Hi everyone!
I know that some of you would love an opportunity to post math problems on this blog for others to solve. So, I have come up with an idea to run a marathon game.
Rules:
1. Look at the posted problem. If you know how to do it, solve it and post your solution, not just your answer.
I know that some of you would love an opportunity to post math problems on this blog for others to solve. So, I have come up with an idea to run a marathon game.
Rules:
1. Look at the posted problem. If you know how to do it, solve it and post your solution, not just your answer.
2. After getting a confirmation that your solution is correct, you will be able to post your own problem for others to solve. (Only the first person with the correct solution will be eligible to post another problem.)
3. At any given time, only 1 problem should be "active".
4. The problem you post should be of your own challenge level.
5. The marathon will run in this chain.
Enjoy!!! This marathon may run for a couple of weeks.
3. At any given time, only 1 problem should be "active".
4. The problem you post should be of your own challenge level.
5. The marathon will run in this chain.
Enjoy!!! This marathon may run for a couple of weeks.
Let's start with an easy problem:
Kareena and a friend order one pizza that is half-pepperoni and half-vegetarian. Kareena eats 1/3 of the pepperoni part and 1/4 of the vegetarian part. What fraction of the pizza did Kareena eat? Express your answer as a common fraction.
HAPPY PROBLEM SOLVING!
Vishal, please provide me with your parents' email for weekly updates and parental contact. For some reason, the messages are unable to reach you with the email I have now.
ReplyDeleteWhat Kareena eats:
ReplyDeletePepperoni 1/3*1/2= 1/6
Vegetarian 1/4*1/2= 1/8
1/6+1/8 =
4/24+3/24 =7/24
Kareena ate 7/24 of the pizza.
---------------------------------
If our answer is not correct, can we try it again?
Happy Holidays everyone!
ReplyDeleteGood job, Nidhi. Your answer is correct and your solution is clear.
ReplyDeletePlease post 1 math problem for others to solve.
Well done on taking the lead for the marathon.
The ratio of men to women in Smalltown is 3 : 2. The population of Bigtown is three times as large as that of Smalltown, and the ratio of men to women in Bigtown is 2 : 3. If Smalltown and Bigtown are combined, what is the ratio of men to women in the combination?
ReplyDeleteThanks guys!
Nidhi's problem is now active.
ReplyDeleteEveryone, solve the problem and take lead in the marathon. Let's keep it running!
My mom's email is smodeku@yahoo.com
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteBigtownm = 3 x Smalltown
ReplyDelete2 x 3 = 6
3 x 3 = 9
add them
6( bigtown men) + 3 ( smalltown men) = 9
9( bigtown women) + 2 ( smalltown women) = 11
the ratio is 9:11
Your answer is absolutely correct! Great job showing your solution too!
ReplyDeleteHere is my problem:
ReplyDeleteOf all the coins in Dave's piggy bank, 30% are pennies, 15% are nickels, 25% are dimes and the rest are quarters. If the total value of all the quarters id $60, what is the total value of all the coins in Kelly's piggy bank?
Have fun solving and a happy New Year!
Wait, do you mean Dave's piggy bank? (The last sentence says Kelly's)
ReplyDeleteAm I allowed to post my answer again? (since I already put up a problem)
ReplyDeleteYes Nidhi, I meant Dave's.I will post the problem again.
ReplyDeleteOf all the coins in Dave's piggy bank, 30% are pennies, 15% are nickels, 25% are dimes and the rest are quarters. If the total value of all the quarters is $60, what is the total value of all the coins in Dave's piggy bank?
Sorry for the mistake!Have fun solving
P.S. Nidhi, I think you can answer to my question. It never says you can't in the directions.
Hi guys! I'm having fun in Orlando. You guys are doing a wonderful job!
ReplyDeleteHappy New Year to all of you! Happy Math Marathon to you too! :)
Keep up the good work. Once school starts, we will be preparing and selecting for the MATHCOUNTS school round.
As Vishal mentioned, anyone can solve a problem and then post a new one numerous times, as long as the answer and solution are the first to be verified.
ReplyDeleteHappy New Year Everyone
ReplyDeleteHappy New Year everyone!
ReplyDeleteanswer: $67.10 in Dave's piggy bank
Maya, check over your work. Please post your solution so that we can see your approach.
ReplyDeleteSorry Maya, your answer is wrong.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteSolution:
ReplyDeleteLet the total number of coins be x. $60 in quarters are equal to 240 coins.
# pennies-30/100 of x
#nickles-15/100 of x
# dimes- 25/100 of x
# quarters- 3/10x = 240 (SOLVE TO GET 800 COINS)
Therefore, there are 800 coins in the piggy bank
Then, solve for x in the equation above. Substitute 800 for x and multiply.
answers:
Pennies- $2.40
nickles- $6.00
dimes- $20.00
quarters- $60.00
TOTAL COST: $88.40
Are we the only ones on the blog?
ReplyDeleteLooks like it.
ReplyDeleteMaya, good job. Your second attempt is correct and your solution is very clear.
ReplyDeleteNow you are in lead of the marathon. Please post a problem for others to solve.
Everyone, get active on the blog and have fun running this marathon!
OK. Here's my question:
ReplyDeleteHow many ways are there to rearrange the word MANGO so that the letter N is always the 4th letter?
There are 4! ways
ReplyDelete4 x 3 x 2 x 1 = 24 ways
Correct! Nice solution, too. I didn't want a solution based on a list of possibilities!
ReplyDeleteVishal, you are once again in charge of posting a problem.
ReplyDeleteNice work! All of you are posting problems that are similar in difficulty to MATHCOUNTS school round problems. I'm sure you all will do great on the upcoming test!
Here is my problem:
ReplyDeleteTeddy gave 30 of his stickers to Ronny. He then gave 60% of his remaining stickers to Gary and kept the rest. If Teddy had 32% of his stickers left then, how many stickers did he give away altogether.
P.S. We have a test?
Solution:
ReplyDeleteLet Teddy have x stickers in total.
Teddy gave Ronny-30 stickers
What Teddy has left: (x-30) stickers
Teddy gave Gary-60/100(x-30)
Teddy NOW has 32/100x
EQUATION:
0.6(x-30)+ 0.32x+30=x
0.6x-18.0+.32x+ 30=x
0.6-x + 0.32x+ =-30+18
-0.4x + 0.32x = -12
.08x=12
Therefore, x is equal to 150.
Ronny: 30
Teddy: (150-30) 120
Gary: 60/100 * 120= 72
He gave away 72+30= 102 stickers
ANSWER: 102 stickers
Good job, Maya. I know that you meant to write 32x/100.
ReplyDeleteAnyone who posts a problem, please visit the blog often to confirm the response so that the chain can continue.
Vishal, we will have a MATHCOUNTS school round, where 5 students will be selected to move on to the chapter round at GA Tech.
ReplyDeleteGood job Maya. Thank you for the information Agni.
ReplyDeleteMaya, please post your problem.
ReplyDeleteI hope this post is still continuing. Agni, I was also out of town in Orlando. I couldn't really access the blog through my phone so I couldn't answer the questions. Can you give me some more info. about the MathCounts competitions and tests.
ReplyDeleteHere is the 6th problem for the marathon chain:
ReplyDeleteThe ratio of two numbers is 3 : 7. When 5 is subtracted from each of those numbers, the new ratio is 1 : 3. What is the smaller of those two numbers before 5 is subtracted?
Keep up the good work, guys!
Avanthika, MATHCOUNTS is a nation-wide math competition for middle schoolers. It consists of several rounds and levels. We are beginning with the school round, followed by the chapter round, and then the state round. The top 4 students in each state get selected for the national round.
ReplyDeleteTo find out more about each round in the different levels, visit the MATHCOUNTS website:
www.mathcounts.org
We will be taking the school round sometime in January. Those who advance will take the Chapter Round in February at Georgia Tech.
Maya, for your problem, can the new ratio be equal to 1:3?
ReplyDeleteLet x and y be the variables
ReplyDeletex/y = 3/7
x = 3/7 y
After subtracting 5 from x and y, we get
x-5/y-5 = 1/3
Substitute
(3/7y) - 5 / y - 5 = 1/3
3/7y - 5 = 1/3(y-5)
3/7y - 5 = 1/3y - 5/3
3y/7 - 1/3y = 5 - 5/3
9y/21 - 7y/21 = 15/3 - 5/3
2y/21 = 10/3
y = 10/3 * 21/2
y = 35
3 : 7 = 15 : 35
x = 15
The smaller number is 15
Thnx
Vishal, you are absolutely correct! Go ahead and post your problem.
ReplyDeleteHere is my problem
ReplyDeleteJustin is making snowballs to build a fort. Justin can build 15 snowballs in an hour but 2 snowballs melt every 15 minutes. How long will it take him to build 210 snowballs?
1 hour: +15 - 8 = +7 snowballs
ReplyDelete210/7 = 30 hours
Problem 8
ReplyDeleteHow many square numbers are greater than 1, but smaller than 200, and is a multiple of 3 and 2?
By square numbers, do you mean like the answer of a number squared?
ReplyDeleteA square number is a number that is the square of another integer, as 1 of 1, 4 of 2, 9 of 3, etc.
ReplyDeleteI just went through all the perfect squares to see if they were divisible by 6, which is the least common multiple of 3 and 2.
ReplyDelete4
9
16
25
36
49
64
81
100
121
144
169
196
The only ones divisible by 6 are 36 (6 x 6) and 144 (6 x 24), so
2 numbers
Nice work, Grace. Now you are entitled to post a problem of your choice.
ReplyDeleteProblem 9
ReplyDeleteCandle A can be burnt completely in 3 hours while Candle B can be burnt completely in 4 hours. At what time should the candles be lit simultaneously so that one candle is twice the length of the other after 12 noon?
The candles should be lit at 10 am
ReplyDeleteAgni! Now I can't post a problem!
ReplyDeleteThe candles should be lit at 9:36am. I mainly did trial and error until I found out that it was between 9:30 and 9:45.
ReplyDeleteCandle A takes 3 hours to burn, so if lit at 9am it would burn out exactly at noon. Candle B takes 4 hours so it would need to be lit at 8am to burn out at noon. I call these times "light time for total burn," or "LTFB."
To figure out what fraction of the candle was left, I used this process. When subtracting times, remember that time is a base 60 number system so for example 3:15-1:40=95 minutes.
candle left= (time lit - LTFB) / time taken to burn out completely
So I tried 9:40.
Candle A:
(9:40-9:00)/180 min.
40/180= 2/9
Candle B:
(9:40-8:00)/240 min.
100/240=5/12
I also tried 9:35 and using the same method I discovered Candle A was 28/144 of its original length and Candle B was 57/144. I knew I was very close so I tried 9:36.
Candle A:
(9:36-9:00)/180
36/180=1/5
Candle B:
(9:36-8:00)/240
96/240=2/5
1/5 x 2=2/5, so the answer is 9:36.
Now can I post a problem?
Grace, you did a wonderful job. I thought I would be challenging everybody with this problem!
ReplyDeleteYou've done the problem nicely, but it would be more efficient to solve it with an algebraic equation:
We know that the candle that burns for four hours should be twice as long as the candle that burns for only three hours. We set up an equation, where t represents the time before noon that the candles were lit.
2 × (1 – 1/3(12 – t)) = 1 – (1/4(12 – t))
Simplifying, we have
-6 + 2t/3 = -2 + t/4
5t/12 = 4
5t = 48
t = 9.6 hours ⇒ 9:36
Checking our solution, we have
1/10 + 1/4 + 1/4 = 3/5 ⇒ 2/5
2/15 + 1/3 + 1/3 = 4/5 ⇒ 1/5
2/5 = 2 × 1/5
It works!
Grace, I've been impressed with your work! You are now entitled to post a problem.
ReplyDeleteOnce you post your problem, remember to visit the blog often to confirm any responses to your problem. Keep up with the good work!
I was trying to think of an algebraic equation, but my brain has been really fried from being stuck in my house so long and I just couldn't.
ReplyDeleteIn a car-racing videogame, there are speed boosts every 1/7 of a lap, not including the start or finish line. There are bombs to throw at your opponents every 11/14 of a lap, not including the start or finish line. You must race for 6 laps. How many times you get a speed boost and a bomb simultaneously during the race?
I'll be going to bed somewhere in the next hour or so, so don't expect me to check your answer if you posted late at night.
It will happen 3 times simultaneously.
ReplyDeleteWhen is there going to be a new marathon problem?
ReplyDeleteExplain how you got your answer, Vishal, and maybe you will get to to post the next one.
ReplyDeleteThe equation is 2x/14 = 11y/14
ReplyDeletewhere
x = the number of times you can use speed boosts including start and stop. There are 42 possible speed boosts so x<42.
y = the number of times you can throw bombs. there are 84/11 possible chances so y<84/11 ~ 7 possible times within 6 laps.
The above equation is true at
1) x = 11 and y = 2
2) x = 22 and y = 4
3) x = 33 and y = 6
One way of doing this problem is:
ReplyDeleteLCM(1/7, 11/14) = 22/14
6 ÷ 22/14 = 6 × 7/11 = 42/11
Floor[42/11] = 3
*The floor of a real number is the greatest whole number less than the number itself.
Vishal, go ahead and post your problem for the marathon.
ReplyDeleteHere is my problem:
ReplyDeleteWhat is the sum of the positive multiples of 13 that are less than 300?
Since it is less than 300, I did this.
ReplyDelete300/13 = 23
23 x 13 = 299
Formula for sum of n consecutive number is: n(n + 1)/2
For us nth number is 23.
So, sum = 23 x 12.
Sum of positive multiples of 13 is 13 x 23 x 12 = 3588.
Therefore, the answer to this problem is 3588.
Good job, you can post a problem
ReplyDeleteNeha, I love your logic! However, you could also solve it using another method:
ReplyDeletefloor[300/13] = 23
Thus, we have an arithmetic sequence, where the differnce between consecutive terms is always the same:
13, 26, 39, ... 13 × 22, 13 × 23
The sum of an arithmetic sequence is defined as:
n × (t₁+ t₂)/2
where n represents the number of terms, t₁represents the first term, and t₂ represents the last term.
Plugging in values, we have:
23 × (13 + 299)/2 = 3588.