Solutions

1. 32 years old

   Brian’s age now = b
   Johan’s age now = j

   Translate the first sentence:

   (1)  j = b + 20

   Translate and simplify the second sentence:

   

   The problem says to solve for b, so combine the two equations by substituting the value for j in equation (1) into equation (2) to eliminate j and solve for b.

   
2. 45 pounds

   Let j = Jackie’s weight, and let s = Stella’s weight. Stella’s weight is the Ultimate Unknown: s = ?

   (1) The two dogs weigh a total of 75 pounds:	(2) Stella weighs 15 pounds less than twice Jackie’s weight:
   j + s = 75	s = 2j − 15

   Combine the two equations by substituting the value for s from equation (2) into equation (1):

   

   Find Stella’s weight by substituting Jackie’s weight into equation (1):

   
3. 45 double burgers

   Let s = the number of single burgers purchased.
   Let d = the number of double burgers purchased.

   (1) Caleb bought 50 burgers:	(2) Caleb spent $72.50 in all:
   s + d = 50	s + 1.5d = 72.50

   Combine the two equations through substitution or by subtracting equation (1) from equation (2).

   
4. 40 minutes

   Let x = the number of minutes.
   A call made by United Telephone costs $10.00 plus $0.25 per minute: 10 + 0.25x.
   A call made by Atlantic Call costs $12.00 plus $0.20 per minute: 12 + 0.20x.

   Set the expressions equal to each other:

   
5. 52 inches

   Let x = the 1st piece of ribbon.
   Let y = the 2nd piece of ribbon.

   (1) The ribbon is 70 inches long:	(2) The 1st piece is 5 inches more than 1/4 as long as the 2nd:
   x + y = 70

   Combine the equations by substituting the value of x from equation (2) into equation (1):

   

   Now, because x + y = 70, x = 18. Thus, x < y, so y is the answer.

6. 23

   You are given actual ages for Jayla, therefore, the easiest way to solve the problem is to think about the extreme scenarios. At one extreme, 18-year-old Jayla could have babysat a child of age 9. Because Jayla is now 32, that child would now be 23. At the other extreme, 22-year-old Jane could have babysat a child of age 11. If Jayla is now 32, that child would be 21. You can see that the first scenario yields the oldest possible current age, 23, of a child that Jayla babysat.

7. (A)

   Let A and B denote Aubrey and Brian’s ages today. Then, their ages 10 years ago would be given by A − 10 and B − 10, respectively. Those ages are related by the problem statement as:

   B − 10 = 2(A − 10)

   Expanding and simplifying yields:

   

   Rewrite the quantities in terms of A and B. Twice Aubrey’s age today is 2A and Brian’s age today is B.

   Ten years ago, Brian was twice as
   old as Aubrey.

   Quantity A	Quantity B
   Twice Aubrey’s age today = 2A	Brian’s age today = B

   According to the equation, B is 10 less than 2A. Therefore, Quantity A is greater.

8. (C)

   Let L and W stand for the length and width of the room in feet. Then, from the first relation, you can write this equation:

   (1) L = W + 8

   Moreover, the area of a rectangle is given by length times width, such that:

   (2) LW = 240

   Taken together, you have two equations with two unknowns, and because the question involves the width rather than the length, you can eliminate the length by substituting from equation (1) into equation (2):

   (W + 8)W = 240

   Now expand the product and move everything to the left-hand side, so that you can solve the quadratic equation by factoring it. This gives:

   

   The two solutions are W = −20 and W = 12. A negative width does not make sense, so W must equal 12 feet.

   It is also possible to arrive at the answer by testing the value in Quantity B as the width of the room. Plug in 12 for W in equation (1):

   L = 12 + 8 = 20 feet

   If W = 12 and L = 20, then the area is (20)(12), which equals 240 square feet. Because this agrees with the given fact, you may conclude that 12 feet is indeed the width of the room.

   Either method arrives at the conclusion. Therefore, the two quantities are equal.

9. (B)

   The simplest method for solving a problem like this is to work backwards from the value in Quantity B. Suppose Jaden sold exactly 90 cars. Then, because he met or surpassed his two car minimum each month (which adds up to 24 cars in the entire year), he would have sold another 90 − 24, which equals 66 cars above the minimum.

   The commission he earned on those cars is calculated as follows:

   $500 × 66 = $33,000

   This would put his total yearly income at $30,000 (base salary) + $33,000 (commission), which sums to $63,000. However, you know that Jaden actually earned less than that; therefore, he must have sold fewer than 90 cars.

   Quantity A	Quantity B
   The number of cars Jaden sold last year = less than 90	90

   Therefore, Quantity B is greater.

   The alternative approach is to translate Jaden’s total earnings into an algebraic expression. Suppose Jaden sold N cars. Once again, noting that he met or surpassed his monthly minimum sales, you would need to subtract 24 cars that do not contribute to his bonus from this total, and then solve for N as follows: