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vmelgargalan
1050 + 5x = 1550 + 5y

5x - 5y = 500

(Divide both sides by 5)

x - y = 100

Only two digits were x - y = 100 is 130 and 30

If this is the case it will match numbers in a year or so ..

1050+1050*1.30 and 1550+1550*.3 ....

In my opinion 40 and 30 should be Answer
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vmelgargalan
1050 + 5x = 1550 + 5y

5x - 5y = 500

(Divide both sides by 5)

x - y = 100

Only two digits were x - y = 100 is 130 and 30

If this is the case it will match numbers in a year or so ..

1050+1050*1.30 and 1550+1550*.3 ....

In my opinion 40 and 30 should be Answer


It is said that after 5 years, the number of employees in organization A will be equal to the number of employees in organization B.
If we take 40 and 30 then after 5 years, the result will be :

Org A : 1,050 + (40*5) = 1,250

Org B : 1,550 + (30*5) = 1,700

It doesn't match, so it can't be our answers.

However, if you take 130 and 30 :

Org A : 1,050 + (130*5) = 1,700

Org B : 1,550 + (30*5) = 1,700

It's a match !


Answers : 130 and 30.
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1050+5x=1550+5y
x-y=100

Only 130 and 30 suffice :)
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Whats the OA? Shouldnt the andwer be A>40 and B>30 ?
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Whats the OA? Shouldnt the andwer be A>40 and B>30 ?

OA is updated

Organization A: 130
Organization B: 30

Thank you!
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Why do we solve by means of an arithmetic progression and not a geometric prgression?

What I initially did was 1050*a^5 = 1550*b^5. a and b being A and B respective constant rates.
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is my approach correct ?

Initally they have difference of 1550-1050 = 500 in b/w them and this 500 needs to be covered in 5 yrs and since rates are same so each year they will cover 500/5 = 100 ka difference
now, since A surpasses B afterwards so A's rate is more than B and only 130 and 30 can be valid options since only then the difference of 100 in a year can be covered
A moved ahead by 130, B moved ahead by 30 so overall difference b/w is 100
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djangobackend
is my approach correct ?

Initally they have difference of 1550-1050 = 500 in b/w them and this 500 needs to be covered in 5 yrs and since rates are same so each year they will cover 500/5 = 100 ka difference
now, since A surpasses B afterwards so A's rate is more than B and only 130 and 30 can be valid options since only then the difference of 100 in a year can be covered
A moved ahead by 130, B moved ahead by 30 so overall difference b/w is 100
Yes, Your approach is correct as well, and it works because of the constant additive rates.
To clarify it lets look at this more mathematically:

1050 + 5a = 1550 + 5b
(Here a is the constant additive rate by which org A is growing and likewise for B)

1550-1050 = 5(a-b)
(a-b)=500/5 ---- This is what you arrived at directly
a-b =100

Hope this helps!
Regards.
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Here there is a fault- the question explicitly states the number of member increases by a constant rate(not constant number) so there will a compounding of the number every year. if the question would have said by a constant number then we could use your method............

vmelgargalan
1050 + 5x = 1550 + 5y

5x - 5y = 500

(Divide both sides by 5)

x - y = 100

Only two digits were x - y = 100 is 130 and 30
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I generally struggle with understanding if a question is on linear growth or exponential growth. how do we identify it in such questions?
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Hi AnushkaKala,

This is a great question, and it's exactly the confusion Rogerper123 and Annukiran raised in the thread when they tried a geometric-progression setup (1050·a^5 = 1550·b^5). Rahul885 pointed to the fix, and it's worth spelling out as a general rule you can reuse.

The single strongest clue: the units

Look at what the rate is measured in.

- A rate given as an absolute amount per time - "members per year," "dollars per month," "liters per hour" - means you add the same fixed number each period. That's linear (arithmetic) growth.
- A rate given as a percent - "grows 5% per year," "increases by a factor of 1.1," "doubles every decade" - means you multiply each period. That's exponential (geometric) growth.

In this question the table literally says "Rate of increase (members per year)". That unit - a count of members added each year - locks it as linear. So each year Org A adds the same fixed number, and after 5 years it has 1050 + 5a. No compounding.

Why "constant rate" felt ambiguous

The phrase "constant rate of increase" sounds like it could go either way, and that's the trap. The word "rate" alone doesn't decide it - the units do. Percentages compound; fixed quantities add.

A quick test to build the habit

Read these two versions of the same sentence and sort each one:

- "Membership grows by 40 members each year." - add 40 every year - linear.
- "Membership grows by 4% each year." - multiply by 1.04 every year - exponential.

Same word ("grows"), different unit, different model. Whenever you're unsure, ask: am I told a number to add, or a percent to multiply by? That question alone will settle it almost every time.

Answer: Column 1 (Organization A) = 130; Column 2 (Organization B) = 30

AnushkaKala
I generally struggle with understanding if a question is on linear growth or exponential growth. how do we identify it in such questions?
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AnushkaKala
I generally struggle with understanding if a question is on linear growth or exponential growth. how do we identify it in such questions?

Constant rate can mean either - "100 members per year" or "5% increase in the number of members per year"
That is why it is important to keep an eye on the options. The answers are in terms of 'members per year'. So rate of increase is a simple addition of members year or year.
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