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by Brent@GMATPrepNow » Thu Jan 17, 2019 5:30 am
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188

Here's a solution that uses one variable.

Let x = Jake's present weight in pounds
So, x - 8 = Jake's hypothetical weight IF he were to lose 8 pounds

If Jake loses 8 pounds, he will weigh twice as much as his sister.
In other words, the sister weighs HALF as much as Jake's hypothetical weight of x - 8 pounds
So, (x - 8)/2 = sister's present weight

Together they NOW weigh 278 pounds.
So, Jake's present weight + sister's present weight = 278
So, x + (x - 8)/2 = 278
Eliminate the fraction by multiplying both sides by 2 to get: 2x + (x - 8) = 556
Simplify: 3x - 8 = 556
Add 8 to both sides: 3x = 564
Solve: x = 564/3 = 188

Answer: E

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by fskilnik@GMATH » Thu Jan 17, 2019 8:30 am
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188
Source: Official Guide
$$? = 2M$$
$$\left\{ \matrix{
\,2M - 8 = 2S \hfill \cr
\,2M + S = 278 \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\left\{ \matrix{
\,M - S = 4 \hfill \cr
\,2M + S = 278 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\left( {{2 \over 3}} \right)3M = \left( {{2 \over 3}} \right)\left( {270 + 12} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2M = 2 \cdot 94 = 188$$


We follow the notations and rationale taught in the GMATH method.

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by GMATGuruNY » Thu Jan 17, 2019 10:07 am
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188
ALWAYS KEEP YOUR EYE ON THE ANSWER CHOICES.

If Jake loses 8 pounds, he will weigh twice as much as his sister.
j-8 = 2s
j = 2s + 8 = even + even = even.

Since Jake's weight must be an EVEN VALUE, the correct answer is E.
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by fskilnik@GMATH » Thu Jan 17, 2019 11:35 am
GMATGuruNY wrote:
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188
ALWAYS KEEP YOUR EYE ON THE ANSWER CHOICES.

If Jake loses 8 pounds, he will weigh twice as much as his sister.
j-8 = 2s
j = 2s + 8 = even + even = even.

Since Jake's weight must be an EVEN VALUE, the correct answer is E.
Good idea, but the final check to see whether the *potential* answer satisfies the question stem is necessary.

I explain: with exactly the same question stem, but 366 in the place of 278, we would come to the wrong alternative choice.
(If sister weights 89.5 pounds, twice her weight is 179, Jake could weight 187 pounds and the argument shown in red fails.)

Regards,
Fabio.
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by GMATGuruNY » Thu Jan 17, 2019 12:13 pm
fskilnik@GMATH wrote:
GMATGuruNY wrote:
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188
ALWAYS KEEP YOUR EYE ON THE ANSWER CHOICES.

If Jake loses 8 pounds, he will weigh twice as much as his sister.
j-8 = 2s
j = 2s + 8 = even + even = even.

Since Jake's weight must be an EVEN VALUE, the correct answer is E.
Good idea, but the final check to see whether the *potential* answer satisfies the question stem is necessary.

I explain: with exactly the same question stem, but 366 in the place of 278, we would come to the wrong alternative choice.
(If sister weights 89.5 pounds, twice her weight is 179, Jake could weight 187 pounds and the argument shown in red fails.)

Regards,
Fabio.
The sum of Jake's weight and his sister's weight is an INTEGER VALUE.
The answer choices indicate that Jake's weight is an INTEGER VALUE.
Implication:
The sister's weight must also be an integer value.
Thus, the value in red -- 89.5 -- is not a valid option for the sister's weight.
Given the conditions in the problem, Jake's weight must be an EVEN INTEGER.
As a result, the only viable answer choice is E.
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Jake and his sister

by GMATGuruNY » Thu Jan 17, 2019 12:20 pm
fskilnik@GMATH wrote:I explain: with exactly the same question stem, but 366 in the place of 278, we would come to the wrong alternative choice.
(If sister weights 89.5 pounds, twice her weight is 179, Jake could weight 187 pounds and the argument shown in red fails.)

Regards,
Fabio
Here, the value in red is the sum of Jake's weight and TWICE the sister's weight:
187 + 2(89.5) = 366.
But the prompt gives an integer value for the sum of Jake's weight and the sister's ACTUAL weight -- a constraint that forces Jake's weight to be an even integer.
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by fskilnik@GMATH » Thu Jan 17, 2019 1:27 pm
GMATGuruNY wrote: The answer choices indicate that Jake's weight is an INTEGER VALUE.
Implication:
The sister's weight must also be an integer value.
The addition of the arguments above validates the argument.
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by fskilnik@GMATH » Thu Jan 17, 2019 1:30 pm
GMATGuruNY wrote: The sum of Jake's weight and his sister's weight is an INTEGER VALUE.
The answer choices indicate that Jake's weight is an INTEGER VALUE.
Implication:
The sister's weight must also be an integer value.
Given the conditions in the problem, Jake's weight must be an EVEN INTEGER.
As a result, the only viable answer choice is E.
This is perfect.
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by [email protected] » Thu Jan 17, 2019 5:24 pm
Hi All,

This question can be solved with fairly straight-forward Algebra (as some of the other solutions have proven). It can also be solved by TESTing THE ANSWERS and a bit of logic.

We're told that the total weight of Jake and his sister is 278 pounds. We're also told that if Jake lost 8 pounds, then he would weight TWICE as much as his sister. This means that, right now, Jake weighs MORE than TWICE his sister. We're asked for Jake's current weight.

Since Jake weighs MORE than TWICE his sister, his weight is MORE than 2/3 of the 278 pounds. Looking at these answer choices, I would TEST one of the bigger values first... Under normal circumstances, that would be Answer D. With a quick estimate though we can see that 2/3 of 270 pounds would be 180 pounds, but Jake has to weight MORE than that, so I'm going to TEST Answer E first....

If Jake weights 188 pounds...
Jake - 8 = 180 pounds....
Sister = 90 pounds
90 + 180 + 8 = 278 pounds
This is a MATCH for the information in the prompt, so Jake MUST weight 188 pounds.

Final Answer: E

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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:56 pm
BTGmoderatorDC wrote:If Jake loses 8 pounds, he will weigh twice as much as his sister. Together they now weigh 278 pounds. What is Jake's present weight, in pounds?

(A) 131
(B) 135
(C) 139
(D) 147
(E) 188
We let J = Jake's current weight and S = Sister's current weight, in pounds, and create the equations:

J - 8 = 2S

J = 2S + 8 (Equation 1)

and

J + S = 278 (Equation 2)

To solve this equation, we can substitute 2S + 8 from Equation 1 for the variable J in Equation 2:

(2S + 8) + S = 278

3S = 270

S = 90

We now know that the sister weighs S = 90 pounds, and we can plug that value into either equation to determine J. Let's plug 90 for S into equation 2:

J + 90 = 278

J = 188

Answer: E

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