SEE Model Paper 1

n(P ∪ M ∪ W) = 25

Fill the middle first, then the pairs, then each crop alone. Paddy and maize only: 45 − 20 = 25; maize and wheat only: 30 − 20 = 10; paddy and wheat only: 35 − 20 = 15. Paddy only: 110 − (25 + 20 + 15) = 50; maize only: 85 − (25 + 20 + 10) = 30; wheat only: 70 − (15 + 20 + 10) = 25. Outside the circles: 25.

Check: 50 + 30 + 25 + 25 + 10 + 15 + 20 + 25 = 200 ✓

Those that grew exactly two crops (the pairs only):
$$\begin{aligned} n_0(P \cap M) &= 45 - 20 = 25 \\ n_0(M \cap W) &= 30 - 20 = 10 \\ n_0(P \cap W) &= 35 - 20 = 15 \end{aligned}$$
Each crop alone:
$$\begin{aligned} n_0(P) &= 110 - (25 + 20 + 15) = 50 \\ n_0(M) &= 85 - (25 + 20 + 10) = 30 \\ n_0(W) &= 70 - (15 + 20 + 10) = 25 \end{aligned}$$
Exactly one crop: 50 + 30 + 25 = 105. 105 households grew exactly one of the three crops.

What it means: just over half of the 200 households grew only one of these crops.

Paddy with at least one other crop: 25 + 15 + 20 = 60 households (paddy and maize only, paddy and wheat only, all three). Half of 200 is 100, and 60 is only 30% of the households. No, the report is wrong: 60 households, 30%, not more than half.

$$\text{CA} = P\left(1 + \frac{R}{400}\right)^{4T}$$
where P is the principal, R% the yearly rate and T the time in years.

P = 400000, R = 8, T = 2: the rate is 2% a quarter for 8 quarters.
$$\begin{aligned} \text{CA} &= 400000\left(1 + \frac{8}{400}\right)^{8} \\ &= 400000 \times 1.02^8 \\ &= 400000 \times 1.171659381\ldots \\ &\approx 468663.75 \end{aligned}$$
She gets Rs 4,68,663.75.

Finance company, compounded yearly:
$$\text{CA} = 400000\left(1 + \frac{8.2}{100}\right)^2 = 400000 \times 1.170724 = 468289.60$$
Development bank: Rs 4,68,663.75. Difference: 468663.75 − 468289.60 = 374.15.

The development bank's offer gives Rs 374.15 more interest, although its rate is lower: interest added every quarter starts earning interest sooner.

$$V_T = V_0\left(1 - \frac{R}{100}\right)^T$$
where V₀ is the original value and R% the yearly rate of depreciation.

$$V_2 = 250000\left(1 - \frac{20}{100}\right)^2 = 250000 \times 0.64 = 160000$$
Rs 1,60,000.

The old scooter after 3 years:
$$V_3 = 250000 \times 0.8^3 = 250000 \times 0.512 = 128000$$
A new one after 3 years, its price growing at 5% a year:
$$P_3 = 250000\left(1 + \frac{5}{100}\right)^3 = 250000 \times 1.157625 = 289406.25$$
He must add 289406.25 − 128000 = 161406.25. Rs 1,61,406.25.

What it means: the old scooter pays for less than half of the new one, because its value falls while the new price rises.

The bank buys the dollars from Pemba, so it uses its buying rate, Rs 150.40.
$$6000 \times 150.40 = 902400$$
Rs 9,02,400.

Now the bank sells him dollars, at Rs 151.00: 450 × 151 = 67950. Rs 67,950.

If the rupee is revalued by 2%, a dollar buys 2% fewer rupees: the new buying rate is 150.40 × 0.98 = 147.392. Then 6000 × 147.392 = 884352, which is 902400 − 884352 = 18048 less.

He should change the dollars today: waiting would cost him Rs 18,048.

$$l^2 = h^2 + \left(\frac{a}{2}\right)^2$$

$$h = \sqrt{l^2 - \left(\frac{a}{2}\right)^2} = \sqrt{20^2 - 12^2} = \sqrt{256} = 16$$
16 cm.

The four triangular faces of one lantern (its lateral surface):
$$\text{LSA} = 2al = 2 \times 24 \times 20 = 960 \text{ cm}^2$$
For 12 lanterns: 12 × 960 = 11520 cm² = 1.152 m² (as 1 m² = 10,000 cm²).

Cost = 1.152 × 300 = 345.60. Rs 345.60.

$$V = \frac13 \pi r^2 h$$

$$\begin{aligned} V &= \frac13 \pi r^2 h + \frac23 \pi r^3 \\ &= \frac13 \times \frac{22}{7} \times 3^2 \times 4 + \frac23 \times \frac{22}{7} \times 3^3 \\ &= \frac{22}{7}(12 + 18) = \frac{660}{7} \approx 94.29 \end{aligned}$$
About 94.29 cm³.

The box: $\pi r^2 h = \pi \times 9 \times 7 = 63\pi$ cm³. The top: $30\pi$ cm³ (from b). Empty: $63\pi - 30\pi = 33\pi$ cm³.

Fraction empty $= \dfrac{33\pi}{63\pi} = \dfrac{11}{21}$. $\frac{11}{21}$ of the box, a little more than half, is empty.

Four walls: 2h(l + b) = 2 × 4 × (16 + 10) = 208 m². Ceiling: l × b = 16 × 10 = 160 m².

Doors: 2 × 2.5 × 2 = 10 m². Windows: 8 × 1.5 × 1.2 = 14.4 m².

Area to paint = 208 + 160 − 10 − 14.4 = 343.6 m².

Cost = 343.6 × 125 = 42950. Rs 42,950.

Tiles: 16 ÷ 0.5 = 32 along the length and 10 ÷ 0.5 = 20 across, so 32 × 20 = 640 tiles, which fit exactly. Cost = 640 × 120 = 76800.

Carpet: floor = 16 × 10 = 160 m², cost = 160 × 450 = 72000.

The carpet is cheaper, by 76800 − 72000 = Rs 4,800.

$$S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$$
where a is the first term and d the common difference (or $S_n = \frac{n}{2}(a + l)$, with l the last term).

a = 20, d = 4, n = 15:
$$S_{15} = \frac{15}{2}\left[2 \times 20 + 14 \times 4\right] = \frac{15}{2} \times 96 = 720$$
720 students.

720 is less than 900, so no. With n rows, $S_n = \frac{n}{2}[40 + 4(n - 1)] = n(2n + 18)$.
$$\begin{aligned} S_{17} &= 17 \times 52 = 884 \\ S_{18} &= 18 \times 54 = 972 \end{aligned}$$
17 rows seat 884, still short of 900; 18 rows seat 972. 3 more rows are needed (18 rows in all), and then 72 seats are left over.

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Let the breadth be x m; then the length is (x + 7) m. The diagonal and two sides make a right-angled triangle, so
$$\begin{aligned} x^2 + (x + 7)^2 &= 17^2 \\ 2x^2 + 14x + 49 &= 289 \\ x^2 + 7x - 120 &= 0 \end{aligned}$$
Product −120, sum 7: 15 and −8.
$$\begin{aligned} x^2 + 15x - 8x - 120 &= 0 \\ (x + 15)(x - 8) &= 0 \end{aligned}$$
So x = −15 or 8. A breadth can't be negative, so x = 8.

The rectangle is 15 m long and 8 m wide. Check: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓

Lay each panel with its 1 m side along the 15 m length and its 2 m side across the 8 m breadth: 15 panels in a row and 8 ÷ 2 = 4 rows, so 15 × 4 = 60 panels. They cover 60 × 2 = 120 m², the whole 15 × 8 rectangle. 60 panels.

Write $2^{x + 1} = 2 \times 2^x$ and $2^{3 - x} = \dfrac{8}{2^x}$, and let $2^x = a$:
$$\begin{aligned} 2a + \frac{8}{a} &= 17 \\ 2a^2 - 17a + 8 &= 0 \end{aligned}$$
Product 16, sum −17: −16 and −1.
$$\begin{aligned} 2a^2 - 16a - a + 8 &= 0 \\ 2a(a - 8) - 1(a - 8) &= 0 \\ (2a - 1)(a - 8) &= 0 \end{aligned}$$
So $a = \frac12$ or 8. Then $2^x = 2^{-1}$ gives x = −1, and $2^x = 2^3$ gives x = 3.

x = −1 or 3. Check x = 3: $2^4 + 2^0 = 16 + 1 = 17$ ✓

The LCM of the denominators is $(p - q)(p + q) = p^2 - q^2$:
$$\frac{p(p + q) - q(p - q) - 2pq}{p^2 - q^2} = \frac{p^2 + pq - pq + q^2 - 2pq}{p^2 - q^2} = \frac{p^2 - 2pq + q^2}{p^2 - q^2}$$
The top is $(p - q)^2$:
$$\frac{(p - q)^2}{(p - q)(p + q)} = \frac{p - q}{p + q}$$
$\dfrac{p - q}{p + q}$.

Area of parallelogram ABCD = area of parallelogram ABEF: parallelograms on the same base and between the same parallels are equal in area.

Area of ABCD = area of ABEF = 54 cm². Area = base × height, so the height (the distance between the parallels) = 54 ÷ 9 = 6 cm.

The diagonal BD halves parallelogram ABCD: area of $\triangle BCD = \frac12 \times 54 =$ 27 cm².

Proved.

The opposite angles of a cyclic quadrilateral are supplementary: they add up to 180∘°. Here $\angle PQR + \angle PSR = 180°$ and $\angle QPS + \angle QRS = 180°$.

$$\begin{aligned} (3x + 10) + (2x + 20) &= 180 \\ 5x + 30 &= 180 \\ x &= 30 \end{aligned}$$
x = 30. Check: $\angle PQR = 100°$ and $\angle PSR = 80°$, which add up to 180∘° ✓

Draw two circles, of radii 3 cm and 3.5 cm. In each, mark four points A, B, C, D on the circle and join them to make a cyclic quadrilateral ABCD.

Measure the four angles with a protractor:

Conclusion: in both figures each pair of opposite angles adds up to 180∘°, so the opposite angles of a cyclic quadrilateral are supplementary. (Your own measurements may differ by a degree or so.)

Draw △ABC from its three sides, draw ST through A parallel to BC, bisect BC at M, make an angle of 45∘° at B meeting ST at P, and complete the parallelogram PBMQ. It has the same height as the triangle on half its base, so the same area.

The triangle and the parallelogram lie between the same parallels BC and ST, so they have the same height. A parallelogram on the whole base BC would have twice the triangle's area (a triangle is half the parallelogram on the same base between the same parallels). Halving the base halves the parallelogram, so the one on $BM = \frac12 BC$ has exactly the triangle's area.

$\sin 60° = \dfrac{AB}{AC}$ (perpendicular over hypotenuse).

$$AB = AC \sin 60° = 12 \times \frac{\sqrt3}{2} = 6\sqrt3 \approx 10.39$$
About 10.39 m.

The angle of elevation of A from D equals the angle of depression, 30∘° (alternate angles). In △ABD:
$$\tan 30° = \frac{AB}{BD} \quad\Rightarrow\quad BD = \frac{6\sqrt3}{1/\sqrt3} = 6 \times 3 = 18$$
18 m.

From D, 18 m from the wall, at 45∘°: the window's height $= 18 \tan 45° = 18$ m. It is $18 - 6\sqrt3 \approx 18 - 10.39 =$ 7.61 m above A.

$$M_o = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$
where L is the lower limit of the modal class, f₁ its frequency, f₀ and f₂ the frequencies of the classes before and after it, and h the class width.

Cumulative frequencies: 5, 13, 25, 39, 46, 50. N = 50, so Q₁ is the $\frac{N}{4} = 12.5$th worker's wage, in the class 700–800 (the first cf of 12.5 or more is 13). There L = 700, cf = 5, f = 8, h = 100:
$$Q_1 = 700 + \frac{12.5 - 5}{8} \times 100 = 700 + 93.75 = 793.75$$
Q₁ = Rs 793.75: a quarter of the workers earn less than this.

$$\overline{X} = \frac{\sum fm}{N} = \frac{44700}{50} = 894$$
The mean daily wage is Rs 894.

Raising every wage by 10% multiplies every value, so the mean, by 1.1: 1.1 × 894 = 983.40. Adding Rs 80 to every wage adds 80 to the mean: 894 + 80 = 974. He is wrong: the 10% raise costs more, Rs 9.40 a worker a day on average (Rs 470 a day for all 50).

$$P(\overline{A}) = 1 - P(A)$$

The events are independent, so
$$P(\text{both score}) = 0.8 \times 0.7 = 0.56$$
0.56 ($\frac{14}{25}$).

S: scores, M: misses. Rojina: $P(S) = 0.8 = \frac45$, $P(M) = 0.2 = \frac15$. Puja: $P(S) = 0.7 = \frac{7}{10}$, $P(M) = 0.3 = \frac{3}{10}$.

Check: 0.56 + 0.24 + 0.14 + 0.06 = 1 ✓

At least one scores unless both miss: P(at least one) = 1 − P(MM) = 1 − 0.2 × 0.3 = 1 − 0.06 = 0.94. 0.94, so the coach is right: they fail to score at all only about 6 times in 100.