Linear Equations and Inequalities in One Variable — Check Your Progress

Solve the following equation:

$\dfrac{3}{5}x + 2 = \dfrac{2}{3}x$

Answer

$\dfrac{3}{5}x + 2 = \dfrac{2}{3}x$

L.C.M. of 5 and 3 = 15

Multiplying both sides by 15, we get:

$\Rightarrow \Big[\dfrac{3}{5}x + 2\Big] \times 15 = \dfrac{2}{3}x \times 15 \\[1em] \Rightarrow \dfrac{3}{5}x \times 15 + 2 \times 15 = 10x \\[1em] \Rightarrow 9x + 30 = 10x \\[1em] \Rightarrow 30 = 10x - 9x \\[1em] \Rightarrow x = 30$

Hence, $\bm{x = 30}$.

Solve the following equation:

$\dfrac{x}{2} - \dfrac{3x}{4} + \dfrac{x}{6} = \dfrac{-2}{3}$

Answer

$\dfrac{x}{2} - \dfrac{3x}{4} + \dfrac{x}{6} = \dfrac{-2}{3}$

L.C.M. of 2, 4, 6 and 3 = 12

Multiplying both sides by 12, we get:

$\Rightarrow \Big[\dfrac{x}{2} - \dfrac{3x}{4} + \dfrac{x}{6}\Big] \times 12 = \dfrac{-2}{3} \times 12 \\[1em] \Rightarrow \dfrac{x}{2} \times 12 - \dfrac{3x}{4} \times 12 + \dfrac{x}{6} \times 12 = - 8 \\[1em] \Rightarrow 6x - 9x + 2x = - 8 \\[1em] \Rightarrow - x = - 8 \\[1em] \Rightarrow x = 8$

Hence, $\bm{x = 8}$.

Solve the following equation:

$5x - 3(4x - 3) = 2(4 - 5x) + 10$

Answer

Solving,

$\Rightarrow 5x - 3(4x - 3) = 2(4 - 5x) + 10 \\[1em] \Rightarrow 5x - 12x + 9 = 8 - 10x + 10 \\[1em] \Rightarrow - 7x + 9 = 18 - 10x \\[1em] \Rightarrow - 7x + 10x = 18 - 9 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = \dfrac{9}{3} \\[1em] \Rightarrow x = 3$

Hence, $\bm{x = 3}$.

Solve the following equation:

$\dfrac{3x + 2}{5x + 4} = \dfrac{3}{4}$

Answer

$\dfrac{3x + 2}{5x + 4} = \dfrac{3}{4}$

By cross multiplication, we get

$\Rightarrow 4(3x + 2) = 3(5x + 4) \\[1em] \Rightarrow 12x + 8 = 15x + 12 \\[1em] \Rightarrow 8 - 12 = 15x - 12x \\[1em] \Rightarrow 3x = - 4 \\[1em] \Rightarrow x = - \dfrac{4}{3}$

Hence, $\bm{x = -\dfrac{4}{3}}$.

Solve the following equation:

$\dfrac{2x - 3}{5x - 4} = \dfrac{2x + 1}{5x + 20}$

Answer

$\dfrac{2x - 3}{5x - 4} = \dfrac{2x + 1}{5x + 20}$

By cross multiplication, we get

$\Rightarrow (2x - 3)(5x + 20) = (2x + 1)(5x - 4) \\[1em] \Rightarrow 10x^{2} + 40x - 15x - 60 = 10x^{2} - 8x + 5x - 4 \\[1em] \Rightarrow 10x^{2} + 25x - 60 = 10x^{2} - 3x - 4 \\[1em] \Rightarrow 25x + 3x = - 4 + 60 \\[1em] \Rightarrow 28x = 56 \\[1em] \Rightarrow x = \dfrac{56}{28} \\[1em] \Rightarrow x = 2$

Hence, $\bm{x = 2}$.

Solve the following equation:

$\dfrac{2x + 5}{2} - \dfrac{5x}{x - 1} = x$

Answer

$\dfrac{2x + 5}{2} - \dfrac{5x}{x - 1} = x$

L.C.M. of 2 and $(x - 1)$ = $2(x - 1)$

Multiplying both sides by $2(x - 1)$, we get:

$\Rightarrow \Big[\dfrac{(2x + 5)}{2} - \dfrac{5x}{(x - 1)}\Big] \times 2(x - 1) = x \times 2(x - 1) \\[1em] \Rightarrow \dfrac{(2x + 5)}{2} \times 2(x - 1) - \dfrac{5x}{(x - 1)} \times 2(x - 1) = 2x(x - 1) \\[1em] \Rightarrow (x - 1)(2x + 5) - 10x = 2x(x - 1) \\[1em] \Rightarrow 2x^{2} + 5x - 2x - 5 - 10x = 2x^{2} - 2x \\[1em] \Rightarrow 2x^{2} - 7x - 5 = 2x^{2} - 2x \\[1em] \Rightarrow - 7x + 2x = 5 \\[1em] \Rightarrow - 5x = 5 \\[1em] \Rightarrow x = - \dfrac{5}{5} \\[1em] \Rightarrow x = - 1$

Hence, $\bm{x = -1}$.

The sum of three consecutive multiples of 11 is 363. Find these multiples.

Answer

Let the three consecutive multiples of 11 be $11x, 11x + 11$ and $11x + 22$.

According to the given condition,

$\Rightarrow 11x + (11x + 11) + (11x + 22) = 363 \\[1em] \Rightarrow 33x + 33 = 363 \\[1em] \Rightarrow 33x = 363 - 33 \\[1em] \Rightarrow 33x = 330 \\[1em] \Rightarrow x = \dfrac{330}{33} \\[1em] \Rightarrow x = 10$

So, the multiples are 11 × 10 = 110, 110 + 11 = 121 and 110 + 22 = 132.

Hence, the required multiples are 110, 121 and 132.

Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Answer

Let the smaller number be $x$.

As one number exceeds the other by 15, the greater number = $x + 15$

According to the given condition,

$\Rightarrow x + (x + 15) = 95 \\[1em] \Rightarrow 2x + 15 = 95 \\[1em] \Rightarrow 2x = 95 - 15 \\[1em] \Rightarrow 2x = 80 \\[1em] \Rightarrow x = \dfrac{80}{2} \\[1em] \Rightarrow x = 40$

So, the greater number = 40 + 15 = 55

Hence, the required numbers are 55 and 40.

One-half of a number is equal to one-third of its succeeding number. Find the first number.

Answer

Let the first number be $x$.

Then, its succeeding number = $x + 1$

According to the given condition,

⇒ $\dfrac{1}{2}x = \dfrac{1}{3}(x + 1)$

By cross multiplication, we get

$\Rightarrow 3x = 2(x + 1) \\[1em] \Rightarrow 3x = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 \\[1em] \Rightarrow x = 2$

Hence, the first number = 2.

The numerator of a rational number is 8 less than its denominator. If the numerator is increased by 2 and denominator is decreased by 1, the number obtained is $\dfrac{1}{2}$. Find the number.

Answer

Let the denominator of the rational number be $x$.

Then, its numerator = $x - 8$

So, the rational number = $\dfrac{x - 8}{x}$

On increasing the numerator by 2 and decreasing the denominator by 1, we get $\dfrac{x - 8 + 2}{x - 1} = \dfrac{x - 6}{x - 1}$

According to the given condition,

⇒ $\dfrac{x - 6}{x - 1} = \dfrac{1}{2}$

By cross multiplication, we get

$\Rightarrow 2(x - 6) = x - 1 \\[1em] \Rightarrow 2x - 12 = x - 1 \\[1em] \Rightarrow 2x - x = - 1 + 12 \\[1em] \Rightarrow x = 11$

So, the numerator = 11 - 8 = 3

Hence, the required rational number = $\mathbf{\dfrac{3}{11}}$.

The present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years later the sum of their ages will be 54 years. What are their present ages?

Answer

Let the present ages of Rohit and Mayank be $11x$ years and $8x$ years respectively.

After 8 years, Rohit's age = $(11x + 8)$ years and Mayank's age = $(8x + 8)$ years

According to the given condition,

$\Rightarrow (11x + 8) + (8x + 8) = 54 \\[1em] \Rightarrow 19x + 16 = 54 \\[1em] \Rightarrow 19x = 54 - 16 \\[1em] \Rightarrow 19x = 38 \\[1em] \Rightarrow x = \dfrac{38}{19} \\[1em] \Rightarrow x = 2$

So, Rohit's present age = 11 × 2 = 22 years

and Mayank's present age = 8 × 2 = 16 years

Hence, the present ages of Rohit and Mayank are 22 years and 16 years respectively.

A father's age is 3 times the sum of ages of his two sons. Five years later he will be twice the sum of ages of his two sons. Find the present age of the father.

Answer

Let the sum of the present ages of the two sons be $x$ years.

Then, the present age of the father = $3x$ years

Five years later, each son will be 5 years older, so the sum of the ages of the two sons = $(x + 10)$ years

and the age of the father = $(3x + 5)$ years

According to the given condition,

$\Rightarrow 3x + 5 = 2(x + 10) \\[1em] \Rightarrow 3x + 5 = 2x + 20 \\[1em] \Rightarrow 3x - 2x = 20 - 5 \\[1em] \Rightarrow x = 15$

So, the present age of the father = 3 × 15 = 45 years

Hence, the present age of the father = 45 years.

The digits of a two digit number differ by 7. If the digits are interchanged and the resulting number is added to the original number we get 121. Find the original number.

Answer

Let the digit at unit's place be $x$ and the digit at ten's place be $y$.

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

According to the given condition,

$\Rightarrow (10y + x) + (10x + y) = 121 \\[1em] \Rightarrow 11x + 11y = 121 \\[1em] \Rightarrow 11(x + y) = 121 \\[1em] \Rightarrow x + y = \dfrac{121}{11} \\[1em] \Rightarrow x + y = 11 \quad \text{...(1)}$

Also given, the digits differ by 7.

Case 1 : Let $x - y = 7$ ...(2)

Adding equations (1) and (2), we get

$\Rightarrow 2x = 18 \\[1em] \Rightarrow x = 9$

Substituting $x = 9$ in equation (1), we get $y = 11 - 9 = 2$

So, the number = 10 × 2 + 9 = 29

Case 2 : Let $y - x = 7$ ...(3)

Adding equations (1) and (3), we get

$\Rightarrow 2y = 18 \\[1em] \Rightarrow y = 9$

Substituting $y = 9$ in equation (1), we get $x = 11 - 9 = 2$

So, the number = 10 × 9 + 2 = 92

Hence, the original number is 29 or 92.

The ten's digit of a two digit number exceeds its unit's digit by 5. When digits are reversed, the new number added to the original number becomes 99. Find the original number.

Answer

Let the digit at unit's place be $x$.

As the ten's digit exceeds the unit's digit by 5, the digit at ten's place = $x + 5$

So, the original number = $10(x + 5) + x$

On reversing the digits, the new number = $10x + (x + 5)$

According to the given condition,

$\Rightarrow [10(x + 5) + x] + [10x + (x + 5)] = 99 \\[1em] \Rightarrow 10x + 50 + x + 10x + x + 5 = 99 \\[1em] \Rightarrow 22x + 55 = 99 \\[1em] \Rightarrow 22x = 99 - 55 \\[1em] \Rightarrow 22x = 44 \\[1em] \Rightarrow x = \dfrac{44}{22} \\[1em] \Rightarrow x = 2$

So, the digit at ten's place = 2 + 5 = 7

and the original number = 10 × 7 + 2 = 72

Hence, the original number = 72.

Sonia withdrew ₹ 80,000 from a bank. She asked the cashier to give her ₹ 500 and ₹ 200 currency notes in return. She got 250 currency notes in all. Find the number of each kind of currency notes.

Answer

Let the number of ₹ 500 notes be $x$.

As she got 250 currency notes in all, the number of ₹ 200 notes = $250 - x$

Value of ₹ 500 notes = ₹ $500x$

Value of ₹ 200 notes = ₹ $200(250 - x)$

According to the given condition,

$\Rightarrow 500x + 200(250 - x) = 80000 \\[1em] \Rightarrow 500x + 50000 - 200x = 80000 \\[1em] \Rightarrow 300x = 80000 - 50000 \\[1em] \Rightarrow 300x = 30000 \\[1em] \Rightarrow x = \dfrac{30000}{300} \\[1em] \Rightarrow x = 100$

So, the number of ₹ 200 notes = 250 - 100 = 150

Hence, Sonia got 100 notes of ₹ 500 and 150 notes of ₹ 200.

Ajay covers a distance of 240 km in $4\dfrac{1}{4}$ hours. Some part of the journey was covered at the speed of 45 km/h and the remaining at 60 km/h. Find the distance covered by him at the rate of 60 km/h.

Answer

Let the distance covered at the rate of 60 km/h be $x$ km.

Then, the distance covered at the rate of 45 km/h = $(240 - x)$ km

We know that,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken to cover $(240 - x)$ km at 45 km/h = $\dfrac{240 - x}{45}$ hours

Time taken to cover $x$ km at 60 km/h = $\dfrac{x}{60}$ hours

Total time = $4\dfrac{1}{4}$ hours = $\dfrac{17}{4}$ hours

According to the given condition,

⇒ $\dfrac{240 - x}{45} + \dfrac{x}{60} = \dfrac{17}{4}$

L.C.M. of 45, 60 and 4 = 180

Multiplying both sides by 180, we get:

$\Rightarrow \Big[\dfrac{(240 - x)}{45} + \dfrac{x}{60}\Big] \times 180 = \dfrac{17}{4} \times 180 \\[1em] \Rightarrow \dfrac{(240 - x)}{45} \times 180 + \dfrac{x}{60} \times 180 = 765 \\[1em] \Rightarrow 4(240 - x) + 3x = 765 \\[1em] \Rightarrow 960 - 4x + 3x = 765 \\[1em] \Rightarrow 960 - x = 765 \\[1em] \Rightarrow x = 960 - 765 \\[1em] \Rightarrow x = 195$

Hence, the distance covered at the rate of 60 km/h = 195 km.

If $x$ ∈ {even integers}, represent the solution set of the inequation $-5 \le x \lt 5$ on a number line.

Answer

$- 5 \le x \lt 5$

∵ $x$ is an even integer.

The even integers which are greater than or equal to - 5 and less than 5 are - 4, - 2, 0, 2 and 4.

Hence, solution set = {-4, -2, 0, 2, 4}.

The solution set is shown by thick dots on the number line.

Solve the following inequality and graph its solution on a number line:

$-\dfrac{1}{4} \le \dfrac{1}{2} - \dfrac{x}{3} \lt 2, x \in I$

Answer

$-\dfrac{1}{4} \le \dfrac{1}{2} - \dfrac{x}{3} \lt 2$

Subtracting $\dfrac{1}{2}$ from each term, we get

$\Rightarrow -\dfrac{1}{4} - \dfrac{1}{2} \le - \dfrac{x}{3} \lt 2 - \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{-1 - 2}{4} \le - \dfrac{x}{3} \lt \dfrac{4 - 1}{2} \\[1em] \Rightarrow -\dfrac{3}{4} \le - \dfrac{x}{3} \lt \dfrac{3}{2}$

Multiplying each term by -3 and reversing the symbol of inequality, we get

$\Rightarrow \dfrac{9}{4} \ge x \gt - \dfrac{9}{2} \\[1em] \Rightarrow - \dfrac{9}{2} \lt x \le \dfrac{9}{4} \\[1em] \Rightarrow - 4.5 \lt x \le 2.25$

∵ $x$ is an integer.

Hence, solution set = {-4, -3, -2, -1, 0, 1, 2}.

The solution set is shown by thick dots on the number line.

A man covers a distance of 24 km in $3\dfrac{1}{2}$ hours partly on foot at the speed of 4.5 km/h and partly on bicycle at the speed of 10 km/h. Find the distance covered on foot.

Answer

Let the distance covered on foot be $x$ km.

Then, the distance covered on bicycle = $(24 - x)$ km

We know that,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken to cover $x$ km on foot = $\dfrac{x}{4.5}$ hours = $\dfrac{2x}{9}$ hours

Time taken to cover $(24 - x)$ km on bicycle = $\dfrac{24 - x}{10}$ hours

Total time = $3\dfrac{1}{2}$ hours = $\dfrac{7}{2}$ hours

According to the given condition,

⇒ $\dfrac{2x}{9} + \dfrac{24 - x}{10} = \dfrac{7}{2}$

L.C.M. of 9, 10 and 2 = 90

Multiplying both sides by 90, we get:

$\Rightarrow \Big[\dfrac{2x}{9} + \dfrac{(24 - x)}{10}\Big] \times 90 = \dfrac{7}{2} \times 90 \\[1em] \Rightarrow \dfrac{2x}{9} \times 90 + \dfrac{(24 - x)}{10} \times 90 = 315 \\[1em] \Rightarrow 20x + 9(24 - x) = 315 \\[1em] \Rightarrow 20x + 216 - 9x = 315 \\[1em] \Rightarrow 11x = 315 - 216 \\[1em] \Rightarrow 11x = 99 \\[1em] \Rightarrow x = \dfrac{99}{11} \\[1em] \Rightarrow x = 9$

Hence, the distance covered on foot = 9 km.

A person preparing a medicine wants to convert 15% alcohol solution into 32% alcohol solution. Find how much pure alcohol he should mix in 400 mL of 15% alcohol solution to obtain required solution?

Answer

Let the quantity of pure alcohol to be mixed be $x$ mL.

Quantity of alcohol in 400 mL of 15% alcohol solution = $\dfrac{15}{100} \times 400 = 60$ mL

On mixing $x$ mL of pure alcohol,

quantity of alcohol in the new solution = $(60 + x)$ mL

and quantity of the new solution = $(400 + x)$ mL

As the new solution is a 32% alcohol solution,

⇒ $\dfrac{60 + x}{400 + x} = \dfrac{32}{100}$

By cross multiplication, we get

$\Rightarrow 100(60 + x) = 32(400 + x) \\[1em] \Rightarrow 6000 + 100x = 12800 + 32x \\[1em] \Rightarrow 100x - 32x = 12800 - 6000 \\[1em] \Rightarrow 68x = 6800 \\[1em] \Rightarrow x = \dfrac{6800}{68} \\[1em] \Rightarrow x = 100$

Hence, 100 mL of pure alcohol should be mixed.

Rahul covers a distance from P to Q on bicycle at 10 km/h and returns back at 9 km/h. Anuj covers the distance from P to Q and Q to P both at 12 km/h. On calculation we find that Anuj took 10 minutes less than Rahul. Find the distance between P and Q.

Answer

Let the distance between P and Q be $x$ km.

We know that,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken by Rahul = $\dfrac{x}{10} + \dfrac{x}{9}$ hours

Time taken by Anuj = $\dfrac{x}{12} + \dfrac{x}{12} = \dfrac{x}{6}$ hours

Given, Anuj took 10 minutes less than Rahul.

10 minutes = $\dfrac{10}{60}$ hours = $\dfrac{1}{6}$ hours

According to the given condition,

⇒ $\dfrac{x}{10} + \dfrac{x}{9} - \dfrac{x}{6} = \dfrac{1}{6}$

L.C.M. of 10, 9 and 6 = 90

Multiplying both sides by 90, we get:

$\Rightarrow \Big[\dfrac{x}{10} + \dfrac{x}{9} - \dfrac{x}{6}\Big] \times 90 = \dfrac{1}{6} \times 90 \\[1em] \Rightarrow \dfrac{x}{10} \times 90 + \dfrac{x}{9} \times 90 - \dfrac{x}{6} \times 90 = 15 \\[1em] \Rightarrow 9x + 10x - 15x = 15 \\[1em] \Rightarrow 4x = 15 \\[1em] \Rightarrow x = \dfrac{15}{4} \\[1em] \Rightarrow x = 3.75$

Hence, the distance between P and Q = 3.75 km.

Solve: $\dfrac{x - b - c}{a} + \dfrac{x - c - a}{b} + \dfrac{x - a - b}{c} = 3$

Answer

$\dfrac{x - b - c}{a} + \dfrac{x - c - a}{b} + \dfrac{x - a - b}{c} = 3$

Subtracting 3 from both the sides, we get

$\Rightarrow \dfrac{x - b - c}{a} + \dfrac{x - c - a}{b} + \dfrac{x - a - b}{c} - 3 = 3 - 3 \\[1em] \Rightarrow \left(\dfrac{x - b - c}{a} - 1\right) + \left(\dfrac{x - c - a}{b} - 1\right) + \left(\dfrac{x - a - b}{c} - 1\right) = 3 - 3 \\[1em] \Rightarrow \dfrac{x - b - c - a}{a} + \dfrac{x - c - a - b}{b} + \dfrac{x - a - b - c}{c} = 0 \\[1em] \Rightarrow (x - a - b - c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = 0$

As $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ne 0$, we get

$\Rightarrow x - a - b - c = 0 \\[1em] \Rightarrow x = a + b + c$

Hence, $\bm{x = a + b + c}$.