CBSE Class 8 Mathematics Linear Equations In One Variable Worksheet Set 02

Question. Which of the following are the consecutive multiples of 3 whose sum is 51 ?
(A) 24, 27
(B) 20, 31
(C) 40, 11
(D) 25, 26
Answer : A

Question. If 2 - (x/3) = 4 , what is the value of'x'?
(A) -6
(B) 6
(C) -4
(D) 4
Answer : A

Question. The digits in the tens place and units place of a two digit number are t and u respectively. If 1 is placed between them, what is the value of the three d igit number so formed ?
(A) 1 OOu + t + 1 0
(B) 1 OOt + 1 0 + u
(C) 1 Ot + u + 1
(D) 1 OOt + 1 Ou + 1
Answer : B

Question. How much pure alcohol should be added to 400 ml of strength 15% to make its strength 32% ?
(A) 50 m/
(B) 75 m/
(C) 100m/
(D) 150m/
Answer : C

Question. Which of the following is the solution of 2x/5 = 10/3 ?
(A) X= 25/3
(B) X= 35/3
(C) X= 45/2
(D) X= 55/2
Answer : A

Question. What is the solution of (x + 4)² - (x - 5)² = 9 ?
(A) x = 1
(B) x = 2
(C) X = 3
(D) X = 4
Answer : A

Question. There are 3 bags of flour. Bag B weighs thrice as much as bag A and bag C weighs 4 kg less than bag B. If bag A weighs 6 kg, how heavy is bag C ?
(A) 52 kg
(B) 18 kg
(C) 22 kg
(D) 14 kg
Answer : D

Question. One number is 3 times another number. If 15 is added to both the numbers, one of the new numbers becomes twice that of the other new number. Find the numbers.
(A) 15 and 60
(B) 30 and 45
(C) 15 and 30
(D) 15 and 45
Answer : D

Question.  The width of Sudha's garden is 2/3 of its length. If its perimeter is 40 m, find its dimensions.
(A) length = 12 m and breadth = 8 m
(B) length = 8 m and breadth = 12 m
(C) length = 12 m and breadth = 6 m
(D) length = 8 m and breadth = 4 m
Answer : A

Question. A boy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum.
He does 24 sums and obtains 37 marks.
What was the number of correct sums ?
(A) 20
(B) 17
(C) 31
(D) 19
Answer : B

Question. If 9- 7x = 5- 3x, what is the value of'x'?
(A) 0
(B) 1/4
(C) 1/2
(D) 1
Answer : D

Question. The sum of two numbers is 35 and their difference is 13. What are the numbers ?
(A) 10,25
(B) 22, 13
(C) 24, 11
(D) 20,15
Answer : C

Question. Arif, Ajay and Madhav secured m + 20, 2m - 17 and m - 5 marks respectively in an English test. What is their average marks?
(A) 4m-8/3
(B) 4m- 2/3
(C) 4m + 8/3
(D) 4m+ 2/3
Answer : B

Question. Three numbers are such that the second is twice the first and the third is thrice the first. If their sum is 1 2 , what is the value of the largest number?
(A) 4
(B) 2
(C) 6
(D) 8
Answer : C

Question. If a number increased by 8% of itself gives 135, what is the number ?
(A) 112
(B) 100
(C) 125
(D) 75
Answer : C

Question. When 75% of a number is added to 75 the result is the number again. What is the number?
(A) 150
(B) 300
(C) 100
(D) 450
Answer : B

Question. Subtracting 5 from thrice x results in 4. 
What is the value of 15 subtracted from nine times 'x ' ?
(A) 15
(B) 3
(C) 27
(D) 12
Answer : D

Question. X is a two-digit number. Y is the number obtained on reversing the digits of X.
Which of the following is true ?
(A) X+ Y is divisible by 10.
(B) X - Y is divisible by 6.
(C) X - Y is divisible by 9.
(D) X + Y is divisible by 8.
Answer : C

LINEAR EQUATIONS

Question 1. Solve for x :
i. 4x - 7 - (x + 4) = 3x + 4 - (2x - 1)
ii. \(\frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8\)
Answer: Part (i):
Given equation: \(4x - 7 - (x + 4) = 3x + 4 - (2x - 1)\)
Remove brackets: \(4x - 7 - x - 4 = 3x + 4 - 2x + 1\)
Combine like terms on both sides: \(3x - 11 = x + 5\)
Transpose x terms to the left and constants to the right: \(3x - x = 5 + 11\)
Simplify: \(2x = 16\)
Divide by 2: \(x = 8\)

Part (ii):
Given equation: \(\frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8\)
Expand the numerator: \(\frac{34 - 17x - 5x - 60}{1 - 7x} = 8\)
Simplify numerator: \(\frac{-22x - 26}{1 - 7x} = 8\)
Cross-multiply: \(-22x - 26 = 8(1 - 7x)\)
Expand right side: \(-22x - 26 = 8 - 56x\)
Transpose terms: \(56x - 22x = 8 + 26\)
Simplify: \(34x = 34\)
Divide by 34: \(x = 1\)
In simple words: Open the brackets carefully, collect all variable terms on one side, and isolate the variable x.

Exam Tip: Be extra careful with negative signs outside parentheses when expanding terms like \(-(x + 4)\) or \(-5(x + 12)\).

Question 2. The sum of three consecutive even numbers is 30. Find the numbers.
Answer: Let the three consecutive even numbers be \(x\), \(x + 2\), and \(x + 4\).
According to the question, their sum is 30:
\(x + (x + 2) + (x + 4) = 30\)
Combine like terms:
\(3x + 6 = 30\)
Transpose 6 to the right side:
\(3x = 30 - 6\)
\(3x = 24\)
Divide by 3:
\(x = \frac{24}{3} = 8\)
Therefore, the three numbers are:
First number: \(x = 8\)
Second number: \(x + 2 = 8 + 2 = 10\)
Third number: \(x + 4 = 8 + 4 = 12\)
Check: \(8 + 10 + 12 = 30\).
In simple words: Consecutive even numbers differ by 2, so represent them as x, x + 2, and x + 4, then form an equation and solve.

Exam Tip: Consecutive even and odd numbers both increase by 2, so always use \(x\), \(x + 2\), and \(x + 4\) for both cases.

Question 3. The sum of three consecutive odd numbers is 63. Find the numbers.
Answer: Let the three consecutive odd numbers be \(x\), \(x + 2\), and \(x + 4\).
According to the question, their sum is 63:
\(x + (x + 2) + (x + 4) = 63\)
Combine like terms:
\(3x + 6 = 63\)
Transpose 6 to the right side:
\(3x = 63 - 6\)
\(3x = 57\)
Divide by 3:
\(x = \frac{57}{3} = 19\)
Therefore, the three numbers are:
First number: \(x = 19\)
Second number: \(x + 2 = 19 + 2 = 21\)
Third number: \(x + 4 = 19 + 4 = 23\)
Check: \(19 + 21 + 23 = 63\).
In simple words: Even and odd consecutive numbers both increment by 2. Solve for x and find the sequence.

Exam Tip: Always verify your final answer by adding the numbers together to check if they match the given sum.

Question 4. The sum of two twin primes is 60. Find the two prime numbers.
Answer: Let the two twin prime numbers be \(x\) and \(x + 2\).
According to the question, their sum is 60:
\(x + (x + 2) = 60\)
Combine like terms:
\(2x + 2 = 60\)
Transpose 2:
\(2x = 60 - 2\)
\(2x = 58\)
Divide by 2:
\(x = 29\)
Therefore, the first prime number is 29, and the second prime number is \(29 + 2 = 31\).
Check: 29 and 31 are both prime numbers and twin primes (difference of 2), and their sum is \(29 + 31 = 60\).
In simple words: Twin primes are pairs of prime numbers that have a difference of 2. Set up the equation \(x + x + 2 = 60\) and solve.

Exam Tip: Remember that twin primes are prime numbers with a difference of 2 (like 11 and 13, or 29 and 31).

Question 5. The measures of the angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.
Answer: Let the measures of the angles of the triangle be \(1x\), \(2x\), and \(3x\).
We know that the sum of all angles in a triangle is \(180^\circ\):
\(1x + 2x + 3x = 180^\circ\)
Combine like terms:
\(6x = 180^\circ\)
Divide by 6:
\(x = \frac{180^\circ}{6} = 30^\circ\)
Now, find the individual angles:
First angle = \(1x = 1 \times 30^\circ = 30^\circ\)
Second angle = \(2x = 2 \times 30^\circ = 60^\circ\)
Third angle = \(3x = 3 \times 30^\circ = 90^\circ\)
In simple words: When angles are given in a ratio, use a variable x to represent them, add them to equal \(180^\circ\), and solve for x.

Exam Tip: Always include the degree symbol (\(^\circ\)) in your final answer when finding angle measures.

Question 6. The numerator of a fraction is 3 less than its denominator. If we add 1 to both numerator and denominator, it becomes equal to \(\frac{4}{5}\). Find the fraction.
Answer: Let the denominator of the fraction be \(x\).
Since the numerator is 3 less than the denominator, the numerator is \(x - 3\).
The original fraction is \(\frac{x - 3}{x}\).
According to the question, if we add 1 to both numerator and denominator, the fraction becomes \(\frac{4}{5}\):
\(\frac{(x - 3) + 1}{x + 1} = \frac{4}{5}\)
Simplify the numerator:
\(\frac{x - 2}{x + 1} = \frac{4}{5}\)
Cross-multiply:
\(5(x - 2) = 4(x + 1)\)
Expand both sides:
\(5x - 10 = 4x + 4\)
Transpose terms:
\(5x - 4x = 4 + 10\)
\(x = 14\)
Thus, the denominator is 14, and the numerator is \(14 - 3 = 11\).
Therefore, the fraction is \(\frac{11}{14}\).
In simple words: Express numerator in terms of denominator, set up the modified fraction equation, cross-multiply, and solve.

Exam Tip: Write down the fraction clearly at the end by substituting the value of x back into your assumed expression.

Question 7. Renu’s mother is four times as old as Renu. After 5 years her mother will be three times as old as she will then be. Find their present ages.
Answer: Let Renu's present age be \(x\) years.
Then, her mother's present age is \(4x\) years.
After 5 years:
Renu's age will be \(x + 5\) years.
Mother's age will be \(4x + 5\) years.
According to the question, after 5 years, the mother will be three times as old as Renu:
\(4x + 5 = 3(x + 5)\)
Expand the right side:
\(4x + 5 = 3x + 15\)
Transpose terms:
\(4x - 3x = 15 - 5\)
\(x = 10\)
Therefore:
Renu's present age \(x = 10\) years.
Mother's present age \(4x = 4 \times 10 = 40\) years.
In simple words: Add 5 to both ages for the future condition, set up the equation according to the multiple given, and solve for x.

Exam Tip: In age problems, remember to add (or subtract) the given number of years to both persons' ages before setting up the equation.

Question 8. The sum of four consecutive multiples of 7 is 70. Find these multiples.
Answer: Let the four consecutive multiples of 7 be \(7x\), \(7(x + 1)\), \(7(x + 2)\), and \(7(x + 3)\), or simply let them be represented as \(7x\), \(7x + 7\), \(7x + 14\), and \(7x + 21\).
According to the question, their sum is 70:
\(7x + (7x + 7) + (7x + 14) + (7x + 21) = 70\)
Combine like terms:
\(28x + 42 = 70\)
Transpose 42:
\(28x = 70 - 42\)
\(28x = 28\)
Divide by 28:
\(x = 1\)
Now, find the four multiples:
First multiple: \(7x = 7 \times 1 = 7\)
Second multiple: \(7x + 7 = 7 + 7 = 14\)
Third multiple: \(7x + 14 = 7 + 14 = 21\)
Fourth multiple: \(7x + 21 = 7 + 21 = 28\)
Check: \(7 + 14 + 21 + 28 = 70\).
In simple words: Consecutive multiples of 7 increase by 7 each time. Set them up as \(7x, 7x+7, 7x+14, 7x+21\) and solve.

Exam Tip: Factoring out 7 makes the equation \(7(x + x+1 + x+2 + x+3) = 70\) very easy to solve as \(4x + 6 = 10\).

Question 9. The sum of two numbers is 50. If the larger number is divided by the smaller number we get 4. Find the numbers.
Answer: Let the smaller number be \(x\).
Since the sum of the two numbers is 50, the larger number is \(50 - x\).
According to the question, when the larger number is divided by the smaller number, the result is 4:
\(\frac{50 - x}{x} = 4\)
Multiply both sides by \(x\):
\(50 - x = 4x\)
Transpose \(-x\) to the right side:
\(50 = 4x + x\)
\(50 = 5x\)
Divide by 5:
\(x = \frac{50}{5} = 10\)
Therefore:
Smaller number = \(10\)
Larger number = \(50 - 10 = 40\)
Check: \(40 \div 10 = 4\).
In simple words: If the smaller number is x, the larger is 50 - x. Set up the division equation equal to 4 and solve for x.

Exam Tip: Assigning variables to the smaller number often keeps expressions simpler and avoids negative coefficients.

Question 10. The perimeter of a triangle is 49 cm. One side is 7 cm longer than another side and 5 cm shorter than the third side. Find the sides.
Answer: Let the first side be \(x\).
According to the question:
- One side is 7 cm longer than the first side, so the second side is \(x + 7\).
- The same side is 5 cm shorter than the third side, which means the third side is 5 cm longer than the second side: \((x + 7) + 5 = x + 12\).
Alternatively, let the second side be \(x\). Then the first side is \(x - 7\), and the third side is \(x + 5\).
Let us use the first side as \(x\):
Side 1 = \(x\)
Side 2 = \(x + 7\)
Side 3 = \(x + 12\)
The perimeter of a triangle is the sum of all three sides, which is given as 49 cm:
\(x + (x + 7) + (x + 12) = 49\)
Combine like terms:
\(3x + 19 = 49\)
Transpose 19:
\(3x = 49 - 19\)
\(3x = 30\)
Divide by 3:
\(x = 10\)
Therefore, the lengths of the three sides are:
First side = \(10\) cm
Second side = \(10 + 7 = 17\) cm
Third side = \(10 + 12 = 22\) cm
Check: \(10 + 17 + 22 = 49\) cm.
In simple words: Express all three sides in terms of a single variable x, add them up to equal the perimeter 49, and solve.

Exam Tip: Read carefully to check whether the comparisons are relative to the first side or another specific side.

Question 11. In a quadrilateral ABCD, \(\angle A = (2x + 4)^\circ\), \(\angle B = (2x - 13)^\circ\), \(\angle C = (3x + 11)^\circ\) and \(\angle D = (4x - 5)^\circ\). Find the measures of the angles.
Answer: We know that the sum of all interior angles in a quadrilateral is \(360^\circ$\):
\(\angle A + \angle B + \angle C + \angle D = 360^\circ\)
Substitute the given expressions:
\((2x + 4) + (2x - 13) + (3x + 11) + (4x - 5) = 360\)
Combine like \(x\) terms and constant terms:
\((2x + 2x + 3x + 4x) + (4 - 13 + 11 - 5) = 360\)
\(11x - 3 = 360\)
Transpose -3 to the right side:
\(11x = 360 + 3\)
\(11x = 363\)
Divide by 11:
\(x = \frac{363}{11} = 33\)
Now, substitute \(x = 33\) into each angle expression:
\(\angle A = 2(33) + 4 = 66 + 4 = 70^\circ\)
\(\angle B = 2(33) - 13 = 66 - 13 = 53^\circ\)
\(\angle C = 3(33) + 11 = 99 + 11 = 110^\circ\)
\(\angle D = 4(33) - 5 = 132 - 5 = 127^\circ\)
Check: \(70^\circ + 53^\circ + 110^\circ + 127^\circ = 360^\circ\).
In simple words: The sum of angles in a quadrilateral is \(360^\circ\). Add all four algebraic expressions, solve for x, and substitute back.

Exam Tip: Always state the geometric property used (sum of angles of a quadrilateral is \(360^\circ\)) at the beginning of your solution.

Question 12. Half of the number of boys of Class 8 B went to the football ground to play. One-fourth of the number of boys went to the Library to take books. Remaining 10 boys went to the 3rd Language room. Find the number of boys of Class 8 B.
Answer: Let the total number of boys in Class 8 B be \(x\).
- Number of boys who went to the football ground = \(\frac{1}{2}x\)
- Number of boys who went to the library = \(\frac{1}{4}x\)
- Remaining boys who went to the 3rd Language room = \(10\)
The sum of all these groups equals the total number of boys \(x\):
\(\frac{1}{2}x + \frac{1}{4}x + 10 = x\)
Find a common denominator to add the fractions (LCM of 2 and 4 is 4):
\(\frac{2x}{4} + \frac{x}{4} + 10 = x\)
\(\frac{3x}{4} + 10 = x\)
Transpose \(\frac{3x}{4}\) to the right side:
\(10 = x - \frac{3x}{4}\)
\(10 = \frac{4x - 3x}{4}\)
\(10 = \frac{x}{4}\)
Multiply both sides by 4:
\(x = 40\)
Therefore, the total number of boys in Class 8 B is 40.
In simple words: Set the sum of the fractional parts and remaining boys equal to the total variable x, then solve the linear equation.

Question 13. In \(\triangle ABC\), \(\angle A = \angle C\) and \(\angle B = \angle A + \angle C\). Find the measures of the angles of the triangle.
Answer: We are given that \(\angle A = \angle C\).
We are also given that \(\angle B = \angle A + \angle C\).
Since \(\angle A = \angle C\), we can substitute \(\angle C\) with \(\angle A\) in the expression for \(\angle B\):
\(\angle B = \angle A + \angle A = 2\angle A\)
We know that the sum of the angles in any triangle is \(180^\circ\):
\(\angle A + \angle B + \angle C = 180^\circ\)
Substitute \(\angle B = 2\angle A\) and \(\angle C = \angle A\):
\(\angle A + 2\angle A + \angle A = 180^\circ\)
Combine like terms:
\(4\angle A = 180^\circ\)
Divide by 4:
\(\angle A = \frac{180^\circ}{4} = 45^\circ\)
Since \(\angle C = \angle A\):
\(\angle C = 45^\circ\)
Since \(\angle B = 2\angle A\):
\(\angle B = 2 \times 45^\circ = 90^\circ\)
Therefore, the measures of the angles are \(\angle A = 45^\circ\), \(\angle B = 90^\circ\), and \(\angle C = 45^\circ\).
In simple words: Express all angles in terms of \(\angle A\), use the angle sum property of triangles (\(180^\circ\)) to find \(\angle A\), and then find the others.

Exam Tip: Substituting all variables into a single common variable (like \(\angle A\)) is a powerful method to solve triangle angle problems.