Age Word Problems<\/p>\n
\u00a0<\/strong><\/p>\n Solution<\/p>\n Let x = Sid’s age now\u00a0\u00a0 (smaller number)<\/p>\n 4x = father’s age now<\/p>\n The problem tells you about their ages at another time. Five years ago your age would be 5 less than your age now.<\/p>\n So<\/p>\n x – 5 = Sid’s age 5 years ago<\/p>\n 4x – 5 = father’s age 5 years ago<\/p>\n Five years ago he was seven times as old (as he was)<\/p>\n 4x – 5 = 7(x – 5)<\/p>\n 4x – 5 = 7x – 35<\/p>\n 4x = 7x – 30<\/p>\n -3x = -30<\/p>\n Divide both sides of the equation by 3<\/p>\n x = 10<\/p>\n So,<\/p>\n Sid is 10 and the father is 40.<\/p>\n <\/p>\n Solution<\/p>\n Let x = Costello’s age now\u00a0\u00a0 (smaller number)<\/p>\n x + 6 = Abbott’s age now<\/p>\n Then<\/strong><\/p>\n x – 6 = Costello’s age 6 years ago<\/p>\n (x + 6) – 6 = Abbott’s age 6 years ago<\/p>\n Six years ago she was twice as old as him.<\/p>\n Equation<\/strong><\/p>\n (x + 6) – 6 = 2(x – 6)<\/p>\n x = 2x – 12<\/p>\n -x = -12<\/p>\n x = 12 and x + 6 = 18<\/p>\n So,<\/p>\n Costello is 12 and Abbott is 18.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let x = Willie’s age now.<\/p>\n Let x + 26 = Father’s age now.<\/p>\n In 10 Years<\/p>\n x + 10 = Willie’s age in 10 years<\/p>\n (x + 26) + 10 = Father’s age in 10 years<\/p>\n Equation<\/p>\n x + 10 + (x + 26) + 10 = 80<\/p>\n 2x + 46 = 80<\/p>\n x = 17 and x + 26 = 43<\/p>\n So,<\/p>\n Willie is 17 now.<\/p>\n Willie’s father is 43 now.<\/p>\n <\/p>\n Solution<\/p>\n Let x = Inda Cates age now.<\/p>\n Let 2x = Tess T. Fye’s age now.<\/p>\n Then<\/p>\n x – 8 = 8 subtracted from Inda Cate’s age.<\/p>\n 2x + 4 = 4 added to Tess T. Fye’s age.<\/p>\n Equation:<\/strong><\/p>\n 2x +\u00a0 4 = 4(x – 8)<\/p>\n 2x + 4 = 4x – 32<\/p>\n -2x = -36<\/p>\n x = 18<\/p>\n 2x = 36<\/p>\n So,<\/strong><\/p>\n Tess T. Fye’s age is 36 years.<\/p>\n Inda Cate’s age is 18 years.<\/p>\n Solution<\/p>\n Let x = son’s age now.<\/p>\n Let 4x = Father’s age now.<\/p>\n In 3 Years:<\/strong><\/p>\n x + 3 = son’s age in 3 years<\/p>\n 4x + 3 = father’s age in 3 years<\/p>\n In 3 years, the father will be three times as old as his son.<\/p>\n Equation:<\/strong><\/p>\n 4x + 3 = 3(x + 3)<\/p>\n 4x + 3 = 3x + 9<\/p>\n x = 6 and 4x = 24<\/p>\n So,<\/strong><\/p>\n The father’s age now is 24 years.<\/p>\n The son’s age now is 6 years.<\/p>\n Solution Let w + 8 Robin’s age now.<\/p>\n Then<\/p>\n w – 20 = Willie’s age 20 years ago.<\/p>\n (w + 8) – 20 = Robin’s age 20 years ago.<\/p>\n Equation<\/p>\n (w + 8) – 20 = 3(w – 20)<\/p>\n w -12 = 3w – 60<\/p>\n -2w = -48<\/p>\n w = 24 and w + 8 = 32<\/p>\n So,<\/p>\n Willy is 24 and Robin is 32.<\/p>\n Solution<\/p>\n Doug’s age = x<\/p>\n Bill’s age = 2x<\/p>\n Bill Dupp is twice as old as Doug Upp.<\/p>\n Then<\/strong><\/p>\n 2x – 16 = Bill’s current age<\/p>\n x + 16 = Doug’s current age<\/p>\n If 16 is added to Doug’s age and 16 is subtracted from Bill’s age they will be equal.<\/p>\n Equation<\/strong><\/p>\n 2x – 16 = x + 16<\/p>\n 2x = x + 32<\/p>\n x = 32<\/p>\n So,<\/strong><\/p>\n Doug Dupps is 32 and Bill Dupp is 64.<\/p>\n <\/p>\n Solution<\/p>\n Let x\u00a0= Hobbes\u2019s age now.<\/p>\n x\u00a0– 4 = Calvin\u2019s age now.<\/p>\n In 2 years time, Hobbes’s age is\u00a0x\u00a0+\u00a02<\/p>\n In 2 years time, Calvin’s age is (x – 4) + 2<\/p>\n The problem tells you about their ages at another time. Two years later your age would be 2 more than your age now.<\/p>\n Equation<\/p>\n x + 2 = 2[(x – 4) + 2]<\/p>\n x + 2 = 2x – 8 + 4<\/p>\n x + 2 = 2x – 4<\/p>\n x = 6 and x – 4 = 2<\/p>\n So,<\/p>\n So Calvin’s age\u00a0is 2 and Hobbes\u2019s age is 6.<\/p>\n <\/p>\n Solution<\/p>\n Let x = Jewell’s age The problem talked about their age at another time which was ten years ago. So, that would be less than 10 because it was ten years ago. Ten years ago the sum of their ages was 46.<\/p>\n Equation<\/strong><\/p>\n x – 10 + 2x – 10 = 46<\/p>\n 3x – 20 = 46<\/p>\n 3x = 66<\/p>\n Answer<\/strong><\/p>\n x = 22<\/p>\n So,\u00a0<\/strong><\/p>\n Jewell’s age is 22 and Justin’s is 44.<\/p>\n Solution<\/p>\n Now<\/p>\n Let the Florentine statue = x<\/p>\n Let the Roman statue = 3x<\/p>\n In 100 years<\/p>\n The age of the Florentine statue = x + 100<\/p>\n The age of the roman statute =\u00a0 3x + 100<\/p>\n Equation<\/p>\n 3x + 100 = 2(x + 100)<\/p>\n 3x + 100 = 2x + 200<\/p>\n x = 100<\/p>\n 3x = 300<\/p>\n So<\/p>\n The age of the Florentine statue is 100 now.<\/p>\n The age of the roman statute is 300 now.<\/p>\n Solution<\/p>\n Now<\/p>\n Let Butch’s age= X+20<\/p>\n Let Janet’s age= X<\/p>\n 10 Years From Now<\/p>\n Let Butch’s age = X + 30<\/p>\n Let Janet’s age = X + 10<\/p>\n Equation<\/p>\n x + 30 = 2(x + 10)<\/p>\n x + 30 = 2x + 20<\/p>\n 10 = x<\/p>\n So,<\/p>\n Butch Err’s age now is 10 years.<\/p>\n Janet’s age now is 30 years.<\/p>\n <\/p>\n Solution<\/p>\n Let x = Fran Tick’s age now.<\/p>\n Let 3x = Moe Tell’s age now.<\/p>\n Then<\/p>\n Moe’s age = 3x – 20<\/p>\n Fran’s age = x + 20<\/p>\n If 20 is added to Fran’s age and 20 is subtracted from Moe’s then their ages will be equal.<\/p>\n Equation<\/p>\n 3x – 20 = x + 20<\/p>\n 3x= x + 40<\/p>\n 2x = 40<\/p>\n x = 20 and 3x = 60<\/p>\n So,<\/p>\n Moe Tell’s age is 60 and Fran Tick’s age is 20 years old.<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Let a = Andy’s age now.<\/p>\n Let k = Kate’s age now.<\/p>\n Then<\/p>\n a = 2k<\/p>\n <\/p>\n (a+6) + (k+6) = 60<\/p>\n <\/p>\n Substitute a for 2k<\/p>\n <\/p>\n Equation<\/p>\n <\/p>\n (2k + 6) + (k+6) = 60<\/p>\n <\/p>\n Combine Like Terms<\/p>\n <\/p>\n 3k + 12 = 60<\/p>\n <\/p>\n 3k = 48<\/p>\n <\/p>\n k = 16<\/p>\n <\/p>\n a = 32<\/p>\n <\/p>\n So,<\/p>\n <\/p>\n Andy is 32 years old and Kate is 16 years old.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let w = Mrs. Wang’s age now.<\/p>\n Let d = daughter’s age now.<\/p>\n w = d + 23<\/p>\n d + 5 + w + 5 = 63 or d + w + 10 = 63 or d + w = 53<\/p>\n Solve w = d + 23<\/p>\n d + (d + 23) = 53<\/p>\n 2d + 23 = 53<\/p>\n 2d = 30<\/p>\n d = 15 and w = 38<\/p>\n So<\/p>\n The daughter is 15 and Mrs. Wang is 38.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let Matthew’s age = 3j<\/p>\n Let Jenny’s age = j<\/p>\n 7 Years From Now<\/p>\n 3j + 7 = 2(j + 7)<\/p>\n Equation<\/p>\n 3j + 7 = 2(j + 7)<\/p>\n 3j + 7 = 2j + 14<\/p>\n j + 7 = 14<\/p>\n j = 7<\/p>\n So,<\/p>\n Matthew’s age is 21 and Jenny’s age= 7<\/p>\n Solution<\/p>\n Let j = Juan’s age now.<\/p>\n Let s = sister’s age now.<\/p>\n Then<\/p>\n j\u00a0 = s + 8<\/p>\n <\/p>\n j + 3 = 2(s + 3)<\/p>\n <\/p>\n Substitute s + 8 for j<\/p>\n <\/p>\n Equation<\/p>\n <\/p>\n (s + 8) + 3 = 2(s + 3)<\/p>\n <\/p>\n s + 11 = 2s + 6<\/p>\n <\/p>\n -s + 11 = 6<\/p>\n <\/p>\n -s = -5<\/p>\n <\/p>\n s = 5<\/p>\n <\/p>\n So,<\/p>\n <\/p>\n Juan is 13<\/p>\n <\/p>\n His sister is 5<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let m = Melissa\u2019s age now<\/p>\n Let j = Joyce\u2019s age now<\/p>\n<\/a> j = m + 24<\/p>\n j + 2 = 3 (m + 2)<\/p>\n Substitute\u00a0 m + 24 for j<\/p>\n<\/a> (m + 24) + 2 = 3 (m + 2)<\/p>\n m + 22 = 3m + 2<\/p>\n -2m = -20<\/p>\n m = 10<\/p>\n So,<\/p>\n Joyce is 34 years old<\/p>\n Melissa is 10 years old<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Let t = Tom\u2019s age now.<\/p>\n Let j = Jerry\u2019s age now.<\/p>\n Then<\/p>\n t = j + 4<\/p>\n t + 9 = 5(j + 9)<\/p>\n Substitute (j + 4) for t.<\/p>\n Equation<\/p>\n (j + 4) – 9 = 5 (j – 9)<\/p>\n j – 5 = 5j – 45<\/p>\n j = 5j – 40<\/p>\n – 4j = – 40<\/p>\n – j = – 10<\/p>\n j = 10<\/p>\n So,<\/p>\n t = 14<\/p>\n j = 10<\/p>\n Tome is 14 and Jerry is 10 now.<\/p>\n <\/p>\n Solution<\/p>\n Let k = Kathy\u2019s age now.<\/p>\n Let b = Bill\u2019s age now.<\/p>\n Then<\/p>\n k + 6 = b<\/p>\n 2(k – 12) = b – 12<\/p>\n Substitute (k + 6) for b.<\/p>\n Equation<\/p>\n <\/p>\n 2(k – 12) = (k + 6) – 12<\/p>\n 2k – 24 = k – 6<\/p>\n k – 24 = -6<\/p>\n k = 18<\/p>\n So,<\/p>\n k = 18<\/p>\n b = 24<\/p>\n Kathy is 18 and Bill is 24 now.<\/p>\n 12 years ago Kathy was 6 and Bill was 12<\/p>\n <\/p>\n Solution<\/p>\n Let g = Dr. Garcia\u2019s age now.<\/p>\n Let s = son\u2019s age now.<\/p>\n Then<\/p>\n 2s = g<\/p>\n g – 20 = 4(s-20)<\/p>\n Substitute 2s for g.<\/p>\n Equation<\/p>\n 2s – 20 = 4(s-20)<\/p>\n 2s – 20 = 4s-80<\/p>\n -20 = 2s – 80<\/p>\n 60 = 2s<\/p>\n s = 30<\/p>\n So,<\/p>\n s = 30<\/p>\n g = 60<\/p>\n The son is 30 now and Dr. Garcia is 60 now.<\/p>\n 20 years ago the son was 10 and Dr. Garcia was 40<\/p>\n Solution<\/p>\n Let k = Mr. Klinker’s age now.<\/p>\n Let d = Daughter’s age now.<\/p>\n Then<\/p>\n k + x = 2( d + x)<\/p>\n Replace k with 35 and d with 10<\/p>\n Equation<\/p>\n 35 + x = 20 + 2x<\/p>\n 35 + x = 20 + 2x<\/p>\n 35 = 20 + x<\/p>\n 15 = x<\/p>\n x = 15<\/p>\n So,<\/p>\n In 15 years, Mr. Klinker will be 50, and his daughter will be 25, thus making Mr. Klinker twice his daughter’s age.<\/p>\n <\/p>\n Solution<\/p>\n Let g = George’s age now.<\/p>\n Let m = mother’s age now.<\/p>\n Then<\/p>\n g + x = 3(m + x)<\/p>\n Replace g with 7, and m with 37.<\/p>\n Equation<\/p>\n 37 + x = 3(7+x)<\/p>\n 37 + x = 21 + 3x<\/p>\n 37 = 21 + 2x<\/p>\n 16 = 2x<\/p>\n x = 8<\/p>\n So,<\/p>\n In 8 years, George’s mother will be 3 times older than him.<\/p>\n <\/p>\n Solution<\/p>\n Let p = Pete’s age now.<\/p>\n Let g = grandfather’s age now.<\/p>\n Then<\/p>\n 6(p – x) = g – x<\/p>\n Replace p with 14, and g with 54.<\/p>\n Equation 84 – 6x = 54 – x<\/p>\n 84 = 54 + 5x<\/p>\n 30 = 5x<\/p>\n x = 6<\/p>\n So,<\/p>\n 6 years ago Pete was 8 and the grandfather was 48.<\/p>\n <\/p>\n Solution<\/p>\n Let d = Dorothy’s age now<\/p>\n Let r = Rita’s age now<\/p>\n Then<\/p>\n d + 14 = r<\/p>\n r – 10 = 3(d – 10)<\/p>\n Replace r with d + 14<\/p>\n Equation<\/p>\n r – 10 = 3(d – 10)<\/p>\n d + 14 – 10 = 3(d – 10)<\/p>\n d + 4 = 3d – 30<\/p>\n 4 = 2d – 30<\/p>\n 2d = 34<\/p>\n d = 17<\/p>\n So,<\/p>\n Dorothy is 17 and Rita is 31.<\/p>\n <\/p>\n 10 years ago Dorothy was 7 and Rita was 21.<\/p>\n <\/p>\n Solution<\/p>\n Let f = Ms. Ford’s age now<\/p>\n Let l = Ms. Lincoln’s age now<\/p>\n Then<\/p>\n 2(f –\u00a0 x) = l – x<\/p>\n Replace f with 48, and l with 35.<\/p>\n Equation<\/p>\n f –\u00a0 x = 2(l – x)<\/p>\n 48 –\u00a0 x = 2(35 – x)<\/p>\n 48 – x = 70 – 2x<\/p>\n 48 + x = 70<\/p>\n <\/p>\n x = 22<\/p>\n <\/p>\n So,<\/p>\n In 22 years ago , Ms. Ford’s was 26 and Ms. Lincoln’s was 13.<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let s = Steve’s age now<\/p>\n Let j = Janis’s age now<\/p>\n s = 5j<\/p>\n Then (In 12 Years)<\/p>\n (s + 12) = 2(j + 12)<\/p>\n Solve<\/p>\n s = 5j<\/p>\n (s + 12) = 2(j + 12)<\/p>\n (5j + 12) = 2(j + 12)<\/p>\n 5j + 12 = 2j + 24<\/p>\n 3j = 12<\/p>\n j = 4 so s = 32<\/p>\n So,<\/p>\n Janis’s age is 4 and Steve’s age is 32.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let m = Mary’s age now.<\/p>\n Let t = Toni’s age now.<\/p>\n Let s = Sam’s age now.<\/p>\n t + 4 = m<\/p>\n s = 2m<\/p>\n m + t + s = 8t<\/p>\n Replace m with (t + 4)<\/p>\n Replace s with 2m and then replace the m in 2m with (t + 4)<\/p>\n Equation<\/p>\n (t + 4) + t + 2m = 8t<\/p>\n (t + 4) + t + 2(t+ 4) = 8t<\/p>\n t + 4 + t + 2t + 8 = 8t<\/p>\n 4t + 12 = 8t<\/p>\n 4t = 12<\/p>\n t = 3<\/p>\n So,<\/p>\n Toni is 3<\/p>\n Mary is 7<\/p>\n Sam is 14<\/p>\n 3 + 7 + 14 = 8(3)<\/p>\n 24 = 24<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let l = Larry’s age now<\/p>\n Let s = sister’s age now<\/p>\n s + 8 = l<\/p>\n 2(s + 3) = l + 3<\/p>\n Replace l with s + 8<\/p>\n 2(s + 3) = l + 3<\/p>\n 2(s + 3) = (s + 8) + 3<\/p>\n 2s + 6 = s + 11<\/p>\n s + 6 = 11<\/p>\n s = 5<\/p>\n So,<\/p>\n Larry is 13 and his sister is 5<\/p>\n In 3 years Larry will be 16 and his sister will be 8.<\/p>\n <\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let b = Barry’s age now.<\/p>\n Let s = sister’s age now<\/p>\n b = 8 + s<\/p>\n Then (In 3 Years)<\/p>\n (b + 3) = 2(s + 3)<\/p>\n Solve<\/strong><\/p>\n b = 8 + s<\/p>\n (b + 3) = 2(s + 3)<\/p>\n s + 11 = 2s + 6<\/p>\n 11 = s + 6<\/p>\n s = 5<\/p>\n So,<\/p>\n Barry is 13 and his sister is 5.<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let j = Jennifer’s age now.<\/p>\n Let s = Sue’s age now<\/p>\n j = s + 6<\/p>\n Then (In 4 Years)<\/p>\n j + 4 = 2(s – 5)<\/p>\n Solve<\/strong><\/p>\n (s + 6) + 4 = 2(s – 5)<\/p>\n s + 10 = 2s – 10<\/p>\n 10 = s – 10<\/p>\n 20 = s<\/p>\n s = 20<\/p>\n So,<\/p>\n Sue is 20 and Jennifer is 26.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let a = Adam’s age now.<\/p>\n Let e = Eve’s age now<\/p>\n a + 5 = e<\/p>\n Then (In 1 Years)<\/p>\n 3(a – 4) = e + 1<\/p>\n Solve<\/strong><\/p>\n 3(a – 4) = e + 1<\/p>\n Replace e with (a + 5)<\/p>\n 3(a – 4) = a + 5 + 1<\/p>\n 3a – 12 = a + 6<\/p>\n 2a – 12 = 6<\/p>\n 2a = 18<\/p>\n a = 9<\/p>\n So,<\/p>\n Adam is 9 and Eve is 14<\/p>\n In one year Eve will be 15 and 4 years ago Adam was 5.<\/p>\n <\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let j = Jack’s age now.<\/p>\n Let x = Jill’s age now<\/p>\n j = 2x<\/p>\n Then (In 2 Years)<\/p>\n j + 2 = 4(x – 9)<\/p>\n Solve<\/strong><\/p>\n j + 2 = 4(x – 9)<\/p>\n (2x) + 2 = 4(x – 9)<\/p>\n 2x + 2 = 4x – 36<\/p>\n 2 = 2x – 36<\/p>\n 38 = 2x<\/p>\n x = 19<\/p>\n So,<\/p>\n Jill is 19 and Jack is 38.<\/p>\n <\/p>\n Solution<\/p>\n Now<\/p>\n Let k = Katie’s age now.<\/p>\n Let a = Anne’s age now<\/p>\n k – 4 = 2(a – 4)<\/p>\n k = a + 6<\/p>\n Solve<\/strong><\/p>\n k – 4 = 2(a – 4)<\/p>\n Replace k with (a + 6)<\/p>\n (a + 6) – 4 = 2(a – 4)<\/p>\n a + 2 = 2a – 8<\/p>\n 2 = a – 8<\/p>\n a = 10<\/p>\n So,<\/p>\n Anne is 10 and Katie is 16<\/p>\n In 6 years Anne will be 16 and Katie is 16 now.<\/p>\n 4 years ago Anne was 6 and Katie was 12<\/p>\n <\/p>\n 34. Five years ago. Tom was one third as old as his father was then. In 5 years Tom will be half as old as his father will be then. Find their ages now.<\/p>\n Solution<\/p>\n Now<\/p>\n Let t = Tom’s age now.<\/p>\n Let f = Father’s age now<\/p>\n 3(t – 5) = f – 5<\/p>\n 2(t + 5) = f + 5<\/p>\n Solve<\/strong><\/p>\n 2 equations<\/strong><\/p>\n 3(t – 5) = f – 5<\/p>\n 2(t + 5) = f + 5<\/p>\n First equation<\/strong><\/p>\n 3(t – 5) = f – 5<\/p>\n 3t – 15 = f – 5<\/p>\n Second equation<\/strong><\/p>\n 2(t + 5) = f + 5<\/p>\n 2t + 10\u00a0 = f + 5<\/p>\n Solve second equation for f<\/strong><\/p>\n 2t + 10\u00a0 = f + 5<\/p>\n 2t + 5 = f<\/p>\n Substatute f form second equation for f in first equation<\/strong><\/p>\n 3t – 15 = (2t + 5) – 5<\/p>\n Solve<\/strong><\/p>\n 3t – 15 = 2t<\/p>\n t = 15<\/p>\n So,<\/strong><\/p>\n Tom is 15 and his Father is 35.<\/p>\n Five years ago Tom was 10 and his Father was 30. Tom will be 1\/3 his father’s age.<\/p>\n In 5 years Tom will be 20 and his Father will be 40. Tom will be half his father’s a<\/p>\n <\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" Age Word Problems \u00a0 Sid Upp\u2019s father is four times as old as Tip. Five years ago he was seven times as old. Find the absolute value of the difference of the ages now. Solution Let x = Sid’s age now\u00a0\u00a0 (smaller number) 4x = father’s age now The problem tells you about their ages … Continue reading Age Word Problems<\/span> \n
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\nd + w + = 53<\/p>\n\n
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\n6(14 – x) = 54 – x<\/p>\n\n
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