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

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a,b,c
t = 2,4,5
r = 10%
amount = a(1+1/5) = b(1+2/5) = c(1+1/2)
6a/5 = 4b/5 = 3c/2
a+b+c = 1860
5c/4 + 15c/14 + c = 1860
93c/28 = 1860
c = 560

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17 Jul 2026, 23:36

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Let the common amt. received at maturity be x

Arman's share = x/1.2 = 5x/6
Bela's share = x/1.4 = 5x/7
Cora's share = x/1.5 = 2x/3

So,
5x/6 + 5x/7 + 2x/3 = 1860
35x + 30x + 28x = 1860*42
93x = 78,120
x = 840
Cora's original share = 2x/3 = 560

Ans : C

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17 Jul 2026, 23:45

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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?

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

SI = P*R*T/100
1860 is to be divided among A,B,C
interest of a,b,c 10% for each
time a,b,c is 2,4,5 years respecively
value of a = 1.2x , b = 1.4y , c= 1.5z
let final value be n
1.2x=1.4y=1.5z=n
x= 5n/6
y= 5n/7
z= 2n/3
x+y+z = 1860
solve in terms of n we get
5n/6+ 5n/7 + 2n/3 = 1860
n = 840
deposit value of Cora = 2*840/ 3 = 560
560 ; option C is correct

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18 Jul 2026, 00:17

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say A= x
B = y
C = z

each with 10% interest
A for 2 yrs
B for 4 yrs
C for 5 yrs.

total money at the end of their respective years is same for all.

total for A = x+ 0.2x= 1.2x
B = y+0.4y = 1.4y
C= z+0.5z = 1.5z

1.2x=1.4y=1.5z

12x=14y=15z
x= 15z/12
y= 15z/14

x+y+z = 1860
15z/12 + 15z/14 + z = 1860

upon solving this we get, z= 560

Choice C

Bunuel

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Simple Interest: Amount = Principal × (1 + rt)

r = 10% = 0.1

Let amount be x.

Then,

Arman's share:

P(1 + 0.1 × 2) = x

1.2P = x

----> P = x/1.2 = 5x/6

Bela's share:

1.4P = x

----> P = x/1.4 = 5x/7

Cora's share:

1.5P = x

----> P = x/1.5 = 2x/3

The total original amount is $1,860.

----> 5x/6 + 5x/7 + 2x/3 = 1860

LCM = 42.

35x + 30x + 28x
---------------- = 1860
42

93x/42 = 1860

31x/14 = 1860

31x = 26,040

x = 840

Cora's original share: 2x/3

= 2(840)/3

= 560

Answer: C

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18 Jul 2026, 00:57

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simple interest = Prt
total amount = Prt + P = P(1+rt)
A: A(1+0.1*2) = 1.2A
B: B(1+0.1*4) = 1.4B
C: C(1+0.1*5) = 1.5C
we are given that the amounts are equal
1.2A=1.4B=1.5C
12A=14B=15C
A+B+C=1860
A=15C/12, B=15C/14
15C/12 + 15C/14 +C = 1860
279C = 84*1860
C = 560

Ans: C

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18 Jul 2026, 01:28

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A+B+C=1860
simple interest= A=P(1+rt)
r=0.10,
let x=common final value
Arman(2 years):
x=A(1+0.10*2)=1.2A
A=X/1.2
Bela(4 years)
x=B(1+0.1*4)=1.4B
B=X/1.4
Cora(5 years)
x=C(1+0.1*5)=1.5C
C=X/1.5

X/1.2+X/1.4+X/1.5=1860
X(5/6+5/7+2/3)=1860
X(35/42+30/42+28/42)=1860
X(93/42)=1860
X=1860*42/93=20*42=840

C=840/1.5=560

ANS: C. 560

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18 Jul 2026, 02:24

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Lets say F is the final value
Let A for Arman, B for Bela and C for Cora

so F = A(1+0.1x2) = 1.2A, A = F/1.2
Same for Bela dn Cora

Total share is F/1.2 + F/1.4 + F/1.5 = 1860

Convert the decimals into fractions

F = 840

C=840/1.5 = 560

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18 Jul 2026, 02:34

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Total amount= $1860
A’s share time= 2 yrs
B’s share time= 4 yrs
C’s share time= 5 yrs
Three deposits have same value at end of their respective deposit periods.
C’s original share= ?
Simple Interest= P*R*T/100
Total Amount= P+PRT/100 = P(1+0.RT)
A’s share= A(1+0.10*2)=1.2A
B’s share= B(1+0.10*4)=1.4B
C’s share= C(1+0.10*5)=1.5C
All deposits have same values at their end period = M
1.2A=M ; A=M/1.2
1.4B=M ; B=M/1.4
1.5C=M ; C=M/1.5
Total value of all three shares = 1860
A+B+C=1860
(10M/12)+(10M/14)+(10M/15)=1860
10M((1/12)+(1/14)+(1/15))=1860
M((35+30+28)/420)=186
M(93/420)=186
M=186*420/93=840

C’s share= M/1.5 =840/1.5=560

C. 560

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18 Jul 2026, 03:05

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Using SI formula we get -
1.2A = 1.4B = 1.5C lets suppose = k
Take lcm & solve,
So k = 840, also sum = 1860
Thus c = 560 upon solving
OPTION C

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18 Jul 2026, 03:12

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Let us consider A, B and C as the principal amounts divided among Arman, Bela nad Cora respectively. It is given A+B+C = 1860
Also it is given that amount A is deposited for 2 years, amount B for 4 years and amount C for 5 years at 10 percent Simple Interest respectively.

So the amount for Arman at the end of 2 years = A + (A*10*2)/100 = 6A/5
The amount for Bela at the end of 4 years = B + (B*10*4)/100 = 7B/5
The amount for Cora at the end of 5 years = C + (C*10*5)/100 = 3C/2

It is given in the question that their individual shares at the end of their respective deposit periods are equal and hence we can equate

6A/5 = 3C/2 --> A =5C/4
7B/5 = 3C/2. --> B = 15C/14

Since, A + B + C = 1860
5C/4 + 15C/14 + C = 1860
LCM of 4 and 14 is 28 and hence the eqn becomes --> 93C/28 = 1860

On solving this we get C= 560 dollars (Option C is the right answer)

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Ans Choice C: Given that A + B + C = 1860 ;
Arman deposit after 2yrs @10% SI = 1.2A
Bela deposit after 4yrs @10% SI = 1.4B
Cora deposit after 5yrs @10% SI = 1.5C
We are given that these are equal. i.e 1.2A = 1.4B = 1.5C

putting back in the original eq and solving for C,

15C/12 + 15C/14 + C = 1860;

we get C=560.

Bunuel

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18 Jul 2026, 03:23

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Let the original shares be Arman=a, Bela=b and Cora = c
So total earning for each will be:
(1+0.1x2)a=1.2a
(1+0.1x4)b= 1.4b
(1+0.1x5)c= 1.5c
1.2a=1.4b=1.5c=k
1.2a=k so a=k/1.2, b= k/1.4. c=k/1.5
k/1.2+k/1.4+k/1.5=1860
93k/42=1860
K=840
c= k/1.5= 840/1.5=560
Ans C

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18 Jul 2026, 04:16

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Let principle amount with Arman, bela, cora be a, b and c respectively.
a+b+c = 1860. This is our eqn 1

Amount with Arman at the end = a + (a*0.1*2) = 1.2a
Similarly, amount with Bela at the end = 1.4b and amount with Cora at the end = 1.5c

Since all three amounts at the end are equal, 1.2a= 1.5c or a=1.5c/1.2
Similarly, 1.4b=1.5c or b = 1.5c/1.4

On substituting these values into our eqn 1 and solving, we get, c =560

Bunuel

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18 Jul 2026, 04:31

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Share of A,B,C after deposits complete considering original amounts are a,b,c

A = 1.2a
B = 1.4b
C = 1.5c

given a + b + b = 1860 & 1.2a = 1.4b = 1.5c = d (deposit at the end)

So we know c = d/1.5, a = d/1.2, b = d/1.4
d/1.2 + b/1.4 + c/1.5. = 1860 => d = 840

So C's amount is 840/1.5 = 560

Ans - C

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18 Jul 2026, 04:34

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let Arman (a), Bela (b), Cora (c) be respective shares:
a+b+c=1860
1.2a=1.4b=1.5c as post interest tenure total value is same
1.5c/1.2 + 1.5c/1.4 + c =1860
==> c= 560

Bunuel

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18 Jul 2026, 04:36

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Formula: A=P(1+rt)
where A=Final amount, P=Principal/Origial Deposit, r=annual interest rate, and t=years

Then, x=final amount is similiar in A, B, and C

Arman: x = A (1+0.1*2) = 1.2A
Bela: x = B (1+0.1*4) = 1.4B
Cora: x = C (1+0.1*5) = 1.5B

1.2A = 1.4B = 1.5C ... (1)
A+B+C=1860 ... (2)

Then we acan find C
- 1.2A = 1.5C -> A= 1.5/1.2 C
- 1.4B = 1.5C -> B = 1.5/1.4 C

A+B+C = 1.5/1.2 C + 1.5/1.4 C + C = 1860
C = 560 (C)

Bunuel

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	r	t	p	i	Amount
A	10%	2	a	0.2a	1.2a
B	10%	4	b	0.4b	1.4b
C	10%	5	c	0.5c	1.5c

1.2a=1.4b=1.5c

a/b=14/12=7:6=35:30
b/c=15/14=30:28
a:b:c= 35:30:28

.: 35x+30x+28x=1860
.: 93x=1860
.: x=20

.:c=28*20= 560

.: Ans is C.

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18 Jul 2026, 05:27

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Answer: C

Create equation a + b + c = 1860

Notice the total value of the shares are equal, not simply the interest rates, so we have the equation:
Or; 1.2a = 1.4b = 1.5c

Lets' rewrite each of a and be in terms of c and plug into the first equation. Personally I found it easier to convert to fractions.

a = 1.4b/1.2
b = 1.5c/1.4

1.4 cancels out in a and we have a = 1.5c/1.2

c + 1.5c/1.4 + 1.5c/1.2 = 1860

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

c + 15c/14 + 5c/4 = 1860

1860(28)/93 = c

c = 560

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18 Jul 2026, 05:33

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IMO : C
A = P (1+rt)
r = 10% = 0.1
A (1+0.1 .20 = 1.2A = X ==> A = 5X/6
similarly, after aclculation,
B = 5X/7
C = 2X/3
A+B+C = 1860
after solving====> C = 2/3 *840 = 560