Bunuel
Tough and Tricky questions: Word Problems.
2 fewer columns of tiles and 4 more rows of tiles. After the change in layout, the patio will still have 70 tiles, and it will still be rectangular. How many rows are in the tile patio before the change in layout?
A. 5
B. 6
C. 10
D. 13
E. 28
Kudos for a correct solution. OFFICIAL SOLUTION:(C) The fastest way to solve this problem is by plugging in the answer choices. If there were 10 rows to to start, then there must have been 7 columns to get 70. If you add 4 to 10 and subtract 2 from 7, you get 5 and 14, which multiplies out to 70.
Another way to solve this problem is by using logic. If a rectangular patio is composed of 70 square tiles, then the original layout, in rows by columns, must be either 70 × 1, 35 × 2, 14 × 5, 10 × 7, 7 × 10, 5 × 14, 2 × 35, or 1 × 70.
Since removing 2 tiles from a row is the same as removing 2 columns, the only way to get a rectangle by removing 2 columns and adding 4 rows is by starting with a 10×7 patio and ending up with a 14×5 patio. So there must be 10 rows.
The correct answer is C.
Alternative Method (Algebraic Solution):This problem can also be solved algebraically. Let R be the number of rows and C be the number of columns in the tile patio before the change in shape. The patio has 70 tiles, so RC = 70. After the change in layout, there will be 4 more rows, and 2 fewer tiles in each row, and the same total of 70 tiles, so we know that:
(R + 4)(C – 2) = 70.
Using FOIL on this equation, we get:
RC – 2R + 4C – 8 = 70.
Since RC = 70, we can substitute into the second equation to get:
70 – 2R + 4C – 8 = 70.
Combining terms, we get:
-R + 2C – 4 = 0.
R = 2C – 4. Let’s plug it in RC = 70.
(2C – 4)C = 70
C² – 2C – 35 = 0
The solutions are C = -5 and C = 7. C must be positive, so C = 7 is the only option. So R = 10. The correct answer is C.