Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Set 12.2 Solutions

Question 1. Mother is 25 years older than her son. Find son's age, if after 8 years ratio of son's age to mother's age will be \(\frac{4}{9}\).

Answer:
Let the son's present age be x years.
\( \therefore \) Mother's present age = (x + 25) years
After 8 years,
Son's age = (x + 8) years
Mother's age = (x + 25 + 8) = (x + 33) years
Since, the ratio of the son's age to mother's age after 8 years is \( \frac{4}{9} \).

\( \therefore \frac{x+8}{x+33} = \frac{4}{9} \)

\( \therefore 9 (x + 8) = 4 (x + 33) \)

\( \therefore 9x + 72 = 4x + 132 \)

\( \therefore 9x - 4x = 132 - 72 \)

\( \therefore 5x = 60 \)

\( \therefore x = \frac{60}{5} \)

\( \therefore x = 12 \)
\( \therefore \) Son's present age is 12 years.
In simple words: We set up an equation based on the given ages and their ratio after 8 years. Solving the equation \( \frac{x+8}{x+33} = \frac{4}{9} \) gives the son's current age.

🎯 Exam Tip: Ensure to correctly formulate the algebraic equation from the word problem, paying attention to age changes over time and ratio definitions.

Question 2. The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent to \(\frac{1}{2}\). Find the fraction.
Answer:
Let the numerator of the fraction be x.
The denominator of a fraction is greater than its numerator by 12.
\( \therefore \) Denominator of the fraction = (x + 12)

\( \therefore \) The required fraction = \( \frac{x}{x+12} \)
For the new fraction,
numerator is decreased by 2.
\( \therefore \) The new numerator = (x - 2)
Also, denominator is increased by 7.
\( \therefore \) The new denominator = (x + 12) + 7
= (x + 19)
Since, the new fraction is equivalent to \( \frac{1}{2} \)

\( \therefore \frac{x-2}{x+19} = \frac{1}{2} \)

\( \therefore 2(x - 2) = 1(x + 19) \)

\( \therefore 2x - 4 = x + 19 \)

\( \therefore 2x - x = 19 + 4 \)

\( \therefore x = 23 \)

\( \therefore \) The required fraction = \( \frac{x}{x+12} = \frac{23}{23+12} = \frac{23}{35} \)

\( \therefore \) The required fraction is \( \frac{23}{35} \).
In simple words: We define the numerator as 'x' and the denominator as 'x+12'. We then form a new fraction based on given conditions and equate it to \( \frac{1}{2} \) to solve for 'x' and find the original fraction.

🎯 Exam Tip: Clearly define variables for the numerator and denominator, then systematically apply the given conditions to form and solve the equation.

Question 3. The ratio of the weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Answer:
Let the weight of zinc in the brass utensil be x gm.
Since, the ratio of the weights of copper to zinc in brass is 13:7.

\( \therefore \frac{\text{Weight of copper}}{\text{Weight of zinc}} = \frac{13}{7} \)

\( \therefore \frac{\text{Weight of copper}}{x} = \frac{13}{7} \)

\( \therefore \) Weight of copper in the brass utensil = \( (\frac{13}{7}x) \) gm
The weight of the brass utensil = 700 gm

\( \therefore \frac{13}{7}x + x = 700 \)

\( \therefore 13x + 7x = 700 \times 7 \)

\( \therefore 13x + 7x = 4900 \)

\( \therefore 20x = 4900 \)

\( \therefore x = \frac{4900}{20} \)

\( \therefore x = 245 \)
\( \therefore \) The weight of zinc in the brass utensil is 245 gm.
In simple words: We set up a proportion based on the given ratio of copper to zinc. We then express the total weight as the sum of copper and zinc weights and solve for the weight of zinc.

🎯 Exam Tip: When dealing with ratios and total quantities, express individual parts in terms of a common variable and form an equation based on the total.

Question 4. Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Answer:
Let the three consecutive whole numbers be (x - 1), x and (x + 1).
\( \therefore \) Sum of the three numbers
= (x - 1) + x + (x + 1)
= 3x
Given that, the sum of the three numbers is greater than 45 and less than 54.
When the sum of the three numbers is 45,
3x = 45

\( \therefore x = \frac{45}{3} \)

\( \therefore x = 15 \)
When the sum of the three numbers is 54,

\( \therefore 3x = 54 \)

\( \therefore x = \frac{54}{3} \)

\( \therefore x = 18 \)
\( \therefore \) the value of x is greater than 15 and less than 18.
\( \therefore \) the value of x is either 16 or 17
Case I:
If the value of x is 16, then the three consecutive whole numbers are
(16 - 1), 16,(16 + 1) i.e., 15, 16, 17
Case II:
If the value of x is 17, then the three consecutive whole numbers are (17 - 1), 17, (17 + 1) i.e., 16, 17, 18.
\( \therefore \) The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.
In simple words: We represent three consecutive whole numbers as \(x-1\), \(x\), and \(x+1\). We then use the given condition that their sum is between 45 and 54 (exclusive) to find the possible integer values for \(x\).

🎯 Exam Tip: For consecutive whole numbers, represent them as \(x-1\), \(x\), and \(x+1\) to simplify calculations. Solve inequalities by considering the boundary conditions.

Question 5. In a two-digit number, digit at the ten's place is twice the digit at unit's place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Answer:
Let the digit at unit's place be x.
The digit at the ten's place is twice the digit at unit's place.
\( \therefore \) The digit at ten's place = 2x

Since, the sum of the original number and the new number is 66.
\( \therefore 21x + 12x = 66 \)

\( \therefore 33x = 66 \)

\( \therefore x = \frac{66}{33} \)

\( \therefore x = 2 \)

\( \therefore \) Original number = 21x = 21 \( \times \) 2 = 42
\( \therefore \) the original number is 42.
In simple words: We represent the two-digit number using variables for its units and tens digits based on the given relationship. Then, we form an equation by adding the original number and the number with interchanged digits, solving for the unknown digit.

🎯 Exam Tip: Remember that a two-digit number with tens digit 'a' and units digit 'b' is represented as \(10a + b\). Clearly define the digits before setting up the equation.

Question 6. Some tickets of Rs. 200 and some of Rs. 100, of a drama in a theatre were sold. The number of tickets of Rs. 200 sold was 20 more than the number of tickets of Rs. 100. The total amount received by the theatre by sale of tickets was Rs. 37000. Find the number of Rs. 100 tickets sold.
Answer:
Let the number of tickets sold of Rs. 100 be x.
The number of tickets of Rs. 200 sold was 20 more than the number of tickets of Rs. 100.
\( \therefore \) Number of tickets sold of Rs. 200 = (x + 20)
\( \therefore \) Total amount received by the theatre through the sale of tickets = 100 \( \times \) x + 200 \( \times \) (x + 20)
= 100x + 200x + 4000
= 300x + 4000
Since, the total amount received by the theatre through the sale of tickets = Rs. 37000
\( \therefore 300x + 4000 = 37000 \)

\( \therefore 300x = 37000 - 4000 \)

\( \therefore 300x = 33000 \)

\( \therefore x = \frac{33000}{300} \)

\( \therefore x = 110 \)
\( \therefore \) 110 tickets of Rs. 100 were sold.
In simple words: We define the number of Rs. 100 tickets as 'x' and use it to express the number of Rs. 200 tickets. Then, we set up an equation where the total revenue from both types of tickets equals the given total amount, solving for 'x'.

🎯 Exam Tip: Be careful to distinguish between the number of tickets and the total value contributed by each type of ticket when forming the sum equation.

Question 7. Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Answer:
Let the three consecutive natural numbers be (x - 1), x and (x + 1).
Here, the smallest number is (x - 1) and the greatest number is (x + 1).
Since, five times the smallest number is 9 more than four times the greatest number.
\( \therefore 5 \times (x - 1) = [4 \times (x + 1)] + 9 \)

\( \therefore 5x - 5 = 4x + 4 + 9 \)

\( \therefore 5x - 5 = 4x + 13 \)

\( \therefore 5x - 4x = 13 + 5 \)

\( \therefore x = 18 \).

\( \therefore \) the three numbers are (18 - 1), 18, (18 + 1)
i. e., 17, 18, 19
\( \therefore \) The three consecutive natural numbers are 17,18 and 19.
In simple words: We represent the three consecutive natural numbers as \(x-1\), \(x\), and \(x+1\). An equation is formed by setting five times the smallest number equal to nine more than four times the greatest number, which allows us to find 'x' and thus the numbers.

🎯 Exam Tip: Clearly identify the smallest and greatest numbers when working with consecutive numbers and translate the 'is more than' or 'is less than' phrases into correct algebraic operations.

Question 8. Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs. 54. Then he sold the bicycle to Nikhil for Rs. 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Answer:
Let the cost price at which Raju purchased the bicycle be Rs. x.
Since, Raju sold the bicycle at 8% profit to Amit.
\( \therefore \) Selling price of bicycle for Raju = x + 8% of x

\( = x + \frac{8}{100} \times x \)

\( = \frac{100x+8x}{100} \)

\( = \frac{108x}{100} \)
Since, Amit spent Rs. 54 on repairing the bicycle and then sold it to Nikhil for Rs. 1134, at no loss and no profit.
\( \therefore \) Selling price of bicycle + repairing cost = Rs. 1134

\( \therefore \frac{108x}{100} + 54 = 1134 \)

\( \therefore \frac{108x}{100} = 1134 - 54 \)

\( \therefore \frac{108x}{100} = 1080 \)

\( \therefore 108x = 1080 \times 100 \)

\( \therefore 108x = 108000 \)

\( \therefore x = \frac{108000}{108} \)

\( \therefore x = 1000 \)

\( \therefore \) The cost price of the bicycle at which Raju purchased it is Rs. 1000.
In simple words: We represent Raju's original cost price as 'x'. We calculate Raju's selling price to Amit with an 8% profit. Since Amit sold it with no profit or loss after repairs, Amit's total cost (purchase price + repair cost) equals his selling price. This relationship forms an equation to find 'x'.

🎯 Exam Tip: Clearly differentiate between cost price, selling price, and repair costs. Remember that 'no loss and no profit' means the selling price equals the total cost incurred by the seller.

Question 9. A cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Answer:
Let the number of runs required by the cricket player to score in the third match be x.
Number of runs scored by the player in first match = 180
Number of runs scored in second match = 257
\( \therefore \) Total runs scored by the player = 180 + 257 + x = 437 + x
Average of runs in the three matches = \( \frac{437+x}{3} \)
Since, the average of runs should be 230.

\( \therefore \frac{437+x}{3} = 230 \)

\( \therefore 437 + x = 230 \times 3 \)

\( \therefore 437 + x = 690 \)

\( \therefore x = 690 - 437 \)

\( \therefore x = 253 \)
\( \therefore \) The cricket player should score 253 runs in the third match.
In simple words: We let 'x' be the runs in the third match. The average is the total runs divided by 3 matches. We set this average equal to 230 and solve the resulting equation for 'x'.

🎯 Exam Tip: The average of a set of numbers is their sum divided by the count of numbers. Use this definition to set up the equation for the unknown score.

Question 10. Sudhir's present age is 5 more than three times the age of Viru. Anil's age is half the age of Sudhir. If the ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5:6, then find Viru's age.
Answer:
Let Viru's present age be x years.
Sudhir's present age is 5 more than three times the age of Viru.
\( \therefore \) Sudhir's present age = (3x + 5) years
Anil's age is half the age of Sudhir.

\( \therefore \) Anil's present age = \( (\frac{3x+5}{2}) \) years
Since, the ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5:6.

\( \therefore \frac{(3x+5)+x}{3 \times (\frac{3x+5}{2})} = \frac{5}{6} \)

\( \therefore \frac{4x+5}{\frac{3(3x+5)}{2}} = \frac{5}{6} \)

\( \therefore 6 \times (4x+5) = (\frac{3}{2}) (3x+5) \times 5 \)

\( \therefore 24x + 30 = \frac{9x+15}{2} \times 5 \)

\( \therefore 24x + 30 = \frac{45x+75}{2} \)

\( \therefore 2 \times (24x + 30) = 45x + 75 \)

\( \therefore 48x + 60 = 45x + 75 \)

\( \therefore 48x - 45x = 75 - 60 \)

\( \therefore 3x = 15 \)

\( \therefore x = \frac{15}{3} \)

\( \therefore x = 5 \)
\( \therefore \) Viru's present age is 5 years.
In simple words: We express the ages of Sudhir and Anil in terms of Viru's age 'x'. Then, we use the given ratio of the sum of Sudhir's and Viru's ages to three times Anil's age to form and solve a complex algebraic equation for 'x'.

🎯 Exam Tip: Break down complex age word problems by defining each person's age relative to a single variable. Be meticulous with algebraic simplification, especially when dealing with fractions in equations.

Maharashtra Board Class 8 Maths Chapter 12 Equations In One Variable Practice Set 12.2 Intext Questions And Activities

Question 1. Write correct numbers in the boxes given. (Textbook pg. no. 78) length is 3 times the breadth
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत दिखाया गया है। आयत की परिधि 40 सेमी है और चौड़ाई को 'x' के रूप में दर्शाया गया है। यह दर्शाता है कि लंबाई चौड़ाई की 3 गुना है, इसलिए लंबाई '3x' है।
Answer:
Perimeter of the rectangle = 40
\( \therefore \) 2(3x + 1x) = 40

\( \therefore \) 2 \( \times \) 4x = 40

\( \therefore \) 8x = 40

\( \therefore \) x = 5
\( \therefore \) Breadth of rectangle = 5 cm and Length of rectangle = 15 cm
In simple words: We're given a rectangle's perimeter and that its length is three times its breadth. By representing breadth as 'x', we form an equation using the perimeter formula \(2(\text{length} + \text{breadth}) = \text{perimeter}\) to find 'x' (breadth) and then the length.

🎯 Exam Tip: Remember the formula for the perimeter of a rectangle, \(P = 2(l + b)\), and correctly substitute the given relationships between length and breadth.