Tommy invited some friends from school and several cousins to see a movie for his ninth birthday party. Each school fired is 9 years old. Each cousin is 11 years old. None of his cousins attend his school. If the sum of the ages of Tommy, his friends, and his cousins is 122, how many of Tommy's school friends and how many cousins attended the movie?
Please, don't tell me the answer, just tell me how to solve it. I don't want to cheat.
Another:
Find the greatest whole number that meets all the following conditions:
a. It is greater than 100.
b. It is less than 200.
c. It is 20 greater when rounded to the nearest 100 than when rounded to the nearest 10.
Once again, don't tell me the answer, just tell me how to solve it. After all, what would I learn if I cheated? This question might be on my SAT's, and then I wouldn't know how to work it out. So remember, just tell me how to solve it!
Please and Thank You!!
^.^
2007-02-07 10:40:57 · 7 answers · asked by Ironica 2 in Education & Reference ➔ Homework Help
And it's a worksheet, so I can't look in the back of the book.
2007-02-07 10:54:34 · update #1
1) (9s + 9) + 11c = 122
becomes
9s + 11c = 113
s = # students
c= # cousins
The extra nine is Tommy's age
both s and c have to be a real integer (i.e. 1, 2, 3, 4, 5, etc)
Plug in numbers for c until 113 - 11c is a number divisible by 9
2) Because of item c, it has to be greater that 150 since anything less that 150 would round down and cause it to be less when rounded to the nearest 100.
2007-02-07 11:00:11 · answer #1 · answered by mstafa_isu 2 · 0⤊ 0⤋
For the first one:
Tommy = 9, Friends = 9, Cousins = 11, and Total = 122
Tommy's age + Age of x amount of Friends + Age of y amount of Cousins = 122
9 + 9x + 11y = 122
9x + 11y = 113
Without more info you need to create a table so the most amount of cousins are = 113/11 or 10.27 but we can't have a fraction of a person so 10 cousins.
Put 10 into the equation for y and if x ends up a fraction then go to 9 cousins, then 8, 7, 6, 5, 4, 3, 2, 1, or 0 until the answer is whole for x.
For the second one (answer is here since it is just thinking quickly):
We want the greatest number that fits the following:
100 < x < 200 so we know a range of numbers
x = 20 more rounded to the 100s then rounded to the 10s
Since we want the highest we immediately know to start at the top. 190 is rounded to 200 and 190 so 10 more (200 - 190) and 180 is rounded to 200 and 180 so 20 more (200 - 180). thus the answer is between 190 and 180. 185 is rounded to 200 and 190 which is 10 while 184 is 200 and 180 and 20 more. So the answer is 184.
2007-02-07 19:15:58 · answer #2 · answered by jstout 3 · 0⤊ 0⤋
If the combined ages are 122 and assuming Tommy is also nine, subtract nine from 122 until you get a number divisible exactly by 11.
Second one: for it to be 20 greater when rounded to the nearest 100 than to the nearest 10, are you shooting for a number at least 20 greater than 100 or a number at least 20 less than 200? There will be more than one number that can be rounded to the nearest hundred and to the nearest ten and give the difference you want. You want the largest of them.
You can mail me to check your answers or to clarify.
2007-02-07 19:11:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
First question: You need to have variables for th kids who are nine (including tommy) and the kids are are eleven. There may be more than one answer- set the equation with variable equal to 122. YOu are probably going to have to fill out a table with possible answers to the variables. Obviously you answer will need to be whole numbers
Second question:
This is also a thinking question. You can use one variable and the rules of rounding. Parts a and b limit your number. Part c is the important part when rounding. take for example 39- what number does it round to nearest hundreds? what number does it round to for the nearest 10's?
2007-02-07 18:49:26 · answer #4 · answered by Turtle 2 · 0⤊ 0⤋
Thanks for wanting to solve it yourself. So many people just post their homework here.
I only looked at the first one, but you have x children @ 9 yrs old, another x children @ 11 yrs old, and tommy (x).
9x + 11x + x (tommy) = 122
Work it out from there...
2007-02-07 18:51:19 · answer #5 · answered by T J 6 · 0⤊ 1⤋
9x + 11y = 122 (a)
x + 11 = 122 (b)
multiply row b by -9
9x + 11y = 122
-9x -9y = -1098
x cancel out
take away numbers in row b from row a
2y = -976
976/2=488
2007-02-07 18:56:53 · answer #6 · answered by 100%IrishXOX 3 · 0⤊ 2⤋
check the solutions in the back of the book!
2007-02-07 18:46:25 · answer #7 · answered by pro@answering 1 · 0⤊ 0⤋