NCERT Solutions Class 7 Mathematics Ganita Prakash 2 Chapter 07 Finding the Unknown

Question 1. Solve these equations and check the solutions. (a) 3x - 10 = 35 (b) 5s = 3s (c) 3u - 7 = 2u + 3 (d) 4(m + 6) - 8 = 2m - 4 (e) u/15 = 6
Answer:
(a) Begin with 3x - 10 = 35. Add 10 to both sides to get 3x = 45. Divide both sides by 3, which gives x = 15. To check: when x = 15, the left side is 3(15) - 10 = 45 - 10 = 35, and the right side is 35. Since both sides match, the solution is verified.

(b) Start with 5s = 3s. Subtract 3s from both sides, so 5s - 3s = 0, which means 2s = 0. Divide both sides by 2 to get s = 0. To check: when s = 0, the left side is 5(0) = 0 and the right side is 3(0) = 0. Both sides are equal, confirming the solution.

(c) Begin with 3u - 7 = 2u + 3. Subtract 2u from both sides, giving u - 7 = 3. Add 7 to both sides to get u = 10. To check: when u = 10, the left side is 3(10) - 7 = 30 - 7 = 23, and the right side is 2(10) + 3 = 20 + 3 = 23. Both sides equal 23, so the solution is correct.

(d) Start with 4(m + 6) - 8 = 2m - 4. Expand the left side using the distributive property: 4m + 24 - 8 = 2m - 4, which simplifies to 4m + 16 = 2m - 4. Subtract 2m from both sides to get 2m + 16 = -4. Subtract 16 from both sides, so 2m = -20. Divide both sides by 2 to get m = -10. To check: when m = -10, the left side is 4(-10 + 6) - 8 = 4(-4) - 8 = -16 - 8 = -24, and the right side is 2(-10) - 4 = -20 - 4 = -24. Both sides match, confirming the solution.

(e) Start with u/15 = 6. Multiply both sides by 15 to get u = 90. To check: when u = 90, the left side is 90/15 = 6 and the right side is 6. Both sides are equal, so the solution is verified.
In simple words: To solve an equation, move all numbers to one side and all letters to the other. Do the same operation on both sides, and then simplify to find the answer. Always check by putting your answer back into the original equation.

Exam Tip: Always perform the same operation on both sides of the equation to keep it balanced. Checking your solution by substituting it back into the original equation catches careless errors.

Question 2. Frame an equation that has no solution.
Answer: One example of an equation with no solution is u + 6 = u + 2. When you subtract u from both sides, you get 6 = 2, which is a false statement. Since this statement can never be true, no value of u will make this equation true. An equation has no solution when simplifying it leads to a false statement rather than a valid number for the unknown.
In simple words: An equation has no solution when you get a false statement like 6 = 2. No number can make the equation work.

Exam Tip: Watch for equations that simplify to false statements with no unknown present - these always have no solution, not all values as solutions.

Question 1. Write 5 equations whose solution is x = -2.
Answer:
(i) Start with x = -2. Add 3 to both sides: x + 3 = 1. Multiply both sides by 2: 2(x + 3) = 2.

(ii) Start with x = -2. Multiply both sides by 3: 3x = -6. Add 2 to both sides: 3x + 2 = -4.

(iii) Start with x = -2. Divide both sides by 4: \( \frac{x}{4} = -\frac{1}{2} \).

(iv) Start with x = -2. Multiply both sides by 5: 5x = -10. Add 12 to both sides: 5x + 12 = 2.

(v) Start with x = -2. Subtract 7 from both sides: x - 7 = -9.
In simple words: To create an equation with a given solution, start with that solution and use different operations (add, subtract, multiply, divide) to build different equations that all work out to the same answer.

Exam Tip: Always verify your created equations by substituting the solution back in to make sure they work correctly.

Question 2. Find the value of each unknown: (a) 2y = 60 (b) -8 = 5x - 3 (c) -53w = -15 (d) 13 - z = 8 (e) k + 8 = 12 - k (f) 7m = m - 3 (g) 3n = 10 + n
Answer:
(a) Divide both sides by 2: y = 30.

(b) Add 3 to both sides: -5 = 5x. Divide both sides by 5: x = -1.

(c) Divide both sides by -53: \( w = \frac{15}{53} \).

(d) Subtract 13 from both sides: -z = -5. Divide both sides by -1: z = 5.

(e) Add k to both sides: 2k + 8 = 12. Subtract 8 from both sides: 2k = 4. Divide both sides by 2: k = 2.

(f) Subtract m from both sides: 6m = -3. Divide both sides by 6: \( m = -\frac{1}{2} \).

(g) Subtract n from both sides: 2n = 10. Divide both sides by 2: n = 5.
In simple words: Use addition or subtraction to move all numbers and variables to separate sides, then use multiplication or division to isolate the unknown.

Exam Tip: When a variable appears on both sides, collect all variable terms on one side first, then solve.

Question 3. I am a 3-digit number. My hundred's digit is 3 less than my ten's digit. My ten's digit is 3 less than my unit's digit. The sum of all three digits is 15. Who am I?
Answer: Let the ten's digit be x. Then the hundred's digit is x - 3 and the unit's digit is x + 3. The sum of all three digits equals 15, so (x - 3) + x + (x + 3) = 15. This simplifies to 3x = 15. Dividing both sides by 3 gives x = 5. Therefore, the ten's digit is 5, the hundred's digit is 5 - 3 = 2, and the unit's digit is 5 + 3 = 8. The 3-digit number is 258.
In simple words: Write each digit as an expression using one variable, then use the fact that they add to 15 to find that variable's value.

Exam Tip: When describing digits of a number using algebra, express each digit in terms of one variable to create a single equation.

Question 4. The weight of a brick is 1 kg more than half its weight. What is the weight of the brick?
Answer: Let the weight of the brick be x kg. According to the problem, x = x/2 + 1. Subtract x/2 from both sides: x - x/2 = 1, which gives (2x - x)/2 = 1, so x/2 = 1. Multiply both sides by 2: x = 2. Therefore, the weight of the brick is 2 kg.
In simple words: If the brick weighs more than half of itself by exactly 1 kg, then the other half must also weigh 1 kg, making the total weight 2 kg.

Exam Tip: When a quantity is described in terms of its own parts, set up the equation carefully by expressing each part using the variable.

Question 5. One quarter of a number increased by 9 gives the same number. What is the number?
Answer: Let the number be x. According to the problem, x/4 + 9 = x. Subtract x/4 from both sides: 9 = x - x/4. This becomes 9 = (4x - x)/4, so 9 = 3x/4. Multiply both sides by 4/3: x = 9 × (4/3) = 12. Therefore, the number is 12.
In simple words: If you take one quarter of a number and add 9, you get back the original number. This means three quarters of the number equals 9, so the number is 12.

Exam Tip: Use fractions carefully when working with "parts" of a number; converting to a common denominator helps avoid errors.

Question 6. Given 4k + 1 = 13, find the values of: (a) 8k + 2 (b) 4k (c) k (d) 4k - 1 (e) -k - 2
Answer: First, solve for k. From 4k + 1 = 13, subtract 1 from both sides to get 4k = 12. Divide both sides by 4: k = 3.

(a) 8k + 2 = 8(3) + 2 = 24 + 2 = 26.

(b) 4k = 4(3) = 12.

(c) k = 3.

(d) 4k - 1 = 4(3) - 1 = 12 - 1 = 11.

(e) -k - 2 = -3 - 2 = -5.
In simple words: Find the value of k first, then use it to calculate each expression by substituting k = 3.

Exam Tip: Once you find the main variable, use its value systematically to evaluate all requested expressions without re-solving.

Question 1. Fill in the blanks with integers. (a) 5 × ____ - 8 = 37 (b) 37 - (33 - ____) = 35 (c) -3 × (-11 + ____) = 45
Answer:
(a) Let the unknown be x. So 5x - 8 = 37. Add 8 to both sides: 5x = 45. Divide by 5: x = 9. Check: 5(9) - 8 = 45 - 8 = 37. ✓

(b) Let the unknown be x. So 37 - (33 - x) = 35. Expand: 37 - 33 + x = 35, which gives 4 + x = 35. Subtract 4 from both sides: x = 31. Check: 37 - (33 - 31) = 37 - 2 = 35. ✓

(c) Let the unknown be y. So -3(-11 + y) = 45. Divide both sides by -3: -11 + y = -15. Add 11 to both sides: y = -4. Check: -3(-11 + (-4)) = -3(-15) = 45. ✓
In simple words: Treat the blank as an unknown variable, set up the equation, and solve step by step using inverse operations.

Exam Tip: Always verify your answer by substituting it back into the original equation to check that both sides are equal.

Question 2. Ranju is a daily wage labourer. She earns Rs 750 a day. Her employer pays her in 50 and 100-rupee notes. If Ranju gets an equal number of 50 and 100 rupee notes, how many notes of each does she have?
Answer: Let the number of Rs 50 notes be x. Since Ranju gets an equal number of each type, she also gets x notes of Rs 100. The total amount is Rs 750, so 50x + 100x = 750. This simplifies to 150x = 750. Divide both sides by 150: x = 5. Therefore, Ranju has 5 notes of Rs 50 and 5 notes of Rs 100.
In simple words: If she gets the same count of each note type, the total value of all notes must add up to 750, which lets you find how many of each she receives.

Exam Tip: When a problem states "equal number," treat that as a key constraint that simplifies setting up your equation.

Question 3. In the given 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? Write an equation to describe this problem.
Answer: Let u represent the number of blue dots hidden by each black blob. The picture shows 3 black blobs and 4 blue dots visible. The total number of dots is 25, so the equation is 3u + 4 = 25. Subtract 4 from both sides: 3u = 21. Divide both sides by 3: u = 7. Each black blob hides 7 blue dots.
In simple words: Set up an equation by counting visible dots and hidden dots, then solve to find how many dots each blob covers.

Exam Tip: Always define your variable clearly (what does u represent?) and write the equation before solving.

Question 4. Here are machines that take an input, perform an operation on it, and send out the result as an output. Find the inputs in the following cases.
Answer:
(a) (i) The input is 9. Working: 9 + 3 = 12, then 12 × 4 = 48, then 48 - 5 = 43.

(ii) The input is 17. Working: 17 + 3 = 20, then 20 × 4 = 80, then 80 - 5 = 75.

(b) (i) Let the unknown number be x. The machine takes x, multiplies by 3 and also adds 3 to get two results, then subtracts the second from the first: x × 3 - (x + 3) = 63. This simplifies to 3x - x - 3 = 63, so 2x - 3 = 63. Add 3 to both sides: 2x = 66. Divide by 2: x = 33.

(ii) Let the unknown number be y. Using the same machine, y × 3 - (y + 3) = 227. This simplifies to 2y - 3 = 227. Add 3: 2y = 230. Divide by 2: y = 115.
In simple words: Work backwards from the output by reversing each operation, or set up an equation matching the sequence of operations shown in the machine.

Exam Tip: When working with machines or sequence diagrams, trace each operation step by step to build your equation correctly.

Question 5. What are the inputs to these machines?
Answer:
(i) Let the unknown number be a. The machine divides by 3 twice, giving (a ÷ 3) ÷ 3 = 5. Work backwards: multiply by 3, so a ÷ 3 = 15. Multiply by 3 again: a = 45.

(ii) Let the unknown number be b. The machine subtracts 4 twice, giving (b - 4) - 4 = -11. Simplify: b - 8 = -11. Add 8 to both sides: b = -3.
In simple words: When a machine repeats an operation, apply the inverse operations in reverse order to find the input from the output.

Exam Tip: Reverse the operations in the opposite order - if the machine adds then multiplies, undo by dividing then subtracting.

Question 6. A taxi driver charges a fixed fee of Rs 800 per day plus Rs 20 for each kilometre travelled. If the total cost for a taxi ride is Rs 2200, determine the number of kilometres travelled.
Answer: Let the distance travelled be x km. The total cost equals the fixed fee plus the variable charge: 800 + 20x = 2200. Subtract 800 from both sides: 20x = 1400. Divide both sides by 20: x = 70. The taxi travelled 70 kilometres.
In simple words: Set up an equation with a fixed part (800) plus a variable part (20 times the distance), then solve for the distance.

Exam Tip: Clearly separate fixed charges from variable charges when setting up word problem equations.

Question 7. The sum of two numbers is 76. One number is three times the other number. What are the numbers?
Answer: Let one number be x. Then the other number is 3x. Since their sum is 76, we have x + 3x = 76. This simplifies to 4x = 76. Divide both sides by 4: x = 19. The first number is 19 and the second number is 3 × 19 = 57.
In simple words: If one number is a multiple of the other, express both using the same variable, then use their sum to find each one.

Exam Tip: When two unknowns are related by a ratio or multiple, express them both in terms of a single variable.

Question 8. The figure shows the diagram for a window with a grill. What is the gap between the two rods in the grill?
Answer: The total height of the window including the frame is 34 cm. Excluding the frame (removing 3 cm from the top and 3 cm from the bottom), the usable height is 34 - 3 - 3 = 28 cm. The grill has 5 rods, each 2 cm wide, for a total of 5 × 2 = 10 cm. Let x be the height of each gap. There are 6 gaps between and around the 5 rods. The sum of all gap heights is 6x. According to the problem, 10 + 6x = 28. Subtract 10 from both sides: 6x = 18. Divide by 6: x = 3. The gap between two rods is 3 cm.
In simple words: Add up all the space used by the rods, subtract from the total usable space, and divide by the number of gaps to find each gap's size.

Exam Tip: When working with repeated structures, count both the items and the spaces between them carefully.

Question 9. In a restaurant, a fruit juice costs Rs 15 less than a chocolate milkshake. If 4 fruit juices and 7 chocolate milkshakes cost Rs 600, find the cost of the fruit juice and milkshake.
Answer: Let the cost of a chocolate milkshake be Rs x. Then the cost of a fruit juice is Rs (x - 15). The total cost of 4 fruit juices and 7 milkshakes is Rs 600, so 4(x - 15) + 7x = 600. Expand: 4x - 60 + 7x = 600. Combine like terms: 11x - 60 = 600. Add 60 to both sides: 11x = 660. Divide by 11: x = 60. The cost of 1 chocolate milkshake is Rs 60, and the cost of 1 fruit juice is Rs 60 - 15 = Rs 45.
In simple words: Express one item's cost in terms of the other, then use the total cost to find both prices.

Exam Tip: When one item's cost is described relative to another, use that relationship to write all costs in terms of a single variable.

Question 10. Given 28p - 36 = 98, find the value of 14p - 19 and 28p - 38.
Answer: From 28p - 36 = 98, add 36 to both sides: 28p = 134. Divide both sides by 2: 14p = 67. Now find each requested value:

(i) 14p - 19 = 67 - 19 = 48.

(ii) 28p - 38 = 134 - 38 = 96.
In simple words: Once you find the value of 28p, use it to calculate related expressions quickly without finding p itself.

Exam Tip: When asked for multiple expressions, look for shortcuts - often you can find a related expression's value without solving for the variable.

Question 11. The steps to solve three equations are shown below. Identify and correct any mistakes.
Answer:
(a) The original equation is 6x + 9 = 66. The error occurs in the first step: the constant should be moved to one side and variable terms to the other before dividing out the coefficient. The correct solution is: 6x + 9 = 66 → 6x = 66 - 9 → 6x = 57 → x = 57/6 = 19/2 = 9.5.

(b) The steps shown are: 14y + 24 = 36 → 7y + 12 = 18 → 7y = 6 → y = 6/7. There is no error; dividing the entire equation by 2 in the first step is a valid approach.

(c) The original equation is 4x - 5 = 9x + 8. There are two errors: first, when moving -5 to the other side, it should become +5, not disappear. Second, the final answer is stated as x = -5/3, but it should be x = -13/5 = -2.6. The correct solution is: 4x - 5 = 9x + 8 → 4x = 9x + 8 + 5 → 4x = 9x + 13 → 4x - 9x = 13 → -5x = 13 → x = -13/5.
In simple words: Always move constants and variables to opposite sides before isolating the variable. Pay careful attention to signs when moving terms, especially negative numbers.

Exam Tip: Common mistakes include forgetting to change the sign when moving a term to the other side, and combining steps incorrectly. Check each operation separately.

Question 12. Find the measures of the angles of these triangles.
Answer:
(i) In triangle ABC, sides AB and AC are equal. By the property of isosceles triangles, angles opposite to equal sides are also equal, so angle B = angle C. Given that angle B = y + 15, we also have angle C = y + 15. The sum of angles in a triangle is 180°, so y + (y + 15) + (y + 15) = 180°. Simplify: 3y + 30 = 180°. Subtract 30: 3y = 150°. Divide by 3: y = 50°. Therefore, angle A = 50°, angle B = 50° + 15° = 65°, and angle C = 65°.

(ii) In triangle ABC, the sum of angles is 180°. So x + (x - 10) + (x + 10) = 180°. Simplify: 3x = 180°. Divide by 3: x = 60°. Therefore, angle A = 60°, angle B = 60° - 10° = 50°, and angle C = 60° + 10° = 70°.
In simple words: Use the angle sum property of triangles (all three angles add to 180°) and the property that equal sides have equal opposite angles to set up and solve equations for unknown angles.

Exam Tip: Always remember that the sum of angles in any triangle is 180°, and in isosceles triangles, the angles opposite the equal sides are themselves equal.

Question 13. Write 4 equations whose solution is u = 6.
Answer:
(i) Start with u = 6. Multiply both sides by 2/3: (2/3)u = 4.

(ii) Start with u = 6. Add 7 to both sides: u + 7 = 13.

(iii) Start with u = 6. Multiply both sides by 2: 2u = 12. Add 3 to both sides: 2u + 3 = 15.

(iv) Start with u = 6. Multiply both sides by 3: 3u = 18. Subtract 5 from both sides: 3u - 5 = 13.
In simple words: To create equations with a specific solution, start with that solution and apply operations like add, subtract, multiply, or divide to form new equations.

Exam Tip: Always verify your created equations by substituting the solution back to confirm they are correct.

Question 14. The Bakhshali Manuscript (300 CE) mentions the following problem. The amount given to the first person is not known. The second person is given twice as much as the first. The third person is given thrice as much as the second, and the fourth person four times as much as the third. The total amount distributed is 132. What is the amount given to the first person?
Answer: Let the amount given to the first person be x. The second person receives 2x. The third person receives 3(2x) = 6x. The fourth person receives 4(6x) = 24x. The total amount is 132, so x + 2x + 6x + 24x = 132. Combine: 33x = 132. Divide by 33: x = 4. The amount given to the first person is 4 units.
In simple words: Express each person's share as a multiple of the first person's amount, add them all up, and use the total to find the first person's share.

Exam Tip: When amounts are related as multiples of each other, express all amounts using the same variable based on one person's share.

Question 15. The height of a giraffe is two and a half metres more than half its height. How tall is the giraffe?
Answer: Let the height of the giraffe be x m. According to the problem, x = (1/2)x + 2.5. Subtract (1/2)x from both sides: x - (1/2)x = 2.5. This gives (1/2)x = 2.5. Multiply both sides by 2: x = 5. The height of the giraffe is 5 metres.
In simple words: If the whole height is half itself plus 2.5 m, then the other half must also be 2.5 m, making the total 5 m.

Exam Tip: When a quantity is described as a fraction of itself plus some amount, isolate that fraction to find the whole easily.

Question 16. Two separate figures are given below. Each figure shows the first few positions in a sequence of arrangements made with sticks. Identify the pattern and answer the following questions for each figure: (a) How many squares are in position number 11 of the sequence? (b) How many sticks are needed to arrange position number 11 of the sequence? (c) Can an arrangement in this sequence be made using exactly 85 sticks? If yes, which position number will it correspond to? (d) Can an arrangement in this sequence be made using exactly 150 sticks? If yes, which position number will it correspond to?
Answer:
(a) In the arrow pattern, position 1 has 4 squares (1 + 3), position 2 has 7 squares (1 + 3 + 3), position 3 has 10 squares (1 + 3 + 3 + 3). The pattern shows that position n has 1 + 3n squares. For position 11: 1 + 3(11) = 1 + 33 = 34 squares.

(b) In the staircase pattern, the number of sticks follows the formula 13 + 9(n - 1). For position 11: 13 + 9(11 - 1) = 13 + 9(10) = 13 + 90 = 103 sticks.

(c) To find if 85 sticks is possible, set 13 + 9(n - 1) = 85. Simplify: 13 + 9n - 9 = 85, so 4 + 9n = 85. Subtract 4: 9n = 81. Divide by 9: n = 9. Yes, position 9 uses exactly 85 sticks.

(d) To find if 150 sticks is possible, set 13 + 9(n - 1) = 150. Simplify: 13 + 9n - 9 = 150, so 4 + 9n = 150. Subtract 4: 9n = 146. Divide by 9: n = 146/9 ≈ 16.2. Since n is not a whole number, no position in the sequence uses exactly 150 sticks.
In simple words: Identify the pattern by examining the first few positions, express it as a formula, then use algebra to answer questions about specific positions or stick counts.

Exam Tip: Always check that your answer makes sense - for example, the position number should be a whole number, and the calculated quantity should match the formula.

Question 17. A number increased by 36 is equal to ten times itself. What is the number?
Answer: Let the number be x. According to the problem, x + 36 = 10x. Subtract x from both sides: 36 = 9x. Divide by 9: x = 4. The number is 4.
In simple words: If a number plus 36 equals ten times that number, then nine times that number must equal 36, making the number 4.

Exam Tip: When a number is described as being related to itself through multiplication or addition, collect all variable terms on one side first.

Question 18. Solve these equations: (a) 5(r + 2) = 10 (b) -3(u + 2) = 2(u - 1) (c) 2(7 - 2n) = -6 (d) 2(x - 4) = -16 (e) 6(x - 1) = 2(x - 1) - 4 (f) 3 - 7s = 7 - 3s (g) 2x + 1 = 6 - (2x - 3) (h) 10 - 5x = 3(x - 4) - 2(x - 7)
Answer:
(a) Divide both sides by 5: r + 2 = 2. Subtract 2: r = 0.

(b) Expand the left side: -3u - 6 = 2u - 2. Add 3u to both sides: -6 = 5u - 2. Add 2: -4 = 5u. Divide by 5: u = -4/5.

(c) Divide both sides by 2: 7 - 2n = -3. Subtract 7: -2n = -10. Divide by -2: n = 5.

(d) Divide both sides by 2: x - 4 = -8. Add 4: x = -4.

(e) Expand: 6x - 6 = 2x - 2 - 4, so 6x - 6 = 2x - 6. Subtract 2x: 4x - 6 = -6. Add 6: 4x = 0. So x = 0.

(f) Rearrange: 3 - 7 = -3s + 7s, so -4 = 4s. Divide by 4: s = -1.

(g) Expand the right side: 2x + 1 = 6 - 2x + 3, so 2x + 1 = 9 - 2x. Add 2x: 4x + 1 = 9. Subtract 1: 4x = 8. Divide by 4: x = 2.

(h) Expand: 10 - 5x = 3x - 12 - 2x + 14, so 10 - 5x = x + 2. Subtract x and add 5x: 10 = 6x + 2. Subtract 2: 8 = 6x. Divide by 6: x = 4/3.
In simple words: Expand any brackets using the distributive property, move all variable terms to one side and constants to the other, then isolate the variable.

Exam Tip: Always expand brackets fully before collecting like terms - careless errors in expansion are a common source of mistakes.

Question 20. There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?
Answer: Let the number of children be x and the number of donkeys be y. Since each has one head, x + y = 28, so y = 28 - x. Children have 2 feet each and donkeys have 4 feet each, so 2x + 4y = 80. Substitute y = 28 - x: 2x + 4(28 - x) = 80. Expand: 2x + 112 - 4x = 80. Simplify: -2x + 112 = 80. Subtract 112: -2x = -32. Divide by -2: x = 16. Therefore, there are 16 children and 28 - 16 = 12 donkeys.
In simple words: Set up two equations using the fact that heads count all animals while feet depend on each type. Use substitution to reduce to one variable and solve.

Exam Tip: In problems with multiple unknowns, count what you know about each type of object to create all the equations you need.