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ꯄꯨꯀꯆꯦꯜ ꯁꯦꯡꯗꯕꯥ ꯗꯤꯂꯔ ꯑꯃꯅꯥ ꯃꯍꯥꯛꯀꯤ ꯄꯣꯠꯂꯃꯁꯤꯡ ꯃꯃꯜ ꯍꯣꯡꯕꯥ ꯃꯃꯂꯗꯥ ꯌꯣꯜꯂꯤ ꯍꯥꯌꯅꯥ ꯂꯥꯑꯣꯊꯣꯀꯏ ꯑꯗꯨꯕꯨ $950$ gꯒꯤ ꯑꯔꯨꯝꯕꯥ ꯑꯃꯥ $1$ kgꯒꯤ ꯃꯍꯨꯠꯇꯥ ꯁꯤꯖꯤꯟꯅꯩ꯫ ꯃꯍꯥꯛꯀꯤ ꯒꯦꯏꯟ ꯆꯥꯗꯥ ꯀꯌꯥꯅꯣ?

A$5\%$
B$5\dfrac{5}{19}\%$
C$5.5\%$
D$10\%$
Answer & Solution
Correct answer: B. $5\dfrac{5}{19}\%$
**ꯄ꯭ꯔꯤꯟꯁꯤꯄꯜ.** ꯁꯤ.ꯄꯤ.ꯗꯥ ꯌꯣꯜꯂꯤꯉꯩꯗꯥ ꯗꯤꯂꯔꯅꯥ ꯑꯔꯨꯝꯕꯥ ꯑꯗꯨ ꯁꯦꯃꯗꯣꯀꯄꯥ ꯃꯇꯃꯗꯥ, $\text{Gain}\% = \dfrac{\text{Error}}{\text{True Value} - \text{Error}} \times 100$. **ꯑꯦꯞꯂꯥꯏ ꯇꯧꯕꯥ.** ꯑꯁꯣꯌꯕꯥ $= 1000 - 950 = 50$ g, ꯑꯆꯨꯝꯕꯥ ꯚꯦꯜꯌꯨ $= 1000$ g. ꯒꯦꯏꯟ $\% = \dfrac{50}{950} \times 100 = \dfrac{100}{19} \approx 5\tfrac{5}{19}\%$. **ꯀꯔꯤꯒꯤ ꯑꯣꯄꯁꯟ A ꯑꯁꯤ ꯑꯔꯥꯅꯕꯥ ꯑꯣꯏꯔꯤꯕꯅꯣ.** ꯀꯔꯤꯒꯨꯝꯕꯥ ꯒꯦꯏꯟ ꯑꯗꯨ *ꯑꯆꯨꯝꯕꯥ ꯑꯔꯨꯝꯕꯥ* (50/1000)ꯗꯥ ꯂꯩꯔꯕꯗꯤ $5\%$ ꯑꯁꯤ ꯄꯥꯎꯈꯨꯝ ꯑꯣꯏꯒꯅꯤ꯫ ꯒꯦꯏꯟ ꯑꯁꯤ *ꯃꯍꯥꯛꯅꯥ ꯇꯁꯦꯡꯅꯥ ꯄꯤꯔꯤꯕꯥ* ꯑꯗꯨꯗꯥ ꯂꯩ, ꯃꯁꯤ $950$ gꯅꯤ — ꯗꯦꯅꯣꯃꯤꯅꯦꯇꯔ ꯑꯁꯤ ꯑꯆꯨꯝꯕꯅꯤ − ꯑꯁꯣꯏꯕꯥ, ꯑꯆꯨꯝꯕꯥ ꯅꯠꯇꯦ꯫
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