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# There are two packets – x and y. ‘x’ contains only 25 cents coins and

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Joined: 19 Oct 2018
Posts: 1151
Location: India
There are two packets – x and y. ‘x’ contains only 25 cents coins and  [#permalink]

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17 Jun 2019, 17:30
7
00:00

Difficulty:

65% (hard)

Question Stats:

43% (02:28) correct 57% (03:50) wrong based on 21 sessions

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There are two packets – x and y. ‘x’ contains only 25 cents coins and ‘y’ contains only 50 cents coins. Total amount in both the packets put together is 80 Dollars. For every 4 coins transferred from x to y, 3 coins are transferred from y to x. It is also known that only 25 cents coins are transferred from x and only 50 cents coins are transferred from y. After ‘n’ such transfers, the total number of coins in both the packets becomes equal. Which of following can be a value of n?

A. 8
B. 12
C. 16
D. 20
E. 24
Senior Manager
Joined: 31 May 2018
Posts: 453
Location: United States
Concentration: Finance, Marketing
There are two packets – x and y. ‘x’ contains only 25 cents coins and  [#permalink]

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17 Jun 2019, 20:19
1
nick1816 wrote:
There are two packets – x and y. ‘x’ contains only 25 cents coins and ‘y’ contains only 50 cents coins. Total amount in both the packets put together is 80 Dollars. For every 4 coins transferred from x to y, 3 coins are transferred from y to x. It is also known that only 25 cents coins are transferred from x and only 50 cents coins are transferred from y. After ‘n’ such transfers, the total number of coins in both the packets becomes equal. Which of following can be a value of n?

A. 8
B. 12
C. 16
D. 20
E. 24

let there be a 25 cent coin and b 50 cent coin
since the total amount in both packets are given

0.25a + 0.50b = 80
a + 2b = 320.......(1)
now
box (X).................box(Y)
a coins................b coins (For every 4 coins transferred from x to y, 3 coins are transferred from y to x and after n such transfer)
a-4n...................b+4n ( transfer from x to y)
a-4n+3n..............b+4n-3n (transfer from y to x)

a-4n+3n=b+4n-3n (given)
a-n=b+n
a-b = 2n.............(2)

solving equation (1) and (2)
we get a = $$\frac{320+4n}{3}$$

a = integer (number of coins)
so 320 + 4n must be multiple of 3
only option c satisfy
n = 16
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Re: There are two packets – x and y. ‘x’ contains only 25 cents coins and  [#permalink]

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17 Jun 2019, 21:25
1
Let the number of coins in two packets - x and y be a and b.

Now 0.25a + 0.5b = $80 ______________(1) Each time 4 coins are being taken from packet x and 3 coins are being taken from packet y. This operation is performed for n times. a-4n --> b+4n a-4n+3n --> b+4n-3n After n operations, this count became equal in both packets. a - n = b + n a - b = 2n __________(2) multiplying eq (2) by 0.5 0.5a - 0.5b = n adding this with eq (1) 0.5a - 0.5b = n 0.25a + 0.5b = 80 --------------------------- 0.75a = 80 + n ----------------------- Now a has to be a positive integer as the number of coins cannot be any other value apart from a positive integer. From options, only option C for n = 16, a value will be a positive integer. Only for n = 16, a = 128 and b = 96. These values satisfy the eq (1) andd hence n = 16. OPTION: C Manager Joined: 07 May 2018 Posts: 61 Re: There are two packets – x and y. ‘x’ contains only 25 cents coins and [#permalink] ### Show Tags 20 Jun 2019, 13:03 eswarchethu135 wrote: Let the number of coins in two packets - x and y be a and b. Now 0.25a + 0.5b =$80 ______________(1)

Each time 4 coins are being taken from packet x and 3 coins are being taken from packet y. This operation is performed for n times.

a-4n --> b+4n
a-4n+3n --> b+4n-3n

After n operations, this count became equal in both packets.

a - n = b + n

a - b = 2n __________(2)

multiplying eq (2) by 0.5

0.5a - 0.5b = n

0.5a - 0.5b = n
0.25a + 0.5b = 80
---------------------------
0.75a = 80 + n
-----------------------

Now a has to be a positive integer as the number of coins cannot be any other value apart from a positive integer.

From options, only option C for n = 16, a value will be a positive integer. Only for n = 16, a = 128 and b = 96. These values satisfy the eq (1) andd hence n = 16.

OPTION: C

Could you explain why the answer is 16 for n

You do get positive integers with Options A & B
Re: There are two packets – x and y. ‘x’ contains only 25 cents coins and   [#permalink] 20 Jun 2019, 13:03
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