GSEB Class 7 Maths Solutions Chapter 4 Simple Equations Exercise 4.4

Question 1. Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number, you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \( \frac{5}{2} \) of the number, the result is 23.
Answer:
(a) Let the unknown number be \( x \).
Eight times this number equals \( 8x \).
As per the given situation, we have:
\( 8x + 4 = 60 \)
Moving 4 to the right side of the equation:
\( 8x = 60 - 4 \)
\( 8x = 56 \)
Dividing both sides by 8, we obtain:
\( \frac{8x}{8} = \frac{56}{8} \)
\( x = 7 \)
So, the missing number is 7.
In simple words: First, you choose a letter for the unknown number. Then, you write down the math problem exactly as the words describe it. Finally, you solve the equation to find out what that unknown number is.

Exam Tip: Always define your variable clearly at the start of your solution. Double-check your arithmetic after isolating the variable.

Answer:
(b) Let the number be \( x \).
One-fifth of the number is \( \frac{1}{5}x \).
According to the problem, we get:
\( \frac{1}{5}x - 4 = 3 \)
Moving -4 to the right side of the equation:
\( \frac{1}{5}x = 3 + 4 \)
\( \frac{x}{5} = 7 \)
Multiplying both sides by 5, we have:
\( \frac{x}{5} \times 5 = 7 \times 5 \)
\( x = 35 \)
Thus, the missing number is 35.
In simple words: To find the number, turn the words into a math sentence. Then, move numbers around to get the letter by itself on one side and the answer on the other.

Exam Tip: Remember that "one-fifth of a number" means dividing the number by 5 or multiplying it by \( \frac{1}{5} \).

Answer:
(c) Let the number be \( x \).
Three-fourths of the number is \( \frac{3}{4}x \).
As per the given condition, we have:
\( \frac{3}{4}x + 3 = 21 \)
Moving 3 to the right side of the equation:
\( \frac{3}{4}x = 21 - 3 \)
\( \frac{3}{4}x = 18 \)
Multiplying both sides by \( \frac{4}{3} \) to isolate \( x \):
\( \frac{3}{4}x \times \frac{4}{3} = 18 \times \frac{4}{3} \)
\( x = 6 \times 4 \)
\( x = 24 \)
Therefore, the missing number is 24.
In simple words: Set up the equation from the words, then move the constant term to the other side. Finally, multiply by the inverse of the fraction to get the number.

Exam Tip: To clear a fraction from a variable, multiply both sides of the equation by the reciprocal of that fraction.

Answer:
(d) Let the number be \( x \).
Twice the number is \( 2x \).
According to the condition, we have:
\( 2x - 11 = 15 \)
Moving -11 to the right side of the equation:
\( 2x = 15 + 11 \)
\( 2x = 26 \)
Dividing both sides by 2, we get:
\( \frac{2x}{2} = \frac{26}{2} \)
\( x = 13 \)
Hence, the missing number is 13.
In simple words: If you subtract a number from two times another number, you get a result. To find that number, add the subtracted value to the result, then divide by two.

Exam Tip: Be careful with signs when transposing terms. A negative term becomes positive when moved to the other side.

Answer:
(e) Let Munna have \( x \) notebooks.
Thrice the number of notebooks is \( 3x \).
Now, as per the condition, we have:
\( 50 - 3x = 8 \)
Moving 50 to the right side of the equation:
\( -3x = 8 - 50 \)
\( -3x = -42 \)
Dividing both sides by (-3), we get:
\( \frac{-3x}{-3} = \frac{-42}{-3} \)
\( x = 14 \)
Thus, the number of notebooks Munna has is 14.
In simple words: When you take three times a number of notebooks from 50, and you get 8, it means that three times the number of notebooks is 42. Divide 42 by 3 to find the actual number of notebooks.

Exam Tip: When dividing by a negative number, ensure both the numerator and denominator's signs are handled correctly to get a positive result if applicable.

Answer:
(f) Let the number be \( x \).
Then, according to the condition, we have:
\( \frac{x + 19}{5} = 8 \)
Multiplying both sides by 5, we have:
\( \frac{x + 19}{5} \times 5 = 8 \times 5 \)
\( x + 19 = 40 \)
Moving 19 to the right side of the equation:
\( x = 40 - 19 \)
\( x = 21 \)
Thus, the missing number is 21.
In simple words: If you add 19 to a number and then divide the whole thing by 5 to get 8, first multiply 8 by 5. Then, subtract 19 from that result to find the original number.

Exam Tip: Always perform operations in reverse order of PEMDAS/BODMAS when solving equations (undo division, then undo addition/subtraction).

Answer:
(g) Let the number be \( x \).
\( \frac{5}{2} \) of the number is \( \frac{5}{2}x \).
According to the condition, we have:
\( \frac{5}{2}x - 7 = 23 \)
Moving -7 from the left side to the right side of the equation:
\( \frac{5}{2}x = 23 + 7 \)
\( \frac{5}{2}x = 30 \)
Dividing both sides by \( \frac{5}{2} \) (which is the same as multiplying by \( \frac{2}{5} \)), we have:
\( x = 30 \times \frac{2}{5} \)
\( x = \frac{60}{5} \)
\( x = 12 \)
Thus, the missing number is 12.
In simple words: If taking 7 away from five-halves of a number gives 23, first add 7 to 23. Then, to find the number, multiply that sum by the inverse of five-halves, which is two-fifths.

Exam Tip: To divide by a fraction, multiply by its reciprocal. This is a common step when solving equations involving fractions.

Question 2. Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? {Remember, the sum of three angles of a triangle is 180°.)
(c) Sachin scored twice as many runs as Rahul. Together their runs fell two short of a double century. How many runs did each one score?
Answer:
(a) Let the lowest marks scored be \( x \).
Then, twice the lowest marks is \( 2x \).
According to the condition given, we have:
[Twice the lowest marks] \( + 7 = \) [Highest marks]
\( 2x + 7 = 87 \)
Moving 7 from the left side to the right side:
\( 2x = 87 - 7 \)
\( 2x = 80 \)
Dividing both sides by 2, we get:
\( \frac{2x}{2} = \frac{80}{2} \)
\( x = 40 \)
Therefore, the lowest marks are 40.
In simple words: The highest score is 87. It's also 7 more than twice the lowest score. So, twice the lowest score must be 80. Then, the lowest score is half of 80, which is 40.

Exam Tip: Read word problems carefully to identify the "is" (equals sign), "twice" (multiply by 2), and "plus" (addition) keywords for setting up the equation correctly.

Answer:
(b) Let the base angle be \( x^\circ \).
Since it's an isosceles triangle, the other base angle is also \( x^\circ \).
The vertex angle is given as \( 40^\circ \).
The sum of the angles in a triangle is \( x^\circ + x^\circ + 40^\circ \).
This sum is equal to \( 2x^\circ + 40^\circ \).
According to the condition (sum of angles in a triangle is \( 180^\circ \)), we have:
\( 2x^\circ + 40^\circ = 180^\circ \)
Moving \( 40^\circ \) from the left side to the right side:
\( 2x^\circ = 180^\circ - 40^\circ \)
\( 2x^\circ = 140^\circ \)
Dividing both sides by 2, we have:
\( \frac{2x^\circ}{2} = \frac{140^\circ}{2} \)
\( x^\circ = 70^\circ \)
So, the base angles of the triangle are \( 70^\circ \) each.
In simple words: In an isosceles triangle, the two base angles are the same. If the top angle is 40 degrees and all angles add up to 180 degrees, then the two base angles together must make 140 degrees. Divide 140 by 2 to get 70 degrees for each base angle.

Exam Tip: Always remember the fundamental property that the sum of interior angles in any triangle is \( 180^\circ \).

Answer:
(c) Let the number of runs scored by Rahul be \( x \).
Since Sachin scored twice as many runs as Rahul, Sachin's runs are \( 2x \).
The sum of their runs is \( x + 2x = 3x \).
According to the condition, their combined runs were two short of a double century.
A double century is \( 2 \times 100 = 200 \) runs.
So, the sum of their runs \( = 200 - 2 \)
\( 3x = 198 \)
Dividing both sides by 3, we have:
\( \frac{3x}{3} = \frac{198}{3} \)
\( x = 66 \)
Thus, Rahul scored 66 runs.
And Sachin's runs \( = 2x = 2 \times 66 = 132 \) runs.
Therefore, Rahul scored 66 runs and Sachin scored 132 runs.
In simple words: Rahul and Sachin scored runs, with Sachin scoring double Rahul's runs. Their total score was 2 short of 200, making it 198 runs. Since Sachin scored twice as much, their combined score is three times Rahul's score. Divide 198 by 3 to find Rahul's score, then double that for Sachin's score.

Exam Tip: Translate "double century" into a numerical value first (200), and "two short of" means subtracting 2 from that value. This helps in forming the equation accurately.

Question 3. Solve the following:
(i) Irfan says that he has 1 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. What is Laxmi's age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruits trees planted if the number of non-fruit trees planted was 77?
Answer:
(i) Let the number of marbles Parmit has be \( x \).
Five times \( x \) is \( 5x \).
Irfan has 1 marble more than five times Parmit's marbles, so Irfan has \( 5x + 1 \) marbles. (Note: The OCR says "1 marbles more" but the solution uses "7", so I'm using "7" to be consistent with the provided solution steps, assuming the question had a typo or was reworded.)
Irfan has 37 marbles.
According to the condition, we have:
\( 5x + 7 = 37 \)
Moving 7 from the left side to the right side:
\( 5x = 37 - 7 \)
\( 5x = 30 \)
Dividing both sides by 5, we have:
\( \frac{5x}{5} = \frac{30}{5} \)
\( x = 6 \)
Therefore, Parmit has 6 marbles.
In simple words: If Irfan has 37 marbles, and that's 7 more than five times Parmit's marbles, then five times Parmit's marbles must be 30. Divide 30 by 5 to find out Parmit has 6 marbles.

Exam Tip: Be careful with phrases like "more than" and "times." "Five times x plus 7" translates to \( 5x + 7 \), not \( 5(x+7) \).

Answer:
(ii) Let Laxmi's age be \( x \) years.
Three times Laxmi's age is \( 3x \).
Laxmi's father is 4 years older than three times Laxmi's age, so his age is \( 3x + 4 \).
Laxmi's father is 49 years old.
According to the condition, we have:
\( 3x + 4 = 49 \)
Moving 4 from the left side to the right side:
\( 3x = 49 - 4 \)
\( 3x = 45 \)
Dividing both sides by 3, we have:
\( \frac{3x}{3} = \frac{45}{3} \)
\( x = 15 \)
Therefore, Laxmi's age is 15 years.
In simple words: Laxmi's father is 49 years old, which is 4 years more than three times Laxmi's age. So, three times Laxmi's age must be 45. Divide 45 by 3 to find that Laxmi is 15 years old.

Exam Tip: When dealing with age problems, clearly define the age of each person using variables and set up the equation based on their relationship.

Answer:
(iii) Let the number of fruit trees be \( x \).
Three times the number of fruit trees is \( 3x \).
The number of non-fruit trees is 2 more than three times the number of fruit trees, so it is \( 2 + 3x \).
The number of non-fruit trees planted was 77.
According to the condition, we have:
\( 2 + 3x = 77 \)
Moving 2 from the left side to the right side:
\( 3x = 77 - 2 \)
\( 3x = 75 \)
Dividing both sides by 3, we have:
\( x = \frac{75}{3} \)
\( x = 25 \)
Therefore, the number of fruit trees is 25.
In simple words: If 77 non-fruit trees were planted, and this number is 2 more than three times the fruit trees, then three times the fruit trees must be 75. Dividing 75 by 3 reveals there were 25 fruit trees planted.

Exam Tip: Always clearly identify what each variable represents to avoid confusion, especially in problems with multiple quantities like fruit and non-fruit trees.

Question 4. Solve the following riddle:
I am a number, Tell my identity!
Take me seven times over And add a fifty!
To reach a triple century You still need forty!
Answer:
Let the missing number be \( x \).
"Take me seven times over" means \( 7x \).
"And add a fifty" means \( 7x + 50 \).
A "triple century" is \( 3 \times 100 = 300 \).
"To reach a triple century You still need forty!" means the current value is \( 300 - 40 \).
So, according to the riddle, we have:
\( 7x + 50 = 300 - 40 \)
\( 7x + 50 = 260 \)
Moving 50 from the left side to the right side:
\( 7x = 260 - 50 \)
\( 7x = 210 \)
Dividing both sides by 7, we have:
\( \frac{7x}{7} = \frac{210}{7} \)
\( x = 30 \)
Therefore, the missing number is 30.
In simple words: The riddle describes a number. If you multiply it by 7 and add 50, you get a value that is 40 less than 300. This means \( 7x + 50 \) equals 260. To find \( x \), subtract 50 from 260, which gives 210. Then, divide 210 by 7, and the number is 30.

Exam Tip: Break down riddles or complex word problems into smaller, manageable phrases. Translate each phrase into a mathematical expression to build the full equation step-by-step.