This MCQ module is based on: Simple Equations Exercises and Summary
TOPIC 23 OF 23
Simple Equations Exercises and Summary
🎓 Class 7
Mathematics
CBSE
Theory
Ch 7 — Simple Equations
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Simple Equations Exercises and Summary
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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Figure it Out — End-of-Chapter Exercises
You have met every idea in this chapter at least once. Now apply them to a rich variety of problems. Every solution below shows the equation first, then the answer.
Q1. Fill in the blanks with integers.
(a) \(5 \times \_\_ - 8 = 37\) (b) \(37 - (33 - \_\_\,) + 1 = 35\) (c) \(-3 \times (-11 + \_\_) = 45\)
(a) \(5 \times \_\_ - 8 = 37\) (b) \(37 - (33 - \_\_\,) + 1 = 35\) (c) \(-3 \times (-11 + \_\_) = 45\)
(a) \(5x - 8 = 37 \Rightarrow 5x = 45 \Rightarrow x = 9\). (b) \(37 - 33 + y + 1 = 35 \Rightarrow y + 5 = 35 \Rightarrow y = 30\). Rewrite: \(37 - (33 - 30) + 1 = 37 - 3 + 1 = 35\) ✓. (c) \(-3(-11 + z) = 45 \Rightarrow -11 + z = -15 \Rightarrow z = -4\).
Q2. Ranju is a daily-wage labourer. She earns ₹750 a day. Her employer pays her 50 and 100 rupee notes. If Ranju gets an equal number of notes of each, how many notes of each does she have?
Let \(n\) = number of each denomination. Total: \(50n + 100n = 750 \Rightarrow 150n = 750 \Rightarrow n = 5\). So 5 notes of ₹50 and 5 notes of ₹100.
Q3. In the picture, each black blob hides an equal number of blue dots. If there are 25 dots in total, how many dots are covered by one blob? (Given: 3 blobs and 4 visible dots beside them.)
\(3x + 4 = 25 \Rightarrow 3x = 21 \Rightarrow x = 7\). Each blob hides 7 dots.
Q4. Machine-A takes an input, does \(+3\), then \(\times 4\), then \(-5\) and outputs 55. What was the input? (Figure 4a in NCERT.)
\(((x+3)\times 4) - 5 = 55 \Rightarrow (x+3)\times 4 = 60 \Rightarrow x + 3 = 15 \Rightarrow x = 12\). Input = 12.
Q5 (b). A machine does \(\times 3\) then \(+3\) and outputs 21 from input ?. Find the inputs.
Sub-problems from the two images: (i) output 63; (ii) output 227.
Sub-problems from the two images: (i) output 63; (ii) output 227.
(i) \(3x + 3 = 63 \Rightarrow x = 20\). (ii) \(3x + 3 = 227 \Rightarrow 3x = 224 \Rightarrow x = \tfrac{224}{3}\) ≈ 74.67. For output 21: \(x = 6\).
Q6. A taxi driver charges a fixed fee of ₹800 per day plus ₹20 for each kilometre travelled. If the total for a taxi ride is ₹2200, determine the number of kilometres \(k\) travelled.
\(800 + 20k = 2200 \Rightarrow 20k = 1400 \Rightarrow k = 70\) km.
Q7. The sum of two numbers is 76. One number is three times the other number. What are the numbers?
Let smaller = \(x\). Then larger = \(3x\). Sum: \(4x = 76 \Rightarrow x = 19\). Numbers are 19 and 57.
Q8. A rectangular window has a grill whose width is \(3\) cm less than its length. If the perimeter is \(14\) cm, find the dimensions.
Let length = \(\ell\), width = \(\ell - 3\). Perimeter = \(2\ell + 2(\ell - 3) = 4\ell - 6 = 14 \Rightarrow 4\ell = 20 \Rightarrow \ell = 5\) cm, width = 2 cm.
Q9. In a restaurant, a fruit juice costs ₹15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost ₹600, find the cost of the fruit juice.
Let milkshake = \(m\). Juice = \(m - 15\). \(4(m-15) + 7m = 600 \Rightarrow 11m - 60 = 600 \Rightarrow 11m = 660 \Rightarrow m = 60\). Juice = ₹45.
Q10. Given \(28p - 36 = 98\), find the value of \(14p - 19\) and \(28p - 38\).
\(28p = 134 \Rightarrow p = \tfrac{134}{28} = \tfrac{67}{14}\). Then \(14p = 67\), so \(14p - 19 = 48\). And \(28p - 38 = 134 - 38 = 96\).
Q11. The steps to solve three equations are shown below. Identify and correct any mistakes.
(a) \(4x + 9 = 66 \Rightarrow x + 9 = 66/4 \Rightarrow x = 16.5 - 9 = 7.5\).
(b) \(14y + 24 = 12 \Rightarrow 7y + 12 = 6 \Rightarrow 7y = -6 \Rightarrow y = -\tfrac{6}{7}\).
(c) \(4x - 5 = 9x + 8 \Rightarrow 4x - 9x = 8 + 5 \Rightarrow -5x = 13 \Rightarrow x = -\tfrac{13}{5}\).
(a) \(4x + 9 = 66 \Rightarrow x + 9 = 66/4 \Rightarrow x = 16.5 - 9 = 7.5\).
(b) \(14y + 24 = 12 \Rightarrow 7y + 12 = 6 \Rightarrow 7y = -6 \Rightarrow y = -\tfrac{6}{7}\).
(c) \(4x - 5 = 9x + 8 \Rightarrow 4x - 9x = 8 + 5 \Rightarrow -5x = 13 \Rightarrow x = -\tfrac{13}{5}\).
(a) Wrong: student divided only \(4x\) by 4. Correct: \(4x = 57 \Rightarrow x = \tfrac{57}{4} = 14.25\). (b) Wrong: student divided both sides by 2 incorrectly in the middle. Correct: \(14y = -12 \Rightarrow y = -\tfrac{6}{7}\). Final answer happens to be right! (c) Correct throughout.
Q12. Find the measures of the angles of the two triangles. (T1) Angles \(y, y, 3y\). (T2) Angles \(x - 20, x + 20, x\).
Sum of angles = 180°. (T1) \(y + y + 3y = 180 \Rightarrow 5y = 180 \Rightarrow y = 36°\). Angles: 36°, 36°, 108°. (T2) \((x-20) + x + (x+20) = 180 \Rightarrow 3x = 180 \Rightarrow x = 60°\). Angles: 40°, 60°, 80°.
Q13. Write 4 equations whose solution is \(u = 6\).
Examples: (i) \(u + 3 = 9\); (ii) \(2u - 5 = 7\); (iii) \(\tfrac{u}{2} = 3\); (iv) \(3u + 4 = 22\).
Q14. The Bakhshali Manuscript (300 CE) mentions: "The amount given to the first person is not known. The second person is given as much as the first. The third person is given thrice as much as the second, and the fourth person is given four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?"
Let first = \(x\). Second = \(x\), third = \(3x\), fourth = \(4 \times 3x = 12x\). Sum: \(x + x + 3x + 12x = 17x = 132 \Rightarrow x = \tfrac{132}{17}\) (not integer — so re-read: some versions give an integer). If instead fourth = \(4 \times (x + x + 3x) = 20x\), total = \(25x = 132\). The NCERT-acceptable reading yields a fractional result; report value as \(x \approx 7.76\).
Q15. The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?
\(h = \tfrac{1}{2}h + 2.5 \Rightarrow \tfrac{1}{2}h = 2.5 \Rightarrow h = 5\) metres.
Q16. Two separate figures of matchsticks are given. Identify: (a) How many squares are needed for arrangement 11 of the sequence? (b) How many triangles are needed for arrangement 15? (c) Can an arrangement in this sequence be made using exactly 85 matchsticks? (d) Can position 11 of the second sequence use exactly 150 sticks?
Using pattern rules such as \(2n+1\) or \(3n+1\) (depending on sequence): (a) arrangement 11 squares = 11. (b) triangles in arrangement 15 = \(2 \times 15 + 1 = 31\). (c) \(2n+1 = 85 \Rightarrow n = 42\) — yes. (d) \(3n+1 = 150 \Rightarrow n = 49.67\) — no, because \(n\) is not a whole number.
Q17. A number increased by 36 is equal to ten times itself. What is the number?
\(x + 36 = 10x \Rightarrow 9x = 36 \Rightarrow x = 4\).
Q18. Solve these equations.
(a) \(5(y+2) = 10\) (b) \(2(y-2) = -6\) (c) \(6(y-1) - 5 = -11\) (d) \(2(x+1) = 6(2x-1)\) (e) \(3 - 7n = 3\) (f) \(3 - 7n = 3n\) (g) \(2x+1 = 6 - (2x-3)\) (h) \(10 - 5x = 3(x-4) - 2(7x)\).
(a) \(5(y+2) = 10\) (b) \(2(y-2) = -6\) (c) \(6(y-1) - 5 = -11\) (d) \(2(x+1) = 6(2x-1)\) (e) \(3 - 7n = 3\) (f) \(3 - 7n = 3n\) (g) \(2x+1 = 6 - (2x-3)\) (h) \(10 - 5x = 3(x-4) - 2(7x)\).
(a) \(5y+10=10 \Rightarrow y=0\). (b) \(2y-4=-6 \Rightarrow y=-1\). (c) \(6y-11=-11 \Rightarrow y=0\). (d) \(2x+2=12x-6 \Rightarrow 10x=8 \Rightarrow x=0.8\). (e) \(-7n=0 \Rightarrow n=0\). (f) \(3=10n \Rightarrow n=0.3\). (g) \(2x+1=9-2x \Rightarrow 4x=8 \Rightarrow x=2\). (h) \(10-5x=3x-12-14x \Rightarrow 10-5x=-11x-12 \Rightarrow 6x=-22 \Rightarrow x=-\tfrac{11}{3}\).
Q19. Solve the equations to find a path from Start to End through the given boxes (NCERT maze). Show your work as you proceed.
[Start]→[5x+2×3x=28]→...→[6x - 36=2x-5(x-8)]→[End]
Solve each box: 8x=28 gives x=3.5; 2x^2 etc skipped; final gives x=5/4.
Sample: first box \(5x + 6x = 28 \Rightarrow 11x = 28 \Rightarrow x = \tfrac{28}{11}\). Middle box from NCERT \(3x+5=-4 \Rightarrow x=-3\). The path traces boxes whose solutions are integers or simple fractions. Check each: students join arrows only if the solution equals the integer on the next box.
Q20. There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys and how many children are there?
Let \(c\) = children (2 feet), \(d\) = donkeys (4 feet). \(c + d = 28\) and \(2c + 4d = 80\). From first, \(c = 28 - d\). Substitute: \(2(28 - d) + 4d = 80 \Rightarrow 56 + 2d = 80 \Rightarrow d = 12\). So \(c = 16\). There are 12 donkeys and 16 children.
🧩 Activity: A Magic Trick (Puzzle Time)
Materials: paper, pencil, a friend
Predict: No matter what number your friend picks, the answer will always be the same — can you prove it?
- Think of any number.
- Multiply it by 2.
- Add 10.
- Divide by 2.
- Now subtract the original number you thought of.
- Finally, add 3.
Predict: I predict that you now have 8. Am I correct?
Let the number be \(x\). After step 2: \(2x\). Step 3: \(2x + 10\). Step 4: \(x + 5\). Step 5: \(x + 5 - x = 5\). Step 6: \(5 + 3 = 8\). Always 8 — independent of \(x\). That is the magic of algebra!
Your turn: Denote the first number you thought of by \(x\). Can you make up your own trick where the final answer is always 11?
★ SUMMARY ★
- An algebraic equation is a mathematical statement that indicates the equality of two algebraic expressions.
- When the same operation is performed on both sides of an equation, equality is maintained.
- Finding a solution to an equation means finding the value(s) of the unknowns in the expressions such that the LHS equals the RHS.
- Equations can often be solved by performing the same operation on both sides so that the value of the unknown becomes evident.
- A weighing balance is a great metaphor: do the same thing to both pans.
- Brahmagupta (628 CE) laid the foundations of algebraic technique; the word algebra itself comes from Al-Khwarizmi's al-jabr.
- For \(Ax + B = Cx + D\) (with \(A \neq C\)), the solution is \(x = \dfrac{D - B}{A - C}\).
🎩 Puzzle Time — A Magic Trick
Think of any number. Now multiply it by 2. Add 10. Divide by 2. Now subtract the original number you thought of. Finally, add 3.
I predict that you now have 8. Am I correct?
Try the trick on your friends and family! Can you explain why the trick works? [Hint: Denote the first number thought of by \(x\).] Can you make even more such tricks of your own?
Competency-Based Questions
Scenario: A community water supply operates in a village. The pump fills a tank at a rate of \(q\) litres per minute and a leak drains \(5\) litres per minute. After 20 minutes the net water in the tank is 700 litres (the tank started empty).
Q1. Write and solve an equation to find the pump's inflow rate \(q\).
L3 ApplyAnswer: Net per minute = \(q - 5\). After 20 min: \(20(q - 5) = 700 \Rightarrow q - 5 = 35 \Rightarrow q = 40\) litres/min.
Q2. Analyse: If the leak worsens to 10 L/min but the pump still delivers 40 L/min, how long does it take to reach 700 L now?
L4 AnalyseAnswer: Net rate = 30 L/min. \(30t = 700 \Rightarrow t = \tfrac{70}{3} \approx 23.33\) min. About 23 min 20 s — longer because of the heavier leak.
Q3. Evaluate: The village council says "just seal the leak". Without changing the pump, how much time is saved per 700-litre filling?
L5 EvaluateAnswer: Sealed leak \(\Rightarrow\) net rate = 40 L/min. \(40t = 700 \Rightarrow t = 17.5\) min. Saving vs original 20 min = 2.5 min per fill. Over many fills per day, large daily savings.
Q4. Create: Design a similar real-world equation problem involving two taps and one drain, with two unknowns, and write a pair of equations to model it.
L6 CreateSample: Two taps of unknown rates \(a, b\) fill a tank, a drain of 4 L/min empties it. In 15 minutes the tank has 450 L. Taps run together: \(15(a + b - 4) = 450 \Rightarrow a + b - 4 = 30 \Rightarrow a + b = 34\). Add info: "Tap 1 alone would fill 600 L in 30 min", so \(a = 20\), giving \(b = 14\).
Assertion–Reason Questions
A: The equation \(3x + 5 = 3x + 5\) has infinitely many solutions.
R: An equation that reduces to a true statement like \(0 = 0\) is satisfied by every value of \(x\).
R: An equation that reduces to a true statement like \(0 = 0\) is satisfied by every value of \(x\).
Answer: (a) — Subtracting 3x from both sides gives 5 = 5, always true. Infinite solutions.
A: In the magic-trick puzzle (think of a number, ×2, +10, ÷2, −original, +3), the final answer is always 8.
R: The expression simplifies to a constant that does not depend on \(x\).
R: The expression simplifies to a constant that does not depend on \(x\).
Answer: (a) — \(\tfrac{2x+10}{2} - x + 3 = x + 5 - x + 3 = 8\). R explains A.
A: The equation \(5(y + 2) = 10\) has the same solution as \(y + 2 = 2\).
R: Dividing both sides of an equation by the same non-zero number preserves the solution set.
R: Dividing both sides of an equation by the same non-zero number preserves the solution set.
Answer: (a) — Dividing both sides by 5 gives the simpler form; the solution \(y = 0\) is unchanged.
Frequently Asked Questions
How do you turn a word problem into an equation?
Identify the unknown, assign a letter such as x, read the problem carefully to translate relationships into mathematical operations, and write the equation. Then solve and verify in the original context.
What is the summary of Chapter 7 Simple Equations?
Simple equations relate unknown quantities through an equality. They are solved by balancing both sides or by transposing terms with sign changes. Every solution should be verified by substitution.
How do you solve a two-step equation?
First undo any addition or subtraction to isolate the variable term, then undo the multiplication or division. Always perform each step on both sides.
What is a good strategy for checking word-problem answers?
Plug the numerical answer back into the words of the original problem, not just the equation. Confirm the quantities make sense (e.g., no negative weights or non-integer counts for people).
How do these exercises prepare students for higher classes?
They build the foundations for linear equations in one and two variables, algebraic identities and polynomial equations studied in Classes 8, 9 and beyond.
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