This problem can be solved as a simple word problem in which we must convert words to math. Before we create our equations, we want to define some variables.

J = Jake’s current weight, in pounds

S = Sister’s current weight, in pounds

We are told that “If Jake loses 8 pounds, he will weigh twice as much as his sister." We put this into an equation:

J – 8 = 2S

We can isolate J by adding 8 to 2S:

J = 2S + 8 (Equation 1)

Next, we are told that “Together they now weigh 278 pounds.” We can also put this into an equation.

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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Video solution from Quant Reasoning:

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Let J be Jake's present weight and S be his sister's weight, then we can construct two linear equations:
J=2S+8;
J+S=278.

Subtract one from another: S=278-2S-8 --> S=90 --> J=188.

Answer: E.
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(x-8) = 2s

s+x = 278 --> s= 278-x

x-8 = 2(278-x)

3x=564

x=188

..................... Jack ......................... Sister

Now ................ x ................................. y

8 yrs before....... (x-8)

2y = x - 8

x = 2y+8

Given that 2y + 8 + y = 270

y = 90

x = 188

Answer = E

Hi All,

This question can be solved with fairly straight-forward Algebra (as 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:

GMAT assassins aren't born, they're made,
Rich
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You can also test the answers.
When you realize that when you choose 188 for Jake's weight you get from the equations, a nice round number for his sister age and the work involved is fast:
J-8=2S, where J stands for Jake and S for his sister --> 188-8=2S --> S=90.
Now you can plug in the obtained value in the second equation: J+S= 278 --> As we have chosen 188 for Jake and got 90 pounds for his sister--> 188+90=278. This is a match. Option E is the correct answer.

let Jack's Weight be J
His sister weight be S
J-8 = 2S ----1
J+S = 278 ---2
Then,
J= 2S+8
2S+8+S = 278
S= 270/3
S=90
then J = 188
Answer is E

One of my favorite method to attempt questions is by using as low level an approach as possible:

- On this question we know that 278 is the current sum of Jake's and his sister weights

- We also know that 270 is the sum of their weights if Jake loses 8 pounds

- Since 270 equals 2 parts from Jake's weight and 1 part from his sister weight so we can have following ratio

- Jake: 2 parts of 270 (180) and Sister: 1 part of 270 (90)

- So Jake's current weight will be those 2 parts (180) plus the weight that he has not lost yet (8) = 188 (option E)

- In fact we don't even need to calculate 188. Since we know that Jake's weight is at least 180 and since none of the other options is anywhere close to 180, so our answer is automatically option - E. Moreover, if you are pressed for time then you don't even have to consider 8. You can use the approximation technique to divide 278 into 3 equal parts which will be about ~93 and so Jake's weight is approximately 186 (and so only option E works)

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

Cheers,
Brent
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\(? = 2M\)

\(\left\{ \matrix{
\,2M - 8 = 2S \hfill \cr
\,2M + S = 278 \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\left\{ \matrix{
\,M - S = 4 \hfill \cr
\,2M + S = 278 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\left( {{2 \over 3}} \right)3M = \left( {{2 \over 3}} \right)\left( {270 + 12} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2M = 2 \cdot 94 = 188\)


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

Regards,
Fabio.
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I loved it how FANewJersey has put this question in, option E is indeed pretty far from others and if hard pressed on time such a trick can indeed be really handy

All you need to do for this one is find out the first equation and start plugging in answer choices. So:

J-8=2(S)

A - 123 = 2(S)
B - 127 = 2(S)
C - 131 = 2(S)
D - 149 = 2(S)
E - 180 = 2(S)

If you were to go on to find what S=, you will notice only one is even (E).

Because the weight together is an integer, you can stop here and grab E as the final answer.

Even-8=Even
Even+2*Even= 278 (Even)

Well, we need an Even. Only E.

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Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart
Is it the right way...?

Let Jake's weight is 11 (odd number). After losing 8 pounds, his new weight would be 3. This 3 pounds is the twice of his sister. So, his sister's weight is \(\frac{3}{2}=1.5\) pound.
Their present weight is 11+1.5 pounds=12.5 pounds (this is fraction value). The question prompt gave their total weight is EVEN (278). So, Jake's weight can't be odd (say 11). The weight of Jake must be EVEN. In the answer option, every answer option is ODD without choice E. So, choice E is the correct choice.

Or,
We can cross out choices A, B and C easily.
A, B and C:
Half of 278 is 139, which is choice C. Jake's weight has to be a lot more than half of 278. Because, together they weigh 278, but Jake is more than twice as much as his sister. So, he needs to be a lot more than half of 278. The correct choice MUST be greater than 139.

Thanks__

Hi Asad,

YES - the Number Properties that you've described could be used to correctly answer this question. This goes to show how important it is to pay attention to how the answers are written. There are plenty of circumstances (in BOTH the Quant and Verbal sections), in which you can use the 5 answer choices "against" the prompt to logically eliminate the wrong answers and zero-in on the correct one.

GMAT assassins aren't born, they're made,
Rich
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If Jake loses 8 pounds, his weight will become twice that of her sister's i.e. an even number. Now out of all the 5 options, only option E is even, subtracting 8 from which will leave an even number (rest all answer options are odd, subtracting 8 from which will leave an odd number, which cannot be twice of a number). Hence correct answer option is E.