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Re: The ratio of the number of plants that Keisha has to the number [#permalink]
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HolaMaven wrote:
The ratio of the number of plants that Keisha has to the number of plants Sam has is 1:5. After Sam gives Keisha 5 plants, the ratio of plants Keisha has to the plants Sam has will be 2:7. How many more plants will Sam have than Keisha after 5 plants are given?

A. 30
B. 45
C. 50
D. 60
E. 75




let the common ratio be x, so \(\frac{k}{s} = \frac{x}{5x}\)
After s gives 5 to k, ratio becomes \(\frac{x+5}{5x-5}=\frac{2}{7}.....7x+35=10x-10.....3x=45...x=15\)

so the difference prior to exchange of 5 is (5-1)*x=4*15=60..
now when s gives 5 to k, difference reduces by 5+5=10...60-10=50
OR
number = 15*1:15*4=15:75
after exchange of 5 it becomes = 15+5:75-5=20:70 so difference = 70-20=50

C
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The ratio of the number of plants that Keisha has to the number [#permalink]
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HolaMaven wrote:
The ratio of the number of plants that Keisha has to the number of plants Sam has is 1:5. After Sam gives Keisha 5 plants, the ratio of plants Keisha has to the plants Sam has will be 2:7. How many more plants will Sam have than Keisha after 5 plants are given?

A. 30
B. 45
C. 50
D. 60
E. 75

Use x as a multiplier with the original ratio. Add and subtract 5 from that ratio, and set LHS equal to new ratio. Solve for x to yield how many plants each person had originally. Then calculate totals after the exchange.

Ratio of K's plants to S's plants:\(\frac{K}{S}\) = \(\frac{(1)x}{5x}\)

S gives K five plants. Account for it twice. The total quantities of both people have changed: \(\frac{x + 5}{5x -\\
5}\)

. . . and after S gives K the plants, the new ratio is 2:9

\(\frac{x + 5}{5x -\\
5}\) = \(\frac{2}{9}\)

10x - 10 = 7x + 35
3x = 45
x = 15 --> the multiplier for the original ratio quantities

So K had x = 15 plants
And S had 5x = 75 plants

K now has 5 more = 20
S now has 5 fewer = 70

How many more plants does S have after five are given? 70 - 20 = 50

Answer C
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Re: The ratio of the number of plants that Keisha has to the number [#permalink]
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