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

1 2 3 4

GoodKey

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17 Jul 2026, 11:34

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A + B + C = 1860

Since the final value of each deposit is same across ABC
Amount = P + SI
SI = PRT/100

Therefore, 1.2A = 1.4B = 1.5C
B = 1.5C/1.4
A = 3C/4

Substituting the values in the first equation
3C/4 + 15C/14 + C = 1860
93C/28 = 1860
C = 560 OPTION C

TanviNagrani

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The correct answer is C. 560

A=P(1+rt), r=10%=0.1
For A (Arman) t=2
A*(1+0.1*2)=1.2A

For B (Bela), t=4
B*(1+0.1*4)=1.4B

For C (Cora), t=5
C*(1+0.1*5)=1.5C

1.2A=1.4B=1.5C, from the question.
12A=14B=15C

A=15C/12=5C/4
B=15C/14

putting these into the equation
A+B+C= 1860 (from the question)
5C/4+15C/14+C=1860
(35C+30C+28C)/28=1860
C(35+30+28)/28=1860
C=(1860*28)/93
C=20*28=560

Hence our answer.

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17 Jul 2026, 11:44

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inference: at the end, everyone must have the same share and A+B+C=1860
Ask: Find C

1.2A = 1.5C -> div by 3 -> 4A = 5C
1.4B = 1.5C ---------------> 14B = 15C

so i'll multiply A+B+C = 1860 with 4x14 on both sides

4 x 14 (A+B+C) = 1860 X 4 X 14
14 X 4A + 4 X 14B + 4 X 14 X C = 1860 X 4 X 14

Substitute 4A = 5C & 14B = 15 C

14 X 5C + 4 X 15C + 4 X 14 C = 1860 X 4 X 14
DIV BY 2
7 X 5C + 2 X 15C + 4 X 7C = 1860 X 4 X 7
C ( 35+30+28)
93 C = 1860 X 4 X 7

93 X 2 = 186

C = 20 X 4 X 7 -> 80 X 7 -> 560

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17 Jul 2026, 11:46

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i am going with option c.
1/1.2:1/1.4:1/1.5 = 175:150:140
175+150+140=465
1860/465=4
c's share = 140x4=560
a incorrect - too small, final amount would be less than others
b incorrect - does not produce equal final amount
d incorrect - too large, final amount would exceed others
e incorrect - also too large, final amount would exceed others.

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17 Jul 2026, 12:14

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Let A be armans share, B be Belas share and C be coras share

Since it is simple interest 10% than value is 1.1 x number of years. Final values of all three people are equal. FV

Arman 2 years. 1.2A = final value FV
Bela. 4 years. 1.4B =FV
Cora. 5 years. 1.5C = FV

A = FV/1.2 = 5FV/6
B = FV/1.4 = 5FV/7
C = FV/1.5 = 2V/3

we know A + B + C = 1860

then 5FV/6 + 5FV/7 + 2FV/3 = 1860
Common denominator is 42

therefore (35FV + 30FV + 28 FV) /42 = 1860

93 FV /42 = 1860
31FV/14 = 1860
FV = 1860 x 14/31 = 840

Cora original share is

C = 2FV/3 = 2(840)/3 = 560

ANSWER C $560

ShreyaMittal1609

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My ans is C.) 560

Let shares be a, b, and c

Amounts at the end of 2,4 and 5 years (a+ a*2*10/100) = 1.2a=1.4b=1.5c or 12a=14b=15c

Also a+b+c= 1860
Now replace a and b with c

15c/12+15c/14+c=1860
93c= 620*84
c=560

Bunuel

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

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If we consider the common maturity amount is M.
1860 to be divided amongst 3 students.
Arman has Principal: M/1.2 because 10% over 2 years is 20%
Bella has Principal : M/1.4 because 10% over 4 years is 40%
Cora has principal: M/1.5 because 10% over 5 years is 50%

so (M/1.2) + (M/1.4) + (M/1.4) =1860
Solving this gives M = 840.
So Cora's Original Share: 840/1.5 = 560

IMO ans is C.

Bunuel

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A amount = a + 0.2a =1.2a
B amount = b+ 0.4b=1.4b
C amount= c+0.5c =1.5c

1.2a=1.4b=1.5c
a+b+c=1860

15c/12 + 15c/14 + c=1860
c=560

Ans C

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17 Jul 2026, 13:03

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x + y + z = 1860

1.2x = 1.4x = 1.5z

i.e 12x = 14x = 15z , LCM of 12 , 14 and 15 is = 2^2*3*5*7

Divide each element by the LCM we get the ratio as

Hence ratio of x : y : z as is 35 : 30 : 28 ---> 93k

Hence value of z = 1860/93*28 = 560

I ll go with C

Bunuel

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amount for arman= 1.2A
amount for bela= 1.4B
Amount for Cora= 1.5C
B=15/14C
A=5/4C

5/4 c+ 15/14 c+ c= 1860
c= 560

Bunuel

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Since here, the final amount is supposed to be the same, hence A:B:C - 1/1.2:1/1.4:1/1.5 - SIMPLIFY THIS - 35:30:28. nOW, 1860*28/93 as Cora is 28 parts of 93, ans is 560%. Which is C option.

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Total amount held=$1860
Let Arman, Bela and Cora' share be a b and c respectively
It is shared that the at the rate of 10% annual simple interest their respective shares are deposited for 2,4 and 5 years such that the final amount is same
total Share of A= a+(a*10*2)/100= 12a/10, similarly for B= 14b/10 and for C= 15c/10
12a/10=14b/10=15c/10 =Sum
Let the sum be a common multiple of the sum held by each : Sum= 4*3*5*7 S
a:b:c= 35:30:28
Then c= 1860*28/(28+35+30)=560
Option C is correct

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Bunuel

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Make equations and write in terms of Cora(C).
A+2A/10 = B + 4B/10 = C + 5C/10
Express A and B in terms of C
A = 5C/4 B = 15C/14
C + 5c/4 + 15c/14 = 1860
C = 560
Option C

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I used ratios to solve this question, you could use other methods as well

As given, 1.2a=1.4b=1.5c

12a=14b=15c

to find the ratios of each, we could take two parts together,

12a=14b,
a/b=7/6

14b=15c,
b/c=15/14

Now we could find a:b:c :: 105:90:84
simplified a:b:c :: 35:30:28
the total of a+b+c is of the form 93k (35+30+28)

And given total of a+b+c=1860
so k = 20

so the value for c =28k = 28*20 = 560
IMO C

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C, 56-

Simple interest so each share just gets a multiplier. Arman 1.2, Bela 1.4, Cora 1.5, All three end up equal, so the shares run inverse to those: 1/1.2 : 1/1.4 : 1/1.5, which is 5/7 :5/7 : 2/3

Clear the fraction over 42 and its 35 : 30 :28. Sum is 93 parts and 1860/93 = 20 per part.

Cora gets 28 times 20 = 560.

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The value of each persons share is equal to their share * (1+rate* # of years). All values are the same so I will set it to V

V = Arman's share * (1 + .1*2) = A(1.2)
V= Bella * (1+ .1*4)= B(1.4)
V = Cora*(1 + .1*5) = C(1.5)

We can rewrite each equation to find shares in terms of values

A = V/1.2
B = V/1.4
C = V/1.5

Then we can make the equation

total shares = V/1.2 + V/1.4 + v/1.5 =1860 then solve for V

V = 840

C= V/1.5 = 840/1.5 = $560.

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Let common maturity valuse for all 3 deposits be M

Using simple interest formula (Interest = 10%):

Arman (2 years) -> M = 1.2P, so P = 5M/6
Bela (4 years) -> M = 1.4P, so P = 5M/7
Cora (5 years) -> M = 1.5P, so P = 2M/3

Their original deports add up to $1860: 5M/6 + 5M/7 + 2M/3 = 1860

Simplifying: 31M/14 = 1860 --> M = 840

Cora's original deposit = (2/3) * 840 = 560

Answer: Option C: 560

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Let the final value of each deposits be A.
& intial share of Arman, Bela, & cora are Pa, Pb, Pc.

Given,
Rate of interest, r = 10 %

So, original shares,

Arman's share in 2 years, Pa = V /(1+0.2) = V / 1.2
Bela's share in 4 years, Pb = V ( 1+ 0.4 = V/1.4
Cora's share in 5 years, Pc = V ( 1+0.5) = V/1.5

As total distributed amount is 1860,

V/1.2 +V/1.4 + V /1.5 = 1860

V ( 5/6 + 5/7 + 2/3 ) = 1860
V (31/14) = 1860
V = 840

Cora's initial share, Pc = 840/1.5 = 560

Answer : C ( 560)

Bunuel

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Let the principal amounts deposited for Arman, Bela & Cora be A,B & C respectively
Interest amounts for them respectively will be

Arman = A*10/100*2 = 2A/10
Bela = B*10/100*4 = 4B/10
Cora = C*10/100*5 = 5C/10

Maturity amounts then become

Arman = 1+2A/10 = 12A/10
Bela = 1+4B/10 = 14B/10
Cora = 1+5C/10 = 15C/10

As per question, all these amounts are equal and A+B+C = 1860

12A/10=15C/10 or A=5C/4
14B/10=15C/10 or B=15C/14
15C/10=12A/10 or C=4A/5

Replacing the values, 5C/4+15C/14+C = 1860
93C/28 = 1860, C=560. Answer C

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For this question we have:
a+b+c= 1860

And the deposits at the end for all three are the same.
So,
Deposit of A: a + a*10/100*2 = 1.2a
Deposit of B: b + b*10/100*4 = 1.4b
Deposit of C: c + c*10/100*5 = 1.5c

1.2a = 1.4b = 1.5c

Now we have to find c.

Let's consider c =560
1.5*c = 3*280 (this goes by 1.2a and 1.4b as well)
b= 600
a= 700.

To test it out: 1.4*600 = 840
1.2* 700 = 840
1.5* 560 = 840.
Other options were giving values in decimal which would not allow the a+b+c = 1860.