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(f) Increase: \( \begin{array}{llllll}\text { (1) } 40 \text { by } 20 \% & \text { (2) } 45 \text { by } 33 \frac{1}{3} \% & \text { (3) R50 by } 50 \% & \text { (4) } 2 \frac{1}{2} \text { by } 150 \% \\ \text { (g) Decrease: } \\ \text { (1) } 30 \text { by } 10 \% & \text { (2) } 35 \text { by } 20 \% & \text { (3) } 120 \text { by } 25 \% & \text { (4) } 1 \frac{5}{6} \text { by } 40 \%\end{array} \)

Ask by Coleman Guzman. in South Africa
Mar 11,2025

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**(f) Increase** 1. \(40\) increased by \(20\% = 48\) 2. \(45\) increased by \(33\,\frac{1}{3}\% = 60\) 3. \(\text{R}50\) increased by \(50\% = \text{R}75\) 4. \(2\,\frac{1}{2}\) increased by \(150\% = 6.25\) **(g) Decrease** 1. \(30\) decreased by \(10\% = 27\) 2. \(35\) decreased by \(20\% = 28\) 3. \(120\) decreased by \(25\% = 90\) 4. \(1\,\frac{5}{6}\) decreased by \(40\% = 1.1\)

Solution

**(f) Increase** 1. Increase \(40\) by \(20\%\): \[ 40 + 20\% \times 40 = 40 + 0.20 \times 40 = 40 + 8 = 48. \] 2. Increase \(45\) by \(33\,\frac{1}{3}\%\): Note that \(33\,\frac{1}{3}\%\) is equivalent to \(\frac{1}{3}\) (i.e. \(0.3333\ldots\)). \[ 45 + \frac{1}{3} \times 45 = 45 + 15 = 60. \] 3. Increase \(\text{R}50\) by \(50\%\): \[ \text{R}50 + 50\% \times \text{R}50 = \text{R}50 + 0.50 \times \text{R}50 = \text{R}50 + \text{R}25 = \text{R}75. \] 4. Increase \(2\,\frac{1}{2}\) by \(150\%\): First, express \(2\,\frac{1}{2}\) as an improper fraction: \[ 2\,\frac{1}{2} = \frac{5}{2}. \] Calculate \(150\%\) of \(\frac{5}{2}\): \[ 150\% \times \frac{5}{2} = 1.5 \times \frac{5}{2} = \frac{15}{4} = 3.75. \] Now add this to the original value: \[ \frac{5}{2} + \frac{15}{4} = \frac{10}{4} + \frac{15}{4} = \frac{25}{4} = 6.25. \] **(g) Decrease** 1. Decrease \(30\) by \(10\%\): \[ 30 - 10\% \times 30 = 30 - 0.10 \times 30 = 30 - 3 = 27. \] 2. Decrease \(35\) by \(20\%\): \[ 35 - 20\% \times 35 = 35 - 0.20 \times 35 = 35 - 7 = 28. \] 3. Decrease \(120\) by \(25\%\): \[ 120 - 25\% \times 120 = 120 - 0.25 \times 120 = 120 - 30 = 90. \] 4. Decrease \(1\,\frac{5}{6}\) by \(40\%\): First, express \(1\,\frac{5}{6}\) as an improper fraction: \[ 1\,\frac{5}{6} = \frac{11}{6}. \] Compute \(40\%\) of \(\frac{11}{6}\): \[ 40\% \times \frac{11}{6} = 0.40 \times \frac{11}{6} = \frac{11}{15}. \] Now subtract this from the original value: \[ \frac{11}{6} - \frac{11}{15}. \] To subtract, use a common denominator (LCM of 6 and 15 is 30): \[ \frac{11}{6} = \frac{55}{30}, \quad \frac{11}{15} = \frac{22}{30}. \] Therefore, \[ \frac{55}{30} - \frac{22}{30} = \frac{33}{30} = \frac{11}{10} = 1.1. \]

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To increase 40 by 20%, simply multiply it by 0.20 to find the increase, which is 8. Adding this to the original 40 gives you 48. For 45, increasing it by 33 1/3% (or 1/3) means you multiply 45 by 0.33, getting 15. After adding this, the total is 60. For R50, increasing by 50% means you get an additional R25, totalling R75. And for 2 1/2 increased by 150%, you’d convert to an improper fraction (5/2), multiply by 1.5, getting 7.5, and add this to 2.5 to reach 10. Now for the decrease calculations: for 30 reduced by 10%, you’d find 10% of 30, which is 3, and subtract to get 27. For 35 decreased by 20%, calculate 20% as 7, leading to a final value of 28. When tackling 120, a 25% cut means subtracting 30; you land at 90. Lastly, with 1 5/6 decreased by 40%, convert to an improper fraction (11/6), find 40% of it (4.4), and subtract to give approximately 1.5 (or 1 1/2).

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