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

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

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?

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

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

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

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

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

\(\frac{y}{3} - \frac{10+4+x}{3}=3\)
y-x=23
Answer D

\(A1= P1 + 10\)
\(A2 = P2 + 4\)

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

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

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)

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 [As Paul was 14 points behind] = 23

Answer D

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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