Problem: In light of the original problem, suppose for each 1 meter of sliding, Isaac now gains 5 times the speed instead of 3 times the speed. In this case, how much speed does he gain after sliding 1 meter?

InternLM-Math Base 20B:

  Reasoning: Isaac gains 5 times the speed for each 1 meter of sliding.
Isaac gains 5 times the speed for each 1 meter of sliding.
### Answer: 5

Expected answer: 5

Correct? Yes

Problem: In a further experiment, a third rocket with the same mass of 8 is tested, but this time it is equipped with an improved fuel system that requires less fuel. Now, this rocket requires only 5 units of fuel per mass. How much fuel is this third rocket going to need?

InternLM-Math Base 20B:

  Reasoning: The mass of the rocket is 8, so the fuel needed for the rocket is 8*5 = <<8*5=40>>40.
### Answer: 40

Expected answer: 40

Correct? Yes

Problem: In addition to planting seeds in the plots, Theresa wanted to give 7 gift bags each containing as many seeds as the number she planted in each plot. How many seeds should Theresa set aside for the gift bags?

InternLM-Math Base 20B:

  Reasoning: If Theresa wanted to give 7 gift bags each containing as many seeds as the number she planted in each plot, then she should set aside 7*8 = <<7*8=56>>56 seeds for the gift bags.
### Answer: 56

Expected answer: 56

Correct? Yes

Problem: In a change of plans, Theresa divided the 80 seeds evenly between only 2 plots instead of 10. This meant each plot now received a different number of seeds than before. If she later planted an extra plot with the same amount of seeds as in each of the new plots, how many seeds did she use for the extra plot?

InternLM-Math Base 20B:

  Reasoning: If Theresa divided the 80 seeds evenly between only 2 plots instead of 10, then she planted 80/2 = <<80/2=40>>40 seeds in each plot.
If she later planted an extra plot with the same amount of seeds as in each of the new plots, then she used 40 seeds for the extra plot.
### Answer: 40

Expected answer: 40

Correct? Yes

Problem: Remembering his magical spell, Captain Blackbeard starts to second guess himself. What if his magical spell had only doubled the amount of treasure chests instead of tripling? How many treasure chests would be in the cave in that case, after the spell is cast?

InternLM-Math Base 20B:

  Reasoning: 11*2 = <<11*2=22>>22
### Answer: 22

Expected answer: 22

Correct? Yes

Problem: If instead the baby giraffe grows by 4 feet every year, how tall would the giraffe be after 7 years?

InternLM-Math Base 20B:

  Reasoning: 4 × 7 = <<4*7=28>>28
The giraffe would be 28 feet tall after 7 years.
### Answer: 28

Expected answer: 28

Correct? Yes

Problem: After examining his ropes, Pete figures out that he needs to use all of his rope for a project. If he combines all 4 pieces of his 8 foot long ropes, how many feet of rope will he have for his project?

InternLM-Math Base 20B:

  Reasoning: Pete has 4 pieces of rope. Each piece of rope is 8 feet long. So, Pete has 4*8 = <<4*8=32>>32 feet of rope in total.
### Answer: 32

Expected answer: 32

Correct? Yes

Problem: Upon rechecking his inventory, Pete realizes he actually has 12 pieces of rope not 4. If each of these 12 ropes are 8 feet each, how much total length of rope does Pete have now?

InternLM-Math Base 20B:

  Reasoning: Pete has 12 pieces of rope. Each piece of rope is 8 feet long. So, Pete has 12*8 = <<12*8=96>>96 feet of rope in total.
### Answer: 96

Expected answer: 96

Correct? Yes

Problem: Earlier, we learned the Marine Institute distributed 81 starfish equally across 9 tanks. But what if they had only 12 starfish instead? How many starfish would equal distribution across the 9 tanks result in then?

InternLM-Math Base 20B:

  Reasoning: 12/9 = <<12/9=1.3333333333333333>>1.3333333333333333
### Answer: 1.3333333333333333

Expected answer: 4/3

Correct? Yes

Problem: After finding toys in the sandbox for two days, Lisa decides to bury and later dig up the exact square of the number of toys she found on the second day. How many toys will Lisa bury and dig up if she follows this plan?

InternLM-Math Base 20B:

  Reasoning: Lisa will bury and dig up <<2^2=4>>4 toys.
### Answer: 4

Expected answer: 4

Correct? Yes

Problem: Instead of finding twice as many toys as the previous day, Lisa now finds ten times as many toys as the previous day. How many toys will she find on the second day following this new rule?

InternLM-Math Base 20B:

  Reasoning: Lisa found <<2^10=1024>>1024 toys on the second day.
### Answer: 1024

Expected answer: 10

Correct? No

Problem: Given that the robot processed information to match the system's speed, it needed to multiply this new processing speed by itself to calculate its efficiency rate. What is the robot's efficiency rate?

InternLM-Math Base 20B:

  Reasoning: 1*1 = <<1*1=1>>1.
The robot's efficiency rate is 1.
### Answer: 1

Expected answer: 1

Correct? Yes

Problem: If the robot's processing speed were slightly faster and instead was able to process information in 12 microseconds which still needed to be divided by 11 to match the system's speed, how quickly could the robot process information to match the system's speed then?

InternLM-Math Base 20B:

  Reasoning: 12/11 = <<12/11=1.0909090909090909>>1.0909090909090909.
The robot could process information in 1.0909090909090909 microsecond to match the system's speed.
### Answer: 1.0909090909090909

Expected answer: 12/11

Correct? Yes

Problem: Each of these tomatoes will produce 3 new seeds of its own. Using all the seeds from the tomatoes that Nina grows, how many new tomato plants could she potentially grow?

InternLM-Math Base 20B:

  Reasoning: Nina will produce 6*3 = <<6*3=18>>18 tomatoes.
### Answer: 18

Expected answer: 54

Correct? No

Problem: In the above problem, the alien learned another new word. If they transmit each word 11 times per minute and the alien can now transmit two words, how many transmissions will the alien be able to make after 1 minute?

InternLM-Math Base 20B:

  Reasoning: The alien can transmit 11 words per minute. After 1 minute, they will be able to transmit 11 words.
### Answer: 11

Expected answer: 22

Correct? No

Problem: In the previous problem, instead of transmitting each word 11 times per minute, suppose the alien can only transmit each word 3 times per minute. How many transmissions will the alien be able to make after 1 minute if they learn a new word?

InternLM-Math Base 20B:

  Reasoning: The alien can transmit 3 words per minute. After 1 minute, they will be able to transmit 3 words.
### Answer: 3

Expected answer: 3

Correct? Yes

Problem: If the unicorn maintains the same speed, how many miles would it travel in 3 hours?

InternLM-Math Base 20B:

  Reasoning: The unicorn travels 5*3 = <<5*3=15>>15 miles in 3 hours.
### Answer: 15

Expected answer: 15

Correct? Yes

Problem: If instead the unicorn traveled 35 miles in 2 hours, how many miles would the unicorn travel in one hour?

InternLM-Math Base 20B:

  Reasoning: The unicorn travels 35/2 = <<35/2=17.5>>17.5 miles in one hour.
### Answer: 17.5

Expected answer: 35/2

Correct? Yes

Problem: Following from the previous problem, suppose Spectra, another superhero, has 5 times the strength of Jetstream. What is the strength of Spectra?

InternLM-Math Base 20B:

  Reasoning: Jetstream has 15 times the strength of himself. Spectra has 5 times the strength of Jetstream. So, Spectra has 15 × 5 = <<15*5=75>>75 times the strength of himself.
### Answer: 75

Expected answer: 75

Correct? Yes

Problem: Jason had made one necklace requiring 6 beads, but now he wants to make a total of 3 necklaces. How many beads will he need in total to make these 3 necklaces?

InternLM-Math Base 20B:

  Reasoning: 1 necklace requires 6 beads.
6/1 = <<6/1=6>>6 beads are needed to make 1 necklace.
6*3 = <<6*3=18>>18 beads are needed to make 3 necklaces.
### Answer: 18

Expected answer: 18

Correct? Yes

Problem: In the original problem, Jason made one necklace using 6 beads. But what if he wanted to use a new design that requires 8 times as many beads as his original design? How many beads would Jason need to create this new necklace?

InternLM-Math Base 20B:

  Reasoning: 1 necklace requires 6 beads.
6/1 = <<6/1=6>>6 beads are needed to make 1 necklace.
### Answer: 6

Expected answer: 48

Correct? No

Problem: After successfully unlocking the puzzle box with the code obtained by multiplying 2 by 3, a new challenge appears on the screen inside the box. It tells you to multiply the previous code by 12 to get the final secret code to complete the puzzle. What is the final code?

InternLM-Math Base 20B:

  Reasoning: 6*12 = <<6*12=72>>72.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: After completing the previous challenge, another puzzle appears on the next screen inside the box. It instructs you to multiply 2 by 7 instead of 3. What number will you get by following this new instruction for the puzzle?

InternLM-Math Base 20B:

  Reasoning: 2*7 = <<2*7=14>>14.
### Answer: 14

Expected answer: 14

Correct? Yes

Problem: If Peter then puts all of the money from the one wallet into a piggy bank, how much money will be in the piggy bank?

InternLM-Math Base 20B:

  Reasoning: Peter has 54 dollars in the wallet. If he puts all the money from the one wallet into a piggy bank, he will have 54 dollars in the piggy bank.
### Answer: 54

Expected answer: 54

Correct? Yes

Problem: If Peter originally had 12 dollars in each of the 6 wallets instead of 9, how much money would be in the one wallet then?

InternLM-Math Base 20B:

  Reasoning: Peter has 6 wallets and in each wallet, he has 12 dollars. So, he has 6*12 = <<6*12=72>>72 dollars altogether.
If he stacks all the money in one wallet, he will have 72 dollars in that wallet.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: Now consider that the butterfly only flaps its wings 9 times in one second. How many times would it flap its wings in 4 seconds?

InternLM-Math Base 20B:

  Reasoning: 9 × 4 = <<9*4=36>>36.
The butterfly flaps its wings 36 times in 4 seconds.
### Answer: 36

Expected answer: 36

Correct? Yes

Problem: If the plane's speed was instead 11 miles per minute, how many miles would the plane cover in the same time period of 6 minutes?

InternLM-Math Base 20B:

  Reasoning: 11 × 6 = <<11*6=66>>66
The plane would cover 66 miles in 6 minutes.
### Answer: 66

Expected answer: 66

Correct? Yes

Problem: In a different scenario, Josephine originally had 9 pairs of ice skates instead of 27 that she wanted to distribute equally among the 9 stores. How many pairs of ice skates would each store receive in this case?

InternLM-Math Base 20B:

  Reasoning: Josephine will give 9/9 = <<9/9=1>>1 pair of ice skates to each store.
### Answer: 1

Expected answer: 1

Correct? Yes

Problem: Billy decides to double his collection of kites. After doubling the initial amount, how many kites will he have in total?

InternLM-Math Base 20B:

  Reasoning: Billy has 2 × 27 = <<2*27=54>>54 kites in total.
### Answer: 54

Expected answer: 54

Correct? Yes

Problem: Billy initially thought he had 9 sets of kites, but he actually had 11 sets. Given that each set contains 9 kites, how many kites does he really have in total?

InternLM-Math Base 20B:

  Reasoning: Billy has 11 × 9 = <<11*9=99>>99 kites in total.
### Answer: 99

Expected answer: 99

Correct? Yes

Problem: Continuing the fairy tale, the magical creature took the outcome from her previous spell and multiplied it by 5. What is the new number after this spell?

InternLM-Math Base 20B:

  Reasoning: 16*5 = <<16*5=80>>80
### Answer: 80

Expected answer: 80

Correct? Yes

Problem: Continuing the magical story, suppose the magical creature had initially started with the number 11 instead of 4, and had multiplied this number by 4. What would have been the outcome of the first spell in that case?

InternLM-Math Base 20B:

  Reasoning: 11*4 = <<11*4=44>>44
### Answer: 44

Expected answer: 44

Correct? Yes

Problem: At first, there were 12 rooms instead of 3 to distribute the 27 chairs into. How many chairs would each room have in this case?

InternLM-Math Base 20B:

  Reasoning: 27/12 = <<27/12=2.25>>2.25
Each room would have 2.25 chairs.
### Answer: 2.25

Expected answer: 9/4

Correct? Yes

Problem: Now, imagine if the store decided to stock only one toy car per shelf instead, distributing 8 toy cars uniformly over 8 shelves. How many toy cars would be there per shelf?

InternLM-Math Base 20B:

  Reasoning: 8/8 = <<8/8=1>>1 toy car is on each shelf.
### Answer: 1

Expected answer: 1

Correct? Yes

Problem: In addition to the 3 pizzas, the restaurant also wants to make a special pizza that takes the same amount of time as the 3 pizzas combined. How long will it take to make the special pizza?

InternLM-Math Base 20B:

  Reasoning: 3 full pizzas have 3*3 = <<3*3=9>>9 slices.
Each slice takes 6 minutes to cook.
Therefore, it will take 9*6 = <<9*6=54>>54 minutes to cook all the slices.
### Answer: 54

Expected answer: 54

Correct? Yes

Problem: The restaurant has revised its pizza cooking time. Now, each slice of pizza takes only 1 minute to cook but a full pizza still has 3 slices. Now, how long will it take to make 3 full pizzas?

InternLM-Math Base 20B:

  Reasoning: 3 full pizzas have 3*3 = <<3*3=9>>9 slices.
Each slice takes 1 minute to cook.
Therefore, it will take 9*1 = <<9*1=9>>9 minutes to cook all the slices.
### Answer: 9

Expected answer: 9

Correct? Yes

Problem: Considering the previous problem, what if Arthur only drove his car for 2 hours each day for that same 4-day period? How many total hours did Arthur spend driving his car over this period?

InternLM-Math Base 20B:

  Reasoning: Arthur drove his car 2*4 = <<2*4=8>>8 hours over this period.
### Answer: 8

Expected answer: 8

Correct? Yes

Problem: Following the previous problem, if you were given the task to make two blankets with the same size and design, how many yards of yarn would you need in total?

InternLM-Math Base 20B:

  Reasoning: You would need 30*2 = <<30*2=60>>60 yards of yarn.
### Answer: 60

Expected answer: 60

Correct? Yes

Problem: Following the initial problem, consider if each blanket requires six yards of yarn for just one section, but the new design demands that a blanket is made up of 12 such sections. How many yards of yarn would you now need to create one blanket with this new design?

InternLM-Math Base 20B:

  Reasoning: Each blanket requires 6*12 = <<6*12=72>>72 yards of yarn.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: Following the previous scenario, consider that the bus was actually traveling at a speed of 10 miles per hour, not 7. How many miles would the bus have travelled after 6 hours in this case?

InternLM-Math Base 20B:

  Reasoning: 10 × 6 = <<10*6=60>>60.
The bus would have travelled 60 miles after 6 hours.
### Answer: 60

Expected answer: 60

Correct? Yes

Problem: Superhero Starlight, after completing her special power-up training, can now increase her initial power level by three times. If her initial power level was the product of multiplying 3 by 5, what is her new power level after the power-up training?

InternLM-Math Base 20B:

  Reasoning: 3*5 = <<3*5=15>>15
3*15 = <<3*15=45>>45
### Answer: 45

Expected answer: 45

Correct? Yes

Problem: Considering that Superhero Starlight initially had a power level that was the product of multiplying 3 by 5, let's imagine if her initial power level was actually the product of multiplying 3 by 6. What would her initial power level be in this scenario?

InternLM-Math Base 20B:

  Reasoning: 3*6 = <<3*6=18>>18
### Answer: 18

Expected answer: 18

Correct? Yes

Problem: Now consider that Jessica has another backpack, this one has a side that measures 4 inches. What is the area of one side of this backpack, again assuming it is a perfect square?

InternLM-Math Base 20B:

  Reasoning: The area of a square is the length of one side squared.
The area of one side of Jessica's backpack is 4^2 = <<4^2=16>>16 square inches.
### Answer: 16

Expected answer: 16

Correct? Yes

Problem: Suppose Captain Sparrow had initially found 36 gold doubloons, but this time he wants to divide them equally among 7 crew members, not 3. How many gold doubloons will each crew member get now?

InternLM-Math Base 20B:

  Reasoning: 36/7 = <<36/7=5>>5
Each crew member will get 5 gold doubloons.
### Answer: 5

Expected answer: 36/7

Correct? No

Problem: In the previous problem, the robot factory increased its output to twice its original amount, making 10 robots each hour. If the factory were to then multiply this new hourly output by the same amount they were originally making each hour (5 robots), how many robots would the factory make every hour?

InternLM-Math Base 20B:

  Reasoning: The factory produces 10*5 = <<10*5=50>>50 robots each hour.
### Answer: 50

Expected answer: 50

Correct? Yes

Problem: In the previous problem, the robot factory was originally making 5 robots each hour. However, if the factory were to increase its output to six times the original amount instead, how many robots would it produce each hour?

InternLM-Math Base 20B:

  Reasoning: The factory produces 5*6 = <<5*6=30>>30 robots each hour.
### Answer: 30

Expected answer: 30

Correct? Yes

Problem: If one bike wheel still has a radius of 81/9 units, but now the height of a certain bike rack is only 6 times this radius, what is the height of the bike rack?

InternLM-Math Base 20B:

  Reasoning: The height of the bike rack is 6 × 81/9 = <<6*81/9=54>>54 units.
### Answer: 54

Expected answer: 54

Correct? Yes

Problem: Following on from the previous problem, if Jamie continued swinging at the same pace for 3 more minutes, how many times would he have swung in total over those extra minutes?

InternLM-Math Base 20B:

  Reasoning: Jamie swung 11 times in every minute and he spent 3 minutes swinging. So, he swung 11*3 = <<11*3=33>>33 times in total over those extra minutes.
### Answer: 33

Expected answer: 33

Correct? Yes

Problem: After his first session, Jamie took a break and then returned to the swing. This second time, he swung 6 times per minute for one minute. How many times did he swing in that minute?

InternLM-Math Base 20B:

  Reasoning: Jamie swung 6 times in every minute and he spent 1 minute swinging. So, he swung 6 times in total.
### Answer: 6

Expected answer: 6

Correct? Yes

Problem: Let's say instead of initially having 9 robot toys and buying the same amount again, James originally had 9 robot toys, but this time, he decided to buy 5 more sets of 9 toys each. How many robot toys does James have now?

InternLM-Math Base 20B:

  Reasoning: James has 9+5*9 = <<9+5*9=54>>54 robot toys now.
### Answer: 54

Expected answer: 54

Correct? Yes

Problem: The farmer from the previous problem sells each egg for 1 dollar. How much money does the farmer make in a week by selling all the eggs?

InternLM-Math Base 20B:

  Reasoning: 11 × 7 × 1 = <<11*7*1=77>>77
### Answer: 77

Expected answer: 77

Correct? Yes

Problem: In the previous problem, if the farmer only had 2 chickens instead of 11, and each chicken still lays 7 eggs in a week, how many eggs would the chickens lay in total in a week?

InternLM-Math Base 20B:

  Reasoning: 2 × 7 = <<2*7=14>>14
### Answer: 14

Expected answer: 14

Correct? Yes

Problem: The enchanted tree that has been growing in the forest doesn't stop there. Overnight, it grows half of its daily growth rate, which is 4 cm. If this overnight height is multiplied by 10, how long is the final growth of the tree over these 10 nights?

InternLM-Math Base 20B:

  Reasoning: The tree grows 4/2 = <<4/2=2>>2 centimeters each night.
### Answer: 20

Expected answer: 20

Correct? Yes

Problem: The enchanted tree that was previously growing at an overnight rate of 4 cm changes its growth pattern. Now, it starts to grow at a rate of 10 centimeters each day, but it still grows only half of this amount each night. How many centimeters does the tree now grow overnight?

InternLM-Math Base 20B:

  Reasoning: The tree grows 10/2 = <<10/2=5>>5 centimeters each night.
### Answer: 5

Expected answer: 5

Correct? Yes

Problem: After having divided the mice among the 11 cages, the scientist realized that she made a mistake and originally there were only 77 mice but some of them were counted twice. After correcting for this mistake, she realized that she actually only had 77 / 2 mice. How many mice did the scientist have originally?

InternLM-Math Base 20B:

  Reasoning: 77/2 = <<77/2=38.5>>38.5 mice were originally in the lab.
### Answer: 38.5

Expected answer: 77/2

Correct? Yes

Problem: Sally realizes that she made a mistake in her calculations for making earrings. Instead of needing 35 beads for a set of earrings, she only needs 35 beads for a display of five earrings. Therefore, she recalculates and finds that to make each earring she only needs to divide this new amount by five. If she wants to make just 2 earrings, how many beads will she need now?

InternLM-Math Base 20B:

  Reasoning: Sally needs 35/5 = <<35/5=7>>7 beads to make each earring.
Sally needs 2*7 = <<2*7=14>>14 beads to make 2 earrings.
### Answer: 14

Expected answer: 14

Correct? Yes

Problem: In a new scenario, Brenda has 9 apples but this time she only wants to divide them evenly among her 2 closest friends. Given this change, how many apples will each friend receive now?

InternLM-Math Base 20B:

  Reasoning: 9/2 = <<9/2=4.5>>4.5
Each friend will get 4.5 apples.
### Answer: 4.5

Expected answer: 9/2

Correct? Yes

Problem: After completing his daring stunt, the tightrope walker decides to challenge himself further and wants to perform the same stunt on a tightrope that's only 1 foot off the ground. Now, how far would he need to walk considering the tightrope is still five times the height above the ground?

InternLM-Math Base 20B:

  Reasoning: The tightrope is 5*1 = <<5*1=5>>5 feet above the ground.
The stuntman will need to walk 5 feet to complete his stunt.
### Answer: 5

Expected answer: 5

Correct? Yes

Problem: The length of one side of a square is equal to the result of dividing 30 by 6. The total area of the square is found by multiplying the length of one side by itself. Now, if we multiply this area by 3, what will be the new value?

InternLM-Math Base 20B:

  Reasoning: The length of one side of the square is 30/6 = <<30/6=5>>5.
The area of the square is 5*5 = <<5*5=25>>25.
The new value is 25*3 = <<25*3=75>>75.
### Answer: 75

Expected answer: 75

Correct? Yes

Problem: In the original problem, the length of one side of the square was found by dividing 30 by 6. Now consider a new scenario where the length of the square is instead found by dividing number 12 by 6. What would be the total area of this new square?

InternLM-Math Base 20B:

  Reasoning: The length of one side of the square is 12/6 = <<12/6=2>>2.
The area of the square is 2*2 = <<2*2=4>>4.
### Answer: 4

Expected answer: 4

Correct? Yes

Problem: If the first star, instead of emitting particles for 8 seconds, only emits particles for 5 seconds, how many particles will then it emit?

InternLM-Math Base 20B:

  Reasoning: 7 × 5 = <<7*5=35>>35.
The star emits 35 particles in total.
### Answer: 35

Expected answer: 35

Correct? Yes

Problem: The shoe factory increased its production time from five hours to ten hours. Now, how many shoes will the factory produce if it's still able to create 24 shoes every three hours?

InternLM-Math Base 20B:

  Reasoning: The factory can create 24/3 = <<24/3=8>>8 shoes in one hour.
The factory can create 8*10 = <<8*10=80>>80 shoes in ten hours.
### Answer: 80

Expected answer: 80

Correct? Yes

Problem: Considering the same unique species of butterfly, but now in a different region it only lives for 5 days each week instead of per month. How many days total does the butterfly species live in a 7-week span?

InternLM-Math Base 20B:

  Reasoning: The butterfly species lives 5 days in each week. There are 7 weeks in a 7-week span. So, the butterfly species lives 5*7 = <<5*7=35>>35 days in a 7-week span.
### Answer: 35

Expected answer: 35

Correct? Yes

Problem: In fairy land, change the original number of fairies an ordinary mushroom can feed from 10 to 11. With this change, how many fairies can a magic mushroom, which can provide food as 4 ordinary mushrooms, feed for a day?

InternLM-Math Base 20B:

  Reasoning: 4 ordinary mushrooms can feed 4*11 = <<4*11=44>>44 fairies for a day.
Therefore, a magic mushroom can feed 44 fairies for a day.
### Answer: 44

Expected answer: 44

Correct? Yes

Problem: If the speed of the river were instead 4 kilometers per hour for 2 hours, how far would the river flow within these 2 hours?

InternLM-Math Base 20B:

  Reasoning: 4*2 = <<4*2=8>>8 kilometers.
### Answer: 8 kilometers

Expected answer: 8

Correct? Yes

Problem: Following up on the last problem: If Ava decides to triple the height of the slide that is double her height, how high will the new slide be?

InternLM-Math Base 20B:

  Reasoning: The new slide is 2*3*12 = <<2*3*12=72>>72 feet tall.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: In the previous problem, Ava was initially considering a slide double her height when she was 12 feet tall. However, let's say Ava is actually 11 feet tall. In that case, how high would the slide be if it was still double her height?

InternLM-Math Base 20B:

  Reasoning: The slide is 2*11 = <<2*11=22>>22 feet tall.
### Answer: 22

Expected answer: 22

Correct? Yes

Problem: What if the tap was turned on for 7 minutes instead of 9 minutes? How much water would the tap fill in the tank in that case?

InternLM-Math Base 20B:

  Reasoning: The tank will have 2*7 = <<2*7=14>>14 liters of water after 7 minutes.
### Answer: 14

Expected answer: 14

Correct? Yes

Problem: Fred received 11 letters in one week instead of 12. If he still reads them evenly over 3 days, how many letters does he read per day now?

InternLM-Math Base 20B:

  Reasoning: Fred reads 11/3 = <<11/3=3.6666666666666665>>3.6666666666666665 letters per day.
### Answer: 3.6666666666666665

Expected answer: 11/3

Correct? Yes

Problem: After the shifts mentioned in the previous problem, the toy car factory decides to change its production rate. Now, instead of producing 8 cars per hour as before, it produces 5 cars per hour. If the factory continues to run for 12 hours a day, how many cars will it produce in one day now?

InternLM-Math Base 20B:

  Reasoning: The factory produces 5 cars per hour, and it runs for 12 hours a day. Therefore, the factory produces 5*12 = <<5*12=60>>60 cars per day.
### Answer: 60

Expected answer: 60

Correct? Yes

Problem: If instead Baby Lily originally got 44 toys but this time, she divided them evenly among her 11 friends, each friend would get 'i' number of toys. If we multiply 'i' by 5, how many toys would that represent?

InternLM-Math Base 20B:

  Reasoning: Each friend would get 44/11 = <<44/11=4>>4 toys.
If we multiply 4 by 5, we get 4*5 = <<4*5=20>>20.
### Answer: 20

Expected answer: 20

Correct? Yes

Problem: If each of Mike's children decided to save the money they received from their father every month for a year, how much money would each child have saved by the end of the year?

InternLM-Math Base 20B:

  Reasoning: 3*12 = <<3*12=36>>36
Each child would have saved $36 by the end of the year.
### Answer: 36

Expected answer: 36

Correct? Yes

Problem: If Mike had $11 instead of $9 and he still decided to divide it equally among his 3 children, how much money would each child receive now?

InternLM-Math Base 20B:

  Reasoning: 11/3 = <<11/3=3.6666666666666665>>3.6666666666666665
Each child would recieve $3.67.
### Answer: 3.67

Expected answer: 11/3

Correct? No

Problem: If instead the cow produced 72 liters of milk in 10 days, and you still wanted to find out how much milk it would produce in 12 days, how much would that be?

InternLM-Math Base 20B:

  Reasoning: 72/10 = <<72/10=7.2>>7.2 liters of milk per day.
7.2*12 = <<7.2*12=86.4>>86.4 liters of milk in 12 days.
### Answer: 86.4

Expected answer: 432/5

Correct? Yes

Problem: If the scale of the map is adjusted so that instead of 30 miles being represented by 10 inches, it is now represented by 12 inches, how many miles would 1 inch represent on the new map?

InternLM-Math Base 20B:

  Reasoning: 1 inch represents 30/12 = <<30/12=2.5>>2.5 miles.
### Answer: 2.5

Expected answer: 5/2

Correct? Yes

Problem: If the truck with all the tires fully loaded is carrying cargo weighing the same as the total load the tires can handle, what would be the total weight being carried by the truck?

InternLM-Math Base 20B:

  Reasoning: 8*8 = <<8*8=64>>64
### Answer: 64

Expected answer: 64

Correct? Yes

Problem: In the previous scenario, each tire was capable of carrying a load of 8 tons. But what if each tire was only capable of carrying a load of 3 tons? How much total load could the truck carry then when all tires are fully loaded?

InternLM-Math Base 20B:

  Reasoning: 8*3 = <<8*3=24>>24
### Answer: 24

Expected answer: 24

Correct? Yes

Problem: If instead you started with just 1 fruit, how many cups of juice would you be able to make?

InternLM-Math Base 20B:

  Reasoning: 1*1 = <<1*1=1>>1
You would be able to make 1 cup of juice.
### Answer: 1

Expected answer: 1

Correct? Yes

Problem: Sara decided to duplicate her complete set of colored pencils and she bought exactly the same amount as she had. Now, how many colored pencils does Sara have in total?

InternLM-Math Base 20B:

  Reasoning: Sara has 25 colored pencils, and she bought 25 more. So, she has 25+25 = <<25+25=50>>50 colored pencils in total.
### Answer: 50

Expected answer: 50

Correct? Yes

Problem: Sara suddenly realized that she had miscalculated; she actually has 11 of each type of colored pencil, not 5. With this correction, how many colored pencils does Sara really have?

InternLM-Math Base 20B:

  Reasoning: Sara has 5 colors, and she has 11 of each type. So, she has 5*11 = <<5*11=55>>55 colored pencils in total.
### Answer: 55

Expected answer: 55

Correct? Yes

Problem: You have decided to play 8 rounds of the same game. If you continue to use the same 12-letter word in each round, how many points will you accumulate in total from all the rounds?

InternLM-Math Base 20B:

  Reasoning: 8 × 12 = <<8*12=96>>96
### Answer: 96

Expected answer: 96

Correct? Yes

Problem: You decide to change your strategy in the next game and instead use a 9-letter word. If each letter still corresponds to 1 point, how many points will you get in this game?

InternLM-Math Base 20B:

  Reasoning: 9 × 1 = <<9*1=9>>9
### Answer: 9

Expected answer: 9

Correct? Yes

Problem: Jake decided to not drink any of the water just yet and stored all the bottles in his garage. After leaving the bottles in the garage untouched, how many bottles does Jake still have?

InternLM-Math Base 20B:

  Reasoning: Jake still has 49 bottles of water.
### Answer: 49

Expected answer: 49

Correct? Yes

Problem: Considering from the previous question, what if each piano only had 3 legs instead of 4. With 2 pianos in the music room, how many legs would there be in total?

InternLM-Math Base 20B:

  Reasoning: 3*2 = <<3*2=6>>6
### Answer: 6

Expected answer: 6

Correct? Yes

Problem: Following the previous scenario, if there are 11 keyrings in total, how many keys does each person get?

InternLM-Math Base 20B:

  Reasoning: Each person gets 11*2/2 = <<11*2/2=11>>11 keys.
### Answer: 11

Expected answer: 11

Correct? Yes

Problem: In the previous scenario, suppose now that each keyring holds 3 keys, but still has to be split amongst two people. How many keys does each person get now?

InternLM-Math Base 20B:

  Reasoning: Each person gets 3/2 = <<3/2=1.5>>1.5 keys.
### Answer: 1.5

Expected answer: 3/2

Correct? Yes

Problem: Javier decided to pack more bottles into his crates. Now, each crate contains 8 bottles instead of 6. How many bottles does Javier have now with his 9 crates?

InternLM-Math Base 20B:

  Reasoning: Javier has 9 × 8 = <<9*8=72>>72 bottles.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: Following the previous question: if the exchange rate changes and now 1 euro equals 1 dollar, how much will the shirt cost in dollars now?

InternLM-Math Base 20B:

  Reasoning: 1 euro = 1 dollar, so 10 euros = 10 dollars.
The shirt costs 10 dollars.
### Answer: 10

Expected answer: 10

Correct? Yes

Problem: After the price changes, the shirt now costs 9 dollars. If 1 dollar equals 1 euro, how much does the shirt cost in euros now?

InternLM-Math Base 20B:

  Reasoning: 1 dollar = 1 euro, so 9 dollars = 9 euros.
The shirt now costs 9 euros.
### Answer: 9

Expected answer: 9

Correct? Yes

Problem: Suppose the original size of the beach ball was actually 5 feet in diameter, not 3 feet as previously stated. If it is still being blown up to be twice its original size, what will be the diameter of the beach ball after it is blown up now?

InternLM-Math Base 20B:

  Reasoning: The diameter of the beach ball after it is blown up is 5*2 = <<5*2=10>>10 feet.
### Answer: 10

Expected answer: 10

Correct? Yes

Problem: The farmer then decides to combine all the tomatoes from the 4 boxes back into one large container. How many tomatoes will there be in the large container?

InternLM-Math Base 20B:

  Reasoning: 6*4 = <<6*4=24>>24
There will be 24 tomatoes in the large container.
### Answer: 24

Expected answer: 24

Correct? Yes

Problem: In a different scenario, if the farmer had decided to equally divide the 24 tomatoes between only 2 boxes instead of 4, how many tomatoes would have been in each box then?

InternLM-Math Base 20B:

  Reasoning: 24/2 = <<24/2=12>>12
There would have been 12 tomatoes in each box.
### Answer: 12

Expected answer: 12

Correct? Yes

Problem: In the previous situation, John multiplied the distance he traveled with 6. But this time, he decides to ride his scooter for a bit longer. Instead of multiplying by 6, he now multiplies the same initial half of 16, distance "B", by 8 to find out the new total distance "X" he traveled back and forth. What is the new total distance "X" he traveled?

InternLM-Math Base 20B:

  Reasoning: B = 16/2 = <<16/2=8>>8
X = 8*B = 8*<<8>>8 = <<8*8=64>>64
### Answer: 64

Expected answer: 64

Correct? Yes

Problem: Now, imagine Lisa spent 5 days over the course of a week doing the same routine of watching 2 episodes back-to-back of her favorite TV show, which is 8 minutes per episode. How long in total did she spend watching her TV show over those 5 days?

InternLM-Math Base 20B:

  Reasoning: Lisa spent 5*16 = <<5*16=80>>80 minutes watching her favorite TV show over those 5 days.
### Answer: 80

Expected answer: 80

Correct? Yes

Problem: In a twist, one day Lisa decided to watch a different TV show, which is 12 minutes long per episode. If she watches 2 episodes of this show back-to-back without any breaks, how long will she be watching this different TV show in total?

InternLM-Math Base 20B:

  Reasoning: Lisa will be watching the different TV show for 12+12 = <<12+12=24>>24 minutes in total.
### Answer: 24

Expected answer: 24

Correct? Yes

Problem: Consider the starting point of the previous problem where Jane had 1 blanket and John had 3 times as many blankets as Jane. Now, suppose that Jane initially has 6 blankets instead of 1. In this new scenario, how many blankets does John have initially?

InternLM-Math Base 20B:

  Reasoning: John has 3*6 = <<3*6=18>>18 blankets.
### Answer: 18

Expected answer: 18

Correct? Yes

Problem: If John originally had $10 instead of $28, and he still wanted to divide it equally among his 7 children, how much would each child receive now?

InternLM-Math Base 20B:

  Reasoning: John will divide $10 by 7.
Each child will receive $1.42.
### Answer: 1.42

Expected answer: 10/7

Correct? No

Problem: Continuing from the previous situation, after selling 1 scooter a day for 8 days, Kate decides to hold a grand sale for 8 more days where she sells the same number of scooters she sold in the previous 8 days every day. How many additional scooters will she sell during this grand sale?

InternLM-Math Base 20B:

  Reasoning: Kate sells 8 scooters every day during the grand sale. She will sell 8*8 = <<8*8=64>>64 additional scooters during the grand sale.
### Answer: 64

Expected answer: 64

Correct? Yes

Problem: Let's revise the original situation. Instead of selling 1 scooter a day, let's say Kate was actually selling 7 scooters a day for 8 days. How many scooters did she sell in this period?

InternLM-Math Base 20B:

  Reasoning: Kate sells 7 scooters every day. She will sell 7*8 = <<7*8=56>>56 scooters after 8 days.
### Answer: 56

Expected answer: 56

Correct? Yes

Problem: Now imagine that the tree actually produces 12 seeds each year instead of 10. How many seeds does the tree produce in 1 year?

InternLM-Math Base 20B:

  Reasoning: 1 tree produces 12 seeds each year.
1 tree produces 12 seeds in 1 year.
### Answer: 12

Expected answer: 12

Correct? Yes

Problem: Following the same scenario where the friends were divided into groups, if each group decided to buy 7 pizzas, how many pizzas would be bought in total?

InternLM-Math Base 20B:

  Reasoning: 4*7 = <<4*7=28>>28
There will be 28 pizzas bought in total.
### Answer: 28

Expected answer: 28

Correct? Yes

Problem: Continuing from our previous scenario, let's say that instead of having 12 friends, there were actually only 6 friends to begin with. If they still want to divide themselves into groups with 3 friends each, how many group will there be?

InternLM-Math Base 20B:

  Reasoning: 6/3 = <<6/3=2>>2
There will be 2 groups.
### Answer: 2

Expected answer: 2

Correct? Yes

Problem: Continuing from the previous situation, Emma and her friends decide to make their sandcastles stronger. They pour a layer of seawater on each shovel used in their sandcastles. If each shovel full of sand requires 1 bottle of seawater, how many bottles of seawater will they need in total?

InternLM-Math Base 20B:

  Reasoning: Emma and her friends use 72 shovels of sand in total. So they need 72 bottles of seawater in total.
### Answer: 72

Expected answer: 72

Correct? Yes

Problem: Following the previous situation with Emma and her friends building sandcastles, imagine that they had originally started with 5 sandcastles instead of 6. Each of these 5 sandcastles still uses 4 buckets of sand and each of these buckets contains 3 shovels of sand. How many shovels of sand would they have used then?

InternLM-Math Base 20B:

  Reasoning: Each sandcastle uses 4 buckets of sand, and they build 5 sandcastles. So they use 4*5 = <<4*5=20>>20 buckets of sand in total.
Each bucket of sand has 3 shovels full of sand in it. So they use 20*3 = <<20*3=60>>60 shovels of sand in total.
### Answer: 60

Expected answer: 60

Correct? Yes

Problem: After reconsidering, Jenna decided to only make 7 bracelets instead of 8. Now, how many beads will Jenna need to make the bracelets?

InternLM-Math Base 20B:

  Reasoning: Jenna needs 12 × 7 = <<12*7=84>>84 beads in total.
### Answer: 84

Expected answer: 84

Correct? Yes

Problem: If Penny started using 9 milliliters of toothpaste for each brushing instead of 11, how many times could she brush her teeth with the new tube of toothpaste?

InternLM-Math Base 20B:

  Reasoning: Penny can brush her teeth 66/9 = <<66/9=7>>7 times with the new tube of toothpaste.
### Answer: 7

Expected answer: 22/3

Correct? No

Problem: If Vanessa only had 4 bunches of grapes and each bunch had 5 grapes instead of 11, how many grapes would Vanessa have in total then?

InternLM-Math Base 20B:

  Reasoning: 4*5 = <<4*5=20>>20
### Answer: 20

Expected answer: 20

Correct? Yes

Problem: Mary used her 6 yards of fabric to make a skirt. Each skirt she makes, requires 6 yards of fabric. She decides to make as many skirts as skirts she already made (i.e., she wants to multiply the number of skirts by the number of skirts she already made). How many skirts will Mary have made in total?

InternLM-Math Base 20B:

  Reasoning: Mary will have made 1*1 = <<1*1=1>>1 skirts in total.
### Answer: 1

Expected answer: 1

Correct? Yes

Problem: Mary now has a new length of fabric that's 7 yards long. Keeping in mind that each skirt requires 6 yards of fabric, how many skirts can Mary make with the new fabric she has?

InternLM-Math Base 20B:

  Reasoning: Mary can make 7/6 = <<7/6=1.1666666666666667>>1.1666666666666667 skirts with the new fabric she has.
### Answer: 1.1666666666666667

Expected answer: 7/6

Correct? Yes

Problem: Following up from the previous question, if Jordan decides to again double the total number of cards he has, how many cards will Jordan end up with at that point?

InternLM-Math Base 20B:

  Reasoning: Jordan has 18*2 = <<18*2=36>>36 cards.
### Answer: 36

Expected answer: 36

Correct? Yes

Problem: Following up from the previous question, if originally Jordan had 5 card packs instead of 9, and he still decided to double the total number of his cards, how many cards would Jordan have then?

InternLM-Math Base 20B:

  Reasoning: Jordan has 5*2 = <<5*2=10>>10 cards.
### Answer: 10

Expected answer: 10

Correct? Yes

Problem: In the magical world, a donezal is a unit of time equal to 10 wizard staff channeling periods. If every wizard staff channeling period lasts as long as the spell in the previous problem, how much time does one donezal represent?

InternLM-Math Base 20B:

  Reasoning: 10 × 10 = <<10*10=100>>100 minutes.
### Answer: 100

Expected answer: 100

Correct? Yes

Problem: In a new experiment, the wizard decided to divide the same 60-minute spell into only 4 equal periods before the wizard staff needs to take a break. How long does each period of casting last now?

InternLM-Math Base 20B:

  Reasoning: 60/4 = <<60/4=15>>15 minutes.
### Answer: 15

Expected answer: 15

Correct? Yes

Problem: In the previous problem, the drummer was hitting his drum 11 times per minute. But what if his rhythm slows down and he now hits the drum only 2 times per minute. How many times will he hit the drum in 8 minutes then?

InternLM-Math Base 20B:

  Reasoning: 2*8 = <<2*8=16>>16
### Answer: 16

Expected answer: 16

Correct? Yes

Problem: Now, John is reconsidering his packing strategy. Instead of packing two pairs of socks per pair of pants, he thinks he only needs to pack two socks for every pair of pants. If he sticks to this new strategy, how many pairs of pants should he pack for his 4 pairs of socks?

InternLM-Math Base 20B:

  Reasoning: John needs 4/2 = <<4/2=2>>2 pairs of pants.
### Answer: 2

Expected answer: 2

Correct? Yes

Problem: If, instead of 8 pens per pack, there were 6 pens per pack, how many pens would John have in total with his 5 packs?

InternLM-Math Base 20B:

  Reasoning: John has 5 packs of pens. Each pack contains 6 pens. So, John has 5*6 = <<5*6=30>>30 pens in total.
### Answer: 30

Expected answer: 30

Correct? Yes

Problem: Suppose there are more guards available and Captain Blackbeard decided to split the total time into 12 shifts instead of 6. How many hours will each guard now have to protect the ship under this new arrangement?

InternLM-Math Base 20B:

  Reasoning: 60/12 = <<60/12=5>>5 hours.
Each guard will have to protect the ship for 5 hours.
### Answer: 5

Expected answer: 5

Correct? Yes

Problem: Going further from the previous problem, John now realizes he needs twice as much rope! How much will it cost him to buy 12 meters of rope in all?

InternLM-Math Base 20B:

  Reasoning: John needs a 12-meter long rope.
The cost of 1 meter of rope is $5.
The cost of 12 meters of rope is 12*5 = <<12*5=60>>60 dollars.
### Answer: 60

Expected answer: 60

Correct? Yes

Problem: After considering his needs again, John realized that he actually needs 7 meters of rope, not 6. How much will it cost him to buy this new length of rope?

InternLM-Math Base 20B:

  Reasoning: John needs a 7-meter long rope.
The cost of 1 meter of rope is $5.
The cost of 7 meters of rope is 7*5 = <<7*5=35>>35 dollars.
### Answer: 35

Expected answer: 35

Correct? Yes

Problem: In the previous scenario, Amelia was distributing gifts among 7 of her friends. But what if Amelia instead planned to distribute her 42 gifts equally among only 6 friends instead? How many gifts would each friend receive in this case?

InternLM-Math Base 20B:

  Reasoning: Amelia will distribute 42/6 = <<42/6=7>>7 gifts to each of her 6 best friends.
### Answer: 7

Expected answer: 7

Correct? Yes

Problem: If the mother bear had only 40 honey pots and wanted to evenly distribute them among her 10 cubs instead, how many honey pots would each cub receive now?

InternLM-Math Base 20B:

  Reasoning: 40/10 = <<40/10=4>>4
Each cub would receive 4 honey pots.
### Answer: 4

Expected answer: 4

Correct? Yes

Problem: In the previous situation, what if instead of eating 4kg of food per hour, the elephant ate only 3kg of food per hour? How much food would the elephant eat in 4 hours then?

InternLM-Math Base 20B:

  Reasoning: 3*4 = <<3*4=12>>12
### Answer: 12

Expected answer: 12

Correct? Yes

Problem: In addition to the 3-headed dragon, another dragon of the same kind joins it. Between the two dragons, how many eyes are there in total?

InternLM-Math Base 20B:

  Reasoning: 2*3*8 = <<2*3*8=48>>48
### Answer: 48

Expected answer: 48

Correct? Yes

Problem: Suppose that each dragon instead had 11 eyes per head, not 8. How many eyes would one three-headed dragon have then?

InternLM-Math Base 20B:

  Reasoning: 3*11 = <<3*11=33>>33
### Answer: 33

Expected answer: 33

Correct? Yes

Problem: Following from the last problem, Charlie then takes the amount of chocolates he has and decides to multiply it by 9 for a birthday party. How many chocolates does Charlie have now?

InternLM-Math Base 20B:

  Reasoning: Charlie has 4*9 = <<4*9=36>>36 chocolates after multiplying the amount by 9.
### Answer: 36

Expected answer: 36

Correct? Yes

Problem: Following from the previous problems, let's consider a new scenario. This time, suppose Charlie only had 2 chocolates and again decided to share half of them with his sister. Then, he counts his remaining chocolates and decides to double the amount. How many chocolates does Charlie have now in this scenario?

InternLM-Math Base 20B:

  Reasoning: Charlie has 2/2 = <<2/2=1>>1 chocolate left after sharing half of them with his sister. Then he has 1*2 = <<1*2=2>>2 chocolates after doubling the amount.
### Answer: 2

Expected answer: 2

Correct? Yes

Problem: Suppose the number of walls in the castle was incorrectly counted before, and there are actually 10 walls instead of 6. If the castle still has a total of 36 towers, how many towers would be on each wall in that case?

InternLM-Math Base 20B:

  Reasoning: There are 36/10 = <<36/10=3.6>>3.6 towers on each wall.
### Answer: 3.6

Expected answer: 18/5

Correct? Yes