Mathematics: Unit-wise Subjective Question Bank

A Mathematics answer is marked in steps, and the steps are worth more than the final number.

A two-mark question expects the formula and the substitution; a four-mark question expects the formula, the substitution, the working and the answer with its unit.

Write every step, even the obvious one, because a marker can award a step you wrote and cannot award a step you did in your head.

How a four-mark solution is marked

One mark for the correct formula or construction, one for correct substitution, one for correct simplification and one for the final answer with its unit. A wrong arithmetic slip with correct method still earns three of the four — but only if the steps are on the page.

6.1 Sets

1. If A = \{1,2,3,4,5\} and B = \{4,5,6,7\}, find A \cup B, A \cap B and n(A \cup B). [2]

A \cup B = \{1,2,3,4,5,6,7\}; A \cap B = \{4,5\}; n(A \cup B) = 7.

2. In a class of 40 students, 25 play football, 18 play volleyball and 8 play both. Find how many play at least one game and how many play neither. [4]

n(F \cup V) = n(F) + n(V) - n(F \cap V) = 25 + 18 - 8 = 35 play at least one game. Neither = 40 - 35 = 5 students. (A Venn diagram with 17, 8 and 10 in the three regions and 5 outside earns the method marks.)

6.2 Whole, Rational and Irrational Numbers

1. Write any two differences between a rational and an irrational number with one example of each. [2]

A rational number can be written as \frac{p}{q} with integers p and q \neq 0, and its decimal form terminates or recurs — for example \frac{3}{4} = 0.75.

An irrational number cannot be written as such a fraction and its decimal neither terminates nor recurs — for example \sqrt{5} = 2.2360\ldots

2. Simplify: \frac{3}{4} + \frac{5}{6} - \frac{2}{3}. [2]

LCM of 4, 6 and 3 is 12: \frac{9}{12} + \frac{10}{12} - \frac{8}{12} = \frac{11}{12}.

3. Insert any two rational numbers between \frac{1}{3} and \frac{1}{2}. [4]

Write both with a common denominator large enough to leave room: \frac{1}{3} = \frac{12}{36} and \frac{1}{2} = \frac{18}{36}. Two numbers between them are \frac{14}{36} = \frac{7}{18} and \frac{15}{36} = \frac{5}{12}.

(Checking: 0.333 < 0.388 < 0.416 < 0.5.) Infinitely many such numbers exist, which is why the rationals are said to be dense.

6.3 Ratio, Proportion and the Unitary Method

1. Divide Rs 1,750 among A, B and C in the ratio 2:3:5. [2]

Total parts = 10; one part = \frac{1750}{10} = 175. A = 350, B = 525, C = 875 rupees.

2. If 15 workers can dig a canal in 24 days, how many workers are needed to dig it in 18 days? [2]

Work is inversely proportional to time: 15\times24 = x\times18, so x = \frac{360}{18} = 20 workers.

3. A pipe fills a tank in 6 hours and another in 4 hours. If both are opened together, how long will the tank take to fill? [4]

In one hour the first fills \frac{1}{6} of the tank and the second \frac{1}{4}. Together in one hour: \frac{1}{6} + \frac{1}{4} = \frac{2+3}{12} = \frac{5}{12}. Time = \frac{12}{5} = 2.4 hours = 2 hours 24 minutes.

6.4 Profit and Loss

1. An article costing Rs 1,500 is sold at a profit of 18 %. Find the selling price. [2]

Profit = \frac{18}{100}\times1500 = 270; S.P. = 1500 + 270 = Rs 1,770.

2. A shopkeeper marks a jacket at Rs 4,000, allows 15 % discount and then charges 13 % VAT. Find the amount the customer pays. [4]

Discount = 15 % of 4000 = 600; price after discount = 3400. VAT = 13 % of 3400 = 442. Amount paid = 3400 + 442 = Rs 3,842. (Note the order: discount is on the marked price, VAT on the discounted price.)

3. By selling a bicycle for Rs 6,300 a trader loses 10 %. At what price should it be sold to gain 12 %? [4]

Let the cost price be x. Then 0.90x = 6300, so x = 7000. For a 12 % gain, S.P. = 1.12\times7000 = Rs 7,840.

6.5 Simple Interest

1. Find the simple interest on Rs 12,000 for 3 years at 9 % per annum. [2]

I = \frac{PTR}{100} = \frac{12000\times3\times9}{100} = Rs 3,240.

2. A sum of Rs 8,000 amounts to Rs 10,400 in 4 years at simple interest. Find the rate. [4]

I = 10400 - 8000 = 2400. R = \frac{100I}{PT} = \frac{100\times2400}{8000\times4} = \frac{240000}{32000} = 7.5 % per annum.

3. In what time will Rs 15,000 earn Rs 4,500 as simple interest at 10 % per annum? [2]

T = \frac{100I}{PR} = \frac{100\times4500}{15000\times10} = 3 years.

6.6 Area and Volume

1. Find the area of a circle of radius 14 cm, taking \pi = \frac{22}{7}. [2]

A = \pi r^2 = \frac{22}{7}\times14\times14 = 616 cm².

2. A room is 6 m long, 4 m wide and 3 m high. Find the cost of painting its four walls and ceiling at Rs 90 per m². [4]

Four walls = 2h(l+b) = 2\times3\times(6+4) = 60 m². Ceiling = l\times b = 24 m². Total area = 84 m²; cost = 84\times90 = Rs 7,560.

3. A cylindrical water tank of radius 1.4 m and height 2 m is full. Find the volume of water in litres, taking \pi = \frac{22}{7}. [4]

V = \pi r^2 h = \frac{22}{7}\times1.4\times1.4\times2 = 12.32 m³. 1 m³ = 1000 litres, so the tank holds 12,320 litres.

6.7 Indices and Algebra

1. Simplify: \frac{x^5 \times x^3}{x^4}. [2]

\frac{x^{5+3}}{x^4} = x^{8-4} = x^4.

2. Factorise: (i) x^2 - 25 (ii) x^2 + 7x + 12. [2]

(i) (x+5)(x-5). (ii) (x+3)(x+4), since 3+4 = 7 and 3\times4 = 12.

3. Simplify: \frac{x^2-9}{x^2+x-12}. [4]

Numerator = (x+3)(x-3). Denominator = (x+4)(x-3). Cancelling the common factor (x-3): \frac{x+3}{x+4}, valid for x \neq 3 and x \neq -4.

4. Solve: \frac{2x-1}{3} - \frac{x+2}{4} = 1. [4]

Multiply through by 12: 4(2x-1) - 3(x+2) = 12. 8x - 4 - 3x - 6 = 12 \Rightarrow 5x - 10 = 12 \Rightarrow 5x = 22, so x = \frac{22}{5} = 4.4.

6.8 Equation and Graph

1. Draw the graph of y = 2x - 1 for x = -1, 0, 1, 2 (give the table of values and describe the line). [4]

Table: x = -1 \to y = -3; x = 0 \to y = -1; x = 1 \to y = 1; x = 2 \to y = 3. Plotting (-1,-3), (0,-1), (1,1), (2,3) gives a straight line rising to the right, cutting the y-axis at -1 with slope 2 (for every 1 unit right, 2 units up).

2. The sum of two numbers is 45 and their difference is 11. Find the numbers. [4]

Let the numbers be x and y with x > y. x + y = 45 and x - y = 11. Adding: 2x = 56, so x = 28; then y = 45 - 28 = 17.

Check 28 + 17 = 45 and 28 - 17 = 11. The numbers are 28 and 17.

6.9 Lines, Angles and Plane Figures

1. Two angles of a triangle are 48° and 67°. Find the third angle and the exterior angle at that vertex. [2]

Third angle = 180° - (48° + 67°) = 65°. Its exterior angle = 180° - 65° = 115° (which equals 48° + 67°, as the exterior-angle property requires).

2. Find each interior angle of a regular polygon of 9 sides. [2]

Interior angle = \frac{(n-2)180°}{n} = \frac{7\times180°}{9} = 140°.

3. In a parallelogram ABCD, \angle A = 65°. Find the other three angles, giving reasons. [4]

Opposite angles of a parallelogram are equal, so \angle C = \angle A = 65°. Co-interior angles between the parallel sides are supplementary, so \angle B = 180° - 65° = 115°. Then \angle D = \angle B = 115°.

Check 65 + 115 + 65 + 115 = 360°.

6.10 Congruency and Similarity

1. State the SAS test of congruency and illustrate it with a pair of triangles. [2]

If two sides and the included angle of one triangle equal the corresponding two sides and included angle of another, the triangles are congruent. In \triangle ABC and \triangle PQR, if AB = PQ, AC = PR and \angle A = \angle P, then \triangle ABC \cong \triangle PQR.

2. In two similar triangles the sides of the first are 6 cm, 8 cm and 10 cm, and the shortest side of the second is 9 cm. Find the other two sides of the second triangle. [4]

Scale factor = \frac{9}{6} = 1.5. Other sides = 8\times1.5 = 12 cm and 10\times1.5 = 15 cm. (Check: 9:12:15 = 6:8:10, so the triangles are similar.)

6.11 Solid Objects, Tessellation and Scale Drawing

1. Verify Euler's relation for a triangular prism. [2]

A triangular prism has F = 5, V = 6, E = 9. Then F + V - E = 5 + 6 - 9 = 2, which verifies the relation.

2. Explain why a regular pentagon does not tessellate while a regular hexagon does. [4]

For a tessellation, whole polygons must fill the 360° round a point with no gap and no overlap, so the interior angle must divide 360° exactly.

A regular hexagon has interior angle 120°, and 3\times120° = 360° — three hexagons meet exactly, as in a honeycomb.

A regular pentagon has interior angle 108°; three give 324° (a gap of 36°) and four give 432° (an overlap), so no arrangement closes.

3. A map is drawn to the scale 1:25{,}000. Two villages are 7.2 cm apart on the map. Find the real distance in kilometres, and the map distance for a 5 km road. [4]

Real distance = 7.2\times25{,}000 = 180{,}000 cm = 1.8 km. For 5 km = 500{,}000 cm: map distance = \frac{500000}{25000} = 20 cm.

6.12 Coordinates and Transformation

1. Plot A(2,3), B(2,-1) and C(-3,3) and state which two of them lie on a line parallel to the y-axis. [2]

A and B both have x = 2, so AB is parallel to the y-axis. (A and C both have y = 3, so AC is parallel to the x-axis.)

2. Find the image of the triangle with vertices P(1,2), Q(4,2), R(1,5) under (i) reflection in the y-axis (ii) translation \binom{2}{-3}. [4]

(i) Reflection in the y-axis, (x,y)\to(-x,y): P'(-1,2), Q'(-4,2), R'(-1,5). (ii) Translation (x,y)\to(x+2, y-3): P''(3,-1), Q''(6,-1), R''(3,2).

Both images are congruent to the original, since reflection and translation preserve lengths and angles.

6.13 Bearing

1. A village B is 6 km due east of A, and C is 6 km due north of B. Find the bearing of B from A and of A from B. [4]

Bearings are measured clockwise from north. Due east is 090°, so the bearing of B from A is 090°. The bearing of A from B is the back bearing, 090° + 180° = 270° (due west), which agrees with A lying west of B.

6.14 Statistics

1. Find the mean, median and mode of: 5, 7, 7, 9, 12, 14, 14, 14, 18. [4]

Sum = 100, n = 9, so mean = \frac{100}{9} = 11.11 (2 d.p.). Ordered already; the 5th value is the median = 12. 14 occurs three times, so the mode = 14.

2. The marks of 40 students are given: 0–20: 6, 20–40: 10, 40–60: 14, 60–80: 7, 80–100: 3. Find the mean using mid-values. [4]

Mid-values 10, 30, 50, 70, 90. \sum fx = 6(10) + 10(30) + 14(50) + 7(70) + 3(90) = 60 + 300 + 700 + 490 + 270 = 1820. Mean = \frac{1820}{40} = 45.5 marks.

3. In a survey of 72 families, 24 use firewood, 30 use gas, 12 use electricity and 6 use biogas for cooking. Find the angle of each sector for a pie chart. [4]

Each angle = \frac{\text{families}}{72}\times360° = \text{families}\times5°. Firewood = 120°; gas = 150°; electricity = 60°; biogas = 30°. Check: 120 + 150 + 60 + 30 = 360°.

6.15 Extended model answers (8 to 10 marks)

Long problems in Mathematics are marked for method, not only for the answer. Write the formula, substitute with units, calculate in clear steps and check the result. The three solutions below show the full layout.

Extended questions

1. A shopkeeper buys a television for Rs 30,000 and marks it at Rs 40,000. He allows a 10 percent discount and charges 13 percent VAT. Find the selling price, the profit percent and the bill amount. Also find the simple interest on Rs 50,000 at 8 percent for 3 years. [10]
2. A cylindrical water tank has radius 3.5 m and height 4 m. Find its capacity in litres, its curved surface area and its total surface area. Take \pi = \frac{22}{7}. [10]
3. The cost of 3 pens and 2 copies is Rs 90, and the cost of 2 pens and 3 copies is Rs 85. Find the price of one pen and one copy, and verify the answer. [10]

Model answer D1: discount, VAT, profit and simple interest (10 marks)

Introduction. In trade, the marked price (MP) is the price printed on the goods. The selling price (SP) is the price actually paid after discount, and VAT is a tax added on the selling price. In Nepal the VAT rate is 13 percent.

Profit or loss is always calculated on the cost price (CP).

Step 1: discount. Discount = 10\% \text{ of } 40{,}000 = \frac{10}{100}\times 40{,}000 = \text{Rs } 4{,}000.

Step 2: selling price before VAT. SP = MP - \text{discount} = 40{,}000 - 4{,}000 = Rs 36,000.

Step 3: profit and profit percent. Profit = SP - CP = 36{,}000 - 30{,}000 = \text{Rs } 6{,}000. Profit percent = \frac{6{,}000}{30{,}000}\times 100 = 20 percent.

Step 4: VAT and bill amount. VAT = 13\% \text{ of } 36{,}000 = \frac{13}{100}\times 36{,}000 = \text{Rs } 4{,}680. Bill amount = 36{,}000 + 4{,}680 = Rs 40,680.

Item	Working	Amount (Rs)
Marked price	given	40,000
Discount (10 percent)	40,000 × 10/100	4,000
Selling price	40,000 − 4,000	36,000
VAT (13 percent)	36,000 × 13/100	4,680
Bill amount	36,000 + 4,680	40,680

Step 5: simple interest. I = \frac{P\,T\,R}{100} = \frac{50{,}000\times 3\times 8}{100} = Rs 12,000. Amount = 50{,}000 + 12{,}000 = Rs 62,000.

Check. Profit percent 20 percent of Rs 30,000 is Rs 6,000, which agrees with SP − CP. VAT is paid by the customer, so it is not part of the shopkeeper's profit.

Conclusion. The customer pays Rs 40,680, the shopkeeper earns a profit of 20 percent, and Rs 4,680 goes to the government as VAT. Always check the bill in a shop: the VAT is added after the discount.

Model answer D2: cylindrical water tank (10 marks)

Introduction. A cylinder has two equal circular faces and one curved surface. For a tank of radius r and height h, the volume is V = \pi r^2 h, the curved surface area is CSA = 2\pi r h and the total surface area is TSA = 2\pi r(r + h). One cubic metre holds 1,000 litres.

Given: r = 3.5 m, h = 4 m, \pi = \frac{22}{7}.

Step 1: capacity.

V = \frac{22}{7}\times 3.5\times 3.5\times 4 = \frac{22}{7}\times 12.25\times 4 = 22\times 1.75\times 4 = 154\ \text{m}^3.

Capacity = 154\times 1000 = 154,000 litres.

Step 2: curved surface area.

CSA = 2\times\frac{22}{7}\times 3.5\times 4 = 2\times 22\times 0.5\times 4 = 88 m².

Step 3: total surface area. The area of one circular face is \pi r^2 = \frac{22}{7}\times 12.25 = 38.5 m². Two faces = 77 m².

TSA = CSA + 2\pi r^2 = 88 + 77 = 165 m².

Quantity	Formula	Result
Volume	πr²h	154 m³
Capacity	volume × 1000	154,000 L
Curved surface area	2πrh	88 m²
Area of two ends	2πr²	77 m²
Total surface area	2πr(r + h)	165 m²

Check. Using the shortcut, TSA = 2\pi r(r+h) = 2\times\frac{22}{7}\times 3.5\times 7.5 = 22\times 7.5 = 165 m², which matches.

Application. If a school needs 4,000 litres of water every day, this tank lasts \frac{154{,}000}{4{,}000} = 38.5 days. If painting the outer surface of a closed tank costs Rs 40 per m², the cost is 165\times 40 = Rs 6,600.

Conclusion. The tank holds 154,000 litres and needs 165 m² of metal sheet. The same three formulas solve any cylinder problem, from a drum of oil to a water tower.

Model answer D3: simultaneous equations (10 marks)

Introduction. Two equations with two unknown quantities are called simultaneous equations. They are solved together by elimination or substitution. Let the price of one pen be Rs p and the price of one copy be Rs c.

Formation of equations.

3p + 2c = 90 \quad\cdots(1)

2p + 3c = 85 \quad\cdots(2)

Solution by elimination. To remove c, make its coefficient equal. Multiply (1) by 3 and (2) by 2.

9p + 6c = 270 \quad\cdots(3)

4p + 6c = 170 \quad\cdots(4)

Subtract (4) from (3): 5p = 100, so p = 20.

Put p = 20 in (1): 3(20) + 2c = 90, so 60 + 2c = 90, 2c = 30 and c = 15.

Verification.

Equation	Left side	Right side	Result
(1) 3p + 2c	3(20) + 2(15) = 60 + 30 = 90	90	true
(2) 2p + 3c	2(20) + 3(15) = 40 + 45 = 85	85	true

Answer. One pen costs Rs 20 and one copy costs Rs 15.

Conclusion. Elimination removes one unknown at a time. Always verify in both original equations, because a slip in one step gives a wrong answer that still satisfies only one equation.