Tutors Answer Your Questions about Age Word Problems (FREE)
Question 454963: I need help on this question. I know the answer but i dont understand how to get to it...thanks if you could reply this
THE QUESTION IS:
When i was 14 my mother was 41, and she is now twice as old as am. how old am i?
the answer is 27 i know but how do u get to it...thanks, RACHEL
Click here to see answer by amoresroy(333)  |
Question 454963: I need help on this question. I know the answer but i dont understand how to get to it...thanks if you could reply this
THE QUESTION IS:
When i was 14 my mother was 41, and she is now twice as old as am. how old am i?
the answer is 27 i know but how do u get to it...thanks, RACHEL
Click here to see answer by Alan3354(30993)  |
Question 455614: I have this problem. Kirsten is 4 times Claires age. Emily is 6yrs older than Claire and 6yrs younger than Zack. if anna is twice as old as Claire and the total of their ages is 81, How hod is Kirsten? I can't even get my equation started.
Click here to see answer by richwmiller(9143)  |
Question 455774: The age of Betty's mother is 7 years more than twice Betty's age. The difference in their ages is 21 years. How old is Betty? Show the linear equation in one variable that you used then show each step in solving for your answer.
Click here to see answer by richwmiller(9143)  |
Question 455538:
my moms age in years is the same as my brothers age in months.
My mothers age is 6 times my sisters age.
My age is the sum of my brother and sisters ages.
My fathers age in years is the number of seasons I've been alive.
My mother and father are the same age.
Click here to see answer by richwmiller(9143)  |
Question 456929: A family determined the average cost of maintaining and operating the family car to be about $.30 per mile. On one trip, the family drove at an average rate of 50 miles per hour for a total of 6.5 hours. On the second trip, they drove at an average rate of 55 miles per hour for a total of 6 hours. which trip cost more? how much more?
Click here to see answer by richwmiller(9143)  |
Question 458420: Determine the dollar amount for the cement needed for a sidewalk that is 4 inches thick, 3 feet wide, and 40 feet long. The cement is sold for $76 per cubic yard.(Calculate the problem using fractions to avoid rounding errors caused by repeating decimals.)
Click here to see answer by richwmiller(9143)  |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030
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