In Mr Tan's P$6$ class, the number of girls is $\dfrac{3}{5}$ the number of boys. After $8$ new girls join and $2$ boys transfer out, the ratio of girls to boys becomes $10 : 9$. How many boys were ORIGINALLY in the class?
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Worked Solution
Step 1: Translate the fraction into ratio parts. 'Girls is $\\dfrac{3}{5}$ the number of boys' means girls : boys $= 3 : 5$. So boys are the LARGER quantity. Let boys $= 5k$ and girls $= 3k$.\n
Step 2: Apply the changes. After $8$ girls join, girls $= 3k + 8$. After $2$ boys leave, boys $= 5k - 2$.\n
Step 3: Set up the new ratio. $(3k + 8) : (5k - 2) = 10 : 9$, i.e. $\\dfrac{3k + 8}{5k - 2} = \\dfrac{10}{9}$.\n
Step 4: Cross-multiply.\n $9(3k + 8) = 10(5k - 2)$\n $27k + 72 = 50k - 20$.\n
Step 5: Solve.\n $50k - 27k = 72 + 20$\n $23k = 92$\n $k = 4$.\n
Step 6: Boys originally $= 5k = 5 \\times 4 = 20$.\nCheck: girls originally $= 3 \\times 4 = 12$. After $8$ girls join: $20$ girls. After $2$ boys leave: $18$ boys. New ratio $20 : 18 = 10 : 9$. ✓
Answer: 20
Correct answer: 20
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