CBSE Class 8 Mathematics Linear Equations In One Variable Worksheet Set 01

Question. In an examination a student was asked to find 3/14 th of a certain number. By mistake, he found 3/4 of it. His answer was 150 more than the correct answer. The number given him is :-
(A) 290
(B) 280
(C) 240
(D) 180
Answer: B

Question. If 10 is added to four times a certain number the result is 5 less than five times the number. The number is :-
(A) 10
(B) 15
(C) 20
(D) 25
Answer: B

Question. There were only two candidates in an election. One got 62% voltes and was elected by a margin of 144 votes. The total number of voters were :-
(A) 500
(B) 600
(C) 700
(D) 800
Answer: B

Question. A number consists of two digits whose sum is 8. If 18 is added to the number its digits are interchanged. Find the number :-
(A) 53
(B) 35
(C) 92
(D) 63
Answer: B

Question. A student has to secure 40% marks to pass. He got 40 marks and failed by 40 marks. The maximum number of marks is :-
(A) 160
(B) 180
(C) 200
(D) 320
Answer: C

Question. If a number increased by 8% of itself gives 135, then that number is :-
(A) 112
(B) 100
(C) 125
(D) none of these
Answer: C

Question. If 3.1/x X 3.3/4 = 12.1/2, then the value of 'x' is :-
(A) 1
(B) 1/3
(C) 2
(D) 3
Answer: D

Question. A number 351 is divided into two parts in th ratio 2 : 7. Find the product of the number :-
(A) 20,294
(B) 21,294
(C) 25,295
(D) 31,294
Answer: B

Question. If 2/3 of a number is 20 less than the original number, then the number is :-
(A) 60
(B) 40
(C) 80
(D) 120
Answer: A

Question. If the sum of three consecutive even numbers is 234, then the smallest among them is :-
(A) 76
(B) 78
(C) 80
(D) None of these
Answer: A

Question. By selling a bicycle for Rs 1, 885, a man gains 16%. At what price did he buy the bicycle ?
(A) Rs. 1, 625
(B) Rs. 1, 825
(C) Rs. 2, 000
(D) None of these
Answer: A

Question. 3/4 rd of a number when multiplied by 3/4 th of the same number make 338. The number is :-
(A) 18
(B) 24
(C) 36
(D) 26
Answer: D

Question. The ratio of two numbers is 3 : 8 and their difference is 115. The largest number is :-
(A) 69
(B) 115
(C) 184
(D) 230
Answer: C

Question. Twenty years ago, my age was 1/3 rd of what it is now. My present age is :-
(A) 66 years
(B) 30 years
(C) 33 years
(D) 36 years
Answer: B

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

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

Question. The solution of x-5/2 - x-3/5 = 1/2 is :-
(A) x = 7
(B) x = 9
(C) x = 8
(D) x = 5
Answer: C

Question. X and Y together can do a piece of work in 8 days, which X alone can do in 12 days. In how many days can Y do the same work alone ?
(A) 12 days
(B) 24 days
(C) 36 days
(D) 16 days
Answer: B

Question. Two angles in a triangle are in the ratio 4 : 5. If the sum of these angles is equal to the third angle, then the third angle is :-
(A) 180°
(B) 40°
(C) 50°
(D) 90°
Answer: D

Question. A man is 5 years older than his wife and the wife is now thrice as odd as their daughter, who is 10 years old. How old was the man when his daughter was born ?
(A) 20 years
(B) 23 years
(C) 25 years
(D) 30 years
Answer: C

Question. The prices of a scooter and cycle are in the ratio 9 : 5. If a scooter costs Rs. 4, 200 more than a cycle. The price of cycle is :-
(A) Rs. 5, 250
(B) Rs. 5, 200
(C) Rs. 5,000
(D) Rs. 4, 800
Answer: A

Question. A boat goes downstream and covers the distance between two ports in 4 hrs while it covers the same distance upstream in 5 hrs. If the speed of the stream is 2 kmph, the speed fo the boat in still water is :-
(A) 15 km/hr
(B) 20 km/hr
(C) 24 km/hr
(D) 18 km/hr 
Answer: D

Question. If the digit 1 is placed after a two digit number whose tens digit is 't' and units digit is 'u', the new number is :-
(A) 10t + u + 1
(B) 100t + 10u + 1
(C) t + u + 1
(D) none of these
Answer: B

Question. The ratio of number of males to number of females in a club are 7 : 4. If there are 84 males in the club, the total number of members in the club are :-
(A) 126
(B) 132
(C) 136
(D) 148
Answer: B

Question. The solution of 2x+2/3x-1 = 3/2 is :-
(A) x = 1
(B) x = – 1
(C) x = 2
(D) x = – 3
Answer: A

Mathematics Revision Assignment

Question 1. Name the property used in each of the following:
(a) \(\frac{6}{7} + \left(\frac{-6}{7}\right) = 0 = \left(\frac{-6}{7}\right) + \frac{6}{7}\)
(b) \(\frac{3}{4} + \left(-\frac{5}{6} + \frac{7}{3}\right) = \left(\frac{3}{4} + -\frac{5}{6}\right) + \frac{7}{3}\)
(c) \(\left(\frac{-2}{7}\right) \times 1 = \left(\frac{-2}{7}\right) = 1 \times \left(\frac{-2}{7}\right)\)
(d) \(\frac{5}{7} \times \frac{2}{3} = \frac{2}{3} \times \frac{5}{7}\)
(e) \(\left(-\frac{9}{13} + \frac{6}{7}\right) \times \frac{3}{5} = -\frac{9}{13} \times \frac{3}{5} + \frac{6}{7} \times \frac{3}{5}\)
Answer: (a) Additive Inverse Property
(b) Associative Property of Addition
(c) Multiplicative Identity Property
(d) Commutative Property of Multiplication
(e) Distributive Property of Multiplication over Addition
In simple words: These are fundamental algebraic rules that show how numbers behave when added or multiplied.

Exam Tip: Memorize the standard names of properties: Commutative (order changes), Associative (grouping changes), Distributive (spreading multiplication over addition), Identity (leaving the number unchanged), and Inverse (combining to get the identity).

Question 2. Solve: (a) \(\frac{3}{7} + \frac{-5}{11} + \frac{-5}{14} + \frac{3}{7}\)
Answer: Group the like denominators or take the LCM of denominators 7, 11, 14, and 7.
LCM of 7, 11, and 14 is \(2 \times 7 \times 11 = 154\).
Rearrange terms:
\(= \left(\frac{3}{7} + \frac{3}{7}\right) + \left(\frac{-5}{11}\right) + \left(\frac{-5}{14}\right)\)
\(= \frac{6}{7} - \frac{5}{11} - \frac{5}{14}\)
Convert to common denominator 154:
\(= \frac{6 \times 22}{154} - \frac{5 \times 14}{154} - \frac{5 \times 11}{154}\)
\(= \frac{132 - 70 - 55}{154}\)
\(= \frac{132 - 125}{154} = \frac{7}{154} = \frac{1}{22}\)
In simple words: Combine fractions by finding a common denominator and adding or subtracting their numerators.

Exam Tip: Always look for terms with identical denominators to simplify first using commutative and associative properties.

Question 3. Find the multiplicative inverse:
(a) \(-2x \times \frac{-3}{5}\) (assuming the expression is to simplify first: \(\frac{6x}{5}\), so its inverse is \(\frac{5}{6x}\); or treating as two separate items or standard product. Let us evaluate as a single rational expression \(-2 \times \left(-\frac{3}{5}\right) = \frac{6}{5}\))
(b) \(-\frac{5}{8} \times \frac{16}{15}\)
(c) \(-\frac{4}{11} \times \frac{22}{-6}\)
(d) \(\frac{-4}{-13}\)
Answer:
(a) Expression value = \(-2 \times \left(-\frac{3}{5}\right) = \frac{6}{5}\). Multiplicative inverse = \(\frac{5}{6}\).
(b) Expression value = \(-\frac{5}{8} \times \frac{16}{15} = -\frac{1}{1} \times \frac{2}{3} = -\frac{2}{3}\). Multiplicative inverse = \(-\frac{3}{2}\).
(c) Expression value = \(-\frac{4}{11} \times \frac{22}{-6} = -\frac{4}{1} \times \frac{2}{-6} = \frac{-8}{-6} = \frac{4}{3}\). Multiplicative inverse = \(\frac{3}{4}\).
(d) Expression value = \(\frac{-4}{-13} = \frac{4}{13}\). Multiplicative inverse = \(\frac{13}{4}\).
In simple words: The multiplicative inverse (reciprocal) of a number is what you multiply it by to get 1. Simply flip the numerator and denominator.

Exam Tip: Always simplify the given expression to a single fraction before finding its reciprocal.

Question 4. Solve using distributive property:
(a) \(\frac{2}{7} \times \left(\frac{7}{16} + \frac{21}{4}\right)\)
(b) \(-\frac{5}{4} \times \left(\frac{8}{5} + \frac{16}{5}\right)\)
Answer:
(a) Using distributive property \(a(b + c) = ab + ac\):
\(= \left(\frac{2}{7} \times \frac{7}{16}\right) + \left(\frac{2}{7} \times \frac{21}{4}\right)\)
\(= \frac{1}{8} + \frac{3}{2}\)
LCM of 8 and 2 is 8.
\(= \frac{1}{8} + \frac{12}{8} = \frac{13}{8}\)

(b) Using distributive property:
\(= \left(-\frac{5}{4} \times \frac{8}{5}\right) + \left(-\frac{5}{4} \times \frac{16}{5}\right)\)
\(= -2 + (-4) = -6\)
In simple words: Multiply the outer term by each term inside the parentheses separately before adding.

Exam Tip: Distributing first often cancels out common factors in numerators and denominators, making calculations much easier than adding inside the brackets first.

Question 5. Find 5 rational number between:
(a) \(-\frac{1}{2}\) and \(\frac{1}{4}\)
(b) \(\frac{1}{2}\) and \(\frac{1}{5}\)
(c) \(\frac{2}{3}\) and \(\frac{3}{4}\)
Answer:
(a) Make denominators equal (LCM of 2 and 4 is 4):
\(-\frac{1}{2} = -\frac{2}{4}\) and \(\frac{1}{4}\)
Multiply numerator and denominator by 6 to get enough space:
\(-\frac{12}{24}\) and \(\frac{6}{24}\)
Five rational numbers: \(-\frac{11}{24}, -\frac{10}{24}, -\frac{9}{24}, -\frac{8}{24}, -\frac{7}{24}\).

(b) Make denominators equal (LCM of 2 and 5 is 10):
\(\frac{1}{2} = \frac{5}{10}\) and \(\frac{1}{5} = \frac{2}{10}\)
Multiply by 6:
\(\frac{30}{60}\) and \(\frac{12}{60}\)
Five rational numbers: \(\frac{13}{60}, \frac{14}{60}, \frac{15}{60}, \frac{16}{60}, \frac{17}{60}\).

(c) Make denominators equal (LCM of 3 and 4 is 12):
\(\frac{2}{3} = \frac{8}{12}\) and \(\frac{3}{4} = \frac{9}{12}\)
Multiply by 6:
\(\frac{48}{72}\) and \(\frac{54}{72}\)
Five rational numbers: \(\frac{49}{72}, \frac{50}{72}, \frac{51}{72}, \frac{52}{72}, \frac{53}{72}\).
In simple words: Convert the fractions to have the same denominator, then scale them up so you can pick numbers in between.

Exam Tip: If you need \(n\) rational numbers, multiply both fractions' numerators and denominators by \(n + 1\) (or higher) to ensure sufficient gap.

Question 6. Represent the following on number line:
(a) \(\left(-1 \times \frac{8}{2}\right)\) (Note: simplifying gives -4)
(b) \(\frac{4}{5} \times \frac{15}{16}\)
Answer:
(a) Value = \(-4\).
Construction: Draw a straight horizontal line, mark integer points ..., -5, -4, -3, -2, -1, 0, 1, 2, ..., and highlight the point at -4.

(b) Value = \(\frac{4}{5} \times \frac{15}{16} = \frac{3}{4}\).
Construction: Draw a number line between 0 and 1. Divide the segment into 4 equal parts (fourths). Mark the third division point, which represents \(\frac{3}{4}\).
In simple words: Simplify the expression first, then plot the resulting fraction or integer on a standard numbered line.

Exam Tip: Always reduce the fraction to its simplest form before deciding how to divide the segments on the number line.

Question 7. Solve and verify:
(a) \(5x - (3x - 1) = x - 4\)
(b) \(\frac{3}{5}x + 1 = \frac{2}{5} - 3x\)
(c) \(9.8x - 13.8 = 2.7 - \frac{1}{5}x\)
(d) \(5x + 2 - 3(x + 2) = -2(x + 1) + 5(x - 2)\)
(e) \(3m + \frac{1}{2} = \frac{3}{8} + m\)
(f) \(x + 2(x - 1) = \frac{5}{2}\)
(g) \(0.3(6 + m) = 0.4(8 - m)\)
Answer:
(a) \(5x - 3x + 1 = x - 4\)
\(2x + 1 = x - 4\)
\(2x - x = -4 - 1\)
\(x = -5\)
Verification:
LHS = \(5(-5) - (3(-5) - 1) = -25 - (-15 - 1) = -25 - (-16) = -25 + 16 = -9\)
RHS = \(-5 - 4 = -9\). (LHS = RHS verified)

(b) \(\frac{3}{5}x + 3x = \frac{2}{5} - 1\)
\(\frac{18}{5}x = -\frac{3}{5}\)
\(x = -\frac{3}{5} \times \frac{5}{18} = -\frac{3}{18} = -\frac{1}{6}\)
Verification:
LHS = \(\frac{3}{5}\left(-\frac{1}{6}\right) + 1 = -\frac{1}{10} + 1 = \frac{9}{10}\)
RHS = \(\frac{2}{5} - 3\left(-\frac{1}{6}\right) = \frac{2}{5} + \frac{1}{2} = \frac{9}{10}\). (LHS = RHS verified)

(c) \(9.8x + 0.2x = 2.7 + 13.8\)
\(10.0x = 16.5\)
\(x = 1.65\)

(d) \(5x + 2 - 3x - 6 = -2x - 2 + 5x - 10\)
\(2x - 4 = 3x - 12\)
\(-4 + 12 = 3x - 2x\)
\(x = 8\)

(e) \(3m - m = \frac{3}{8} - \frac{1}{2}\)
\(2m = \frac{3}{8} - \frac{4}{8} = -\frac{1}{8}\)
\(m = -\frac{1}{16}\)

(f) \(x + 2x - 2 = \frac{5}{2}\)
\(3x = \frac{5}{2} + 2 = \frac{9}{2}\)
\(x = \frac{9}{6} = \frac{3}{2}\)

(g) \(1.8 + 0.3m = 3.2 - 0.4m\)
\(0.3m + 0.4m = 3.2 - 1.8\)
\(0.7m = 1.4\)
\(m = 2\)
In simple words: Collect all variable terms on one side and constants on the other, then isolate the variable.

Exam Tip: Always substitute your final answer back into both sides of the original equation to verify your work during exams.

Question 8. Reema's mother is four times as old as Reema. After 5 years, her mother will be 3 times as old as she will be then. Find their present ages.
Answer:
Let Reema's present age be \(x\) years.
Then Reema's mother's present age = \(4x\) years.
After 5 years:
Reema's age = \(x + 5\)
Mother's age = \(4x + 5\)
According to the question:
\(4x + 5 = 3(x + 5)\)
\(4x + 5 = 3x + 15\)
\(4x - 3x = 15 - 5\)
\(x = 10\)
Reema's present age = 10 years.
Mother's present age = \(4 \times 10 = 40\) years.
In simple words: Set up an equation based on their ages after 5 years and solve for Reema's age.

Exam Tip: Remember to add 5 to both ages when dealing with "after 5 years" conditions.

Question 9. 70 coins of 10p and 50p are kept in a purse. If the total value of the money in the purse is Rs. 19. Find the number of each type of coins.
Answer:
Let the number of 10p coins be \(x\).
Then the number of 50p coins = \(70 - x\).
Total value = Rs. 19 = 1900 paise (or convert value in rupees: \(0.10x + 0.50(70 - x) = 19\)).
\(0.10x + 35 - 0.50x = 19\)
\(-0.40x = 19 - 35\)
\(-0.40x = -16\)
\(x = \frac{16}{0.40} = 40\)
Number of 10p coins = 40.
Number of 50p coins = \(70 - 40 = 30\).
In simple words: Use the total number of coins to express one type in terms of the other, then equate total monetary value to Rs. 19.

Exam Tip: Always keep units consistent (either convert everything to rupees or everything to paise).

Question 10. One fifth of a number increased by 5 is equal to 4 less than the one-fourth of that number. Find the number.
Answer:
Let the number be \(x\).
According to the question:
\(\frac{x}{5} + 5 = \frac{x}{4} - 4\)
Rearranging terms:
\(5 + 4 = \frac{x}{4} - \frac{x}{5}\)
\(9 = \frac{5x - 4x}{20}\)
\(9 = \frac{x}{20}\)
\(x = 9 \times 20 = 180\)
In simple words: Translate the sentence directly into a mathematical equation and solve for \(x\).

Exam Tip: "4 less than one-fourth" means \(\frac{x}{4} - 4\), not \(4 - \frac{x}{4}\).

Question 11. Manav has a small box with Rs. 740 in coins of denominations Rs. 1, Rs. 2 and Rs. 5. If the number of coins of denominations Rs. 1, Rs. 2 are 3 times the no. of coins of Re. 1 and the no. of coins of denomination of Rs. 5 are twice the no. of coins of Rs. 2, how many coins of each denomination does Manav have?
Answer:
Let the number of Re. 1 coins be \(x\).
Number of Rs. 2 coins = \(3x\).
Number of Rs. 5 coins = \(2 \times (\text{number of Rs. 2 coins}) = 2(3x) = 6x\).
Total value calculation:
Value of Re. 1 coins = \(1 \times x = x\)
Value of Rs. 2 coins = \(2 \times 3x = 6x\)
Value of Rs. 5 coins = \(5 \times 6x = 30x\)
Total money = \(x + 6x + 30x = 37x\)
Given total money = Rs. 740.
\(37x = 740\)
\(x = \frac{740}{37} = 20\)
Number of Re. 1 coins = 20.
Number of Rs. 2 coins = \(3 \times 20 = 60\).
Number of Rs. 5 coins = \(6 \times 20 = 120\).
In simple words: Express all coin counts in terms of Re. 1 coins, calculate total value, and solve for \(x\).

Exam Tip: Be careful with reading the wording: Rs. 5 coins are twice the number of Rs. 2 coins, so multiply Rs. 2 count by 2.

Question 12. Sum of three consecutive multiples of 7 is 777. Find these multiples.
Answer:
Let the three consecutive multiples of 7 be \(7x, 7(x + 1),\) and \(7(x + 2)\) (or \(x, x + 7, x + 14\)).
Using \(x, x+7, x+14\):
\(x + (x + 7) + (x + 14) = 777\)
\(3x + 21 = 777\)
\(3x = 777 - 21\)
\(3x = 756\)
\(x = \frac{756}{3} = 252\)
The three multiples are:
First number = 252
Second number = \(252 + 7 = 259\)
Third number = \(252 + 14 = 266\)
In simple words: Consecutive multiples differ by 7, so let them be \(x, x+7,\) and \(x+14\).

Exam Tip: Always check your answer by adding the three resulting numbers to verify their sum equals 777.