✅ STEP-BY-STEP SOLUTION:

Step 1: Identify the pattern
The instalments form an Arithmetic Progression (AP) with:
First term (a) = ₹1000
Common difference (d) = ₹100
Total loan amount = ₹1,18,000

Step 2 (i): Amount in 30th instalment

Using formula:$$ a_n = a + (n-1)d$$

$$a_{30} = 1000 + (30-1) \times 100$$

$$a_{30} = 1000 + 29 \times 100$$

$$a_{30} = 1000 + 2900 = ₹3900$$

✅ Answer (i): $${₹3900}$$

Step 3 (ii): Amount in last instalment (40th)

$$a_{40} = 1000 + (40-1) \times 100$$

$$a_{40} = 1000 + 39 \times 100$$

$$a_{40} = 1000 + 3900 = ₹4900$$

✅ Answer (ii): $${₹4900}$$

Step 4 (iii)(a): Amount paid after 30th instalment

First find total amount paid in 30 instalments using sum formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$S_{30} = \frac{30}{2}[2 \times 1000 + (30-1) \times 100]$$

$$S_{30} = 15[2000 + 2900]$$

$$S_{30} = 15 \times 4900 = ₹73,500$$

Amount still to be paid = Total loan – Amount paid in 30 instalments

= ₹1,18,000 – ₹73,500 = ₹44,500

✅ Answer (iii)(a): $${₹44,500}$$

Step 5 (iii)(b): Ratio of 10th instalment to last instalment (OR)

10th instalment: $$a_{10} = 1000 + (10-1) \times 100 $$

$$= 1000 + 900 $$

$$= ₹1900$$

Last instalment (40th): $$a_{40} = ₹4900$$ (from above)

Ratio = 1900 : 4900 = 19 : 49

✅ Answer (iii)(b): $${19 : 49}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Understand the situation
Let height of building = h meters.
At second position (angle 60°), car is 25 m away from building.
So, distance from building at second position = 25 m.

Step 2 (i): Find height of building using second position

In right triangle at second position:

$$\tan 60^\circ = \frac{h}{25}$$

$$\sqrt{3} = \frac{h}{25}$$

$$h = 25\sqrt{3} meters$$

✅ Answer (i): $${25\sqrt{3} \text{ m}}$$

Step 3: Find distance of first position from building

At first position (angle 30°):

$$\tan 30^\circ$$

$$= \frac{h}{d} $$where d

= distance of first position from building$$\frac{1}{\sqrt{3}}$$

$$= \frac{25\sqrt{3}}{d}$$

$$d = 25\sqrt{3} \times \sqrt{3}$$

$$= 25 \times 3 = 75 m$$

Step 4 (ii): Distance between two positions

Distance = First position distance – Second position distance

= 75 m – 25 m = 50 m

✅ Answer (ii): $${50 \text{ m}}$$

Step 5 (iii)(a): Total time to reach foot of building

Speed of car = Distance covered / Time taken

Car covers 50 m in 6 seconds, so speed = $$\frac{50}{6}$$

$$= \frac{25}{3} m/s$$

Total distance from first position to building = 75 m

$$Time = \frac{\text{Distance}}{\text{Speed}}$$

$$= \frac{75}{\frac{25}{3}}$$

$$= 75 \times \frac{3}{25} $$

$$= 3 \times 3 = 9 seconds$$

✅ Answer (iii)(a): $${9 \text{ seconds}}$$

Step 6 (iii)(b): Distance of observer from car at 60° (OR)

At 60° angle, we have a right triangle with:

Height = $$25\sqrt{3} m$$, base = 25 m

Distance from observer to car = hypotenuse

Using Pythagoras theorem:

$$d = \sqrt{(25\sqrt{3})^2 + 25^2}$$

$$= \sqrt{1875 + 625}$$

$$= \sqrt{2500} = 50 m$$

✅ Answer (iii)(b): $${50 \text{ m}}$$

✅ STEP-BY-STEP SOLUTION:

Step 1 (i): Diagonal of cube with edge 6 cm
Formula for diagonal of cube = \(a\sqrt{3}\) where a = edge length
Diagonal = \(6\sqrt{3}\) cm

✅ Answer (i): $${6\sqrt{3} \text{ cm}}$$

Step 2 (ii): Volume of cube with edge 7 cm

Volume of cube = $$a^3$$

Volume = $$7^3 = 343 cm³$$

✅ Answer (ii): $${343 \text{ cm}^3}$$

Step 3 (iii)(a): Curved surface area of hemisphere (ice-cream)

Radius r = 7 cm

Curved surface area of hemisphere = $$2\pi r^2$$

Using $$\pi = \frac{22}{7}$$:

$$CSA = 2 \times \frac{22}{7} \times 7^2$$

$$= 2 \times \frac{22}{7} \times 49$$

$$= 2 \times 22 \times 7$$

$$= 44 \times 7 = 308 cm²$$

✅ Answer (iii)(a): $${308 \text{ cm}^2}$$

Step 4 (iii)(b): Surface area of cuboid formed by joining two cubes (OR)

Two cubes of edge 4 cm joined end-to-end form a cuboid of dimensions:

Length = 4 + 4 = 8 cm

Breadth = 4 cm

Height = 4 cm

Surface area of cuboid = 2(lb + bh + hl)

$$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$$

= 2(32 + 16 + 32)

$$= 2 \times 80 = 160 cm²$$

✅ Answer (iii)(b): $${160 \text{ cm}^2}$$