Linear Equations - Exercise 14(B)

Three-sevenths of a number is 12. Find the number.

Answer

Let the required number be x.

Then, three-sevenths of this number = 12

$\therefore \dfrac{3}{7}x = 12 \\[1em] 3x = 12 \times 7 \\[1em] 3x = 84 \\[1em] x = \dfrac{84}{3} \\[1em] = 28$

Hence, the required number is 28.

A number increased by 9 gives 43. Find the number.

Answer

Let the required number be x.

Then, the number increased by 9 = 43

∴ x + 9 = 43

⇒ x = 43 - 9

⇒ x = 34

Hence, the required number is 34.

A number diminished by 11 gives 57. Find the number.

Answer

Let the required number be x.

Then, the number diminished by 11 = 57

∴ x - 11 = 57

⇒ x = 57 + 11

⇒ x = 68

Hence, the required number is 68.

Thrice a number increased by 6 equals 39. Find the number.

Answer

Let the required number be x.

Then, thrice the number (3x) increased by 6 = 39

∴ 3x + 6 = 39

⇒ 3x = 39 - 6

⇒ 3x = 33

⇒ x = $\dfrac{33}{3}$

⇒ x = 11

Hence, the required number is 11.

Three-fourths of a number exceeds its one-third by 15. Find the number.

Answer

Let the required number be x.

Three-fourths of the number = $\dfrac{3}{4}x$

One-third of the number = $\dfrac{1}{3}x$

According to the question, the difference between them is 15.

$\therefore \dfrac{3}{4}x - \dfrac{1}{3}x = 15 \\[1em] \Rightarrow \dfrac{9x - 4x}{12} = 15 \\[1em] \Rightarrow \dfrac{5x}{12} = 15 \\[1em] \Rightarrow 5x = 15 \times 12 \\[1em] \Rightarrow 5x = 180 \\[1em] \Rightarrow x = \dfrac{180}{5} \\[1em] \Rightarrow x = 36$

Hence, the required number is 36.

A number when divided by 4 is reduced by 21. Find the number.

Answer

Let the required number be x.

When the number is divided by 4, the result is equal to 21 less than the original number.

$\therefore \dfrac{x}{4} = x - 21 \\[1em] \Rightarrow x = 4(x - 21) \\[1em] \Rightarrow x = 4x - 84 \\[1em] \Rightarrow 84 = 4x - x \quad \text{[Transposing +x to RHS]} \\[1em] \Rightarrow 84 = 3x \\[1em] \Rightarrow x = \dfrac{84}{3} \\[1em] \Rightarrow x = 28$

Hence, the required number is 28.

A number is as much greater than 36 as is less than 86. Find the number.

Answer

Let the required number be x.

"As much greater than 36" means: (x - 36)

"As is less than 86" means: (86 - x)

Since these two differences are equal:

∴ x - 36 = 86 - x

⇒ x + x = 86 + 36 $\quad$ [Transposing -36 to RHS and -x to LHS]

⇒ 2x = 122

⇒ x = $\dfrac{122}{2}$

⇒ x = 61

Hence, the required number is 61.

A number exceeds its four-sevenths by 18. Find the number.

Answer

Let the required number be x.

Four-sevenths of the number = $\dfrac{4}{7}x$

According to the given condition we have:

$\phantom{=} x - \dfrac{4}{7}x = 18 \\[1em] \Rightarrow \dfrac{7x - 4x}{7} = 18 \\[1em] \Rightarrow \dfrac{3x}{7} = 18 \\[1em] \Rightarrow 3x = 18 \times 7 \\[1em] \Rightarrow 3x = 126 \\[1em] \Rightarrow x = \dfrac{126}{3} \\[1em] \Rightarrow x = 42$

Hence, the required number is 42.

A number exceeds 20% of itself by 40. Find the number.

Answer

Let the required number be x.

20% of the number = $\dfrac{20}{100}x = \dfrac{1}{5}x$

According to the question x exceeds 20% of itself by 40:

$\therefore x - \dfrac{1}{5}x = 40 \\[1em] \Rightarrow \dfrac{5x - x}{5} = 40 \\[1em] \Rightarrow \dfrac{4x}{5} = 40 \\[1em] \Rightarrow 4x = 40 \times 5 \\[1em] \Rightarrow 4x = 200 \\[1em] \Rightarrow x = \dfrac{200}{4} \\[1em] \Rightarrow x = 50$

Hence, the required number is 50.

If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.

Answer

Let the required number be x.

"10 added to four times the number" = 4x + 10

"5 less than five times the number" = 5x - 5

According to the given condition we have:

4x + 10 = 5x - 5

⇒ 4x - 5x = - 5 - 10 $\quad$ [Transposing +10 to RHS and +5x to LHS]

⇒ -x = -15

⇒ x = 15

Hence, the required number is 15.

One fourth of a number is increased by 7 and the result is multiplied by 3. Thus, we obtain 36. Find the number.

Answer

Let the required number be x.

One-fourth of the number increased by 7 is $\Big(\dfrac{x}{4} + 7\Big)$.

According to the given condition we have:

$\phantom{=} 3 \times \Big(\dfrac{x}{4} + 7\Big) = 36 \\[1em] \Rightarrow \dfrac{x}{4} + 7 = \dfrac{36}{3} \\[1em] \Rightarrow \dfrac{x}{4} + 7 = 12 \\[1em] \Rightarrow \dfrac{x}{4} = 12 - 7 \quad \text{[Transposing +7 to RHS]} \\[1em] \Rightarrow \dfrac{x}{4} = 5 \\[1em] \Rightarrow x = 5 \times 4 \\[1em] \Rightarrow x = 20$

Hence, the required number is 20.

The sum of two consecutive odd numbers is 56. Find the numbers.

Answer

Let the two consecutive odd numbers be x and x + 2.

The sum of two consecutive odd numbers is 56:

∴ x + (x + 2) = 56

⇒ 2x + 2 = 56

⇒ 2x = 56 - 2 $\quad$ [Transposing +2 to RHS]

⇒ 2x = 54

⇒ x = $\dfrac{54}{2}$

⇒ x = 27

The numbers are 27 and 27 + 2 = 29.

Hence, the required numbers are 27 and 29.

The sum of three consecutive even numbers is 48. Find the numbers.

Answer

Let the three consecutive even numbers be x, x + 2, and x + 4.

The sum of three consecutive even numbers is 48:

∴ x + (x + 2) + (x + 4) = 48

⇒ 3x + 6 = 48

⇒ 3x = 48 - 6 $\quad$ [Transposing +6 to RHS]

⇒ 3x = 42

⇒ x = $\dfrac{42}{3}$

⇒ x = 14

The numbers are 14, 14 + 2 = 16, and 14 + 4 = 18.

Hence, the required numbers are 14, 16 and 18.

One of the two numbers exceeds the other by 9. Four times the smaller added to five times the larger gives 108. Find the numbers.

Answer

Let the smaller number be x.

Then the larger number = x + 9.

Four times the smaller added to five times the larger gives 108:

∴ 4(x) + 5(x + 9) = 108

⇒ 4x + 5x + 45 = 108

⇒ 9x + 45 = 108

⇒ 9x = 108 - 45 $\quad$ [Transposing +45 to RHS]

⇒ 9x = 63

⇒ x = $\dfrac{63}{9}$

⇒ x = 7

Smaller number = 7, Larger number = 7 + 9 = 16.

Hence, the required numbers are 16 and 7.

In a class of 40 students, the number of girls is three-fifths of the number of boys. Find the number of boys in the class.

Answer

Let the number of boys be x.

Then the number of girls = $\dfrac{3}{5}x$.

Total students = 40.

Boys + Girls = Total students

$\therefore x + \dfrac{3}{5}x = 40 \\[1em] \Rightarrow \dfrac{5x + 3x}{5} = 40 \\[1em] \Rightarrow \dfrac{8x}{5} = 40 \\[1em] \Rightarrow 8x = 40 \times 5 \\[1em] \Rightarrow 8x = 200 \\[1em] \Rightarrow x = \dfrac{200}{8} \\[1em] \Rightarrow x = 25$

Hence, the number of boys in the class is 25.

The length of a rectangular park is three times its breadth. If the perimeter of the park is 192 metres, find the dimensions of the park.

Answer

Let the breadth of the rectangular park be x metres.

Then, the length of the park = 3x metres.

Perimeter of the park = 192 metres

We know the formula:

Perimeter of a rectangle = 2(Length + Breadth)

∴ 192 = 2(3x + x)

⇒ 192 = 6x + 2x

⇒ 192 = 8x

⇒ x = $\dfrac{192}{8}$

⇒ x = 24

∴ Breadth = x = 24 m

And

Length = 3x = 3 × 24 m = 72 m

Hence, the dimensions of the park are : Length = 72 m and Breadth = 24 m.

Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.

Answer

Let the length of the third side be x metres.

Then, each of the two equal sides = (2x - 5) metres.

Perimeter of the triangle = 55 metres.

We know the formula:

Perimeter of a triangle = Sum of all three sides

∴ 55 = (2x - 5) + (2x - 5) + x

⇒ 55 = 2x + 2x + x - 5 - 5

⇒ 55 = 5x - 10

⇒ 55 + 10 = 5x $\quad$ [Transposing -10 to LHS]

⇒ 65 = 5x

⇒ x = $\dfrac{65}{5}$

⇒ x = 13

∴ Third side = x = 13 m

Each equal side = (2x - 5) m = (2 x 13 - 5) m = (26 - 5) m= 21 m

Hence, the lengths of the sides of the triangle are 13 m, 21 m and 21 m.

Two supplementary angles differ by 44°. Find the angles.

Answer

Let the required angles be x° and (x + 44)°.

Since the sum of supplementary angles is 180°, we have:

x + (x + 44) = 180

⇒ 2x + 44 = 180

⇒ 2x = 180 - 44 $\quad$ [Transposing +44 to RHS]

⇒ 2x = 136

⇒ x = $\dfrac{136}{2}$

⇒ x = 68

⇒ (x + 44)° = (68 + 44)° = 112°.

Hence, the required angles are 68° and 112°.

The total cost of 3 tables and 2 chairs is ₹ 8745. If a table costs ₹ 40 more than a chair, find the price of each.

Answer

Let the price of a chair be ₹ x.

Then, the price of a table = ₹ (x + 40).

Total cost = Cost of 3 tables + Cost of 2 chairs

Substituting the values in above, we get:

₹ 8745 = 3(x + 40) + 2x

⇒ ₹ 8745 = 3x + 120 + 2x

⇒ ₹ 8745 = 5x + 120

⇒ ₹ 8745 - 120 = 5x $\quad$ [Transposing +120 to LHS]

⇒ ₹ 8625 = 5x

⇒ x = ₹ $\dfrac{8625}{5}$

⇒ x = ₹ 1725

Price of a chair = ₹ x = ₹ 1725.

Price of a table = ₹ (x + 40) = ₹ (1725 + 40) = ₹ 1765.

Hence, the price of a chair is ₹ 1725 and price of a table is ₹ 1765.

The denominator of a fraction is 3 more than the numerator. If 2 is added to the numerator and 5 is added to the denominator, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Answer

Let the numerator of the required fraction be x.

Then, its denominator = (x + 3).

The fraction is $\dfrac{x}{x + 3}$.

According to the question, if 2 is added to the numerator and 5 is added to the denominator, the fraction becomes $\dfrac{1}{2}$.

$\therefore \dfrac{x + 2}{(x + 3) + 5} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x + 2}{x + 8} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x + 2) = 1(x + 8) \\[1em] \text{[By cross-multiplication]} \\[1em] \Rightarrow 2x + 4 = x + 8 \\[1em] \Rightarrow 2x - x = 8 - 4 \\[1em] \text{[Transposing +4 to RHS and +x to LHS]} \\[1em] \Rightarrow x = 4$

Numerator = x = 4

Denominator = (x + 3) = 4 + 3 = 7.

Hence, the required fraction is $\dfrac{4}{7}$.

A man is twice as old as his son. 20 years ago, the age of the man was 12 times the age of the son. Find their present ages.

Answer

Let the son's present age be x years.

Then, the man's present age = 2x years.

20 years ago:

Son's age = (x - 20) years.

Man's age = (2x - 20) years.

According to the question, 20 years ago, the age of the man was 12 times the age of the son:

∴ 2x - 20 = 12(x - 20)

⇒ 2x - 20 = 12x - 240

⇒ 2x - 12x = -240 + 20 $\quad$ [Transposing -20 to RHS and +12x to LHS]

⇒ -10x = -220

⇒ x = $\dfrac{-220}{-10}$

⇒ x = 22

Son's present age = x = 22 years.

Man's present age = 2x years = (2 × 22) years = 44 years.

Hence, the present age of the man is 44 years and his son's age is 22 years.

A man is 28 years older than his son. After 10 years, he will be thrice as old as his son. Find their present ages.

Answer

Let the son's present age be x years.

Then, the man's present age = (x + 28) years.

After 10 years:

Son's age = (x + 10) years.

Man's age = (x + 28 + 10) years = (x + 38) years.

According to the question, after 10 years, the man will be thrice as old as his son:

∴ x + 38 = 3(x + 10)

⇒ x + 38 = 3x + 30

⇒ x - 3x = 30 - 38 $\quad$ [Transposing +38 to RHS and +3x to LHS]

⇒ -2x = -8

⇒ x = $\dfrac{-8}{-2}$

⇒ x = 4

Son's present age = x years = 4 years.

Man's present age = (x + 28) years = (4 + 28) years = 32 years.

Hence, the man's present age is 32 years and his son's age is 4 years.

Sunita is 24 years older than her daughter Kavita. 6 years ago, Sunita was thrice as old as Kavita. Find their present ages.

Answer

Let Kavita's present age be x years.

Then, Sunita's present age = (x + 24) years.

6 years ago:

Kavita's age = (x - 6) years.

Sunita's age = (x + 24 - 6) years = (x + 18) years.

According to the question, 6 years ago, Sunita was thrice as old as Kavita:

∴ x + 18 = 3(x - 6)

⇒ x + 18 = 3x - 18

⇒ 18 + 18 = 3x - x $\quad$ [Transposing +x to RHS and -18 to LHS]

⇒ 36 = 2x

⇒ x = $\dfrac{36}{2}$

⇒ x = 18

Kavita's present age = x years = 18 years.

Sunita's present age = (x + 24) years = (18 + 24) years = 42 years.

Hence, Sunita's present age is 42 years and Kavita's present age is 18 years.

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer

Let the first part be x.

Then, the second part = (184 - x).

According to the question, one-third of the first part exceeds one-seventh of the second part by 8:

$\therefore \dfrac{1}{3}x - \dfrac{1}{7}(184 - x) = 8 \\[1em] \Rightarrow \dfrac{7x - 3(184 - x)}{21} = 8 \\[1em] \Rightarrow 7x - 552 + 3x = 8 \times 21 \\[1em] \Rightarrow 10x - 552 = 168 \\[1em] \Rightarrow 10x = 168 + 552 \\[1em] \text{[Transposing -552 to RHS]} \\[1em] \Rightarrow 10x = 720 \\[1em] \Rightarrow x = \dfrac{720}{10} \\[1em] \Rightarrow x = 72$

First part = x = 72.

Second part = (184 - x) = (184 - 72) = 112.

Hence, the two parts are 72 and 112.

A sum of ₹ 500 is in the form of denominations of ₹ 5 and ₹ 10. If the total number of notes is 90, find the number of notes of each type.

Answer

Let the number of ₹ 5 notes be x.

Then, the number of ₹ 10 notes = (90 - x).

Total value of ₹ 5 notes = 5x

Total value of ₹ 10 notes = 10 × (90 - x) = 900 - 10x

According to the question, the total sum is ₹ 500:

∴ 5x + (900 - 10x) = 500

⇒ 5x + 900 - 10x = 500 $\quad$ [Removing brackets]

⇒ 5x - 10x = 500 - 900 $\quad$ [Transposing +900 to RHS]

⇒ -5x = -400

⇒ x = $\dfrac{-400}{-5}$

⇒ x = 80

Number of ₹ 5 notes = x = 80.

Number of ₹ 10 notes = (90 - x) = (90 - 80) = 10.

Hence, number of ₹ 5 notes = 80 and number of ₹ 10 notes = 10.

There are some 50 paisa and some 25 paisa coins in a bag. If the total number of coins is 30 and their total value is ₹ 11, find the number of coins of each kind.

Answer

First, convert the total value to paisa:

₹ 1 = 100 paisa

∴ ₹ 11 = (11 × 100) paisa = 1100 paisa.

Let the number of 50 paisa coins be x.

Then, the number of 25 paisa coins = (30 - x).

Value of 50 paisa coins = 50x

Value of 25 paisa coins = 25 × (30 - x) = 750 - 25x

According to the question, total value is ₹ 11:

∴ 50x + (750 - 25x) = 1100

⇒ 50x + 750 - 25x = 1100 $\quad$ [Removing brackets]

⇒ 25x = 1100 - 750 $\quad$ [Transposing +750 to RHS]

⇒ 25x = 350

⇒ x = $\dfrac{350}{25}$

⇒ x = 14

Number of 50 paisa coins = x = 14.

Number of 25 paisa coins = (30 - x) = (30 - 14) = 16.

Hence, the number of 50 paisa coins = 14 and the number of 25 paisa coins = 16.

A labourer is engaged for 20 days on the condition that he will receive ₹ 280 for each day he works and will be fined ₹ 60 for each day he is absent. If he receives ₹ 2540 in all, for how many days did he remain absent?

Answer

Let the number of days the labourer was absent be x.

Then, the number of days he worked = (20 - x).

Amount earned for working = 280 x (20 - x) = 5600 - 280x

Amount fined for being absent = 60 × x = 60x

Total amount received = Earnings - Fines

∴ (5600 - 280x) - 60x = 2540

Remove brackets:

⇒ 5600 - 280x - 60x = 2540

⇒ 5600 - 340x = 2540

Transpose -340x to RHS and +2540 to LHS:

⇒ 5600 - 2540 = 340x

⇒ 3060 = 340x

⇒ x = $\dfrac{3060}{340}$

⇒ x = 9

Hence, the labourer remained absent for 9 days.