a man is 32 years older than his son. 10 years ago he was 3 times as old as his son was then. find their ages?
411481
Simple word problem.\n\nStep 1: Assign known quantities to variables:\n\nM = age of the man\nS = age of his son\n\nStep 2: Translate sentences in the word problem into equations:\n\nM = S + 32\nM - 10 = 3 * (S - 10)\n\nStep 3: Solve multiple equations:\n\nM - 10 = 3 * (S - 10)\n(S + 32) - 10 = 3 * (S - 10)\nS + 22 = 3S - 30\n52 = 2S\nS = 26\nM = 26 + 32 = 58\n\nThe man is currently 58\nHis son is currently 26
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He was 84 years old (the math of the years is 85, but he died in March and his birthday was in December). Click the links below to learn more. Way too much info to copy and paste. :o)
738815
Ahmed is seven years old and his father is forty-nine years old.\n\nGood Luck!!!!
1111510
John is 13\n\nMary is 8
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J=2B\nG=B+7\n(J-4)+(B-4)+(G-4)=55\nJ+B+G-12=55\nJ+B+G=67\n2B+B+B+7=67\n4B=67-7\n4B=60\nB=15\nsub J=30\nG=22\ncheck 4 yearsago the sum of their ages\n=11+26+18=55 checks\nso the present ages are\nJohn=39\nGreg=22\nBill=15
670672
Was 40 and 10. Now 45 and 15.\n\nYou do not need the information that the combined age was 50 in 2001 to be given you, to be able to solve this.\n\nM = 3S\nM-5 = 4 (S-5) = 4S - 20\nM = 4S - 15\nSubtract 1st line from 3rd line\n0 = S - 15\nS = 15\nSubstitute in 1st line to get \nM = 45
504307
52x \n\nthere are 52 weeks in 1 year.  so you multiply 52 (the # of weeks by X the # of years
184815
Hi, \nFirstly, read it carefully and determine what is required.\nThen, note down whatever information is given. Like distance, dad's age whatever.\nThen, find the "missing link". Its something which is essential for the completion of the problem.\nTo complete the above mentioned step, u need to have a clear memory of all the formulas. It would best if you knew how to deduce the formula, cuz then you would have a firm base.\nThen, just use the formula. Substitute the values, and find out the result.
942741
A HR for 20 year old = 70\n\nB A=R/(0.7*220)\n\nC 9 months old
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rememer the following values:\n99.99%=13 half lives(appr.)\n99.9%=10 half lives\n90%=3 half lives(appr.)\nhence now the answer is 10*30=300  years(appr.)
919110
"Q: 11) When three times a number, x, is added to a second number, y, the total is 33. The\nfirst number added to three times the second number is 19.\na)Form two equations connecting x and y.\nb)Find the two numbers."\n\neq 1: 3x + y = 33\neq 2: x + 3y = 19\n\nmultiply eq 2 by 3:\n3x + 9y = 57\nthen, subtr from eq 1:\n-8y = -24\ny = 3\nsolve for x using eq 2:\nx + 9 = 19\nx = 10\n\n"\nQ:11) Ahmed is x years old and his father is x2 years old. The sum of their ages is 56\nyears. Form an equation in x and solve it to find the age of each"\n\nx + 2x = 56\n3x = 56\nx = 18-2/3 years old for Ahmed\n2x = 37-1/3 years old for his father\n\nHope this helps!
51849
9,2,2\nbcoz 6+6+1=9+2+2=13\nThis is the only conflicting triple. as the eldest is one  so their age is 9,2,2
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20 years\n\n4 scores and 7 years ago = 87 years
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let the age of tony be x and sarah be y\nthen,\nfour years ago,\n(x-4)=5(y-4)\nx-4=5y-20\nx-5y= -16......................(1)\nafter two years,\n(x+2)=2(y+2)\nx+2=2y+4\nx-2y=2..........................(2)\nby subtracting eqn. (1) from (2),\n x -2y=2\n-x+5y=16\n--------------------\n3y=18\ny=6\nput value of y in eqn (2)\nx-2*6=2\nx=2+12\nx=14\ntherefore, the age of tony and sarah now are 14 and 6 respectively.
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Sydney's age = x\nDevaughn's age = 3x\n\nx + 3x = 92\n4x = 92\nx = 23\n\nSydney is 23\nDevaughn is 69
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Dayum that b i t c h is old, son!
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Here is a claim of 10 consecutive sons:\n\nhttp://famteam.com/newfamily.html