The United States mint produces coins in 1-cent, 5-cent, 10-cent, 25-c

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Ih the highest denominations is 50 cent in the jar, then maximum possible number of coins jar is 1+50= 51. Don't check for cases when there is 50 cent coin in the jar.

We'll check the same thing with 25 cent coin.

If there is a 25 cent coin in the jar, then maximum number of coin jar can have is 1+75 = 76 (III)

We don't have to check for any other 25 cent cases, as other 2 options are greater than 76. (We can reject option A and B here)


If the highest denominations is 10 cent in the jar, then maximum possible number of coins is 1+90= 91 (I)

Now we have to check only 1 case, whether 86 coins in jar is possible.

If there is a 25 cent coin in the jar then maximum number of coin jar can have is 2+80 = 82.

Hence if there are 86 coins in the jar, then there is either 0 or 1 10-cent coin in the jar.

First case- There is 1 10-cent in jar

x+y+1= 86....(1)

x+5y+10=100...(2)

No integral solution is possible

Second case- There is 0 10-cent in jar

x+y= 86....(1)

x+5y=100...(2)

No integral solution is possible

Hence there can never be 86 coins in the jar.


D

I am not a fan of algebra and only a few GMAT questions NEED algebraic equations to be solved. Instead of looking for algebraic ways in logical solutions, go the other way around - look for logical solution in algebraic questions. For that, GMAT will reward you.

Here, if you want to look at it algebraically, see this:
Try to use integer solutions to an equation in a limited way (considering only 1c and 5c coins).

x + 5y = 100

First simple solution is 100, 0 (you take 100 coins of 1c and 0 coins of 5c)

Next few integer sols:

95, 1 (Total 96 coins)
90, 2 (Total 92 coins)
85, 3 (Total 88 coins)
80, 4 (Total 84 coins)
75, 5 (Total 80 coins)
70, 6 (Total 76 coins)
...
fewer and fewer coins going down...

You will not get 81 coins.

Algebraic approach:
Let A=# of 1c coins, B=# of 5c coins, C=# of 10c coins, D=# of 25c coins, E=# of 50c coins.

(1) A+5B+10C+25D+50E=100 (Cent value of coins)

I.
(2) A+B+C+D+E=91 (Total number of coins)
Subtract equations (1)-(2): 4B+9C+24D+49E=9
Equation will work if B=0, C=1, D=0, E=0; i.e. one 10c coin. Can then solve for A using equation (2), A+0+1+0+0=91, A = 90 (# of 1c coins).

II.
(2) A+B+C+D+E=81 (Total number of coins)
Subtract equations (1)-(2): 4B+9C+24D+49E=19
From a quick glance, no combination of B,C,D,E will fit into the equation. Thus, eliminate II.

III.
(2) A+B+C+D+E=76 (Total number of coins)
Subtract equations (1)-(2): 4B+9C+24D+49E=24
Equation can work if B=0, C=0, D=1, E=0; i.e. one 25c coin. Can then solve for A using equation (2), A+0+0+1+0=76, A = 75 (# of 1c coins).
Equation can also work if B=6, C=0, D=0, E=0; i.e. six 5c coins. Can then solve for A using equation (2), A+6+0+0+0=76, A = 70 (# of 1c coins).

We can tell very quickly that 91 coins and 76 are possible.

I. 91 coins = 90 pennies, 1 dime.
III. 76 coins = 75 pennies, 1 25-cent

Now, lets take a look at (II) 81.

81 coins:

80 pennies, 2 10-cent. 82 > 81. Not possible.
75 pennies, 5 nickels. 80 coins < 81. Not possible.

By testing values we can see that it's not possible to get 100 cents with 81 coins. Answer is D.

Hello,

Shouldn't it be 50 instead of 10 in the first line ?

Algebraically, this looks tough to solve - may be because i see 5 variables in a + 5b + 10c + 25d + 50e. I could reach out to D and E but faltered as for a moment i thought all types of coins are required to solve. One way to solve can be to check minimum and maximum number of coins that sum exactly 100 cents.
The minimum number of coins would be e = 2(2 50-cents coins).
The maximum number of coins would be a = 100(100 1-cents coins).

As options available are nearer to 100, most likely we should solve keeping 1-cents coins at maximum possible numbers. The 1-cents coins need to be in multiples of 5 as we can see from the equation 'a + 5(b + 2c + 5c + 10e)'.
So, lowering down the number of coins from 100 the next lowest value is
96 [a(95) + 5b(1) = 100]
92 [a(90) + 5b(2)]
91 [a(90) + 10c(1)]
88 [a(85) + 5b(3)]
87 [a(85) + 5b(1) + 10c(1)]
84 [a(80) + 5b(4)]
83 [a(80) + 5b(2) + 10c(1)]
82 [a(80) + 10c(2)]
80 [a(75) + 5b(5)]
78 [a(75) + 5b(1) + 10c(2)]
76 [a(75) + 25d(1)]

As we can see 81 is not possible, only 91 and 76 are possible.

Looking back at the approach i see we can check for highest number of coins - a combination of largest possible number of 1-cents coins and 1 coin of any other - but this might no click at first under exam condition.
1-cent coins(a) = 100 coins
1-cent(a) + 5-cent coins(b) = 96
1-cent(a) + 10-cent(c) = 91
1-cent(a) + 25-cent(d) = 76
1-cent(a) + 50-cent(e) = 51

Finally, had we had coins of denominations 15-cent and 20-cents we would had 86 and 81 respectively.
Answer D.

Back of the envelope thought: Algebraic equation actually helps to form a base to start with.
Hope this helps.

i agree with some of the people here, the question indirectly states(THESE COINS that we have to use all the coins, it becomes lengthy to calculate then.

Hi Experts -- chetan2u, Bunuel, VeritasKarishma, generis

I tried doing trial and error and after 2 minutes of trying different cases -- i was able to get 76 but wasn't able to get 91 or 81, so i guessed and moved on

Just wondering, what is the concept being tested in this problem specifically...is it divisibility ?

How would you solve this using logic / reasoning using theory we have learnt as part of the strategy guides ?

Also, i tried to find a similar problems to this -- is this a similar problem / theory being tested on your view ?

https://gmatclub.com/forum/if-x-and-y-a ... 05722.html

Hi VeritasKarishma,

very nice explanation. I have just one curiosity. Is there a way to represent what you said by an integral equation?



Please let me know if we could represent the situation algebraically, it would help.

Thanks

what is the best way to solve this kind of question instead of assuming values?

Hi experts,

I have a query regarding the language of the question.

The question stem says, "If a jar contains exactly 100 cents worth of THESE coins".

Doesn't that mean all the denominations have to be there to make, to sum up to exactly 100 cents?
The wording is so confusing.

Hi Experts,

I do not agree with the solution methods provided. The Q quality is poor. It says If a jar contains exactly 100 cents worth of these coins, which of the following could be the total number of coins in the jar?. So, the question should have been framed a atleast any coins or the total value of the coins changed to smaller values. I do not see why such a question is being tested in Gmat...

Thanks,

The question is fine.

"100 cents worth of these coins" means the worth of the coins in the jar is 100 c.
This is the same as saying "total value of the coins in the jar is 100 c"
GMAC tests its questions thoroughly before making them live so any question with ambiguous language (as per test-taker feedback through experimental questions) is rooted out. So just trust the process and do your best.

Given: The United States mint produces coins in 1-cent, 5-cent, 10-cent, 25-cent, and 50-cent denominations.

Asked: If a jar contains exactly 100 cents worth of these coins, which of the following could be the total number of coins in the jar?

x1 + 5x2 + 10x3 + 25x4 + 50x5 = 100
x1 + x2 + x3 + x4 + x5 = ?


x5=0;x4=0;x3=1;x2=0;x1=90
There could be 91 coins in the jar.


x5=0;x4=0;x3=2;x2=0;x1=80; Total coins = 82
x5=0;x4=0;x3=1;x2=2;x1=80; Total coins = 83
x5=0;x4=1;x3=0;x2=0;x1=75; Total coins = 76
There could NOT be 81 coins in the jar.


x5=0;x4=1;x3=0;x2=0;x1=75
There could be 76 coins in the jar.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

IMO D

The number of minimum coins used to make 100 cents is 2 coins of 50 cents each.
The number of maximum coins used to make 100 cents is 100 coins of 1 cent each.
So, the number of coins could be any number from 2-100.

You can think of this:100 coins of 1 cent are required to make 100 cents.
5 coins of 1 cent each can be replaced with 1 coin of 5 cents so the total possible coins could be 96(100-4), 92(100-4-4), and so on.
Similarly,10 coins of 1 cent each can be replaced with 1 coin of 10 cents, hence the number of possible coins could be 91(100-9),82(100-9-9), and so on. So,91 is a possible option.
Similarly,25 coins of 1 cent each can be replaced with 1 coin of 25 cents, hence the number of possible coins could be 76(100-24),52(100-48), and so on. So,76 is a possible option.

There is no combination of 81 coins that could make 100 cents.

Hence, the correct answer is D.