there are 7 children standing in a line ,not all of whom have the same number of cakes with them, if the first child distributes his cakes to the remaining six children such that he doubles their respective no.of cakes ,then he will be left with four cakes. instead ,if the second child takes away 2 cakes from each of the remaining six ,then he will be left with 3 cakes less than the no.of cakes that the 1 child initially had.what is the total no.of cakes that are there with the 3 child to seventh child
Let the number of cakes be a,b,c,d,e,f,g for the 7 children
a-(b+c+d+e+f+g) = 4
b+(2*6) = a-3
we want c+d+e+f+g. Call it x
a-b-x = 4
b+12 = a-3
x = (a-b)-4
a-b = 15
So, x = 11
Well, it seems like these children are really into cake business! Let's break it down.
We know that if the first child distributes his cakes to the remaining six children, he will be left with four cakes. So we can assume that initially, the first child had x cakes (where x is some number).
If he doubles the number of cakes for each of the remaining six children, then each child will get 2x cakes (since their respective number of cakes doubles). And the first child will be left with 4 cakes, so we can set up the equation:
x - 6(2x) = 4
Simplifying this, we get:
x - 12x = 4
-11x = 4
x = -4/11
Hmm, it looks like we've gotten a negative number for the number of cakes the first child initially had. That doesn't seem right, does it? My apologies, it seems there might be a mistake in the information provided. Let's try another approach, using the information about the second child.
According to the second child, if they take away 2 cakes from each of the remaining six children, they will be left with 3 cakes less than the number of cakes the first child initially had. So let's assume the first child originally had y cakes.
If the second child takes away 2 cakes from each of the remaining six children, then each of them will have y - 2 cakes. And the second child will be left with 3 cakes less than the first child initially had, so we can set up the equation:
y - 2 - 3 = y
Simplifying this, we get:
-5 = 0
Wait a minute, that equation doesn't make any sense! It seems like we've encountered another mistake in the information provided.
My apologies for the confusion. It seems like the information is contradictory or incomplete. Without the correct information, I'm afraid I can't determine the total number of cakes with the 3rd to 7th child.
Let's break down the information provided step-by-step:
Step 1: The first child distributes his cakes to the remaining six children, doubling their respective number of cakes, and is left with four cakes.
Let's denote the number of cakes the first child initially had as "x".
After distributing his cakes, the remaining six children each have 2x cakes, and the first child is left with 4 cakes.
Step 2: The second child takes away 2 cakes from each of the remaining six children and is left with 3 cakes less than the number of cakes the first child initially had.
This means that the second child has (x-3) cakes after taking away 2 cakes from each of the remaining six children.
Now, let's find the value of "x" using both the steps:
In step 1:
The remaining six children each have 2x cakes, so the total number of cakes they have is 6 * 2x = 12x.
Adding the 4 cakes that the first child is left with, we have 12x + 4.
In step 2:
The second child has (x-3) cakes.
According to the given information, the second child is left with 3 cakes less than the first child initially had.
So, we have (x-3) = x - 3.
Simplifying further, we get x = x - 3 + 3, which further simplifies to x = x.
Since we haven't obtained a specific value for "x" in step 2, we need to analyze further.
Let's assume that the second child initially had more cakes than the first child.
Since the second child takes away 2 cakes from each of the remaining six children, the number of cakes they have after this step is (2x - 12).
But according to the given information, the second child is left with 3 cakes less than the first child initially had.
So, we have (2x - 12) = x - 3.
Simplifying further, we get x = 9.
So, the initial number of cakes the first child had, denoted by "x", is 9.
Now, let's find the number of cakes the remaining three children (child 3 to child 7) have:
In step 1:
As the first child distributes his cakes, each of the remaining six children now has 2x = 2 * 9 = 18 cakes.
In step 2:
The second child takes away 2 cakes from each of the remaining six children, so they are left with (18 - 2) = 16 cakes each.
The second child, using the same calculation as before, has (x - 3) = 9 - 3 = 6 cakes.
Therefore, the total number of cakes that the third child to the seventh child have is:
5 children * 16 cakes (each) + 6 cakes (child 2) = 86 cakes.
To solve this problem, let's break it down step by step and calculate the number of cakes with each child.
Let's assume the number of cakes with the 1st child is x.
According to the problem, if the 1st child distributes his cakes to the remaining six children in such a way that he doubles their respective number of cakes, he will be left with 4 cakes. This means, each of the six remaining children will have (x/2) + 4 cakes.
So, the number of cakes with the 1st child after distribution = 4.
Now, let's analyze the second part of the problem. It says that if the second child takes away 2 cakes from each of the remaining six children, he will be left with 3 cakes less than the number of cakes that the 1st child initially had.
So, the number of cakes with each of the remaining six children after the second child takes away 2 cakes from them = (x/2) + 4 - 2 = (x/2) + 2.
According to the problem, the second child will be left with 3 cakes less than what the 1st child initially had. So, the number of cakes with the second child = x - 3.
Now, we know that the number of cakes with the second child = (x/2) + 2.
Equating these two expressions, we can solve for x:
x - 3 = (x/2) + 2
Simplifying the equation, we get:
2x - 6 = x + 4
Bringing x terms to one side and constant terms to the other side:
2x - x = 4 + 6
x = 10
Therefore, there are 10 cakes with the 1st child.
Now, let's calculate the number of cakes with the 3rd child to the 7th child:
3rd child = (x/2) + 4 = (10/2) + 4 = 9 cakes
4th child = (x/2) + 4 = (10/2) + 4 = 9 cakes
5th child = (x/2) + 4 = (10/2) + 4 = 9 cakes
6th child = (x/2) + 4 = (10/2) + 4 = 9 cakes
7th child = (x/2) + 4 = (10/2) + 4 = 9 cakes
Therefore, the total number of cakes with the 3rd child to the 7th child is 9 + 9 + 9 + 9 + 9 = 45 cakes.