MHB Solve for J and B: Junhao & Bala Stamps Problem

  • Thread starter Thread starter Johnx1
  • Start date Start date
  • Tags Tags
    Fractions
Click For Summary
Junhao and Bala collect stamps, with the relationship between their collections defined by two equations. Junhao has 76 more stamps than Bala, leading to the equation J = 76 + B. Additionally, 1/3 of Junhao's stamps equals 3/5 of Bala's, resulting in J = (9/5)B. By substituting the first equation into the second, both J and B can be solved. This method clarifies the approach to finding the number of stamps each collector has.
Johnx1
Messages
48
Reaction score
0
Junhao and Bala both collect stamps. 1/3 of Junhao's stamps is equal to 3/5 of Bala's stamps. Junhao has 76 more stamps than Bala. How many stamps does each of them have?

My answer:

Number of stamps Junhao have =J
Number of stamps Bala have = B

We know that Junhao has 76 more stamps than Bala => Junhao = 76 + Bala.

I'm not sure if this is a correct way to do it, but we know 1/3 of Junao's stamp equal to 3/5 of Bala's stamps.
so => (1/3)J = (3/5)BI got stuck and couldn't figure out a clear way to do this.
 
Mathematics news on Phys.org
Johnx said:
Number of stamps Junhao have =J
Number of stamps Bala have = B

We know that Junhao has 76 more stamps than Bala => Junhao = 76 + Bala.

Why don't you change the equation to following using J and B

$$J = 76 + B$$

and you already have the following equation,

$$ \frac{J}{3} = \frac{3B}{5} \implies J = \frac{9B}{5}$$

now you have two unknown variables and two equations. So you can find the values for J and B by substituting the J on the first equation from the second
 
BAdhi said:
$$ \frac{J}{3} = \frac{3B}{5} \implies J = \frac{9B}{5}$$
Thank you for this. I didn't see it this way.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K