please if anyone could help me with this
question:
1. A class paid $20 for a cake and $4 per child for one slice of cheese pizza.
they paid $140.
how many children are in the class?
2. Rajie want to divide 24 peanuts, 64 raisins, and 56 apricots evenly into packets (with 0 food/things left over. no i did not copy right i put this in my own words. no i did not use a name i know of, and YES this is a real question).
what is the greatest number of packets he can make?
3. a number has.... remainder 2 when it's divided by 3.
remainder 4 when it's divided into 5.
what is the number?
4. Make as man three digit numbers as you can using the digits 5, 1, and 0 (use each of the digits only once).
which of the numbers are divisible by...
a) 2 c) 10 or....
b) 5 d) 3 ??
PLEASE IF ANYONE COULD HELP?!?!?!? I could really use the help!! A.S.A.P please!!
After subtracting the $20 from the $140 we would know how much we paid for the pizza.
How many slices do we have if each slice costs $4 ?
2. This is like finding the LCM of 24, 64, and 56
At this level of arithmetic, you must have learned how to find that.
e.g.
24 = 2*2*2*3
64 = 2*2*2*2*2*2
56 = 2*2*2*7
3.
numbers which have a remainder of 2 when divided by 3:
2 5 8 11 14 17 20 23 26 29 32 35 ...
numbers which have a remainder of 4 when divided by 5:
4 9 14 19 24 29 34 39 ...
What do you think?
4.
can't use the 0 at the front
number of possible numbers
= 2*2*1
= 4
Gee, there are only 4 of them, why not just list them, and then you can answer the last part.
#1: 20 + 4x = 140
#2: GCD(24,64,56) = 8
#3: I suspect you meant that there is a remainder of 4 when it is divided by 5. If sol, just list the numbers with these two properties:
2,5,8,11,14,17,...
4,9,14,19,...
14 = 4*3+2 = 2*5+4
#4. There are only 4 such numbers: 105,150,501,510
for #2, go with Steve's answer.
I should have called it GCD instead of LCM
Sure! I'll be happy to help you with these questions. Let's go through each question step by step:
1. To find the number of children in the class, we need to set up an equation. The class paid $20 for the cake and $4 per child for one slice of cheese pizza. Since they paid a total of $140, we can write the equation as:
20 + 4x = 140
where x represents the number of children in the class.
To solve this equation, we first subtract 20 from both sides:
4x = 120
Then, we divide both sides by 4:
x = 30
So, there are 30 children in the class.
2. To find the greatest number of packets Rajie can make, we need to find the greatest common divisor (GCD) of the numbers of peanuts, raisins, and apricots. The GCD will represent the maximum number of packets that can be evenly divided.
The numbers given are: 24 peanuts, 64 raisins, and 56 apricots.
To find the GCD, we can list the prime factors of each number and identify the common factors.
24 = 2^3 * 3
64 = 2^6
56 = 2^3 * 7
By looking at the common factors, we see that the greatest power of 2 that appears in all three numbers is 2^3. Therefore, the greatest number of packets Rajie can make is 2^3, which is equal to 8 packets.
3. To find the number with the given remainders, we can use the Chinese remainder theorem. In this case, we know that the number has a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5.
To find the number, we can use the following equation:
x ≡ 2 (mod 3)
x ≡ 4 (mod 5)
By solving this system of congruences, we can find the unique solution for x. In this case, the number is 14.
4. To make three-digit numbers using the digits 5, 1, and 0 (using each digit only once), we can list all possible combinations:
105, 150, 501, 510
Now, let's check which numbers are divisible by the given numbers:
a) To check divisibility by 2, we look at the last digit. In this case, 150, 510, and 510 are divisible by 2.
b) To check divisibility by 5, we look at the last digit. In this case, 150 is divisible by 5.
c) To check divisibility by 10, we look at the last two digits. None of the numbers listed are divisible by 10.
d) To check divisibility by 3, we can sum up the digits and check if the sum is divisible by 3. In this case, 150 and 510 are divisible by 3.
So, the numbers that satisfy the conditions are:
a) 150, 510
b) 150
c) None
d) 150, 510
I hope this helps! If you have any more questions, feel free to ask.