Final Exam Model Paper 3

E = {2, 4, 6, 8, 10, 12, 14}, so n(E) = 7.

P = {2, 3, 5, 7, 11, 13}, S = {1, 4, 9}, E = {2, 4, 6, 8, 10, 12, 14}. No number is in all three sets, and none is in both P and S; 4 is in S and E, and 2 in P and E.

Left side: P ∩ E = {2}, so
$$\overline{P \cap E} = \{1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$$
Right side: P = {1, 4, 6, 8, 9, 10, 12, 14, 15} and E = {1, 3, 5, 7, 9, 11, 13, 15}, so
$$\overline{P} \cup \overline{E} = \{1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$$
Both sides are the same set. So P ∩ E = P ∪ E.

$P \cap S = \varnothing$: P and S are disjoint sets. The reason: a square number other than 1 is k × k with k > 1, so it has the factor k and isn't prime; and 1 isn't a prime number either.

The social security tax.

Yearly income = 13 × 110000 = 1430000. Deposits in the trust $= 12 \times \frac{10}{100} \times 110000 = 12 \times 11000 = 132000$.
$$\text{Taxable income} = 1430000 - 132000 = 1298000$$
Rs 12,98,000.

Rs 12,98,000 = Rs 10,00,000 + Rs 2,98,000.
$$\begin{aligned} \text{Tax} &= \tfrac{1}{100} \times 1000000 + \tfrac{10}{100} \times 298000 \\ &= 10000 + 29800 = 39800 \end{aligned}$$
Relief: $\frac{10}{100} \times 39800 = 3980$. She pays 39800 − 3980 = 35820.

Rs 35,820.

A bonus is money that an organization pays its employees once a year, on top of their salary, as a part of the profit it made that year.

Bonus fund $= \frac{10}{100} \times 7200000 = 720000$; each = 720000 ÷ 48 = Rs 15,000.

New profit $= 7200000 - \frac{20}{100} \times 7200000 = 5760000$; bonus fund = 576000; each of 50 employees gets 576000 ÷ 50 = 11520.

Each bonus falls by 15000 − 11520 = Rs 3,480.

9 units is less than the minimum 10 units, so the water charge is the minimum, Rs 100. Bill = 100 + 50 = Rs 150.

72 units = 20 + 10 + 20 + 22, in the 51–100 block, so the minimum charge is Rs 75.
$$\begin{aligned} \text{Energy charge} &= 20 \times 3 + 10 \times 6.50 + 20 \times 8 + 22 \times 9.50 \\ &= 60 + 65 + 160 + 209 = 494 \\ \text{Bill} &= 75 + 494 = 569 \end{aligned}$$
Rs 569.

Day: 14 + 39 × 9 = 14 + 351 = 365. Night: 21 + 58.50 × 9 = 21 + 526.50 = 547.50.

547.50 ÷ 365 = 1.5. Yes, he is right: the night fare, Rs 547.50, is exactly 1.5 times the day fare, Rs 365, because every night rate is 50% more than the day rate.

20 dhur.

$s = \dfrac{130 + 140 + 150}{2} = \dfrac{420}{2} =$ 210 ft.

$$\begin{aligned} \text{Area} &= \sqrt{s(s - a)(s - b)(s - c)} \\ &= \sqrt{210 \times 80 \times 70 \times 60} \\ &= \sqrt{70560000} = 8400 \end{aligned}$$
8,400 ft². In kattha: 8400 ÷ 3645 = 2.3045…: 2 kattha, and 0.3045 × 20 = 6.09 dhur.

8,400 ft², that is 2 kattha 6.09 dhur.

1 dhur = 3645 ÷ 20 = 182.25 ft², so the neighbor's plot is 2 × 3645 + 5 × 182.25 = 7290 + 911.25 = 8201.25 ft².

Rajkumar's field is bigger, by 8400 − 8201.25 = 198.75 ft² (about 1.09 dhur).

$$A = 4\pi r^2$$

The water rises by as much as the stone's volume, spread over the tin's base.
$$\text{Stone} = \frac43 \pi r^3 = \frac43 \times \frac{22}{7} \times 3^3 = \frac{792}{7} \approx 113.14 \text{ cm}^3$$
$$\text{Base of the tin} = \pi R^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \approx 113.14 \text{ cm}^2$$
$$\text{Rise} = \frac{792/7}{792/7} = 1$$
The water rises by 1 cm. (Both are $36\pi$: $\frac43\pi \times 27 = 36\pi$ and $\pi \times 36 = 36\pi$.)

$$\text{Cross-section} = \frac12 \times 3 \times 0.5 = 0.75 \text{ m}^2$$
$$\text{Volume} = 0.75 \times 1.2 = 0.9 \text{ m}^3$$
0.9 m³ of concrete.

Cost = 0.9 × 15000 = 13500. No, Rs 11,250 is not enough (Rs 2,250 short).

With Rs 11,250 they can buy 11250 ÷ 15000 = 0.75 m³. Width $= \dfrac{\text{volume}}{\text{cross-section}} = \dfrac{0.75}{0.75} =$ 1 m.

The common difference is the fixed number that each term exceeds the term before it by: d = t₂ − t₁ = t₃ − t₂ = ⋯.

From the 4th row to the 9th row there are 5 steps:
$$t_9 - t_4 = 5d \quad\Rightarrow\quad 41 - 26 = 5d \quad\Rightarrow\quad d = 3$$
3 seats more in each row.

t₄ = a + 3d: 26 = a + 9, so a = 17.
$$t_{20} = a + 19d = 17 + 19 \times 3 = 74$$
17 seats in the first row and 74 in the 20th.

$$a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$$

First expression:
$$27a^3 + 1 = (3a)^3 + 1^3 = (3a + 1)(9a^2 - 3a + 1)$$
Second expression:
$$\begin{aligned} 81a^4 + 9a^2 + 1 &= (9a^2 + 1)^2 - 18a^2 + 9a^2 \\ &= (9a^2 + 1)^2 - (3a)^2 \\ &= (9a^2 + 3a + 1)(9a^2 - 3a + 1) \end{aligned}$$
The factor in both: HCF $= 9a^2 - 3a + 1$.

Division law, then power law:
$$x^{\frac{a - b}{ab}} \times x^{\frac{b - c}{bc}} \times x^{\frac{c - a}{ca}}$$
Each index splits: $\dfrac{a - b}{ab} = \dfrac{1}{b} - \dfrac{1}{a}$, $\dfrac{b - c}{bc} = \dfrac{1}{c} - \dfrac{1}{b}$, $\dfrac{c - a}{ca} = \dfrac{1}{a} - \dfrac{1}{c}$.

Multiplication law:
$$x^{\frac1b - \frac1a + \frac1c - \frac1b + \frac1a - \frac1c} = x^0 = 1$$
1.

Let Sujata be x years old and her mother y years old:
$$\begin{aligned} y &= 4x \quad \text{(i)} \\ y + 5 &= 3(x + 5) \quad \text{(ii)} \end{aligned}$$
Put (i) in (ii): 4x + 5 = 3x + 15, so x = 10 and y = 40.

Sujata is 10 years old and her mother 40. Check: after 5 years, 45 = 3 × 15 ✓

After t years: 40 + t = 2(10 + t), so 40 + t = 20 + 2t and t = 20. After 20 years, when they are 30 and 60.

No, never again: the difference in their ages stays 30 years, so the ratio keeps falling towards 1 as both grow older (four times now, three times after 5 years, twice after 20).

Figures: draw two triangles ABC of different sizes, for example with AB = 3 cm, BC = 4 cm, CA = 5 cm, and AB = 4 cm, BC = 6 cm, CA = 7 cm.

Table: measure the sides and add them in pairs:

Conclusion: the sum of any two sides of a triangle is greater than the third side.

In △ADE and △ABC: ∠ADE = ∠ABC (corresponding angles, $DE \parallel BC$) and ∠A is common. So $\triangle ADE \sim \triangle ABC$, and their corresponding sides are in the same ratio:
$$\frac{AD}{AB} = \frac{DE}{BC} \quad\Rightarrow\quad \frac{4}{4 + 6} = \frac{5}{BC} \quad\Rightarrow\quad BC = \frac{5 \times 10}{4} = 12.5$$
BC = 12.5 cm.

The diagonals of a parallelogram bisect each other: AO = OC and BO = OD.

AO = OC: 2x + 1 = x + 4, so x = 3. Then AO = OC = 7 cm and AC = 14 cm.

Draw AB = 6.5 cm. At B, construct $\angle ABX = 60^\circ$ with a compass and cut BC = 4 cm on BX. At C, construct $\angle BCY = 120^\circ$ on the same side as A (co-interior with 60∘°, so $CY \parallel AB$), and cut CD = 3.5 cm on CY. Join AD.

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

$\angle OMA = 90^\circ$. The line joining the center of a circle and the midpoint of a chord is perpendicular to the chord.

$AM = \frac{48}{2} = 24$ cm. In the right-angled △OMA, the radius OA is the hypotenuse:
$$OA = \sqrt{OM^2 + AM^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$
The radius is 25 cm.

The distance of CD from the center: $\sqrt{25^2 - 7^2} = \sqrt{576} = 24$ cm. AB is nearer (7 cm against 24 cm): the longer chord is nearer the center.

Each student's size is written separately, as collected: raw data, an individual series.

$$\sum fX = 4 \times 1 + 5 \times 4 + 6 \times 8 + 7 \times 5 + 8 \times 2 = 4 + 20 + 48 + 35 + 16 = 123$$
$$\overline{X} = \frac{123}{20} = 6.15$$
The mean size is 6.15.

The mode, size 6: it is the size most students wear (8 of 20). The mean, 6.15, isn't a size shoes are made in, and it says nothing about which size is most needed.

$$\text{Empirical probability} = \frac{\text{number of times the event happened}}{\text{total number of trials}}$$

Trials: 200 seeds; the event happened 170 times.
$$P(\text{germinates}) = \frac{170}{200} = \frac{17}{20} = 0.85$$
$\frac{17}{20}$ (0.85).

$$P(\text{does not germinate}) = 1 - \frac{17}{20} = \frac{3}{20}$$
Expected number: $\frac{3}{20} \times 1000 = 150$. About 150 seeds.

No. An empirical probability comes from a test, so it only estimates what to expect: the 1,000 seeds may give a few more or a few fewer than 850, though the number is likely to be near it.

$\sin 30^\circ =$ $\dfrac12$.

$$\left(\frac{\sqrt3}{2}\right)^2 + \left(\frac{\sqrt3}{2}\right)^2 = \frac34 + \frac34 = \frac32$$
$\frac32$.

$$\sin 30^\circ = \frac{BK}{AK} \quad\Rightarrow\quad BK = 60 \times \frac12 = 30$$
30 m.

$\sin\theta = \dfrac{30\sqrt3}{60} = \dfrac{\sqrt3}{2} = \sin 60^\circ$, so $\theta = 60^\circ$.