CBSE Class 7 Mathematics Simple Equations Worksheet Set 03

I. SOLVE

Question I(i). 7y + 5 = 19
Answer: \( y = 2 \)
To solve this equation, first move 5 to the other side by subtracting it from 19:
\( 7y = 19 - 5 \)
\( 7y = 14 \)
Now, divide both sides by 7 to find y:
\( y = \frac{14}{7} \)
\( y = 2 \)
In simple words: Subtract 5 from 19 to get 14, and then divide 14 by 7 to get 2.

Exam Tip: Always perform the inverse operation to move numbers to the other side of the equals sign.

Question I(ii). \( \frac{3k + 9}{6} = 5 \)
Answer: \( k = 7 \)
First, multiply both sides by 6 to remove the fraction:
\( 3k + 9 = 5 \times 6 \)
\( 3k + 9 = 30 \)
Next, subtract 9 from both sides:
\( 3k = 30 - 9 \)
\( 3k = 21 \)
Finally, divide both sides by 3 to find k:
\( k = \frac{21}{3} \)
\( k = 7 \)
In simple words: Multiply 5 by 6 to get 30, then subtract 9 to get 21, and divide by 3 to get 7.

Exam Tip: When a whole side is divided by a number, multiply by that number first to clear the fraction.

Question I(iii). a + 5 = 11
Answer: \( a = 6 \)
Subtract 5 from both sides of the equation:
\( a = 11 - 5 \)
\( a = 6 \)
In simple words: Take 5 away from 11 to find that a is equal to 6.

Exam Tip: Simple equations can be checked quickly by plugging your answer back in to see if both sides are equal.

Question I(iv). 12a + 4 = 40
Answer: \( a = 3 \)
First, subtract 4 from both sides:
\( 12a = 40 - 4 \)
\( 12a = 36 \)
Now, divide both sides by 12:
\( a = \frac{36}{12} \)
\( a = 3 \)
In simple words: Subtract 4 from 40 to get 36, then divide 36 by 12 to get 3.

Exam Tip: Remember to carry out additions or subtractions before doing multiplications or divisions when solving linear equations.

Question I(v). \( \frac{2x}{4} = 11 \)
Answer: \( x = 22 \)
First, simplify the fraction on the left:
\( \frac{x}{2} = 11 \)
Now, multiply both sides by 2 to find x:
\( x = 11 \times 2 \)
\( x = 22 \)
In simple words: Multiply 11 by 4 to get 44, and then divide by 2 to find 22.

Exam Tip: You can either simplify the fraction first or multiply both sides directly; both methods yield the correct answer.

Question I(vi). \( \frac{3m}{7} = 9 \)
Answer: \( m = 21 \)
First, multiply both sides by 7 to eliminate the denominator:
\( 3m = 9 \times 7 \)
\( 3m = 63 \)
Next, divide both sides by 3 to solve for m:
\( m = \frac{63}{3} \)
\( m = 21 \)
In simple words: Multiply 9 by 7 to get 63, then divide 63 by 3 to get 21.

Exam Tip: Keep your calculations clean by writing down each step on a new line.

Question I(vii). 2a – 4 = 6
Answer: \( a = 5 \)
First, add 4 to both sides of the equation:
\( 2a = 6 + 4 \)
\( 2a = 10 \)
Next, divide both sides by 2:
\( a = \frac{10}{2} \)
\( a = 5 \)
In simple words: Add 4 to 6 to get 10, then divide 10 by 2 to get 5.

Exam Tip: Changing a minus sign to a plus sign when shifting terms across the equals sign is crucial.

Question I(viii). 4 (x + 3 ) = 20
Answer: \( x = 2 \)
First, divide both sides by 4 to open the brackets:
\( x + 3 = \frac{20}{4} \)
\( x + 3 = 5 \)
Now, subtract 3 from both sides:
\( x = 5 - 3 \)
\( x = 2 \)
In simple words: Divide 20 by 4 to get 5, then subtract 3 to get 2.

Exam Tip: Dividing by the coefficient outside the bracket is often faster than expanding the bracket.

Question I(ix). 8x - 5 = 27
Answer: \( x = 4 \)
First, add 5 to both sides:
\( 8x = 27 + 5 \)
\( 8x = 32 \)
Next, divide both sides by 8:
\( x = \frac{32}{8} \)
\( x = 4 \)
In simple words: Add 5 to 27 to get 32, then divide 32 by 8 to get 4.

Exam Tip: Make sure you double-check simple addition and division to avoid making small errors.

Question I(x). \( \frac{2m}{7} + 6 = 10 \)
Answer: \( m = 14 \)
First, subtract 6 from both sides:
\( \frac{2m}{7} = 10 - 6 \)
\( \frac{2m}{7} = 4 \)
Next, multiply both sides by 7:
\( 2m = 4 \times 7 \)
\( 2m = 28 \)
Finally, divide both sides by 2:
\( m = \frac{28}{2} \)
\( m = 14 \)
In simple words: Subtract 6 from 10 to get 4, multiply by 7 to get 28, and divide by 2 to get 14.

Exam Tip: Isolate the fraction term first before multiplying by the denominator.

Question I(xi). 3 ( 2x - 1 ) + 5 = 14
Answer: \( x = 2 \)
First, subtract 5 from both sides:
\( 3(2x - 1) = 14 - 5 \)
\( 3(2x - 1) = 9 \)
Next, divide both sides by 3:
\( 2x - 1 = \frac{9}{3} \)
\( 2x - 1 = 3 \)
Now, add 1 to both sides:
\( 2x = 3 + 1 \)
\( 2x = 4 \)
Finally, divide both sides by 2:
\( x = \frac{4}{2} \)
\( x = 2 \)
In simple words: Take 5 from 14 to get 9, divide by 3 to get 3, add 1 to get 4, then divide by 2 to get 2.

Exam Tip: Solving step-by-step from outside to inside the bracket keeps your work clean and correct.

Question I(xii). 2z – ( 7 – 5z ) – 21 = 0
Answer: \( z = 4 \)
First, open the brackets carefully by changing the signs inside:
\( 2z - 7 + 5z - 21 = 0 \)
Combine the like terms together:
\( (2z + 5z) + (-7 - 21) = 0 \)
\( 7z - 28 = 0 \)
Now, add 28 to both sides:
\( 7z = 28 \)
Finally, divide both sides by 7:
\( z = \frac{28}{7} \)
\( z = 4 \)
In simple words: Combine the z terms to get 7z, group the numbers to get -28, move 28 to the other side, and divide by 7 to get 4.

Exam Tip: Be extra careful with minus signs in front of brackets because they flip the signs inside.

Question I(xiii). \( \frac{8x}{3} + 4 = 12 \)
Answer: \( x = 3 \)
First, subtract 4 from both sides:
\( \frac{8x}{3} = 12 - 4 \)
\( \frac{8x}{3} = 8 \)
Next, multiply both sides by 3:
\( 8x = 8 \times 3 \)
\( 8x = 24 \)
Finally, divide both sides by 8:
\( x = \frac{24}{8} \)
\( x = 3 \)
In simple words: Take 4 away from 12 to get 8, multiply by 3 to get 24, then divide by 8 to get 3.

Exam Tip: Check your final value by placing it back into the fraction to ensure the equality holds true.

Question I(xiv). 3 ( y – 1 ) = 2 ( 2y – 6 )
Answer: \( y = 9 \)
First, expand both sides by multiplying through the brackets:
\( 3y - 3 = 4y - 12 \)
Now, move the y terms to one side and the constant numbers to the other:
\( 12 - 3 = 4y - 3y \)
\( 9 = y \)
\( y = 9 \)
In simple words: Expand the brackets to get 3y - 3 and 4y - 12, then rearrange the parts to find that y equals 9.

Exam Tip: Moving the smaller variable term to the side of the larger one prevents negative coefficients.

Question I(xv). 5 – 6 ( x - 2 ) + 13 = 6
Answer: \( x = 4 \)
First, open the bracket by expanding:
\( 5 - 6x + 12 + 13 = 6 \)
Combine the constant numbers together:
\( (5 + 12 + 13) - 6x = 6 \)
\( 30 - 6x = 6 \)
Now, rearrange the terms to solve for x:
\( 30 - 6 = 6x \)
\( 24 = 6x \)
Finally, divide by 6:
\( x = \frac{24}{6} \)
\( x = 4 \)
In simple words: Simplify the numbers to get 30 - 6x = 6, then subtract 6 from 30 to get 24, and divide by 6 to get 4.

Exam Tip: Be careful to apply the negative sign to all terms inside the parentheses when expanding.

II. Set up equations and solve.

Question II(i). Add 7 to 3times a number, you get 34. Find the number.
Answer: \( x = 9 \)
Let the unknown number be \( x \).
According to the statement, 7 added to 3 times the number is equal to 34:
\( 3x + 7 = 34 \)
To solve this equation, subtract 7 from both sides:
\( 3x = 34 - 7 \)
\( 3x = 27 \)
Now, divide by 3 to find the number:
\( x = \frac{27}{3} \)
\( x = 9 \)
The required number is 9.
In simple words: Write the equation as 3x + 7 = 34, then subtract 7 to get 27, and divide by 3 to get 9.

Exam Tip: Translate words like "times" into multiplication and "get" into an equals sign when setting up simple equations.

Question II(ii). Subtract 23 from thrice a number the result is 7.find the number.
Answer: \( x = 10 \)
Let the unknown number be \( x \).
The statement tells us that subtracting 23 from 3 times the number yields 7:
\( 3x - 23 = 7 \)
Add 23 to both sides to isolate the variable term:
\( 3x = 7 + 23 \)
\( 3x = 30 \)
Now, divide by 3 to solve for x:
\( x = \frac{30}{3} \)
\( x = 10 \)
The required number is 10.
In simple words: Set up the equation 3x - 23 = 7, add 23 to 7 to get 30, then divide by 3 to get 10.

Exam Tip: Pay attention to the order of subtraction; "subtract from" means the subtracted value comes after the variable term.

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Question II(iii). Anmol’s mother is 4 years more than 3times Anmol’s age. Find Anmol’s age if mother is 40 years old.
Answer: Anmol's age is 12 years old.
Let Anmol's age be \( x \) years.
The mother's age is 4 years more than 3 times Anmol's age, which is \( 3x + 4 \) years.
We are given that the mother is 40 years old:
\( 3x + 4 = 40 \)
To solve for x, subtract 4 from both sides:
\( 3x = 40 - 4 \)
\( 3x = 36 \)
Now, divide by 3:
\( x = \frac{36}{3} \)
\( x = 12 \)
Anmol's age is 12 years.
In simple words: Set up the equation 3x + 4 = 40, subtract 4 to get 36, then divide 36 by 3 to find Anmol's age of 12.

Exam Tip: Define the variable clearly first (e.g., let Anmol's age be x) to avoid getting confused between the ages.

Question II(iv). Tony scored twice as many runs as Kohli. Together they scored 6 short of triple century. How many runs did each score?
Answer: Tony scored 196 runs and Kohli scored 98 runs.
Let Kohli's score be \( x \) runs.
Then, Tony's score is \( 2x \) runs.
Together, their total score is \( x + 2x = 3x \).
A triple century is 300 runs. "6 short of triple century" means:
\( 300 - 6 = 294 \) runs.
Set up the equation:
\( 3x = 294 \)
Divide both sides by 3 to find x:
\( x = \frac{294}{3} \)
\( x = 98 \)
So, Kohli's score is 98 runs.
Tony's score is:
\( 2 \times 98 = 196 \) runs.
In simple words: Kohli scored 98 runs and Tony scored 196 runs. Together they scored 294 runs, which is 6 runs less than 300.

Exam Tip: Always state the final scores of both individuals separately to secure full marks on word problems.

Question II(v). The three angles of a triangle are in the ratio 2:3:5. Find the measure of each angle. Classify the triangle
Answer: The angles are \( 36^\circ \), \( 54^\circ \), and \( 90^\circ \); the triangle is a right-angled triangle.
Let the three angles of the triangle be \( 2x \), \( 3x \), and \( 5x \).
The sum of the interior angles of any triangle is \( 180^\circ \):
\( 2x + 3x + 5x = 180 \)
\( 10x = 180 \)
Divide both sides by 10 to find x:
\( x = 18 \)
Now, calculate the measure of each angle:
First angle: \( 2 \times 18 = 36^\circ \)
Second angle: \( 3 \times 18 = 54^\circ \)
Third angle: \( 5 \times 18 = 90^\circ \)
Since the largest angle is \( 90^\circ \), the triangle is classified as a right-angled triangle.
In simple words: Add the ratio parts to get 10, divide 180 by 10 to get 18, and multiply each ratio by 18 to find the angles.

Exam Tip: Always use the angle sum property of a triangle (which is 180 degrees) to set up ratio equations involving angles.