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m=18(k)+12
n=24(r)+14

Both m and n must also be a mulitiple of 6. Therefore on diving each by 6, remainder is 0 for m and 2 for n and m+n must give a
remainder of 2 on division by 6. Thus only 50 and 92 fit the bill.
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deeksha6
m=18(k)+12
n=24(r)+14

Both m and n must also be a mulitiple of 6. Therefore on diving each by 6, remainder is 0 for m and 2 for n and m+n must give a
remainder of 2 on division by 6. Thus only 50 and 92 fit the bill.

Hi deeksha,
You may be correct in your observation, but there is more to it..
If I give you say 14,20,32, they will fit in in your approach but will not be correct...
Thus any of these choices used would have got you to a wrong answer

Now WHY?
m+n= 18a+24b+26... where m=18a+12 and n=24b+14...
What Does 18a+24b+26 mean
1) it has to be more than 26..
2) x and y will be integers so next number will be 26+18*1+24*0=26+18=44..m=12+18=30 and n=14..
Next 26+18*0+24*1=26+24=50..m is 12 and n 14+24=38
Next 26+18*2+24*0=26+36=62.. m is 12+18*2=48..n is 14
And so on
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m = 12, 30, 48, 66, ....
n = 14, 38, 62, 86, .......

12+38 = 50
30+62 = 92

Ans: D
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m = 18p+12, n= 24q+14
m+n = 18p+24q+26 = 6(some quotient)+2, so choices should leave remainder when divided by 6
therefore answer is D) I ad III
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Bunuel
The variables m and n are positive integers. When m is divided by 18, there is a remainder of 12, and when n is divided by 24, there is a remainder of 14. Which of the following are possible values of m + n?

I. 50
II. 70
III. 92

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

m+n=18q+24p+26
(m+n)/2=9q+12p+13
I. 25-13=12=9q+12p
possible: q=0; p=1
II. 35-13=22=9q+12p
not possible
III. 46-13=33=9q+12p
possible: q=1; p=2
D
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Bunuel
The variables m and n are positive integers. When m is divided by 18, there is a remainder of 12, and when n is divided by 24, there is a remainder of 14. Which of the following are possible values of m + n?

I. 50
II. 70
III. 92

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

(Divisor*quotient) + remainder = dividend

18*x + 12=m
24*w +14=n

m+n= 18x+24w+26

for m+n =50,
50=18x+24w+26
This can be the case when w=1 and x=0. So possible

for m+n=70
70=18x+24w+26. not possible for any combination of x and w.

for m+n=92,
92=18x+24w+26
This can be the case when w=2 and x=1. so possible

Hence D.
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Bunuel
The variables m and n are positive integers. When m is divided by 18, there is a remainder of 12, and when n is divided by 24, there is a remainder of 14. Which of the following are possible values of m + n?

I. 50
II. 70
III. 92

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

Think of it in terms of divisibility fundamentals

" When m is divided by 18, there is a remainder of 12" - When m balls are divided into groups of 18, we are left with 12 balls

"when n is divided by 24, there is a remainder of 14" - When n balls are divided into groups of 24, we are left with 14 balls

So m + n balls would mean some groups of 18 balls, some groups of 24 balls and a group of 12+14 = 26 balls
The groups of 18 balls can be split into 3 groups of 6 balls (HCF of 18 and 24) each and the groups of 24 balls can be split into 4 groups of 6 balls each too. As for the 26 balls, we can make 4 groups of 6 balls each and still have 2 balls leftover.
So m + n would be 2 more than a multiple of 6. Only I and III satisfy this.

Answer (D)
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