If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then

Question

If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200

Bunuel · Accepted Answer

Square k + m = 20;

(k + m)^2 = 20^2;

k^2 + 2km + m^2 = 400.

Since given that k^2 + m^2 = 289, then 289 + 2km = 400.

km = 111/2 = 55.5.

Answer: A.

P.S. Is this question really from GMAT Prep?

BrentGMATPrepNow · Answer

Bunuel's solution is perfect. 
I just want to point out that there's no shortage of questions that test the concepts tested here.

Important expansions:   (x + y)² = x² + 2xy + y²   and   (x - y)² = x² - 2xy + y²   
So, if you see questions that include information about x² + y² and 2xy (or just xy), then you should consider their relationship with the above expansions.

Cheers,
Brent

mvictor · Answer

clearly, I need to brush up my quant skills...I took into consideration that k and m are integers...damn!

Abhishek009 · Answer

( k + m ) ^2= k^2 + m^2 + 2km

So,  ( 20 ) ^2= 289 + 2km

Or,  400 = 289 + 2km

Or,  2km = 111

Or,  km = 55.5

Thus, km must be between 20 and 60, answer must be option (A) between 20 and 60

chrtpmdr · Answer

Different year - same problem.

It's funny how the test makers can correctly measure the probability of test takers to fall into certain traps. 
Because after you see it the light bulb goes on, but when you actually in the "zone" and calculating you don't even look left or right and watch for signs.

I did not even consider that the numbers might not be integers. Beginner mistake!

BrentGMATPrepNow · Answer

GIVEN: k + m = 20 and  k² + m² = 289

Take:  k + m = 20 
Square both sides to get: (k + m)² = 20²
Expand and simplify the left side: k² + 2km + m² = 400
Rewrite as: (  k² + m²  ) + 2km= 400
Substitute to get: (  289  ) + 2km= 400
Subtract 289 from both sides to get: 2km = 111
Divide both sides by 2 to get: km = 55.5

Answer: A

Cheers,
Brent

ScottTargetTestPrep · Answer

Solution:

Since k + m = 20, we have:

(k + m)^2 = 20^2

k^2 + m^2 + 2km = 400

Since k^2 + m^2 = 289, we have:

289 + 2km = 400

2km = 111

km = 55.5

Answer: A

hiranmay · Answer

If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60 -->   correct :  k^2 + m^2 = 289  & k + m = 20 =>   (k + m)^2 = (20)^2 => k^2+m^2+2km = 400 => 289+2km=400=> 2km=111 => km = 55.5 , 55.5 is between 20 and 60, so km is between 20 and 60 
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200

MathRevolution · Answer

k + m = 20

On squaring both the sides, we get

(k + m )^2  = 400

On expanding,  k^2  +  m^2  + 2km = 400

Given:  k^2  +  m^2  = 289

=> 289 + 2km = 400

=> 2km = 400-289 = 111

=> km =  \frac{111 }{ 2}  = 55.5

It lies between 20 and 60.

Answer A

Kinshook · Answer

Asked: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(k + m)^2 = 400 = k^2 + m^2 + 2km = 289 + 2km
2km = 400 - 289 = 111
km = 55.55

IMO A

800GMAT2019 · Answer

How come this is not solvable by this manner?
(20-M)^2 + M^2 = 289
...
2M^2-40M+111 = 0 
Factoring this into form (2M... ) (M....)

Or this method?

Plugging this into Quadratic Equation ----> 2M^2-40M+111 = 0

The crucial step of converting k + m = 20 into this --> (k + m)^2 = 20^2; never occurred to me. How would I know next time NOT to take the two steps I took above and instead identify this key step which makes the numbers work out nicely? Perhaps there was a clue in the problem I should have identified? Thank you!

PS91 · Answer

k+m = 20 
(k+m)^2 = 400
k^2 + m^2 + 2km = 400 
2km = 111
km = 55.5 
20<55.5<60 
So, A

Basshead · Answer

Given: k + m = 20 and k^2 + m^2 = 289, what is km?

Notice  289 is the square of 17 ; thus, it's not possible for k and m to be integers.

Square k + m = 20 to get k^2 + 2km + m^2 = 400.

We're given k^2 + m^2 is 289:

400 - 289 = 111

2km = 111
km = 55.5

Answer is A.

TheBipedalHorse · Answer

k and m are numbers such that k + m = 20 and k^2 + m^2 = 289

Helpful to know - 289 is basically 17^2

k + m = 20
square both sides
k^2 + m^2 + 2km = 20^2
or
17^2 + 2km = 20^2
or
KM = (0.5)(20^2 - 17^2) =  (0.5)(37)(3)

egmat · Answer

Hey there! I see you're working through this algebraic problem involving the sum and sum of squares. Let's tackle this together - it's actually quite elegant once you see the key insight.

Here's how to approach this problem:

Let me show you the strategic way to think about this. When you have two numbers with a known sum and sum of squares, there's a beautiful algebraic relationship we can use.

Step 1: Recognize what we have 
We're given:
-  k + m = 20 
-  k^2 + m^2 = 289 
- We need to find  km

Step 2: Apply the key insight 
Notice how if we square the sum  (k + m) , we get something useful. When you expand  (k + m)^2 , you get:
 (k + m)^2 = k^2 + 2km + m^2

This can be rearranged as:
 (k + m)^2 = k^2 + m^2 + 2km

Step 3: Substitute and solve 
Now let's plug in what we know:
-  (k + m)^2 = 20^2 = 400 
-  k^2 + m^2 = 289

So:  400 = 289 + 2km

Solving for  km :
 2km = 400 - 289 = 111 
 km = 55.5

Step 4: Find the right range 
Since  km = 55.5 , and  20 < 55.5 < 60 , the answer is  (A) between 20 and 60 .

The beauty of this approach is that you never need to find the individual values of k and m - the product emerges directly from the given information!

You can check out the  step-by-step solution on Neuron by e-GMAT  to master this algebraic pattern systematically and learn how to recognize when this technique applies to other symmetric function problems. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here , where you'll find comprehensive explanations and practice quizzes to build consistent accuracy with these algebraic manipulation questions.

If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then

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