# Consumer Arithmetic

• Sep 5th 2009, 09:26 PM
Consumer Arithmetic
(Nod) Hello everyone! I was just wondering if someone could explain these questions to me (and how to solve them). I seem to be having a memory block. Thanks in advance.

1. Claudia, to save for her trip to the Gold Coast, puts her tax return cheque of $890 into an account paying 8% p.a. How long (in years, correct to 1 decimal place) will her money take to accumulate to$1300?

2. After investing $6000 into a savings account paying 7% interest p.a., Ian received$6238. For how many days (correct to the nearest day) was the money invested?

(Doh)
• Sep 5th 2009, 10:49 PM
Quote:

(Nod) Hello everyone! I was just wondering if someone could explain these questions to me (and how to solve them). I seem to be having a memory block. Thanks in advance.

1. Claudia, to save for her trip to the Gold Coast, puts her tax return cheque of $890 into an account paying 8% p.a. How long (in years, correct to 1 decimal place) will her money take to accumulate to$1300?

2. After investing $6000 into a savings account paying 7% interest p.a., Ian received$6238. For how many days (correct to the nearest day) was the money invested?

(Doh)

These questions don't state whether it's Simple or Compound Interest. Here's the first one done in the two different ways:

1) Simple Interest: $I=\frac{PRT}{100}$

$I= 1300 -890 = 410=\frac{890\times8\times T}{100}$

$\Rightarrow T=\frac{410\times100}{890\times8}=5.8$ years, to 1 d.p.

Compound Interest: $A=P\Big(1+\frac{R}{100}\Big)^n$, where the interest is compounded yearly

$1300 = 890(1+0.08)^n$

$\Rightarrow 1.08^n=\frac{1300}{890}=1.4607$

$\Rightarrow n\log1.08=\log1.4607$

$\Rightarrow n = \frac{\log1.4607}{\log1.08}=4.9$ years to 1 d.p.

• Sep 5th 2009, 11:16 PM
Wilmer

Future Value - Free Math Help
• Sep 5th 2009, 11:40 PM
2. After investing $6000 into a savings account paying 7% interest p.a., Ian received$6238. For how many days (correct to the nearest day) was the money invested?

I'm assuming it's Simple Interest, as 5.8 is the answer obtained by the teacher.

Thus, this would be the second solution:

I= 6238 - 6000
= 238.

238= 6000 x 0.08 x T / 100
238= 480 x T / 100

T=238 x 100 / 6000 x 7
= 56.7
= 57

Therefore, the money was invested for 57 days.

Am I right?

Thanks again.
• Sep 6th 2009, 12:13 AM
Quote:

2. After investing $6000 into a savings account paying 7% interest p.a., Ian received$6238. For how many days (correct to the nearest day) was the money invested?

I'm assuming it's Simple Interest, as 5.8 is the answer obtained by the teacher.

Thus, this would be the second solution:

I= 6238 - 6000
= 238.

238= 6000 x 0.08 x T / 100 This should be 6000 x 7 x T /100
238= 480 x T / 100

T=238 x 100 / 6000 x 7 Correct!
= 56.7 No. 0.566...
= 57

Therefore, the money was invested for 57 days.

Am I right?

Thanks again.

See my comments in red. The time is 0.5666... years. Multiply by 365 to get the number of days, which is 207.

• Sep 6th 2009, 12:16 AM
Oh, sorry. So, it was 207 days.
• Sep 6th 2009, 12:22 AM
Thus, this would be the second solution:

I= 6238 - 6000
= 238.

238= 6000 x 7 x T / 100
238= 420 x T

T=238 x 100 / 6000 x 7
= 0.566.

0.566 x 365 = 207

Therefore, the money was invested for 207 days.