Subjects algebra

Percentage Problems E769D2

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Use the AI math solver

1. **Problem 4a:** What percent of 200 is 11? The formula for percentage is: $$\text{Percentage} = \frac{\text{Part}}{\text{Base}} \times 100$$ Here, Part = 11, Base = 200. Calculate: $$\frac{11}{200} \times 100 = 5.5\%$$ 2. **Problem 4b:** 0.1 is 20% of what number? Let the number be $x$. Then: $$0.1 = 20\% \times x = \frac{20}{100} \times x = 0.2x$$ Solve for $x$: $$x = \frac{0.1}{0.2}$$ Show cancellation: $$x = \frac{0.1}{\cancel{0.2}} \times \frac{\cancel{1}}{1} = 0.5$$ 3. **Problem 4c:** Find 250% of 250. $$250\% = \frac{250}{100} = 2.5$$ Calculate: $$2.5 \times 250 = 625$$ 4. **Problem 4d:** 20% of what number is 7? Let the number be $y$. $$20\% \times y = 7$$ $$\frac{20}{100} y = 7$$ Solve for $y$: $$y = \frac{7}{\frac{20}{100}} = \frac{7}{0.2}$$ Show cancellation: $$y = \frac{7}{\cancel{0.2}} \times \frac{\cancel{1}}{1} = 35$$ 5. **Problem 5:** Justin earned 63 out of 75 points, Stephanie earned 63 out of 70. Find difference in percent scores. Justin's percentage: $$\frac{63}{75} \times 100 = 84\%$$ Stephanie's percentage: $$\frac{63}{70} \times 100 = 90\%$$ Difference: $$90\% - 84\% = 6\%$$ 6. **Problem 6:** Fire department received 71 false alarms out of 639. What percent were not false? Number not false: $$639 - 71 = 568$$ Percent not false: $$\frac{568}{639} \times 100 \approx 88.89\%$$ 7. **Problem 7:** Six middle schools sent bands, which is 30% of all middle schools. Find total number. Let total schools be $N$. $$30\% \times N = 6$$ $$\frac{30}{100} N = 6$$ Solve for $N$: $$N = \frac{6}{0.3} = 20$$ 8. **Problem 8a:** 29% + 25% = 54% prefer pepperoni or cheese. If 590 students prefer these, total surveyed is: Let total be $T$. $$54\% \times T = 590$$ $$0.54 T = 590$$ $$T = \frac{590}{0.54} \approx 1092.59 \approx 1093$$ 8b. If 2000 students surveyed, how many answered "Other" (6%)? $$6\% \times 2000 = 0.06 \times 2000 = 120$$ 9. **Problem 9:** Joyce's age is 25% of Marta's, Marta's is 50% of Cesar's. Joyce is 11. Let Cesar's age be $C$. Marta's age: $$M = 0.5 C$$ Joyce's age: $$J = 0.25 M = 11$$ Substitute $M$: $$0.25 \times 0.5 C = 11$$ $$0.125 C = 11$$ Solve for $C$: $$C = \frac{11}{0.125} = 88$$ Find $M$: $$M = 0.5 \times 88 = 44$$ **Final answers:** 4a: 5.5% 4b: 0.5 4c: 625 4d: 35 5: 6% 6: 88.89% 7: 20 8a: 1093 8b: 120 9: Marta 44, Cesar 88
29%25%10%30%6%