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Math Word Problems and Solutions - Distance, Speed, Time

Problem 1 A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? Click to see solution Solution: Let $x$ be the number of kilograms he sold in the morning.Then in the afternoon he sold $2x$ kilograms. So, the total is $x + 2x = 3x$. This must be equal to 360. $3x = 360$ $x = \frac{360}{3}$ $x = 120$ Therefore, the salesman sold 120 kg in the morning and $2\cdot 120 = 240$ kg in the afternoon.

Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 2 kg more than Peter. Together the three of them picked 26 kg of chestnuts. How many kilograms did each of them pick? Click to see solution Solution: Let $x$ be the amount Peter picked. Then Mary and Lucy picked $2x$ and $x+2$, respectively. So $x+2x+x+2=26$ $4x=24$ $x=6$ Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.

Problem 3 Sophia finished $\frac{2}{3}$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book? Click to see solution Solution: Let $x$ be the total number of pages in the book, then she finished $\frac{2}{3}\cdot x$ pages. Then she has $x-\frac{2}{3}\cdot x=\frac{1}{3}\cdot x$ pages left. $\frac{2}{3}\cdot x-\frac{1}{3}\cdot x=90$ $\frac{1}{3}\cdot x=90$ $x=270$ So the book is 270 pages long.

Problem 4 A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then? Click to see solution Solution: If each of $6$ tractors ploughed $120$ hectares a day and they finished the work in $4$ days, then the whole field is: $120\cdot 6 \cdot 4 = 720 \cdot 4 = 2880$ hectares. Let's suppose that each of the four tractors ploughed $x$ hectares a day. Therefore in 5 days they ploughed $5 \cdot 4 \cdot x = 20 \cdot x$ hectares, which equals the area of the whole field, 2880 hectares. So, we get $20x = 2880$ $ x = \frac{2880}{20} = 144$. Hence, each of the four tractors would plough 144 hectares a day.

Problem 5 A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose? Click to see solution Solution: Let $x$ be the number he chose, then $2\cdot x - 138 = 102$ $2x = 240$ $x = 120$

Problem 6 I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose? Click to see solution Solution: Let $x$ be the number I chose, then $\frac{x}{5}-154=6$ $\frac{x}{5}=160$ $x=800$

Problem 8 One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by 18 cm 2 . Find the lengths of all sides. Click to see solution Solution: Let $x$ be the length of the longer side $x \gt 3$, then the other side's length is $x-3$ cm. Then the area is S 1 = x(x - 3) cm 2 . After we increase the lengths of the sides they will become $(x +1)$ and $(x - 3 + 1) = (x - 2)$ cm long. Hence the area of the new rectangle will be $A_2 = (x + 1)\cdot(x - 2)$ cm 2 , which is 18 cm 2 more than the first area. Therefore $A_1 +18 = A_2$ $x(x - 3) + 18 = (x + 1)(x - 2)$ $x^2 - 3x + 18 = x^2 + x - 2x - 2$ $2x = 20$ $x = 10$. So, the sides of the rectangle are $10$ cm and $(10 - 3) = 7$ cm long.

Problem 9 The first year, two cows produced 8100 litres of milk. The second year their production increased by 15% and 10% respectively, and the total amount of milk increased to 9100 litres a year. How many litres were milked from each cow each year? Click to see solution Solution: Let x be the amount of milk the first cow produced during the first year. Then the second cow produced $(8100 - x)$ litres of milk that year. The second year, each cow produced the same amount of milk as they did the first year plus the increase of $15\%$ or $10\%$. So $8100 + \frac{15}{100}\cdot x + \frac{10}{100} \cdot (8100 - x) = 9100$ Therefore $8100 + \frac{3}{20}x + \frac{1}{10}(8100 - x) = 9100$ $\frac{1}{20}x = 190$ $x = 3800$ Therefore, the cows produced 3800 and 4300 litres of milk the first year, and $4370$ and $4730$ litres of milk the second year, respectively.

Problem 10 The distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C at 12 pm, and by that time the express train stopped at at intermediate station for 10 min and the freight train stopped for 5 min. Find: a) The distance between stations C and B. b) The time when the freight train left station B. Click to see solution Solution a) Let x be the distance between stations B and C. Then the distance from station C to station A is $(148 - x)$ km. By the time of the meeting at station C, the express train travelled for $\frac{148-x}{80}+\frac{10}{60}$ hours and the freight train travelled for $\frac{x}{36}+\frac{5}{60}$ hours. The trains left at the same time, so: $\frac{148 - x}{80} + \frac{1}{6} = \frac{x}{36} + \frac{1}{12}$. The common denominator for 6, 12, 36, 80 is 720. Then $9(148 - x) +120 = 20x +60$ $1332 - 9x + 120 = 20x + 60$ $29x = 1392$ $x = 48$. Therefore the distance between stations B and C is 48 km. b) By the time of the meeting at station C the freight train rode for $\frac{48}{36} + \frac{5}{60}$ hours, i.e. $1$ hour and $25$ min. Therefore it left station B at $12 - (1 + \frac{25}{60}) = 10 + \frac{35}{60}$ hours, i.e. at 10:35 am.

Problem 11 Susan drives from city A to city B. After two hours of driving she noticed that she covered 80 km and calculated that, if she continued driving at the same speed, she would end up been 15 minutes late. So she increased her speed by 10 km/hr and she arrived at city B 36 minutes earlier than she planned. Find the distance between cities A and B. Click to see solution Solution: Let $x$ be the distance between A and B. Since Susan covered 80 km in 2 hours, her speed was $V = \frac{80}{2} = 40$ km/hr. If she continued at the same speed she would be $15$ minutes late, i.e. the planned time on the road is $\frac{x}{40} - \frac{15}{60}$ hr. The rest of the distance is $(x - 80)$ km. $V = 40 + 10 = 50$ km/hr. So, she covered the distance between A and B in $2 +\frac{x - 80}{50}$ hr, and it was 36 min less than planned. Therefore, the planned time was $2 + \frac{x -80}{50} + \frac{36}{60}$. When we equalize the expressions for the scheduled time, we get the equation: $\frac{x}{40} - \frac{15}{60} = 2 + \frac{x -80}{50} + \frac{36}{60}$ $\frac{x - 10}{40} = \frac{100 + x - 80 + 30}{50}$ $\frac{x - 10}{4} = \frac{x +50}{5}$ $5x - 50 = 4x + 200$ $x = 250$ So, the distance between cities A and B is 250 km.

Problem 12 To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3 days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced. Find how many parts the company made and how many days this took. Click to see solution Solution: Let $x$ be the number of days the company worked. Then 25x is the number of parts they planned to make. At the new production rate they made: $3\cdot 25 + (x - 3)\cdot 30 = 75 + 30(x - 3)$ Therefore: $25 x = 75 + 30(x -3) - 100$ $25x = 75 +30x -90 - 100$ $190 -75 = 30x -25$ $115 = 5x$ $x = 23$ So the company worked 23 days and they made $23\cdot 25+100 = 675$ pieces.

Problem 13 There are 24 students in a seventh grade class. They decided to plant birches and roses at the school's backyard. While each girl planted 3 roses, every three boys planted 1 birch. By the end of the day they planted $24$ plants. How many birches and roses were planted? Click to see solution Solution: Let $x$ be the number of roses. Then the number of birches is $24 - x$, and the number of boys is $3\times (24-x)$. If each girl planted 3 roses, there are $\frac{x}{3}$ girls in the class. We know that there are 24 students in the class. Therefore $\frac{x}{3} + 3(24 - x) = 24$ $x + 9(24 - x) = 3\cdot 24$ $x +216 - 9x = 72$ $216 - 72 = 8x$ $\frac{144}{8} = x$ $x = 18$ So, students planted 18 roses and 24 - x = 24 - 18 = 6 birches.

Problem 14 A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find: a) The distance the car has covered. b) The time that took it to get from C to B. Click to see solution Solution: From the statement of the problem we don't know if the 15 min stop in town C was planned or it was unexpected. So we have to consider both cases. A The stop was planned. Let us consider only the trip from C to B, and let $x$ be the number of hours the driver spent on this trip. Then the distance from C to B is $S = 40\cdot x$ km. If the driver could use the initial route, it would take him $x - \frac{30}{60} = x - \frac{1}{2}$ hours to drive from C to B. The distance from C to B according to the initially itinerary was $(x - \frac{1}{2})\cdot 32$ km, and this distance is $28$ km shorter than $40\cdot x$ km. Then we have the equation $(x - 1/2)\cdot 32 + 28 = 40x$ $32x -16 +28 = 40x$ $-8x = -12$ $8x = 12$ $x = \frac{12}{8}$ $x = 1 \frac{4}{8} = 1 \frac{1}{2} = 1 \frac{30}{60} =$ 1 hr 30 min. So, the car covered the distance between C and B in 1 hour and 30 min. The distance from A to B is $3\cdot 32 + \frac{12}{8}\cdot 40 = 96 + 60 = 156$ km. B Suppose it took $x$ hours for him to get from C to B. Then the distance is $S = 40\cdot x$ km. The driver did not plan the stop at C. Let we accept that he stopped because he had to change the route. It took $x - \frac{30}{60} + \frac{15}{60} = x - \frac{15}{60} = x - \frac{1}{4}$ h to drive from C to B. The distance from C to B is $32(x - \frac{1}{4})$ km, which is $28$ km shorter than $40\cdot x$, i.e. $32(x - \frac{1}{4}) + 28 = 40x$ $32x - 8 +28 = 40x$ $20= 8x$ $x = \frac{20}{8} = \frac{5}{2} = 2 \text{hr } 30 \text{min}.$ The distance covered equals $ 40 \times 2.5 = 100 km$.

Problem 15 If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially? Click to see solution Solution: Let $x$ be the number of days in the initial plan. Therefore, the whole field is $120\cdot x$ hectares. The farmer had to work for $x + 2$ days, and he ploughed $85(x + 2)$ hectares, leaving $40$ hectares unploughed. Then we have the equation: $120x = 85(x + 2) + 40$ $35x = 210$ $x = 6$ So the farmer planned to have the work done in 6 days, and the area of the farm field is $120\cdot 6 = 720$ hectares.

Problem 16 A woodworker normally makes a certain number of parts in 24 days. But he was able to increase his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also he made 80 extra parts. How many parts does the woodworker normally makes per day and how many pieces does he make in 24 days? Click to see solution Solution: Let $x$ be the number of parts the woodworker normally makes daily. In 24 days he makes $24\cdot x$ pieces. His new daily production rate is $x + 5$ pieces and in $22$ days he made $22 \cdot (x + 5)$ parts. This is 80 more than $24\cdot x$. Therefore the equation is: $24\cdot x + 80 = 22(x +5)$ $30 = 2x$ $x = 15$ Normally he makes 15 parts a day and in 24 days he makes $15 \cdot 24 = 360$ parts.

Problem 17 A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker. Click to see solution Solution: Let x km/hr be the initial speed of the biker, then his speed during the second part of the trip is x + 2 km/hr. Half the distance between two cities equals $2\frac{30}{60} \cdot x$ km and $2\frac{20}{60} \cdot (x + 2)$ km. From the equation: $2\frac{30}{60} \cdot x = 2\frac{20}{60} \cdot (x+2)$ we get $x = 28$ km/hr. The intial speed of the biker is 28 km/h. Half the distance between the two towns is $2 h 30 min \times 28 = 2.5 \times 28 = 70$. So the distance is $2 \times 70 = 140$ km.

Problem 18 A train covered half of the distance between stations A and B at the speed of 48 km/hr, but then it had to stop for 15 min. To make up for the delay, it increased its speed by $\frac{5}{3}$ m/sec and it arrived to station B on time. Find the distance between the two stations and the speed of the train after the stop. Click to see solution Solution: First let us determine the speed of the train after the stop. The speed was increased by $\frac{5}{3}$ m/sec $= \frac{5\cdot 60\cdot 60}{\frac{3}{1000}}$ km/hr = $6$ km/hr. Therefore, the new speed is $48 + 6 = 54$ km/hr. If it took $x$ hours to cover the first half of the distance, then it took $x - \frac{15}{60} = x - 0.25$ hr to cover the second part. So the equation is: $48 \cdot x = 54 \cdot (x - 0.25)$ $48 \cdot x = 54 \cdot x - 54\cdot 0.25$ $48 \cdot x - 54 \cdot x = - 13.5$ $-6x = - 13.5$ $x = 2.25$ h. The whole distance is $2 \times 48 \times 2.25 = 216$ km.

Problem 19 Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together. For how many days have each of them worked and what percentage of the job have each of them completed? Click to see solution Solution: First we will find the daily productivity of every worker. If we consider the whole job as unit (1), Elizabeth does $\frac{1}{15}$ of the job per day and Tony does $75\%$ of $\frac{1}{15}$, i.e. $\frac{75}{100}\cdot \frac{1}{15} = \frac{1}{20}$. Suppose that Tony worked alone for $x$ days. Then he finished $\frac{x}{20}$ of the total job alone. Working together for 6 days, the two workers finished $6\cdot (\frac{1}{15}+\frac{1}{20}) = 6\cdot \frac{7}{60} = \frac{7}{10}$ of the job. The sum of $\frac{x}{20}$ and $\frac{7}{10}$ gives us the whole job, i.e. $1$. So we get the equation: $\frac{x}{20}+\frac{7}{10}=1$ $\frac{x}{20} = \frac{3}{10}$ $x = 6$. Tony worked for 6 + 6 = 12 days and Elizabeth worked for $6$ days. The part of job done is $12\cdot \frac{1}{20} = \frac{60}{100} = 60\%$ for Tony, and $6\cdot \frac{1}{15} = \frac{40}{100} = 40\%$ for Elizabeth.

Problem 20 A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule. a) What is the area of the field? b) In how many days did the farmer get the job done? c) In how many days did the farmer plan to get the job done? Click to see solution Solution: First of all we will find the new daily productivity of the farmer in hectares per day: 25% of 120 hectares is $\frac{25}{100} \cdot 120 = 30$ hectares, therefore $120 + 30 = 150$ hectares is the new daily productivity. Lets x be the planned number of days allotted for the job. Then the farm is $120\cdot x$ hectares. On the other hand, we get the same area if we add $120 \cdot 2$ hectares to $150(x -4)$ hectares. Then we get the equation $120x = 120\cdot 2 + 150(x -4)$ $x = 12$ So, the job was initially supposed to take 12 days, but actually the field was ploughed in 12 - 2 =10 days. The field's area is $120 \cdot 12 = 1440$ hectares.

Problem 21 To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by $33 \times \frac{1}{3}\%$, and finished the work 1 day earlier than it was planned. A) What is the area of the grass field? B) How many days did it take to mow the whole field? C) How many days were scheduled initially for this job? Hint : See problem 20 and solve by yourself. Answer: A) 120 hectares; B) 7 days; C) 8 days.

Problem 22 A train travels from station A to station B. If the train leaves station A and makes 75 km/hr, it arrives at station B 48 minutes ahead of scheduled. If it made 50 km/hr, then by the scheduled time of arrival it would still have 40 km more to go to station B. Find: A) The distance between the two stations; B) The time it takes the train to travel from A to B according to the schedule; C) The speed of the train when it's on schedule. Click to see solution Solution: Let $x$ be the scheduled time for the trip from A to B. Then the distance between A and B can be found in two ways. On one hand, this distance equals $75(x - \frac{48}{60})$ km. On the other hand, it is $50x + 40$ km. So we get the equation: $75(x - \frac{48}{60}) = 50x + 40$ $x = 4$ hr is the scheduled travel time. The distance between the two stations is $50\cdot 4 +40 = 240$ km. Then the speed the train must keep to be on schedule is $\frac{240}{4} = 60$ km/hr.

Problem 23 The distance between towns A and B is 300 km. One train departs from town A and another train departs from town B, both leaving at the same moment of time and heading towards each other. We know that one of them is 10 km/hr faster than the other. Find the speeds of both trains if 2 hours after their departure the distance between them is 40 km. Click to see solution Solution: Let the speed of the slower train be $x$ km/hr. Then the speed of the faster train is $(x + 10)$ km/hr. In 2 hours they cover $2x$ km and $2(x +10)$km, respectively. Therefore if they didn't meet yet, the whole distance from A to B is $2x + 2(x +10) +40 = 4x +60$ km. However, if they already met and continued to move, the distance would be $2x + 2(x + 10) - 40 = 4x - 20$km. So we get the following equations: $4x + 60 = 300$ $4x = 240$ $x = 60$ or $4x - 20 = 300$ $4x = 320$ $x = 80$ Hence the speed of the slower train is $60$ km/hr or $80$ km/hr and the speed of the faster train is $70$ km/hr or $90$ km/hr.

Problem 24 A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by $\frac{50}{9}$ m/sec, it will arrive in town B 30 min earlier than scheduled. Find: A) The distance between the two towns; B) The bus's scheduled time of arrival in B; C) The speed of the bus when it's on schedule. Click to see solution Solution: First we will determine the speed of the bus following its increase. The speed is increased by $\frac{50}{9}$ m/sec $= \frac{50\cdot60\cdot60}{\frac{9}{1000}}$ km/hr $= 20$ km/hr. Therefore, the new speed is $V = 50 + 20 = 70$ km/hr. If $x$ is the number of hours according to the schedule, then at the speed of 50 km/hr the bus travels from A to B within $(x +\frac{42}{60})$ hr. When the speed of the bus is $V = 70$ km/hr, the travel time is $x - \frac{30}{60}$ hr. Then $50(x +\frac{42}{60}) = 70(x-\frac{30}{60})$ $5(x+\frac{7}{10}) = 7(x-\frac{1}{2})$ $\frac{7}{2} + \frac{7}{2} = 7x -5x$ $2x = 7$ $x = \frac{7}{2}$ hr. So, the bus is scheduled to make the trip in $3$ hr $30$ min. The distance between the two towns is $70(\frac{7}{2} - \frac{1}{2}) = 70\cdot 3 = 210$ km and the scheduled speed is $\frac{210}{\frac{7}{2}} = 60$ km/hr.

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HIX Tutor’s AI Math Problem Solver Features at a Glance

1. What is Math AI?

A Math AI is an artificial intelligence-powered tool designed to solve complex mathematical problems efficiently and accurately. By utilizing advanced algorithms and computational power, Math AI can provide step-by-step solutions, offer insights into problem-solving strategies, and enhance our overall understanding of various mathematical concepts.

2. What math subjects does the AI math problem solver cover?

Our AI math solver tool is trained to solve a wide range of math subjects, including but not limited to, algebra, geometry, calculus, trigonometry, calculus, and more.

3. How long does this math AI take to get an answer to my math question?

HIX Tutor's math AI Solver is available 24/7 and delivers correct, step-by-step solutions to maths questions almost immediately after you submit the query.

4. What's the best Math AI solver?

HIX Tutor is the best Math AI that offers comprehensive solutions for solving complex mathematical problems. With its advanced features, such as step-by-step explanations, personalized learning paths, and interactive problem-solving tools, HIX Tutor aims to help students and professionals better understand mathematical concepts and improve their problem-solving skills.

5. How does HIX Tutor's AI for maths work?

HIX Tutor's math AI helper utilizes advanced algorithms and deep learning techniques to analyze and understand math problems. It breaks down the problem into steps, applies relevant mathematical concepts, and provides detailed explanations for each step of the solution.

6. Is HIX Tutor's mathematics AI solver accurate?

Yes, our math AI solving tool is designed to deliver accurate solutions. It has been trained on a vast dataset of mathematical problems to ensure its proficiency in providing reliable answers. However, it's important to note that while our tool strives for accuracy, occasional errors or misunderstandings may occur.

7. Is HIX Tutor's math AI solver replace the need for a maths tutor?

While our maths AI tool is designed to provide comprehensive support and step-by-step solutions, it is not intended to replace the guidance and expertise of a human maths tutor. It can be a valuable tool to supplement your learning and provide quick answers, but for in-depth understanding and tailored guidance, a maths tutor may still be beneficial in certain situations.

8. Is this math AI free to use?

Yes, you can try Math AI Solver at no cost. Once you’ve reached your free question limit, you’ll need to purchase an affordable subscription.

Discover Frequently Asked Math Questions and Their Answers

• How do you write the quadratic function #y=x^2+14x+11# in vertex form?
• How many points does #y=-2x^2+x-3# have in common with the vertex and where is the vertex in relation to the x axis?
• How do you solve #4x^4 - 16x^2 + 15 = 0#?
• How do you solve #2x^2+3x-2=0#?
• How do you solve #7(x-4)^2-2=54# using any method?
• How do you solve #x^2 + 5x + 6 = 0# algebraically?
• How do you use factoring to solve this equation #3x^2/4=27#?
• What is the vertex of # y = (1/8)(x – 5)^2 - 3#?
• How do you solve #| x^2+3x-2 | =2#?
• How do you solve #2x²+3x=5 # using the quadratic formula?
• How do you find the derivative of #y=tan(3x)# ?
• How do you differentiate #f(x)= 1/ (lnx)# using the quotient rule?
• How do you differentiate #(3+sin(x))/(3x+cos(x))#?
• What is the derivative of this function #sin^-1(x/4)#?
• What is the derivative of this function #y=sin^-1(2x)#?
• What is the derivative of this function #arcsec(x^3)#?
• What is the derivative of #y=sin(tan2x)#?
• How do you differentiate #cos(pi*x^2)#?
• What is the derivative of #f(x)=(x^2-4)ln(x^3/3-4x)#?
• What is the derivative of #y=3sin(x) - sin(3x)#?
• A triangle has corners at #(5 ,1 )#, #(2 ,9 )#, and #(4 ,3 )#. What is the area of the triangle's circumscribed circle?
• How can we find the area of irregular shapes?
• A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/3 #, and the triangle's area is #24 #. What is the area of the triangle's incircle?
• An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,1 )# to #(8 ,5 )# and the triangle's area is #27 #, what are the possible coordinates of the triangle's third corner?
• A triangle has corners at #(7 , 9 )#, #(3 ,7 )#, and #(4 ,8 )#. What is the radius of the triangle's inscribed circle?
• Circle A has a center at #(2 ,3 )# and a radius of #1 #. Circle B has a center at #(0 ,-2 )# and a radius of #4 #. Do the circles overlap? If not, what is the smallest distance between them?
• A parallelogram has sides A, B, C, and D. Sides A and B have a length of #2 # and sides C and D have a length of # 7 #. If the angle between sides A and C is #pi/4 #, what is the area of the parallelogram?
• What is a quadrilateral that is not a parallelogram and not a trapezoid?
• Your teacher made 8 triangles he need help to identify what type triangles they are. Help him?: 1) #12, 16, 20# 2) #15, 17, 22# 3) #6, 16, 26# 4) #12, 12, 15# 5) #5,12,13# 6) #7,24,25# 7) #8,15,17# 8) #9,40,41#
• A triangle has corners A, B, and C located at #(3 ,5 )#, #(2 ,9 )#, and #(4 , 8 )#, respectively. What are the endpoints and length of the altitude going through corner C?
• What is the GCF of the set #64, 16n^2, 32n#?
• How do you write the reciprocal number of 5?
• Jeanie has a 3/4 yard piece of ribbon. She needs one 3/8 yard piece and one 1/2 yard piece. Can she cut the piece of ribbon into the two smaller pieces? Why?
• How do you find the GCF of #25k, 35j#?
• How do you write 132/100 in a mixed number?
• How do you evaluate the power #2^3#?
• How do you simplify #(4^6)^2 #?
• How do you convert 3.2 tons to pounds?
• How do you solve #\frac { 5} { 8} + \frac { 3} { 2} ( 4- \frac { 1} { 4} ) - \frac { 1} { 8}#?
• What are some acronyms for PEMDAS?
• How do you find all the asymptotes for function #y=(3x^2+2x-1)/(x^2-4 )#?
• How do you determine whether the graph of #y^2+3x=0# is symmetric with respect to the x axis, y axis or neither?
• How do you determine whether the graph of #y^2=(4x^2)/9-4# is symmetric with respect to the x axis, y axis, the line y=x or y=-x, or none of these?
• How do you find the end behavior of #-x^3+3x^2+x-3#?
• How do you find the asymptotes for #(2x^2 - x - 38) / (x^2 - 4)#?
• How do you find the asymptotes for #f(x) = (x^2) / (x^2 + 1)#?
• How do you find the vertical, horizontal and slant asymptotes of: #(3x-2) / (x+1)#?
• How do you find the Vertical, Horizontal, and Oblique Asymptote given #s(t)=(8t)/sin(t)#?
• How do you find vertical, horizontal and oblique asymptotes for #(x^3+1)/(x^2+3x)#?
• How do you find vertical, horizontal and oblique asymptotes for #y = (4x^3 + x^2 + x + 5 )/( x^2 + 3x)#?
• What is a pooled variance?
• What is the mean, mode median and range of 11, 12, 13, 12, 14, 11, 12?
• What is the z-score of sample X, if #n = 81, mu= 43, St. Dev. =90, and E[X] =57#?
• The camera club has five members, and the mathematics club has eight. There is only one member common to both clubs. In how many ways could a committee of four people be formed with at least one member from each club?
• How many permutations are there of the letter in the word baseball?
• How do you evaluate 6p4?
• What is the probability of #X= 6# successes, using the binomial formula?
• A lottery has a $100 000 first prize, a $25 000 second prize, and five $500 third prizes. A total of 50000 tickets are sold. What is the probability of winning a prize in this lottery?
• When a event is reported, the probability that is a negative event is 30%. What is the probability that 3 out of 5 reported events are negative?
• What is the median of 5, 19, 2, 28, 25?
• How do you solve this trigonometric equation?
• What is the frequency of #f(theta)= sin 3 t - cos2 t #?
• How do you evaluate #Sin(pi/2) + 6 cos(pi/3) #?
• How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=9, b=40, c=41?
• How do you express (5*pi)/4 into degree?
• Solve for θ on the interval [90°,180°]:2tanθ +19 = 0?
• Prove that #((cos(33^@))^2-(cos(57^@))^2)/((sin(10.5^@))^2-(sin(34.5^@))^2)= -sqrt2# ?
• A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/3# and the angle between sides B and C is #pi/6#. If side B has a length of 26, what is the area of the triangle?
• How do you write the following in trigonometric form and perform the operation given #(sqrt3+i)(1+i)#?
• A triangle has sides A, B, and C. The angle between sides A and B is #(2pi)/3#. If side C has a length of #32 # and the angle between sides B and C is #pi/12#, what is the length of side A?

Trust HIX Tutor’s Math AI for Reliable Math Solutions

Don’t waste hours slaving over complex equations or confusing mathematical word problems. Math AI Solver is the ultimate solution for completing math assessments quickly and efficiently. Get started now.

Math Word Problems

Welcome to the math word problems worksheets page at Math-Drills.com! On this page, you will find Math word and story problems worksheets with single- and multi-step solutions on a variety of math topics including addition, multiplication, subtraction, division and other math topics. It is usually a good idea to ensure students already have a strategy or two in place to complete the math operations involved in a particular question. For example, students may need a way to figure out what 7 × 8 is or have previously memorized the answer before you give them a word problem that involves finding the answer to 7 × 8.

There are a number of strategies used in solving math word problems; if you don't have a favorite, try the Math-Drills.com problem-solving strategy:

• Question : Understand what the question is asking. What operation or operations do you need to use to solve this question? Ask for help to understand the question if you can't do it on your own.
• Estimate : Use an estimation strategy, so you can check your answer for reasonableness in the evaluate step. Try underestimating and overestimating, so you know what range the answer is supposed to be in. Be flexible in rounding numbers if it will make your estimate easier.
• Strategize : Choose a strategy to solve the problem. Will you use mental math, manipulatives, or pencil and paper? Use a strategy that works for you. Save the calculator until the evaluate stage.
• Calculate : Use your strategy to solve the problem.
• Evaluate : Compare your answer to your estimate. If you under and overestimated, is the answer in the correct range. If you rounded up or down, does the answer make sense (e.g. is it a little less or a little more than the estimate). Also check with a calculator.

Most Popular Math Word Problems this Week

Arithmetic Word Problems

• Addition Word Problems One-Step Addition Word Problems Using Single-Digit Numbers One-Step Addition Word Problems Using Two-Digit Numbers
• Subtraction Word Problems Subtraction Facts Word Problems With Differences from 5 to 12
• Multiplication Word Problems One-Step Multiplication Word Problems up to 10 × 10
• Division Word Problems Division Facts Word Problems with Quotients from 5 to 12
• Multi-Step Word Problems Easy Multi-Step Word Problems

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Probability Problem Solver

This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of artificial intelligence large language models to parse and generate natural language answers. This creates a math problem solver that's more accurate than ChatGPT, more flexible than a math calculator, and provides answers faster than a human tutor.

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Problem Solver Subjects

Our math problem solver that lets you input a wide variety of probability math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects.

• Math Word Problems
• Pre-Algebra
• Geometry Graphing
• Trigonometry
• Precalculus
• Finite Math
• Linear Algebra

Here are example math problems within each subject that can be input into the calculator and solved. This list is constanstly growing as functionality is added to the calculator.

Basic Math Solutions

Below are examples of basic math problems that can be solved.

• Long Arithmetic
• Rational Numbers
• Operations with Fractions
• Ratios, Proportions, Percents
• Measurement, Area, and Volume
• Factors, Fractions, and Exponents
• Unit Conversions
• Data Measurement and Statistics
• Points and Line Segments

Math Word Problem Solutions

Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

Pre-Algebra Solutions

Below are examples of Pre-Algebra math problems that can be solved.

• Variables, Expressions, and Integers
• Simplifying and Evaluating Expressions
• Solving Equations
• Multi-Step Equations and Inequalities
• Ratios, Proportions, and Percents
• Linear Equations and Inequalities

Algebra Solutions

Below are examples of Algebra math problems that can be solved.

• Algebra Concepts and Expressions
• Points, Lines, and Line Segments
• Simplifying Polynomials
• Factoring Polynomials
• Linear Equations
• Absolute Value Expressions and Equations
• Radical Expressions and Equations
• Systems of Equations
• Quadratic Equations
• Inequalities
• Complex Numbers and Vector Analysis
• Logarithmic Expressions and Equations
• Exponential Expressions and Equations
• Conic Sections
• Vector Spaces
• 3d Coordinate System
• Eigenvalues and Eigenvectors
• Linear Transformations
• Number Sets
• Analytic Geometry

Trigonometry Solutions

Below are examples of Trigonometry math problems that can be solved.

• Algebra Concepts and Expressions Review
• Right Triangle Trigonometry
• Radian Measure and Circular Functions
• Graphing Trigonometric Functions
• Simplifying Trigonometric Expressions
• Verifying Trigonometric Identities
• Solving Trigonometric Equations
• Complex Numbers
• Analytic Geometry in Polar Coordinates
• Exponential and Logarithmic Functions
• Vector Arithmetic

Precalculus Solutions

Below are examples of Precalculus math problems that can be solved.

• Operations on Functions
• Rational Expressions and Equations
• Polynomial and Rational Functions
• Analytic Trigonometry
• Sequences and Series
• Analytic Geometry in Rectangular Coordinates
• Limits and an Introduction to Calculus

Calculus Solutions

Below are examples of Calculus math problems that can be solved.

• Evaluating Limits
• Derivatives
• Applications of Differentiation
• Applications of Integration
• Techniques of Integration
• Parametric Equations and Polar Coordinates
• Differential Equations

Statistics Solutions

Below are examples of Statistics problems that can be solved.

• Algebra Review
• Average Descriptive Statistics
• Dispersion Statistics
• Probability
• Probability Distributions
• Frequency Distribution
• Normal Distributions
• t-Distributions
• Hypothesis Testing
• Estimation and Sample Size
• Correlation and Regression

Finite Math Solutions

Below are examples of Finite Math problems that can be solved.

• Polynomials and Expressions
• Equations and Inequalities
• Linear Functions and Points
• Systems of Linear Equations
• Mathematics of Finance
• Statistical Distributions

Linear Algebra Solutions

Below are examples of Linear Algebra math problems that can be solved.

• Introduction to Matrices
• Linear Independence and Combinations

Chemistry Solutions

Below are examples of Chemistry problems that can be solved.

• Unit Conversion
• Atomic Structure
• Molecules and Compounds
• Chemical Equations and Reactions
• Behavior of Gases
• Solutions and Concentrations

Physics Solutions

Below are examples of Physics math problems that can be solved.

• Static Equilibrium
• Dynamic Equilibrium
• Kinematics Equations
• Electricity
• Thermodymanics

Geometry Graphing Solutions

Below are examples of Geometry and graphing math problems that can be solved.

• Step By Step Graphing
• Linear Equations and Functions
• Polar Equations

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