Final Exam Model Paper 2

$$n(F \cup V) = n(F) + n(V) - n(F \cap V)$$

Football only: 35 − 10 = 25; volleyball only: 28 − 10 = 18; both: 10; neither: 7.

$$n(F \cup V) = n(F) + n(V) - n(F \cap V) = 35 + 28 - 10 = 53$$
53 students like at least one of the games, and 7 like neither:
$$n(U) = n(F \cup V) + n(\overline{F \cup V}) = 53 + 7 = 60$$
60 students were asked. Check from the diagram: 25 + 10 + 18 + 7 = 60 ✓

Only football: 25 students. Volleyball (with or without football): 28 students. No: 25 is less than 28, so fewer students like only football than like volleyball.

VAT is a tax on the value added to goods and services at each stage of making and selling them; the buyer pays it on the price, and the seller passes it to the government. In Nepal it is 13%.

$$\text{Discount} = \frac{15}{100} \times 48000 = 7200$$
$$\text{Price after discount} = 48000 - 7200 = 40800$$
Rs 40,800.

VAT is added after the discount:
$$\text{VAT} = \frac{13}{100} \times 40800 = 5304$$
$$\text{She pays } 40800 + 5304 = 46104$$
Rs 46,104.

What it means: still Rs 1,896 less than the marked price, as the discount is bigger than the tax.

The shareholders of the company, each in proportion to the number of shares they own.

Dividend per share $= \dfrac{20000000}{5000000} = 4$: Rs 4 on a share of Rs 100, that is $\frac{4}{100} \times 100\% =$ 4% of the face value.

This year's dividend is $\frac{2}{8} = 25\%$ of the profit. Next year: $\frac{25}{100} \times 100000000 = 25000000$; per share 25000000 ÷ 5000000 = 5.

Bimala: 1200 × 5 = Rs 6,000.

21 + 58.50 × 4 = 21 + 234 = Rs 255.

80,000 liters = 80 units: 56 units at the minimum charge and 80 − 56 = 24 units more.
$$\begin{aligned} \text{Water charge} &= 3960 + 24 \times 71 = 3960 + 1704 = 5664 \\ \text{Bill} &= 5664 + \tfrac{50}{100} \times 5664 = 5664 + 2832 = 8496 \end{aligned}$$
Rs 8,496.

Both taxes together make the bill TC × 1.13 × 1.13 = 1.2769 × TC.
$$\text{TC} \le \frac{750}{1.2769} = 587.36\ldots$$
The TC is a whole number of rupees, so at most Rs 587. Calls = 175 + (587 − 200) = 562 calls.

Check: 562 calls cost 587 × 1.2769 = 749.54; one more call would cost 588 × 1.2769 = 750.82, over Rs 750.

$$\text{Area of four walls} = 2h(l + b)$$
where l, b and h are the room's length, breadth and height.

2 × 3.5 × (8 + 6) = 7 × 14 = 98 m².

Ceiling = 8 × 6 = 48 m². Door = 2 × 1.2 = 2.4 m²; windows = 3 × 1.5 × 1.2 = 5.4 m².
$$\text{Area to paint} = 98 + 48 - 2.4 - 5.4 = 138.2 \text{ m}^2$$
$$\text{Cost} = 138.2 \times 160 = 22112$$
Rs 22,112.

Two coats: 2 × 138.2 = 276.4 m². Tins: 276.4 ÷ 40 = 6.91. A part of a tin can't be bought, and 6 tins would cover only 240 m², so 7 tins must be bought.

$$V = \frac23 \pi r^3$$

$$\text{Pot} = \pi r^2 h = \frac{22}{7} \times 21^2 \times 28 = 38808 \text{ cm}^3$$
$$\text{Bowl} = \frac23 \pi r^3 = \frac23 \times \frac{22}{7} \times 10.5^3 = 2425.5 \text{ cm}^3$$
$$\text{Bowls} = \frac{38808}{2425.5} = 16$$
16 bowls.

$$\text{Cross-section} = \frac12(1.2 + 0.8) \times 0.6 = 0.6 \text{ m}^2$$
$$\text{Volume} = 0.6 \times 3 = 1.8 \text{ m}^3$$
0.6 m², and 1.8 m³ (1,800 liters) of water.

Time to fill = 1800 ÷ 12 = 150 minutes = 2 hours 30 minutes, so the trough is full at 8:30.

No: at 8:15 it is not yet full; it needs 15 minutes more (it holds 135 × 12 = 1620 liters then).

$$t_n = ar^{n - 1}$$
where a is the first term and r the common ratio.

$r = \dfrac{t_2}{t_1} = \dfrac{4}{8} =$ $\dfrac12$.

a = 8, $r = \frac12$, $t_n = 0.25 = \frac14$:
$$\begin{aligned} 8 \times \left(\tfrac12\right)^{n - 1} &= \tfrac14 \\ \left(\tfrac12\right)^{n - 1} &= \tfrac{1}{32} = \left(\tfrac12\right)^5 \end{aligned}$$
n − 1 = 5, n = 6. After the 6th bounce.

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

First expression:
$$27y^3 - 8 = (3y)^3 - 2^3 = (3y - 2)(9y^2 + 6y + 4)$$
Second expression:
$$\begin{aligned} 81y^4 + 36y^2 + 16 &= (9y^2 + 4)^2 - 72y^2 + 36y^2 \\ &= (9y^2 + 4)^2 - (6y)^2 \\ &= (9y^2 + 6y + 4)(9y^2 - 6y + 4) \end{aligned}$$
The LCM takes every factor, the shared one once:
LCM $= (3y - 2)(9y^2 + 6y + 4)(9y^2 - 6y + 4)$.

Multiplication law: $3^{n + 2} = 3^n \times 3^2$ and $3^{n + 1} = 3^n \times 3$:
$$\frac{3^n \times 9 - 3^n}{3^n \times 3 + 3^n}$$
Take out the common factor $3^n$:
$$\frac{3^n(9 - 1)}{3^n(3 + 1)} = \frac{8}{4}$$
= 2.

Let there be x hens and y goats:
$$\begin{aligned} x + y &= 46 \quad \text{(i)} \\ 2x + 4y &= 128 \quad \text{(ii)} \end{aligned}$$
(ii) − 2 × (i): 2y = 128 − 92 = 36, so y = 18 and x = 46 − 18 = 28.

28 hens and 18 goats. Check: 2 × 28 + 4 × 18 = 56 + 72 = 128 ✓

Selling k hens takes away k heads and 2k legs:
$$\begin{aligned} 128 - 2k &= 3(46 - k) \\ 128 - 2k &= 138 - 3k \\ k &= 10 \end{aligned}$$
He sold 10 hens. Check: 18 hens and 18 goats have 36 heads and 36 + 72 = 108 = 3 × 36 legs ✓

Figures: draw two triangles ABC with sides of different lengths, for example BC = 6 cm, CA = 5 cm, AB = 4 cm, and BC = 8 cm, CA = 6 cm, AB = 5 cm.

Table: measure the angles opposite the longest and the shortest side:

(Measured angles may differ by a degree or so.)

Conclusion: in a triangle, the angle opposite the longer side is greater than the angle opposite the shorter side.

The exterior angle BAE equals the two opposite interior angles: $x + 2x = 120^\circ$, so $x = 40^\circ$, and $\angle ABC = 40^\circ$. $\angle BAC = 180^\circ - 120^\circ = 60^\circ$ (a linear pair).

∠BAC:∠ABC = 60:40 = 3 : 2.

The diagonals of a rectangle are equal and bisect each other: PR = QS, and PO = OR = QO = OS.

PR = QS: 4y − 2 = 2y + 8, so 2y = 10 and y = 5. Then PR = 18 cm, and $OP = \frac12 PR =$ 9 cm.

Draw a rough sketch first. Draw AB = 5.5 cm. With center A and radius 6.5 cm, and with center B and radius 4.5 cm, draw arcs that cut at C; join AC and BC. On the other side of AC from B, draw an arc of radius 4 cm from A and one of radius 5 cm from C; they cut at D. Join AD and CD.

ABCD is the quadrilateral. See it step by step in Construct it: a quadrilateral.

OM = ON: equal chords of a circle are equidistant from the center.

The perpendicular from the center bisects the chord, so MB = 8 cm. Join OB, a radius: in the right-angled △OMB,
$$OM = \sqrt{OB^2 - MB^2} = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$$
OM = 6 cm.

16 cm. EF is as far from the center as AB (6 cm), and chords which are equidistant from the center of a circle are equal, so EF = AB = 16 cm. (Or: half of EF is $\sqrt{10^2 - 6^2} = 8$ cm.)

Each age is given with its frequency: a discrete series.

$$\text{Median} = \left(\frac{N + 1}{2}\right)\text{th item} = \left(\frac{25 + 1}{2}\right)\text{th item} = 13\text{th item}$$
In the cf column, 15 is the first number of 13 or more, so the 13th player is 16. The median age is 16 years.

$Q_1 = \left(\frac{N + 1}{4}\right)$th item = 6.5th item: the 6th and 7th players are both 15 (the cf 8 covers items 4 to 8), so Q₁ = 15.

$Q_3 = \left(\frac{3(N + 1)}{4}\right)$th item = 19.5th item: the 19th and 20th are both 17 (items 16 to 21), so Q₃ = 17.

The organizers are right: the middle half of the players lie between Q₁ = 15 and Q₃ = 17 years.

0. (For example, the probability of getting a 7.)

S = {1, 2, 3, 4, 5, 6}, n(S) = 6. The prime numbers are {2, 3, 5}, so n(E) = 3.
$$P(\text{prime}) = \frac36 = \frac12$$
$\frac12$. (1 is not a prime number.)

The multiples of 3 are {3, 6}: $P(\text{multiple of 3}) = \frac26 = \frac13$.
$$P(\text{not a multiple of 3}) = 1 - \frac13 = \frac23$$
$\frac23$.

Greater than 4: {5, 6}, probability $\frac26 = \frac13$. Even: {2, 4, 6}, probability $\frac36 = \frac12$. No: an even number is more likely, as it has more favorable outcomes.

PQ is opposite θ (the perpendicular) and PR is the hypotenuse: $\sin\theta = \dfrac{p}{h} =$ $\dfrac{PQ}{PR}$.

$PR = \sqrt{PQ^2 + QR^2} = \sqrt{64 + 225} = \sqrt{289} =$ 17 cm.

$\cos\theta = \frac{QR}{PR} = \frac{15}{17}$ and $\sin\theta = \frac{8}{17}$:
$$\frac{1 - \frac{15}{17}}{\frac{8}{17}} = \frac{\frac{2}{17}}{\frac{8}{17}} = \frac28 = \frac14$$
$\frac14$.

$\tan 30^\circ = \dfrac{PQ}{QR}$, so $PQ = 15 \times \dfrac{1}{\sqrt3} = \dfrac{15}{\sqrt3} = 5\sqrt3 \approx 8.66$. $5\sqrt3$ cm, about 8.66 cm.