I see the image you uploaded with several math questions. Let's start solving some of them:

10. Simplify:

• $\frac{0.00625}{0.0025} = 2.5$
• $\frac{0.792}{1.98} = 0.4$

11. Evaluate:

(i) $0.64 \times 500 \div 0.16$ $0.64 \times 500 = 320 \quad \text{and} \quad 320 \div 0.16 = 2000$ So, the answer is 2000.

(ii) $65.1 \div (1.25 + 3.4)$ $1.25 + 3.4 = 4.65 \quad \text{and} \quad 65.1 \div 4.65 \approx 14$ So, the answer is 14.

12. Saurabh buys toothbrushes:

(i) He pays ₹43.95, and if he gives ₹50: $50 - 43.95 = 6.05$ So, he will get ₹6.05 back.

(ii) If each toothbrush costs ₹14.65, the number of toothbrushes he bought is: $\frac{43.95}{14.65} = 3$ So, he bought 3 toothbrushes.

13. Largest square in a circle with diameter 5.2 cm:

The side $s$ of the largest square that can be inscribed in a circle is given by: $s = \frac{d}{\sqrt{2}} = \frac{5.2}{\sqrt{2}} \approx 3.677 \, \text{cm}$ So, the side length of the square is approximately 3.68 cm.

14. Ms. Gupta’s cloth:

Total cloth = 24.36 m
Used for 9 dresses = $9 \times 2.14 = 19.26 \, \text{m}$
Remaining cloth = $24.36 - 19.26 = 5.1 \, \text{m}$

So, she had 5.1 m of cloth left for 3 shirts.

15. Converting km to miles:

1.6 km = 1 mile
How many miles are there in 135.28 km? $\frac{135.28}{1.6} = 84.55 \, \text{miles}$ So, there are 84.55 miles in 135.28 km.

16. MCQ:

$1.6 \times 2.13 = 3.408$ So, the correct answer is (c) 3.408.

17. MCQ:

$0.05 \times 5 - 0.005 \times 5 = 0.25 - 0.025 = 0.225$ So, the correct answer is (b) 0.225.

Would you like more details on any of these solutions?

Here are 5 related questions:

1. How can you find the area of the square in question 13?
2. How would you calculate the exact total cost of 5 toothbrushes at ₹14.65 each?
3. What is the general formula for evaluating expressions like in question 11?
4. How can you convert between different units of length in the metric and imperial systems?
5. What is the quickest way to check your result in simplifications like in question 10?

Tip: Always double-check your calculations, especially when dealing with decimal numbers, to avoid rounding errors.