The Tomato And The Bean

R

RainyScotland

New member
Joined
May 16, 2009
Messages
4
Tom's Dad sowed some tomato seed in February. He gave Tom one of the tomato plants in a pot.

At the beginning of May Tom put his tomato plant outside. On the same day he sowed a bean in another pot.

Ten days later the bean plant was just 1 cm (centimetre) above the soil surface. Tom measured his tomato plant which was already 38 cm tall.

Each evening Tom measured his two plants.

On the evening of the next day the little bean plant had grown another 2 cm so it was 3 cm high. Each day it continued to grow double the amount it had grown the day before.

The tomato plant grew at a steady 5 cm a day.


After how many days were the two plants the same height when Tom measured them in the evening? How high were they?
 
RainyScotland said:
Tom's Dad sowed some tomato seed in February. He gave Tom one of the tomato plants in a pot.

At the beginning of May Tom put his tomato plant outside. On the same day he sowed a bean in another pot.

Ten days later the bean plant was just 1 cm (centimetre) above the soil surface. Tom measured his tomato plant which was already 38 cm tall.

Each evening Tom measured his two plants.

On the evening of the next day the little bean plant had grown another 2 cm so it was 3 cm high. Each day it continued to grow double the amount it had grown the day before.

The tomato plant grew at a steady 5 cm a day.


After how many days were the two plants the same height when Tom measured them in the evening? How high were they?
Click to expand...

Hint:

The tomato plant is growing by arithmetic progression

and

the bean plant is growing by geometric progression.

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
Well, I got that one was an arithmetic and one was a geometric sequence. My problem is I guess I don't understand what the question wants you to find because the plants are never the same height...
 
How do you know?

The bean plant will start smaller but surpass the tomato plant.
 
Yes, I figured that.. The tomato seed starts at 38 and then grows 5cm so, its then 43, then 48. That means that d is equal to 5.

The bean seed starts at 1cm and then it grows to be 3cm and from then on it doubles itself. So, from 3 to 6 to 12.
Therefore, the common ratio, r, is equal to 2.

What I don't understand is how do I use this information to find the anwser...
 
RainyScotland said:
Yes, I figured that.. The tomato seed starts at 38 and then grows 5cm so, its then 43, then 48. That means that d is equal to 5.
>The bean seed starts at 1cm and then it grows to be 3cm and from then on it doubles itself>----No, I don't think that's what it says!! So, from 3 to 6 to 12.
Therefore, the common ratio, r, is equal to 2.

What I don't understand is how do I use this information to find the anwser...
Click to expand...

Well, not exactly, on the bean plant....

From how I read the problem, the bean plant grew 1 cm the first day, from 1 cm to 3 cm (that's 2 cm, right?) on the second day, and each consecutive day it grows double what it did the day before. On day 2, the bean plant grew 2^2 inches. So, on day 3, the bean plant would grow 2^3 inches. Be aware that this growth rate cannot continue very far!!

So...on day 1 after growth was observed, the bean plant was 1 cm.

The next day, it was 3 cm.

The day after that, it would be 3 cm + 4 cm, or 7 cm....

And the day after THAT, the bean plant would grow 2*3 cm, or 8 cm, and its height would be 7 cm + 8 cm, or 15 cm

You can make a table of the height of the two plants:

Tomato: 38, 43, 48, 53, 58, 63, 68, 73, 78, .....
Bean: 1, 3, 7, 15, 31, ......


After "n" days, the height of the tomato plant would be 38 + 5n

The height of the bean plant would be 1 + 2^n

When the two plants are the same height,

38 + 5n = 1 + 2^n
 
Make a table:
Code: 
days             Tomato                         Bean
1                   38                               1
2                   38 + (2-1)*5                 1 + 2^(2-1)
3                   38 + (3-1) *5                1+ 2^(3-1)
4                   38 + (4-1) *5                1 + 2^(4-1)
5                   38 + (5-1)*5                 1 + 2^(5-1)
.
.
.

Suppose the "n" th day - their heights are equal.

What are the expressions of their height on the "n" th day?

Equate those and solve for "n".
 
Mrspi, shouldn't your "38 + 5n = 1 + 2^n" be 2^n - 1 = 38 + 5(n-1) : which simplifies to 2^n = 5n + 34

Subhotosh, shouldn't your "1 + 2^(2-1)" be 2^(2-1) - 1 ?
 
1st column = n, 2nd = tomato, 3rd = bean:
Code: 
1:         38        1    1
2:    5    43        2    3
3:    5    48        4    7
4:    5    53        8   15
5:    5    58       16   31
6:    5    63       32   63
Notice the bean doubles its previous day

Pick n=4:
tomato: 38 + 5(n-1) = 38 + 5(3) = 38 + 15 = 53
bean: 2^n - 1 = 2^4 - 1 = 16 - 1 = 15

Try n = 6:
tomato: 38 + 5(n-1) = 38 + 5(5) = 38 + 25 = 63
bean: 2^n - 1 = 2^6 - 1 = 64 - 1 = 63

Clearly, happens when: 2^n - 1 = 38 + 5(n-1)

Now go make a bean and tomato sandwich :wink:
 
The best way to solve for 'n' would be to solve by making a chart (or graph). It does not have a closed form analytical solution.
 
Rains, what Subhotosh is telling you is it's impossible to directly solve an equation in form:
a^x + bx + c = 0 ; graphing or iteration is needed...
 
You must log in or register to reply here.
Top