Problem: Given the previous scenario but this time, let's say that the full-grown giraffe has a different height of 78 feet. If a baby giraffe is still 7.34 feet shorter than this full-grown giraffe, then how tall is the baby giraffe now?

InternLM-Math Base 20B:

  Reasoning: The height of the baby giraffe is 78-7.34 = <<78-7.34=70.66>>70.66 feet.
### Answer: 70.66

Expected answer: 70.6600000000000

Correct? Yes

Problem: The scooter store decided to multiply the discount you received, $j, by 15.1 to determine how much they'd donate to a local charity based on your purchase. How much money would be donated to the charity as a result of your scooter purchase?

InternLM-Math Base 20B:

  Reasoning: The amount donated to the charity is 15.1 * $j = <<15.1*20.6=312.66>>$312.66.
### Answer: $312.66

Expected answer: 311.060000000000

Correct? No

Problem: If instead the scooter was initially priced at $66 before being discounted to $39.3, how much would be the discount on the scooter?

InternLM-Math Base 20B:

  Reasoning: The difference between the original price and the sale price is the discount. The discount is $66 - $39.3 = <<66-39.3=26.7>>$26.7.
### Answer: $26.7

Expected answer: 26.7000000000000

Correct? Yes

Problem: In a different playground, there was also a slide that was initially 71.36 meters long. However, safety concerns caused the officials to reduce the length by 24 meters instead of the original 50.0 meters. What is the new length of this slide?

InternLM-Math Base 20B:

  Reasoning: The new length of the slide is 71.36 - 24 = <<71.36-24=47.36>>47.36 meters.
### Answer: 47.36

Expected answer: 47.3600000000000

Correct? Yes

Problem: The farmer decides to make tomato sauce and uses up 81.0 kg of the tomatoes he collected. After processing those tomatoes into sauce, how much weight in tomatoes does he have left?

InternLM-Math Base 20B:

  Reasoning: 143.89-81.0 = <<143.89-81.0=62.89>>62.89
### Answer: 62.89

Expected answer: 62.8900000000000

Correct? Yes

Problem: If instead of 57.19 kg, the farmer actually collected 84 kg of tomatoes on the second day, what would be the new total weight of tomatoes he collected over the two days?

InternLM-Math Base 20B:

  Reasoning: 86.7+84 = <<86.7+84=170.7>>170.7
### Answer: 170.7

Expected answer: 170.700000000000

Correct? Yes

Problem: Following his earlier sales, the farmer sold an additional 4.8 batches of tomatoes. Including the new sales, how many total batches of tomatoes has the farmer sold now?

InternLM-Math Base 20B:

  Reasoning: 38.6+4.8 = <<38.6+4.8=43.4>>43.4 batches of tomatoes were sold.
### Answer: 43.4

Expected answer: 43.4000000000000

Correct? Yes

Problem: After recalculating, it turns out the farmer had actually sold 83 kilograms of tomatoes, not 111.94 kilograms, still in batches of 2.9 kilograms each. How many batches of tomatoes did he actually sell?

InternLM-Math Base 20B:

  Reasoning: 83/2.9 = <<83/2.9=28.6>>28.6 batches of tomatoes were sold.
### Answer: 28.6

Expected answer: 28.6206896551724

Correct? Yes

Problem: Instead of the original price being $75.0, assume that the original price of the skirt was $49. How much would the skirt cost after the first discount of $37.5?

InternLM-Math Base 20B:

  Reasoning: The skirt's new price is $49 - $37.5 = <<49.0-37.5=11.5>>$11.5.
### Answer: $11.5

Expected answer: 11.5000000000000

Correct? Yes

Problem: In addition to the existing wheat and corn fields, the farm has a newly acquired barley field that spans 7.16 acres. What is the total acreage of wheat, corn, and barley fields altogether on the farm now?

InternLM-Math Base 20B:

  Reasoning: 35.9+11.5+7.16 = <<35.9+11.5+7.16=54.56>>54.56
### Answer: 54.56

Expected answer: 54.5600000000000

Correct? Yes

Problem: Suppose instead of the 35.9 acres of wheat in the original calculation, the farm actually has 93 acres of wheat fields. Taking the updated acreage of wheat fields into account and keeping the 11.5 acres of corn fields the same, what would then be the total acreage of wheat and corn fields on the farm?

InternLM-Math Base 20B:

  Reasoning: 93+11.5 = <<93+11.5=104.5>>104.5
### Answer: 104.5

Expected answer: 104.500000000000

Correct? Yes

Problem: Following the previous problem, after filling each bucket, 2.4 liters of water is lost due to leakage. What would be the revised amount of water in each bucket?

InternLM-Math Base 20B:

  Reasoning: 8.2-2.4 = <<8.2-2.4=5.8>>5.8
There are 5.8 liters of water in each bucket.
### Answer: 5.8

Expected answer: 5.80000000000000

Correct? Yes

Problem: Following the previous problem, suppose that instead of 65.9 buckets the water was divided into 33 buckets. How many liters of water would have been in each bucket under this new arrangement?

InternLM-Math Base 20B:

  Reasoning: 540.38/33 = <<540.38/33=16.4>>16.4
There are 16.4 liters of water in each bucket.
### Answer: 16.4

Expected answer: 16.3751515151515

Correct? Yes

Problem: After increasing its speed, the UFO maintained its new speed for 52.9 hours. How many miles did the UFO travel at its new speed?

InternLM-Math Base 20B:

  Reasoning: The UFO traveled 97.46*52.9 = <<97.46*52.9=5163.634>>5163.634 miles at its new speed.
### Answer: 5163.634

Expected answer: 5155.63400000000

Correct? No

Problem: Suppose that the UFO was originally moving at a speed of 94.83 mph, but then it increased its speed by 72 mph instead of 2.63 mph. What would its new speed be now?

InternLM-Math Base 20B:

  Reasoning: The UFO's new speed would be 94.83+72 = <<94.83+72=166.83>>166.83 miles per hour.
### Answer: 166.83

Expected answer: 166.830000000000

Correct? Yes

Problem: After swimming 95.34 kilometers to the east and another 41.6 kilometers in the same direction, the Penguin continued its journey and swam an additional 97.85 kilometers. What is the total distance that the Penguin swam?

InternLM-Math Base 20B:

  Reasoning: The Penguin swam 95.34+41.6+97.85 = <<95.34+41.6+97.85=234.79>>234.79 kilometers.
### Answer: 234.79

Expected answer: 234.790000000000

Correct? Yes

Problem: Suppose instead of swimming 95.34 kilometers to the east initially, the penguin only swam 12 kilometers before continuing for another 41.6 kilometers in the same direction. How much distance would the penguin have covered in this scenario?

InternLM-Math Base 20B:

  Reasoning: The penguin would have swum 12+41.6 = <<12+41.6=53.6>>53.6 kilometers.
### Answer: 53.6

Expected answer: 53.6000000000000

Correct? Yes

Problem: In the park, there is another identical slide next to the first one which also gets extended by the same 9.86 feet. After this extension, what is the total length of the two slides put together?

InternLM-Math Base 20B:

  Reasoning: 2*86.42+2*9.86 = <<2*86.42+2*9.86=192.56>>192.56
### Answer: 192.56

Expected answer: 192.560000000000

Correct? Yes

Problem: Suppose the park decided to make an alteration to the plan and instead wanted to extend the slide by 81 feet rather than 9.86 feet. How long would the slide be after this extension?

InternLM-Math Base 20B:

  Reasoning: 86.42+81 = <<86.42+81=167.42>>167.42
### Answer: 167.42

Expected answer: 167.420000000000

Correct? Yes

Problem: After buying the ice skates and skate guards, you decide to take advantage of a special offer that gives you a discount of $46.83. What would be your new total cost after this discount is applied?

InternLM-Math Base 20B:

  Reasoning: Your new total cost after the discount is applied is $136.26-$46.83 = <<136.26-46.83=89.43>>89.43.
### Answer: 89.43

Expected answer: 89.4300000000000

Correct? Yes

Problem: Suppose instead of the original pair of ice skates costing $81.66, you found a different pair on sale for only $25. If you still buy the skate guards for an additional $54.6, what would be the total cost of these discounted ice skates combined with the skate guards?

InternLM-Math Base 20B:

  Reasoning: The total cost of the discounted ice skates combined with the skate guards is $25+$54.6 = <<25+54.6=79.6>>79.6.
### Answer: 79.6

Expected answer: 79.6000000000000

Correct? Yes

Problem: If originally, the farmer had harvested 36.0 kilograms of strawberries and now instead, he sold 12 kilograms of strawberries. How many kilograms of strawberries does he have left now?

InternLM-Math Base 20B:

  Reasoning: The farmer has 36.0-12 = <<36.0-12=24>>24 kilograms of strawberries left.
### Answer: 24

Expected answer: 24.0000000000000

Correct? Yes

Problem: Consider the same scenario but with a different original cost. If the pillow originally cost $72 instead of $22.88 and its price is reduced to $14.2 after a sale, what would be the amount of the discount now?

InternLM-Math Base 20B:

  Reasoning: The difference between the original price and the discounted price is the amount of the discount.
72-14.2 = <<72-14.2=57.8>>57.8
### Answer: 57.8

Expected answer: 57.8000000000000

Correct? Yes

Problem: In the initial situation, the bunch of bananas weighed 79 grams instead of 77.8 grams. After removing the peel, which weighed 7.0 grams as before, what would be the new weight of the bananas this time?

InternLM-Math Base 20B:

  Reasoning: 79-7.0 = <<79-7.0=72>>72
### Answer: 72

Expected answer: 72.0000000000000

Correct? Yes

Problem: Following up on our previous question about the book with 15.18 words per page and 24.0 pages, if 45.7 words were removed from the book for editing, how many words does the book contain now?

InternLM-Math Base 20B:

  Reasoning: The book has 364.32-45.7 = <<364.32-45.7=318.62>>318.62 words in total.
### Answer: 318.62

Expected answer: 318.620000000000

Correct? Yes

Problem: Let's revisit the book situation again. What if, initially, each page in the book had 30 words instead of 15.18, and the book still had 24.0 pages. How many words would the book contain in that case?

InternLM-Math Base 20B:

  Reasoning: The book would have 30*24.0 = <<30*24.0=720>>720 words in total.
### Answer: 720

Expected answer: 720.000000000000

Correct? Yes

Problem: After the kids added sand to their sandcastle, they decided to make it even bigger. They multiplied the original weight of the sand by 46.6. What is the total weight of the sand in the sandcastle after their addition?

InternLM-Math Base 20B:

  Reasoning: 62.78*46.6 = <<62.78*46.6=2917.948>>2917.948
### Answer: 2917.948

Expected answer: 2925.54800000000

Correct? No

Problem: Suppose the retailer has recalculated his inventory and found that they have made an error in their initial calculation. In actuality, it only takes 27 meters of material to make one pair of socks instead of 98.6 meters. How many pairs of socks can they now make with the corrected amount of material?

InternLM-Math Base 20B:

  Reasoning: 5600.48/27 = <<5600.48/27=207.8>>207.8
The retailer can make 207.8 pairs of socks with the corrected amount of material.
### Answer: 207.8

Expected answer: 207.425185185185

Correct? Yes

Problem: As a follow-up to the previous problem, suppose that a space probe needs to use 50.03 kilograms of fuel for every million kilometers it travels towards the sun. If it starts its journey on the second day, how many kilograms of fuel will it need to reach the Sun?

InternLM-Math Base 20B:

  Reasoning: The space probe needs 71.39*50.03 = <<71.39*50.03=3577.0117>>3577.0117 kilograms of fuel to reach the Sun.
### Answer: 3577.0117

Expected answer: 3571.64170000000

Correct? No

Problem: As a follow up to the previous problem, suppose that instead of the initial distance being 56.14 million kilometers, the Earth was initially 60 million kilometers far from the Sun. After increasing by 15.25 million kilometers, what would be the distance from Earth to the Sun on the following day, in millions of kilometers?

InternLM-Math Base 20B:

  Reasoning: The distance to the Sun on the second day is 60+15.25 = <<60+15.25=75.25>>75.25 million kilometers.
### Answer: 75.25

Expected answer: 75.2500000000000

Correct? Yes

Problem: Suppose that for each degree drop in temperature, the cloud loses 97.29 liters of water vapor. How much water vapor would the cloud lose given the temperature change from the previous question?

InternLM-Math Base 20B:

  Reasoning: 35.43*97.29 = <<35.43*97.29=3450.8147>>3450.8147
### Answer: 3450.8147

Expected answer: 3446.98470000000

Correct? No

Problem: Suppose that instead of dropping from 92.5 degrees Fahrenheit, the temperature in the cloud layer had dropped from 87 degrees Fahrenheit. How much would the temperature change be in this situation?

InternLM-Math Base 20B:

 Reasoning: 87-57.07 = <<87-57.07=29.93>>29.93
### Answer: 29.93

Expected answer: 29.9300000000000

Correct? Yes

Problem: After the giraffe walks 42.53 miles to the north in the morning and 28.9 miles to the east in the afternoon, it then retraces 60.5 miles back towards its starting point. How many miles is the giraffe from its starting point now?

InternLM-Math Base 20B:

  Reasoning: The giraffe is 71.43-60.5 = <<71.43-60.5=10.93>>10.93 miles from its starting point.
### Answer: 10.93

Expected answer: 10.9300000000000

Correct? Yes

Problem: If the giraffe had walked 42.53 miles to the north in the morning and then 57 miles to the east in the afternoon instead of 28.9, how many miles in total would the giraffe have walked?

InternLM-Math Base 20B:

  Reasoning: The total distance the Giraffe walked is 42.53+57 = <<42.53+57=99.53>>99.53 miles.
### Answer: 99.53

Expected answer: 99.5300000000000

Correct? Yes

Problem: A train ticket costs $14.15 and the cost of a meal on the train is $95.07. After your meal, you decide to buy a cool souvenir train mug for $4.02. How much would it cost in total for the train ticket, meal, and souvenir mug?

InternLM-Math Base 20B:

  Reasoning: The total cost is $14.15+$95.07+$4.02 = <<14.15+95.07+4.02=113.24>>$113.24.
### Answer: $113.24

Expected answer: 113.240000000000

Correct? Yes

Problem: The train ticket originally costs $14.15, and in this follow-up question, we're considering what happens if the meal cost is changed to $59. Now, how much is the total cost for the ticket and the meal?

InternLM-Math Base 20B:

  Reasoning: The total cost is $14.15+$59 = <<14.15+59=73.15>>$73.15.
### Answer: $73.15

Expected answer: 73.1500000000000

Correct? Yes

Problem: Following the viewing of the galaxy, the scientist decides to amplify her binocular lens, multiplying the viewing scope by 5.21 times. What is the new approximate distance in light years that she can view through each scope after amplification?

InternLM-Math Base 20B:

  Reasoning: 50.6*5.21 = <<50.6*5.21=263.9>>263.9
### Answer: 263.9

Expected answer: 262.584000000000

Correct? No

Problem: Following her studies, the scientist decides to adjust her binoculars to view the galaxy through only 65 different scopes instead of the previous 85.3. If the total distance remains the same, 4299.12 light years, what is the new approximate distance in light years that she sees through each scope?

InternLM-Math Base 20B:

  Reasoning: 4299.12/65 = <<4299.12/65=66.3>>66.3
### Answer: 66.3

Expected answer: 66.1403076923077

Correct? Yes

Problem: Originally, the water company decided to divide the 1774.46 cubic meters of water between 28.9 storage tanks. However, it later decides to evenly distribute the water among 46 tanks instead. How many cubic meters of water will each tank contain under this new distribution plan?

InternLM-Math Base 20B:

  Reasoning: 1774.46/46 = <<1774.46/46=38.4>>38.4
Each tank will contain 38.4 cubic meters of water.
### Answer: 38.4

Expected answer: 38.5752173913043

Correct? Yes

Problem: After the modifications, the star's magnitude brightness is multiplied by a factor of 96.1 due to interference from the earth's atmosphere. What is the value of the star's final observed brightness after being adjusted for atmospheric interference?

InternLM-Math Base 20B:

  Reasoning: The star's final observed brightness is 101.16*96.1 = <<101.16*96.1=9725.976>>9725.976 units.
### Answer: 9725.976

Expected answer: 9721.47600000000

Correct? No

Problem: Suppose the star's observed magnitude brightness was modified by 54 units instead of 4.7 units, keeping its intrinsic brightness the same at 96.46 units. What would be the total magnitude brightness of the star then when these modifications are added?

InternLM-Math Base 20B:

  Reasoning: The total magnitude brightness of the star is 96.46+54 = <<96.46+54=150.46>>150.46 units.
### Answer: 150.46

Expected answer: 150.460000000000

Correct? Yes

Problem: If a smaller blanket has a width of 94.35 feet, what would be the total width in feet if you laid the large and small blankets side by side?

InternLM-Math Base 20B:

  Reasoning: The total width of the blankets is 3.8+94.35 = <<3.8+94.35=98.15>>98.15 feet.
### Answer: 98.15

Expected answer: 98.1500000000000

Correct? Yes

Problem: If the large blanket actually had a total area of 86 square feet instead of 175.56 while keeping the length identical at 46.2 feet, what would be the new width in feet?

InternLM-Math Base 20B:

  Reasoning: The width of the blanket is 86/46.2 = <<86/46.2=1.86>>1.86 feet.
### Answer: 1.86

Expected answer: 1.86147186147186

Correct? Yes

Problem: After successfully handling the small fire, the firefighter then moves on to control a medium fire, using an additional 46.61 liters of water. How many liters of water has the firefighter used in total?

InternLM-Math Base 20B:

  Reasoning: The firefighter used 123.63+46.61 = <<123.63+46.61=170.24>>170.24 liters of water in total.
### Answer: 170.24

Expected answer: 170.240000000000

Correct? Yes

Problem: Suppose the firefighter initially used 92 liters of water to extinguish the small fire instead of 37.7 liters, but still uses an additional 85.93 liters to cool down the area. How many liters of water has the firefighter used in total now?

InternLM-Math Base 20B:

  Reasoning: The firefighter used 92+85.93 = <<92+85.93=177.93>>177.93 liters of water in total.
### Answer: 177.93

Expected answer: 177.930000000000

Correct? Yes

Problem: Following the last snow season, the weather department is predicting that the upcoming season will have double the number of snowfalls as the previous season. If the previous season had a certain number of snowfalls, each measuring 38.3 inches for a total of 509.39 inches, how many snowfalls are expected in the upcoming winter season?

InternLM-Math Base 20B:

  Reasoning: 509.39/38.3 = <<509.39/38.3=13.2>>13.2
There were 13 snowfalls during the previous season.
### Answer: 13

Expected answer: 26.6000000000000

Correct? No

Problem: In retrospect, the Met Office revised their measurements for the previous winter season. Instead of a total of 509.39 inches of snowfall, there were only 83 inches, still with each snowfall measuring 38.3 inches. With that in mind, how many snowfalls were there truly last winter season?

InternLM-Math Base 20B:

  Reasoning: 83/38.3 = <<83/38.3=2.2>>2.2
There were 2 snowfalls during the season.
### Answer: 2

Expected answer: 2.16710182767624

Correct? Yes

Problem: Following from the previous question, let's consider a smaller clock that initially measures 6 cm in height. If it also has a base of 1.58 cm that is removed, what would be the height of the main body of this smaller clock without the base?

InternLM-Math Base 20B:

  Reasoning: The main body of the smaller clock is 6-1.58 = <<6-1.58=4.42>>4.42 cm tall.
### Answer: 4.42

Expected answer: 4.42000000000000

Correct? Yes

Problem: Following up from the previous hot chocolate problem: after adjusting your recipe, you realize that you actually need 8 ounces of cocoa powder per cup instead of the originally planned 3.91 ounces. If you are making 12.7 cups of hot chocolate, how much cocoa will you need now?

InternLM-Math Base 20B:

  Reasoning: 8 × 12.7 = <<8*12.7=101.6>>101.6
You will need 101.6 ounces of cocoa powder.
### Answer: 101.6

Expected answer: 101.600000000000

Correct? Yes

Problem: Following up on the previous problem, the toy car runs on a battery that drains power proportional to the change in its speed per second. The rate is 24.8 mAh for each cm/s change in speed. How much power would the battery have drained in 1 second due to the speed change?

InternLM-Math Base 20B:

  Reasoning: The change in the toy car's speed is 70.9 cm/s.
The battery drains 24.8 mAh for each cm/s change in speed.
The battery drains 24.8 * 70.9 = <<24.8*70.9=1788.32>>1788.32 mAh for the change in speed.
### Answer: 1788.32

Expected answer: 1758.32000000000

Correct? No

Problem: In the previous scenario, if the toy car's speed after 1 second was modified to 29 cm/s instead of 28.0 cm/s, what would be the new change in the toy car's speed?

InternLM-Math Base 20B:

  Reasoning: The toy car's speed changed from 98.9 cm/s to 29.0 cm/s.
The change in the toy car's speed is 98.9 - 29.0 = <<98.9-29.0=69.9>>69.9 cm/s.
### Answer: 69.9

Expected answer: 69.9000000000000

Correct? Yes

Problem: Following the previous question, the lion hunter then sees a third pride that has a total of 3.72 lions (including cubs as half). How many lions (taking cubs as half each) did the lion hunter count across all three prides?

InternLM-Math Base 20B:

  Reasoning: The lion hunter counted 25.5+29.9+3.72 = <<25.5+29.9+3.72=59.12>>59.12 lions in total.
### Answer: 59.12

Expected answer: 59.1200000000000

Correct? Yes

Problem: Following the previous story, what if the lion hunter actually saw 76 lions in the first pride instead of 25.5 that he initially counted? When he then counted the second pride with 29.9 lions (including cubs as half), how many lions (considering cubs as half each) did the lion hunter count in total?

InternLM-Math Base 20B:

  Reasoning: The lion hunter counted 76+29.9 = <<76+29.9=105.9>>105.9 lions in total.
### Answer: 105.9

Expected answer: 105.900000000000

Correct? Yes

Problem: A brown bear ate 50.1 pounds of fish in the morning. Then, it ate another 21 pounds in the afternoon instead of 65.7 pounds. How many total pounds of fish did the brown bear consume in one day now?

InternLM-Math Base 20B:

  Reasoning: The brown bear consumed 50.1+21 = <<50.1+21=71.1>>71.1 pounds of fish in one day.
### Answer: 71.1

Expected answer: 71.1000000000000

Correct? Yes

Problem: After calculating the flour used to make the cakes, suppose this amount is then mixed with 74.2 grams of sugar per kilogram of the flour. How many grams of sugar would be needed in total?

InternLM-Math Base 20B:

  Reasoning: The bakery used 10.47 kilograms of flour to make the cakes. Therefore, the amount of sugar needed is 10.47*74.2 = <<10.47*74.2=778.57>>778.57 grams.
### Answer: 778.57

Expected answer: 776.874000000000

Correct? No

Problem: Suppose the bakery initially started with 68 kilograms of flour instead of 62.06 kilograms. How many kilograms of flour would have been used to make the cakes then?

InternLM-Math Base 20B:

  Reasoning: The bakery would have used 68-51.59 = <<68-51.59=16.41>>16.41 kilograms of flour to make the cakes.
### Answer: 16.41

Expected answer: 16.4100000000000

Correct? Yes

Problem: After adding the value of the gold and the pearl, the mermaid decides to give $28.05 worth of her treasure to her best sea creature friend. How much is the value of the remaining treasures after giving away part of it?

InternLM-Math Base 20B:

  Reasoning: The value of the remaining treasures is 98.69-28.05 = <<98.69-28.05=70.64>>70.64.
### Answer: 70.64

Expected answer: 70.6400000000000

Correct? Yes

Problem: Suppose instead of the gold being worth $86.7, it was actually worth $71. Now how much is the total value of the treasure the mermaid found?

InternLM-Math Base 20B:

  Reasoning: The total worth of the treasure is 71+11.99 = <<71+11.99=82.99>>82.99.
### Answer: 82.99

Expected answer: 82.9900000000000

Correct? Yes

Problem: The astronaut originally was at a distance of 61 meters from the spaceship before he moved an additional 85.98 meters further out into space. What is the total distance the astronaut moved away from the spaceship given this new initial distance?

InternLM-Math Base 20B:

  Reasoning: The astronaut moved 61+85.98 = <<61+85.98=146.98>>146.98 meters away from the spaceship.
### Answer: 146.98

Expected answer: 146.980000000000

Correct? Yes

Problem: From the previous problem, we knew a city bus has a capacity of 68.0 passengers with 10.66 seats reserved for specific riders. Now, suppose there are an additional 73.24 available seats added to the bus after an upgrade. How many total seats are now available for general riders after the upgrade?

InternLM-Math Base 20B:

  Reasoning: 10.66+73.24 = <<10.66+73.24=83.9>>83.9 seats are available for general riders.
### Answer: 83.9

Expected answer: 130.580000000000

Correct? No

Problem: From the previous problem, a city bus was upgraded to have more seats for general riders. However, now consider that originally, the bus only had a capacity of 28 passengers instead of 68.0, with the same 10.66 seats reserved for specific riders. How many seats would then have been available for general riders before the upgrade?

InternLM-Math Base 20B:

  Reasoning: 28.0-10.66 = <<28.0-10.66=17.34>>17.34 seats were available for general riders before the upgrade.
### Answer: 17.34

Expected answer: 17.3400000000000

Correct? Yes

Problem: In the previous scenario, the ballerina danced her main routine 72.0 times. However, let's consider what would happen if she only performed it 25 times instead. How much total time would she spend if she only performed her main routine 25 times?

InternLM-Math Base 20B:

  Reasoning: 93.5 × 25 = <<93.5*25=2337.5>>2337.5 seconds.
### Answer: 2337.5 seconds

Expected answer: 2337.50000000000

Correct? Yes

Problem: The chocolate maker originally thought they had purchased 23.76 kg of milk chocolate, but realized later that they had actually bought 91 kg of milk chocolate. Given that they still purchased 49.8 kg of dark chocolate as initially thought, what would be the new total weight in kg of the chocolate purchased?

InternLM-Math Base 20B:

  Reasoning: 91+49.8 = <<91+49.8=140.8>>140.8
### Answer: 140.8

Expected answer: 140.800000000000

Correct? Yes

Problem: After the tuning slips, the frequency a piano plays notes at is further reduced by 39.2 Hz. What frequency does the piano play notes at now?

InternLM-Math Base 20B:

  Reasoning: The piano now plays notes at 52.9-39.2 = <<52.9-39.2=13.7>>13.7 Hz.
### Answer: 13.7

Expected answer: 13.7000000000000

Correct? Yes

Problem: Following up on the previous scenario, the toy factory also made an additional $74.2 from selling accessories for the toy cars on the same day. What is the total amount of money the factory made from selling both the toy cars and the accessories?

InternLM-Math Base 20B:

  Reasoning: The factory made $489.6 from selling toy cars and $74.2 from selling accessories. The total amount of money the factory made is $489.6 + $74.2 = <<489.6+74.2=563.8>>563.8 dollars.
### Answer: 563.8

Expected answer: 563.494000000000

Correct? Yes

Problem: Let's revisit the scenario at the toy factory. Instead of selling 10.2 toy cars, they actually sold 63 toy cars in a day, with each car still priced at $47.97. How much total revenue did the factory now make from selling toy cars?

InternLM-Math Base 20B:

  Reasoning: 63 cars were sold, each for $47.97. So 63 cars were sold for 63*47.97 = <<63*47.97=3030.11>>3030.11 dollars.
### Answer: 3030.11

Expected answer: 3022.11000000000

Correct? No

Problem: Following up from the previous question where the superhero could fly a certain distance in an hour, let's assume he has a battery pack that allows him to fly an additional 81.04 miles per hour. Adding this extra distance to the previously calculated hourly flight distance, how many miles can the superhero now travels in an hour with his improved abilities?

InternLM-Math Base 20B:

  Reasoning: The superhero can fly 35.0+81.04 = <<35.0+81.04=116.04>>116.04 miles in one hour.
### Answer: 116.04

Expected answer: 116.040000000000

Correct? Yes

Problem: Following the UFO's descent from an altitude of 72.18 km to 70.7 km, it then flew an additional distance calculated by multiplying the distance it descended by 37.2. What is this additional distance flown by the UFO?

InternLM-Math Base 20B:

  Reasoning: 1.48*37.2 = <<1.48*37.2=55.056>>55.056
The additional distance flown by the UFO is 55.056 km.
### Answer: 55.056

Expected answer: 55.0560000000000

Correct? Yes

Problem: Following the UFO's additional movement, suppose it originally descended from an altitude of 90 km instead of 72.18 km to the same final altitude of 70.7 km. How much would it have descended then?

InternLM-Math Base 20B:

  Reasoning: 90-70.7 = <<90-70.7=19.3>>19.3
The UFO would have descended by 19.3 km.
### Answer: 19.3

Expected answer: 19.3000000000000

Correct? Yes

Problem: If the region continued to receive the same amount of rain per day for the following 6.43 days, how many additional inches of rain would the region receive?

InternLM-Math Base 20B:

  Reasoning: 3.6*6.43 = <<3.6*6.43=23.148>>23.148 additional inches of rain would fall.
### Answer: 23.148

Expected answer: 23.1480000000000

Correct? Yes

Problem: If instead of 11.5 days, the region originally received the total of 41.4 inches of rain over a period of 21 days, how many inches of rain would the region have received each day?

InternLM-Math Base 20B:

  Reasoning: 41.4/21 = <<41.4/21=1.97>>1.97 inches of rain fell each day.
### Answer: 1.97

Expected answer: 1.97142857142857

Correct? Yes

Problem: If the optimal work rate for such robots is 43.9 units per hour, by how much was this robot's work rate short of the optimal rate?

InternLM-Math Base 20B:

  Reasoning: The robot's work rate was 43.9-16.6 = <<43.9-16.6=27.3>>27.3 units per hour short of the optimal rate.
### Answer: 27.3

Expected answer: 27.3000000000000

Correct? Yes

Problem: Suppose the robot completed the work over 76 hours instead of 84.5 hours, while still accomplishing 1402.7 units of work. What would then be the unit of work completed per hour in this case?

InternLM-Math Base 20B:

  Reasoning: The robot completed 1402.7/76 = <<1402.7/76=18.4>>18.4 units of work per hour.
### Answer: 18.4

Expected answer: 18.4565789473684

Correct? Yes

Problem: A competitor's toy car runs at a speed of 66.1 meters per minute. By how much is this faster than our toy car that runs at a speed determined by traveling a total distance of 5460.48 meters over 86.4 minutes of battery life?

InternLM-Math Base 20B:

  Reasoning: The competitor's toy car is 66.1-63.5 = <<66.1-63.5=2.6>>2.6 meters per minute faster.
### Answer: 2.6

Expected answer: 2.90000000000000

Correct? Yes

Problem: If the battery life of our toy car was instead 41 minutes while still covering the distance of 5460.48 meters, how would that alter the toy car's speed in meters per minute?

InternLM-Math Base 20B:

  Reasoning: The toy car's speed is 5460.48/41 = <<5460.48/41=132.9>>132.9 meters per minute.
### Answer: 132.9

Expected answer: 133.182439024390

Correct? No

Problem: The robot mentioned before that travels a total distance of 1149.72 meters over a time span of 15.6 minutes has an energy consumption rate of 25.58 joules per meter. How much total energy does the robot consume in joules during this span of time?

InternLM-Math Base 20B:

  Reasoning: The robot consumes 1149.72*25.58 = <<1149.72*25.58=29,383.5376>>29,383.5376 joules in total.
### Answer: 29,383.5376

Expected answer: 1885.24600000000

Correct? No

Problem: In the same scenario as before, let's consider if the robot traveled a lesser distance of 46 meters over the same time span of 15.6 minutes. How many meters would the robot travel per minute then?

InternLM-Math Base 20B:

  Reasoning: The robot travels 46/15.6 = <<46/15.6=2.9>>2.9 meters per minute.
### Answer: 2.9

Expected answer: 2.94871794871795

Correct? Yes

Problem: In light of a new sleeping bag design, the manufacturer has decided to adjust the material requirements. Now each sleeping bag will use 94 square feet of material instead of 51.2. Given the same total amount of material available as before, 2631.68 square feet, how many of these new sleeping bags can now be produced?

InternLM-Math Base 20B:

  Reasoning: The manufacturer can produce 2631.68/94 = <<2631.68/94=28>>28 sleeping bags with the available material.
### Answer: 28

Expected answer: 27.9965957446808

Correct? No

Problem: Continuing from the previous scenario, suppose the plane increases its speed again by 89.64 km/hr. What is the total speed of the plane now, considering all three speed rates?

InternLM-Math Base 20B:

  Reasoning: 144.44+89.64 = <<144.44+89.64=234.08>>234.08
### Answer: 234.08

Expected answer: 234.080000000000

Correct? Yes

Problem: In the original scenario, assume instead that the computer game consumes 2714.46 MBs of data over 4 hours of gameplay. In this case, how many MBs of data would the game consume per hour?

InternLM-Math Base 20B:

  Reasoning: 2714.46/4 = <<2714.46/4=678.6>>678.6 MBs of data per hour.
### Answer: 678.6

Expected answer: 678.615000000000

Correct? Yes

Problem: In a revision of the previous scenario, let's say the kite actually traveled 7102.88 meters over the course of 83 minutes instead of 86.2 minutes. In this case, how many meters would the kite have traveled per minute?

InternLM-Math Base 20B:

  Reasoning: 7102.88/83 = <<7102.88/83=85.5>>85.5
The kite traveled 85.5 meters per minute.
### Answer: 85.5

Expected answer: 85.5768674698795

Correct? Yes

Problem: Follow up question to previous problem: After finding out how many liters of juice each kilogram of oranges produces, if we have 4.8 kilograms of oranges, how many liters of juice can we make from them?

InternLM-Math Base 20B:

  Reasoning: 4.8*43.9 = <<4.8*43.9=211.72>>211.72
We can make 211.72 liters of juice from 4.8 kilograms of oranges.
### Answer: 211.72

Expected answer: 210.240000000000

Correct? No

Problem: Follow-up to the previous problem: Instead of using 18.1 kilograms of oranges, what if the juice stand used 38 kilograms of oranges to make 792.78 liters of juice? In this case, how many liters of juice would be produced by each kilogram of oranges?

InternLM-Math Base 20B:

  Reasoning: 792.78/38 = <<792.78/38=20.8>>20.8
Each kilogram of oranges would produce 20.8 liters of juice.
### Answer: 20.8

Expected answer: 20.8626315789474

Correct? Yes

Problem: In addition to the skateboard part and the sticker, you also plan on buying a helmet that costs $84.8. What would the total cost be for all three items?

InternLM-Math Base 20B:

  Reasoning: The total cost is $52.9+$95.23+$84.8 = <<52.9+95.23+84.8=242.93>>242.93.
### Answer: 242.93

Expected answer: 232.930000000000

Correct? No

Problem: Suppose instead of the skateboard part costing $52.9, it actually cost $33. How much would the total cost be if you purchased this skateboard part and the $95.23 sticker?

InternLM-Math Base 20B:

  Reasoning: The total cost is $33+$95.23 = <<33+95.23=128.23>>128.23.
### Answer: 128.23

Expected answer: 128.230000000000

Correct? Yes

Problem: In the same scenario, suppose instead that the elephant's original weight was 49 tons before it started losing weight and reached 42.8 tons. How much weight did the elephant lose in this case?

InternLM-Math Base 20B:

  Reasoning: The elephant lost 49-42.8 = <<49-42.8=6.2>>6.2 tons.
### Answer: 6.2

Expected answer: 6.20000000000000

Correct? Yes

Problem: After packaging together the two types of cheese, the food manufacturer realizes that the package needs to maintain a standard weight of 72.5 grams. How much does the manufacturer need to subtract from the total weight of the combination of the two cheeses to meet this standard?

InternLM-Math Base 20B:

  Reasoning: 184.3-72.5 = <<184.3-72.5=111.8>>111.8
### Answer: 111.8

Expected answer: 111.800000000000

Correct? Yes

Problem: Let's reconsider the problem by supposing that the weight of the first type of cheese was originally 15 grams instead of 92.6 grams. What would then be the total weight of the two types of cheese when packaged together?

InternLM-Math Base 20B:

  Reasoning: 15+91.7 = <<15+91.7=106.7>>106.7
### Answer: 106.7

Expected answer: 106.700000000000

Correct? Yes

Problem: On a farm, there are 77.66 acres of land. If 31.94 acres are used for growing corn, how many acres are not used for growing corn? If each of the unused acres can yield 6.5 bushels of wheat, how many bushels of wheat can be grown in total?

InternLM-Math Base 20B:

  Reasoning: 77.66 - 31.94 = <<77.66-31.94=45.72>>45.72
### Answer: 45.72

Expected answer: 297.180000000000

Correct? No

Problem: In the same farm scenario, suppose there had originally been 81 acres of land instead of 77.66. If still 31.94 acres are used for growing corn, how many acres would now be left unused?

InternLM-Math Base 20B:

  Reasoning: 81 - 31.94 = <<81-31.94=49.06>>49.06
### Answer: 49.06

Expected answer: 49.0600000000000

Correct? Yes

Problem: Continuing from the initial scenario, suppose the rocket initially only rose to an altitude of 43.2 km and needed to ascend an additional 15 km to deliver the satellite into its required position. What would be the rocket's altitude when it has ascended the additional distance?

InternLM-Math Base 20B:

  Reasoning: The rocket needs to ascend 15 km to reach the necessary position. Therefore, the rocket's altitude will be 43.2+15 = <<43.2+15=58.2>>58.2 km when it has ascended the additional distance.
### Answer: 58.2

Expected answer: 58.2000000000000

Correct? Yes

Problem: The skateboard travels a total of 3769.24 meters in 48.2 minutes. If it traveled at a constant speed, how many meters does it cover in one minute? After subtracting 58.8 meters from this distance, how far has the skateboard traveled?

InternLM-Math Base 20B:

  Reasoning: 3769.24/48.2 = <<3769.24/48.2=78.2>>78.2
The skateboard covers 78.2 meters in one minute.
### Answer: 78.2

Expected answer: 19.4000000000000

Correct? No

Problem: A skateboard travels a total of 63 meters in 48.2 minutes. If it traveled at a constant speed, how many meters does it cover in one minute?

InternLM-Math Base 20B:

  Reasoning: 63/48.2 = <<63/48.2=1.3>>1.3
The skateboard covers 1.3 meters in one minute.
### Answer: 1.3

Expected answer: 1.30705394190871

Correct? Yes

Problem: If each marshmallow in both bags was replicated the exact number of times equal to the total of marshmallows in the two bags, how many marshmallows would there then be in total?

InternLM-Math Base 20B:

  Reasoning: 14.0*20.9 = <<14.0*20.9=293.6>>293.6
### Answer: 293.6

Expected answer: 436.810000000000

Correct? No

Problem: If there were 90 marshmallows in the first bag instead of 14.0, and 6.9 marshmallows in the second bag, how many marshmallows would there be in total?

InternLM-Math Base 20B:

  Reasoning: 90+6.9 = <<90+6.9=96.9>>96.9
### Answer: 96.9

Expected answer: 96.9000000000000

Correct? Yes

Problem: After the Triceratops dinosaur, the heaviest dinosaur is a type that is 59.6 times the weight of the Triceratops. How heavy is this dinosaur?

InternLM-Math Base 20B:

  Reasoning: The heaviest dinosaur weighs 162.1*59.6 = <<162.1*59.6=9638.16>>9638.16 tons.
### Answer: 9638.16

Expected answer: 9661.16000000000

Correct? No

Problem: Suppose instead of the Triceratops being 69.3 tons heavier than the Stegosaurus, it was only 26 tons heavier. How heavy would the Triceratops be then?

InternLM-Math Base 20B:

  Reasoning: A Triceratops dinosaur weighs 92.8+26 = <<92.8+26=118.8>>118.8 tons.
### Answer: 118.8

Expected answer: 118.800000000000

Correct? Yes

Problem: After using the oxygen from the supplementary bottle, the firefighter finds another backup oxygen source that adds an additional 21.46 liters of oxygen. How much total oxygen does he now have for the mission, including the backup source?

InternLM-Math Base 20B:

  Reasoning: The firefighter has 119.4+21.46 = <<119.4+21.46=140.86>>140.86 liters of oxygen for this mission.
### Answer: 140.86

Expected answer: 140.860000000000

Correct? Yes

Problem: Originally, the firefighter's oxygen tank did not start with 97.3 liters, but with 36 liters of oxygen instead. After adding the 22.1 liters of oxygen from the supplementary bottle, how much total oxygen does he have for the mission at that point?

InternLM-Math Base 20B:

  Reasoning: The firefighter has 36+22.1 = <<36+22.1=58.1>>58.1 liters of oxygen for this mission.
### Answer: 58.1

Expected answer: 58.1000000000000

Correct? Yes

Problem: In the previous problem, the scooter initially traveled at a constant speed of 95.1 kilometers per hour. What if instead, the scooter initially traveled at a lower speed of 27 kilometers per hour. What will be the total distance traveled in 4.8 hours at this reduced speed?

InternLM-Math Base 20B:

  Reasoning: 27 × 4.8 = <<27*4.8=130.8>>130.8
The scooter will travel 130.8 kilometers.
### Answer: 130.8

Expected answer: 129.600000000000

Correct? No

Problem: The length of the chain on one side of a swing is 89.5 cm. The length of the chain on the other side of the swing is 70.77 cm, and there is an additional 3.5 cm chain used to attach the swing to the frame. What is the total length of the chain used for the swing, including the attachment?

InternLM-Math Base 20B:

  Reasoning: The total length of the chain is 89.5+70.77+3.5 = <<89.5+70.77+3.5=153.77>>153.77 cm.
### Answer: 153.77

Expected answer: 163.770000000000

Correct? No

Problem: Previously, the length of the chain on one side of a swing was 89.5 cm. However, after some adjustments, the new length of the chain on this side is now 13 cm. The length of the chain on the other side remains the same at 70.77 cm. What is the total length of the chain used for the swing now?

InternLM-Math Base 20B:

  Reasoning: The total length of the chain is 13+70.77 = <<13+70.77=83.77>>83.77 cm.
### Answer: 83.77

Expected answer: 83.7700000000000

Correct? Yes

Problem: The candy store manager realized that they had made an error in their original calculation, and the store was actually initially stocked with 30.83 pounds of chocolate candies but ended up with 15 pounds after some sales. How many pounds of candies were really sold?

InternLM-Math Base 20B:

  Reasoning: 30.83 - 15 = <<30.83-15=15.83>>15.83 pounds of candies were sold.
### Answer: 15.83

Expected answer: 15.8300000000000

Correct? Yes

Problem: Continuing from the previous question, you decide to sell the watch you bought for a profit, marking up the price you paid by 15.28 times. How much are you selling the watch for?

InternLM-Math Base 20B:

  Reasoning: You are selling the watch for $38.40 * 15.28 = <<38.4*15.28=585.632>>$585.63.
### Answer: 585.63

Expected answer: 586.752000000000

Correct? No

Problem: In the previous scenario, what if the discount on the watch was instead $36.00 instead of $19.96? How much would you have paid for the watch in this case?

InternLM-Math Base 20B:

  Reasoning: You paid $58.36 - $36.00 = <<58.36-36.00=22.36>>$22.36 for the watch.
### Answer: 22.36

Expected answer: 22.3600000000000

Correct? Yes

Problem: Reconsidering the two comets, suppose this time the first comet covered a distance of 61 million kilometers instead of 32.52 million kilometers, and the second traveled the same distance as before, i.e. 27.3 million kilometers. Combining the two, what is the total distance both comets traveled now?

InternLM-Math Base 20B:

  Reasoning: 61+27.3 = <<61+27.3=88.3>>88.3
### Answer: 88.3

Expected answer: 88.3000000000000

Correct? Yes

Problem: Suppose the wizard initially started with a different quantity of the first ingredient, having 51 units instead of 56.9 units. How many units would there be in total in the potion then?

InternLM-Math Base 20B:

  Reasoning: 51+59.42 = <<51+59.42=110.42>>110.42
### Answer: 110.42

Expected answer: 110.420000000000

Correct? Yes