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GMAT Club World Cup 2026 (Day 10): A school sets aside $1,860 to be

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GMAT Club Official Solution:

A school sets aside $1,860 to be divided among three students: Arman, Bela, and Cora. Each student’s share will be deposited at 10% annual simple interest. Arman’s share will be deposited for 2 years, Bela’s share for 4 years, and Cora’s share for 5 years. If the shares are chosen so that the three deposits have the same value at the end of their respective deposit periods, how much is Cora’s original share?

A. $520
B. $540
C. $560
D. $600
E. $620

Let the common final value of each deposit be F.

At simple interest, final value = principal * (1 + rt), where r is the annual interest rate as a decimal and t is the number of years.

• Arman’s share is deposited for 2 years, so his deposit grows by a factor of 1 + 0.10 * 2 = 1.2 = 6/5.

So Arman’s original share is F/(6/5) = 5F/6.

• Bela’s share is deposited for 4 years, so her deposit grows by a factor of 1 + 0.10 * 4 = 1.4 = 7/5.

So Bela’s original share is F/(7/5) = 5F/7.

• Cora’s share is deposited for 5 years, so her deposit grows by a factor of 1 + 0.10 * 5 = 1.5 = 3/2.

So Cora’s original share is F/(3/2) = 2F/3.

Since the total original amount is $1,860:

5F/6 + 5F/7 + 2F/3 = 1860

35F/42 + 30F/42 + 28F/42 = 1860

93F/42 = 1860

F = 840

Cora’s original share is:

2/3 * F = 2/3 * 840 = 560

Answer: C.

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Given or let A+B+C = 1860

for Arman = A+SIA = A + 20A/100 = 6A/5
Bela = B+SIB = B + 40B/100 = 7B/5
Cora = C + SIC = C + 50C/100 = 3C/2

given 6A/5 = 7B/5 = 3C/2 = x

A = 5x/6
B = 5x/7
C = 2x/3

x [5/6+5/7+2/3] = 1860
x [ 93/42] = 93 *20
x = 840

therefore C = (840*2)/3 = 560

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Answer: C) 560

Let's let their shares equal a, b, and c, respectively.

a(1+0.1*2) = b(1+0.1*4) = c(1+0.1*5) = x, where x is the value of their deposits at the end of their respective periods
1.2a = 1.4b = 1.5c
a = x/1.2, b = x/1.4, c = x/1.5
a+b+c = 1860
x/1.2 + x/1.4 + x/1.5 = 1860
x = 840

c = 840/1.5 = 560

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Armans Total amount after 2 years = 1 +0.2x

B's = 1+ 0.4y

C's = 1 + 0.5z

1.2x = 1.4y = 1.5z

let each final value b K

K/1.2 + K/.1.4 + K/1.5 = 1860

1.2 = 6/5 1.4 = 7/5 1.5 = 3/2

5/6 : 5:7 / 2:3 = 35:30:28 (Taking lcm)

28*20 = 560

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A school sets aside $1,860 to be divided among three students: Arman, Bela, and Cora.
Each student’s share will be deposited at 10% annual simple interest.
Arman’s share will be deposited for 2 years, Bela’s share for 4 years, and Cora’s share for 5 years.

If the shares are chosen so that the three deposits have the same value at the end of their respective deposit periods, how much is Cora’s original share?

Let Arman's, Bela's and Cora's shares be a, b & c respectively

a + b + c = 1860

At the end of respective deposit periods : -
Arman's value = a(1+2*10%) = 1.2a
Bela's value = b(1+4*10%) = 1.4b
Cora's value = c(1+5*10%) = 1.5c

1.2a = 1.4b = 1.5c
12a = 14b = 15c = k

a = k/12; b = k/14; c = k/15

k/12 + k/14 + k/15 = 1860
k(35+30+28)/2^2*3*5*7 = 1860
93k/420 = 1860
k = 420*1860/93 = 420*20 = 8400

c = k/15 = 560

Cora's original share = $560

IMO C

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A school sets aside $1,860 to be divided among three students: Arman, Bela, and Cora. Each student’s share will be deposited at 10% annual simple interest. Arman’s share will be deposited for 2 years, Bela’s share for 4 years, and Cora’s share for 5 years. If the shares are chosen so that the three deposits have the same value at the end of their respective deposit periods, how much is Cora’s original share?

Total value= Principal+Interest
Armans share= A(1+0.10*2)=1.20A
Belas share= B(1+0.10*4)=1.40B
Coras share= C(1+0.10*5)=1.50C
Future values of all are equal
1.2A=1.4B=1.5C=k
Sum of all 3 shares= 1860
A+B+C=1860
(5k/6)+(5k/7)(2k/3)=1860
k=840
Coras original share= 2k/3 =2(840)/3=560

C

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We are given that,

The total amount that is split between Arman, Bela and Cora is $1860. Let their respective shares be a, b and c
Thus,
a+b+c = 1860

Now, their shares are deposited for 2, 4 and 5 years respectively at 10% simple interest annually.
Thus, Arman's share after 2 years = a (original share) + 0.1*2*a (simple interest for two years @ 10% annual) = 1.2a
Bela's share = b + 0.1*4*b = 1.4b
Cora's share = c + 0.1*5*c = 1.5c

We are given that at the end of the deposit periods, their shares are same.
Thus, 1.2a = 1.4b = 1.5c
1.2a = 1.5c
a = (1.5/1.2)*c = 5c/4
Also, 1.4b = 1.5c
b = (1.5/1.4)*c = 15c/14

Thus, the original equation can be expressed as
5c/4 + 15c/14 + c = 1860

(70c + 60c + 56c)/56 = 1860
Thus, 186c = 1860*56
c = 560

Thus Cora's original share was $560. Answer is (C)

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Arman's deposits at the end of 2 years (A) = $P_1 + (P_1 * 0.1 * 2) = 1.2P_1$
Bela's deposits at the end of 4 years (B) = $P_2 + (P_2 * 0.1 * 4) = 1.4P_2$
Cora's deposits at the end of 5 years (C) = $P_3 + (P_3 * 0.1 * 5) = 1.5P_3$

Given that, A = B = C, so $1.2P_1 = 1.5P_3$
$P_1 = 5/4P_3$

Similarly, $1.4P_2 = 1.5P_3$
$P_2 = 5/14P_3$

Also, given that $P_1 + P_2 + P_3 = 1860$
Substituting values of $P_1$ and $P_2$

$5/4P_3 + 5/14P_3 + P_3 = 1860$
$P_3 = 560$

IMO, the answer is C.

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Given:
Total amount $1860
To be divided among three friends Aman(a), Bela(b), Cora(c)
Each person individual share amount will be paid 10% annual simple interest rate

Time period of deposit: a- 2 years, b=4 years, c=5 years

At end of their respective time period each will have same amount, How much was C's share?

a+ b+ c=1860

a(1+10*2/100)= b(1+10*4/100)= c(1+10*5/100) =k
6/5*a= 7/5*b= 3/2*c =k

a=5/6* k, b= 5/7*k , c= 2/3*k

5/6*k+ 5/7*k+2/3*k= 1860
( (35+30+28)/42)*k=1860
93/42*k= 1860
k= 1860*42/93 = 840

a=700
b= 600
c=560

so C's share or Cora's share is 560

Option- C: answer

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Simple interest: Final amount = Principal (1+rt)

* Arman(2yr) = A = f/1.2 = 5f/6
* Bela (4 yr) = B = f/1.4 = 5f/7
* Cora (5 yr) = C = f/1.5 = 2f/3
Original amount = 1860

5f/6 + 5f/7 + 2f/3 = 1860
35f+30f+28f / 42 = 1860
93 f /42 = 1860
31f /14 = 1860
f = 840
Cora original share c = 2f/3 = 560
Answer: C

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Here's my solution for this question

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Bunuel

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Correct Answer C

Amount = Principal (1+ Rate * No of years)

Let principal for Aman, Bella, Cora be A,B,C

Aman = 1.2A
Bella = 1.4B
Cora = 1.5C

1.2A = 1.4B = 1.5C

Aman in terms of Cora is equal A = 1.5C/1.2
Bella in terms of Cora is equal B = 1.5C/1.4

A + B + C = 1860

1.5C/1.2 + 1.5C/1.4 + C = 1860

C = 560

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A = P(1+rt/100)
r = 10
P = a,b,c
t = 2,4,5
A = 6a/5 = 7b/5 = 3c/2
also, a+b+c = 1860
as we need c, putting everything as c, we get
5c/4+15c/14+c = 1860
93c/28 = 1860
c = 560 Ans.

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Arman: A*1.2
Bela:B*1.4
Cora: C*1.5

since the final values are equal
1.2A=1.4B=1.5C
12A=14B=15C

Ratio parts

A=420/12=35
B=420/14=30
C=420/15=28

35+30+28=93 PARTS

1 parts=1860/93=20
since cora has 28 parts

cora's share =28*20=560

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Here's how I thought about this question:

Let's set student A's original share is A,
student B's original share is B,
student C's original share is C.

1,860=A+B+C

Total money for the 1st year, under the condition of annual simple interest:
A: A+A*10%
=A(1+10%)

Total money for the 2nd year, under the condition of annual simple interest:
A: (A+A*10%)+A*10%
=A+2*A*10%
=A(1+20%)
=1.2A

Therefore, we can infer:
B's total money for 4 years will be: B+4*B*10%
=B(1+40%)
=1.4B

and
C's total money for 5 years will be: C+5*C*10%
=C(1+50%)
=1.5C

The premise in the question says that these three savings have the same value at the end of their respective deposit periods,
so we can come up this: 1.2A=1.4B=1.5C
, and the question wants to know C, so we can infer:
A=1.5C/1.2
B=1.5C/1.4

Plus, the premise in the question says these three students' original money are from 1860,
so we can list the line as: A+B+C=1,860, which is:
(1.5C/1.2)+(1.5C/1.4)+C =1,860
=> (10.5C+9C+8.4C)/8.4 =1,860
=>27.9C=1,860*8.4
=>27.9C=15,624
=>C=560

Therefore, the answer of this question is C.560.

If anything above is incorrect, pls let me know.

Thank you!

Bunuel

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Bunuel

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Person	Arman	Bela	Cora
Principal	A	B	C
Rate	10%	10%	10%
Time (yrs)	2	4	5
Amount	Am	Am	Am

A + B +C = 1860 (i)
Also amount Am is same for all 3 deposits. Thus
A + (A * 10*2/100) = B + (B *10*4/100) = C + (C*10*5/100)
A + A/5 = B + 2B/5 = C + C/2
6A/5 = 7B/5 = 3C/2
Multiply by 10
12A = 14B = 15C
A = 15C/12 = 5C/4
B = 15C/14
Put values of A and B in eq (i)
5C/4 + 15C/14 + C = 1860
93C/28 = 1860
C = 560
Answer - (C)

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IMO C
Let Arman, Bela & Cora original shar e be a, b, c resp
Then a+b+c=1860--------(1)
Also, a+2a/10= b+4b/10 = c+5c/10 = SI equal as per Q
Then 6a/5 = 7b/5 =3c/2 =k(say)
Thn , a= 5k/6, b= 5k/7, c=2k/3, substituting, a,b, c in (1)
we get k= 20x42
Then c = 2k/3 =560 Hence Cora Original share is $560

Bunuel

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Arman's share = A
Bela's share = B
Clara's share = C

A(1+2*0.1) = B(1+4*0.1) = C(1+5*0.1)
1.2A=1.4B=1.5C

12A=14B=15C
LCM of 12, 14 and 15 = 420
hence A:B:C = (420/12) : (420/14) : (420/15) = 35:30:28 = 35x : 30x : 28x

hence A + B + C = 1860 = 35x +30x +28x = 93 x then x = 1860/93x = 20

C=28x = 28*20 =560

Correct answer is C=560

Bunuel

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Let the shares of Arman be a, Bela be b, and Cora be c.

Then, given a+b+c = 1860 -- eq1

now,
Values for each deposit
Arman - a*10*2/100 + a = \frac{6a}{5}
Bela - b*10*4/100 +b = \frac{7y}{5}
Cora - c*10*5/100 +c = \frac{3c}{2}

Given that the value of each deposit is the same,
so, \frac{6a}{5 = 7b/5 = 3c/2}

Rewriting a and b in terms of c, and putting that in eq 1, we get
c + \frac{5c}{4 + 15c/14} = 1860
Solving for c, we get c = 560

Bunuel

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