Jeff plans to lose weight by reducing his daily calorie intake each we

Question

Jeff plans to lose weight by reducing his daily calorie intake each week. How many weeks does it take for his calorie intake to fall below half of its original level

(1) Jeff currently consumes 2,500 calories per day.
(2) Jeff plans to reduce his daily calorie intake by 10 percent each week.

chetan2u · Answer

Hi,
Statement one just gives us intake , nothing about reduction. Insufficient
Statement two tells us the % reduction each week. Since the answer is also in terms of %, we donot require the value per se.
We can find when this reduction will lead to 50% reduction from 100%. Sufficient
Ans B

EMPOWERgmatRichC · Answer

Hi 23a2012,

With certain DS questions, you have to pay careful attention to what is specifically ASKED for. The nature of the question might allow you more than one way to approach the work.

Here, we're asked how many weeks it will take Jeff to reduce his calorie intake by HALF.

Fact 1: Jeff currently consumes 2,500 calories per day.

This tells us nothing about how many calories he will reduce each week, so we cannot answer the question.
Fact 1 is INSUFFICIENT.

Fact 2: Jeff plans to reduce his daily calorie intake by 10 percent each week.

While we're not told Jeff's starting number of calories, THAT information is actually irrelevant here. We know that he's going to reduce his number of calories by 10% each week, so we COULD figure out how many weeks it takes to get that number down to HALF (meaning 50%). Here's how you could do the work, if you needed to...

By reducing the number by 10% each week, that is mathematically equivalent to multiplying the prior week's number by 0.9

Start = X
Week 1 = .9X
Week 2 = .9(.9X) = .81X
Week 3 = .9(.81X) = .729X
Week 4 = .9(.729X) = .6561X
Week 5 = .9(.6561X) = .59049X
Week 6 = .9(.59049X) = .531441X
Week 7 = .9(.531441X) = .4782969X

As you can see, the math IS rather tedious, but we don't actually have to do the math to answer this question. We know that we WOULD eventually get to the answer and THAT knowledge is enough.
Fact 2 is SUFFICIENT

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

AkshdeepS · Answer

What I did wrong with this question is that I thought we need initial calorie value which is to be reduced. Also, I referred both statements together when evaluated B.

ScottTargetTestPrep · Answer

Statement One Alone:

Jeff currently consumes 2,500 calories per day.

Statement one is not sufficient to answer the question since we don’t know how much he plans to reduce his daily calorie intake.

Statement Two Alone:

Jeff plans to reduce his daily calorie intake by 10 percent each week.

Statement two is sufficient to answer the question. For example, if his daily calorie intake in the first week is 2000 calories, then his daily calorie intake in the second week is 2000 x 0.9 = 1800 calories. His daily calorie intake in the third week is 1800 x 0.9 = 1620 calories. Therefore, by continuing this process, we can find the number of weeks it takes for his calorie intake to fall below half its original level. (Note: We don’t have to find the actual number of weeks; we only need to know it’s possible to find it.)

Answer: B

mangamma · Answer

Official explanation 
SOurce is veritas prep

Solution: B.

In order to solve, we need to know what the calorie reductions will be. Statement 2 says that Jeff will reduce his daily calorie intake by 10 percent each week. If we say that week 1 is x, then week 2 is .9x, week 3 is .81x, and so on. We can figure out when his calorie intake is less than .5x, so Statement 2 is sufficient. Since the calorie reductions are done by percent, we don’t need to know actual calorie values. Statement 1 tells us his current consumption level, but we have nothing about the calorie reductions. Insufficient, so the correct answer is B.

coled90 · Answer

Since it is not important here, we do not need to find the value.... but for those curious how to do it:

If  L_{new} = \frac{1}{2} * L_{original} = 0.9^n * L_{original} 
 \frac{1}{2} * L_{original} = 0.9^n * L_{original} 
-> the  L_{original}  cancels out
result is:  \frac{1}{2}= 0.9^n  
Now solve for  n

(would not be expected to do this on the test, just to know it COULD be done)
take the Natural Logarithm of each side, to undo the exponential:
  ln(\frac{1}{2}) = ln(0.9^n)  
use logarithm rules to pull out the  n  to the front of the right side:
  ln(\frac{1}{2}) = n*ln(0.9)  
  n = \frac{ln(0.5)}{ln(0.9)}  
 n= 6.58 weeks

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