Ada and Paul received their scores on three tests. On the fi

Question

Ada and Paul received their scores on three tests. On the first test, Ada's score was 10 points higher than Paul's score. On the second test, Ada's score was 4 points higher than Paul's score. If Paul 's average (arithmetic mean) score on the three tests was 3 points higher than Ada's average score on the three tests, then Paul's score on the third test was how many points higher than Ada's score?

(A) 9
(B) 14
(C) 17
(D) 23
(E) 25

(A) 9
(B) 14
(C) 17
(D) 23
(E) 25

Bunuel · Accepted Answer

SOLUTION

Ada and Paul received their scores on three tests. On the first test, Ada's score was 10 points higher than Paul's score. On the second test, Ada's score was 4 points higher than Paul's score. If Paul 's average (arithmetic mean) score on the three tests was 3 points higher than Ada's average score on the three tests, then Paul's score on the third test was how many points higher than Ada's score?

(A) 9
(B) 14
(C) 17
(D) 23
(E) 25

(A) 9
(B) 14
(C) 17
(D) 23
(E) 25

Paul 's average score on the three tests was 3 points higher than Ada's average score on the three tests, means that Paul scored 3*3 = 9 points more than Ada.

On the first two tests, Ada scored 10 + 4 = 14 points more than Paul, thus Paul's score on the third test was 9 + 14 = 23 points more than that of Ada's.

Answer: D.

Bigstein09 · Answer

I solved by picking numbers.

1st exam - Ada outscores Paul by 10.

Paul - 90
Ada - 100

2nd exam - Ada outscores Paul by 4

Paul - 90
Ada - 94

3rd Exam - We need to figure out by how many points did Paul outscore Ada.

Paul - 90
Ada - ??

If Paul's average for the 3 exams is 90, then Ada's average must be 87. Since there are 3 exams, in order to average 87 Ada needs 261 total points (87*3). To get Ada's score for the third exam, subtract the total of her first two exams from the total points needed.

261-194=67, therefore:

Paul - 90
Ada - 67

Paul's score is 23 points higher. Answer D.

BrentGMATPrepNow · Answer

Here's a slightly different approach.

Let A, B, C = Ada's 3 test scores respectively
Let X, Y, Z = Paul's 3 test scores respectively

Paul's average score on the three tests was 3 points higher than Ada's average score on the three tests  
In other words, Paul's average score - Ada's average score = 3
Or, we can write: (X+Y+Z)/3 - (A+B+C)/3 = 3
Multiply both sides by 3 to get:   (X + Y + Z) - (A + B + C) = 9

On the first test, Ada's score was 10 points higher than Paul's score.  
We can plug in some nice numbers that satisfy this condition. 
Let's say that A = 10 and X = 0

On the second test, Ada's score was 4 points higher than Paul's score.  
Let's say that B = 4 and Y = 0

When we plug these values into   (X + Y + Z) - (A + B + C) = 9  , we get:
(0 + 0 + Z) - (10 + 4 + C) = 9
Simplify: Z - C - 14 = 9
Simplify: Z - C = 23

Since Z-C represents (Paul's 3rd test score) - (Ada's 3rd test score), we can see that the correct answer is D

Cheers,
Brent

WoundedTiger · Answer

Ada and Paul received their scores on three tests. On the first test, Ada's score was 10 points higher than Paul's score. On the second test, Ada's score was 4 points higher than Paul's score. If Paul 's average (arithmetic mean) score on the three tests was 3 points higher than Ada's average score on the three tests, then Paul's score on the third test was how many points higher than Ada's score?

(A) 9
(B) 14
(C) 17
(D) 23
(E) 25

Let Paul's score in test 1,2 and 3 be x,y and z

So Ada' score in test 1 and 2 will be x+10 ,y+4 . Let Ada's score in 3rd test be a

So as per the question 
(x+10+y+4+a)/3+3= (x+y+z)/3------> (x+y+14+a)/3+3= (x+y+z)/3------> (14+a)/3+3= z/3 or z = (14+a)+9 or z=a+23

Ans D

650 level is okay

Posted from my mobile device

jmyer028 · Answer

For this one, I used Smart Numbers...

Ada's first two test scores = 10 + 4 (sum to 14)

Paul's first two test scores = 0 + 0 (sum to 0)

To keep the math simple, I made the average of Ada's test a multiple of 3... so her last score was a 4, giving an average of 6. For Paul to have an average 3 points higher, his average has to be 9... meaning his last score is 27.

27 - 4 = 23.

Answer is D

XuanPhong · Answer

P(avg) = (T1+T2+T3)/3
P(avg) = 3
T1 = -10
T2 = -4
T3 = ?

3 = (-10 + (-4) + T3)/3

3*3 = -14 + T3
9 = -14 + T3
T3 = 23

PareshGmat · Answer

Ada ........... Paul

10 ................ 0 ........... First Test

4 .................. 0 ............. Second Test

0 ................... x ............... Third Test (Say x is score by Paul in third test)

\frac{14}{3} + 3 = \frac{x}{3}

\frac{23}{3} = \frac{x}{3}

x = 23

Answer = D

sameersanjeev · Answer

Bunuel you are the God of quant man! I looked at the official explanation I was like what the hell! So many variables and equations and then I look at your solution and the problem becomes so so simple!! Thanks a ton! :D

BrainLab · Answer

\frac{y}{3} - \frac{10+4+x}{3}=3

y - x = 23

Answer: D.

Salsanousi · Answer

A1= P1 + 10 
 A2 = P2 + 4

\frac{P1 + P2 + P3}{3}= 3 + \frac{A1 + A2 + A3}{3}

Multiply by 3 in both sides we get
P1 + P2 + P3 = 9 + A1 + A2 + A3

P1 + P2 + P3 = 9 + P1 + 10 + P2 + 4 + A3

P1 and P2 cancel

We are interested in P3 - A3 = 9 + 10 + 4 = 23

Answer choice D

mozerng · Answer

Ada scored 10+4+x Lets look for number that makes the sum devisable by 3 -> 1
so Ada scored 10+4+1=15 -> 15:3 = 5(average score)

Paul scored 0+0+x
We know Paul´s average score was 3 higher -> 5+3=8

8x3=24(Paul´s 3rd score)
Since Ada scored 1 in her 3rd score we know that Paul scored 24-1 higher -> 23 (D)

MathRevolution · Answer

First test: Ada = 10 , Paul : 0

Second test: Ada = 4 , Paul : 0

Overall Paul is 14 points behind.

Average of all three tests: Paul: 3 Ada 0

To make an average of '3' tests higher by 3 points, an increase of 3 * 3 = 9 should be there.

Also, 14 + 9   = 23

Answer D

Bambi2021 · Answer

Number line approach:

X-------------------------------------(-3)--------0------------(+4)---------------(+10)

x needs to make -3 the average of three points.

The distance from -3 is 7 + 13 = 20.

The distance in the other direction must also be 20 from -3 = -23.

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