DAY 394
At an outdoor party, six places are set at each table. There are six chairs at each table, and all the tables and chairs have four legs each. There are a total of 196 legs. How many people are invited to the party?
Answer:
42 ........... Each table and its six chairs = 28 legs (7 x4) ........... 196/28 = 7 tables ........... 7 x 6 = 42
Another way to solve this would be 196/4 = 49. This means there are a combined total of 49 tables and chairs, but we don't know how many of each. Since there are 6 chairs at each table, guessing and checking gives us 7 x 6 = 42 ......... 42 chairs and 7 tables.
Monday, 30 April 2012
DAY 393
At a certain high school, there are twice as many girl graduates as boy graduates. If 3/4 of the girl graduates and 5/6 of the boy graduates go on to college, what fraction of the schools graduates does this represent?
Answer:
7/9 ............. Let x = boy graduates and 2x = girl graduates ........... 3x = total number of graduates. 3/4(2x) + 5/6(x) = 7/3(x) = graduates who go on to college .......... 7/3(x)/3x = 7/9
At a certain high school, there are twice as many girl graduates as boy graduates. If 3/4 of the girl graduates and 5/6 of the boy graduates go on to college, what fraction of the schools graduates does this represent?
Answer:
7/9 ............. Let x = boy graduates and 2x = girl graduates ........... 3x = total number of graduates. 3/4(2x) + 5/6(x) = 7/3(x) = graduates who go on to college .......... 7/3(x)/3x = 7/9
Saturday, 28 April 2012
DAY 392
Before you is a bowl filled with $50, $20, $10, and $5 bills. You are blindfolded and told you may pull out the bills one at a time, but you must stop when you have four bills of the same denomination. What is the most money you could pull out of the bowl before you had to stop?
Answer:
$305 .......... You could pull out three $50's, three $20's, three $10's, three $5's, and another $50. Another way of thinking of it is: $50 + $20 + $10 + $5 = $85 ............... $85 x 3 = $255 ............... $255 + $50 = $305.
Before you is a bowl filled with $50, $20, $10, and $5 bills. You are blindfolded and told you may pull out the bills one at a time, but you must stop when you have four bills of the same denomination. What is the most money you could pull out of the bowl before you had to stop?
Answer:
$305 .......... You could pull out three $50's, three $20's, three $10's, three $5's, and another $50. Another way of thinking of it is: $50 + $20 + $10 + $5 = $85 ............... $85 x 3 = $255 ............... $255 + $50 = $305.
Tuesday, 24 April 2012
DAY 391
Which three consecutive, even numbers, when multiplied together, will give the product 8xxxx2 ? The x's represent four digits, though not necessarily the same ones.
Answer:
884,352
Since 90 cubed is 729,000 and the product of the three numbers is over 800,000, all three numbers must begin with 9. Since they are even, the numbers must be either 90, 92, and 94, or 92, 94, and 96, or 94, 96, and 98. The product must end with 2. Looking at the second digits of 90, 92, and 94 - 0 x 2 x 4 = 0 - ending digit 0. Looking at the second digits of 92, 94, and 96 - 2 x 4 x 6 = 48 - ending digit 8. Looking at the second digits in 94, 96, and 98 ------- 4 x 6 x 8 = 192 - ending digit 2, which is what is required. So 94 x 96 x 98 = 884,352.
Which three consecutive, even numbers, when multiplied together, will give the product 8xxxx2 ? The x's represent four digits, though not necessarily the same ones.
Answer:
884,352
Since 90 cubed is 729,000 and the product of the three numbers is over 800,000, all three numbers must begin with 9. Since they are even, the numbers must be either 90, 92, and 94, or 92, 94, and 96, or 94, 96, and 98. The product must end with 2. Looking at the second digits of 90, 92, and 94 - 0 x 2 x 4 = 0 - ending digit 0. Looking at the second digits of 92, 94, and 96 - 2 x 4 x 6 = 48 - ending digit 8. Looking at the second digits in 94, 96, and 98 ------- 4 x 6 x 8 = 192 - ending digit 2, which is what is required. So 94 x 96 x 98 = 884,352.
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