If x, y, and z are positive integers and 3x = 4y = 7z, then the least

Question

If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is

A. 33
B. 40
C. 49
D. 61
E. 84

https://lh3.googleusercontent.com/_NgVz7MmLvZg/Tb8Wu-Y4hPI/AAAAAAAAAEE/6notAbHoQrU/1.jpg

Bunuel · Accepted Answer

Notice that since 3, 4, and 7 are co-prime (they don't share any common factor but 1), then:

x must be a multiple of 4 and 7, so the least value of x is 28 (LCM of 4 and 7);
y must be a multiple of 3 and 7, so the least value of y is 21;
z must be a multiple of 3 and 4, so the least value of z is 12;

So, the least possible value of x + y + z is 28+21+12=61.

Answer: D.

KarishmaB · Answer

When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'.

3x = 4y = 7z = k

x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7} 
Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.

x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61

Spidy001 · Answer

given 3x=4y=7z

x+y+z in terms of x

= x+(3x/4)+(3x/7) = 61x/28

now checking with each of the answers and see which value gives a minimum integer value.

A x = 28*33/61 , not an integer

B,C can be ruled out similarly.

D is minimum value as x = 61*28/61 = 28

Answer is D.

pike · Answer

3x = 4y = 7z

LCM = 3 * 4 * 7 = 84
84 / 3 = 28
84 / 4 = 21
84 / 7 = 12

28 + 21 + 12 = 61

hamm0 · Answer

The LCM of 3, 4, and 7 = 84. (Easy to figure out - since 3, 4, and 7 are prime, the LCM is 3*4*7)

3(x) = 84 ->  x = 28 
4(y) = 84 ->  y = 21 
7(z) = 84 ->  z = 12

28+21+12 =   61  .

Hope this helps.

WoundedTiger · Answer

We know x,y and z are postive integers so based on given information, we can say
x= 4/3y and z= 4/7y

then x+y+z= 4/3y+y+4/7y-------> (4/3y+4/7y)+ y ------> 40y/21 +y -----> 61 y/21

61y/21= Sum of positive integers x,y and z. This sum also is a postive integer. Let us say it as Integer "a"

therefoee 61y/21= a now this implies y= a*21/61 ----For y to be an integer, we see that "a" has to be multiple of 61 and the smallest multiple of 61 is 1

Hence ans should D i.e 61

PareshGmat · Answer

LCM of 3, 4 & 7 is 84

Least value of x+y+z

= \frac{84}{3} + \frac{84}{4} + \frac{84}{7}

= 28 + 21 + 12 = 61

Answer = D

energetics · Answer

1) We should minimize z while keeping 7z divisible by a multiple of 3 and 4 (which means z must be even, since 7 is odd and we need divisibility by an even number)

2) 7z must be divisible by 12 because 3x and 4y will have 12 as their LCM. First multiple of 7 divisible by 12 is 84, which is trap AC E) --> we need x+y+z NOT the LCM of all 3

3) 84/3 = 28, 84/4 = 21, 84/7 = 12 
28 + 21 + 12 = 61, D)

pandajee · Answer

VeritasKarishma 
Ma'm, why have you replaced "K" by 3x4x7 (confused by this equality-- 3x=4y=7z=k)

KarishmaB · Answer

You have three different terms, all equal to each other. We bring it equal to k to bring everything in terms of a single variable k. Then we have x, y and z, all in terms of k and we can easily find their ratio. 
x = k/3
y = k/4
z = k/7

Since x, y and z all need to be integer, k/3 should be an integer so k should have 3 as a factor. k/4 should be an integer so k should have 4 as a factor. k/7 should be an integer so k should have 7 as a factor. 
So k should have 3, 4 and 7 as factors so minimum value of k = 3*4*7 
So min value of x = 28, of y = 21 and of z = 12

Alternatively, we can focus on one pair of equation at a time:
3x = 4y
x/y = 4/3 = 28/21

4y = 7z
y/z = 7/4 = 21/12

x : y: z = 28:21:12

Since x, y and z are integers, least value of their sum = 28 + 21 + 12 = 61

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