Questions requiring help in deriving the method of doing

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ctmckoay

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(1) On the 1st of Jan this year, Edu-kingdom puts 160 cents in a piggy bank. Every year, the amount which is put in the bank is doubled. If there is $49.60 in the Edu-kingdom bank account, how many years has it been since the first deposit has been made ?

Ans : 4

(2) Travelling by train from location A to B, I passed a sign saying "Location B - 280 km". After 18 more kilometres, I passed another sign saying "Location A 172 kilometres". How far is it by train from location A to B ?

Ans : 434km

(3) The pages of a book are numbered, 1,2,3.....In total, it takes 852 digist to number all the pages of the book. What is the number of the last page ?

Ans : 320

(4) Suppose that x is an odd integer, and that the sum .. (Anyone can help on this..as in how do i show a grade 5 the simple way to get the answer x ?

1 + 3 + 5 + 7 + ....+ (x-2) + x = 400.
What is x ?

Ans : 39

Appreciate your help.
 
I see substantial (basic) algebra intended to do these, instead of just arithmetic. Can these be moved
to "Beginning Algebra" instead?
 
Hello, ctmckoay!

We can baby-talk our way through #3 . . .

(3) The pages of a book are numbered 1, 2, 3, ...
In total, it takes 852 digist to number all the pages of the book.
What is the number of the last page ?
. . Ans: 320
Click to expand...

\(\displaystyle \text{Pages 1 to 9 have nine one-digit numbers.}\)
. . \(\displaystyle \text{They use 9 digits.}\)

\(\displaystyle \text{Pages 10 to 99 have ninety two-digit numbers.}\)
. . \(\displaystyle \text{They use 180 digits.}\)

\(\displaystyle \text{Page 100 to }n\text{ have }n-99\text{ three-digit numbers.}\)
. . \(\displaystyle \text{They use }3(n-99)\text{ digits.}\)


\(\displaystyle \text{The total number of digits is }852\!:\;\;3(n-99) + 180 + 9 \:=\:852\)

. . \(\displaystyle 3(n-99) \:=\:663 \quad\Rightarrow\quad n - 99 \:=\:221 \quad\Rightarrow\quad \boxed{n \:=\:320}\)
 
Hello, ctmckoay!

Did you make a sketch for #2 ?

(2) Travelling by train from location A to B, I passed a sign saying "Location B - 280 km".
After 18 more kilometres, I passed another sign saying "Location A 172 kilometres".
How far is it by train from location A to B ?
. . Answer : 434 km
Click to expand...
Code: 
                : - - -  280  - - - :
    A *---------*-------*-----------* B
                X   18  Y
      : - - -  172  - - :

\(\displaystyle \text{At }X\text{, it is 280 km to }B\!:\;X\!B = 280 \quad\Rightarrow\quad Y\!B = 280 - 18 \:=\:262\)

\(\displaystyle \text{At }Y\text{, it is 172 km from }A\!:\;AY = 172 \quad\Rightarrow\quad AX \:=\:172 - 18 \:=\:154\)


\(\displaystyle \text{Therefore: }\:AB \;=\;AX + XY + YB \;=\;154 + 18 + 262 \;=\;434\text{ km.}\)
 
Hi mmm4444bot,

For Q1, I've tried your method but it gives me 30 years to accumulate $49.60. The problem is the answer is 4 years. So honestly I'm not sure where I went wrong. The question is as written and I've just reproduced it in the post only.

Help.
 
Hi soroban,

Thanks for your feedback and reply. I have to be honest, I only managed to do a mental map and did not draw it out. Will do so the next time. Thanks for giving me the light.

Appreciate it.
 
Re:

mmm4444bot said:
ctmckoay said:
Every year, the amount which is put in the bank is doubled.

This sentence could be better worded. The kingdom does not put money in the bank every year.


I'm thinking that the following is what they mean.

First, convert 160 cents to $1.60, and work with dollars instead of cents.

On January 1, 2010, the kingdom deposits $1.60 into a bank account.

Every year, the bank adds $1.60 (interest payment) to the kingdom's account.

How many years will it take for the account balance to reach $49.60 ?

Here is the scenario:

Code: 
January 1, 2010     Kingdom's Deposit: $1.60     Balance: $1.60

January 1, 2011     Interest Payment:  $1.60     Balance: $3.20

January 1, 2012     Interest Payment:  $1.60     Balance: $4.80

January 1, 2013     Interest Payment:  $1.60     Balance: $6.40

January 1, 2014     Interest Payment:  $1.60     Balance: $8.00

January 1, 2015     Interest Payment:  $1.60     Balance: $9.60

et cetera

Here is the "pattern":

After zero years, the balance is 1.6 + (0)(1.6) dollars.

After one year, the balance is 1.6 + (1)(1.6) dollars.

After two years, the balance is 1.6 + (2)(1.6) dollars.

After three years, the balance is 1.6 + (3)(1.6) dollars.

After four years, the balance is 1.6 + (4)(1.6) dollars.

After five years, the balance is 1.6 + (5)(1.6) dollars.

et cetera

Therefore, we can write a mathematical model:

After t years, the balance (B) is modeled by the equation B = 1.6 + (t)(1.6)

Does this give you any ideas about how to find t (the number of years) that it takes for the balance (B) to reach 49.6 dollars ?

If you still need help, please ask specific questions or share what you've done thus far.

Cheers ~ Mark 8-)

Click to expand...
Click to expand...

I may have missed something terribly significant...but after reading the original problem SEVERAL times (very carefully), I see nothing about interest!!
It's a piggy bank!!!!!


The first year, $1.60 is put into the piggy bank.

It sits there. No interest. No withdrawals (that are mentioned).

Ok...the next year, the amount put into the piggy bank is DOUBLED. So, 2*($1.60) is added, and the amount in the piggy bank after 2 years is

$1.60 + 2*($1.60)


The third year, the amount put into the piggy bank is doubled again. So, this year, 2*2*($1.60) is added, and the amount in the piggy bank is

$1.60 + 2*($1.60) + 2*2*($1.60)

Hmmmmm.....looks to me like we have a geometric sequence going, where the first term is $1.60 and the common ratio is 2.

What is the formula for the SUM of a geometric sequence with first term a1, common ratio r, and n terms?

S[sub:29e9viw5]n[/sub:29e9viw5] = (a[sub:29e9viw5]1[/sub:29e9viw5] - a[sub:29e9viw5]1[/sub:29e9viw5]*r[sup:29e9viw5]n[/sup:29e9viw5]) / (1 - r)

We know that the ending amount, s[sub:29e9viw5]n[/sub:29e9viw5], is 49.60.

We know that the beginning amount, a[sub:29e9viw5]1[/sub:29e9viw5], is $1.60.

We know that r = 2.

We are looking for n.

Substitute the values we KNOW into the formula, and solve the resulting equation for n.

I did that, and I got n = 4 (so it takes FOUR years for the amount in the piggy bank to reach $49.60.
 
ctmckoay said:
For Q1, I've tried your method but it gives me 30 years to accumulate $49.60. The problem is the answer is 4 years.
Click to expand...

Darn! I totally missed the word "piggy", and then got confused about money doubling every year. Please excuse me.

Every year, the kingdom puts twice as much money into a piggy bank as they did the year before.

See Mrs. Pi's reply. Her reading skills are much better than mine!

 
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