DAV Class 8 Maths Chapter 9 Brain Teasers
Linear Equations in One Variable Brain Teasers
1. A. Tick (✓) the correct option.
(a)
If \( \dfrac{5x}{3} - 4 = \dfrac{2x}{5} \), then the numerical value of \( (2x - 7) \) is— \(
\begin{aligned}
(i)&\ \frac{19}{13} \\
(ii)&\ \frac{-13}{19} \\
(iii)&\ 0 \\
(iv)&\ \frac{13}{19} \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \frac{5x}{3} - 4 &= \frac{2x}{5} \\[6pt] \frac{5x}{3} - \frac{2x}{5} &= 4 \\[6pt] \frac{25x - 6x}{15} &= 4 \\[6pt] 19x &= 4 \times 15 \\[6pt] 19x &= 60 \\[6pt] x &= \frac{60}{19} \\ \\ \text{Numerical value of } & (2x - 7) \\[6pt] &= 2\left(\frac{60}{19}\right) - 7 \\[6pt] &= \frac{120}{19} - 7 \\[6pt] &= \frac{120 - 133}{19} \\[6pt] &= \frac{-13}{19} \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{\dfrac{-13}{19}} \)
(b)
A number 351 is divided into two parts in the ratio 2 : 7. The product of the numbers is— \(
\begin{aligned}
(i)&\ 20294 \\
(ii)&\ 21294 \\
(iii)&\ 25295 \\
(iv)&\ 31294 \\
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Let the two parts be } & 2x \text{ and } 7x \\[6pt] 2x + 7x &= 351 \\[6pt] 9x &= 351 \\[6pt] x &= \frac{351}{9} \\[6pt] x &= 39 \\ \\ \text{First part} &= 2 \times 39 = 78 \\ \text{Second part} &= 7 \times 39 = 273 \\ \\ \text{Product} &= 78 \times 273 \\ &= 21294 \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{21294} \)
(c)
Value of \( x \) for which \( \dfrac{2x - 3}{3x + 2} = \dfrac{-2}{3} \) is— \(
\begin{aligned}
(i)&\ \frac{5}{10} \\[6pt]
(ii)&\ \frac{5}{11} \\[6pt]
(iii)&\ \frac{11}{5} \\[6pt]
(iv)&\ \frac{5}{12} \\[6pt]
\end{aligned}
\)
Solution
\[ \begin{aligned} \frac{2x - 3}{3x + 2} &= \frac{-2}{3} \\[6pt] 3(2x - 3) &= -2(3x + 2) \\[6pt] 6x - 9 &= -6x - 4 \\[6pt] 6x + 6x &= -4 + 9 \\[6pt] 12x &= 5 \\[6pt] x &= \frac{5}{12} \end{aligned} \]Answer \( {\color{orange}(iv)}\ \color{red}{\dfrac{5}{12}} \)
(d)
Solution of a linear equation in one variable is always— \(
\begin{aligned}
(i)&\ \text{a natural number} \\
(ii)&\ \text{a whole number} \\
(iii)&\ \text{a real number} \\
(iv)&\ \text{an integer} \\
\end{aligned}
\)
Solution
A linear equation in one variable can have any real number as its solution (including fractions, decimals, integers, etc.).
Answer \( {\color{orange}(iii)}\ \color{red}{\text{a real number}} \)
(e)
The perimeter of a rectangle is numerically equal to its area. If the width of a rectangle is \( 2\dfrac{3}{4}
\) cm, then its length is— \(
\begin{aligned}
(i)&\ \frac{11}{3} \text{ cm} \\[6pt]
(ii)&\ \frac{22}{3} \text{ cm} \\[6pt]
(iii)&\ 11 \text{ cm} \\[6pt]
(iv)&\ 10 \text{ cm} \\[6pt]
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Let length} &= x \\[6pt] \text{Breadth } &= 2\frac{3}{4} \implies \frac{11}{4} \text{ cm} \\[6pt] \text{Perimeter} &= \text{Area} \\[6pt] 2(l + b) &= l \times b \\[6pt] 2\left(x + \frac{11}{4}\right) &= x \times \frac{11}{4} \\[6pt] 2x + \frac{11}{2} &= \frac{11x}{4} \\[6pt] \frac{11}{2} &= \frac{11x}{4} - 2x \\[6pt] \frac{11}{2} &= \frac{11x - 8x}{4} \\[6pt] \frac{11}{2} &= \frac{3x}{4} \\[6pt] 3x \times 2 &= 11 \times 4 \\[6pt] 6x &= 44 \\[6pt] x &= \frac{44}{6} \implies \frac{22}{3} \text{ cm} \\[6pt] Length &= \frac{22}{3} \text{ cm} \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{\dfrac{22}{3} \text{ cm}} \)
B. Answer the following questions.
(a) Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30.
Solution
\[ \begin{aligned} \text{Let the number be } & x \\[6pt] \frac{x}{5} + 30 &= \frac{x}{4} - 30 \\[6pt] 30 + 30 &= \frac{x}{4} - \frac{x}{5} \\[6pt] 60 &= \frac{5x - 4x}{20} \\[6pt] 60 &= \frac{x}{20} \\[6pt] x &= 60 \times 20 \\[6pt] x &= 1200 \end{aligned} \]Answer The number is \( \color{red} 1200 \)
(b) Raman had 30 flowers. He offered some flowers in a temple and found that the ratio of the number of remaining flowers to that of flowers in the beginning is 2 : 5. Find the number of flowers offered by him in the temple.
Solution
\[ \begin{aligned} \text{Let flowers offered in the temple} & = x \\[6pt] \text{Total flowers} & = 30 \\[6pt] \text{Remaining flowers} & = 30 - x \\[6pt] \textbf{According to the} & \textbf{ question} \\[6pt] (30 - x) : x &= 2 : 5 \\[6pt] \frac{30 - x}{30} &= \frac{2}{5} \\[6pt] 5(30 - x) &= 2 \times 30 \\[6pt] 150 - 5x &= 60 \\[6pt] - 5x &= 60 - 150 \\[6pt] - 5x &= -90 \\[6pt] x &= \frac{90}{5} \\[6pt] x &= 18 \end{aligned} \]Answer Number of flowers offered by Raman in the temple is \( \color{red} 18 \)
(c) The tens and ones digits of a two-digit number are the same. When the number is added to its reverse, the sum is 88. What is the number?
Solution
\[ \begin{aligned} \text{Let the ones place} & = x \\[6pt] \text{Let the tens place} & = x \\[6pt] \text{Original number} &= 10x + x \implies 11x \\[6pt] \text{Reverse number} &= 10x + x \implies 11x \\[6pt] \textbf{According to the} & \textbf{ question} \\[6pt] 11x + 11x &= 88 \\[6pt] 22x &= 88 \\[6pt] x &= \frac{88}{22} \\[6pt] x &= 4 \\ \\ \text{Required number} &= 11x \\[6pt] &= 11(4) \\[6pt] &= 44 \end{aligned} \]Answer The number is \( \color{red} 44 \)
(d) Solve for \( x \): \( \dfrac{x^{2} - 9}{x + 3} = \dfrac{4}{7} \)
Solution
\[ \begin{aligned} \frac{x^{2} - 9}{x + 3} &= \frac{4}{7} \\[6pt] \frac{x^{2} - 3^2}{x + 3} &= \frac{4}{7} \\[6pt] \frac{(x - 3)(x + 3)}{(x + 3)} &= \frac{4}{7} \\[6pt] x - 3 &= \frac{4}{7} \\[6pt] x &= \frac{4}{7} + 3 \\[6pt] x &= \frac{4 + 21}{7} \\[6pt] x &= \frac{25}{7} \end{aligned} \]Answer \( \color{red}{x = \dfrac{25}{7}} \)
(e) Rohan brought a pack of 100 chocolates on his birthday to distribute in his class. After giving two chocolates to each student and his class teacher, he is left with ten chocolates. How many students are there in the class?
Solution
\[ \begin{aligned} \text{Let the no. of persons} & = x \\[6pt] \text{Total chocolates} & = 100 \\[6pt] \text{Remaining chocolates} & = 10 \\[6pt] \textbf{According to the} & \textbf{ question} \\[6pt] 2x + 10 &= 100 \\[6pt] 2x &= 100 - 10 \\[6pt] 2x &= 90 \\[6pt] x &= \frac{90}{2} \\[6pt] x &= 45\\[6pt] \text{No. of students} & = 45 - 1 \\[6pt] \text{No. of students} & = 44\end{aligned} \]Answer Number of students in the class is \( \color{red} 44 \)
2. Solve the following equations.
(i) \( \dfrac{2z+7}{3z+8} = \dfrac{1}{4} \)
Solution
\[ \begin{aligned} \frac{2z + 7}{3z + 8} &= \frac{1}{4} \\[6pt] 4(2z + 7) &= 1(3z + 8) \\[6pt] 8z + 28 &= 3z + 8 \\[6pt] 8z - 3z &= 8 - 28 \\[6pt] 5z &= -20 \\[6pt] z &= \frac{-20}{5} \\[6pt] \color{green} z &= \color{green} -4 \end{aligned} \]Answer \( \color{red} z = -4 \)
(ii) \( \dfrac{3+2y}{2+5y} = \dfrac{7}{12} \)
Solution
\[ \begin{aligned} \frac{3 + 2y}{2 + 5y} &= \frac{7}{12} \\[6pt] 12(3 + 2y) &= 7(2 + 5y) \\[6pt] 36 + 24y &= 14 + 35y \\[6pt] 36 - 14 &= 35y - 24y \\[6pt] 22 &= 11y \\[6pt] y &= \frac{22}{11} \\[6pt] \color{green} y &= \color{green} 2 \end{aligned} \]Answer \( \color{red} y = 2 \)
(iii) \( \dfrac{2x-1}{1+5x} = \dfrac{1}{2} \)
Solution
\[ \begin{aligned} \frac{2x - 1}{1 + 5x} &= \frac{1}{2} \\[6pt] 2(2x - 1) &= 1(1 + 5x) \\[6pt] 4x - 2 &= 1 + 5x \\[6pt] 4x - 5x &= 1 + 2 \\[6pt] -x &= 3 \\[6pt] \color{green} x &= \color{green} -3 \end{aligned} \]Answer \( \color{red} x = -3 \)
(iv) \( \dfrac{5k-7}{3k-9} = -\dfrac{6}{7} \)
Solution
\[ \begin{aligned} \frac{5k - 7}{3k - 9} &= \frac{-6}{7} \\[6pt] 7(5k - 7) &= -6(3k - 9) \\[6pt] 35k - 49 &= -18k + 54 \\[6pt] 35k + 18k &= 54 + 49 \\[6pt] 53k &= 103 \\[6pt] \color{green} k &= \color{green} \frac{103}{53} \end{aligned} \]Answer \( \color{red} k = \dfrac{103}{53} \)
(v) \( \dfrac{\frac{3}{7}z - \frac{1}{2}}{z - \frac{1}{4}} - \dfrac{3}{14} = \dfrac{1}{7} \)
Solution
\[ \begin{aligned} \frac{\frac{3}{7}z - \frac{1}{2}}{z - \frac{1}{4}} - \frac{3}{14} &= \frac{1}{7} \\[6pt] \frac{\frac{3}{7}z - \frac{1}{2}}{z - \frac{1}{4}} &= \frac{1}{7} + \frac{3}{14} \\[6pt] \frac{\frac{3}{7}z - \frac{1}{2}}{z - \frac{1}{4}} &= \frac{2 + 3}{14} \\[6pt] \frac{\frac{3z}{7} - \frac{1}{2}}{z - \frac{1}{4}} &= \frac{5}{14} \\[6pt] \frac{\frac{6z - 7}{14}}{\frac{4z - 1}{4}} &= \frac{5}{14} \\[6pt] \frac{6z - 7}{14} \times \frac{4}{4z - 1} &= \frac{5}{14} \\[6pt] \frac{4(6z - 7)}{4z - 1} &= 5 \\[6pt] 24z - 28 &= 20z - 5 \\[6pt] 24z - 20z &= 28 - 5 \\[6pt] 4z &= 23 \\[6pt] \color{green} z &= \color{green} \frac{23}{4} \end{aligned} \]Answer \( \color{red} z = \dfrac{23}{4} \)
(vi) \( \dfrac{p - \frac{2}{5}}{p + \frac{2}{5}} = 5 \)
Solution
\[ \begin{aligned} \frac{p - \frac{2}{5}}{p + \frac{2}{5}} &= 5 \\[6pt] p - \frac{2}{5} &= 5(p + \frac{2}{5}) \\[6pt] p - \frac{2}{5} &= 5p + 5 \times \frac{2}{5} \\[6pt] p - \frac{2}{5} &= 5p + 2 \\[6pt] p - 5p &= 2 + \frac{2}{5} \\[6pt] -4p &= \frac{10 + 2}{5} \\[6pt] -4p &= \frac{12}{5} \\[6pt] p &= \frac{12}{5 \times (-4)} \\[6pt] p &= \frac{3}{5 \times (-1)} \\[6pt] \color{green} p &= \color{green} -\frac{3}{5} \end{aligned} \]Answer \( \color{red} p = -\dfrac{3}{5} \)
(vii) \( \dfrac{(x+1)(4x-3) - 4x^2 + 5}{4x+1} = 2 \)
Solution
\[ \begin{aligned} \frac{(x+1)(4x-3) - 4x^2 + 5}{4x+1} &= 2 \\[6pt] \frac{(4x^2 - 3x + 4x - 3) - 4x^2 + 5}{4x+1} &= 2 \\[6pt] \frac{4x^2 + x - 3 - 4x^2 + 5}{4x+1} &= 2 \\[6pt] \frac{x + 2}{4x+1} &= 2 \\[6pt] x + 2 &= 2(4x + 1) \\[6pt] x + 2 &= 8x + 2 \\[6pt] x - 8x &= 2 - 2 \\[6pt] -7x &= 0 \\[6pt] \color{green} x &= \color{green} 0 \end{aligned} \]Answer \( \color{red} x = 0 \)
(viii) \( \dfrac{x^2 - (x+2)(x-2)}{x+3} = \dfrac{1}{2} \)
Solution
\[ \begin{aligned} \frac{x^2 - (x+2)(x-2)}{x+3} &= \frac{1}{2} \\[6pt] \frac{x^2 - (x^2 - 4)}{x + 3} &= \frac{1}{2} \\[6pt] \frac{x^2 - x^2 + 4}{x+3} &= \frac{1}{2} \\[6pt] \frac{4}{x+3} &= \frac{1}{2} \\[6pt] 4 \times 2 &= 1 \times (x+3) \\[6pt] 8 &= x + 3 \\[6pt] 8 - 3 &= x \\[6pt] \color{green} x &= \color{green} 5 \end{aligned} \]Answer \( \color{red} x = 5 \)
3. Find the positive value of the variable for which the given equation is satisfied.
(i) \( \dfrac{1-x^2}{1+x^2} = \dfrac{-4}{5} \)
Solution
\[ \begin{aligned} \frac{1 - x^2}{1 + x^2} &= \frac{-4}{5} \\[6pt] 5(1 - x^2) &= -4(1 + x^2) \\[6pt] 5 - 5x^2 &= -4 - 4x^2 \\[6pt] 5 + 4 &= 5x^2 - 4x^2 \\[6pt] 9 &= x^2 \\[6pt] x &= \sqrt{9} \\[6pt] \color{green} x &= \color{green} 3 \end{aligned} \]Answer \( \color{red} x = 3 \)
(ii) \( \dfrac{3z^2+7}{4+z^2} = 2 \)
Solution
\[ \begin{aligned} \frac{3z^2 + 7}{4 + z^2} &= 2 \\[6pt] 3z^2 + 7 &= 2(4 + z^2) \\[6pt] 3z^2 + 7 &= 8 + 2z^2 \\[6pt] 3z^2 - 2z^2 &= 8 - 7 \\[6pt] z^2 &= 1 \\[6pt] z &= \sqrt{1} \\[6pt] \color{green} z &= \color{green} 1 \end{aligned} \]Answer \( \color{red} z = 1 \)
4. Five years ago, a mother was seven times as old as her daughter. Five years hence, she will be three times as old as her daughter. Find their present ages.
Solution
| 5 years ago (Past) | Present | 5 years after (Future) | |
|---|---|---|---|
| Daughter | \( x \) | \( x +5 \) | \( x + 5 + 5\) \( \implies x +10 \) |
| Mother | \( 7x \) | \( 7x + 5 \) | \( 7x + 5 + 5\) \( \implies 7x +10 \) |
Answer Present ages: Mother = \( \color{red} 40 \) years, Daughter = \( \color{red} 10 \) years
5. The digit at ones place of a 2-digit number is four times the digit at tens place. The number obtained by reversing the digits exceeds the given number by 54. Find the given number.
Solution
\[ \begin{aligned} \text{Let the digit at tens place be } & x \\[6pt] \text{Digit at ones place} &= 4x \\ \text{Original number} &= 10(x) + 4x = 14x \\ \\ \textbf{Reversing the digits} \\[6pt] \text{Tens digit} &= 4x \\ \text{Ones digit} &= x \\[6pt] \text{Reversed number} &= 10(4x) + x = 41x \\ \\ \textbf{According to the} & \textbf{ question} \\[6pt] 41x - 14x &= 54 \\ 27x &= 54 \\ x &= 2 \\ \\ \text{Digit at tens place} &= 2 \\ \text{Digit at ones place} &= 4(2) = 8 \\[6pt] \textbf{Original number} &= 28 \end{aligned} \]Answer The given number is \( \color{red} 28 \)
6. The sum of the digits of a 2-digit number is 10. The number obtained by interchanging the digits exceeds the original number by 36. Find the original number.
Solution
\[ \begin{aligned} \text{Let the digit at tens place be } & x \\[6pt] \text{The digit at ones place} &= (10 - x) \\ \\ \text{Original number} &= 10(x) + (10 - x) \\[6pt] &= 10x + 10 - x = 9x + 10 \\ \\ \text{New number} &= 10(10 - x) + x \\[6pt] &= 100 - 10x + x = 100 - 9x \\ \\ \textbf{According to the} & \textbf{ question} \\[6pt] (100 - 9x) - (9x + 10) &= 36 \\[6pt] 100 - 9x - 9x - 10 &= 36 \\[6pt] 90 - 18x &= 36 \\[6pt] 90 - 36 &= 18x \\[6pt] 54 &= 18x \\[6pt] x &= \frac{54}{18} = 3 \\ \\ \text{Original number} &= 9(3) + 10 \\ &= 27 + 10 \\ &= 37 \end{aligned} \]Answer The original number is \( \color{red} 37 \)
7. The sum of two consecutive multiples of 6 is 66. Find these multiples.
Solution
\[ \begin{aligned} \text{Let the two consecutive multiples of 6 be } & 6x \text{ and } 6(x + 1) \\ \\ 6x + 6(x + 1) &= 66 \\[6pt] 6x + 6x + 6 &= 66 \\[6pt] 12x + 6 &= 66 \\[6pt] 12x &= 66 - 6 \\[6pt] 12x &= 60 \\[6pt] x &= \frac{60}{12} = 5 \\ \\ \text{First multiple} &= 6(5) = 30 \\[6pt] \text{Second multiple} &= 6(5 + 1) = 36 \end{aligned} \]Answer The multiples are \( \color{red} 30 \) and \( \color{red} 36 \)
8. The numerator of a rational number is 3 less than its denominator. If the numerator is increased by 1 and the denominator is increased by 3, the number becomes \( \dfrac{1}{2} \). Find the rational number.
Solution
\[ \begin{aligned} \text{Let the denominator be } & x \\[6pt] \text{Numerator} &= (x - 3) \\ \\ \text{Rational number} &= \frac{x - 3}{x} \\ \\ \text{New Rational number} &= \frac{x - 3 +1}{x + 3} \implies \frac{x - 2}{x + 3} \\ \\ \textbf{According to the} & \textbf{ question} \\[6pt] \frac{x - 2}{x + 3} &= \frac{1}{2} \\[6pt] 2(x - 2) &= 1(x + 3) \\[6pt] 2x - 4 &= x + 3 \\[6pt] 2x - x &= 3 + 4 \\[6pt] x &= 7 \\ \\ \text{Rational number} &= \frac{x - 3}{x} \implies \frac{4}{7} \end{aligned} \]Answer The rational number is \( \color{red} \dfrac{4}{7} \)
9. A race-boat covers a distance of 60 km downstream in one and a half hour. It covers this distance upstream in 2 hours. The speed of the race-boat in still water is 35 km/hr. Find the speed of the stream.
Solution
\[ \begin{aligned} \text{Distance } (d) &= 60 \text{ km} \\ \\ \text{Downstream time} &= 1 \frac{1}{2} \text{ hours} \implies \frac{3}{2} \text{ hours} \\[6pt] \text{Speed downstream} &= 60 \div \frac{3}{2} \\[6pt] &= 60 \times \frac{2}{3} \\[6pt] &= 40 \text{ km/hr} \\ \\ \text{Let speed of stream be } & x \text{ km/hr} \\[6pt] \text{Speed in still water } &= 35 \text{ km/hr} \\ \\ \text{Speed downstream} &= 35 + x \\[6pt] 40 &= 35 + x \\[6pt] x &= 40 - 35 \\[6pt] \text{Speed of stream}&= 5 \text{ km/hr} \end{aligned} \]Answer Speed of the stream is \( \color{red} 5 \ km/hr \)
10. The sides (other than hypotenuse) of a right triangle are in the ratio 3 : 4. A rectangle is described on its hypotenuse, the hypotenuse being the longer side of the rectangle. The breadth of the rectangle is \( \dfrac{4}{5} \) of its length. Find the shortest side of the right triangle if the perimeter of the rectangle is 180 cm.
Solution
\[ \begin{aligned} \text{Let the sides of the right triangle be } & 3x \text{ and } 4x \\ \\ \text{Pythagoras theorem}\\ (Hypotenuse)^2 &= (3x)^2 + (4x)^2 \\ (Hypotenuse)^2 &= 9x^2 + 16x^2 \\ (Hypotenuse)^2 &= 25x^2 \\ Hypotenuse &= 5x \\[6pt] l &= 5x \\[6pt] b &= \frac{4}{5} \times 5x \\[6pt] b&= 4x \\ \\ \text{Perimeter of rectangle} &= 2(l + b) \\[6pt] 180 &= 2(5x + 4x) \\[6pt] 180 &= 2(9x) \\[6pt] 180 &= 18x \\[6pt] x &= \frac{180}{18} \\[6pt] x &= 10 \ cm \\ \\ \text{Shortest side of triangle} &= 3x \\ &= 3(10) \\ &= 30 \text{ cm} \end{aligned} \]Answer The shortest side of the triangle is \( \color{red} 30 \) cm
11. Divide 243 into three parts such that half of the first part, one-third of the second part and one-fourth of the third part are all equal.
Solution
\[ \begin{aligned} \text{Let the three parts be } & x, y \text{ and } z \\[6pt] x + y + z &= 243 \\ \\ \frac{x}{2} &= \frac{y}{3} = \frac{z}{4} = k \\ \\ \implies x &= 2k, \ y = 3k, \ z = 4k \\ \\ \textbf{According to the} & \textbf{ question} \\[6pt] 2k + 3k + 4k &= 243 \\[6pt] 9k &= 243 \\[6pt] k &= \frac{243}{9} = 27 \\ \\ \text{First part } (x) &= 2 \times 27 = 54 \\[6pt] \text{Second part } (y) &= 3 \times 27 = 81 \\[6pt] \text{Third part } (z) &= 4 \times 27 = 108 \end{aligned} \]Answer The three parts are \( \color{red} 54, 81 \) and \( \color{red} 108 \)
12. A purse has only two-rupee and five-rupee coins. The sum of the coins is 36 and the total value of the coins is ₹ 84. Find the number of five-rupee coins.
Solution
\[ \begin{aligned} \text{Total number of coins} &= 36 \\[6pt] \text{Let the no. of five-rupee coins be } &= x \\ \text{No. of two-rupee coins} &= (36 - x) \\ \\ \text{Value of five-rupee coins} &= 5 \times x \\ & \implies 5x \\[6pt] \text{Value of two-rupee coins} &= 2 \times (36 - x) \\ &\implies 72 - 2x \\ \\ \textbf{According to the} & \textbf{ question} \\[6pt] \text{Total value} &= 84 \\ 5x + 72 - 2x &= 84 \\ 3x + 72 &= 84 \\ 3x &= 84 - 72 \\ 3x &= 12 \\ x &= 4 \end{aligned} \]Answer The number of five-rupee coins is \( \color{red} 4 \)