Selling price of a toy car is Rs 540. If the profit made by shopkeeper is 20%, what is the cost price of this toy? Simplify: 25*5^2*t^8 / (10^3*t^4) Find any three rational numbers... Selling price of a toy car is Rs 540. If the profit made by shopkeeper is 20%, what is the cost price of this toy? Simplify: 25*5^2*t^8 / (10^3*t^4) Find any three rational numbers between 1/4 and 1/2 Read more

Steps to Solve

1. Find the cost price of the toy car Let the cost price be $x$. The selling price is given as Rs 540, and the profit is 20%. The selling price is the cost price plus the profit, so we have: $x + 0.20x = 540$

2. Solve the equation to find the cost price Combine the terms with $x$: $1.20x = 540$ Divide both sides by 1.20 to find $x$: $x = \frac{540}{1.20} = 450$

3. Simplify the algebraic expression The expression is $\frac{25 \times 5^2 \times t^8}{10^3 \times t^4}$. We can rewrite this as: $\frac{25 \times 25 \times t^8}{1000 \times t^4}$

4. Simplify the numbers $25 \times 25 = 625$ and $1000 = 10^3$ So we have: $\frac{625 \times t^8}{1000 \times t^4}$ $\frac{625}{1000} = \frac{5}{8}$

5. Simplify the variable term $\frac{t^8}{t^4} = t^{8-4} = t^4$

6. Combine the simplified terms $\frac{5}{8} t^4$

7. Find three rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$ First, find a common denominator for the fractions. The least common denominator for 4 and 2 is 4, so we can express the fractions as $\frac{1}{4}$ and $\frac{2}{4}$. To find rational numbers between these, we can use a larger denominator, for example multiplying both fractions by $\frac{4}{4}$ which gives $\frac{4}{16}$ and $\frac{8}{16}$. Three rational numbers between these are $\frac{5}{16}$, $\frac{6}{16}$, $\frac{7}{16}$. These simplify to $\frac{5}{16}, \frac{3}{8}, \frac{7}{16}$.