CBSE Class 7 Mathematics Simple Equations Assignment Set 02

M. C. Q.
Choose the correct option from the given options.

Question 1. Which of the following is not a linear equation in \(x\)?
(i) \(x^2 = 3x + 5\)
(ii) \(7x = 5\)
(iii) \(x = 2x + 5\)
(iv) \(x - 1 = 0\)
Answer: (i) \(x^2 = 3x + 5\)
A linear equation is an algebraic equation in which each term has an exponent of 1 and the highest degree of the variable is 1. In option (i), the variable \(x\) has an exponent of 2 (\(x^2\)), which makes it a quadratic equation, not a linear equation.
In simple words: Linear equations have the variable with a power of 1, but \(x^2 = 3x + 5\) has a power of 2.

Exam Tip: Always check the highest power of the variable. If the degree is greater than 1, it is not a linear equation.

Question 2. Which of the following is not the correct solution of equation \(3x - 2 = 10\)?
(i) \(x = 4\)
(ii) \(x = 2\)
(iii) \(2x = 8\)
(iv) \(3x = 12\)
Answer: (ii) \(x = 2\)
Solving the equation \(3x - 2 = 10\):
\(3x = 10 + 2\)
\(3x = 12\), which is option (iv)
\(x = \frac{12}{3} = 4\), which is option (i)
Checking option (iii) \(2x = 8\): dividing by 2 gives \(x = 4\), which is also valid.
Substituting \(x = 2\) into the original equation: \(3(2) - 2 = 6 - 2 = 4 \neq 10\). Thus, \(x = 2\) is not correct.
In simple words: Plugging \(x = 2\) into the equation does not give 10 on the left side.

Exam Tip: Verify a solution by substituting the value back into the original equation to check if the left-hand side equals the right-hand side.

Question 3. Which of the following is not a correct statement of \(3p - 2 = 7\)?
(i) 3p is 7 more than 2
(ii) difference of 3p and 2 is 7 if \(3p > 2\)
(iii) 2 subtracted from 3p gives 7
(iv) 2 is less than 3 times p gives 7.
Answer: (i) 3p is 7 more than 2
The equation \(3p - 2 = 7\) means that when 2 is subtracted from \(3p\), the result is 7 (or \(3p\) is 2 more than 7, i.e., \(3p = 7 + 2 = 9\)). Statement (i) incorrectly claims that \(3p\) is 7 more than 2, which would be written as \(3p = 2 + 7\).
In simple words: The equation means \(3p\) minus 2 equals 7, so \(3p\) is actually 9, not 7 more than 2.

Exam Tip: Read word statements carefully and translate subtraction as "less than" or "subtracted from" and equals as "gives" or "is".

Question 4. The solution of \(\frac{3x}{4} = 9\) is
(i) 36
(ii) 12
(iii) 6
(iv) 1
Answer: (ii) 12
Given equation: \(\frac{3x}{4} = 9\)
Multiply both sides by 4:
\(3x = 9 \times 4\)
\(3x = 36\)
Divide both sides by 3:
\(x = \frac{36}{3} = 12\)
In simple words: Multiply 9 by 4 to get 36, then divide by 3 to get 12.

Exam Tip: Clear denominators first by multiplying both sides by the denominator to simplify solving.

Question 5. The solution of \(\frac{1}{5}(2x - 1) = 3\) is
(i) 8
(ii) 16
(iii) 24
(iv) 32
Answer: (i) 8
Given equation: \(\frac{1}{5}(2x - 1) = 3\)
Multiply both sides by 5:
\(2x - 1 = 3 \times 5\)
\(2x - 1 = 15\)
Add 1 to both sides:
\(2x = 15 + 1\)
\(2x = 16\)
Divide by 2:
\(x = \frac{16}{2} = 8\)
In simple words: Multiply 3 by 5, add 1, and divide by 2 to find \(x\).

Exam Tip: Deal with fractions outside brackets first by multiplying through by the denominator.

Question 6. The solution of \(\frac{3x - 5}{2} = \frac{2x - 5}{3}\) is
(i) 2
(ii) 1
(iii) 3
(iv) 4
Answer: (ii) 1
Given equation: \(\frac{3x - 5}{2} = \frac{2x - 5}{3}\)
Cross-multiply:
\(3(3x - 5) = 2(2x - 5)\)
Expand both sides:
\(9x - 15 = 4x - 10\)
Rearrange terms (bring \(4x\) to the left and \(-15\) to the right):
\(9x - 4x = 15 - 10\)
\(5x = 5\)
\(x = 1\)
In simple words: Cross-multiply to remove fractions, then group like terms together.

Exam Tip: Cross-multiplication is the quickest way to solve equations where a single fraction equals a single fraction.

Question 7. If 3 is added to a number, the result is 15. The number is
(i) 15
(ii) 14
(iii) 12
(iv) 13
Answer: (iii) 12
Let the number be \(x\).
According to the question: \(x + 3 = 15\)
Subtract 3 from both sides:
\(x = 15 - 3\)
\(x = 12\)
In simple words: Subtract 3 from 15 to find the original number.

Question 8. Anil's father is 8 times as old as Anil. If sum of the two ages is 72 years, the age of Anil is
(i) 11 yrs
(ii) 10 yrs
(iii) 9 yrs
(iv) 8 yrs
Answer: (iv) 8 yrs
Let Anil's age be \(x\) years.
Then Anil's father's age = \(8x\) years.
Sum of their ages = \(x + 8x = 72\)
\(9x = 72\)
\(x = \frac{72}{9} = 8\)
In simple words: Divide the total sum of ages (72) by \(1 + 8 = 9\) to get Anil's age.

Exam Tip: In age problems involving multiples, express both ages in terms of a single variable.

Question 9. Sum of two consecutive numbers is 29. One of them is
(i) 14
(ii) 28
(iii) 20
(iv) 25
Answer: (i) 14
Let the two consecutive numbers be \(x\) and \(x + 1\).
Their sum: \(x + (x + 1) = 29\)
\(2x + 1 = 29\)
\(2x = 29 - 1\)
\(2x = 28\)
\(x = 14\)
The other number is \(14 + 1 = 15\). Since 14 is among the options, it is the correct choice.
In simple words: Two numbers that follow each other and add up to 29 are 14 and 15.

Exam Tip: Consecutive numbers are represented as \(x\) and \(x + 1\).

Question 10. A man is 7 times as old as his son. If he is 56 years old, the age of son is
(i) 10 yrs
(ii) 8 yrs
(iii) 9 yrs
(iv) 6 yrs
Answer: (ii) 8 yrs
Let the son's age be \(y\) years.
Father's age = \(7y\)
Given father's age = 56 years
\(7y = 56\)
\(y = \frac{56}{7} = 8\)
In simple words: Divide the father's age (56) by 7 to find the son's age.

Exam Tip: When one person's age is a multiple of another, divide the given age by that multiplier.