What is the value of 2x + 2y ?

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Re: What is the value of 2x + 2y ?

by Brent@GMATPrepNow » Sun Jun 14, 2020 7:22 am

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BTGModeratorVI wrote:
Sat Jun 13, 2020 5:34 am
What is the value of 2x + 2y ?

(1) 3x + 5y = 60
(2) 5x + 3y = 68

Answer: C
Source: GMAT prep
Target question: What is the value of 2x + 2y ?

Statement 1: 3x + 5y = 60
Since there are infinitely many solutions to the equation 3x + 5y = 60, it seems likely that statement 1 it's not sufficient.
To verify this, let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two cases:
Case a: x = 20 and y = 0. In this case, the answer to the target question is 2x + 2y = 2(20) + 2(0) = 40
Case b: x = 0 and y = 12. In this case, the answer to the target question is 2x + 2y = 2(0) + 2(12) = 24
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 5x + 3y = 68
There are several values of x and y that satisfy statement 2. Here are two cases:
Case a: x = 0 and y = 20. In this case, the answer to the target question is 2x + 2y = 2(0) + 2(20) = 40
Case b: x = 12 and y = 0. In this case, the answer to the target question is 2x + 2y = 2(12) + 2(0) = 24
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that 3x + 5y = 60
Statement 2 tells us that 5x + 3y = 68
If we ADD the two equations we get: 8x + 8y = 128
Divide both sides of the equation by 4 to get: 2x + 2y = 32
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Re: What is the value of 2x + 2y ?

by deloitte247 » Sun Jun 14, 2020 10:59 am

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$$Statement\ 1\ =>\ 3x\ +\ 5y\ =\ 60$$
$$\frac{3x}{3}=\frac{60-5y}{3}$$
$$x=\frac{60-5}{3}$$
The value of x and y remains unknown. Statement 1 is NOT SUFFICIENT

$$Statement\ 2\ =>\ 5x\ +\ 3y\ =\ 68$$
$$\frac{3y}{3}=\frac{68-5x}{3}$$
The value of x and y remains unknown. Statement 2 is NOT SUFFICIENT

Combining both statements together =>
$$3x+5y=60.......eqn\ 1$$
$$5x+3y=68.......eqn\ 2$$
$$from\ eqn\ 1;\ \ x=\frac{60-5y}{3}$$
$$substituting\ x\ in\ eqn\ 2$$
$$5\left(\frac{60-5y}{3}\right)+3y=68$$
$$\frac{300-25y}{3}+3y=68$$
$$\frac{300-25y+9y}{3}=68$$
$$300-25y+9y=204$$
$$-16y=204-300$$
$$\frac{-16y}{-16}=\frac{-96}{-16}$$
$$y=6$$
$$then\ from\ equation\ 1\ substitute\ y$$
$$3x+5\left(6\right)=60$$
$$\frac{3x}{3}=\frac{60-30}{3}=\frac{30}{3}$$
$$x=10$$
$$\sin ce\ y=6\ and\ x=10$$
$$2x+2y=>2\left(10\right)+2\left(6\right)=>20+12=32$$

Both statements combined together ARE SUFFICIENT
Answer = C