1. Solve the following equations and check your answers.
(1). Adding 4 to twice a number yields \( \frac{25}{6}\) Find the number.
Solution:
\begin{align*} \text{Let the } & \text{number be } x \\ 2x + 4 &= \frac{25}{6} \\ 2x &= \frac{25}{6} - 4 \\ 2x &= \frac{25 - 24}{6} \\ 2x &= \frac{1}{6} \\ x &= \frac{1}{6} \times \frac{1}{2} \\ &\boxed{x = \frac{1}{12}} \\ \text{Check} \\ \text{For } x &= \frac{1}{12} \\ &= 2x + 4 \\ &= 2\left(\frac{1}{12}\right) + 4 \\ &= \frac{1}{6} + 4 \\ &= \frac{1}{6} + \frac{24}{6} \\ &= \frac{25}{6} \\ \end{align*}
The number is \({\boxed{\frac{1}{12}}}\).(2). A number when added to its two-thirds is equal to \(55\). Find the number.
Solution:
\begin{align*} \text{Let the number } & \text{be } x \\ x + \frac{2}{3}x &= 55 \\ \frac{3x + 2x}{3} &= 55 \\ \frac{5x}{3} &= 55 \\ x &= \cancelto{11}{55} \times \frac{3}{\cancel5} \\ &\boxed{x = 33} \\ \text{Check} \\ \text{For } x &= 33 \\ &= x + \frac{2}{3}x \\ &= 33 + \frac{2}{\cancel3}(\cancelto{11}{33}) \\ &= 33 + 22 \\ &= 55 \\ \end{align*}
The number is \({\boxed{33}}\).(3). A number when multiplied by 4 exceeds itself by \(45\). Find the number.
Solution:
\begin{align*} \text{Let the number } & \text{be } x \\ 4x &= x + 45 \\ 4x - x &= 45 \\ 3x &= 45 \\ x &= \frac{45}{3} \\ &\boxed{x = 15} \\ \text{Check} \\ \text{For } x &= 15 \\ 4x &= x + 45 \\ 4(15)&= 15 + 45 \\ 60 &= 60 \\ \implies 4x &\text{ exceeds } x \text{ by } 45 \\ \end{align*}
The number is \({\boxed{15}}\).(4). Find a number which when multiplied by 8 and then reduced by 9 is equal to \(47\).
Solution:
\begin{align*} \text{Let the } & \text{number be } x \\ 8x - 9 &= 47 \\ 8x &= 47 + 9 \\ 8x &= 56 \\ x &= \frac{56}{8} \\ &\boxed{x = 7} \\ \text{Check} \\ \text{For } x &= 7 \\ &= 8x - 9 \\ &= 8(7) - 9 \\ &= 56 - 9 \\ &= 47 \\ \end{align*}
The number is \({\boxed{7}}\).(5). The sum of two numbers is \(72\). If one of the number is 6 more than the other, find the number.
Solution:
\begin{align*} \text{Let one number be: } & x \\ \text{ Other Number be: } & x + 6 \\ \text{According to question,} \\ x + (x + 6) &= 72 \\ 2x + 6 &= 72 \\ 2x &= 72 - 6 \\ 2x &= 66 \\ x &= \frac{66}{2} \\ &\boxed{x = 33} \\ \text{The other number is } & x + 6 \\ &= 33 + 6 \\ &\boxed{= 39} \\ \text{Check} \\ \text{Sum of the numbers: } & 33 + 39 \\ &= 72 \\ \text{Difference: } & 39 - 33 \\ &= 6 \\ \end{align*}
The numbers are \({\boxed{33}}\) and \({\boxed{39}}\).(6). The sum of two number is \(99\). If one exceeds the other by 9, find the numbers.
Solution:
\begin{align*} \text{Let one number be: } & x \\ \text{ Other Number be: } & x + 6 \\ \text{According to the problem,} \\ x + (x + 9) &= 99 \\ 2x + 9 &= 99 \\ 2x &= 99 - 9 \\ 2x &= 90 \\ x &= \frac{90}{2} \\ &\boxed{x = 45} \\ \text{So the other number is } & x + 9 \\ &= 45 + 9 \\ &\boxed{= 54} \\ \text{Check} \\ \text{Sum of the numbers: } & 45 + 54 \\ &= 99 \\ \text{Difference: } & 54 - 45 \\ &= 9 \\ \end{align*}
The numbers are \({\boxed{45}}\) and \({\boxed{54}}\).(7). One number is 10 more than the other. If their sum is \(52\), find the numbers.
Solution:
\begin{align*} \text{Let the first number be } & x \\ \text{Other number be } & x + 10 \\ \text{According to the problem,} \\ x + (x + 10) &= 52 \\ 2x + 10 &= 52 \\ 2x &= 52 - 10 \\ 2x &= 42 \\ x &= \frac{42}{2} \\ &\boxed{x = 21} \\ \text{So the second number is } & x + 10 \\ &= 21 + 10 \\ &\boxed{= 31} \\ \text{Check} \\ \text{Sum of the numbers: } & 21 + 31 \\ &= 52 \\ \text{Difference: } & 31 - 21 \\ &= 10 \\ \end{align*}
The numbers are \({\boxed{21}}\) and \({\boxed{31}}\).(8). The length of a rectangle is 20 cm more than its breadth. If the perimeter is \(100\) cm, find the dimension of the rectangle.
Solution:
\begin{align*} \text{Let the breadth } &= x \\ \text{Length } &= x + 20 \\ \text{Perimeter } &= 100 \\ \text{Perimeter of the rectangle = } & 2 \times (\text{L} + \text{B}) \\ 2 \times (\text{L} + \text{B}) &= 100 \\ 2 \times (x + 20 + x) &= 100 \\ 2 \times (2x + 20) &= 100 \\ 2x + 20 &= \frac{100}{2} \\ 2x + 20 &= 50 \\ 2x &= 50 - 20 \\ 2x &= 30 \\ x &= \frac{30}{2} \\ x &= 15 \\ &\boxed{Breadth = 15 \text{ cm}} \\ \text{Length } &= x + 20 \\ &= 15 + 20 \\ &\boxed{Length = 35 \text{ cm}} \\ \text{Check} \\ \text{Perimeter: } & 2 \times (\text{length} + \text{breadth}) \\ &= 2 \times (35 + 15) \\ &= 2 \times (50) \\ &= 100 \text{ cm} \\ \end{align*}
The dimensions are \({\boxed{L = 35 \,cm}}\) and \({\boxed{B = 15 \, cm}}\).(9) The length of a rectangle is three times its width. It the perimeter is \(84\) m, find the length of the rectangle.
Solution:
\begin{align*} \text{Let the breadth } &= x \\ \text{Length } &= 3x \\ \text{Perimeter } &= 84 \\ \text{Perimeter of the rectangle = } & 2 \times (\text{L} + \text{B}) \\ 2 \times (\text{L} + \text{B}) &= 84 \\ 2 \times (x + 3x) &= 84 \\ 2 \times (4x) &= 84 \\ 4x &= \frac{84}{2} \\ 4x &= 42 \\ x &= \frac{42}{4} \\ x &= 10.5 \, cm\\ &\boxed{Breadth = 10.5 \text{ cm}} \\ \text{Length } &= 3x \\ &= 3 \times 10.5 \\ &\boxed{Length = 31.5 \text{ cm}} \\ \text{Check} \\ \text{Perimeter: } & 2 \times (\text{length} + \text{breadth}) \\ &= 2 \times (31.5 + 10.5) \\ &= 2 \times (42) \\ &= 84 \text{ cm} \\ \end{align*}
The dimensions are \({\boxed{L = 31.5 \,cm}}\) and \({\boxed{B = 10.5 \, cm}}\).(10) Two equal sides of an isoscles triangle are each 2 cm more than thrice the third side. If the perimeter of triangle is \(67\) cm, find the lengths of its sides.
Solution:
\begin{align*} \text{Let the the third side be } &= x \text{ cm} \\ \text{Equal sides of isoscles triangle } &= (3x + 2) \text{ cm} \\ \text{Perimeter } &= 67 \text{ cm}\\ \text{Perimeter} &= \text{Sum of all sides } \\ x + (3x + 2) + (3x + 2) &= 67 \\ x + 3x + 2 + 3x + 2 &= 67 \\ 7x + 4 &= 67 \\ 7x &= 67 - 4 \\ 7x &= 63 \\ x &= \frac{63}{7} \\ &\boxed{x = 9 \text{ cm}} \\ \text{Length of the two equal sides } &= 3x + 2 \\ &= 3(9) + 2 \\ &\boxed{= 29 \text{ cm each}} \\ \text{Check} \\ \text{Perimeter} \\ &= 9 + 29 + 29 \\ &= 9 + 58 \\ &= 67 \text{ cm} \\ \end{align*}
The length of the sides are \({\boxed{9 \,cm, 29 \,cm, 29 \,cm}}\).(11) Length of a rectangle is 16 cm less than twice its breadth. If the perimeter of the rectangle is \(100\) cm, find its length and breadth.
Solution:
\begin{align*} \text{Let breadth } & = x \text{ cm} \\ \text{Length } &= 2x - 16 \text{ cm } \\ \text{Perimeter } &= 100 \\ \text{Perimeter of the rectangle } &= 2 \times (\text{L} + \text{B}) \\ 2 \times (\text{L} + \text{B}) &= 100 \\ 2 \times (2x - 16 + x) &= 100 \\ 3x - 16 &= \frac{100}{2} \\ 3x - 16 &= 50 \\ 3x &= 50 + 16 \\ 3x &= 66 \\ x &= \frac{66}{3} \\ x &= 22 \\ &\boxed{x = 22 \text{ cm}} \\ \text{Length } &= 2x - 16 \\ &= 2(22) - 16 \\ &= 44 - 16 \\ &\boxed{l = 28 \text{ cm}} \\ \text{Verification} \\ \text{Perimeter} &= 2(l + b) \\ &= 2(28 + 22) \\ &= 2 \times 50 \\ &= 100 \text{ cm} \\ \end{align*}
The dimensions are \({\boxed{L = 28 \,cm}}\) and \({\boxed{B = 22 \, cm}}\).(12) Find two consecutive positive integers whose sum is \(63\).
Solution:
\(\text{Let the two consecutive positive integers be } x \text{ and } x+1\)\begin{align*} \text{According to question }\\ \text{Sum } &= 63 \\ x + (x + 1) &= 63 \\ x + x + 1 &= 63 \\ 2x + 1 &= 63 \\ 2x &= 63 - 1 \\ 2x &= 62 \\ x &= \frac{62}{2} \\ x &= 31 \\ \boxed{ \text{First integer } = 31} \\ \text{Second integer } &= x + 1 \\ &= 31 + 1 \\ &= 32 \\ \boxed{ \text{Second integer } = 32} \\ \text{Check} \\ \text{The sum of the two integers } &= 31 + 32 \\ & = 63 \\ \end{align*}
The two consecutive positive integers are \(\boxed{31} \) and \(\boxed{32} \)(13) A sum of Rs \(800\) is in the form of denominations of Rs 10 and Rs 20. If the total number of notes is 50, find the number of notes of each type.
Solution:
\begin{align*} \text{Number of Rs 10 notes} &= x \\ \text{Number of Rs 20 notes} &= 50 - x \\ \\ \text{Value of Rs 10 notes} &= 10 \times x \\ \text{Value of Rs 20 notes} &= 20 \times (50 - x) \\ \\ \text{Total money} &= 10 \times x + [20 \times (50 - x)] \\ &= 10x + (20 \times 50 - 20 \times x) \\ &= 10x + 1000 - 20x \\ &= 1000 - 10x \\ \\ \text{According to the equation} \\ \text{Money} &= 800 \\ 1000 - 10x &= 800 \\ - 10x &= 800 - 1000 \\ \cancel- 10x &= \cancel- 200 \\ 10x &= 200 \\ x &= \frac{200}{10} \\ x &= 20 \\ \boxed{ \text{Rs 10 notes } = 20} \\ \text{Rs 20 notes } &= 50 - x \\ \text{Rs 20 notes } &= 50 - 20 \\ \boxed{ \text{Rs 20 notes } = 30} \\ \text{Check} \\ \text{Total Money } &= (10 \times 20) + (20 \times 30) \\ &= (200) + (600) \\ & = \text{Rs } 800 \\ \\ \end{align*}
There are \(\boxed{20 } \) ₹10 notes and \(\boxed{30 } \) ₹20 notes.(14) In a class of 49 students, number of girls is \( \frac{2}{5}\) of the boys. Find the number of boys in the class.
Solution:
\begin{align*} \text{Let the number of boys} &= x \\ \text{Let the number of girls} &= \frac{2}{5}x \\ \\ \text{Total students} &= 49 \\ \text{Boys + Girls} &= 49 \\ x + \frac{2}{5}x &= 49 \\ \frac{5x + 2x}{5} &= 49 \\ \frac{7}{5}x &= 49 \\ x &= \cancelto{7}{49} \times \frac{5}{\cancel7} \\ x &= 7 \times 5 \\ x &= 35 \\ \boxed{ \text{Boys} = 35} \\ \text{Girls} &= \frac{2}{5}x \\ &= \frac{2}{\cancel5} \times \cancelto{7}{35} \\ &= 2 \times 7 \\ \boxed{ \text{Girls} = 14} \\ \\ \text{Check} \\ \text{Total students } &= 15 + 14 \\ &= 49 \\ \end{align*}
There are \({\boxed{35}}\) boys in the class.(15) Leena has \( 117\) rupees in the form of 5 rupee coins and 2 rupee coins. The number of 2 rupee coins is 4 times that of 5 rupee coins. Find the number of coins of each denomination.
Solution:
\begin{align*} \text{Let the number of Rs 5 coins} &= x \\ \text{The number of Rs 2 coins} &= 4x \\ \\ \text{Total value of Rs 5 coins} &= 5x \\ \text{Total value of Rs 2 coins} &= 2 \times 4x \\ &= 8x \\ \\ \text{Total Money} &= 117 \\ \\ \text{According to Question} \\ 5x + 8x &= 117 \\ 13x &= 117 \\ x &= \frac{117}{13} \\ x &= 9 \\ \boxed{ \text{Number of Rs 5 coins} = 9} \\ \\ \text{Number of Rs 2 coins} &= 4x \\ &= 4 \times 9 \\ \boxed{ \text{Number of 2 rupee coins} = 36} \\ \\ \text{Check:} \\ \text{Total value} &= (5 \times 9) + (2 \times 36) \\ &= 45 + 72 \\ &= 117 \\ \end{align*}
There are \(\boxed{9 } \) ₹5 coins and \(\boxed{36 } \) ₹20 coins.(16) A total of Rs \( 80,000\) is to be distributed among 200 persons as prizes. A prize is either of Rs 500 or Rs 100. Find the number of each prize.
Solution:
\( \text{Let the no. of persons who get Rs 500 prizes} = x \)
\( \text{The no. of persons who get Rs 100 prizes} = 200 - x \)
\begin{align*}
\text{Total value of Rs 500 prizes} &= 500x \\
\text{Total value of Rs 100 prizes} &= 100 \times (200 - x) \\ \\
\text{According to the Question} \\ \\
\text{Total Money} &= 80000 \\
500x + 100(200 - x) &= 80,000 \\
500x + 20,000 - 100x &= 80,000 \\
400x + 20,000 &= 80,000 \\
400x &= 80,000 - 20,000 \\
400x &= 60,000 \\
x &= \frac{60,000}{400} \\
x &= 150 \\
\boxed{ \text{Number of Rs 500 prizes} = 150} \\
\\
\text{Number of Rs 100 prizes} &= 200 - x \\
&= 200 - 150 \\
&= 50 \\
\boxed{ \text{Number of Rs 100 prizes} = 50} \\ \\
\text{Check} \\
\text{Total number of prizes} &= x + y \\
&= 150 + 50 \\
&= 200 \\
\\
\text{Total amount distributed} &= (500 \times 150) + (100 \times 50) \\
&= 75,000 + 5,000 \\
&= Rs \, 80,000 \\
\end{align*}
(17) When \( \frac{1}{3}\) is subtracted from a number and the diference is multiplied by 4, the result is 28. Find the number.
Solution:
\begin{align*} \text{Let the number be x} \\ 4 \left( x - \frac{1}{3} \right) &= 28 \\ x - \frac{1}{3} &= \frac{28}{4} \\ x - \frac{1}{3} &= 7 \\ x &= 7 + \frac{1}{3} \\ x &= \frac{21}{3} + \frac{1}{3} \\ x &= \frac{22}{3} \\ \\ \text{Check} \\ &=4 \left( \frac{22}{3} - \frac{1}{3} \right) \\ &=4 \left( \frac{22 - 1}{3} \right) \\ &=4 \times \frac{21}{3} \\ &=4 \times 7 \\ &=28 \\ \end{align*}
The number is \({\boxed{\frac{22}{3}}}\)(18) Sudesh is twice as old as Seema. If six years is subtracted from Seema's age and four years are added to Sudesh's age, Sudesh wil be four times Seema's age. How old were they three years ago?
Solution:
\begin{align*} \text{Let Seema's age be } & = x \\ \text{Then Sudesh's age } & = 2x \\ \\ \text{6 years subtracted from Seema's age will be } & = x - 6 \\ \text{4 years add to Sudesh's age will be } & = 2x + 4 \\ \\ \text{According to the Question} \\ \text{Sudesh will be four times Seema's age} \\ 2x + 4 &= 4(x - 6) \\ 2x + 4 &= 4x - 24 \\ 4 + 24 &= 4x - 2x \\ 28 &= 2x \\ x &= 14 \\ \boxed{\text{Seema's present age} = 14 \text{ years}} \\ \\ \text{Sudesh's age} &= 2x \\ &= 2 \times 14 \\ &= 28 \\ \boxed{\text{Sudesh's present age} = 28 \text{ years}} \\ \\ \text{Three years ago, their ages were:} \\ \text{Seema's age} &= 14 - 3 \\ &= 11 \text{ years} \\ \text{Sudesh's age} &= 2x - 3 \\ &= 28 - 3 \\ &= 25 \text{ years} \\ \end{align*}
Three years ago Seema was \(\boxed{11 \, years } \) and Sudesh was \(\boxed{25 \, years } \) old.(19) The ages of Leena and Heena are in the ratio \( 7 : 5\). Ten years hence, the ratio of their ages wil be \( 9 : 7\). Find their present ages.
Solution:
\begin{align*} \text{Let the present age of Leena } & = 7x \\ \text{Let the present age of Heena } & = 5x \\ \\ \text{Ten years hence}\\ \text{Age of Leena} &= 7x + 10\\ \text{Age of Heena} &= 5x + 10\\ \text{According to the Question} & \\ (7x + 10) : (5x 10) & = 9:7 \\ \frac{7x + 10}{5x + 10} = \frac{9}{7} \\ \text{Cross Multiply} & \\ 7(7x + 10) &= 9(5x + 10) \\ 49x + 70 &= 45x + 90 \\ 49x - 45x &= 90 - 70 \\ 4x &= 20 \\ x &= 5 \\ \text{Present age of Leena} & = 7x \\ & = 7 \times 5 \\ & = 35 \\ \boxed{\text{Present age of Leena} = 35 \text{ years}} \\ \text{Present age of Heena} & = 7x \\ & = 5 \times 5 \\ & = 25 \\ \boxed{\text{Present age of Heena} = 25 \text{ years}} \\ \\ \text{Check} \\ &= \frac{35}{25} \\ &= \frac{7}{5} \\ &= 7:5 \end{align*}
Present ages of Leena is \(\boxed{35 \, years } \) and Heena is \(\boxed{25 \, years } \) old.(20) Vikas is three years older than Deepika. Six years ago, Vikas's age was four times Deepika's age. Find the ages of Deepika and Vikas.
Solution:
\begin{align*} \text{Let Deepika's present age} & = x \text{ years} \\ \text{Vikas's present age} & = x + 3 \text{ years} \\ \text{Six years ago:} & \\ \text{Deepika's age} &= (x - 6) \text{ years} \\ \text{Vikas's age} &= (x + 3) - 6 \\ &= (x - 3) \text{ years} \\ \text{According to the Question:} \\ \text{Vikas's age was four times Deepika's age} \\ (x - 3) &= 4(x - 6) \\ x - 3 &= 4x - 24 \\ 4x - x &= 24 - 3 \\ 3x &= 21 \\ x &= 7 \\ \text{Present age of Deepika} & = x \text{ years} \\ & = 7 \text{ years} \\ \boxed{\text{Present age of Deepika} = 7 \text{ years}} \\ \text{Present age of Vikas} & = x + 3 \\ & = 7 + 3 \\ & = 10 \\ \boxed{\text{Present age of Vikas} = 10 \text{ years}} \\ \end{align*}
Present ages of Vikas is \(\boxed{10 \, years } \) and Deepika is \(\boxed{7 \, years } \) old.