Mr ram has five sons ajay,balu,charan,easwar and ganesh. He had some chocolates with him which he distributed in the following manner. Twice the number of chocolates received by ajay, thrice the number of chocolates received by charan and four times the number of chocolates received by easwar are equal. Five times the number of chocolates received by balu, six times the number of chocolates received by easwar and 8 times the number of chocolates received by ganesh are equal. find the minimum number of chocolates that could have been distributed by Mr.Ram
Please help me solve this.
don't be put off by all the words. Just write each fact down as math.
You have several amounts: a,b,c,e,g.
2a = 3c = 4e
5b = 6e = 8g
Since the LCM(2,3,4) = 12,
2a,3c,4e must be a multiple of 12
Since LCM(5,6,8) = 120,
5b,6e,8g must be a multiple of 120
Now, if 4e is a multiple of 12, e >= 3. if 6e is a multiple of 120, we must have e >= 20.
So, e must be 60. Given that, 4e=240 and 6e=360
2a = 3c = 4e = 240, so
a=120 and c=80
5b = 6e = 8g = 360, so
b=72 and g=45
a+b+c+e+g = 377
Hmm, let me put on my thinking cap! 🤔
First, let's assign variables to the number of chocolates received by each son:
ajay = A
balu = B
charan = C
easwar = E
ganesh = G
According to the problem, the following equations can be formed:
2A = 3C = 4E ---(1)
5B = 6E = 8G ---(2)
To find the minimum number of chocolates distributed, we need to find the common multiples for each set of variables (A, C, E) and (B, E, G).
For the first set (A, C, E):
The least common multiple (LCM) of 2, 3, and 4 is 12.
So, A = 6, C = 4, E = 3.
For the second set (B, E, G):
The LCM of 5, 6, and 8 is 120.
So, B = 24, E = 20, G = 15.
Now, we can find the minimum number of chocolates distributed by adding up all the variables:
A + B + C + E + G = 6 + 24 + 4 + 3 + 15 = 52.
Therefore, the minimum number of chocolates that Mr. Ram could have distributed is 52.
Hope that solves the puzzle with a sweet answer! 🍫😄
Let's solve this step-by-step:
Step 1: Assign variables to the number of chocolates received by each son.
Let's say,
Ajay received A chocolates.
Balu received B chocolates.
Charan received C chocolates.
Easwar received E chocolates.
Ganesh received G chocolates.
Step 2: Set up the given conditions mathematically.
According to the first condition:
2A = 3C = 4E
According to the second condition:
5B = 6E = 8G
Step 3: Simplify the conditions.
From the first condition:
A = (3C) / 2
E = (2A) / 4
E = (2 * (3C/2)) / 4
E = (3C) / 4
Simplifying it further,
A = (3C) / 2
E = (3C) / 4
From the second condition:
B = (6E) / 5
G = (6E) / 8
Simplifying it further,
B = (18C) / 20
G = (27C) / 16
Step 4: Find a common value for all variables.
To find a common value, we need to find the least common multiple (LCM) of all the coefficients.
LCM(2, 3, 4) = 12
LCM(5, 6, 8) = 120
Step 5: Substitute the common value into the equations.
Substituting 12 for A, C, and E:
A = (3C) / 2 becomes 12 = (3C) / 2
12 = (3C) / 2, multiplied by 2 on both sides gives us 24 = 3C
24/3 = C, so C = 8
Substituting 120 for B, E, and G:
B = (18C) / 20 becomes 120 = (18 * 8) / 20
120 = 144 / 20, multiplied by 20 on both sides gives us 2400 = 144
This equation is not valid, so we need to find another solution.
Let's try a different value for C:
C = 40
Substituting 40 for C:
A = (3C) / 2 becomes A = (120) / 2, so A = 60
E = (3C) / 4 becomes E = (120) / 4, so E = 30
Substituting 40, 30, and 120 into the equation for G:
G = (27C) / 16 becomes G = (27 * 40) / 16
G = 1080 / 16, simplified gives G = 67.5
Since the number of chocolates must be a whole number, G cannot be 67.5.
Therefore, there is no valid solution for the given conditions.
To solve this problem, we need to consider the information given and create equations based on the relationships between the number of chocolates received by each son.
Let's assume the number of chocolates received by Ajay, Balu, Charan, Easwar, and Ganesh are represented by variables 'a', 'b', 'c', 'e', and 'g' respectively.
Based on the given conditions, we can create the following equations:
1) Twice the number of chocolates received by Ajay, thrice the number of chocolates received by Charan, and four times the number of chocolates received by Easwar are equal:
2a = 3c
=> 2a - 3c = 0
2) Five times the number of chocolates received by Balu, six times the number of chocolates received by Easwar, and eight times the number of chocolates received by Ganesh are equal:
5b = 6e = 8g
=> 5b - 6e = 0
=> 6e - 8g = 0
Now, we have a system of equations that we can solve simultaneously to find the values of 'a', 'b', 'c', 'e', and 'g'.
To do this, we can use the method of substitution. Let's solve equation 2a - 3c = 0 for 'a':
2a = 3c
=> a = (3c) / 2
Now, substitute this value of 'a' into the equation 5b - 6e = 0:
5b - 6e = 0
=> 5b - 6e = 0
=> 5b - 6e = 0
=> 5b - 6e = 0
=> 5b - 6e = 0
Simplify this equation:
5b - 6e = 0
=> b = (6e) / 5
Substitute the value of 'b' into the equation 6e - 8g = 0:
6e - 8g = 0
=> 6e - 8g = 0
Now, we have two independent equations:
a = (3c) / 2
b = (6e) / 5
Next, we need to find the minimum number of chocolates that could have been distributed by Mr. Ram. To do this, we will assume the minimum values for 'c' and 'e' and calculate the corresponding values of 'a', 'b', and 'g'.
Let's assume 'c' = 2 and 'e' = 5.
Substituting these values into the equations, we get:
a = (3 * 2) / 2
=> a = 3
b = (6 * 5) / 5
=> b = 6
Finally, we calculate the corresponding value of 'g' using b = (6e) / 5:
b = (6 * 5) / 5
=> 6 = 6
So, the minimum number of chocolates that could have been distributed by Mr. Ram is:
a + b + c + e + g
= 3 + 6 + 2 + 5 + 6
= 22
Therefore, Mr. Ram could have distributed a minimum of 22 chocolates to his sons.