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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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1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck.

Let the 1st paycheck be x (integer). Mrs. Rodger got a weekly raise of $ 145. So after completing the 1st week she will get $ (x+145). Similarly after completing the 2nd week she will get $ (x + 145) + $ 145. = $ (x + 145 + 145) = $ (x + 290) So in this way end of every week her salary will increase by $ 145.

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he: (a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents Solution:

Selling price of the first pipe = $1.20

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price - Profit

CP = 1.20 - 20% of CP

CP = 1.20 - 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = \(\frac{1.20}{1.20}\)

Selling price of the Second pipe = $1.20

Let’s try to find the cost price of the second pipe

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP - 0.20CP = 1.20

0.80CP = 1.20

CP = \(\frac{1.20}{0.80}\)

Therefore, total cost price of the two pipes = $1.00 + $1.50 = $2.50

 And total selling price of the two pipes = $1.20 + $1.20 = $2.40

Loss = $2.50 – $2.40 = $0.10

Therefore, Mr. Jones loss 10 cents.

Answer:   (d) 

5. A man has $ 10,000 to invest. He invests $ 4000 at 5 % and $ 3500 at 4 %. In order to have a yearly income of $ 500, he must invest the remainder at: (a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 % Solution: Income from $ 4000 at 5 % in one year = $ 4000 of 5 %. = $ 4000 × 5/100. = $ 4000 × 0.05. = $ 200. Income from $ 3500 at 4 % in one year = $ 3500 of 4 %. = $ 3500 × 4/100. = $ 3500 × 0.04. = $ 140. Total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340. Remaining income amount in order to have a yearly income of $ 500 = $ 500 - $ 340. = $ 160. Total invested amount = $ 4000 + $ 3500 = $7500. Remaining invest amount = $ 10000 - $ 7500 = $ 2500. We know that, Interest = Principal × Rate × Time Interest = $ 160, Principal = $ 2500, Rate = r [we need to find the value of r], Time = 1 year. 160 = 2500 × r × 1. 160 = 2500r 160/2500 = 2500r/2500 [divide both sides by 2500] 0.064 = r r = 0.064 Change it to a percent by moving the decimal to the right two places r = 6.4 % Therefore, he invested the remaining amount $ 2500 at 6.4 % in order to get $ 500 income every year. Answer: (e) 6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was: (a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much Solution: Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr. We know that Speed = Distance/Time. Or, Time = Distance/Speed. So, times taken to covered a distance of 50 miles on his first trip = 50/x hr. And times taken to covered a distance of 300 miles on his later trip = 300/3x hr. = 100/x hr. So we can clearly see that his new time compared with the old time was: twice as much. Answer: (b)

11. Kalin walks at a constant rate of 5/8  kilometers per hour. The beach is 3/4 kilometers from his home. How long does it take Kalin to walk from his home to the beach?

Speed =  \(\frac{5}{8}\) kilometers per hour.

Distance =  \(\frac{3}{4}\) kilometers

Distance =  Speed × Time

or, Time = Distance/Speed

            = (\(\frac{3}{4}\) ÷ \(\frac{5}{8}\)) hours

            = (\(\frac{3}{4}\) × \(\frac{8}{5}\)) hours

            = \(\frac{24}{20}\) hours

            = \(\frac{6}{5}\) hours

            = 1\(\frac{1}{5}\) hours

Answer:  1\(\frac{1}{5}\) hours

12.  The cost price of one refrigerator is $23,547. How many refrigerators can be purchased for $ 588,675 ?

Total amount = $588,675

Cost price of one refrigerator = $23,547.

No. of  refrigerators = $588,675 ÷ $23,547 = 25

13. In an University auditorium each row has 35 seats. Determine the minimum number of rows required to seat 1,575 students at a time.

Total number of seats = 1,575

Each row has 35 seats.

No. of  rows required = 1,575 ÷ 35 = 45

14. What least number must be added to 15896 to get a number exactly divisible by 29?

First divide  15896 by 29.

15896 = 29 × 548 + 4

The remainder is 4

Therefore, we need to add 4 to  15896 it will exactly divisible by 29.

15. Find the largest 6-digit number exactly divisible by 174.

The largest 6-digit number = 999999

Now divide the  largest 6-digit number (999999) by 174

999999 = 174 × 5747 + 21

Remainder = 21

Now subtract the remainder 21 from the  largest 6-digit number 999999.

Therefore, 999999 - 21 =  999978 is the  largest 6-digit number exactly divisible by 174.

Answer: 999978

16. On diving 80,256 by 187, the remainder is 33. Find the quotient.

Dividend = Divisor × Quotient + Remainder

80,256 = 187 × Quotient + 33

187 × Quotient = 80,256 - 33

187 × Quotient =  80223

Quotient  = 80223 ÷ 187 = 429

Answer: 429

17. During prayer time in a school, 35 students stand in each row. Find the minimum number of rows if there are 1,575 students in that school.

Number of students in each row = 35

Number of row =  1,575 ÷ 35 = 45

18. A motorcycle travels 1494 km in 18 hours. What is the average speed of the car?

Distance traveled = 1494 km

Time =  18 hours

Average speed =  Distance traveled ÷ Time 

                      = (1494 ÷ 18) km/hr

                      = 83 km/hr

Answer: 83 km/hr

19. Divide 78938 by 75 and check the result by division algorithm.

By actual division, we have

Here, dividend = 78938,

quotient 1052 and

remainder  38.

   Quotient × Divisor + Remainder = 1052 × 75 + 38

= 78900 + 38

= 78938  = dividend

20. Find the number which when divided by 67 gives 17 as quotient and 11 as remainder.

Here, divisor = 67, quotient = 17 and remainder 11

By division algorithm, we have

              = 67 × 17 + 11

              = 1,139 + 11

              = 1,150

Hence, the required number is 1,150.

Unsolved Questions:

1. Fahrenheit temperature F is a linear function of Celsius temperature C. The ordered pair (0, 32) is an ordered pair of this function because 0°C is equivalent to 32°F, the freezing point of water. The ordered pair (100, 212) is also an ordered pair of this function because 100°C is equivalent to 212° F, the boiling point of water.

2. A sports field is 300 feet long. Write a formula that gives the length of x sports fields in feet. Then use this formula to determine the number of sports fields in 720 feet.

3. A recipe calls for 2 1/2 cups and I want to make 1 1/2 recipes. How many cups do I need?

4. Mario answered 30% of the questions correctly. The test contained a total of 80 questions. How many questions did Mario answer correctly?

5. Mary’s credit card company charges 16% interest on her outstanding credit card balance each month. Her minimum payment is $20 each month. Mary’s credit card bill is $70 in January. Mary only pays the minimum amount each month, and she does not spend any additional money on her credit card. How long, in months, will it take her to pay off her bill from January?

6.  Imagine a can of purple paint that is 3/4 full. This purple paint consists of 40% red paint and 60% blue paint.

Part A : A painter decides to fill the remaining 1/4 of the can with red paint and mixes everything together thoroughly. What percentage of the new mixture is blue paint?

Answer:_________________

Part B: If the painter had chosen to fill the remaining 1/4 of the can with blue paint instead of red paint and mixed it thoroughly, what percentage of the new mixture would be blue paint?

7. Brandon has skittles and M&M's in his candy jar in a ratio of 3:5. His little sister Paige comes home from school one day and when Brandon isn't there she eats 1/3 of the skittles. If there are 56 pieces of candy in the jar after Paige eats the skittles, how many M&M's are in the jar?

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Math word Problems With Answers for Grade 6

Grade 6 maths word problems with answers are presented. Some of these problems are challenging and need more time to solve. Also detailed solutions and full explanations are included.

• Two numbers N and 16 have LCM = 48 and GCF = 8. Find N.
• If the area of a circle is 81pi square feet, find its circumference.
• Find the greatest common factor of 24, 40 and 60.
• In a given school, there are 240 boys and 260 girls. a) What is the ratio of the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school?
• If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend?
• Find k if 64 ÷ k = 4.
• Little John had $8.50. He spent $1.25 on sweets and gave to his two friends $1.20 each. How much money was left?
• What is x if x + 2y = 10 and y = 3?
• A telephone company charges initially $0.50 and then $0.11 for every minute. Write an expression that gives the cost of a call that lasts N minutes.
• A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 180 kilometers?
• A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles?
• A car travels at a speed of 65 miles per hour. How far will it travel in 5 hours?
• A small square of side 2x is cut from the corner of a rectangle with a width of 10 centimeters and length of 20 centimeters. Write an expression in terms of x for the area of the remaining shape.
• A rectangle A with length 10 centimeters and width 5 centimeters is similar to another rectangle B whose length is 30 centimeters. Find the area of rectangle B.
• A school has 10 classes with the same number of students in each class. One day, the weather was bad and many students were absent. 5 classes were half full, 3 classes were 3/4 full and 2 classes were 1/8 empty. A total of 70 students were absent. How many students are in this school when no students are absent?
• The perimeter of square A is 3 times the perimeter of square B. What is the ratio of the area of square A to the area of square B.
• John gave half of his stamps to Jim. Jim gave gave half of his stamps to Carla. Carla gave 1/4 of the stamps given to her to Thomas and kept the remaining 12. How many stamps did John start with?
• Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 60 seconds. If they start rotating now from the same point, when will they be at the same starting point again?
• A segment is 3 units long. It is divided into 9 parts. What fraction of a unit are 2 parts of the segment?
• A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute?
• Carla is 5 years old and Jim is 13 years younger than Peter. One year ago, Peter's age was twice the sum of Carla's & Jim's age. Find the present age of each one of them.
• Linda spent 3/4 of her savings on furniture. She then spent 1/2 of her remaining savings on a fridge. If the fridge cost her $150, what were her original savings?
• The distance bewteen Harry and Kate is 2500 meters. Kate and Harry start walking toward one another and Kate' dog start running back and forth between Harry and Kate at a speed of 120 meters per minute. Harry walks at the speed of 40 meters per minute while Kate walks at the speed of 60 meters per minute. What distance will the dog have travelled when Harry and Kate meet each other?

Answers to the Above Questions

• a) 13:12 b)12:25
• 0.50 + N * 0.11
• 4.5 gallons
• 450 centimeters squared
• 108 cubic centimeters
• 1250 meters/minute
• Carla:5 years, Jim: 6 years, Peter: 19 years.
• 3000 meters

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The Hechinger Report

Covering Innovation & Inequality in Education

Why schools are teaching math word problems all wrong

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CENTRAL FALLS, R.I. — When Natalia Molina began teaching her second grade students word problems earlier this school year, every lesson felt difficult. Most students were stymied by problems such as: “Sally went shopping. She spent $86 on groceries and $39 on clothing. How much more did Sally spend on groceries than on clothing?”

Both Molina, a first-year teacher, and her students had been trained to tackle word problems by zeroing in on key words like “and,” “more” and “total”  — a simplistic approach that Molina said too often led her students astray. After recognizing the word “and,” for instance, they might mistakenly assume that they needed to add two nearby numbers together to arrive at an answer.

Some weaker readers, lost in a sea of text, couldn’t recognize any words at all.

“I saw how overwhelmed they would get,” said Molina, who teaches at Segue Institute for Learning, a predominantly Hispanic charter school in this small city just north of Providence.

So, with the help of a trainer doing work in Rhode Island through a state grant, Molina and some of her colleagues revamped their approach to teaching word problems this winter — an effort that they said is already paying off in terms of increased student confidence and ability. “It has been a game changer for them,” Molina said.

Perhaps no single educational task encompasses as many different skills as the word problem. Between reading, executive functioning, problem solving, computation and vocabulary, there are a lot of ways for students to go wrong. And for that reason, students perform significantly worse overall on word problems compared to questions more narrowly focused on computation or shapes (for example: “Solve 7 + _ = 22” or “What is 64 x 3?”).

If a student excels at word problems, it’s a good sign that they’re generally excelling at school. “Word-problem solving in lower grades is one of the better indicators of overall school success in K-12,” said Lynn Fuchs, a research professor at Vanderbilt University. In a large national survey , for instance, algebra teachers rated word-problem solving as the most important among 15 skills required to excel in the subject.

Teacher takeaways

• Don’t instruct students to focus mainly on “key words” in word problems such as “and” or “more” 
• Mix question types in any lesson so that students don’t assume they just apply the same operation (addition, subtraction) again and again
• Teach students the underlying structure — or schema — of the word problem

Yet most experts and many educators agree that too many schools are doing it wrong, particularly in the elementary grades. And in a small but growing number of classrooms, teachers like Molina are working to change that. “With word problems, there are more struggling learners than non-struggling learners” because they are taught so poorly, said Nicole Bucka, who works with teachers throughout Rhode Island to provide strategies for struggling learners.

Too many teachers, particularly in the early grades, rely on key words to introduce math problems. Posters displaying the terms — sum, minus, fewer, etc. — tied to operations including addition and subtraction are a staple in elementary school classrooms across the country.

Key words can be a convenient crutch for both students and teachers, but they become virtually meaningless as the problems become harder, according to researchers. Key words can help first graders figure out whether to add or subtract more than half of the time, but the strategy rarely works for the multi-step problems students encounter starting in second and third grade. “With multi-step problems, key words don’t work 90 percent of the time,” said Sarah Powell, a professor at the University of Texas in Austin who studies word problems and whose research has highlighted the inefficacy of key words . “But the average kindergarten teacher is not thinking about that; they are teaching 5-year-olds, not 9-year-olds.”

Many teachers in the youngest grades hand out worksheets featuring the same type of word problem repeated over and over again. That’s what Molina’s colleague, Cassandra Santiago, did sometimes last year when leading a classroom on her own for the first time. “It was a mistake,” the first grade teacher said. “It’s really important to mix them up. It makes them think more critically about the parts they have to solve.”

Another flaw with word problem instruction is that the overwhelming majority of questions are divorced from the actual problem-solving a child might have to do outside the classroom in their daily life — or ever, really. “I’ve seen questions about two trains going on the same track,” said William Schmidt, a University Distinguished Professor at Michigan State University. “First, why would they be going on the same track and, second, who cares?”

Schmidt worked on an analysis of about 8,000 word problems used in 23 textbooks in 19 countries. He found that less than one percent had “real world applications” and involved “higher order math applications .”

“That is one of the reasons why children have problems with mathematics,” he said. “They don’t see the connection to the real world … We’re at this point in math right now where we are just teaching students how to manipulate numbers.”

He said a question, aimed at middle schoolers, that does have real world connections and involves more than manipulating numbers, might be: “Shopping at the new store in town includes a 43% discount on all items which are priced the same at $2. The state you live in has a 7% sales tax. You want to buy many things but only have a total of $52 to spend. Describe in words how many things you could buy.”

Schmidt added that relevancy of word problems is one area where few, if any, countries excel. “No one was a shining star leading the way,” he said. 

In her brightly decorated classroom one Tuesday afternoon, Santiago, the first grade teacher, gave each student a set of animal-shaped objects and a sheet of blue paper (the water) and green (the grass). “We’re going to work on a number story,” she told them. “I want you to use your animals to tell me the story.”

“ Once upon a time,” the story began. In this tale, three animals played in the water, and two animals played in the grass. Santiago allowed some time for the ducks, pigs and bears to frolic in the wilds of each student’s desk before she asked the children to write a number sentence that would tell them how many animals they have altogether.

Some of the students relied more on pictorial representations (three dots on one side of a line and two dots on the other) and others on the number sentence (3+2 = 5) but all of them eventually got to five. And Santiago made sure that her next question mixed up the order of operations (so students didn’t incorrectly assume that all they ever have to do is add): “Some more animals came and now there are seven. So how many more came?”

One approach to early elementary word problems that is taking off in some schools, including Segue Institute, has its origins in a special education intervention for struggling math students. Teachers avoid emphasizing key words and ask students instead to identify first the conceptual type of word problem (or schema, as many practitioners and researchers refer to it) they are dealing with: “Total problems,” for instance, involve combining two parts to find a new amount; “change problems” involve increasing or decreasing the amount of something. Total problems do not necessarily involve adding, however.

“The schemas that students learn in kindergarten will continue with them throughout their whole career,” said Powell, the word-problem researcher, who regularly works with districts across the country to help implement the approach. 

In Olathe, Kansas — a district inspired by Powell’s work — teachers had struggled for years with word problems, said Kelly Ulmer, a math support specialist whose goal is to assist in closing academic gaps that resulted from lost instruction time during the pandemic. “We’ve all tried these traditional approaches that weren’t working,” she said. “Sometimes you get pushback on new initiatives from veteran teachers and one of the things that showed us how badly this was needed is that the veteran teachers were the most excited and engaged — they have tried so many things” that haven’t worked.

In Rhode Island, many elementary schools initially used the strategy with students who required extra help, including those in special education, but expanded this use to make it part of the core instruction for all, said Bucka. In some respects, it’s similar to the recent, well publicized evolution of reading instruction in which some special education interventions for struggling readers  — most notably, a greater reliance on phonics in the early grades — have gone mainstream.

There is an extensive research bas e showing that focusing on the different conceptual types of word problems is an effective way of teaching math, although much of the research focuses specifically on students experiencing difficulties in the subject. 

Molina has found asking students to identify word problems by type to be a useful tool with nearly all of her second graders; next school year she hopes to introduce the strategy much earlier.

One recent afternoon, a lesson on word problems started with everyone standing up and chanting in unison: “Part plus part equals total” (they brought two hands together). “Total minus part equals part ” (they took one hand away) .

It’s a way to help students remember different conceptual frameworks for word problems. And it’s especially effective for the students who learn well through listening and repeating. For visual learners, the different types of word problems were mapped out on individual dry erase mats.

The real work began when Molina passed out questions, and the students— organized into the Penguin, Flower Bloom, Red Panda and Marshmallow teams — had to figure out which framework they were dealing with on their own and then work toward an answer. A few months ago, many of them would have automatically shut down when they saw the text on the page, Molina said.

For the Red Pandas, the question under scrutiny was: “The clothing store had 47 shirts. They sold 21, how many do they have now?”

“It’s a total problem,” one student said.

“No, it’s not total,” responded another.

“I think it’s about change,” said a third.

None of the students seemed worried about their lack of consensus, however. And neither was Molina. A correct answer is always nice but those come more often now that most of the students have made a crucial leap. “I notice them thinking more and more,” she said, “about what the question is actually asking.”

This story about word problems was produced by The Hechinger Report , a nonprofit, independent news organization focused on inequality and innovation in education. Sign up for the Hechinger newsletter .

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