Problem: After both Kyle and Lisa's throws, the ball bounced an additional 3/13 yards. Subtract this distance from the total distance the ball traveled when Kyle and Lisa threw it. What is the remaining distance?

Qwen2-Math 72B:

 Reasoning: The ball traveled a total of 6.611111111111111 yards when Kyle and Lisa threw it.
The ball bounced an additional 3/13 = <<3/13=0.23076923076923078>>0.23076923076923078 yards.
The remaining distance is 6.611111111111111-0.23076923076923078 = <<6.611111111111111-0.23076923076923078=6.380341880341881>>6.380341880341881 yards.
### Answer: 6.380341880341881

Expected answer: 1493/234

Correct? Yes

Problem: In their game, suppose that Kyle actually threw the ball for a distance of 15/4 yards, while Lisa still threw the ball for 19/9 yards. How many yards did the ball travel in total now?

Qwen2-Math 72B:

 Reasoning: Kyle threw the ball for a distance of 15/4 = <<15/4=3.75>>3.75 yards.
Lisa threw the ball for a distance of 19/9 = <<19/9=2.111111111111111>>2.111111111111111 yards.
The ball traveled a total of 3.75+2.111111111111111 = <<3.75+2.111111111111111=5.861111111111111>>5.861111111111111 yards.
### Answer: 5.861111111111111

Expected answer: 211/36

Correct? Yes

Problem: In addition to the bananas given to Johnnie the monkey in the morning and afternoon, the zookeeper decides to give Johnnie the same total amount of bananas the next day. How many bananas in total did Johnnie the monkey receive over these two days?

Qwen2-Math 72B:

 Reasoning: Johnnie received 24/7 bananas on the first day.
On the second day, Johnnie received the same amount, which is also 24/7 bananas.
Therefore, Johnnie received a total of 24/7 + 24/7 = <<24/7+24/7=48/7>>48/7 bananas over the two days.
### Answer: 48/7

Expected answer: 48/7

Correct? Yes

Problem: Suppose instead of travelling 4 miles initially, Billy actually traveled 6 miles before rolling back 17/20 of a mile to avoid the pothole. With this change in initial distance, how many miles did Billy ultimately travel forward?

Qwen2-Math 72B:

 Reasoning: Billy rolled back 17/20 of a mile, so he traveled forward 6 - 17/20 = <<6-17/20=5.15>>5.15 miles.
### Answer: 5.15

Expected answer: 103/20

Correct? Yes

Problem: In a change of earlier data, it was discovered that the rocket only traveled 15/3 light years on the first day. However, on the second day, it actually covered 18/7 light years instead of 2/7 light years. With this new information, how far has the rocket traveled in total?

Qwen2-Math 72B:

 Reasoning: The rocket traveled 15/3 + 18/7 = <<15/3+18/7=61/7>>61/7 light years in total.
### Answer: 61/7

Expected answer: 53/7

Correct? No

Problem: Jacob now wants to double the total length of the rope he has. If Jacob starts with the rope that is ((15 / 12) + (17 / 17)) meters long, and then adds another piece of the same length to it, what will be the total length of the combined ropes?

Qwen2-Math 72B:

 Reasoning: Jacob starts with a rope that is ((15 / 12) + (17 / 17)) = <<((15 / 12) + (17 / 17))=2.25>>2.25 meters long.
Jacob adds another piece of the same length to it, so the total length of the combined ropes is 2.25+2.25 = <<2.25+2.25=4.5>>4.5 meters.
### Answer: 4.5

Expected answer: 9/2

Correct? Yes

Problem: Now, suppose Jacob starts with a piece of rope that is 7/12 meters long instead of 15/12 meters. If he still adds the rope piece that is 17/17 meters long to it, what is the total length of the rope now?

Qwen2-Math 72B:

 Reasoning: Jacob starts with a piece of rope that is 7/12 = <<7/12=0.5833333333333334>>0.5833333333333334 meters long.
Jacob adds a piece that is 17/17 = <<17/17=1>>1 meter long to it.
The total length of the rope is 0.5833333333333334+1 = <<0.5833333333333334+1=1.5833333333333334>>1.5833333333333334 meters.
### Answer: 1.5833333333333334

Expected answer: 19/12

Correct? No

Problem: Mark already had 1 full pair of pants and another 13/4 pairs from various incomplete sets. If he goes shopping and buys an additional 13/3 pairs of pants, how many pairs of pants does he have in total now?

Qwen2-Math 72B:

 Reasoning: 13/4 pairs is equivalent to 3 1/4 pairs.
Mark had 1+3 1/4 = <<1+3.25=4.25>>4 1/4 pairs of pants before shopping.
13/3 pairs is equivalent to 4 1/3 pairs.
Mark bought 4 1/3 pairs of pants.
Mark has 4 1/4+4 1/3 = <<4.25+4.333333333333333=8.583333333333334>>8 5/12 pairs of pants in total now.
### Answer: 8 5/12

Expected answer: 103/12

Correct? No

Problem: Instead of starting with 1 full pair of pants and another 13/4 pairs, let's say Mark originally had 1 full pair of pants and 9/4 pairs from various incomplete sets. How many pairs of pants would he then have in total?

Qwen2-Math 72B:

 Reasoning: 9/4 pairs is equivalent to 2 1/4 pairs.
Mark has 1+2 1/4 = <<1+2.25=3.25>>3 1/4 pairs of pants in total.
### Answer: 3 1/4

Expected answer: 13/4

Correct? Yes

Problem: After the logs and twigs burn out, Tom adds 8 pieces of charcoal, each of which burns for 1/5 of an hour. What is the total burn time for the logs, twigs, and charcoal?

Qwen2-Math 72B:

 Reasoning: The charcoal burns for 8*1/5 = <<8*1/5=8/5>>8/5 of an hour.
The total burn time is 19/12 + 8/5 = <<19/12+8/5=133/60>>133/60 of an hour.
### Answer: 133/60

Expected answer: 191/60

Correct? No

Problem: In the original scenario, consider that there were actually 4 logs that each burn for 1/16 of an hour, but Tom only added 15 twigs that each burn for 1/12 of an hour instead. What would be the total burn time for the logs and twigs in this situation?

Qwen2-Math 72B:

 Reasoning: The logs burn for 4*1/16 = <<4*1/16=1/4>>1/4 of an hour.
The twigs burn for 15*1/12 = <<15*1/12=5/4>>5/4 of an hour.
The total burn time is 1/4 + 5/4 = <<1/4+5/4=3/2>>3/2 of an hour.
### Answer: 3/2

Expected answer: 3/2

Correct? Yes

Problem: Now assume the farmer's first apple tree produced 10/9 bushels instead of 10/11 bushels in the morning. With the same amount of apples in the afternoon, how many bushels of apples did this tree produce in one day?

Qwen2-Math 72B:

 Reasoning: The total bushels of apples produced is 10/9 + 17/11 = <<10/9+17/11=277/99>>277/99 bushels.
### Answer: 277/99

Expected answer: 263/99

Correct? No

Problem: If Sam's binoculars instead increased his range of sight by 4/14 of a mile, while Joe's binoculars still increase his range of sight by exactly 1 mile, how much would their combined range increase be when they use the binoculars one after the other?

Qwen2-Math 72B:

 Reasoning: Sam's binoculars increase his range of sight by 4/14 of a mile.
Joe's binoculars increase his range of sight by 1 mile (16/16).
To find the combined range increase, we need to add the two fractions: 4/14 + 16/16.
First, we need to find a common denominator for the two fractions. The least common multiple of 14 and 16 is 112.
Next, we convert the two fractions to have the common denominator of 112: 4/14 = 32/112 and 16/16 = 112/112.
Now, we can add the two fractions: 32/112 + 112/112 = 144/112.
Finally, we simplify the fraction: 144/112 = 1 32/112 = 1 4/14.
### Answer: 1 4/14

Expected answer: 9/7

Correct? Yes

Problem: After mixing the sky blue paint, the painter decides to mix in an additional 12/13 liters of white paint to lighten the shade. How many liters of paint does he have now in total?

Qwen2-Math 72B:

 Reasoning: The painter now has 155/88 + 12/13 = <<155/88+12/13=2015/1144+1056/1144=3071/1144>>3071/1144 liters of paint in total.
### Answer: 3071/1144

Expected answer: 3071/1144

Correct? Yes

Problem: Initially, the painter used 18/11 liters of blue paint and 1/8 liters of white paint. However, what if instead of using 1/8 liters, he used 1/7 liters of white paint to mix the sky blue shade. How many liters of paint would he then have used in total?

Qwen2-Math 72B:

 Reasoning: The painter used 18/11 + 1/7 = <<18/11+1/7=126/77+11/77=137/77>>137/77 liters of paint in total.
### Answer: 137/77

Expected answer: 137/77

Correct? Yes

Problem: Instead of 12/13 of a pound of magical gold coins in the pot, let's say the pot originally contained 17/13 pounds of gold coins. If we still added this amount to the 3/2 pounds of gold coins in the other pot, how many pounds of gold coins are there now?

Qwen2-Math 72B:

 Reasoning: To find the total amount of gold, we need to add the amount of gold in the first pot to the amount of gold in the second pot.
The first pot contains 17/13 pounds of gold coins.
The second pot contains 3/2 pounds of gold coins.
To add these fractions, we need to find a common denominator. The least common multiple of 13 and 2 is 26.
So, we convert 17/13 to 34/26 and 3/2 to 39/26.
Now, we can add the fractions: 34/26 + 39/26 = 73/26.
Therefore, the total amount of gold is 73/26 pounds.
### Answer: 73/26 pounds

Expected answer: 73/26

Correct? Yes

Problem: After watching the original message, Zorg receives an additional part of the message from earth that is 7/8 hours long. What is the total length of all parts of the message that Zorg received?

Qwen2-Math 72B:

 Reasoning: The total length of the original message is 12 hours.
The additional part of the message is 7/8 hours long.
The total length of all parts of the message is 12+7/8 = <<12+7/8=12.875>>12.875 hours.
### Answer: 12.875

Expected answer: 103/8

Correct? Yes

Problem: Upon reviewing the original video message, Zorg realised that the first part of the video message was actually half as long at 4/8 hours instead of 4/4. The second part of the message still remains 11/1 hour long. What is the total length of the message that Zorg has corrected?

Qwen2-Math 72B:

 Reasoning: The first part of the message is 4/8 = <<4/8=0.5>>0.5 hour long.
The second part of the message is 11/1 = <<11/1=11>>11 hours long.
The total length of the message is 0.5+11 = <<0.5+11=11.5>>11.5 hours.
### Answer: 11.5

Expected answer: 23/2

Correct? Yes

Problem: In the same park, there are also 12 frogs evenly distributed across 4 lily pads, meaning each lily pad has 12 / 4 frogs. Adding this to the previous sum of animals in each pond and pool, what is the total number of animals on each respective setting (pond, pool, lily pad) combined?

Qwen2-Math 72B:

 Reasoning: Each lily pad has 12 / 4 = <<12/4=3>>3 frogs.
Adding this to the previous sum of animals in each pond and pool, we have 2.8 + 3 = <<2.8+3=5.8>>5.8 animals on each respective setting (pond, pool, lily pad) combined.
### Answer: 5.8

Expected answer: 29/5

Correct? Yes

Problem: In the same park, let's consider a different scenario. Suppose there are now only 2 ducks equally distributed among 5 ponds. Given that each pond will have 2 / 5 ducks and each small pool still has 8 / 10 fish, what will be the total number of animals across each pond and pool in this new scenario?

Qwen2-Math 72B:

 Reasoning: Each pond will have 2 / 5 = <<2/5=0.4>>0.4 ducks.
Each pool still has 8 / 10 = <<8/10=0.8>>0.8 fish.
Therefore, each pond and pool combined will have 0.4 + 0.8 = <<0.4+0.8=1.2>>1.2 animals.
### Answer: 1.2

Expected answer: 6/5

Correct? Yes

Problem: On a map, 15/11 inches represents the distance between Town A and Town B, and 16/15 inches represents the distance between Town B and Town C. After measuring the distance between Town C and Town D, which was found to be 20/5 inches. What is the total distance between Town A and Town D on the map?

Qwen2-Math 72B:

 Reasoning: The total distance between Town A and Town D on the map is the sum of the distances between Town A and Town B, between Town B and Town C, and between Town C and Town D.
The distance between Town A and Town B is 15/11 inches.
The distance between Town B and Town C is 16/15 inches.
The distance between Town C and Town D is 20/5 inches.
The total distance between Town A and Town D on the map is 15/11 + 16/15 + 20/5 inches.
To add these fractions, we need a common denominator. The least common multiple of 11, 15, and 5 is 165.
So, 15/11 = (15 * 15) / (11 * 15) = 225/165, 16/15 = (16 * 11) / (15 * 11) = 176/165, and 20/5 = (20 * 33) / (5 * 33) = 660/165.
The total distance between Town A and Town D on the map is 225/165 + 176/165 + 660/165 = 1061/165 inches.
### Answer: 1061/165

Expected answer: 1061/165

Correct? Yes

Problem: On the same map, suppose the distance between Town A and Town B is not 15/11 but 15/11 inches, and 16/7 inches represents the distance between Town B and Town C. Now, what would be the total distance on the map from Town A to Town C?

Qwen2-Math 72B:

 Reasoning: The total distance between Town A and Town C on the map is the sum of the distances between Town A and Town B and between Town B and Town C.
The distance between Town A and Town B is 15/11 inches.
The distance between Town B and Town C is 16/7 inches.
The total distance between Town A and Town C on the map is 15/11 + 16/7 inches.
To add these fractions, we need a common denominator. The least common multiple of 11 and 7 is 77.
So, 15/11 = (15 * 7) / (11 * 7) = 105/77 and 16/7 = (16 * 11) / (7 * 11) = 176/77.
The total distance between Town A and Town C on the map is 105/77 + 176/77 = 281/77 inches.
### Answer: 281/77

Expected answer: 281/77

Correct? Yes

Problem: The soap company plans to double their daily production. After producing 13/18 tons of Soap A and 4/6 tons of Soap B each day, how many tons of soap will they produce in total when they double their production?

Qwen2-Math 72B:

 Reasoning: To find the total amount of soap produced each day when the production is doubled, we need to first find the total amount of soap produced each day and then double it.
The amount of Soap A produced is 13/18 tons, and the amount of Soap B produced is 4/6 tons.
To add these fractions, we need to find a common denominator. The least common multiple of 18 and 6 is 18.
So, we convert 4/6 to 12/18.
Now, we can add the fractions: 13/18 + 12/18 = 25/18.
Therefore, the soap company produces 25/18 tons of soap in total each day.
To double this amount, we multiply it by 2: 25/18 * 2 = 50/18.
Therefore, when the soap company doubles their production, they will produce 50/18 tons of soap in total each day.
### Answer: 50/18

Expected answer: 25/9

Correct? Yes

Problem: The soap company decided to increase the production of Soap A to 17/18 tons a day while maintaining the production of Soap B at 4/6 tons a day. How many tons of soap do they produce in total each day after this change?

Qwen2-Math 72B:

 Reasoning: To find the total amount of soap produced each day after the change, we need to add the new amount of Soap A and the amount of Soap B produced.
The new amount of Soap A produced is 17/18 tons, and the amount of Soap B produced is 4/6 tons.
To add these fractions, we need to find a common denominator. The least common multiple of 18 and 6 is 18.
So, we convert 4/6 to 12/18.
Now, we can add the fractions: 17/18 + 12/18 = 29/18.
Therefore, the soap company produces 29/18 tons of soap in total each day after the change.
### Answer: 29/18

Expected answer: 29/18

Correct? Yes

Problem: After tailoring, Mark finds the sleeping bag is still too long for his comfort. He decides to cut off another 13/14 feet from his sleeping bag. What is the length of his sleeping bag now?

Qwen2-Math 72B:

 Reasoning: The length of the sleeping bag after cutting off another 13/14 feet is 101/12 - 13/14 = <<101/12 - 13/14 = 1414/168 - 156/168 = 1258/168 = 629/84>>629/84 feet.
### Answer: 629/84

Expected answer: 629/84

Correct? Yes

Problem: Originally, Mark bought a sleeping bag that was 8/2 feet long instead of 19/2 feet. He still cuts off 13/12 feet from it to fit his height. How long is his sleeping bag now?

Qwen2-Math 72B:

 Reasoning: The length of the sleeping bag after cutting off 13/12 feet is 8/2 - 13/12 = <<8/2 - 13/12 = 48/12 - 13/12 = 35/12>>35/12 feet.
### Answer: 35/12

Expected answer: 35/12

Correct? Yes

Problem: After feeding the monkeys and giraffes, Mary then feeds the elephants 19/11 pounds of food. How many pounds of food in total has she given to the monkeys, giraffes, and elephants at the zoo?

Qwen2-Math 72B:

 Reasoning: Mary has given 1.5 pounds of food in total to the monkeys and giraffes.
Mary feeds the elephants 19/11 = <<19/11=1.7272727272727273>>1.7272727272727273 pounds of food.
Mary has given 1.5+1.7272727272727273 = <<1.5+1.7272727272727273=3.2272727272727273>>3.2272727272727273 pounds of food in total to the monkeys, giraffes, and elephants.
### Answer: 3.2272727272727273

Expected answer: 71/22

Correct? No

Problem: Originally, Mary had given the monkeys 6/20 pounds of food and the giraffes 18/15 pounds of food. But what if she had made a mistake, and she actually gave the giraffes 18/19 pounds of food instead? How many pounds of food in total did she then give to the monkeys and giraffes?

Qwen2-Math 72B:

 Reasoning: Mary feeds 6/20 = <<6/20=0.3>>0.3 pounds of food to the monkeys.
Mary feeds 18/19 = <<18/19=0.9473684210526315>>0.9473684210526315 pounds of food to the giraffes.
Mary has given 0.3+0.9473684210526315 = <<0.3+0.9473684210526315=1.2473684210526315>>1.2473684210526315 pounds of food in total to the monkeys and giraffes.
### Answer: 1.2473684210526315

Expected answer: 237/190

Correct? No

Problem: Instead of collecting a full set of 4/1 cards, what if Albert had only managed to find 4/11 of a full set? Adding this to his initial 20/17 full sets, how many full sets of cards does he have now?

Qwen2-Math 72B:

 Reasoning: Albert has 20/17 + 4/11 = <<20/17+4/11=268/187>>268/187 full sets of cards now.
### Answer: 268/187

Expected answer: 288/187

Correct? No

Problem: Continuing from the prior situation, the turtle then decided to move further, walking an additional 15/20 of a meter towards the same direction. How many meters is the turtle from its original starting point now?

Qwen2-Math 72B:

 Reasoning: The turtle is already 10 meters from its starting point.
It then moves an additional 15/20 = <<15/20=0.75>>0.75 meters in the same direction.
The turtle is now 10 + 0.75 = <<10+0.75=10.75>>10.75 meters from its original starting point.
### Answer: 10.75

Expected answer: 43/4

Correct? Yes

Problem: Reconsider the previous scenario, but this time the turtle returns 15/8 of a meter back towards the starting point instead of the former 15/15 meters. Now, how many meters is the turtle from its starting point after these two movements?

Qwen2-Math 72B:

 Reasoning: The turtle moves 11 meters northward.
Then, it moves 15/8 = <<15/8=1.875>>1.875 meters back towards the starting point.
The turtle is 11 - 1.875 = <<11-1.875=9.125>>9.125 meters from its starting point after these two movements.
### Answer: 9.125

Expected answer: 73/8

Correct? Yes

Problem: After Chloe eats a piece of the chocolate bar, she decides to add a piece from another chocolate bar. The piece from the other bar weighs 4/14 ounces. How many ounces of chocolate does Chloe have now?

Qwen2-Math 72B:

 Reasoning: Chloe has 1 + 4/14 = <<1+4/14=1.2857142857142858>>1.2857142857142858 ounces of chocolate now.
### Answer: 1.2857142857142858

Expected answer: 9/7

Correct? Yes

Problem: Originally, Chloe had a chocolate bar that was 19/2 ounces. What if instead, at the beginning she had eaten a piece that was 9/2 ounces? How many ounces of chocolate would she have had left then?

Qwen2-Math 72B:

 Reasoning: Chloe would have had 19/2 - 9/2 = <<19/2-9/2=5>>5 ounces of chocolate left.
### Answer: 5

Expected answer: 5

Correct? Yes

Problem: In addition to his already expanded backpack, Sammy finds another tiny side pocket that can carry an extra 3/8 of a liter. What is the total capacity of his backpack now?

Qwen2-Math 72B:

 Reasoning: The additional side pocket can hold 3/8 liters.
The total capacity of the backpack will be 38/9 + 3/8 = <<38/9+3/8=323/72>>323/72 liters.
### Answer: 323/72

Expected answer: 331/72

Correct? No

Problem: Instead of the backpack having an initial capacity of 20/9 liters, let's say it could only carry 20/12 liters. If Sammy still adds the same extra pouch of 8/4 liters, what would the total capacity of his backpack be then?

Qwen2-Math 72B:

 Reasoning: The additional pouch can hold 8/4 = <<8/4=2>>2 liters.
The total capacity of the backpack will be 20/12 + 2 = <<20/12+2=34/12>>34/12 liters.
### Answer: 34/12

Expected answer: 11/3

Correct? No

Problem: After counting his newly found gold, Captain Hook discovers an additional 7/11 ounces of gold hidden in a small pouch he had forgotten about. What is the total weight of all the gold Captain Hook now has?

Qwen2-Math 72B:

 Reasoning: Captain Hook has 19/8 + 7/11 = <<19/8+7/11=209/88+56/88=265/88>>265/88 ounces of gold.
### Answer: 265/88

Expected answer: 265/88

Correct? Yes

Problem: In the previous treasure hunt, Captain Hook had originally found 11/8 ounces of gold. But let's say, instead of finding 5/5 ounces of gold as he did in his latest treasure hunt, he discovered 7/5 ounces of gold. How much total weight of gold would Captain Hook have had then?

Qwen2-Math 72B:

 Reasoning: Captain Hook would have had 11/8 + 7/5 = <<11/8+7/5=55/40+56/40=111/40>>111/40 ounces of gold.
### Answer: 111/40

Expected answer: 111/40

Correct? Yes

Problem: Paul realizes that his chair is too tall, so he decides to shorten it by removing a 2 feet section. After doing so, what is the new height of the chair?

Qwen2-Math 72B:

 Reasoning: The original height of the chair was 55/14 feet.
Paul removes a 2 feet section from the chair.
The new height of the chair is 55/14 - 2 = <<55/14-2=55/14-28/14=27/14>>27/14 feet.
### Answer: 27/14

Expected answer: 27/14

Correct? Yes

Problem: John recalls that he actually only had 5/2 shirts to start with before he donated 12/9 of them to charity. How many shirts does he actually have left now?

Qwen2-Math 72B:

 Reasoning: John donated 12/9 = <<12/9=4/3>>4/3 shirts to charity.
John has 5/2 - 4/3 = <<5/2-4/3=7/6>>7/6 shirts left.
### Answer: 7/6

Expected answer: 7/6

Correct? Yes

Problem: If John watched a movie for 10/12 of an hour and then watched a TV show for another 3/8 of an hour, and then repeated this same routine, what is the total time he spent watching in hours?

Qwen2-Math 72B:

 Reasoning: The LCD of 12 and 8 is 24.
10/12 = <<10/12=5/6>>5/6 = <<5/6=20/24>>20/24
3/8 = <<3/8=9/24>>9/24
20/24 + 9/24 = <<20/24+9/24=29/24>>29/24
29/24 = <<29/24=1.2083333333333333>>1.2083333333333333
1.2083333333333333 * 2 = <<1.2083333333333333*2=2.4166666666666665>>2.4166666666666665
### Answer: 2.4166666666666665

Expected answer: 29/12

Correct? Yes

Problem: If instead of watching the movie for 10/12 of an hour, John watched it for 10/7 of an hour, and he still watched the TV show for another 3/8 of an hour, what is the total time he spent watching in hours?

Qwen2-Math 72B:

 Reasoning: The LCD of 7 and 8 is 56.
10/7 = <<10/7=80/56>>80/56
3/8 = <<3/8=21/56>>21/56
80/56 + 21/56 = <<80/56+21/56=101/56>>101/56
101/56 = <<101/56=1.8035714285714286>>1.8035714285714286
### Answer: 1.8035714285714286

Expected answer: 101/56

Correct? Yes

Problem: Mr. Carpenter has now taken on an additional task. After building the chair and doing the upholstery work, he spends another 8/19 hours to apply the varnish. Now, how long does the entire process of making a chair take?

Qwen2-Math 72B:

 Reasoning: To find the total time, we need to add the time taken to build the frame, the time taken to upholster it, and the time taken to apply the varnish.
Time taken to build the frame = 5/3 hours
Time taken to upholster = 10/13 hours
Time taken to apply the varnish = 8/19 hours
Total time = 5/3 + 10/13 + 8/19
To add these fractions, we need a common denominator. The least common multiple of 3, 13, and 19 is 741.
Converting all fractions to have a denominator of 741:
5/3 = (5 * 247) / (3 * 247) = 1235/741
10/13 = (10 * 57) / (13 * 57) = 570/741
8/19 = (8 * 39) / (19 * 39) = 312/741
Total time = 1235/741 + 570/741 + 312/741 = 2117/741
### Answer: 2117/741 hours

Expected answer: 2117/741

Correct? Yes

Problem: In another batch of chair production, Mr. Carpenter managed to reduce his timeframe. Now, he takes only 5/10 hours to build the frame of the chair while the time for upholstery stays the same as 10/13 hours. How long does it take for him to build a chair with the new time for frame construction?

Qwen2-Math 72B:

 Reasoning: To find the total time, we need to add the time taken to build the frame and the time taken to upholster it.
Time taken to build the frame = 5/10 hours
Time taken to upholster = 10/13 hours
Total time = 5/10 + 10/13
To add these fractions, we need a common denominator. The least common multiple of 10 and 13 is 130.
Converting both fractions to have a denominator of 130:
5/10 = (5 * 13) / (10 * 13) = 65/130
10/13 = (10 * 10) / (13 * 10) = 100/130
Total time = 65/130 + 100/130 = 165/130
### Answer: 165/130 hours

Expected answer: 33/26

Correct? Yes

Problem: Suppose instead of the large bead of 15/2 mm diameter, Anna adds a different large bead to her bracelet that is 5/2 mm in diameter. Now, what would be the total added diameter if she adds this new large bead and the small bead to her bracelet?

Qwen2-Math 72B:

 Reasoning: The diameter of the small bead is 8/4 = <<8/4=2>>2 mm.
The diameter of the new large bead is 5/2 = <<5/2=2.5>>2.5 mm.
The total added diameter is 2+2.5 = <<2+2.5=4.5>>4.5 mm.
### Answer: 4.5

Expected answer: 9/2

Correct? Yes

Problem: After further research, it is found that the forest actually had 5/3 acres of pine trees instead of 5/4 acres. Keeping the acreage of oak trees the same, what is the new total acreage of pine and oak trees in the forest?

Qwen2-Math 72B:

 Reasoning: The new total acreage of pine and oak trees in the forest is 5/3 + 1/2 = <<5/3+1/2=13/6>>13/6 acres.
### Answer: 13/6

Expected answer: 13/6

Correct? Yes

Problem: Continuing from the first scenario, suppose that this special tree only produces 10/4 kilograms of fruit in the summer, but it only loses 11/20 kilograms of its fruit in the winter, instead of 12/20 kilograms. What is now the difference in kilograms of the fruit from summer to winter?

Qwen2-Math 72B:

 Reasoning: The tree produces 10/4 = <<10/4=2.5>>2.5 kg of fruit in the summer.
The tree loses 11/20 = <<11/20=0.55>>0.55 kg of fruit in the winter.
The difference in kilograms of the fruit from summer to winter is 2.5 - 0.55 = <<2.5-0.55=1.95>>1.95 kg.
### Answer: 1.95

Expected answer: 39/20

Correct? Yes

Problem: After shortening her routine, the ballerina decides to add a new dance move that lasts 2/20 minutes. How long is her routine now with the new dance move?

Qwen2-Math 72B:

 Reasoning: The ballerina's routine is now 133/80 minutes long. Adding the new dance move, which lasts 2/20 minutes, we need to convert 2/20 to a fraction with a denominator of 80, which is 8/80. Adding 133/80 + 8/80 = 141/80 minutes.
### Answer: 141/80 minutes

Expected answer: 141/80

Correct? Yes

Problem: If originally the ballerina's routine lasted 13/5 minutes but she had only shortened it by 12/16 minutes instead of 15/16 minutes, how long would her routine be then?

Qwen2-Math 72B:

 Reasoning: The ballerina's routine is initially 13/5 minutes long. To shorten it by 12/16 minutes, we subtract 12/16 from 13/5. To do this, we need a common denominator, which is 80. Converting both fractions, we get 208/80 - 12/16 = 208/80 - 60/80 = 148/80 minutes, which simplifies to 37/20 minutes.
### Answer: 37/20 minutes

Expected answer: 37/20

Correct? Yes

Problem: Bob has 11/3 dollars in one pocket of his wallet and 12/19 dollars in another pocket. How much money does he have in total in his wallet? Now, if Bob duplicates the total amount of money in his wallet, how much money will he have then?

Qwen2-Math 72B:

 Reasoning: To find the total amount of money Bob has, we need to add the amounts in each pocket.
First, we need to find a common denominator for the fractions 11/3 and 12/19. The least common multiple of 3 and 19 is 57.
Converting the fractions to have a denominator of 57, we get:
11/3 = (11 * 19) / (3 * 19) = 209/57
12/19 = (12 * 3) / (19 * 3) = 36/57
Now, we can add the fractions:
209/57 + 36/57 = (209 + 36) / 57 = 245/57
Therefore, Bob has a total of 245/57 dollars in his wallet.
Now, if Bob duplicates the total amount of money in his wallet, he will have:
2 * (245/57) = 490/57 dollars.
### Answer: 490/57

Expected answer: 490/57

Correct? Yes

Problem: Suppose Bob starts out with 3/3 (or 1) dollar in one pocket of his wallet instead of 11/3 dollars, and still has 12/19 dollars in another pocket. Now how much money does he have in total in his wallet?

Qwen2-Math 72B:

 Reasoning: To find the total amount of money Bob has, we need to add the amounts in each pocket.
First, we need to find a common denominator for the fractions 3/3 and 12/19. The least common multiple of 3 and 19 is 57.
Converting the fractions to have a denominator of 57, we get:
3/3 = (3 * 19) / (3 * 19) = 57/57
12/19 = (12 * 3) / (19 * 3) = 36/57
Now, we can add the fractions:
57/57 + 36/57 = (57 + 36) / 57 = 93/57
Therefore, Bob has a total of 93/57 dollars in his wallet.
### Answer: 93/57

Expected answer: 31/19

Correct? Yes

Problem: Following the previous problem, if the shark decides to swim an additional 18/9 miles in the evening, what is the total distance the shark would have swam by the end of the day?

Qwen2-Math 72B:

 Reasoning: The shark swims 18/9 = <<18/9=2>>2 miles in the evening.
The shark swims 5/3+2 = <<5/3+2=11/3>>11/3 miles in total by the end of the day.
### Answer: 11/3

Expected answer: 11/3

Correct? Yes

Problem: Continuing from the last problem, suppose the shark had only swam 5/12 miles in the morning instead of 5/5 miles, but still swam the additional 12/18 miles in the afternoon. How many miles would the shark have swam in the morning and afternoon combined?

Qwen2-Math 72B:

 Reasoning: The shark swims 5/12 miles in the morning.
The shark swims 12/18 = <<12/18=2/3>>2/3 miles in the afternoon.
The shark swims 5/12+2/3 = <<5/12+2/3=13/12>>13/12 miles in total.
### Answer: 13/12

Expected answer: 13/12

Correct? Yes

Problem: After typing together, Jim and Mary decide to slow down their pace and reduce their combined typing speed by 2 words every 11 seconds. What becomes their new combined typing speed?

Qwen2-Math 72B:

 Reasoning: Jim and Mary's combined typing speed is 3.5 words per second.
They reduce their speed by 2 words every 11 seconds, which is equivalent to 2/11 words per second.
Their new combined typing speed is 3.5 - 2/11 = <<3.5-2/11=3.1818181818181817>>3.18 words per second.
### Answer: 3.18

Expected answer: 73/22

Correct? No

Problem: Following her performances, the clown took a total break of 13/5 hours. How many hours remained after her performance time is subtracted from her total break time?

Qwen2-Math 72B:

 Reasoning: The clown took a total break of 13/5 hours and performed for 53/40 hours.
To find the remaining time, we need to subtract the performance time from the total break time.
13/5 - 53/40 = <<13/5-53/40=104/40-53/40=51/40>>51/40 hours.
### Answer: 51/40

Expected answer: 51/40

Correct? Yes

Problem: In the previous problem, the clown juggled for 2/10 hours. Now suppose that she actually juggled for 9/10 hours instead. In this scenario, how many hours in total did she perform, if she still rode the unicycle for 9/8 hours?

Qwen2-Math 72B:

 Reasoning: The clown rode a unicycle for 9/8 hours and then juggled for 9/10 hours.
To find the total time she performed, we need to add these two times together.
9/8 + 9/10 = <<9/8+9/10=45/40+36/40=81/40>>81/40 hours.
### Answer: 81/40

Expected answer: 81/40

Correct? Yes

Problem: Following the last scenario, if Timmy later dumps another 16/3 kg of sand into the sandbox, how much sand does the sandbox contain now?

Qwen2-Math 72B:

 Reasoning: Timmy has 4 + 16/3 = <<4+16/3=28/3>>28/3 kg of sand in his sandbox.
### Answer: 28/3

Expected answer: 28/3

Correct? Yes

Problem: In the previous sandbox problem, if Timmy initially dumped 7/2 kg of sand into his sandbox but this time around, he decided to add only 1/20 kg of sand to it instead, how much sand is now in his sandbox?

Qwen2-Math 72B:

 Reasoning: Timmy has 7/2 + 1/20 = <<7/2+1/20=3.55>>3.55 kg of sand in his sandbox.
### Answer: 3.55

Expected answer: 71/20

Correct? Yes

Problem: Originally, the alien UFO traveled 3/10 of a mile, then an additional 16/20 of a mile. But what if, instead, the UFO initially only traveled 3/10 of a mile, before traveling an extra 16/18 of a mile. How much distance did UFO cover in total this time?

Qwen2-Math 72B:

 Reasoning: 3/10 + 16/18 = <<3/10+16/18=107/90>>107/90 miles
### Answer: 107/90

Expected answer: 107/90

Correct? Yes

Problem: In the chair manufacturing firm discussed before, let's say the first machine was originally making 17/14 of a chair every hour instead of 4/14. How many chairs are now made by both machines in one hour?

Qwen2-Math 72B:

 Reasoning: The first machine now makes 17/14 chairs per hour.
The second machine still makes 5 chairs per hour.
Together, the machines make 17/14 + 5 = <<17/14+5=87/14>>87/14 chairs per hour.
### Answer: 87/14

Expected answer: 87/14

Correct? Yes

Problem: After the blue shark and the hammerhead shark swam their respective distances, a great white shark joined them and swam 9/14 miles. What is the total combined distance that the three sharks swam together?

Qwen2-Math 72B:

 Reasoning: The blue shark and the hammerhead shark swam a total of 1.2 miles.
The great white shark swam 9/14 = <<9/14=0.6428571428571429>>0.6428571428571429 miles.
Together, the three sharks swam 1.2 + 0.6428571428571429 = <<1.2+0.6428571428571429=1.8428571428571428>>1.8428571428571428 miles.
### Answer: 1.8428571428571428

Expected answer: 129/70

Correct? Yes

Problem: In the previous situation, let's imagine that the blue shark actually swam for only 2/4 miles and the hammerhead shark swam for 7/11 miles. In this case, how many miles did the blue and hammerhead sharks swim in total?

Qwen2-Math 72B:

 Reasoning: The blue shark swam 2/4 = <<2/4=0.5>>0.5 miles.
The hammerhead shark swam 7/11 = <<7/11=0.6363636363636364>>0.6363636363636364 miles.
Together, they swam 0.5 + 0.6363636363636364 = <<0.5+0.6363636363636364=1.1363636363636365>>1.1363636363636365 miles.
### Answer: 1.1363636363636365

Expected answer: 25/22

Correct? Yes

Problem: After growing 1/10 of its height and an additional 12/6 feet because of the fertilizer, the tree was then pruned, reducing its height by 1 foot. How much is the tree's height now?

Qwen2-Math 72B:

 Reasoning: The tree grew 1/10 + 2 = <<1/10+2=2.1>>2.1 feet in height this month.
After being pruned, the tree's height is reduced by 1 foot.
The tree's height now is 2.1 - 1 = <<2.1-1=1.1>>1.1 feet.
### Answer: 1.1

Expected answer: 11/10

Correct? Yes

Problem: In the previous problem, the tree was initially growing 1/10 of its height every month. Imagine instead that the tree was growing 4/10 of its height every month. If the tree additionally grew 12/6 feet due to a fertilizer, how much did the tree grow in height this particular month?

Qwen2-Math 72B:

 Reasoning: The tree grows 4/10 of its height every month.
This month, it additionally has grown 12/6 = <<12/6=2>>2 feet due to a fertilizer being applied.
The tree has grown 4/10 + 2 = <<4/10+2=2.4>>2.4 feet in height this month.
### Answer: 2.4

Expected answer: 12/5

Correct? Yes

Problem: In addition to the wood Sarah and Alex collected, another camper named Kim contributed 11 pounds of firewood. How many pounds of firewood do they now have in total?

Qwen2-Math 72B:

 Reasoning: To find the total amount of firewood collected, we need to add the amounts collected by Sarah, Alex, and Kim.
Sarah collected 11/20 pounds of firewood.
Alex collected 9/7 pounds of firewood.
Kim contributed 11 pounds of firewood.
Adding these amounts, we get 11/20 + 9/7 + 11 = 257/140 + 11 = 257/140 + 1540/140 = 1797/140 pounds of firewood.
### Answer: 1797/140 pounds of firewood.

Expected answer: 1797/140

Correct? Yes

Problem: Instead of initially collecting 11/20 of a pound of firewood, let's say Sarah collected only half as much, so 10/20 of a pound of firewood, while Alex still collected 9/7 pounds. How many pounds of firewood did they collect in total now?

Qwen2-Math 72B:

 Reasoning: To find the total amount of firewood collected, we need to add the amounts collected by Sarah and Alex.
Sarah collected 10/20 pounds of firewood.
Alex collected 9/7 pounds of firewood.
To add these fractions, we need a common denominator. The least common multiple of 20 and 7 is 140.
Sarah collected (10/20) * (7/7) = 70/140 pounds of firewood.
Alex collected (9/7) * (20/20) = 180/140 pounds of firewood.
Adding these fractions, we get 70/140 + 180/140 = 250/140 pounds of firewood.
### Answer: 250/140 pounds of firewood.

Expected answer: 25/14

Correct? Yes

Problem: After traveling north for a total distance, Captain Bill decided to backtrack and went 20/17 miles south. How many miles away from the original starting point was Captain Bill located then?

Qwen2-Math 72B:

 Reasoning: Captain Bill was 23/5 - 20/17 = <<23/5-20/17=111/85>>111/85 miles away from the original starting point.
### Answer: 111/85

Expected answer: 291/85

Correct? No

Problem: After reconsidering the calculations, Captain Bill realized that he actually sailed 19/18 miles north initially instead of 19/5 miles before changing track and sailing an additional 12/15 miles north. In light of this new information, how many miles north did Captain Bill sail in total now?

Qwen2-Math 72B:

 Reasoning: Captain Bill sailed 19/18 + 12/15 = <<19/18+12/15=23/18>>23/18 miles north in total.
### Answer: 23/18

Expected answer: 167/90

Correct? No

Problem: After making their signature sandwich, the restaurant staff find that they have used 13/9 pounds more meat than they intended to. What is the weight of the meat that should have been in the sandwich?

Qwen2-Math 72B:

 Reasoning: The weight of the meat that should have been in the sandwich is 3.3676470588235294 - 13/9 = <<3.3676470588235294-13/9=2.0776470588235294>>2.0776470588235294 pounds.
### Answer: 2.0776470588235294

Expected answer: 1177/612

Correct? No

Problem: The restaurant realized that the amount of the turkey meat in their signature sandwich should have been 6/4 pounds, not 9/4 pounds. What should be the total weight of the ham and turkey meat in the sandwich, according to this correction?

Qwen2-Math 72B:

 Reasoning: The total weight of the ham and turkey meat in the sandwich, according to this correction, is 19/17 + 6/4 = <<19/17+6/4=1.1176470588235294+1.5=2.6176470588235294>>2.6176470588235294 pounds.
### Answer: 2.6176470588235294

Expected answer: 89/34

Correct? Yes

Problem: Mandy has 9/2 KGs of apples. She uses 1/4 KG of them to make apple pie. Afterwards, she finds 3/19 more KGs of apples in her refrigerator. How many KGs of apples does she have total now?

Qwen2-Math 72B:

 Reasoning: Mandy has 9/2 - 1/4 = <<9/2-1/4=4.25>>4.25 KGs of apples left after making apple pie.
Mandy has 4.25 + 3/19 = <<4.25+3/19=4.421052631578947>>4.421052631578947 KGs of apples total now.
### Answer: 4.421052631578947

Expected answer: 335/76

Correct? No

Problem: In the scenario where Mandy initially had 9/7 KGs of apples instead, and she still uses 1/4 KG of them for her apple pie, how many KGs of apples would she have left?

Qwen2-Math 72B:

 Reasoning: Mandy has 9/7 - 1/4 = <<9/7-1/4=0.9285714285714286>>0.9285714285714286 KGs of apples left.
### Answer: 0.9285714285714286

Expected answer: 29/28

Correct? No

Problem: After measuring his fields, the farmer also remembered that he has an additional patch of land where he grows vegetables, which is 17/11 of an acre. With this additional patch included, what is his total land area in acres?

Qwen2-Math 72B:

 Reasoning: The farmer's total land area is 4.916666666666667 + 17/11 = <<4.916666666666667+17/11=6.56060606060606>>6.56060606060606 acres.
### Answer: 6.56060606060606

Expected answer: 853/132

Correct? No

Problem: Earlier, the farmer had thought that one part of the wheat field was 4 acres. However, he later realized that it was actually 4/9 of an acre. Including the other part that was 11/12 of an acre, what is the corrected total land area in acres used to grow wheat now?

Qwen2-Math 72B:

 Reasoning: The corrected total land area used to grow wheat is 4/9 + 11/12 = <<4/9+11/12=0.4444444444444444+0.9166666666666666=1.3611111111111112>>1.3611111111111112 acres.
### Answer: 1.3611111111111112

Expected answer: 49/36

Correct? Yes

Problem: Jacob bought another 3/4 feet of wire in addition to what was left over from before. How much guitar wire does he have in total now?

Qwen2-Math 72B:

 Reasoning: Jacob has 83/16+3/4 = <<83/16+3/4=95/16>>95/16 feet of wire in total now.
### Answer: 95/16

Expected answer: 95/16

Correct? Yes

Problem: In the previous scenario, what if Jacob initially had 11/2 feet long guitar wire but ended up using 9/16 feet to fix the guitar instead of 5/16 feet. How much guitar wire would be remaining in that case?

Qwen2-Math 72B:

 Reasoning: The remaining wire is 11/2-9/16 = <<11/2-9/16=79/16>>79/16 feet.
### Answer: 79/16

Expected answer: 79/16

Correct? Yes

Problem: Maria initially had 12 candies and 20 stickers. Instead of having 6 friends, let's say she had 13 friends now. If she shares the candies among her 6 friends and the stickers among her 13 friends equally, how many candies and stickers does each friend get in total?

Qwen2-Math 72B:

 Reasoning: Maria gives each of her 6 friends 12/6 = <<12/6=2>>2 candies.
Maria gives each of her 13 friends 20/13 = <<20/13=1.5384615384615385>>1.5384615384615385 stickers.
Each friend gets 2 candies and 1.5384615384615385 stickers in total.
### Answer: 3.5384615384615385

Expected answer: 46/13

Correct? Yes

Problem: A fairy sprinkles 11/12 oz of magic dust over an enchanted field. She then sprinkles an additional 9/10 oz of dust over a grove of enchanted apple trees. Then, she flies over to a nearby forest where she originally had 14/3 oz of magic dust. After using the dust over the field and apple trees, how much magic dust is left for the fairy now?

Qwen2-Math 72B:

 Reasoning: First, we need to find the total amount of magic dust used by the fairy. We already know that she used 109/60 oz of magic dust over the field and apple trees.
Now, we need to subtract this amount from the original amount of magic dust she had, which was 14/3 oz.
To do this, we need to find a common denominator for the two fractions. The least common multiple of 60 and 3 is 60.
So, we convert 14/3 to 280/60.
Now, we can subtract the two fractions: 280/60 - 109/60 = <<280/60-109/60=171/60>>171/60.
To simplify this fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 3.
So, the fairy has 57/20 oz of magic dust left.
### Answer: 57/20 oz

Expected answer: 57/20

Correct? Yes

Problem: A fairy sprinkles 11/12 oz of magic dust over an enchanted field. She then planned to sprinkle an additional 9/10 oz of dust over a grove of enchanted apple trees, but she accidentally sprinkles more, as much as 16/10 oz. What is the total amount of magic dust the fairy used now?

Qwen2-Math 72B:

 Reasoning: To find the total amount of magic dust used, we need to add the three amounts together.
First, we need to find a common denominator for the three fractions. The least common multiple of 12, 10, and 10 is 60.
So, we convert 11/12 to 55/60, 9/10 to 54/60, and 16/10 to 96/60.
Now, we can add the three fractions: 55/60 + 54/60 + 96/60 = <<55/60+54/60+96/60=205/60>>205/60.
To simplify this fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 5.
So, the total amount of magic dust used is 41/12 oz.
### Answer: 41/12 oz

Expected answer: 151/60

Correct? No

Problem: After earning points from using 1 vowel and 1 consonant, you played another round and scored 19/7 points. What is your total score now?

Qwen2-Math 72B:

 Reasoning: The score for the second round is 19/7 = <<19/7=2.7142857142857144>>2.7142857142857144.
The total score is 1.0727272727272727+2.7142857142857144 = <<1.0727272727272727+2.7142857142857144=3.787012987012987>>3.787012987012987.
### Answer: 3.787012987012987

Expected answer: 1458/385

Correct? Yes

Problem: In the same word game, now we assign 5/11 of a point for each vowel used instead of 3/11, while the points given for each consonant remain 8/10. If you use 1 vowel and 1 consonant with this new scoring system, what will your score be?

Qwen2-Math 72B:

 Reasoning: The score for 1 vowel is 5/11 = <<5/11=0.45454545454545453>>0.45454545454545453.
The score for 1 consonant is 8/10 = <<8/10=0.8>>0.8.
The total score is 0.45454545454545453+0.8 = <<0.45454545454545453+0.8=1.2545454545454545>>1.2545454545454545.
### Answer: 1.2545454545454545

Expected answer: 69/55

Correct? No

Problem: You decide to add some more content to your letter. If you add 7/12 of a page, how many pages will the letter be then?

Qwen2-Math 72B:

 Reasoning: 22/9 + 7/12 = <<22/9+7/12=103/36>>103/36 pages.
### Answer: 103/36

Expected answer: 109/36

Correct? No

Problem: What if your original letter was slightly longer, being 15/1 pages in total, but you still removed the same 14/9 pages from it? How many pages would the letter be then?

Qwen2-Math 72B:

 Reasoning: 15/1 - 14/9 = <<15-14/9=121/9>>121/9 pages.
### Answer: 121/9

Expected answer: 121/9

Correct? Yes

Problem: After Anne and Bob painted parts of the table, Carl decided to add to their work by painting another 11/5 of the table. Now, how much of the table has been painted in total?

Qwen2-Math 72B:

 Reasoning: Anne and Bob painted 1 1/20 of the table.
Carl painted 11/5 of the table.
Together, they painted 1 1/20 + 11/5 = <<1 1/20+11/5=1 1/20+22/20=23/20>>23/20 of the table.
Since the table is a whole, we can simplify 23/20 to 1 3/20.
### Answer: 1 3/20

Expected answer: 13/4

Correct? No

Problem: Suppose before Carl added more paint, instead of Bob painting an additional 5/20 of the table, he could only manage to paint 5/14 of it. In this case, how much of the table has been painted by Anne and Bob together?

Qwen2-Math 72B:

 Reasoning: Anne painted 16/20 of the table, and Bob painted 5/14 of the table.
To find the total amount of the table painted, we need to find a common denominator for the fractions.
The least common multiple of 20 and 14 is 140.
So, Anne painted 16/20 = <<16*7/20*7=112/140>>112/140 of the table, and Bob painted 5/14 = <<5*10/14*10=50/140>>50/140 of the table.
Together, they painted 112/140 + 50/140 = <<112/140+50/140=162/140>>162/140 of the table.
Since the table is a whole, we can simplify 162/140 to 1 11/70.
### Answer: 1 11/70

Expected answer: 81/70

Correct? Yes

Problem: The grocery store owner also received an additional batch of yellow apples that weighs 17/15 kg. So, now how many total kilograms of apples are in the store?

Qwen2-Math 72B:

 Reasoning: 17/15 kg of yellow apples is equal to 1.1333333333333333 kg.
The total weight of apples in the store is 3.1+1.1333333333333333 = <<3.1+1.1333333333333333=4.233333333333333>>4.233333333333333 kg.
### Answer: 4.233333333333333

Expected answer: 127/30

Correct? Yes

Problem: Instead of having 17/20 kg of red apples as stated in the original problem, suppose the store only had 3/20 kg of red apples. Along with the 18/8 kg of green apples, how many kilograms of apples would be in the store now?

Qwen2-Math 72B:

 Reasoning: 3/20 kg of red apples is equal to 0.15 kg.
18/8 kg of green apples is equal to 2.25 kg.
The total weight of apples in the store is 0.15+2.25 = <<0.15+2.25=2.4>>2.4 kg.
### Answer: 2.4

Expected answer: 12/5

Correct? Yes

Problem: After the bus has traveled 12/20 of a mile, then an additional 12/12 of a mile, it makes the same journey back. How many miles did the bus travel in total?

Qwen2-Math 72B:

 Reasoning: 12/20 + 12/12 = <<12/20+12/12=1.6>>1.6 miles
1.6 * 2 = <<1.6*2=3.2>>3.2 miles
### Answer: 3.2

Expected answer: 16/5

Correct? Yes

Problem: If instead of driving 12/20 of a mile initially, the bus drives 14/20 of a mile in the beginning and then goes an additional 12/12 of a mile, how many miles did the bus travel in total?

Qwen2-Math 72B:

 Reasoning: 12/12 = 1 mile
14/20 + 12/12 = <<14/20+12/12=1.7>>1.7 miles
### Answer: 1.7

Expected answer: 17/10

Correct? Yes

Problem: Instead of hitting the first note every 17/17 seconds, the drummer decides to change the beat and hit it every 16/17 seconds. They're still hitting the second note every 2/12 seconds though. How long does it now take the drummer to hit the two beats together?

Qwen2-Math 72B:

 Reasoning: The first beat now takes 16/17 = <<16/17=16/17>>16/17 seconds.
The second beat still takes 2/12 = <<2/12=1/6>>1/6 seconds.
The total time for both beats is 16/17 + 1/6 = <<16/17+1/6=103/102>>103/102 seconds.
### Answer: 103/102

Expected answer: 113/102

Correct? No

Problem: Instead of the distance between the first two cities on the map being 13/10 inches, suppose it is 13/7 inches. Keeping the distance from the second city to the third city the same at 14/15 inches, what would then be the total distance between the three cities on the map?

Qwen2-Math 72B:

 Reasoning: The total distance between the three cities on the map would then be 13/7 + 14/15 = <<13/7 + 14/15 = 195/105 + 98/105 = 293/105>>293/105 inches.
### Answer: 293/105 inches

Expected answer: 293/105

Correct? Yes

Problem: After finishing the small art project, Scarlett decides to mix the remaining red paint with 5/7 gallons of yellow paint to make orange paint. How many gallons of paint does she have now?

Qwen2-Math 72B:

 Reasoning: Scarlett has 31/20 + 5/7 = <<31/20+5/7=217/140>>217/140 gallons of paint now.
### Answer: 217/140

Expected answer: 317/140

Correct? No

Problem: Suppose Scarlett initially had 16/10 gallons of red paint. But this time, she uses 2/20 of a gallon for her small project instead. How many gallons of red paint does she have left now?

Qwen2-Math 72B:

 Reasoning: Scarlett will have 16/10 - 2/20 = <<16/10-2/20=30/20>>30/20 gallons of red paint left.
### Answer: 30/20

Expected answer: 3/2

Correct? Yes

Problem: What if the plane initially flew at a lower altitude, only 2 miles high, before it descended 12/15 of a mile? What would be the altitude of the plane after this descent?

Qwen2-Math 72B:

 Reasoning: The plane descends by 12/15 = <<12/15=4/5>>4/5 of a mile.
The new altitude of the plane is 2 - 4/5 = <<2-4/5=6/5>>6/5 miles.
### Answer: 6/5

Expected answer: 6/5

Correct? Yes

Problem: If the colonial fireplace burns firewood at a combined rate of 4/15 cords per hour in the early morning and 14/15 cords per hour in the late night, and during the peak hours, it burns an additional 6/5 cords per hour. Then how much total firewood does the fireplace burn in one hour considering early morning, late night, and peak hours?

Qwen2-Math 72B:

 Reasoning: The total amount of firewood burned in one hour is the sum of the firewood burned in the early morning, the late night, and the peak hours.
The amount of firewood burned in the early morning is 4/15 cords per hour.
The amount of firewood burned in the late night is 14/15 cords per hour.
The amount of firewood burned during the peak hours is 6/5 cords per hour.
The total amount of firewood burned in one hour is 4/15 + 14/15 + 6/5 = <<4/15+14/15+6/5=24/15>>24/15 cords per hour.
### Answer: 24/15

Expected answer: 12/5

Correct? No

Problem: Continuing from the previous scenario, let's suppose the colonial fireplace in the early morning burns firewood at a slightly different rate of 8/15 cords per hour instead of 4/15 cords per hour. The late night burning rate remains the same at 14/15 cords per hour. Given these new circumstances, how much firewood does the fireplace burn in total in one hour split between the early morning and late night?

Qwen2-Math 72B:

 Reasoning: The total amount of firewood burned in one hour is the sum of the firewood burned in the early morning and the late night.
The amount of firewood burned in the early morning is 8/15 cords per hour.
The amount of firewood burned in the late night is 14/15 cords per hour.
The total amount of firewood burned in one hour is 8/15 + 14/15 = <<8/15+14/15=22/15>>22/15 cords per hour.
### Answer: 22/15

Expected answer: 22/15

Correct? Yes

Problem: After traveling ((9 / 20) + (11 / 7)) light years towards earth, the alien continues its journey for an additional 14/2 light years. How many light years in total has the alien traveled?

Qwen2-Math 72B:

 Reasoning: The alien traveled ((9 / 20) + (11 / 7)) + (14 / 2) = <<((9 / 20) + (11 / 7)) + (14 / 2)=1893/140>>1893/140 light years in total.
### Answer: 1893/140

Expected answer: 1263/140

Correct? No

Problem: The alien originally travelled ((9 / 20) + (11 / 7)) light years towards earth. But what if instead, the alien had only travelled (9 / 20) + (11 / 11) light years to start with? How many light years would the alien have travelled then?

Qwen2-Math 72B:

 Reasoning: The alien travelled (9 / 20) + (11 / 11) = <<(9 / 20) + (11 / 11)=29/20>>29/20 light years in total.
### Answer: 29/20

Expected answer: 29/20

Correct? Yes

Problem: After combining the cucumbers, Bob duplicates his total amount. Now, how many cucumbers does he have?

Qwen2-Math 72B:

 Reasoning: Bob has 36/7 cucumbers.
Bob duplicates his cucumbers, so he has 36/7 * 2 = <<36/7*2=72/7>>72/7 cucumbers.
### Answer: 72/7

Expected answer: 72/7

Correct? Yes

Problem: Suppose instead of 1/7 of a cucumber, Bob initially has 17/7 (or about 2 and a half) cucumbers before adding 10/2. How many cucumbers does he have now?

Qwen2-Math 72B:

 Reasoning: Bob has 17/7 cucumbers from the first cucumber.
Bob has 10/2 = <<10/2=5>>5 cucumbers from the second cucumber.
Bob has 17/7+5 = <<17/7+5=52/7>>52/7 cucumbers in total.
### Answer: 52/7

Expected answer: 52/7

Correct? Yes