Gastón took a certain standardized test four times. His score improved

Question

Gastón took a certain standardized test four times. His score improved by s points from his first try to his second; fell by t points from his second try to his third; and improved by u points from his third try to his fourth. If Gastón’s score on his final attempt was N points, which of the following expressions represents the average (arithmetic mean) of all four of Gastón’s scores?

A. \(\frac{N - u + t - s}{4}\)

B. \(N - \frac{3u}{4 }+ \frac{t}{2} - \frac{s}{4}\)

C. \(\frac{N}{4} - \frac{u}{2} + \frac{3t}{4} - s\)

D. \(\frac{N + u - t + s}{4}\)

E. \(N + \frac{3u}{4} - \frac{t}{2} + \frac{s}{4}\)

Show Spoiler

M48-02

Attachment:

GMAT-Club-Forum-qs3j1dew.png [ 19.18 KiB | Viewed 1176 times ]

Show Answer

Official Answer

B

RonPurewal · Accepted Answer

Official Explanation:

Algebraic Approach:

To find the answer algebraically, start from Gaston’s fourth and final score of N points, and find the other three scores by reversing the successive changes one at a time.

4th score: N 
This score is u points higher than Gastón’s third score, so his third score is therefore (N – u) points.

3rd score: N – u 
This score is t points lower than Gastón’s second score, so his second score is therefore (N – u + t) points.

2nd score: N – u + t 
This score is s points higher than Gastón’s first score, so his first score is therefore (N – u + t – s) points.

1st score: N – u + t – s 
The sum of all four of these scores is 4N – 3u + 2t – s points, so the average (arithmetic mean) is therefore  \frac{4N -3u + 2t - s}{4} , or  N - \frac{3u}{4 }+ \frac{t}{2} - \frac{s}{4} , points.

The correct answer is B.

Number-Picking Approach:

Since the values of the variables remain undetermined even when the problem is solved, we can choose our own values. Probably the most straightforward way to do this is to choose Gastón’s four scores directly, rather than to pick numbers directly for the variables (which represent differentials between consecutive scores). We don’t need to choose numbers within the real-life range of any actual standardized test (no need for numbers between 205 and 805!); we can use smaller, ‘friendlier’ numbers instead.

Let’s say
 1st score = 2
2nd score = 3
3rd score = 1
4th score = 4 
so therefore s = 1, t = 2, u = 3, and N = 4; (Note that t is NOT a negative number!)

and the answer to the problem is the average of all four scores, which is (2 + 3 + 1 + 4)/4 = 2.5 points.

Plug these values into   all five   choices—remember coincidences are possible, so you can’t stop at the first value that works—and look for 2.5 to pop out:

A)  \frac{4 - 3 + 2 - 1}{4} = 0.5. Eliminate;

B)  4 - \frac{9}{4} + \frac{2}{2} - \frac{1}{4} = \frac{10}{4} =  2.5 (: Keep!

C)  \frac{4}{4} - \frac{3}{2} + \frac{6}{4} - 1  = 0. Eliminate;

D)  \frac{4 + 3 - 2 + 1}{4}  = 1.5. Eliminate;

E)  4 + \frac{9}{4} - \frac{2}{2} + \frac{1}{4}  = 5.5. Eliminate.

Only choice B survives, so B is the correct answer.

Dereno · Answer

Let the four scores be represented as a,b,c,d respectively.

Given that d = N

Moreover, from a to b , we have +s

From b to c , we have -t

From c to d , we have +u

If,  d = N  ,

c +u = N , then  c = N-u.

Given, b-t = c

b = c+t = N-u+t

b =N - u + t

Given that a +s = b,

Then  a = N-u+t-s

There fore, the individual standardised scores are:

a = N-u+t-s

b =N - u + t

c = N-u

d = N 
 
Arithmetic mean = (a+b+c+d)/4

= (4N -3u + 2t -s)/4

=  N - (3/4)u + (t/2) - (s/4)

B.  N - \frac{3u}{4 }+ \frac{t}{2} - \frac{s}{4}

aditya1818 · Answer

Let First score be f
second score= f+s
third score= f+s-t
fourth score= f+s-t+u =N (as per the question)

But before doing anything, I assumed his first score=100, second score=100, third score=100 and fourth score=100
thus avg score must be 100 as per my above assumption and t,s,u all will be 0 and N will be also equal to 100
This straight away eliminated A, C and D : This was very quick

Now I started with my variables and saw that options doesn't include "f" and I sensed that I need to get rid of "f"
so from f+s-t+u = N, I have substituted "f" in the 1st,2nd and 3rd score which boiled down to as below:
First score = N+t-s-u
second score= N+t-u
third score= N-u
fourth score=N

Now the average clearly from the above is as per option (B)

Dereno · Answer

But before doing anything, I assumed his first score=100, second score=100, third score=100 and fourth score=100 
 thus avg score must be 100 as per my above assumption and t,s,u all will be 0 and N will be also equal to 100 
 This straight away eliminated A, C and D : This was very quick.

During crunch times, this approach might be very effective to narrow down to two options. 😊 nice one.

My mind stuck to the standard algebraic route. If there were numbers my approach could have differed.

aditya1818 · Answer

@   Dereno

You know what makes Quant more interesting is that sometimes you can compare "units" also. Although this questions doesn't help with that but I have come across few questions that might help by simply comparing units.

For eg: Suppose the question mentioned distance, time, Price and all in terms of variables and then question asks find the "xyz distance" and then options will be presented by these variables.

You can boil down or even solve by merely checking the units of the options. Checking whether options (A,B,C,D,E) corresponds to the unit of distance. This is very much intuitive to me but this helps me when all my weapons and reasoning are down. These provides some valuable edge to some lethal questions.

NIKKtheSolver33 · Answer

Let's start by writing the scores, assuming in his first attempt Gastón took a score of A points.
Attempts and scores:
 
1: A
2: A+s
3: (A+s)-t 
4: ((A+s)-t)+u = N

We know that his score on the fourth attempt equals N points.
Looking at the answer choiches we see that we must work using the variable N, and not A. 
Let's equals Gastón scores using the new variable N instead of A.

4: N
3: N-u
2: N-u+t
1: N-u+t-s (Nuts?)

To calculate the arithmetic mean of all his scores we have to sum them and divide by 4.

Total Sum = N + N-u + N-u+t + N-u+t-s = 4N-3u+2t-s
Average = (4N-3u+2t-s) / 4 = N - (3/4)u + (1/2)t - s/4

Answer B.

Edskore · Answer

This is a really clean Algebraic Translation problem — it looks intimidating because of the variable-heavy answer choices, but once you build the scores systematically, it falls into place fast.

Key concept being tested: Algebraic Translation + Arithmetic Mean

The common trap here is expressing all four scores from the first attempt, which creates a messy expression. The smarter move is to work backward from N — the score you're given — since N is your anchor.

Step 1: Name the four scores.
We know the 4th score is N. Work backward:
- Score 4 = N
- Score 3 = N − u (because score improved by u going from 3rd to 4th)
- Score 2 = N − u + t (score fell by t going from 2nd to 3rd)
- Score 1 = N − u + t − s (score improved by s going from 1st to 2nd)

Step 2: Sum the four scores.
Sum = (N − u + t − s) + (N − u + t) + (N − u) + N
= 4N − 3u + 2t − s

Step 3: Divide by 4 to get the mean.
Mean = (4N − 3u + 2t − s) / 4
= N − 3u/4 + t/2 − s/4

That's answer choice B.

Why the trap works: If you start from Score 1 forward, you get a correct but much messier expression — and under time pressure, students make sign errors or lose track of which variable goes where. Anchoring to N and working backward keeps the algebra tight and avoids that.

Takeaway: When a GMAT problem gives you the last value in a sequence, anchoring to it and working backward almost always simplifies the algebra.

---
Kavya | 725 GMAT Focus | Free gamified GMAT prep: edskore.com

Navdeep2000 · Answer

Let x be the initial text score Gastón gets on his first our four tests. 
Accordingly, we can find his next tests score based on the info given to us.

Test 2 :  (x+s) 
Test 3 :  (x+s-t) 
Test 4 :  (x+s-t+u)  which equals N, as per the given

Now, that we have all the test scores, we can solve for x and finally work on his average score.

(x)+(x+s)+ (x+s-t)+(x+s-t+u)/4

Plugging in the value for x, we get this :
 ((N-s+t-u)+(N+t-u)+(N-u)+N)/4

Upon solving this :  (4N-s+2t-3u)/4 , which is the same as writing  (N)+(3u/4)-(t/2)-(s/4) .   Option B

The_ashura · Answer

the scores will be-
4th test=N
3rd test=N-u
2nd test=N-u+t
1st test=N-u+t-s

summing them and dividing by 4, you will get (4N-s+2t-3u)/4 which is also simplified into option B

Gastón took a certain standardized test four times. His score improved

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Gastón took a certain standardized test four times. His score improved

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards