How can i solve these 2 math word problems?

How can i solve these 2 math word problems? Topic: Math thesis paper
June 17, 2019 / By Bellinda
Question: 1. it takes Nana 12 days to type her thesis. a professional typist could type it in 8 days. How long would it take to type the thesis if both worked together? 2. One car travels 30 m.p.h faster than another. In the same time that one travels 200 miles the other goes 320 miles. Find the rate of each car. in problem 2 i know that i have to use the formula D=RxT but i do not what goes where in the formula
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Best Answers: How can i solve these 2 math word problems?

Adriana Adriana | 8 days ago
1) if Nana needs 12 days, that would mean that she finishes 1/12 of the paper in 1 day, similarly it would mean that the pro finishes 1/8. use this formula: 1= x (1/12 + 1/8) and solve for x. the formula means, to finish the whole paper, it would take x days for nana (1/12) and the pro (1/8) working together. you will get 4.8 days. 2) set the speed one one car x, the speed of the other car would be x+30. use the formula D = RxT, you will get 1 equation for each car. car 1: 200 = x *T car 2: 320 = (x+30) *T T is the same for both cars, so solve for T for each equation and set the T's equal to each other. you will get the equation 200/x = 320/(x+30), cross multiply you would get 200(x+30) = 320 x solve for x, you will get x= 50 so car 1 = 50 mph, car 2 = 80 mph
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We found more questions related to the topic: Math thesis paper

i'll do equations only you solve 1. t = trail mix pounds c = cashews pounds t+c = 225 9t + 14c = 225(11) 2. d = rt r = plane speed w = wind speed 1280 = 4(r+w) with wind 1280 = 8(r-w) against wind using top mult by 2 2560 = 8r + 8w 1280 = 8r - 8w elimination (add together) 3840 = 16r 240 = r solving for w (using either equation) w = 80 3. s = senior a = adult s + a = 493 23a + 14s = 8270
answer: a million + sqrt(2) miles permit s1 be the fee of the grasp ans s2 the fee of the practice. The grasp and practice commute in opposite guidelines till the grasp reaches the back of the practice at time t1 at a distance d from the grasp's commencing element. So, d = s1 * t1. The practice's rear is d miles from the grasp's commencing element, so the practice's front must be at a distance of a million - d miles from that element, which became additionally the practice's commencing element. The practice has subsequently traveled a million - d miles, so a million - d = s2 * t1. If we divide d via a million - d we get d/(a million-d) = s1/s2. as quickly as the grasp reaches the tip of the practice, his horse does a piroette and he rides to the front of the practice, accomplishing it because it hits the a million mile mark. The practice could have traveled distance d and the grasp distance a million + d. So s1 * t2 = a million + d and s2 * t2 = d. branch supplies us s1/s2 = (a million+d)/d. d/(a million-d) = s1/s2 = (a million+d)/d d^2 = a million - d^2 2d^2 = a million d^2 = a million/2 d = sqrt(2)/2 The grasp rode distance d, then a million + d for a complete of a million + 2d = a million + sqrt(2)

Adriana Originally Answered: Word problems to solve?
These questions are simple subsitution. Just find the equation and solve. 1) x + y = 55 .05x+.1y=4.50 .05 = value of nickel .1 = value of dime x = number of nickles y = number of dimes. Solve for x and y. x + y = 55 - .5x + y = 45 ( I multiplied every number by 10 in the second equatoin) ____________ .5x = 10 x = 20 20 + y = 55 ( I plugged in x to the first equation) y = 35 2) x = cost of chips y = cost of container 7x + 2y = 13 6x + 3y = 15 Same method as number 2. 3) x = cost of adult y = cost of children 9x + 2y = 525 y = x - 6 Plug in y and solve for x and the plug in x and solve for y.
Adriana Originally Answered: Word problems to solve?
Let Sally work x hours per week therefore karl works x+4 hours per week the equation is 6x=3(x+4) 6x-3x=12 3x=12 x=4 Karl works x+4 hours a week so, Karl works 8(4+4) hours a week.

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