- Mr. Hermann, a stamp collector , sold 2 stamps for $5000 each. Based on his cost , he made a 25% profit on one stamp and lost 20% on the other stamp. What was Mr. Hermann's total loss on the two transactions?
- There are five chidren in the Roberts family. Amy and Ariel are twins. Brett and Bart are also twins. Amy's age plus Brett's age is 25years. Ariel's age plus Charlene's age is 27 years . Bart's age plus Charlene's age is 20 years. What is the sum of the ages of all 5 children in the Roberts fanily?
return = 5000(1.25) + 5000(.8)
= ...
#2
let Amy and Ariel' age be x
let Brett and Bart's age be y
let Charlene's age be z
x+y = 25
x + z = 27
y+z = 20
2nd minus the third
x - y = 7
add that to the first
2x = 32
x=16
in the 1st:
16+y = 27
y = 11
in the third:
11+z = 20
z = 9
so we have 2x + 2y + z
= 32+22+9 = 63
Amy and Ariel are 13 + 13=26.Brett and Bart are 12+12=23.Charlene is 24
To solve the first question, we need to determine the individual cost price for each stamp. Let's assume the cost price for the stamp on which Mr. Hermann made a 25% profit is x dollars.
Since he made a 25% profit on that stamp, the selling price is 1.25x dollars. As mentioned in the problem statement, he sold one of these stamps for $5000, so we equate the selling price to $5000:
1.25x = 5000
Now, let's solve for x.
Dividing both sides of the equation by 1.25:
x = 5000 / 1.25 = $4000
Therefore, the cost price of the stamp on which Mr. Hermann made a 25% profit is $4000.
Now, let's determine the cost price for the other stamp on which Mr. Hermann incurred a loss of 20%. Let's assume the cost price for this stamp is y dollars.
Since he incurred a loss of 20%, the selling price is 0.8y dollars. As mentioned in the problem statement, he sold one of these stamps for $5000, so we equate the selling price to $5000:
0.8y = 5000
Now, let's solve for y.
Dividing both sides of the equation by 0.8:
y = 5000 / 0.8 = $6250
Therefore, the cost price of the stamp on which Mr. Hermann incurred a 20% loss is $6250.
To find out Mr. Hermann's total loss on the two transactions, we subtract the selling prices from the cost prices and add them together:
Total loss = (x - 5000) + (y - 5000)
= (4000 - 5000) + (6250 - 5000)
= -1000 + 1250
= $250
Therefore, Mr. Hermann's total loss on the two transactions is $250.
Now, let's move on to the second question about the ages of the children in the Roberts family.
Let's assign variables to the ages of the children:
- Let Amy's age be A
- Let Ariel's age be A (since they are twins)
- Let Brett's age be B
- Let Bart's age be B (since they are twins)
- Let Charlene's age be C
According to the problem statement, Amy's age plus Brett's age is 25 years, so we have the equation:
A + B = 25
Similarly, Ariel's age plus Charlene's age is 27 years, so we have the equation:
A + C = 27
Also, Bart's age plus Charlene's age is 20 years, so we have the equation:
B + C = 20
Now, we have three equations with three unknowns (A, B, and C). We can solve this system of equations to find the values of A, B, and C.
By subtracting the second equation from the first equation, we get:
(A + B) - (A + C) = 25 - 27
B - C = -2 (Equation 1)
By subtracting the third equation from the second equation, we get:
(A + C) - (B + C) = 27 - 20
A - B = 7 (Equation 2)
Now, we can solve this system of equations.
From Equation 1, we find that B - C = -2, which means B = C - 2 (Equation 3).
Substituting Equation 3 into Equation 2, we get:
A - (C - 2) = 7
A - C + 2 = 7
A - C = 7 - 2
A - C = 5 (Equation 4)
Now, we can substitute the value of A - C from Equation 4 into Equation 1:
B - C = -2
B = C - 2
B = (A - C) - 2
B = 5 - 2
B = 3
Now, we can substitute the values of A and B into any of the original equations to find the value of C. Let's use Equation 3:
B = C - 2
3 = C - 2
C = 3 + 2
C = 5
Now, we have the values of A, B, and C:
A = 5
B = 3
C = 5
To find the sum of the ages of all 5 children in the Roberts family, we add A, A, B, B, and C:
Sum of ages = 5 + 5 + 3 + 3 + 5
= 21
Therefore, the sum of the ages of all 5 children in the Roberts family is 21 years.