A stone weighing 40 kgms was broken into 4 pieces of different weights. Weighings  from 1 kg to 40 kg can be done by using these pieces. Find weight of individual pieces and logical reasoning to arrive at the solution.?
134910
Answer: 1,3,9 and 27 kg\n\nWay of solving:\n\n1) 4 pieces are very small number to weigh such number (40 of different weights. So I just tried to understand a most efficient way of weighing different weights using a small number of pieces.\n\n2) Firstly, I understand the most efficient set of pieces when we put some thing we are weighing on one scale and pieces on another. There are 15 (2^4-1) ways of taking different non-empty subset from set of 4 pieces. So, in most effective set all those subsets must have different weights. Eady to see, that pieces which are powers of 2: 1,2,4,8 kg make such a set (15 different weighings from 1 to 15 kg). \n\n3) Next I guessed that it is possible to put pieces not only on another but also at the same scale as we put some thing we are weighing. So, the weight of little piece can be not only added to the big piece, but also subtract from it, giving more possible combinations of weights.\nThen I guessed that some subset1 without smallest piece (ss1), ss1 + smallest piece (ss1+), another subset without smallest piece (ss2), ss2 with smallest piece on another scale (ss2-) must be consecutive numbers. So, ss1 = ss2 + 3. So, if smallest piece is x kg, the next piece must be 3 * 1st=3x kg.\nThe same situation is with other pieces: 3rd = 3 * 2nd = 9x kg, 4th = 3* 3rd = 27x kg.\nThis set of pieces makes possible 40 different weights.\nAlso, we know that sum of weights of those pieces is 40 kg.\nSo, x + 3x + 9x + 27x = 40kg\nx = 1kg\n\nAnd the final answer is: 1 kg, 3 kg, 9 kg and 27 kg.
441286
find the force of her weight. \nthat should be the answer
580085
B+Y+G+R=16480\n\nNow read the clues:\n\nB(Y+.5G) ----The blue gemstone's cost was equal to the yellow gemstone plus half the cost of the green gemstone\n\nB(3G) ----You paid three times as much for the blue gemstone as you did for the green gemstone.\n\nY(R-G) ---he yellow gemstone's cost was the same as the red gemstone minus the cost of the green gemstone.\n\nR(Y+G) ---The red gemstone cost as much as the yellow and the green gemstones put together. \n\nNow do some math...\n\n2 clues for the Blue stone.  Yellow must cost 2.5G.  B=3G=Y+.5G\n\nNow do substitution for the other gems:\n\nY=R-G ... so ....\n2.5G=R-G ... so ...\nR=3.5G \n\nNow plug them back into the full equation, and solve for the Green gem.\n\nB+Y+G+R=16480\n\n3G+2.5G+G+3.5G=16480\n\n10G=16480\n\nG= 1,648\n\nNow you know the green, and can figure out the others from the clues.\n\nB=3G=4944\nY=2.5G=4120\nR=3.5G=5768\n\nCheck the work:  1648 + 4944 + 4120 + 5768 = 16480\n\nThe red gemstone cost $5768.00
212037
Several people have answered your questions before, and you should use the method that they have used.  If the proportions remain the same.\n\n40/180 = x/900\n\nSolve for x.\n\nDo your own homework, seriously.
60707
stone (st) \na traditional British unit of weight, rarely used in the U.S. Originally the stone varied in size, both from place to place and according to the nature of the item being weighed. A stone of sugar was traditionally 8 pounds, while a stone of wool could be as much as 24 pounds. Eventually the stone was standardized at 14 pounds avoirdupois or approximately 6.350 29 kilograms -- a convenient size because it makes the stone equal to exactly 1/2 quarter or 1/8 hundredweight. Today the stone is used mostly for stating the weight of persons or animals. No -s is added for the plural.\n\nfortnight \na traditional English unit of time equal to 2 weeks or 14 days. The word, a contraction of "fourteen nights," has been used since at least the 1100s.
154221
We think that the thing to do is weigh six balls against six balls, but if you think about it, this would give us no information the only different ball. As we already know that one side will be heavier than the other. \n\nSo that the following plan can be followed, let us number the balls from 1 to 12. For the first weighing let us put on the left pan balls 1,2,3,4 and on the right pan balls 5,6,7,8. \n\nThere are two possibilities. Either they balance, or they don't. If they balance, then the different balls is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different ball is either 11 or 12. \n\nWeigh ball 1 against 11. If they balance, the different ball is number 12. If they do not balance, then 11 is the different ball. \n\nIf 1,2 vs 9,10 do not balance, then the different ball is either 9 or 10. Again, weigh 1 against 9. If they balance, the different ball is number 10, otherwise it is number 9. \n\nThat was the easy part. \n\nWhat if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these balls could be the different ball. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. \n\nSuppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the ball is either 3 or 4. Weigh 4 against 9, a known good ball. If they balance then the different ball is 3, otherwise it is 4. \n\nNow, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy ball, or 1 is a different, light ball. \n\nFor the third weighing, weigh 7 against 8. Whichever side is heavy is the different ball. If they balance, then 1 is the different ball. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy ball or 2 is a light different ball. Weigh 5 against 6. The heavier one is the different ball. If they balance, then 2 is a different light ball.
87
The answer is 9 stones (or maybe more). \n\nAfter adding a look-alike to 9 stones, there will be 10. Name them ABCDEFGHIJ.\n\n"ABC<DEF" means "the stones ABC as a group weigh less than the stones DEF."\n\nWeighing 1. Weigh ABC vs. DEF.\n\nIf ABC<DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC<GHI, the look-alike must be light and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is A.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is B.\n----If ABC=GHI, the look-alike must be heavy and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is E.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is D.\n\nIf ABC=DEF, the look-alike must be among GHIJ.\n--Weighing 2. Weigh G vs. H.\n----If G=H, the look-alike candidates are I and J.\n----Otherwise, the look-alike candidates are G and H.\n----Weighing 3. Weigh one of the two candidates against A.\n------If they differ in weight, the look-alike is that candidate, otherwise it is the other candidate.\n\nIf ABC>DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC>GHI, the look-alike must be heavy and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is B.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is A.\n----If ABC=GHI, the look-alike must be light and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is D.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is E.
89827
1+1=1\n\nOne pile of rocks and another pile of rocks can be combined to make what is just one, albeit larger, single pile of rocks.
443609
Problem setup:  3 pounds is to 5 breasts as 1 lb is to x breasts\n\n3/5=1/x\n\nCross-multiply 3x=5\n\nx=5/3 or 1 and 2/3.
149683
weigh 3 in one and 3 in the other:\n\nif they are equal:\nweigh each of the other 2 in one pan each. One will be lighter, voila\n\nIf they are different:\ntake the lighter set of 3, weigh one in one pan and one in the other and hold one aside.  If one is lighter, voila, if they are equal it's the one set aside.\n\nthat's the whole thing in 2 weighings
388251
Multiply the two numbers and you are done.\n\nSee weight is equal to force.\n\nForce is the rate of change of momentum but since the mass doesn't change the force is then equal to mass times acceleration or F = ma\n\nm, is mass in kg\na is acceleration of gravity in m/s^2\n\nand the units of N, Newtons, is kg-m/s^2\n\nSo your answer is 27 * 9.8 N
413026
A) a grain of sand.\nThink about it this way: one thousand milligrams= 1 gram\n                                          454 grams= 1 pound\nso 454 THOUSAND miligrams=1 pound\n\nhow much would each of those items weigh if you had 454 thousand of them?
483149
1lb    = 0.4536kg\n1.5lb = 0.6804kg\n2lb    = 0.9072kg
644920
On earth, mass and weight are equivalent, but in physics, it's more correct to use mass (since sometimes objects in physics problems are not on earth!)\n\nSo if the object weighs 1 kg, you can use 1 kg as its mass. One newton is the force needed to accelerate 1 kg (mass) at a rate of 1 meter per second^2.
426497
F = GmM/r^2; where G is a constant, m and M are the two masses, and r is the distance between them.\n\nm + M = 4 kg; so m = 4 - M\n\nF = G(4 - M)M/(.25)^2 = 2.5*10^-10; where G = 6.67 × 10−11 N m^2 kg^−2 \n\nUse the above to solve for M, then m = 4 - M to find m.  \n\nAnd there you have it, you do the math.
528670
Neat. You need 1, 3, 9, 27\n\nPut weights on each side:\n\n1: 1 on left, nothing on right\n2: 3 on left, 1 on right\n3: 3 on left\n4: (3,1)()\n5: (9)(3,1)\n6: (9)(3)\n7: (9,1)(3)\n8: (9)(1)\n9: (9)()\n...\n13: (9,3,1)\n14: (27)(9,3,1)\n...\n40: (27,9,3,1)\n\nThe general rule is you need powers of 3, and the reason is that each weight can be counted on the left 0 times, 1 time, or -1 times.\n\nNeat.
391295
Split these two vaiables apart.\n\nDoubling the mass would double the gravitational force, hence the weight also would double. (purely proportional)\n\nIncreasing the distance by 4x would decrease the grav force by 16x. (Inverse square law)\n\nso...put the two factors together...\n\nIf you have 2x, and then 1/16x, then it will decrease by 1/8 of the original weight.\n\n\nm1 * m1 / d^2 = F