DAY 407
Tom and Tina both have some money (in dollars). They each want to buy a gift card, but neither of them has enough money. Tina needs $2 more, and Tom needs $24 more. They decide to combine their money and buy one gift card, but they still don't have enough. How much is the gift card?
Answer:
$25 .................... If Tina only needs $2 more, and they still don't have enough money after Tom gives her what he has, Tom must only have $1.
Saturday, 8 December 2012
Thursday, 25 October 2012
DAY 406
Ten students take a test with a maximum score of 100. The average of the ten scores is 92. What is the lowest score a student could have received?
Answer:
20 ................. For the average to be 92, the ten students must have scored a total of 920 points (920/10 = 92). This is 80 points short of the 1,000 possible points. 1000 - 80 = 20.
Ten students take a test with a maximum score of 100. The average of the ten scores is 92. What is the lowest score a student could have received?
Answer:
20 ................. For the average to be 92, the ten students must have scored a total of 920 points (920/10 = 92). This is 80 points short of the 1,000 possible points. 1000 - 80 = 20.
Thursday, 18 October 2012
Tuesday, 18 September 2012
Friday, 10 August 2012
DAY 403
Fred says to Jack, "I'll bet you $1 that if you give me $2, I'll give you back $3." Should Jack take the bet?
Answer:
No ........... If Jack takes the bet, he gives Fred $2. Fred can then say, "You win. I lose the bet." He then gives Jack $1 - the amount of the bet - but keeps Jack's other dollar.
Friday, 20 July 2012
DAY 402
If 3 cats can catch 3 mice in 3 minutes, how long would it take the 3 cats to catch 100 mice?
Answer:
102 minutes ........... If 3 cats can catch 3 mice in 3 minutes, it takes each cat 3 minutes to catch 1 mouse.
1 cat - 1 mouse - 3 minutes
3 cats - 3 mice - 3 minutes
3 cats - 6 mice - 6 minutes
3 cats - 9 mice - 9 minutes
.
.
.
.
3 cats 99 mice - 99 minutes
(Every 3 minutes, one mouse is caught by each cat. In 99 minutes, each cat catches 33 mice ........ 3 x 33 = 99)
To catch the 100th mouse takes another 3 minutes ........ 99 + 3 = 102 minutes
OR
100 mice/3 cats = 33 - the number of mice caught by each cat
( + remainder 1 mouse)
Since it takes 3 minutes for each cat to catch one mouse,
3 x 33 =99 minutes
+ 3 minutes to catch the remaining mouse = 102 minutes
Wednesday, 20 June 2012
DAY 400
Three children, Peter, Jane, and Eric share a bag of peanuts. There are 770 peanuts in the bag. For every 4 peanuts Peter took, Jane took 3. For every 6 peanuts Peter took, Eric took 7. How many peanuts did Peter get?
Answer:
264 ........... For every 12 peanuts Peter took, Jane took 9 and Eric took 14. This totals 35 peanuts for each round ... 770/35 = 22 rounds ... 12 x 22 = 264
DAY 398
A postman delivers one letter on his first day of work, three letters on his second day of work, five letters on the third day, and so on - delivering two letters more each day than he delivered the previous day. How many total letters has he delivered after fifteen days of work?
Answer:
225
DAY ................ NUMBER OF LETTERS DELIVERED
1 .................................................... 1
2 ................................................... 3
3 ................................................... 5
4 ................................................... 7
5 ................................................... 9
6 .................................................. 11
The total number of letters delivered is the square of the number of days worked. For example, after 4 days, he has delivered 1 + 3 + 5 + 7 = 16 letters.
Saturday, 19 May 2012
Tuesday, 15 May 2012
Tuesday, 1 May 2012
Monday, 30 April 2012
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.
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.
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.
Wednesday, 4 April 2012
Sunday, 18 March 2012
Saturday, 17 March 2012
DAY 388
What would be the product of this multiplication problem?
(x - a)(x - b)(x - c)(x - d) ................ (x - z)
HINT:
You don't need to know the numerical value of any of the variables.
Answer:
Zero ......... The 24th term (x -x) would be zero, thus making the product zero.
(Credit: Andrew Robinson)
Wednesday, 7 March 2012
DAY 387
You have driven the first 10 miles of a 20 mile trip at an average speed of 25 miles per hour. How fast would you have to go in the second half of your trip to average 40 miles per hour over the entire distance?
Answer:
100 miles per hour .......... Using the formula D = RT, it has taken you 24 minutes to drive the first ten miles (10 = 25 T, T = 10/25 or 2/5 hr. = 2/5 x 60 = 24). To average 40 mph over 20 miles would take 30 minutes (20 = 40 T). Therefore you have 6 minutes to drive the remaining 10 miles. 6 minutes is 1/10 of an hour. So 10 = R 1/10, R =100.
Monday, 20 February 2012
DAY 385
Tom: Give me four and we'll each have the same.
Jane: Give me four and I'll have twice as many as you.
How many does each have?
Answer:
Tom has 20 and Jane has 28 ............ If gaining four will make Tom and Jane equal, Jane must have 8 more than Tom.
Let x = Tom
Let x + 8 = Jane
2(x-4) = x + 12
2x - 8 = x + 12
x = 20
x + 8 = 28
OR
Let x = Jane
Let x - 8 = Tom
2(x-12) = x + 4
2x - 24 = x + 4
x = 28
x - 8 = 20
DAY 383
Three men eat three steaks in three days. How many steaks will six men eat in six days?
Answer:
12 ......... If three men can eat three steaks in three days, each man can eat one steak in three days .......... or two steaks in six days. Six men could eat six times that number of steaks in six days, or 12 steaks.
Thursday, 2 February 2012
DAY 382
Consider the digits 1 - 9. They can be arranged in 9! ways. If you were to write down all of these possible arrangements on individual cards and then pick one, what is the possibility that the number you picked would be prime?
Answer:
Zero .......... In any nine-digit number using each of the digits once, the sum of the digits would be 45 (1+2+3+4+5+6+7+8+9 = 45). According to the rules of divisibility, any number in which the digits add up to a multiple of 9, is divisible by 9, and therefore not prime. Since 45 is a multiple of 9, the nine-digit number could not be prime.
Monday, 30 January 2012
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