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3 years ago

13

John works in a car dealership selling new and used cars, He is extremely busy during major holidays selling as many cars as pos

sible. The store’s motto is “Work Hard Earn More”. During the last holiday period, John sold $150,000 worth of vehicles. His commission is 15% of the first $100,000 and 10% on the remaining balance. How much commission did he earn on his total sales?

1 answer:

Talja [164]3 years ago

3 0

$20,000

15% of 100,000$ is 15,000$

10% of the remaining 50,000$ is 5,000$

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ANSWER

1. A

2. C

3. D

4. B

5. E

EXPLANATION

QUESTION A

The given system of equation is :

$x-4y=-8...(1)$

and

$2x-3y=-16...(2)$

Make x the subject in equation (1)

This implies that;

$x=4y-8...(3)$

Put equation(3) into equation (2).

$2(4y-8)-3y=-16$

$8y-16-3y=-16$

Group like terms;

$8y-3y=-16+16$

$5y=0$

$y=0$

Put y=0 into equation(3).

$x=4(0)-8$

$x=-8$

(-8,0)

QUESTION B

First equation: $x-y=-2$

Second equation: $3x-y=0$

Subtract the first equation from the second equation.

$3x-x-y+y=0+2$

Simplify;

$2x=2$

$x=1$

Put x=1 into any of the equations, say the first one.

$1-y=-2$

$-y=-2-1$

$-y=-3$

$y=3$

(1,3)

QUESTION C

First equation: $x+y=-4$

Second equation: $-3x+2y=2$

Make y the subject in the first equation;

$y=-x-4$

Put y=-x-4 into the second equation;

$-3x+2(-x-4)=2$

Expand:

$-3x-2x-8=2$

Group like terms:

$-3x-2x=2+8$

$-5x=10$

$x=-2$

Put x=-2 into $y=-x-4$

This implies that;

$y=--2-4$

$y=-2$

(-2,-2)

QUESTION D

First equation:4x+6y=-12

Second equation: 2x+3y=9

Divide the first equation by 2;

New equation: 2x+3y=-6

Subtract the second equation from the new equation;

$2x-2x+3y-3y=-6-9$

$0=-18$ . This statement is not true.

The system has no solution;

QUESTION E

First equation: $x+y=-1$

Second equation: $4x+4y=-4$

Divide the second equation by 4;

New equation: $x+y=-1$

Subtract the second equation from the new equation;

$x-x+y-y=-1+1$

$\Rightarrow 0=0$ This statement is true.

This system has infinitely many solutions.

3 0

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