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Bunuel
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A children’s gift store sells gift certificates in denominations of $3 09 Mar 2016, 13:00
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B
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75% (hard)

Question Stats:

57% (02:27) correct 43% (02:33) wrong based on 218 sessions

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A children’s gift store sells gift certificates in denominations of $3 and $5. The store sold ‘m’ $3 certificates and ‘n’ $5 certificates worth $93 on a Saturday afternoon. If ‘m’ and ‘n’ are positive integers, how many different values can ‘m’ take?

A. 5
B. 6
C. 7
D. 18
E. 31
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B
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GMAT Focus 1: 735 Q90 V89 DI81
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A children’s gift store sells gift certificates in denominations of $3 09 Mar 2016, 20:43
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Divyadisha
Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

:)
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Hi,
you have to be very careful while reading the Q..
its clearly written that m and n are positive integers, so 93+0 is not a valid solution..
so answer is 7-1= 6..

another way to avoid calculations is --
We can use number properties to home on the answer without doing all calculations..
3m+5n=93=3*31.
lets concentrate on n, as all other terms are multiple of 3
3 and 5 are COPRIME, so n should be a multiple of 3..
so n=3,6,..., lower int value 93/5=18.6=18..
so y can take (18-3)/3 + 1= 5+1=6
ans 6.. B[/b]
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A children’s gift store sells gift certificates in denominations of $3 09 Mar 2016, 20:32
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Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

:)
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A children’s gift store sells gift certificates in denominations of $3 09 Mar 2016, 20:48
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Divyadisha
Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

:)
Show more

Hi Divyadisha ,
The question states that m and n are positive integers , so your last value pair m=31 and n=0 violates this fact .
3m+5n = 93
3*1 + 5*18 = 93
3*6 + 5*15 = 93
3*11 + 5*12 = 93
3*16 + 5*9 = 93
3*21 + 5*6 = 93
3*26 + 5*3 = 93

Number of different values m can take = 6
Answer B
Hope it helps!! :)
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A children’s gift store sells gift certificates in denominations of $3 10 Mar 2016, 09:43
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chetan2u
Divyadisha
Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

:)

Hi,
you have to be very careful while reading the Q..
its clearly written that m and n are positive integers, so 93+0 is not a valid solution..
so answer is 7-1= 6..

another way to avoid calculations is --
We can use number properties to home on the answer without doing all calculations..
3m+5n=93=3*31.
lets concentrate on n, as all other terms are multiple of 3
3 and 5 are COPRIME, so n should be a multiple of 3..
so n=3,6,..., lower int value 93/5=18.6=18..
so y can take (18-3)/3 + 1= 5+1=6
ans 6.. B[/b]
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Thanks for pointing out the mistake. Hope I will not do this silly mistake on the actual day :P
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A children’s gift store sells gift certificates in denominations of $3 10 Mar 2016, 09:44
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Skywalker18
Divyadisha
Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

:)

Hi Divyadisha ,
The question states that m and n are positive integers , so your last value pair m=31 and n=0 violates this fact .
3m+5n = 93
3*1 + 5*18 = 93
3*6 + 5*15 = 93
3*11 + 5*12 = 93
3*16 + 5*9 = 93
3*21 + 5*6 = 93
3*26 + 5*3 = 93

Number of different values m can take = 6
Answer B
Hope it helps!! :)
Show more

Totally agree!! I got to read the question very carefully and not with sleepy head :). Thanks a lot!
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A children’s gift store sells gift certificates in denominations of $3 13 Dec 2019, 06:25
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Bunuel
A children’s gift store sells gift certificates in denominations of $3 and $5. The store sold ‘m’ $3 certificates and ‘n’ $5 certificates worth $93 on a Saturday afternoon. If ‘m’ and ‘n’ are positive integers, how many different values can ‘m’ take?

A. 5
B. 6
C. 7
D. 18
E. 31
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3m+5n=93, 3m=93-5n, m=31-5n/3;
5n/3=integer, so n=multiple(3);
0<5n/3=integer<31, 0<5n≤90, 0<n≤18;
n=[3,6,9,12,15,18]=6

Ans (B)
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A children’s gift store sells gift certificates in denominations of $3 04 Nov 2022, 04:37
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Is the answer 'c' i.e. 7

3m +5n = 93

I kept reducing the 5n factor by 15 as 15 is the factor of 3 and kept adding in the 3m. It gives below combinations:-

3+90
18+75
33+60
48+45
63+30
78+15
93+0

Posted from my mobile device
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A children’s gift store sells gift certificates in denominations of $3 04 Nov 2022, 09:40
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Bunuel
A children’s gift store sells gift certificates in denominations of $3 and $5. The store sold ‘m’ $3 certificates and ‘n’ $5 certificates worth $93 on a Saturday afternoon. If ‘m’ and ‘n’ are positive integers, how many different values can ‘m’ take?

A. 5
B. 6
C. 7
D. 18
E. 31
Show more

We can create the following equation:

3m + 5n = 93

5n = 93 - 3m

5n = 3(31 - m)

n = 3(31 - m)/5

n = Since 5 does not divide 3, we see that 5 must divide 31 - m.

Thus, the possible values of m are 1, 6, 11, 16, 21, 26. Thus, m and n can be a total of 6 values.

Answer: B
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A children’s gift store sells gift certificates in denominations of $3 30 Jan 2024, 21:47
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