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Sub 505 (Easy)|   Algebra|      
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Bunuel
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 17 Feb 2017, 03:05
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B
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87% (02:09) correct 13% (02:51) wrong based on 179 sessions

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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321
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B
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the Updated on: 17 Feb 2017, 03:58
Originally posted by 0akshay0 on 17 Feb 2017, 03:52.
Last edited by 0akshay0 on 17 Feb 2017, 03:58, edited 1 time in total.
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Bunuel
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321
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Since x and y are consecutive odd integers y =x+2
\(x^2 + y^2 = 394\)
\(x^2 + (x+2)^2=394\)
\(x^2 + x^2 + 4x + 4 =394\)
\(x^2 + 2x - 195 = 0\)

x = {-15,13} and corresponding values for y ={-13,15}
so x*y = 15*13 or (-15)*(-13) = 195

Hence Option B is correct
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 17 Feb 2017, 03:54
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x^2 +(X+2)^2=394
x^+ X^2 +4x + 4 =394
x^ +2x -390 =0
(x- 13)(x-15)= 0
so 13 and 15 are x and y
xy = 195

answer B
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 17 Feb 2017, 04:12
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Answer : [B]

y =x +2

therefore, x^2 + (x+2)^2 = 394
which equates to
(x-13)(x+15) = 0

Either x= 13; then y= 15 ; xy = 195

Or x= -15; then y= -13 ; xy = 195
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 18 Feb 2017, 02:29
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Bunuel
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321
Show more

x, y are consecutive odd integers => y=x+2

\(x^2+y^2=394\)
\(x^2+(x+2)^2=394\)
\(x^2+x^2+4x+4=394\)
\(2x(x+2)=394-4\)
\(x(x+2)=\frac{390}{2}\)
\(x(x+2)=195\)
\(xy=195\)

B.
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 18 Feb 2017, 02:44
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Bunuel
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321
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x = 2k-1
y = 2k+1

=> (2k-1)^2 + (2k+1)^2 = 394
solving, k^2 = 49

so, xy = (2k+1)(2k-1) = (2k)^2 - (1)^2 = 195. B
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 23 Feb 2017, 10:30
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Bunuel
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321
Show more

Since x and y are consecutive odd integers, we can say:

x = y + 2

x - y = 2

Squaring both sides of our equation, we have:

(x - y)^2 = 2^2

x^2 + y^2 - 2xy = 4

x^2 + y^2 = 4 + 2xy

Since x^2 + y^2 = 4 + 2xy, we can substitute 4 + 2xy for x^2 + y^2 in the equation x^2 + y^2 = 394 and we have:

4 + 2xy = 394

2xy = 390

xy = 195 (which is the product of x and y)

Answer: B
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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the 06 Aug 2024, 01:10
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My approach to this question is using number properties,
Since the pairs of two consicutive odd numbers can only be
1 - 3
3 - 5
5 - 7
7 - 9
When the squares of the numbers are added, the last digit has to be 4, The 2 possibilities for the unit digit to be 4 are the combination of 3 and 5 and 5 and 7. Try 13*15, and you will get the answer.
I hope this helps!
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