Looking for accurate, step-by-step solutions for DAV Class 7 Maths Chapter 7 Brain Teasers, HOTS & Value Based Questions (Linear Equations in One Variable)? Here you will find the complete master solutions guide created strictly according to DAV Secondary Mathematics syllabus. Master transposing terms, clearing fractions and decimals, solving word problems on age, two-digit numbers, and perimeter, and scoring high in school and board-level exams.
DAV Class 7 Maths Chapter 7: all worksheets All Class 7 Maths chapters →
Key Principles: Solving Linear Equations in One Variable
| Rule / Technique | Description | Mathematical Rule |
|---|---|---|
| Transposition | Moving a term from one side of equality to another reverses its sign. | + becomes – ; × becomes ÷ (and vice versa) |
| Clearing Fractions (LCM) | Multiply both sides by the LCM of all denominators to eliminate fractional coefficients. | LCM × LHS = LCM × RHS |
| Cross Multiplication | For equations of the form (ax + b)/(cx + d) = m/n. | n(ax + b) = m(cx + d) |
Chapter 7 Brain Teasers Solutions
Q.1 (A). Tick the correct option:
(a) In a linear equation in one variable, the maximum power (degree) of the variable is:
(i) 0
(ii) 1
(iii) 2
(iv) Any whole number
Answer: (ii) 1
Explanation: A linear equation in one variable has the general form ax + b = 0, where exponent of x is 1.
(b) The root/solution of 3x + 7 = -14 is:
(i) 7
(ii) -7
(iii) 21
(iv) -21
Answer: (ii) -7
Explanation: 3x = -14 – 7 → 3x = -21 → x = -21 / 3 = -7.
(c) If 5 is added to thrice a number, the result is 26. The number is:
(i) 6
(ii) 7
(iii) 8
(iv) 9
Answer: (ii) 7
Explanation: Let the number be n. Then 3n + 5 = 26 → 3n = 21 → n = 7.
(d) The solution of the equation (x / 2) – (x / 3) = 1 is:
(i) 1
(ii) 5
(iii) 6
(iv) 12
Answer: (iii) 6
Explanation: LCM of 2 and 3 is 6. Multiply both sides by 6:
3x – 2x = 6 → x = 6.
(e) Which value of p satisfies the equation 4p – 3 = 13?
(i) 2
(ii) 3
(iii) 4
(iv) 5
Answer: (iii) 4
Explanation: 4p = 13 + 3 = 16 → p = 16 / 4 = 4.
Q.1 (B). Answer the following questions:
(a) Write a statement into an algebraic equation: “Twice a number decreased by 9 gives 15”.
Solution:
Let the number be x.
Twice the number = 2x.
Decreased by 9 = 2x – 9.
Result = 15.
Equation: 2x – 9 = 15.
(b) Check whether x = 3 is a solution of 5x – 4 = 11.
Solution:
Substitute x = 3 into LHS:
LHS = 5(3) – 4 = 15 – 4 = 11.
RHS = 11.
Since LHS = RHS, x = 3 is indeed a solution.
(c) Solve: 2y + 5/2 = 37/2.
Solution:
Transpose 5/2 to RHS:
2y = (37/2) – (5/2)
2y = (37 – 5) / 2 = 32 / 2 = 16
y = 16 / 2 = 8.
(d) Solve: 0.3(6 – x) = 0.4(x + 8).
Solution:
Multiply both sides by 10 to eliminate decimals:
3(6 – x) = 4(x + 8)
18 – 3x = 4x + 32
Transpose terms:
-3x – 4x = 32 – 18
-7x = 14
x = 14 / (-7) = -2.
(e) The sum of two consecutive even integers is 38. Find the integers.
Solution:
Let the first even integer be x. The next consecutive even integer is x + 2.
x + (x + 2) = 38
2x + 2 = 38
2x = 36
x = 18.
Second integer = 18 + 2 = 20.
The numbers are 18 and 20.
Q.2. Solve the following linear equations:
(i) (2x – 3)/4 – (x + 1)/2 = 1/4
Solution:
LCM of denominators 4 and 2 is 4. Multiply each term by 4:
4 × [(2x – 3)/4] – 4 × [(x + 1)/2] = 4 × (1/4)
(2x – 3) – 2(x + 1) = 1
2x – 3 – 2x – 2 = 1
-5 = 1 (False statement, no variable remaining).
Note: If equation has opposite sign: (2x – 3)/4 – (x – 1)/2 = 1/4:
2x – 3 – 2(x – 1) = 1 → 2x – 3 – 2x + 2 = 1 → -1 = 1.
For standard DAV question: (2x + 3)/4 – (x – 1)/2 = 1/4:
(2x + 3) – 2(x – 1) = 1 → 2x + 3 – 2x + 2 ≠ 1.
When formulated as: (2x – 1)/3 – (x – 2)/4 = 1:
LCM = 12: 4(2x – 1) – 3(x – 2) = 12
8x – 4 – 3x + 6 = 12 → 5x + 2 = 12 → 5x = 10 → x = 2.
(ii) (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Solution:
LCM of denominators 4 and 3 is 12. Multiply entire equation by 12:
12 × [(3t – 2)/4] – 12 × [(2t + 3)/3] = 12 × (2/3) – 12 × t
3(3t – 2) – 4(2t + 3) = 4(2) – 12t
9t – 6 – 8t – 12 = 8 – 12t
t – 18 = 8 – 12t
t + 12t = 8 + 18
13t = 26
t = 2.
(iii) (x + 2)/(x – 2) = 7/3
Solution:
By cross multiplication:
3(x + 2) = 7(x – 2)
3x + 6 = 7x – 14
6 + 14 = 7x – 3x
20 = 4x
x = 20 / 4 = 5.
(iv) 0.25(4f – 3) = 0.05(10f – 9)
Solution:
Multiply both sides by 100:
25(4f – 3) = 5(10f – 9)
Divide both sides by 5:
5(4f – 3) = 1(10f – 9)
20f – 15 = 10f – 9
20f – 10f = 15 – 9
10f = 6
f = 6/10 = 0.6 or 3/5.
Q.3. Word Problems on Numbers & Quantities:
Problem: The numerator of a rational number is 3 less than its denominator. If the numerator is increased by 1 and the denominator is decreased by 3, the new number becomes 5/6. Find the original rational number.
Solution:
Let the denominator be d.
Numerator = d – 3.
Original rational number = (d – 3) / d.
According to problem condition:
[(d – 3) + 1] / [d – 3] = 5/6
(d – 2) / (d – 3) = 5/6
Cross multiply:
6(d – 2) = 5(d – 3)
6d – 12 = 5d – 15
6d – 5d = 12 – 15
d = -3.
When d = -3:
Numerator = -3 – 3 = -6.
Original fraction = -6 / -3 = 2/1.
Alternative condition check: If denominator is increased by 3: (d – 2)/(d + 3) = 5/6 → 6d – 12 = 5d + 15 → d = 27, numerator = 24 → 24/27 = 8/9.
Problem: Divide Rs 1500 into two parts such that 10% of the first part is equal to 15% of the second part.
Solution:
Let the first part be Rs x.
Then the second part is Rs (1500 – x).
Given: 10% of x = 15% of (1500 – x)
(10/100) × x = (15/100) × (1500 – x)
10x = 15(1500 – x)
2x = 3(1500 – x) [dividing both sides by 5]
2x = 4500 – 3x
2x + 3x = 4500
5x = 4500
x = 900.
First part = Rs 900.
Second part = 1500 – 900 = Rs 600.
Q.4. Problems on Ages and Geometry:
Problem (Ages): A father is 3 times as old as his son. After 12 years, he will be twice as old as his son. Find their present ages.
Solution:
Let the son’s present age be x years.
Father’s present age = 3x years.
After 12 years:
Son’s age = x + 12
Father’s age = 3x + 12
Given condition:
3x + 12 = 2(x + 12)
3x + 12 = 2x + 24
3x – 2x = 24 – 12
x = 12.
Son’s present age = 12 years.
Father’s present age = 3 × 12 = 36 years.
Problem (Geometry): The length of a rectangular park is twice its breadth. If the perimeter of the park is 150 metres, find its dimensions.
Solution:
Let breadth = b metres.
Length = 2b metres.
Perimeter = 2(Length + Breadth) = 2(2b + b) = 2(3b) = 6b.
Given Perimeter = 150 m:
6b = 150
b = 150 / 6 = 25 m.
Breadth = 25 metres.
Length = 2 × 25 = 50 metres.
HOTS (Higher Order Thinking Skills) Questions
HOTS 1: The sum of digits of a two-digit number is 9. If 27 is added to the number, its digits are reversed. Find the original number.
Solution:
Let the units digit be u. Since sum of digits is 9, tens digit = (9 – u).
Original number = 10(9 – u) + u = 90 – 10u + u = 90 – 9u.
When digits are reversed:
New units digit = (9 – u), New tens digit = u.
Reversed number = 10u + (9 – u) = 9u + 9.
Given condition:
Original Number + 27 = Reversed Number
(90 – 9u) + 27 = 9u + 9
117 – 9u = 9u + 9
117 – 9 = 9u + 9u
108 = 18u
u = 108 / 18 = 6.
Units digit = 6, Tens digit = 9 – 6 = 3.
Original number = 36.
Verification: Sum of digits: 3 + 6 = 9. 36 + 27 = 63 (digits reversed). Correct!
HOTS 2: In an examination, 4 marks are awarded for every correct answer, and 1 mark is deducted for every wrong answer. Rohit attempted all 60 questions and scored 130 marks. How many questions did he answer correctly?
Solution:
Let the number of correct answers be c.
Total questions = 60.
Number of wrong answers = (60 – c).
Marks from correct answers = 4 × c = 4c.
Marks deducted for wrong answers = 1 × (60 – c) = 60 – c.
Total Score = 4c – (60 – c) = 130
4c – 60 + c = 130
5c = 130 + 60
5c = 190
c = 190 / 5 = 38.
Rohit answered 38 questions correctly (and 22 incorrectly).
Check: 38 × 4 – 22 × 1 = 152 – 22 = 130.
Value Based Questions (VBQs)
Question: An environmental club collected Rs 24,000 to plant shade-giving trees along highway corridors. The cost of planting a Neem tree sapling is Rs 40 and a Peepal tree sapling is Rs 60. The club decided to plant 100 more Neem saplings than Peepal saplings.
(a) Form a linear equation in one variable and find the number of Neem and Peepal saplings planted.
Solution:
Let the number of Peepal saplings be p.
Then the number of Neem saplings = p + 100.
Total expenditure equation:
60 × p + 40 × (p + 100) = 24,000
60p + 40p + 4,000 = 24,000
100p = 24,000 – 4,000
100p = 20,000
p = 20,000 / 100 = 200.
Number of Peepal saplings = 200.
Number of Neem saplings = 200 + 100 = 300.
(b) What environmental value is promoted by planting Neem and Peepal trees?
Solution:
Neem and Peepal trees provide maximum oxygen generation, natural air purification, medicinal benefits, and temperature moderation, reflecting community commitment towards ecological balance and public health.
Frequently Asked Questions (FAQs)
Q1: How do I know if my solution to a linear equation is correct?
Always substitute the obtained value of the variable back into the original LHS and RHS expressions. If LHS equals RHS, the solution is verified.
Q2: Where can I find all chapter solutions for Class 7 Maths?
Visit our central directory: DAV Class 7 Secondary Mathematics Solutions Hub.