Carl invested a certain amount of money for one year, some of which ea

Question

Carl invested a certain amount of money for one year, some of which earned interest at a simple annual interest rate of 2.2 percent and the rest of which earned interest at a simple annual rate of 2.5 percent. If the amount of money that Carl invested for the year earned a total of $680 in interest, which of the following could have been the total amount of money that he invested for the year?

I. $24,708
II. $27,896
III. $31,284

A. None
B. II only
C. III only
D. I and II
E. II and III

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I. $24,708
II. $27,896
III. $31,284

A. None
B. II only
C. III only
D. I and II
E. II and III

https://gmatclub.com/forum/download/file_private.php?id=75858

gmatophobia · Accepted Answer

The minimum principle value is when all the amount is invested at a 2.5% rate.

\frac{p_{	ext{min}} * 2.5 * 1}{100} = 680

p_{	ext{min}} = 680*\frac{100}{2.5} = 27200

The maximum principle value is when all the amount is invested at a 2.2% rate.

\frac{p_{	ext{max}} * 2.2 * 1}{100} = 680

p_{	ext{max}} = 680*\frac{100}{2.2} \approx 30900

Only one value lies in this range i.e. $27,896

Option B

ankitmba95 · Answer

1) 24708*2.5%=617.7 : thus not possible to reach 680

2) 27896 *2.5 =697.4 and 27896*2.2 = 613.7 . Since required value is in between, possible.

3) 31284 * 2.2 = 688.2. Thus both value would be above 680, hence value of 680 not possible.

Hence answer is B.

Edoua · Answer

Let’s have A total amount invested
X amount of A at a 2.2 rate and A-X at 2.5 rate

Interest 1 = X*2.2/100 and interest 2 = (A-X)*2.5/100

Total interest earned = X*2.2/100 + (A-X)*2.5/100 =680

—->> 22X + 25(A-X) =680.000
25 A -3X = 680.000

—->> A = [680.000+3X]/25
——>> A= 27.200 + 3X/25

We noticed that A should be greater than 27.200
We have to consider B and C
But knowing that, X <<<A let’s verify the worst case for X=27200
3*27200/25 = 3264

27200+3264= 30464 == max value

C can’t be our answer

Hence B

Posted from my mobile device

thr3at · Answer

Hello,

I have seen this pattern in Gmat that some questions can be solved by getting the range.
But how can we figure out if this should be done by the range (min and max) or by using equations.

I solved this by using linear equation method. but realised it was time consuming so had to leave the Q after spending 2.5 min.

Thank you for your help

KarishmaB · Answer

The question tests your understanding of weighted averages. Since one sum was invested at 2.2% and the other at 2.5%, overall interest rate would lie between them.

Looking at the given options, say if the amount was 24800 (rounding up for easier calculations),
2.5% of this would be 248*2 + 124 = 496 + 124 = $620 (which is less than $680).
Hence, the amount must be more than $24800 to get $680 as interest.

Say if the amount was 31200, 
2.2% of this would be 312*2 + 62 = $686
So the amount should be less than $31200 to get $680 as interest.

Since option II is easily in the middle of the two, it will work.

Answer (B)

Abha30 · Answer

Why is X=27200 the worst case... can you please explain a bit more in detail

ShuGMATTutor · Answer

Ok lets dumb this down for us. "A problem well defined is halfway through to the solution - Kidlins law"

Visualization  
 The first step is to visualize the problem. Let’s personalize it:  
 Imagine I invested in two accounts and earned a total of $680 in interest.  
 The accounts had interest rates of 2.2% and 2.5%.  
 The question now is to determine how much I could have invested in each account.

Translation  
 If I rethink the problem, they want us to find out the minimum and maximum amounts I might have invested in these two accounts.  
 The lowest total investment would occur if I had invested all my money in the account with a 2.5% interest rate, while the highest total investment would be if I had invested everything in the account with a 2.2% interest rate.  
 You see, if the interest yield is fixed, the lower the interest percentage, the higher the amount invested, right?

Approach  
 Now, all I have to do is some calculations to find out what the maximum and minimum amounts could be.

The minimum invested at 2.5% interest should yield $680. So we can calculate the maximum invested amount - $27,200  
  (You can make the calculations easier - realizing 2.5% is $680 that mean 10% is $680 x 4 and 100% is $680x40 = $27200)

The maximum invested at 2.2% interest should yield $680. So we can calculate it to be around ~ $30,000 
  (check image for estimation) 
 
  Conclusion  
 Thus we can now deduce that only one answer choice fits within our range so we eliminate (I) and (III) and the answer would be (II) only

Shu "TheGMATTutor"
   The GMAT concepts are relatively basic   , but they require a deep understanding to tackle nuanced questions effectively. So  lets dumb it down for us . This is    not just a math and English test   —it is a    deductive test designed to assess your problem-solving skills

SudsMeister · Answer

Let x be the amount invested at 2.2% p.a. and (option given in Q - x) be amount the amount invested at 2.5% p.a.

Option (I): 
x* \frac{2.2}{100}  + (24708 - x)* \frac{2.5}{100}  = 680

\frac{22*x}{1000}  + (24708 - x)* \frac{25}{1000}  = 680

22x + 24708*25 - 25x = 680000

3x = 24708*25 - 680000

Let this be the standard form we will use for options (II) and (III) too.

3x = 25000*25 - 680000
3x = 625000 - 680000

Since x is coming negative, option (I) is not possible as amount cannot be negative.

Option (II): 
3x = 27896*25 - 680000 = 28000* \frac{100}{4}  - 680000 = 700000 - 680000 
3x = 20000
x = 6667

Also, to double check x is less than 27896, so x is a valid amount. Option (II) is possible.

Option (III): 
3x = 31284*25 - 680000 = 31200* \frac{100}{4}  - 680000 = 780000 - 680000
3x = 100000
x = 33333

But x is greater than 31284, so x is not a valid amount. Option (III) is not possible.

Only option (II) is allowed.

Answer: B.

GMATCoachBen · Answer

Bunuel

Suggested question edit: there is a typo "earnead".

Thanks!

Bunuel · Answer

Fixed the typo. Thank you!

GMAT2025champ · Answer

Quite a pain to figure out which linear equation works. But if you do come up with equations fast, and the range approach discussed above by the rest doesn't click fast for you. I thought this might be an alternative.

22a + 25b = 680,000
a + b = one of the answers given

e.g 24,708

b=24,708 -a
substitute in eqn 1
22a + 25 (24,708-a)=680,000
22a-25a=680,000-25(24,708)
-3a=25(4*6800 - 24,708)
-3a=25(27,200-24,708) - you don't have to finish the calculations, as you see the - on the left won't cancel out; 24,708 can't be an answer

b=27,896
-3a=25(4*6800-27,896)
-3a=25(27,200-27,896)
-3a=25*-696 - the negative will cancel out and 6+9+6=27, divisible by 3; 27,896 can be an answer

b=31,284
-3a=25(4*6800-31,284)
-3a=25(27,200-31,284)
-3a=25*-4,084 - the negative will cancel out but 4+8+4=16, is not divisible by 3; 31,284 can't be an answer

Krish13tss · Answer

Answer: (B) II only. 
 The key insight — this is a bounds problem, not a two-equation problem.  Don't try to solve for the two amounts; you have two unknowns and one equation, so it's underdetermined  on purpose . Instead, ask: for a fixed $680 of interest, what totals are even possible?
The blended rate on the whole pile must sit  between 2.2% and 2.5%  (some money at each). Since interest = total × rate → total = 680 ÷ rate, the extremes of the total come from the extremes of the rate:
 
  Everything at the high rate (2.5%):  total = 680 ÷ 0.025 =  $27,200  ← smallest possible total 
  Everything at the low rate (2.2%):  total = 680 ÷ 0.022 =  $30,909  ← largest possible total  
(Higher rate earns $680 on  less  money; lower rate needs  more  money for the same $680.)
So any valid total must land in  $27,200 – $30,909. 
Now just check the three:
 
  I. $24,708  — below $27,200. To earn $680 it'd need a rate of 680/24,708 ≈  2.75% , higher than either rate offered. Impossible. ✗ 
  II. $27,896  — inside the range. Needs 680/27,896 ≈  2.44% , comfortably between 2.2 and 2.5. Possible. ✓ 
  III. $31,284  — above $30,909. Needs 680/31,284 ≈  2.17% , below both rates. Impossible. ✗  
Only II survives →  (B).

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