Do all your work on a separate sheet of paper Leave plenty o

Do all your work on a separate sheet of paper, Leave plenty of space between problems and extremely clear in showing your work. Fold the paper lengthwise with this sheet on the outside. The larger of two numbers is 5 more than twice the smaller. If the smaller is subtracted from the larger, the result is 12. Find the two numbers. A man invests $17,000 in two accounts. One account earns 5% interest per year and the other 6.5%. If his total interest after 1 year is $970, how much did he invest at each rate? How much 25% antifreeze and 50% antifreeze should be combined to give 40 gallons of 30% antifreeze? Use graph paper to graph the solution set. Show how you tested a point to decide on shading. y

Solution

The interest on the 1^st a/c for 1 year is x* 5% = 5x/100 = x/20. The interest on the 2^nd a/c is y*(6.5%) = 6.5y /100 = (13/200)y. Thus, the total interest earned in an year is x/20 + 13y/200. Since we are given that the total interest in an year is $ 970, we have x/20 + 13y/200 = 970 or, on mutiplying both the sides by 200,   10x + 13y = 970* 200 or, 10x + 13y = 194000…(2)

On multiplying both the sides of the 1^st equation by 10, we get 10x + 10 y = 170000 …(3) Now on subtracting the 3^rd equation from the 2^nd equation, we get + 13 y – (10x + 10y) = 194000 – 170000 or, 10x + 13y – 10x – 10 y = 24000 or, 3y = 24000 so that y = 8000. Now, from the 1^st equation, we get x + 8000 = 17000 so that x = 17000 – 8000 = 9000. We can verify this result by substituting these values of x and y in the 2^nd equation.

Thus, the man invests $ 9000 in the account earning 5 % interest and $ 8000 in the account earning 6.5 % interest.

3. Let us assume that x gallons of 25% antifreeze and y gallons of 50% antifreeze are combined to get 40 gallons of 30% antifreeze. Since the total quantity is 40 gallons, we have x + y = 40 …(1)

Also, x gallons of 25 % antifreeze is equivalent to x/4 gallons of 100% antifreeze. Similarly, y gallons of 50 % antifreeze is equivalent to y/2 gallons of 100% anti freeze and 40 gallons of 30 % antifreeze is equivalent to 40*30/100 = 12 gallons of 100% antifreeze. Therefore, we have            x/4 + y/2 = 12. On multiplying both the sides by 4, we get x + 2y = 48 …(2) > now, on subtracting the 1^st equation from the 2^nd equation, we get x + 2y – (x + y) = 48 – 40 or, x + 2y – x – y = 8 or,y = 8. Then from the 1^st equation, we get x + 8 = 40 so that x = 40 – 8 = 32. We can verfy this result by substituting these values of x and y in the 2^nd equation. Thus 32 gallons of 25 % antifrreze has to be combined with 8 gallons of 50 % antifreeze to get 40 gallons of 30 % antifreeze.

6. Let y = ax where a is a constant. Since y = 10 when x = 2, we have 10 = a(2) so that a = 10/2 = 5. Then y = 5x. Now, when x = 6, we have y = 5*6 = 30.

7. Let z = a(y+ x )where a is a constant. Now x = 4 when z = 96 and y = 2. Therefore, 96 = a(2 +4) or, 96 = a (2 + 2) = 4a. Therefore, a = 96/4 = 24. Thus, z = 24( y + x). Now, when z = 108 and y = 1, we have, 108 = 24( 1 +x) or, 1 + x = 108/24 = 4.5 . Therefore, x = 4.5 – 1 = 3.58.

8. Let y = a*1/x² where a is a constant. Since y = 12 when x = 2, we have 12 = a(1/4) or, a = 12 * 4 = 48. Thus y = 48/x². Now, when x = 6, we have y = 48* 1/ 36 = 4/3