leq x \leq 22$

If lengths are taken as whole numbers, the possible lengths are 8 cm to 22 cm.

Solve the double inequality:

$2 \leq 3x - 4 \leq 5$

Add 4 throughout:

$6 \leq 3x \leq 9$

Divide by 3:

$2 \leq x \leq 3$

So the solution is $2 \leq x \leq 3$.

Solve:

$6 \leq -3(2x-4) < 12$

Expand:

$6 \leq -6x + 12 < 12$

Subtract 12 throughout:

$-6 \leq -6x < 0$

Divide by $-6$ and reverse both inequality signs:

$1 \geq x > 0$

So the solution is:

$0 < x \leq 1$.

Solve:

$-3 \leq 4 - \dfrac{7x}{2} \leq 18$

Subtract 4 throughout:

$-7 \leq -\dfrac{7x}{2} \leq 14$

Multiply throughout by $-2/7$ and reverse the inequality signs:

$1 \geq x \geq -4$

So the solution is:

$-4 \leq x \leq 1$.

Solve:

$-15 < \dfrac{3(x-2)}{5} \leq 0$

Multiply throughout by 5:

$-75 < 3(x-2) \leq 0$

Divide by 3:

$-25 < x-2 \leq 0$

Add 2 throughout:

$-23 < x \leq 2$

So the solution is $-23 < x \leq 2$.

Solve:

$-12 < 4 - \dfrac{3x}{-5} \leq 2$

Since $-\dfrac{3x}{-5}=\dfrac{3x}{5}$, we get:

$-12 < 4 + \dfrac{3x}{5} \leq 2$

Subtract 4 throughout:

$-16 < \dfrac{3x}{5} \leq -2$

Multiply by 5:

$-80 < 3x \leq -10$

Divide by 3:

$-\dfrac{80}{3} < x \leq -\dfrac{10}{3}$

Solve:

$7 \leq \dfrac{3x+11}{2} \leq 11$

Multiply throughout by 2:

$14 \leq 3x+11 \leq 22$

Subtract 11:

$3 \leq 3x \leq 11$

Divide by 3:

$1 \leq x \leq \dfrac{11}{3}$

Solve the two inequalities together:

1. $5x+1>-24$

$5x>-25$

$x>-5$

2. $5x-1<24$

$5x<25$

$x<5$

Combining both gives:

$-5 < x < 5$.

Solve both inequalities:

1. $2(x-1)<x+5$

$2x-2<x+5$

$x<7$

2. $3(x+2)>2-x$

$3x+6>2-x$

$4x>-4$

$x>-1$

Combining:

$-1 < x < 7$.

Solve both inequalities:

1. $3x-7>2(x-6)$

$3x-7>2x-12$

$x>-5$

2. $6-x>11-2x$

Add $2x$ to both sides:

$6+x>11$

$x>5$

Combining gives:

$x>5$.

Solve both inequalities:

1. $5(2x-7)-3(2x+3)\leq 0$

$10x-35-6x-9\leq 0$

$4x-44\leq 0$

$4x\leq 44$

$x\leq 11$

2. $2x+19\leq 6x+47$

$19-47\leq 6x-2x$

$-28\leq 4x$

$x\geq -7$

Combining both:

$-7 \leq x \leq 11$.

From the chapter example, if the solution is to be kept between $68^\circ$F and $77^\circ$F, then use

$C=\dfrac{5}{9}(F-32)$.

For $F=68$:

$C=\dfrac{5}{9}(68-32)=\dfrac{5}{9}\cdot 36=20$

For $F=77$:

$C=\dfrac{5}{9}(77-32)=\dfrac{5}{9}\cdot 45=25$

So the range in Celsius is between $20^\circ$C and $25^\circ$C.

Let $x$ litres of 2% solution be added.

The chapter’s worked method gives the condition:

$120 < x < 300$

So the amount of 2% solution must be more than 120 litres but less than 300 litres.

Let the amount of water added be $x$ litres.

Initial acid = $45\%$ of $1125 = 506.25$ litres.

After adding water, acid amount stays the same, but total mixture becomes $1125+x$ litres.

For acid content to be more than $25\%$ but less than $30\%$:

$0.25 < \dfrac{506.25}{1125+x} < 0.30$

From the lower bound:

$506.25 > 0.25(1125+x)$

$506.25 > 281.25 + 0.25x$

$225 > 0.25x$

$x < 900$

From the upper bound:

$506.25 < 0.30(1125+x)$

$506.25 < 337.5 + 0.30x$

$168.75 < 0.30x$

$x > 562.5$

So the amount of water must satisfy:

$562.5 < x < 900$.

For the remaining visible part of the question, the chapter gives the formula

$IQ=\dfrac{MA}{CA}\times 100$

and says to find the range of mental age for 12-year-old children when $80\leq IQ\leq 140$.

So,

$80 \leq \dfrac{MA}{12}\times 100 \leq 140$

Multiply throughout by 12 and divide by 100:

$80\cdot \dfrac{12}{100} \leq MA \leq 140\cdot \dfrac{12}{100}$

$9.6 \leq MA \leq 16.8$

So the mental age lies between 9.6 years and 16.8 years.