GSEB Class 7 Maths Solutions Chapter 4 સાદા સમીકરણ Exercise 4.4

1. Formulate equations based on the given situation and solve them to find the unknown number:

Question (a). If 46 is added to 8 times a number, you get 60.
Answer: Let us suppose the number is \( x \). Eight times this number becomes \( 8x \). When we add 4 to this result, we get \( 8x + 4 \). This final result equals 60.
\( \implies 8x + 4 = 60 \)
\( \implies 8x = 60 - 4 \) (Taking 4 to the right side)
\( \implies 8x = 56 \)
\( \implies \frac{8x}{8} = \frac{56}{8} \) (Dividing both sides by 8)
\( \implies x = 7 \) The number is 7.
In simple words: First, assume the number as 'x'. Multiply it by 8, then add 4. This total should be 60. Solve this simple equation to find 'x'.

Exam Tip: Remember to clearly define your variable (like 'x' for the unknown number) at the beginning of your solution.

Question (b). If 4 is subtracted from one-fifth of a number, you get 3.
Answer: Let the number be \( x \). One-fifth of the number is \( \frac{1}{5}x \). Subtracting 4 from this gives \( \frac{1}{5}x - 4 \). The final result is 3.
\( \implies \frac{1}{5}x - 4 = 3 \)
\( \implies \frac{1}{5}x = 3 + 4 \) (Taking -4 to the right side)
\( \implies \frac{1}{5}x = 7 \)
\( \implies x = 7 \times 5 \) (Multiplying both sides by 5)
\( \implies x = 35 \) The number is 35.
In simple words: Call the number 'x'. Find one-fifth of 'x', then take away 4. This amount should be 3. Figure out what 'x' must be.

Exam Tip: When dealing with fractions in equations, remember to multiply by the reciprocal to isolate the variable.

Question (c). If I take three-fourths of a number and add 3 to it, I get 21.
Answer: Let the number be \( x \). Three-fourths of this number is \( \frac{3}{4}x \). Adding 3 to this result gives \( \frac{3}{4}x + 3 \). This final result equals 21.
\( \implies \frac{3}{4}x + 3 = 21 \)
\( \implies \frac{3}{4}x = 21 - 3 \) (Taking +3 to the right side)
\( \implies \frac{3}{4}x = 18 \)
\( \implies \frac{3}{4}x \times \frac{4}{3} = 18 \times \frac{4}{3} \) (Multiplying both sides by \( \frac{4}{3} \))
\( \implies x = 24 \) The number is 24.
In simple words: Represent the number as 'x'. Calculate three-quarters of 'x' and then add 3. This total should be 21. Solve to discover 'x'.

Exam Tip: Always perform inverse operations to solve for the unknown variable, ensuring you apply them to both sides of the equation.

Question (d). When I subtracted 11 from two times a number, the result was 15.
Answer: Let the number be \( x \). Two times this number is \( 2x \). When we subtract 11 from this result, we get \( 2x - 11 \). This result was 15.
\( \implies 2x - 11 = 15 \)
\( \implies 2x = 15 + 11 \) (Taking -11 to the right side)
\( \implies 2x = 26 \)
\( \implies \frac{2x}{2} = \frac{26}{2} \) (Dividing both sides by 2)
\( \implies x = 13 \) The number is 13.
In simple words: Let 'x' be the number. Double 'x', then take away 11. The final amount is 15. Find the value of 'x'.

Exam Tip: Be careful with signs when moving terms across the equals sign; subtraction becomes addition and vice-versa.

Question (e). Munna subtracted three times the number of notebooks he had from 50, and the result was 8.
Answer: Let us suppose Munna has \( x \) notebooks. Three times the number of these notebooks becomes \( 3x \). When this result is subtracted from 50, we get \( 50 - 3x \). This result equals 8.
\( \implies 50 - 3x = 8 \)
\( \implies -3x = 8 - 50 \) (Taking 50 to the right side)
\( \implies -3x = -42 \)
\( \implies \frac{-3x}{-3} = \frac{-42}{-3} \) (Dividing both sides by -3)
\( \implies x = 14 \) Munna has 14 notebooks.
In simple words: If Munna has 'x' notebooks, multiply 'x' by 3. Subtract this new number from 50. The answer should be 8. Work out how many notebooks Munna has.

Exam Tip: When dividing by a negative number, remember that a negative divided by a negative results in a positive.

Question (f). Ila thought of a number. If she adds 19 to it and divides the sum by 5, she gets 8.
Answer: Let us suppose the number is \( x \). When 19 is added to this number, we get \( x + 19 \). Dividing this result by 5 gives \( \frac{x+19}{5} \). This final result equals 8.
\( \implies \frac{x+19}{5} = 8 \)
\( \implies \frac{x+19}{5} \times 5 = 8 \times 5 \) (Multiplying both sides by 5)
\( \implies x + 19 = 40 \)
\( \implies x = 40 - 19 \) (Taking 19 to the right side)
\( \implies x = 21 \) The number is 21.
In simple words: Let 'x' be the number. Add 19 to 'x', then divide the total by 5. This should give you 8. Find what 'x' is.

Exam Tip: To clear a denominator in an equation, multiply both sides by that denominator.

Question (g). Anwar thinks of a number. He subtracts 7 from \( \frac{5}{2} \) of the number, and the result is 23.
Answer: Let us suppose the number is \( x \). \( \frac{5}{2} \) of this number is \( \frac{5}{2}x \). When he subtracts 7 from this result, we get \( \frac{5}{2}x - 7 \). This result equals 23.
\( \implies \frac{5}{2}x - 7 = 23 \)
\( \implies \frac{5}{2}x = 23 + 7 \) (Taking -7 to the right side)
\( \implies \frac{5}{2}x = 30 \)
\( \implies \frac{5}{2}x \times \frac{2}{5} = 30 \times \frac{2}{5} \) (Multiplying both sides by \( \frac{2}{5} \))
\( \implies x = 12 \) The number is 12.
In simple words: Call the number 'x'. Calculate five-halves of 'x', then take away 7. The answer should be 23. Work out the value of 'x'.

Exam Tip: To eliminate a fractional coefficient, multiply both sides of the equation by its reciprocal.

2. Solve the Following Equations:

Question (a). A teacher told the students in the class that the highest marks obtained by a student are 7 more than twice the lowest marks. If the highest score is 87, what are the lowest marks?
Answer: Let us suppose the lowest marks obtained by a student are \( x \). Twice these marks becomes \( 2x \). Adding 7 to these marks gives \( 2x + 7 \), which represents the highest marks obtained by a student. Also, this result equals 87.
\( \implies 2x + 7 = 87 \)
\( \implies 2x = 87 - 7 \) (Taking 7 to the right side)
\( \implies 2x = 80 \)
\( \implies \frac{2x}{2} = \frac{80}{2} \) (Dividing both sides by 2)
\( \implies x = 40 \) The lowest marks obtained by a student are 40.
In simple words: Let 'x' be the lowest score. The highest score (87) is two times 'x' plus 7. Form an equation and find 'x'.

Exam Tip: Always translate word problems into algebraic expressions step-by-step to avoid errors in forming the equation.

Question (b). In an isosceles triangle, the measures of the two base angles are equal. The vertex angle measures 40°. What will be the measure of the base angles of the triangle? (Recall: The sum of the measures of all three angles of a triangle is 180°.)
Answer: Let us suppose one base angle of the isosceles triangle measures \( x° \). The other base angle of this triangle will also measure \( x° \). The vertex angle of this triangle measures \( 40° \). Now, the total of all three angles in a triangle is \( 180° \).
\( \implies x° + x° + 40° = 180° \)
\( \implies 2x° = 180° - 40° \) (Taking 40° to the right side)
\( \implies 2x° = 140° \)
\( \implies \frac{2x°}{2} = \frac{140°}{2} \) (Dividing both sides by 2)
\( \implies x = 70° \) The measure of the base angles of this isosceles triangle is \( 70° \).
In simple words: In an isosceles triangle, two bottom angles are equal. If the top angle is 40°, and all three angles add up to 180°, find the size of each bottom angle.

Exam Tip: Clearly state the property of triangles being used (sum of angles is 180°) to support your equation and solution.

Question (c). In a match, Sachin's runs are twice Rahul's runs. If their runs are combined, their total runs are 2 less than two centuries. How many runs did both of them score in that match?
Answer: Let us suppose the runs scored by Rahul are \( x \). Sachin scored twice as many runs as Rahul.
\( \implies \) Sachin's runs are \( 2x \). The total of Rahul's and Sachin's runs = \( x + 2x = 3x \). However, this result is 2 less than two centuries, which means \( 200 - 2 = 198 \).
\( \implies 3x = 198 \)
\( \implies \frac{3x}{3} = \frac{198}{3} \) (Dividing both sides by 3)
\( \implies x = 66 \) And \( 2x = 66 \times 2 = 132 \). Rahul's runs are 66 and Sachin's runs are 132.
In simple words: If Rahul scored 'x' runs, Sachin scored '2x' runs. Together, they scored 2 runs less than 200, so 198 runs. Set up an equation \( (x + 2x = 198) \) and solve for 'x' to find each player's score.

Exam Tip: Break down complex word problems into smaller, manageable parts, defining variables for each unknown quantity.

3. Solve the Following:

Question (i). Irfan said that he has marbles more than 3 times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Answer: Let us suppose Parmit has \( x \) marbles. Five times these marbles becomes \( 5x \). Adding 7 to this result gives \( 5x + 7 \), which is equal to the marbles Irfan has. Currently, Irfan has 37 marbles.
\( \implies 5x + 7 = 37 \)
\( \implies 5x = 37 - 7 \) (Taking 7 to the right side)
\( \implies 5x = 30 \)
\( \implies \frac{5x}{5} = \frac{30}{5} \) (Dividing both sides by 5)
\( \implies x = 6 \) Parmit has 6 marbles.
In simple words: If Parmit has 'x' marbles, Irfan has 7 more than 5 times 'x', which totals 37. Form the equation \( 5x + 7 = 37 \) and find 'x'.

Exam Tip: Pay close attention to keywords like "more than," "less than," "times," and "of" when translating word problems into equations.

Question (ii). Lakshmi's father is 49 years old. He is 4 years older than three times Lakshmi's age. What is Lakshmi's age?
Answer: Let us suppose Lakshmi's age is \( x \) years. Three times Lakshmi's age becomes \( 3x \). When we add 4 to this result, we get \( 3x + 4 \). This age is equal to Lakshmi's father's age, which is 49 years.
\( \implies 3x + 4 = 49 \)
\( \implies 3x = 49 - 4 \) (Taking 4 to the right side)
\( \implies 3x = 45 \)
\( \implies \frac{3x}{3} = \frac{45}{3} \) (Dividing both sides by 3)
\( \implies x = 15 \) Lakshmi's age is 15 years.
In simple words: Let Lakshmi's age be 'x'. Her father's age is 4 more than 3 times 'x', and he is 49. Set up the equation \( 3x + 4 = 49 \) and solve for 'x'.

Exam Tip: Clearly state the unit of measurement (e.g., years, marbles, runs) in your final answer for clarity.

Question (iii). The people of Sundargram planted trees in their village garden. Some of those trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. If the number of non-fruit trees is 77, what is the number of fruit trees?
Answer: Let us suppose the number of fruit trees is \( x \). Three times this number becomes \( 3x \). When 2 is added to this result, we get \( 3x + 2 \). This amount is equal to the number of non-fruit trees. However, the number of non-fruit trees is 77.
\( \implies 3x + 2 = 77 \)
\( \implies 3x = 77 - 2 \) (Taking 2 to the right side)
\( \implies 3x = 75 \)
\( \implies \frac{3x}{3} = \frac{75}{3} \) (Dividing both sides by 3)
\( \implies x = 25 \) The number of fruit trees is 25.
In simple words: Let 'x' be the number of fruit trees. The non-fruit trees are two more than three times 'x', and there are 77 non-fruit trees. Form the equation \( 3x + 2 = 77 \) and find 'x'.

Exam Tip: Define variables clearly for different quantities (e.g., fruit trees, non-fruit trees) to avoid confusion in multi-part problems.

4. Solve This Riddle:

Question 4. Tell my identity! Take seven times me. Add fifty. To reach a triple century, you still need forty. What am I?
Answer: Let us suppose the number is \( x \). Seven times this number becomes \( 7x \). A triple century = \( 3 \times 100 = 300 \). According to the riddle: (Seven times the number) + 50 = (Triple century) - 40
\( \implies 7x + 50 = 300 - 40 \)
\( \implies 7x + 50 = 260 \)
\( \implies 7x = 260 - 50 \) (Taking 50 to the right side)
\( \implies 7x = 210 \)
\( \implies \frac{7x}{7} = \frac{210}{7} \) (Dividing both sides by 7)
\( \implies x = 30 \) The number is 30.
In simple words: Let the unknown number be 'x'. The riddle means \( 7x + 50 = 300 - 40 \). Solve this equation to find 'x'.

Exam Tip: For riddles, break down each phrase into a mathematical operation or value to construct the equation accurately.