Final Exam Model Paper 1

C = {1, 2, 3, 6, 9, 18}, so n(C) = 6.

A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {3, 6, 9, 12, 15, 18}, C = {1, 2, 3, 6, 9, 18}. Place the numbers in all three sets first (6 and 18), then those in two (12 in A and B; 3 and 9 in B and C; 2 in A and C), then the rest.

Left side: B ∪ C = {1, 2, 3, 6, 9, 12, 15, 18}, so
$$A \cap (B \cup C) = \{2, 6, 12, 18\}$$
Right side: A ∩ B = {6, 12, 18} and A ∩ C = {2, 6, 18}, so
$$(A \cap B) \cup (A \cap C) = \{2, 6, 12, 18\}$$
Both sides are the same set. So A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

A ∩ B = {6, 12, 18}, and the multiples of 6 from 1 to 18 are also {6, 12, 18}. They are equal sets. The reason: a number that is even (divisible by 2) and a multiple of 3 is divisible by 2 × 3 = 6, and every multiple of 6 is both.

Taxable income is the income of a year on which income tax is charged: the year's income less the deductions the law allows (such as provident fund deposits and insurance premiums).

$$\text{Yearly income} = 13 \times 85000 = 1105000$$
Rs 11,05,000.

Taxable income = 1105000 − 30000 = 1075000: Rs 10,75,000 = Rs 10,00,000 + Rs 75,000.
$$\begin{aligned} \text{Tax} &= \tfrac{1}{100} \times 1000000 + \tfrac{10}{100} \times 75000 \\ &= 10000 + 7500 \\ &= 17500 \end{aligned}$$
He pays Rs 17,500.

What it means: about 1.6% of his income, though his last Rs 75,000 pays 10%.

A commission is the money paid to an agent for buying or selling goods for someone else, usually a percentage of the value of what is sold (or bought).

$$\text{Commission} = \frac{4}{100} \times 360000 = 14400$$
Rs 14,400.

New scheme: $\frac{2}{100} \times 200000 + \frac{6}{100} \times 160000 = 4000 + 9600 = 13600$.

The old 4% scheme pays her more, by 14400 − 13600 = Rs 800. (The new scheme pays more only on bigger sales: above Rs 4,00,000 a month.)

5376 − 5238 = 138 units.

138 units is in the 101–250 block, so the minimum charge is Rs 125. Cut 138 units at the blocks: 138 = 20 + 10 + 20 + 88.
$$\begin{aligned} \text{Energy charge} &= 20 \times 4 + 10 \times 6.50 + 20 \times 8 + 88 \times 9.50 \\ &= 80 + 65 + 160 + 836 = 1141 \\ \text{Bill} &= 125 + 1141 = 1266 \end{aligned}$$
Rs 1,266.

The first 50 units cost 80 + 65 + 160 = 305 in energy charge, and each unit after that Rs 9.50.

Up to 100 units the minimum charge is Rs 100: the bill for 100 units is 100 + 305 + 50 × 9.50 = 880, under Rs 1,000. So they can use more than 100 units, and then the minimum charge is Rs 125. For u units:
$$\begin{aligned} 125 + 305 + 9.50(u - 50) &\le 1000 \\ 9.50(u - 50) &\le 570 \\ u - 50 &\le 60 \end{aligned}$$
At most 110 units (the bill is then exactly Rs 1,000; 111 units would cost Rs 1,009.50). That is 28 units fewer than this month.

$$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}, \quad \text{where } s = \frac{a + b + c}{2}$$
(Heron's formula.)

$$s = \frac{104 + 112 + 120}{2} = \frac{336}{2} = 168$$
168 ft.

$$\begin{aligned} \text{Area} &= \sqrt{168 \times (168 - 104) \times (168 - 112) \times (168 - 120)} \\ &= \sqrt{168 \times 64 \times 56 \times 48} \\ &= \sqrt{28901376} = 5376 \end{aligned}$$
5,376 ft².

A ropani is 5,476 ft², and the plot is 5,376 ft². No, he is not right: the plot is 5476 − 5376 = 100 ft² short of a ropani (about 1 paisa).

$$V = \pi r^2 h$$
where r is the radius of the base and h the height.

Curved surface area $= 2\pi rh$:
$$\begin{aligned} 2 \times \frac{22}{7} \times 0.7 \times h &= 6.6 \\ 4.4h &= 6.6 \\ h &= 1.5 \end{aligned}$$
The tank is 1.5 m high.
$$V = \frac{22}{7} \times 0.7^2 \times 1.5 = \frac{22}{7} \times 0.49 \times 1.5 = 2.31 \text{ m}^3$$
1 m³ = 1,000 liters, so the tank holds 2,310 liters.

Days: 2310 ÷ 330 = 7 days.

The height of an end from its apex to the middle of the 6 m base: $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$ m.

One end $= \frac12 \times 6 \times 4 = 12$ m²; two ends = 2 × 12 = 24 m².

(By Heron's formula: s = 8, $\sqrt{8 \times 3 \times 3 \times 2} = \sqrt{144} = 12$ m².)

Volume of a prism = area of its cross-section × length. The cross-section is the 12 m² end:
$$12 \times L = 300 \quad\Rightarrow\quad L = 25$$
The new roof space is 25 m long, 15 m longer than the old one (which holds only 12 × 10 = 120 m³).

$$t_n = a + (n - 1)d$$
where a is the first term and d the common difference.

d = t₂ − t₁ = 65 − 50 = Rs 15 (she puts in Rs 15 more each week).

a = 50, d = 15, $t_n = 350$:
$$\begin{aligned} 350 &= 50 + (n - 1) \times 15 \\ 15(n - 1) &= 300 \\ n - 1 &= 20 \end{aligned}$$
n = 21. In the 21st week.

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

First expression:
$$8a^3 + 27 = (2a)^3 + 3^3 = (2a + 3)(4a^2 - 6a + 9)$$
Second expression:
$$\begin{aligned} 16a^4 + 36a^2 + 81 &= (4a^2 + 9)^2 - 72a^2 + 36a^2 \\ &= (4a^2 + 9)^2 - (6a)^2 \\ &= (4a^2 + 6a + 9)(4a^2 - 6a + 9) \end{aligned}$$
The factor in both: HCF $= 4a^2 - 6a + 9$.

Division law in each bracket: $\dfrac{x^{a}}{x^{-b}} = x^{a - (-b)} = x^{a + b}$, and so on:
$$(x^{a + b})^{a - b} \times (x^{b + c})^{b - c} \times (x^{c + a})^{c - a}$$
Power law, with $(a + b)(a - b) = a^2 - b^2$:
$$x^{a^2 - b^2} \times x^{b^2 - c^2} \times x^{c^2 - a^2}$$
Multiplication law:
$$x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} = x^0 = 1$$
1.

Let an adult's fare be Rs x and a child's Rs y:
$$\begin{aligned} 3x + 2y &= 4700 \quad \text{(i)} \\ 2x + 3y &= 4050 \quad \text{(ii)} \end{aligned}$$
Multiply (i) by 3 and (ii) by 2, and subtract:
$$\begin{aligned} 9x + 6y &= 14100 \\ 4x + 6y &= 8100 \\ 5x &= 6000 \end{aligned}$$
x = 1200. From (i): 3600 + 2y = 4700, so y = 550.

An adult pays Rs 1,200 and a child Rs 550. Check (ii): 2400 + 1650 = 4050 ✓

New fares: adult $1200 + \frac{10}{100} \times 1200 = 1320$; child 550 + 50 = 600.

For 2 adults and 2 children: 2 × 1320 + 2 × 600 = 2640 + 1200 = 3840.

Yes, Rs 4,000 is enough: the tickets cost Rs 3,840, and Rs 160 is left.

Figures: draw two isosceles triangles ABC with AB = AC, one small and one large (for example AB = AC = 4 cm, BC = 5 cm, and AB = AC = 6 cm, BC = 4 cm). In each, draw the perpendicular bisector of the base BC with a compass: it meets BC at D and goes through A. Join AD.

Table: measure ∠BAD and ∠CAD, and check that D is the middle of BC:

(Measured angles may differ by a degree or so; the two in each row come out equal.)

Conclusion: the perpendicular bisector of the base of an isosceles triangle passes through the vertex and bisects the vertical angle.

$\angle PRQ = 180^\circ - 115^\circ = 65^\circ$ (a linear pair). PQ = PR, so $\angle PQR = \angle PRQ = 65^\circ$ (base angles of an isosceles triangle). The exterior angle PRS equals the two opposite interior angles: $\angle QPR = 115^\circ - 65^\circ = 50^\circ$.

∠QPR:∠PQR = 50:65 = 10 : 13.

The opposite angles of a parallelogram are equal: ∠A = ∠C and ∠B = ∠D.

∠A = ∠C:
$$3x + 10 = 5x - 30 \quad\Rightarrow\quad 2x = 40 \quad\Rightarrow\quad x = 20$$
x = 20, so $\angle A = \angle C = 70^\circ$, and $\angle B = 180^\circ - 70^\circ =$ 110∘° (co-interior angles between $AD \parallel BC$).

Draw PQ = 5 cm. At P, construct $\angle QPX = 60^\circ$ with a compass and cut PS = 5 cm on PX. With centers Q and S and radius 5 cm, draw two arcs that cut each other at R. Join QR and SR.

PQRS is the rhombus (all four sides 5 cm). Check: the diagonal QS is 5 cm too, since △PQS is equilateral. See it step by step in Construct it: a rhombus.

AM = MB. The perpendicular drawn from the center of a circle to a chord bisects the chord.

$AM = MB = \frac{40}{2} = 20$ cm. In the right-angled △OMB, OB = 25 cm is the hypotenuse:
$$OM = \sqrt{OB^2 - MB^2} = \sqrt{25^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15$$
OM = 15 cm.

The deepest point is at the bottom of the pipe, on the line OM produced. From the center to the bottom is a radius, 25 cm, and the water's surface is 15 cm below the center. The water is 25 − 15 = 10 cm deep.

Each value (number of books) is given with its frequency: a discrete series.

$$\overline{X} = \frac{\sum fX}{N} = \frac{73}{30} \approx 2.43$$
About 2.43 books.

More than 2.43 books means 3, 4 or 5 books: 7 + 4 + 3 = 14 students. Half of 30 is 15. No, she is not right: only 14 students, fewer than half, read more than the mean.

1. (For example, the probability that the bead is red, blue or green.)

n(S) = 20 and n(blue) = 8.
$$P(\text{blue}) = \frac{8}{20} = \frac25$$
$\frac25$ (0.4).

$$P(\text{green}) = \frac{7}{20}, \qquad P(\text{not green}) = 1 - \frac{7}{20} = \frac{13}{20}$$
$\frac{13}{20}$ (the 5 red and 8 blue beads: $\frac{13}{20}$ ✓).

Add x red beads:
$$\frac{5 + x}{20 + x} = \frac12 \quad\Rightarrow\quad 10 + 2x = 20 + x \quad\Rightarrow\quad x = 10$$
10 red beads (then 15 of the 30 beads are red).

For θ, BC is the base and AC the hypotenuse: $\dfrac{BC}{AC} = \dfrac{b}{h} =$ cos θ.

AB is the perpendicular for θ: $\tan\theta = \dfrac{p}{b} = \dfrac{AB}{BC} =$ $\dfrac{12}{5}$.

$$\sin 60^\circ = \frac{AB}{AC} \quad\Rightarrow\quad AB = 20 \times \frac{\sqrt3}{2} = 10\sqrt3 \approx 17.32$$
$10\sqrt3$ ft, about 17.32 ft.

AC = 2BC, so $\cos\theta = \dfrac{BC}{AC} = \dfrac12 = \cos 60^\circ$. $\theta = 60^\circ$.