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GMAT Club Olympics 2025 (Day 9): If the positive difference of the

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Bunuel

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If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

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Bunuel

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Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

GMAT Club Official Explanation:

Given that $\sqrt{x} - \sqrt{y} = \sqrt{15 - 10\sqrt{2}}$, we need to find the value of $(x - y)^2$.

Squaring both sides gives:

$x - 2\sqrt{xy} + y = 15 - 10\sqrt{2}$.

Since x and y are integers, we equate rational and irrational parts:

$x + y = 15$ and $2\sqrt{xy} = 10\sqrt{2}$, which implies $xy = 50$.

Now square $x + y = 15$:

$x^2 + 2xy + y^2 = 225$.

Subtract 4xy:

$x^2 - 2xy + y^2 = 225 - 200 = 25$

So, $(x - y)^2 = 25$.

Answer: C.

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Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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. √x2-√y2=√15-10√2 implies √x2=15,√y2=√10√√2 therefore square of the difference is 15-10√2 ,from the choices we take 9, that is the possible solution

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Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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$a^2 - b^2 = \sqrt{(15 - 10[square_root]2})[/square_root]$

squaring both side we get $a^2 + b^2 - 2ab = 15 - 10\sqrt{2}$
Therefore, the sum of $a^2 + b^2 = 15 while ab = 5\sqrt{2}$
Only the value of \sqrt{10} and \sqrt{5} Satisfy these two equations . Hence their positive difference is 5.
and the square of 5 = 25 Hence C option answer.

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10 Jul 2025, 08:15

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positive difference of the square roots of two integers is

√x-√y= $\sqrt{15 - 10\sqrt{2}}$,
square both sides
x+y-2√xy= 15-10√2
we get
x+y = 15 and
√xy= 5√2
xy = 50

(x-y)^2 = (x+y)^2 -4xy
(x-y)^2 = (5√2)^2 -4 * 50
(x-y)^2 = 225-200
(x-y)^2 = 25

OPTION C is correct

Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Let the two numbers be 'x' and 'y', therefore given that,

is √x - √y =√15-10√2, square both sides,

x+y-2√xy = 15 -10√2, comparing terms on both sides we get,

x+y=15 and 2√xy=10√2,

xy must be positive and as their sum is positive, both x and y must be positive,

The above equations hold true only if, x=10,y=5 or x=5 and y=10,

in either case, (x-y)^2 = 25 .... Hence correct answer would be choice C

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So the equation

root of x - root of y =root of (15-10root2)
square both sides and the equation will be
x-2root(xy)+y=15-10root2
Rearrange
x+y-2root(xy)=15-10root2

From this we can understand x+y=15
and 2 root(xy)=10root2
root(xy)=5root2
root(xy)=root5root5root2=root50

So x=10,y=5 (or vice versa) looks good in both the equations
subtract them to get 5 and square it to get 25.

The answer is C which is 25.

Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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$\sqrt{a} - \sqrt{b}$ =$ ((\sqrt{10} - \sqrt{5})^2)^1/2$
a=10
b=5

a-b=10-5
(a-b)^2=25

Ans C

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a^(1/2) -b^(1/2) = (15-10(2^(1/2))^(1/2)
Squaring both sides
a+b-2 (ab)^(1/2) = 15-10(2^(1/2))
Compare both sides
a+b =15 and -2(ab)^(1/2) = -10(2^1/2) ; ab=50

Turn it into an equation
x^2 —15x +50=0
x= (15+5)/2 and x= (15-5)/2
x= 10, 5

a-b = 10-5=5
Square of difference = 5*5=25

C

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root(a) - root(b) = root(15-10*(root(2)))
Square both sides
a + b - 2root(ab) = 15-10*(root(2))
a+b=15
2root(ab) = 10*(root(2))
root(ab) = root(50)
ab=50
a+b=15
Thus a=10 and b=5
So (a-b)^2=25
Answer is C : 25

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Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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We can resolve the equation into a simpler format.

$\sqrt{15 - 10\sqrt{2}}$ => $\sqrt{10 + 5 - 2.\sqrt{50}}$ => $(\sqrt{10} - \sqrt{5})$

Hence, the 2 integers mentioned are 10,5. We need difference square 25.

IMO C

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We begin by designating two integers as X and Y.

√X√Y = √(15-10√2)

Squaring both sides,

X + Y + 2√X√Y = 15 - 10√2

Given that X + Y = 15,
2√X√Y = 10√2, thus XY = 50.

The problem requires finding (X - Y)^2, which equals (X + Y)^2 - 4XY.
Substituting the known values, we get 15^2 - 200 = 25.

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let the two integers be x & y
(x)^1/2-(y)^1/2=(15-10(2)^1/2)^1/2
squaring both sides,
x-y-2(xy)^1/2=15-10(2)^1/2
Since, x & y are integers.
-2(xy)^1/2=-10(2)^1/2
so x-y=15 & square of this would be 225
Option E

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If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

Given:

$\sqrt{a} - \sqrt{b}$ = $\sqrt{15 - 10\sqrt{2}}$

$(\sqrt{a} - \sqrt{b})^2$ = 15 - $ 10\sqrt{2}$

a + b - 2 $ \sqrt{ab} $ = 15 - $ 10\sqrt{2}$

Now, the sum of two integers can only be an integer, so a + b must be 15; a + b = 15

Subsequently, - 2 $ \sqrt{ab} $ = - $ 10\sqrt{2}$

2 $ \sqrt{ab} $ = $ 2\sqrt{2*25}$ = $ 2\sqrt{50}$

ab = 50; a + b = 15; from this we can calculate values of a and b, which would be a = 10 and b = 5;

$ (10-5)^2 $ = 25

A. 9
B. 16
C. 25
D. 100
E. 225

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We are given that the positive difference of their square roots is
√(15−10 √2)
So, ∣ √x − √y ∣= √(15−10√2)

We need to find the square of the difference between these two integers, which is (x−y)^2

Let's simplify the expression √(15−10√2)

This looks like it could be of the form
√{( √a − √b )^2} =∣ √a − √b ∣ = √{a+b − 2√ab}

Comparing 15−10√2 with a+b−2√ab:
We have 2√ab =10√2

=> ab=(5√2)^2
=25×2=50.
And a+b=15.

We need to find two numbers a and b such that their sum is 15 and their product is 50.
By inspection, the numbers are 10 and 5. (10+5=15, 10×5=50).

So,
√{15−10√2} = √{10+5−2√(10×5)} = √{( √10 − √5 )^2} = ∣ 10 − 5 ∣ = √10 − √5 (since √10 > √5 ).

Now we have ∣ √x − √y∣ = √10 − √5

This implies that {√x,√y} must be {√10, √5}.
Therefore, {x,y} must be {10,5}.

We need to find the square of the difference between these two integers, which is (x−y)^2
(x−y)^2 =(10−5)^2 = 5^2 = 25.
Alternatively, (y−x)^2
=(5−10)^2
=(−5)^2 = 25.

The result is 25.

The final answer is 25

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I feel the question could have been written in abetter way , specifically the difference value with proper brackets

Given : sqrt(a) - sqrt(b) = sqrt(15 - 10*sqrt(2))
squaring both sides,
a + b - 2*sqrt(a*b) = 15 - 10*sqrt(2)
If we carefully look, the rhs can also be rewritten as 10 + 5 - 2*sqrt(5*10)
On comparison, a = 10 ; b = 5
So, (a-b)^2 = (10-5)^2 = 25

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Question says:

$(\sqrt{x} - \sqrt{y})$ = $\sqrt{15-10 sqrt(2)}$

$(\sqrt{x} - \sqrt{y})^2$ = $15-10 \sqrt{2}$

Making both sides $(a-b)^2$:
$x+y-2 \sqrt{xy}$ = $15 - 2 \sqrt{25*2}$

So, x+y = 15 and xy=50

We also know, $(x+y)^2 = x^2+y^2+2xy$
$x^2+y^2= (x+y)^2-2xy$
$x^2+y^2= (15)^2-2*50$ =225-100
$x^2+y^2 = 125$

We need to find: $(x-y)^2$ = $x^2 +y^2-2xy$
$(x-y)^2$ = $125-2*50$
$(x-y)^2$ = $125-100$
$(x-y)^2$ = $25$

Hence, Option C) 25

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Let x = the first integer and y = the second integer. Thus:

|√x + √y | = √(15 - 10√2)
Squaring both sides
(√x + √y )^2 = 15 - 10√2
x + y - 2√x√y = 15 - 10√2
Comparing both sides, we have
x + y = 15
2√xy = 10√2 or 2√xy = 2√50
xy = 50

Squaring both sides of x + y = 15
x^2 + y^2 + 2xy = 225
Putting in value of xy = 50 from above
x^2 + y^2 = 225-100 = 125

Now we need (x-y)^2 Or x^2 + y^2 - 2xy
Putting in values from above, x^2 + y^2 = 125 & xy = 50
We have
125 - 2*50 = 25

Ans C

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Bunuel

If the positive difference of the square roots of two integers is $\sqrt{15 - 10\sqrt{2}}$, what is the square of the difference between these two integers?

A. 9
B. 16
C. 25
D. 100
E. 225

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more