Simple Linear Equations — Exercise 14(D)

One-fifth of a number is 5, find the number.

Answer

Given,

One-fifth of a number is 5.

Let the number be $x$.

$\therefore \dfrac{1}{5} \times x = 5 \\[1em] \Rightarrow \dfrac{x}{5} = 5 \\[1em] \Rightarrow x = 25$

Hence, the number is 25.

Six times a number is 72, find the number.

Answer

Given,

Six times a number is 72.

Let the number be $x$.

$\therefore 6x = 72 \\[1em] \Rightarrow x = \dfrac{72}{6} \\[1em] \Rightarrow x = 12$

Hence, the number is 12.

If 15 is added to a number, the result is 69. Find the number.

Answer

Given,

If 15 is added to a number, the result is 69.

Let the number be $x$.

$\therefore x + 15 = 69 \\[1em] \Rightarrow x = 69 - 15 \\[1em] \Rightarrow x = 54$

Hence, the number is 54.

The sum of twice a number and 4 is 80, find the number.

Answer

Given,

The sum of twice a number and 4 is 80.

Let the number be $x$.

$\therefore 2x + 4 = 80 \\[1em] \Rightarrow 2x = 80 - 4 \\[1em] \Rightarrow 2x = 76 \\[1em] \Rightarrow x = \dfrac{76}{2} = 38$

Hence, the number is 38.

The difference between a number and one-fourth of itself is 24, find the number.

Answer

Given,

The difference between a number and one-fourth of itself is 24.

Let the number be $x$.

$\therefore x - \dfrac{1}{4}x = 24 \\[1em] \Rightarrow \dfrac{4x - x}{4} = 24 \\[1em] \Rightarrow \dfrac{3x}{4} = 24 \\[1em] \Rightarrow 3x = 96 \\[1em] \Rightarrow x = 32$

Hence, the number is 32.

Find a number whose one-third part exceeds its one-fifth part by 20.

Answer

Given,

Find a number whose one-third part exceeds its one-fifth part by 20.

Let the number be $x$.

$\therefore \dfrac{x}{3} - \dfrac{x}{5} = 20 \\[1em] \Rightarrow \dfrac{5x - 3x}{15} = 20 \\[1em] \Rightarrow \dfrac{2x}{15} = 20 \\[1em] \Rightarrow 2x = 300 \\[1em] \Rightarrow x = 150$

Hence, the number is 150.

A number is as much greater than 35 as is less than 53. Find the number.

Answer

Given,

A number is as much greater than 35 as is less than 53.

Let the number be $x$.

$\therefore x - 35 = 53 - x \\[1em] \Rightarrow x + x = 53 + 35 \\[1em] \Rightarrow 2x = 88 \\[1em] \Rightarrow x = \dfrac{88}{2} = 44$

Hence, the number is 44.

The sum of two numbers is 18. If one is twice the other, find the numbers.

Answer

Given,

The sum of two numbers is 18. If one is twice the other.

Let the smaller number be $x$, then the other number = $2x$.

$\therefore x + 2x = 18 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6$

∴ Other number = $2x$ = 2 × 6 = 12

Hence, the numbers are 6 and 12.

A number is 15 more than the other. The sum of the two numbers is 195. Find the numbers.

Answer

Given,

A number is 15 more than the other. The sum of the two numbers is 195.

Let the smaller number be $x$, then the greater number = $x$ + 15.

$\therefore x + (x + 15) = 195 \\[1em] \Rightarrow 2x + 15 = 195 \\[1em] \Rightarrow 2x = 180 \\[1em] \Rightarrow x = 90$

∴ Greater number = $x$ + 15 = 90 + 15 = 105

Hence, the numbers are 90 and 105.

The sum of three consecutive even numbers is 54. Find the numbers.

Answer

Given,

The sum of three consecutive even numbers is 54.

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

$\therefore x + (x + 2) + (x + 4) = 54 \\[1em] \Rightarrow 3x + 6 = 54 \\[1em] \Rightarrow 3x = 48 \\[1em] \Rightarrow x = 16$

∴ The numbers are 16, 16 + 2 = 18 and 16 + 4 = 20.

Hence, the numbers are 16, 18 and 20.

The sum of three consecutive odd numbers is 63. Find the numbers.

Answer

Given,

The sum of three consecutive odd numbers is 63.

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

$\therefore x + (x + 2) + (x + 4) = 63 \\[1em] \Rightarrow 3x + 6 = 63 \\[1em] \Rightarrow 3x = 57 \\[1em] \Rightarrow x = 19$

∴ The numbers are 19, 19 + 2 = 21 and 19 + 4 = 23.

Hence, the numbers are 19, 21 and 23.

A man has ₹ $x$ from which he spends ₹ 6. If twice of the money left with him is ₹ 86, find $x$.

Answer

Given,

A man has ₹ $x$ from which he spends ₹ 6. If twice of the money left with him is ₹ 86.

Total money = ₹ $x$

He spends = ₹ 6

Money left with the man = ₹ ($x$ − 6)

$\therefore 2(x - 6) = 86 \\[1em] \Rightarrow x - 6 = 43 \\[1em] \Rightarrow x = 43 + 6 \\[1em] \Rightarrow x = 49$

Hence, $\bm{x}$ = 49.

A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time. Find their present ages.

Answer

Given,

A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time.

Let the present age of the son be $x$ years, then the present age of the man = $4x$ years.

After 20 years:

Age of son = ($x$ + 20) years

Age of man = ($4x$ + 20) years

$\therefore 4x + 20 = 2(x + 20) \\[1em] \Rightarrow 4x + 20 = 2x + 40 \\[1em] \Rightarrow 4x - 2x = 40 - 20 \\[1em] \Rightarrow 2x = 20 \\[1em] \Rightarrow x = 10$

∴ Present age of man = $4x$ = 4 × 10 = 40 years

Hence, the present age of the son is 10 years and that of the man is 40 years.

If 5 is subtracted from three times a number, the result is 16. Find the number.

Answer

Given,

If 5 is subtracted from three times a number, the result is 16.

Let the number be $x$.

$\therefore 3x - 5 = 16 \\[1em] \Rightarrow 3x = 16 + 5 \\[1em] \Rightarrow 3x = 21 \\[1em] \Rightarrow x = 7$

Hence, the number is 7.

Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.

Answer

Given,

Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.

Let the three consecutive natural numbers be $x, x + 1$ and $x + 2$.

$\therefore x + (x + 1) = (x + 2) + 15 \\[1em] \Rightarrow 2x + 1 = x + 17 \\[1em] \Rightarrow 2x - x = 17 - 1 \\[1em] \Rightarrow x = 16$

∴ The numbers are 16, 16 + 1 = 17 and 16 + 2 = 18.

Hence, the numbers are 16, 17 and 18.

The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

Answer

Given,

The difference between two numbers is 7. Six times the smaller plus the larger is 77.

Let the smaller number be $x$, then the larger number = $x$ + 7.

$\therefore 6x + (x + 7) = 77 \\[1em] \Rightarrow 7x + 7 = 77 \\[1em] \Rightarrow 7x = 70 \\[1em] \Rightarrow x = 10$

∴ Larger number = $x$ + 7 = 10 + 7 = 17

Hence, the numbers are 10 and 17.

The length of a rectangular plot exceeds its breadth by 5 metres. If the perimeter of the plot is 142 metres, find the length and the breadth of the plot.

Answer

Given,

The length of a rectangular plot exceeds its breadth by 5 metres. If the perimeter of the plot is 142 metres.

Let the breadth of the plot be $x$ metres, then the length = ($x$ + 5) metres.

Perimeter of rectangle = 2(length + breadth)

Perimeter = 142 meters

$\therefore 2[(x + 5) + x] = 142 \\[1em] \Rightarrow 2(2x + 5) = 142 \\[1em] \Rightarrow 2x + 5 = 71 \\[1em] \Rightarrow 2x = 66 \\[1em] \Rightarrow x = 33$

∴ Length = $x$ + 5 = 33 + 5 = 38 m

Hence, the length of the plot is 38 m and its breadth is 33 m.

The numerator of a fraction is four less than its denominator. If 1 is added to both the numerator and denominator, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Answer

Given,

The numerator of a fraction is four less than its denominator. If 1 is added to both the numerator and denominator, the fraction becomes $\dfrac{1}{2}$.

Let the denominator of the fraction be $x$, then the numerator = $x$ − 4.

∴ Fraction = $\dfrac{x - 4}{x}$

On adding 1 to both the numerator and the denominator,

$\therefore \dfrac{(x - 4) + 1}{x + 1} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x - 3}{x + 1} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x - 3) = x + 1 \\[1em] \Rightarrow 2x - 6 = x + 1 \\[1em] \Rightarrow x = 7$

∴ Numerator = $x$ − 4 = 7 − 4 = 3 and Denominator = 7

Hence, the fraction is $\dfrac{3}{7}$.

A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time. Find their present ages.

Answer

Given,

A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time.

Let the present age of the son be $x$ years, then the present age of the man = $3x$ years.

After 12 years:

Age of son = ($x$ + 12) years

Age of man = ($3x$ + 12) years

$\therefore 3x + 12 = 2(x + 12) \\[1em] \Rightarrow 3x + 12 = 2x + 24 \\[1em] \Rightarrow 3x - 2x = 24 - 12 \\[1em] \Rightarrow x = 12$

∴ Present age of man = $3x$ = 3 × 12 = 36 years

Hence, the present age of the son is 12 years and that of the man is 36 years.

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

Answer

Given,

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

Let the number of ₹ 5 notes be $x$, then the number of ₹ 10 notes = 90 − $x$.

Value of ₹ 5 notes = ₹ $5x$

Value of ₹ 10 notes = ₹ 10(90 − $x$)

$\therefore 5x + 10(90 - x) = 500 \\[1em] \Rightarrow 5x + 900 - 10x = 500 \\[1em] \Rightarrow -5x = 500 - 900 \\[1em] \Rightarrow -5x = -400 \\[1em] \Rightarrow x = 80$

∴ Number of ₹ 10 notes = 90 − 80 = 10

Hence, there are 80 notes of ₹ 5 each and 10 notes of ₹ 10 each.