A customer has taken three loans from the bank and the tenure of each loan is 5 years

[eddy2017]

Did you mean that we could've first have found the weighted mean of those loans?.

Taking the sum of all the weights × the sample
I know the weight could be taken in the form of frequency or could be in the form of percentages?
Is this what you and Dr Khan were referring to ?

 

[JeffM]

Did you mean that we could've first have found the mean of those loans?.

What I mean is that those different means are not arbitrary. Depending on the circumstances, one is usually the most useful.

Let’s take the usual arithmetic mean. in this case we have three rates: 0.14, 0.11, and 0.09. If we take the plain arithmetic mean rate, we get 0.34/3. Is that a good representative rate? Well, SK suggests multiplying that by the overall amount borrowed of 50 lahks, giving us interest of

[math] 50 * \dfrac{0.34}{3} = \dfrac{17.3}{3} \approx 5.7 \text { lahks.}[/math]
Now lets compute the actual interest.

[math]2 * \dfrac{14}{100} + 8 * \dfrac{11}{100} + 40 * \dfrac{9}{100} = \dfrac{28 + 88 + 360}{100} = \dfrac{476}{100} = 4.76 \text { lahks.}[/math]
The straight arithmetic mean rate did not give us anywhere close to a good approximation of the actual interest due.

Let’s try the weighted arithmetic mean rate.

[math]\dfrac{2 * 0.14 + 8 * 0.11 + 40 * 0.09}{2 + 8 + 40} = \dfrac{0.28 + 0.88 + 3.6}{50} = \dfrac{4.76}{50}.[/math]
Let’s multiply that by the overall amount borrowed of 50 lahks.

[math]50 * \dfrac{4.76}{50} = 4.76 \text { lahks}[/math], which is exact.

In other words, when we are averaging coefficients, a weighted arithmetic mean is far more representative that a plain arithmetic mean of the coefficients.

Now i am sure that in context this problem was intended to teach students why the weighted arithmetic mean is sometimes useful. Those of us who have worked with means for decades usually just “see” which one is best in most situations (though I admit that it usually takes me a while to see when it is best to use the harmonic mean).

For a student, the key clue for this problem is likely to come from the topic being covered, which here was almost certainly different types of mean. And of course means are not the only way to measure central tendency.

This is why we want to know what topic gave rise to the exercise. It is to know what sort of clue to provide.

 

[BigBeachBanana]

What I mean is that those different means are not arbitrary. Depending on the circumstances, one is usually the most useful.

Let’s take the usual arithmetic mean. in this case we have three rates: 0.14, 0.11, and 0.09. If we take the plain arithmetic mean rate, we get 0.34/3. Is that a good representative rate? Well, SK suggests multiplying that by the overall amount borrowed of 50 lahks, giving us interest of

[math] 50 * \dfrac{0.34}{3} = \dfrac{17.3}{3} \approx 5.7 \text { lahks.}[/math]
Now lets compute the actual interest.

[math]2 * \dfrac{14}{100} + 8 * \dfrac{11}{100} + 40 * \dfrac{9}{100} = \dfrac{28 + 88 + 360}{100} = \dfrac{476}{100} = 4.76 \text { lahks.}[/math]
The straight arithmetic mean rate did not give us anywhere close to a good approximation of the actual interest due.

Let’s try the weighted arithmetic mean rate.

[math]\dfrac{2 * 0.14 + 8 * 0.11 + 40 * 0.09}{2 + 8 + 40} = \dfrac{0.28 + 0.88 + 3.6}{50} = \dfrac{4.76}{50}.[/math]
Let’s multiply that by the overall amount borrowed of 50 lahks.

[math]50 * \dfrac{4.76}{50} = 4.76 \text { lahks}[/math], which is exact.

In other words, when we are averaging coefficients, a weighted arithmetic mean is far more representative that a plain arithmetic mean of the coefficients.

Now i am sure that in context this problem was intended to teach students why the weighted arithmetic mean is sometimes useful. Those of us who have worked with means for decades usually just “see” which one is best in most situations (though I admit that it usually takes me a while to see when it is best to use the harmonic mean).

For a student, the key clue for this problem is likely to come from the topic being covered, which here was almost certainly different types of mean. And of course means are not the only way to measure central tendency.

This is why we want to know what topic gave rise to the exercise. It is to know what sort of clue to provide.

I want to add to your points. The arithmetic means is a particular case of the weighted-average mean when all the weights are equal.
Second, given this question came from a finance course. I think it is meant to be solved with "finance" formulas i.e. the simple interest equation:
[math]A=P(1+rt)[/math]Future values of the loans:
[math]A_1=200,000(1+.14\cdot 5)\implies A_1=340,000\\ A_2=800,000(1+.11\cdot 5)\implies A_2=1,240,000\\ A_3=4,000,000(1+.09\cdot5)\implies A_3=5,800,000\\ A_1+A_2+A_3=7,380,000[/math]The "overall" interest rate, [imath]r^*[/imath]:
[math]7,380,000=5,000,000(1+r^*\cdot 5) \implies r^*=0.0952[/math]

 

[eddy2017]

What I mean is that those different means are not arbitrary. Depending on the circumstances, one is usually the most useful.

Let’s take the usual arithmetic mean. in this case we have three rates: 0.14, 0.11, and 0.09. If we take the plain arithmetic mean rate, we get 0.34/3. Is that a good representative rate? Well, SK suggests multiplying that by the overall amount borrowed of 50 lahks, giving us interest of

[math] 50 * \dfrac{0.34}{3} = \dfrac{17.3}{3} \approx 5.7 \text { lahks.}[/math]
Now lets compute the actual interest.

[math]2 * \dfrac{14}{100} + 8 * \dfrac{11}{100} + 40 * \dfrac{9}{100} = \dfrac{28 + 88 + 360}{100} = \dfrac{476}{100} = 4.76 \text { lahks.}[/math]
The straight arithmetic mean rate did not give us anywhere close to a good approximation of the actual interest due.

Let’s try the weighted arithmetic mean rate.

[math]\dfrac{2 * 0.14 + 8 * 0.11 + 40 * 0.09}{2 + 8 + 40} = \dfrac{0.28 + 0.88 + 3.6}{50} = \dfrac{4.76}{50}.[/math]
Let’s multiply that by the overall amount borrowed of 50 lahks.

[math]50 * \dfrac{4.76}{50} = 4.76 \text { lahks}[/math], which is exact.

In other words, when we are averaging coefficients, a weighted arithmetic mean is far more representative that a plain arithmetic mean of the coefficients.

Now i am sure that in context this problem was intended to teach students why the weighted arithmetic mean is sometimes useful. Those of us who have worked with means for decades usually just “see” which one is best in most situations (though I admit that it usually takes me a while to see when it is best to use the harmonic mean).

For a student, the key clue for this problem is likely to come from the topic being covered, which here was almost certainly different types of mean. And of course means are not the only way to measure central tendency.

This is why we want to know what topic gave rise to the exercise. It is to know what sort of clue to provide.

Yes, I asked him he confirmed it. It came from a practice lesson on central tendency, especially mean.

 

[Deleted member 4993]

Personal Loan of $ 200,000 at 14% rate of interest.......................... 20000 = 2*10⁴
Car Loan of $ 800,000 at 11% rate of interest .................................... 80000 = 8*10⁴
Home Loan of 400,000 at 9% rate of interest

What is the rate of interest for the overall $ 500,000?>

If the OP was posted correctly , this problem would have been solved in 3 posts.

Interest earned, after 1 year, for Personal Loan @ 14% on ........................... 200,00 →....................= 2800

Interest earned, after 1 year, for Car Loan @ 11% on ......................................800,00 → ...................= 8800

Interest earned, after 1 year, for Home Loan @ 9% on ..................................400,000 →.................. = 36000
-----------------------------------------------------------------------------------------------------------------------
Interest earned , after 1 year, for Total Loan ...................................................... 500,000 →..................... = 47600

What is the rate of interest for the overall $ 500,000?

rate = (interest earned in 1 year)/(total loan) = 47600/500000 = 952 * 10^-4 = 9.52% per year

That statement about 5 years was a red-herring (according to OP). May be the next problem is about "compound interest" - then that will become important.

Eddy, please please please, read (& review & revise) carefully before you hit that post reply button.

The MOST important skill needed for solving (real world) word-problem is:

Read,​

​

read again,​

​

analyze and then​

​

write (and review and revise the written response )​