GRE Math Challenge Question:
Ariel is 3 years older than Elena and Yasemin is 10 years younger than twice Ariel's age
Quantity A Quantity B
Yasemin's age Elena's age
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Explanation:
Let A represent Ariel's age, let E represent Elena's age, and let Y represent Yasemin's age. Ariel is 3 years older than Elana, so A = E + 3. And Yasemin is 10 years younger than twice Ariel's age, so Y = 2A – 10.
Since A = E + 3, if we pick a value for Elena's age, we will know Ariel's age. And, since Y = 2A – 10, once we have Ariel's age, we can also compute Yasemin's age. So, let's pick some values for E and then see how the resulting values for Y compare to E. We need to be sure to pick numbers that will result in all three ages being positive, since negative ages are not possible.
We know that Yasemin is 10 years younger than twice Ariel's age; this means that twice Ariel's age must be greater than 10 in order for Yasemin to have a positive age. Thus, Ariel must be more than 5 years old, so that twice her age is more than 10. For Ariel to be more than 5, Elena must be more than 2, since Ariel is 3 years older than Elena. So, we can pick any age above 2 for Elena.
Let's try E = 5 then A = E + 3 = 5+ 3 = 8
Y = 2A -10 = 2(8) - 10 = 16 - 10 = 6
So when, E = 5 , Y = 6 that is ,hen Elena is 5, Yasemin is 6. In this case, Quantity A is greater.
Now let E = 3 . Then A = E + 3 = 3 + 3 = 6
With A = 6; Y = 2A - 10 = 2(6) - 10 = 12 - 10 = 2
So when E = 3 , Y =2 , . That is, when Elena is 3, Yasemin is 2. In this case, Quantity B is greater.
Since different relationships between the quantities are possible, Choice (D) is correct.
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