A park has 100 rocks. Each rock is either 1 pound, 10 pounds or 50 pounds.?

Question

The total weight of these rocks is 500 pounds. If x is the number of 10-pound rocks, how many different values of x are possible?

Answer 1

Let the no. of 10 pound rocks be x, 1 pound ones be y, and that of 50 pounds be z. Then:
10x + y + 50z = 500
x + y + z = 100
These give:
40x + 49y = 4500 => y= 20*(225-2x)/49
9x + 49z = 400 => z= (400-9x)/49

For y to be a whole number, x=88,39
Out of these, only x = 39 gives a whole number value of z.

Hence, only one value of x is possible, x=39

Hope this helps, and thanks a lot for asking such a great question!

Answer 2

say the number of 1 pound rock is y and that of 50 pound is z.

Therefore 1 y + 10 x + 50 z = 500

or 10 x = 500 - y -50z

or x = (500 -y -50z)/10

or x = 50 - y/10 - 5z
now , y can be a multiple of 10 and that can have a maximum value of 440 , otherwise x will be -tive which is not possible.

Now for z =1, y can have 44 values. so x can have 44 values

Now for z = 2, y can have 39 values , so x will have 39 values

Now for z = 3, y can have 34 values or x can have 34 values

So as similarly , total values of x can be = 44+39+34+29+24+19+14+9+4+1
= 214

Answer 3

Let there be u rocks of 1 pound, x of 10 pounds, and f of 50 pounds. The equations are:

(1) u + 10x + 50f = 500
(2) u + f + x = 100

By subtracting these, 9x + 49f = 400. The quickest method now is to reduce this equation modulo 49 to 9x = 8, which is solved by multiplying by 11 and reducing modulo 49 again, to give x = 39.

Therefore all the possible values of x are 49n + 39, that is to say 39, 88, 137, 186, ... but 88 rocks of 10 pounds are already too heavy, so x = 39 is the one and only answer.

Answer 4

Let x, y and z be the number of 10-pound, 1-pound and fifty pound rocks, respectively. Then given that the total rock weight is 500 pounds, and the number of rocks is 100, we can right this as

10x + y + 50z = 500 (i), and

x + y + z = 100 (ii).

Subtracting (ii) from (i) yields

9x + 49z = 400 (iii)

We proceed by checking values of z for which x is a positive integer. Clearly, z = 1 is one of those values (for which x = 39). Thus we need only check for those values of z = 1 modulo 3 for which x is positive, ie. z = 4 or 7 (as we need divide through by nine). Neither of these produces integers, so we can conclude that z = 1, x = 39 and y = 60.

Hence, there is only one possible value of x, and that value is 39.

Answer 5

If the answer should contain atleast 1 rock of 1 & 50 pounds then the answer if 44

Consider one stone of 50 pounds.total weight of 10 pound stones is 10x. hence there should ten 1 pound stones .

Answer 6

51

Answer 7

given there are x 10 pd rocks
let there are y 1 pd, z 50 pd rocks
we know 10x+y+50z=500-----------------1
x+y+z=100
y=100-x-z--------------------------------------2

substitute 2 in 1we get

10x+100-x-z+50z=500
9x+49z=400-------------------------3
we know x,y,z are whole numbers
in equation 3
z cant be greater than 8 because if z >8then xwill be less than 0 that cant be possible
let z be f(x)
f(x)=(400-9x)/49
but f(x) is a multiple of 1
but only 39 gives a whole number
therefore there is only one value of x ie 39

Answer 8

51.

Answer 9

51 there could be 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49, or 50