1. 7.6 C LASSIC P UZZLES IN T WO V ARIABLES Objective: Solve traditional math puzzles in two variables. Standards Addressed: 2.5.8.A: Select the appropriate methods and strategies to solve problems. 2.8.11.H: Select and use an appropriate strategy to solve systems of equations.

2. Ages Puzzles : The key to solving age puzzles is knowing how to represent a person’s age in the future or the past. If a person’s age is x, then 5 years from now the person’s age will be x + 5. Five years ago, the person’s age was x – 5.

3. E X. 1 A.

4. B. J UAN IS 4 YEARS OLDER THAN HER BROTHER. S IX YEARS AGO, SHE WAS TWICE AS OLD AS HE WAS. H OW OLD IS EACH NOW ? J = B + 4 J – 6 = 2 (B – 6) B + 4 – 6 = 2B – 12 B – 2 = 2B – 12 10 = 1 b Brother = 10 and June = 14.

5. Wind and Current Puzzles : Wind and current puzzles have a common idea. When a plane is traveling with the wind (a tailwind), the speed of the wind is added to the speed of the plane with no wind. When traveling against the wind (a headwind), the speed of the wind is subtracted from the plane’s speed with no wind.

6. E X. 2 A. A PLANE LEAVES N EW Y ORK C ITY AND HEADS FOR C HICAGO, WHICH IS 750 MILES AWAY. T HE PLANE, FLYING AGAINST THE WIND, TAKES 2.5 HOURS TO REACH C HICAGO. A FTER REFUELING, THE PLANE RETURNS TO N EW Y ORK C ITY, TRAVELING WITH THE WIND, IN 2 HOURS. F IND THE SPEED OF THE WIND AND THE SPEED OF THE PLANE WITH NO WIND. 2.5 (x – y) = 750 against the wind 2 (x + y) = 750 with the wind x – y = 300 x + y = 375 2x = 675 X = 337.50 and Y = 37.5

7. B. A RIVERBOAT TRAVELING WITH THE CURRENT TAKES 2.5 HOURS TO MAKE A TRIP OF 45 MILES. T HE RETURN TRIP, AGAINST THE CURRENT, TAKES 2.75 HOURS. F IND THE SPEED OF THE CURRENT AND THE SPEED OF THE BOAT IN STILL WATER. 2.75 (x – y) = 45 against the current 2.5 (x + y) = 45 with the current x – y = 16 and 4/11 x + y = 18 2x = 34 and 4/11 X = 17.1818 and Y =.8181

8. Number-Digit Puzzles : In most number-digit problems the strategy is to write the value of a number in expanded form. You can write 52 as 5(10) + 2. If you reverse the digits in 52, you get 25, which you can write as 2(10) + 5.

9. E X. 3 A.

10. Coin Puzzles : Classic coin problems can usually be set up by using a system of equations – one involving the number of coins and the other involving the monetary value of the coins in cent.

11. E X. 4 A.

12. B. I N THE MATH DEPARTMENT WATER - COOLER FUND THERE ARE 48 QUARTERS AND NICKELS IN ALL, WITH A TOTAL VALUE OF $6.00. F IND HOW MANY QUARTERS AND HOW MANY NICKELS ARE IN THE FUND. QuartersNickelsTotal # of coinsQN48 Value in Cents25Q5N600 Q + N = 48 25Q + 5N = 600 -5Q – 5N = -240 25Q + 5N = 600 20Q = 360 Q = 18 N = 30

13. Chemical-Solution Puzzles : Chemical-solution puzzles involve mixing solutions, typically acids, of different strengths. The strengths of the solutions are expressed as percents. The strategy is to keep track of the total amount of acid, in ounces of grams, in the original solutions and in the final solution.

14. E X. 5 A.

15. B. H OW MANY OUNCES OF A 2% ACID SOLUTION SHOULD BE MIXED WITH A 4% ACID SOLUTION TO PRODUCE 20 OUNCES OF A 3.6% ACID SOLUTION ? 1 st 2 nd 3 rd Amt of solutionXY20 oz % of acid2%4%3.6% Amt of acid.02x.04Y.036(20) X + y = 20.02x +.04y =.036(20) -.02x -.02Y = -.4.02x +.04y =.72.02y =.32 y = 16 of 4% X = 4 of 2%