Chapter 1 · Model TestSolved paper

Model Test Paper – Solved ഉത്തരങ്ങൾ

Every question of the model paper, with a full solution, an animation for every multi-step question, and a marking hint so you can score your own paper honestly.

📖 How marks are usually given

1-mark questions: the correct answer only. Longer questions: about one mark for writing the equations correctly (say what x and y stand for), the rest for the working and the final answer. A correct method with a small arithmetic slip still earns most of the marks – so always write the equations and every step.

SECTION A8 × 1 = 8 marks

If x + y = 10 and x − y = 4, then x = and y = .

(1)

Answer: x = 7, y = 3

Add: 2x = 14 ⇒ x = 7. Subtract: 2y = 6 ⇒ y = 3.

Review: Page 9 · Sum and difference

Two numbers have sum 30 and difference 6. The larger number is .

(1)

Answer: 18

Larger = (sum + difference) ÷ 2 = (30 + 6) ÷ 2 = 18 (the smaller is 12).

Review: Page 9 · Sum and difference

If 2x + 3y = 12 and 2x + y = 8, then 2y = .

(1)

Answer: 4

Subtract: (2x + 3y) − (2x + y) = 2y = 12 − 8 = 4. (So y = 2 and x = 3.)

Review: Page 7 · Elimination

Which pair of equations has the solution x = 2, y = 3?

(a) x + y = 5, x − y = −1(b) x + y = 6, x − y = 1(c) 2x + y = 6, x + y = 5(d) x − y = 1, x + y = 5

(1)

Answer: (a)

2 + 3 = 5 and 2 − 3 = −1. In (b) 2 + 3 ≠ 6; in (c) 2 × 2 + 3 = 7 ≠ 6; in (d) 2 − 3 = −1 ≠ 1. A solution must satisfy both equations.

Review: Page 9 · Solutions of a pair

How many solutions does the single equation x + y = 10 have if x and y may be any numbers?

(a) one(b) two(c) none(d) infinitely many

(1)

Answer: (d) infinitely many

1 and 9, 2 and 8, 2.5 and 7.5, … – one equation in two letters does not fix them. A second, different equation is needed (textbook side box വിവരങ്ങളും പരിഹാരങ്ങളും).

Review: Page 9 · Information and solutions

A two-digit number has tens digit a and units digit b. The number is .

(1)

Answer: 10a + b

For example 47 = 10 × 4 + 7.

Review: Page 4 · Two-digit numbers

3 pens and 2 books cost ₹50; 3 pens and 4 books cost ₹70. One book costs ₹.

(1)

Answer: ₹10

The 2 extra books cost 70 − 50 = ₹20, so one book costs ₹10 (and a pen ₹10 too).

Review: Page 7 · Pens and notebooks

For the pair x + y = 7 and 2x + 2y = 14, which statement is true?

(a) x = 3, y = 4 is the only solution(b) there is no solution(c) x and y cannot be found separately(d) x must equal y

(1)

Answer: (c)

The second equation is just 2 × the first – the same information (textbook side box വ്യത്യസ്തമല്ലാത്ത വിവരങ്ങൾ). Any x and y adding to 7 work.

Review: Page 9 · Same information

SECTION B5 × 2 = 10 marks

The sum of two numbers is 40 and their difference is 10. Find the numbers.

(2)

🎬 Solution

Answer: 25 and 15

x + y = 40, x − y = 10. Adding: 2x = 50 ⇒ x = 25. Subtracting: 2y = 30 ⇒ y = 15. Check: 25 + 15 = 40, 25 − 15 = 10.

Marks: 1 for the two equations (or the bar-model reasoning), 1 for both numbers.

Review: Page 9 · Sum and difference

10.

A table and a chair together cost ₹5000. A table and 3 chairs cost ₹7000. Find the price of each.

(2)

🎬 Solution

Answer: chair ₹1000, table ₹4000

Chair x, table y: x + y = 5000, 3x + y = 7000. Subtract: 2x = 2000 ⇒ x = 1000; y = 5000 − 1000 = 4000.

Marks: 1 for the equations or "2 extra chairs cost ₹2000", 1 for both prices.

Review: Page 5 · Table and chairs

11.

Solve: 3x + 2y = 18, x + 2y = 10.

(2)

🎬 Solution

Answer: x = 4, y = 3

The y-coefficients are equal: subtract, (3x + 2y) − (x + 2y) = 2x = 8 ⇒ x = 4. Then 4 + 2y = 10 ⇒ y = 3. Check: 12 + 6 = 18 ✓.

Marks: 1 for eliminating y (2x = 8), 1 for x = 4, y = 3.

Review: Page 7 · Elimination

12.

Adding 2 to the numerator of a fraction and simplifying gives ½. Adding 1 to its denominator instead gives ⅓. Find the fraction.

(2)

🎬 Solution

Answer: 5/14

Numerator x, denominator y. (x + 2)/y = ½ ⇒ y = 2(x + 2) = 2x + 4. x/(y + 1) = ⅓ ⇒ y + 1 = 3x. Substitute: 2x + 5 = 3x ⇒ x = 5, y = 14. Check: 7/14 = ½ ✓, 5/15 = ⅓ ✓.

Marks: 1 for the two equations y = 2(x + 2), y + 1 = 3x; 1 for 5/14.

Review: Page 5 · The fraction puzzle · Page 6 · Practice set 1, Q6

13.

The perimeter of a rectangle is 30 cm and its length is 3 cm more than its breadth. Find its length and breadth.

(2)

🎬 Solution

Answer: breadth 6 cm, length 9 cm

Perimeter 30 ⇒ length + breadth = 15. With l − b = 3: 2l = 18 ⇒ l = 9, b = 6. Check: 2(9 + 6) = 30 ✓.

Marks: 1 for l + b = 15 and l − b = 3 (or b + (b + 3) = 15), 1 for the sides.

Review: Page 6 · Practice set 1, Q3

SECTION C4 × 3 = 12 marks

14.

2 kg of apples and 3 kg of oranges cost ₹420. 3 kg of apples and 2 kg of oranges cost ₹380. Find the price of 1 kg of each.

(3)

🎬 Solution

Answer: apples ₹60/kg, oranges ₹100/kg

Apple x, orange y: 2x + 3y = 420, 3x + 2y = 380. Add: 5x + 5y = 800 ⇒ x + y = 160. Subtract: y − x = 40. So y = 100, x = 60. Check: 120 + 300 = 420 ✓, 180 + 200 = 380 ✓. (Making coefficients equal also works: × 3 and × 2.)

Marks: 1 for the equations, 1 for x + y = 160 and y − x = 40 (or an elimination step), 1 for both prices.

Review: Page 8 · Practice set 3, Q1

15.

Solve by making the coefficients of one letter equal: 4x + 3y = 24, 3x − 2y = 1.

(3)

🎬 Solution

Answer: x = 3, y = 4

(1) × 2: 8x + 6y = 48. (2) × 3: 9x − 6y = 3. The y-terms are opposite, so add: 17x = 51 ⇒ x = 3. Then 12 + 3y = 24 ⇒ y = 4. Check (2): 9 − 8 = 1 ✓.

Marks: 1 for multiplying correctly, 1 for 17x = 51 (or the matching step for x), 1 for x = 3, y = 4.

Review: Page 7 · Making coefficients equal

16.

The digits of a two-digit number add up to 9. Reversing the digits gives a number 27 more than the original. Find the number.

(3)

🎬 Solution

Answer: 36

Number 10a + b, reversed 10b + a. a + b = 9 and (10b + a) − (10a + b) = 9b − 9a = 27 ⇒ b − a = 3. Sum and difference: b = 6, a = 3. Check: 63 − 36 = 27 ✓.

Marks: 1 for a + b = 9, 1 for 9(b − a) = 27, 1 for 36.

Review: Page 10 · Practice set 4, Q3

17.

₹50000 is invested in two schemes, one at 8% and the other at 6% yearly interest. The total interest after one year is ₹3500. How much was invested in each scheme?

(3)

🎬 Solution

Answer: ₹25000 in each

x at 8%, y at 6%: x + y = 50000, 0.08x + 0.06y = 3500. Subtract 0.06 × (first): 0.02x = 3500 − 3000 = 500 ⇒ x = 25000, y = 25000. Check: 2000 + 1500 = 3500 ✓.

Marks: 1 for the equations, 1 for 0.02x = 500, 1 for both amounts.

Review: Page 6 · Practice set 1, Q7

SECTION D2 × 5 = 10 marks

18.

Five years ago a father was 3 times as old as his son. Five years from now he will be twice as old as his son.

Write two equations for their present ages.
Solve them.
Check your answer with both facts.
Explain why the difference of their ages is the same at every time.

(5)

🎬 Solution

Answer: son 15 years, father 35 years

(i) Father F, son S now: F − 5 = 3(S − 5) and F + 5 = 2(S + 5), that is F = 3S − 10 and F = 2S + 5.

(ii) 3S − 10 = 2S + 5 ⇒ S = 15, F = 35.

(iii) Five years ago: 30 and 10, and 30 = 3 × 10 ✓. In five years: 40 and 20, and 40 = 2 × 20 ✓.

(iv) Every year both become one year older, so the gap F − S = 20 never changes.

Marks: 2 for the two equations, 1 for S = 15, F = 35, 1 for the check, 1 for the explanation.

Review: Page 8 · Practice set 3, Q3 · Page 4 · Ages

19.

The speed of an object after t seconds is u + at metres per second. After 2 seconds its speed is 11 m/s; after 6 seconds it is 23 m/s.

Write two equations in u and a.
Find u and a.
Find its speed after 10 seconds.
The distance travelled in t seconds is ut + ½at². Find the distance travelled in the first 4 seconds.
Explain why the pair x + y = 5, 2x + 2y = 12 has no solution.

(5)

🎬 Solution

Answer: u = 5 m/s, a = 3 m/s per second; 35 m/s; 44 m

(i) u + 2a = 11 and u + 6a = 23.

(ii) Subtract: 4a = 12 ⇒ a = 3; u = 11 − 6 = 5.

(iii) Speed after 10 s = 5 + 3 × 10 = 35 m/s.

(iv) Distance = 5 × 4 + ½ × 3 × 16 = 20 + 24 = 44 m.

(v) Doubling the first equation gives 2x + 2y = 10, but the second says 2x + 2y = 12. The same quantity cannot be both 10 and 12, so the facts contradict each other and no x, y satisfy both.

Marks: 1 for each part.

Review: Page 6 · Practice set 1, Q8 · Page 10 · Practice set 4, Q5 · Page 9 · Impossible data

Marks summary

Section	Questions	Marks	What it tests
A	1 – 8	8	Quick sum–difference, elimination, checking a solution, kinds of pairs, 10a + b
B	9 – 13	10	Forming and solving pairs: numbers, prices, fraction, perimeter
C	14 – 17	12	Elimination with multiplying, digit puzzle, interest
D	18 – 19	10	Ages over time, speed u + at and distance, contradictory data
Total		40

⏱️ Time management (1½ hours = 90 minutes)

Section A: about 10 minutes – one line each; check by putting your answer back.
Section B: about 20 minutes. Always write "let … be x" and the equations first.
Section C: about 30 minutes. Decide early: add/subtract directly, or multiply first?
Section D: about 25 minutes. Answer each part separately and label it (i), (ii), …
Keep 5 minutes to check every answer in both equations.

🎉 Finished the chapter?

Scored 32+? Excellent. Between 24 and 31? Revisit the pages named under the questions you lost marks on. Below 24? Go back to page 5 and work through the section pages again, solving each puzzle all three ways – then retake the test in a week.

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