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Difficulty: Very hardRatio and Proportion Word Problems

At a software company, the ratio of senior developers to junior developers was initially 3:73 : 7. During a quarterly hiring push, 1212 senior developers and 1818 junior developers were hired, but 66 junior developers resigned shortly thereafter. As a result, the ratio of senior developers to junior developers became 4:94 : 9. If a certain number NN of senior developers were subsequently promoted out of the developer pool such that the ratio of senior developers to junior developers became 5:125 : 12, what is the value of NN?

  1. A
    8
  2. 12Answer
  3. C
    15
  4. D
    18
  5. E
    24

Answer

The value of NN is 1212.
By defining the initial counts as 3x3x and 7x7x, accounting for the net additions (+12+12 seniors and +12+12 juniors), and solving 3x+127x+12=49\frac{3x+12}{7x+12} = \frac{4}{9}, we find x=60x = 60. Thus, there are 192192 senior developers and 432432 junior developers after hiring. Setting the final ratio 192−N432=512\frac{192-N}{432} = \frac{5}{12} gives 192−N=180192 - N = 180, which yields N=12N = 12.

Step-by-Step Solution

1
Define initial quantities using a common multiplier xx.
Initial senior developers = 3x3x, Initial junior developers = 7x7x.
The given initial ratio of senior to junior developers is 3:73 : 7.
2
Account for changes in staffing and set up the equation for the new ratio.
Senior developers after hiring = 3x+123x + 12; Net junior developer change = +18−6=+12+18 - 6 = +12, so Junior developers after hiring = 7x+127x + 12. Equation: 3x+127x+12=49\frac{3x + 12}{7x + 12} = \frac{4}{9}.
12 seniors were added, while 18 juniors were hired and 6 resigned.
3
Solve for the multiplier xx.
9(3x+12)=4(7x+12)  ⟹  27x+108=28x+48  ⟹  x=609(3x + 12) = 4(7x + 12) \implies 27x + 108 = 28x + 48 \implies x = 60.
Cross-multiplication yields the value of the algebraic multiplier xx.
4
Calculate post-hiring developer counts.
Senior developers post-hiring = 3(60)+12=1923(60) + 12 = 192; Junior developers post-hiring = 7(60)+12=4327(60) + 12 = 432.
Substitute x=60x = 60 back into the expressions for post-hiring developer counts.
5
Determine NN using the final ratio 5:125 : 12.
192−N432=512  ⟹  192−N=5×36=180  ⟹  N=192−180=12\frac{192 - N}{432} = \frac{5}{12} \implies 192 - N = 5 \times 36 = 180 \implies N = 192 - 180 = 12.
Junior developer count remains unchanged (432432) while senior developer count decreases by NN.

Key Concept

Multi-step algebraic ratio setup and alteration of part-to-part proportions
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Topics

Ratio and Proportion Word Problems

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