For all real numbers x and y, the operation Θ is defined by the equati

Question

For all real numbers x and y, the operation Θ is defined by the equation x Θ y = ax + by + c, where a, b, c are constants. If 3 Θ 5 = 16 and 4 Θ 7 = 30, then what is the value of 1 Θ 1 = ?

A. -15
B. -12
C. -8
D. 0
E. 12

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kanishaksharma · Accepted Answer

3Θ5 = 3a + 5b + c = 16 .....(1)
4Θ7 = 4a + 7b + c = 30 .....(2)
Subtract (1) from (2) 
we get, a + 2b = 14 .....(3)
1 Θ 1 = a + b + c ......(4)
rewrite (1) as 2a + 4b + a + b + c = 16
28 + a + b + c = 16
a + b + c = 16 - 28 = -12

CrackverbalGMAT · Answer

There are times when we get 2 equation in 3 variables, and we need to get a value of a 3rd equation.

In this case 3 Θ 5 = 3x + 5y + c = 16 ...(1) and

4 Θ 7 = 4x + 7y + c = 30 ...(2)

We need to find the value of 1 Θ 1 = x + y + c

The normal method is to do by hit and trial.

We conduct the mathematical operation 3 * Equation 1 - 2 * Equation 2

3 (3x + 5y + c) - 2(4x + 7y + c) = 3* 16 - 2*30

9x + 15y + 3c - 8x - 14y - 2c = 48 - 60

x + y + c = -12

Option B

Arun Kumar

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=j-SI7M-wXmI

Given that x Θ y = ax + by + c, 3 Θ 5 = 16 and 4 Θ 7 = 30 and we need to find the value of 1 Θ 1 = ?

x Θ y = ax + by + c
=> 3 Θ 5 = a*3 + b*5 + c = 16 ...(1)
=> 4 Θ 7 = a*4 + b*7 + c = 30 ...(2)

(2) - (1) we get 
4a - 3a + 7b - 5b = 301-6 = 14
=> a + 2b = 14 ...(3)

(1) - 2*(3) 
3a + 5b + c - 2a - 4b = 16 - (2*14) = -12
=> a + b + c = -12

1 Θ 1  = a + b + c = -12

So,  Answer will be B 
Hope it helps!

Watch the following video to MASTER Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

Oppenheimer1945 · Answer

3a+5b+c=16
4a+7b+c=30
a+b+c=?

Let a+b+c=(3a+5b+c)+k(4a+7b+c)
3+4k=5+7k=1+k
=>3k=-2; k=-2/3

a+b+c=2(3a+5b+c)-3(4a+7b+c)=3(16)-(2)(30)=48-60=-12

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For all real numbers x and y, the operation Θ is defined by the equati

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