InternLM-Math Base 7B on 7.NS.A.3-decimal
Common Core Standard 7.NS.A.3-decimal
This standard covers the following:
Solve real-world and mathematical problems involving the four operations with rational numbers.
Problem: A teacher has a stack of papers that is 12.7 cm high. She adds another stack of papers that is 4.2 cm high to the existing pile. Afterwards, she distributes 2.144 cm worth of papers to her students. How tall is the stack of papers now?
InternLM-Math Base 7B:
| |user| A teacher has a stack of papers that is 12.7 cm high. She adds another stack of papers that is 4.2 cm high to the existing pile. Afterwards, she distributes 2.144 cm worth of papers to her students. How tall is the stack of papers now?
Expected answer: 14.7560000000000
Correct? No
Problem: A dragon flew a total distance of 24.02 km in a day. However, it had to backtrack 1.3 km due to a severe storm. How many kilometers did the dragon travel in the end?
InternLM-Math Base 7B:
Reasoning: The dragon flew 1.3*2 = <<1.3*2=2.6>>2.6 km in the opposite direction. The dragon flew 24.02-2.6 = <<24.02-2.6=21.42>>21.42 km in the end. ### Answer: 21.42
Expected answer: 22.7200000000000
Correct? No
Problem: An elephant weighs 12.1 tons in the morning and gains 15.68 tons throughout the day from eating. What is the elephant's total weight by the end of the day?
InternLM-Math Base 7B:
Reasoning: 12.1+15.68 = <<12.1+15.68=27.78>>27.78 ### Answer: 27.78
Expected answer: 27.7800000000000
Correct? Yes
Problem: The elephant from the previous question then burns off 5.935 tons of weight. What is the elephant's weight now?
InternLM-Math Base 7B:
Reasoning: 27.78-5.935 = <<27.78-5.935=21.845>>21.845 ### Answer: 21.845
Expected answer: 21.8450000000000
Correct? Yes
Problem: A certain breed of mouse can eat up to 29.573 grams of food a day. However, one day it only eats 4.62 grams less than what it usually eats. How much does it eat on that day?
InternLM-Math Base 7B:
Expected answer: 24.9530000000000
Correct? No
Problem: In a zoo, there’s a baby giraffe that is 1.4 meters tall. If it grows another 5.7 meters, how tall will the giraffe be?
InternLM-Math Base 7B:
Answer: 1.4+5.7=<<1.4+5.7=7.1>>7.1 meters
Expected answer: 7.10000000000000
Correct? Yes
Problem: In a zoo, there’s a baby giraffe that is 1.4 meters tall. If it grows another 5.7 meters, it will reach a certain height. Now, if this grown giraffe stretches its neck further and adds another 12.608 meters to its height, how tall will the giraffe be then?
InternLM-Math Base 7B:
Answer: 1.4+5.7+12.608=<<1.4+5.7+12.608=19.808>>19.808 meters
Expected answer: 19.7080000000000
Correct? No
Problem: In another zoo, there is a full-grown giraffe that measures an impressive 23 meters. If the giraffe stretches its neck and grows another 5.7 meters, how tall will the giraffe be then?
InternLM-Math Base 7B:
Answer: 23+5.7=<<23+5.7=28.7>>28.7 meters
Expected answer: 28.7000000000000
Correct? Yes
Problem: An astronaut is floating in space and generates an isotopic energy reading of 24.178 units per second. If he keeps this energy level constant for 1.0 second, how much energy does he generate in total?
InternLM-Math Base 7B:
Reasoning: The astronaut generates 24.178 <<24.178*1.0=24.178>>24.178 units of energy in 1.0 second. ### Answer: 24.178
Expected answer: 24.1780000000000
Correct? Yes
Problem: After generating an isotopic energy reading of 24.178 units for 1.0 second, the astronaut decides to boost his energy output by an additional 4.995 units. What is the new total amount of energy he generates?
InternLM-Math Base 7B:
Reasoning: The astronaut generates 24.178+4.995 = <<24.178+4.995=29.173>>29.173 units of energy in 1.0 second. ### Answer: 29.173
Expected answer: 29.1730000000000
Correct? Yes
Problem: If instead of generating an isotopic energy reading of 24.178 units per second, the astronaut was initially generating 23 units per second for 1.0 second, what would the total energy output be then?
InternLM-Math Base 7B:
Reasoning: The astronaut generates 23 <<23*1.0=23>>23 units of energy in 1.0 second. ### Answer: 23
Expected answer: 23.0000000000000
Correct? Yes
Problem: A treasure chest holds 25.84 kilograms of gold. However, 2.876 kilograms of it is stolen. How much gold is left in the chest?
InternLM-Math Base 7B:
Reasoning: 2.876/25.84 = <<2.876/25.84=0.1111111111111111>>0.1111111111111111 0.1111111111111111*25.84 = <<0.1111111111111111*25.84=2.9999999999999998>>2.9999999999999998 ### Answer: 2.9999999999999998
Expected answer: 22.9640000000000
Correct? No
Problem: A breed of dog grows 15.7 inches tall and a different breed grows 4.9 inches more. How tall is the taller dog?
InternLM-Math Base 7B:
Reasoning: The taller dog is <<15.7+4.9=20.6>>20.6 inches tall. ### Answer: 20.6
Expected answer: 20.6000000000000
Correct? Yes
Problem: In addition to the previous question: The taller dog then bends over to pick up a toy, at which point its height decreases by 2.865 inches. How tall is the dog when it is bending over?
InternLM-Math Base 7B:
Reasoning: The dog is 20.6-2.865=<<20.6-2.865=17.735>>17.735 inches tall when it is bending over. ### Answer: 17.735
Expected answer: 17.7350000000000
Correct? Yes
Problem: If instead of growing 4.9 inches more, the different breed of dog only grew 1 inch more than the 15.7 inches tall dog, how tall would this different breed of dog be?
InternLM-Math Base 7B:
Reasoning: The different breed of dog would be 15.7+1=16.716.7 inches tall. ### Answer: 16.7
Expected answer: 16.7000000000000
Correct? Yes
Problem: A ball weighs 21.985 grams. If another identical ball weighs 25.3 grams but has lost some weight to become equal to the first ball, how much weight has the second ball lost? If we add 8.0 grams to the lost weight, what will be the total?
InternLM-Math Base 7B:
Reasoning: The second ball has lost <<25.3-21.985=3.315>>3.315 grams of weight to become equal to the first ball. The total is 8.0+3.315 = <<8.0+3.315=11.315>>11.315 grams. ### Answer: 11.315
Expected answer: 11.3150000000000
Correct? Yes
Problem: At the beginning of his shift, a firefighter checks the fuel levels for two firetrucks. One truck has 9.155 gallons of fuel and the other has 5.540 gallons of fuel. After refueling both trucks, he adds an additional 9.03 gallons. What is the total amount of fuel in gallons after the firefighter refueled both trucks?
InternLM-Math Base 7B:
Reasoning: The total amount of fuel in the two firetrucks is 9.155+5.540=<<9.155+5.540=14.695>>14.695 gallons. The firefighter refueled 9.03+9.03=<<9.03+9.03=18.06>>18.06 gallons. The total amount of fuel in the two firetrucks is 14.695+18.06=<<14.695+18.06=32.75>>32.75 gallons. ### Answer: 32.75
Expected answer: 23.7250000000000
Correct? No
Problem: A sunflower stem measures 23.57 cm at its longest but due to insufficient sunlight it shrunk to 11.58 cm. How much did the sunflower stem shrink?
InternLM-Math Base 7B:
Reasoning: The sunflower stem shrunk 23.57 – 11.58 = <<23.57 – 11.58=12>>12 cm. ### Answer: 12
Expected answer: 11.9900000000000
Correct? No
Problem: A scented candle is usually priced at $18.77 but today it is discounted by $4.016. What is the current price of the scented candle?
InternLM-Math Base 7B:
Reasoning: The current price of the scented candle is 18.77-4.016 = <<18.77-4.016=14.754>>14.754.
Expected answer: 14.7540000000000
Correct? Yes
Problem: After buying the discounted candle, Helen realized she had a coupon that subtracts another $8.49 from her total payment. How much is the final price she has to pay for the candle after applying the coupon?
InternLM-Math Base 7B:
Reasoning: The final price is 18.77-4.016-8.49 = <<18.77-4.016-8.49=3.254>>3.254.
Expected answer: 6.26400000000000
Correct? No
Problem: Recalling the previous scenario, let's imagine that initially, the scented candle was priced at $18.77, but today it's on a bigger discount of $9 instead of $4.016. What would be the current price of the scented candle with the bigger discount?
InternLM-Math Base 7B:
Reasoning: The current price of the scented candle is 18.77-9 = <<18.77-9=9.77>>9.77.
Expected answer: 9.77000000000000
Correct? Yes
Problem: In a forest, there are 3.9km of northern trails and 7.4km of southern trails. If the total distance of all trails in the forest is 17.9km, how many kilometres of trails are not part of the northern or southern trails?
InternLM-Math Base 7B:
Reasoning: 3.9+7.4= <<3.9+7.4=11.3>>11.3 11.3/17.9= <<11.3/17.9=0.6300>>0.6300 0.6300x17.9= <<0.6300x17.9=11.3>>11.3 ### Answer: 11.3
Expected answer: 6.60000000000000
Correct? No
Problem: A baby weighed 4.8 pounds at birth. After a few weeks, the baby gained 15.67 pounds. The baby's weight increased again by 6.5 pounds a few months later. What is the baby's weight now?
InternLM-Math Base 7B:
Reasoning: The baby's weight is 4.8+15.67+6.5 = <<4.8+15.67+6.5=26.97>>26.97 pounds. ### Answer: 26.97
Expected answer: 26.9700000000000
Correct? Yes
Problem: Previously, the baby's weight increased by 15.67 pounds a few weeks after birth. Suppose instead, the baby gained 11 pounds after a few weeks. The baby's weight still increased by 6.5 pounds a few months later. What would be the baby's weight now in this scenario?
InternLM-Math Base 7B:
Reasoning: The baby's weight is 4.8+11+6.5 = <<4.8+11+6.5=22.3>>22.3 pounds. ### Answer: 22.3
Expected answer: 22.3000000000000
Correct? Yes
Problem: Astronomer Tom has a powerful telescope with a focal length of 22.5 centimeters. He uses an eyepiece that extends the focal length by another 1.832 centimeters. What is the total length of his telescope?
InternLM-Math Base 7B:
Reasoning: 22.5+1.832=<<22.5+1.832=24.332>>24.332 centimeters ### Answer: 24.332
Expected answer: 24.3320000000000
Correct? Yes
Problem: With further calibration, Astronomer Tom realized the eyepiece only extends the telescope's focal length by an additional 2 centimeters, instead of the previous 1.832 centimeters. Considering this, what would be the new total focal length of his telescope?
InternLM-Math Base 7B:
Reasoning: 22.5+2=<<22.5+2=24.5>>24.5 centimeters ### Answer: 24.5
Expected answer: 24.5000000000000
Correct? Yes
Problem: A window has a width of 17.9 inches and an additional length of 1.5 inches for the frame. What is the total width of the window including the frame?
InternLM-Math Base 7B:
Reasoning: 1.5+17.9 = <<1.5+17.9=19.4>>19.4 ### Answer: 19.4
Expected answer: 19.4000000000000
Correct? Yes
Problem: After installing the window, the builder realizes he made an error and the opening for the window is actually 13.48 inches smaller than the total width of the window including the frame. What is the excess length of the window that couldn't fit into the opening?
InternLM-Math Base 7B:
Reasoning: 19.4-13.48 = <<19.4-13.48=5.92>>5.92 ### Answer: 5.92
Expected answer: 5.92000000000000
Correct? Yes
Problem: Suppose the builder made a new window, this time with a width of 28 inches, plus an additional length of 1.5 inches for the frame, how long is the window including the frame this time?
InternLM-Math Base 7B:
Reasoning: 1.5+28 = <<1.5+28=29.5>>29.5 ### Answer: 29.5
Expected answer: 29.5000000000000
Correct? Yes
Problem: Your pen has 5.1ml of ink when it's new. If you add 22.4ml of ink into the pen, how much ink does the pen have in total?
InternLM-Math Base 7B:
Reasoning: 5.1+22.4 = <<5.1+22.4=27.5>>27.5 ### Answer: 27.5
Expected answer: 27.5000000000000
Correct? Yes
Problem: Suppose instead of starting with 5.1ml of ink in your pen, you started with 2ml of ink and then added 22.4ml of ink. How much ink does your pen have in total now?
InternLM-Math Base 7B:
Reasoning: 2+22.4 = <<2+22.4=24.4>>24.4 ### Answer: 24.4
Expected answer: 24.4000000000000
Correct? Yes
Problem: The perimeter of a square is found by adding the lengths of all its sides. If three sides of a square measure 15.4 cm in total, and the length of the remaining side is 1.828 cm, what is the total perimeter of the square?
InternLM-Math Base 7B:
Reasoning: The sum of the three given side lengths is 15.4+15.4+15.4 = <<15.4+15.4+15.4=46.2>>46.2 cm. The sum of all the side lengths of a square is the same as the length of the side times 4. The sum of the three given side lengths is 3/4 of the total perimeter. 3/4 of 46.2 = <<(3/4)*46.2=32.7>>32.7 cm. The total perimeter of the square is 32.7*4 = <<(32.7*4=130.8)>>130.8 cm. ### Answer: 130.8
Expected answer: 17.2280000000000
Correct? No
Problem: In a document, a paragraph has 12.244 words. If we add the number of words in this paragraph to a similar one, then how many words do the two paragraphs have in total?
InternLM-Math Base 7B:
Reasoning: The number of words in the other paragraph is 12.244/2 = <<12.244/2=6.122>>6.122. The number of words in the two paragraphs is 12.244+6.122 = <<12.244+6.122=18.366>>18.366. ### Answer: 18.366
Expected answer: 24.4880000000000
Correct? No
Problem: In a letter writing competition, one of the participants wrote a letter having a length of 29.82 pages. However, the rules require only a maximum of 24.2 pages. How many extra pages did the participant write compared to the maximum limit?
InternLM-Math Base 7B:
Expected answer: 5.62000000000000
Correct? No
Problem: At the zoo, a lion weights 28.32 pounds but loses 21.88 pounds. After gaining 3.4 pounds, what is the lion's new weight?
InternLM-Math Base 7B:
Reasoning: The lion's new weight is 28.32-21.88+3.4 = <<28.32-21.88+3.4= 8.64>>8.64 pounds. ### Answer: 8.64
Expected answer: 9.84000000000000
Correct? No
Problem: Last month, it rained an average of 3.645 inches per day. Today's rainfall is increased by 3.741 inches. How much rain was there in inches today?
InternLM-Math Base 7B:
Reasoning: 3.645+3.741 = <<3.645+3.741=7.386>>7.386 inches of rain. ### Answer: 7.386
Expected answer: 7.38600000000000
Correct? Yes
Problem: Continuing from the last scenario: After the increase in the daily rainfall today, an additional 4.51 inches of rain fell tonight. How much total rainfall was there today in inches?
InternLM-Math Base 7B:
Reasoning: 7.386+4.51 = <<7.386+4.51=11.896>>11.896 inches of rain. ### Answer: 11.896
Expected answer: 11.8960000000000
Correct? Yes
Problem: A window has a length of 1.7 meters and a height of 9.448 meters. A larger window has its length equal to the length and height of the smaller window combined and an extra height of 7.619 meters. What is the total length and height of the larger window?
InternLM-Math Base 7B:
Reasoning: The height of the larger window is 9.448+7.619 = <<9.448+7.619=17.067>>17.067 meters. The length of the larger window is 1.7+9.448+7.619 = <<1.7+9.448+7.619=18.787>>18.787 meters. ### Answer: 18.787, 17.067
Expected answer: 18.7670000000000
Correct? No
Problem: A pair of glasses at an optometrist's store cost $14.8. A special sturdy case for these glasses costs an additional $2.22. How much will it cost to buy the glasses and the case together?
InternLM-Math Base 7B:
Expected answer: 17.0200000000000
Correct? No
Problem: A roller skate normally costs $17.9, but is currently on sale with a discount of $8.069. How much does the sale price of the roller skate cost?
InternLM-Math Base 7B:
Reasoning: The current cost of the roller skate is 17.9-8.069 = <<17.9-8.069=9.831>>9.831. The sale price of the roller skate is $<<9.831>>9.81. ### Answer: 9.81
Expected answer: 9.83100000000000
Correct? Yes
Problem: After finding the sale price of the roller skate, let's say that you also decide to purchase a special set of roller skate guards that cost $3.53. How much would you need to pay in total for the discounted roller skate and the set of roller skate guards?
InternLM-Math Base 7B:
Reasoning: The total cost of the roller skate and the roller skate guards is 9.81+3.53 = <<9.81+3.53=13.34>>13.34. You would need to pay a total of $<<13.34>>$13.34 for the discounted roller skate and the set of roller skate guards. ### Answer: 13.34
Expected answer: 13.3610000000000
Correct? Yes
Problem: Let's consider a different scenario for the roller skate sale. If the discount on the roller skate was instead $6 from its original price of $17.9, how much would the roller skate cost after this new discount?
InternLM-Math Base 7B:
Reasoning: The current cost of the roller skate is 17.9-6 = <<17.9-6=11.9>>11.9. The sale price of the roller skate is $<<11.9>>11.90. ### Answer: 11.9
Expected answer: 11.9000000000000
Correct? Yes
Problem: In a sandbox, average weight of a single toy is 1.18 pounds. If a child put 23 toys in the sandbox, and added a castle mold that weighed 2.04 pounds, what would be the total weight of the items in the sandbox?
InternLM-Math Base 7B:
Reasoning: The total weight of the items in the sandbox is 1.18 ⋅ 23 + 2.04 = <<1.18⋅23+2.04=55.12>>55.12 pounds. ### Answer: 55.12
Expected answer: 29.1800000000000
Correct? No
Problem: At a campfire gathering, each person uses about 1.217 kg of firewood. If there are 18.89 people, approximately how much firewood will be used in total?
InternLM-Math Base 7B:
Reasoning: 1.217 × 18.89 = <<1.217 × 18.89=22.9113>>22.9113 kg.
Expected answer: 22.9891300000000
Correct? No
Problem: A big marshmallow contains 6.1 grams of sugar. A small one has 4.64 grams. What is the total grams of sugar if you were to consume both types of marshmallows?
InternLM-Math Base 7B:
Reasoning: 6.1 + 4.64 = <<6.1+4.64=10.74>>10.74 ### Answer: 10.74
Expected answer: 10.7400000000000
Correct? Yes
Problem: Following the previous question, if you were to consume another marshmallow containing 10.045 grams of sugar, along with the big and small marshmallow, what would be the total grams of sugar you would consume?
InternLM-Math Base 7B:
Reasoning: 10.74+10.045 = <<10.74+10.045=20.785>>20.785 ### Answer: 20.785
Expected answer: 20.7850000000000
Correct? Yes
Problem: Following the previous question, if the amount of sugar in the big marshmallow was actually 6.1 grams and the small marshmallow contained only 2 grams of sugar, what would be the total grams of sugar you would consume from both marshmallows?
InternLM-Math Base 7B:
Reasoning: 6.1 + 2 = <<6.1+2=8.1>>8.1 ### Answer: 8.1
Expected answer: 8.10000000000000
Correct? Yes
Problem: The weight of a toy car and a toy robot are 8.011 and 2.7 kilograms respectively. If we discount the collective weight by 9.526 kilograms, what is the resulting weight?
InternLM-Math Base 7B:
Reasoning: The weight of the car and the robot is 8.011+2.7 = <<8.011+2.7=10.711>>10.711 kilograms. The weight of the car and the robot is 10.711-9.526 = <<10.711-9.526=1.185>>1.185 kilograms. ### Answer: 1.185
Expected answer: 1.18500000000000
Correct? Yes
Problem: Continuing from the previous question, if the resulting weight is translated to Newtons by multiplying each kilogram by 17.5, what will be the total weight in Newtons of the toy car and robot after the discount?
InternLM-Math Base 7B:
Reasoning: The total weight in Newtons of the car and the robot is 1.185*17.5 = <<1.185*17.5=20.63>>20.63. ### Answer: 20.63
Expected answer: 20.7375000000000
Correct? No
Problem: In the previous question, suppose the weight of the toy robot was actually 21 kilograms instead of 2.7 kilograms. After taking into account the 9.526 kilograms discount, what would be the new combined weight of the toy car and robot?
InternLM-Math Base 7B:
Reasoning: The weight of the car and the robot is 8.011+21 = <<8.011+21=29.011>>29.011 kilograms. The weight of the car and the robot is 29.011-9.526 = <<29.011-9.526=19.485>>19.485 kilograms. ### Answer: 19.485
Expected answer: 19.4850000000000
Correct? Yes
Problem: A dog's meal should weigh 28.207 grams. If you already fed your dog 5.801 grams, how much more do you need to feed it if it needs to consume an additional 1.59 grams after this meal?
InternLM-Math Base 7B:
Reasoning: The dog needs to eat <<28.207-5.801-1.59=20.015>>20.015 more grams. ### Answer: 20.015
Expected answer: 23.9960000000000
Correct? No
Problem: A dolphin can leap a maximum of 9.8 feet out of the water. If it jumps an additional 11.484 feet during an extraordinary display, what is the total distance the dolphin covered?
InternLM-Math Base 7B:
Reasoning: 9.8+11.484= <<9.8+11.484=21.284>>21.284 ### Answer: 21.284
Expected answer: 21.2840000000000
Correct? Yes
Problem: After the remarkable display, it was noticed that the dolphin only returned 9.8 feet back to the water level. What was the remaining distance the dolphin covered in mid-air?
InternLM-Math Base 7B:
Reasoning: 11.484-9.8= <<11.484-9.8=1.684>>1.684 ### Answer: 1.684
Expected answer: 11.4840000000000
Correct? No
Problem: In another remarkable leap, instead of rising an additional 11.484 feet, the dolphin soared an extra 20 feet above the initial 9.8 feet. What is the total distance the dolphin covered in this jump?
InternLM-Math Base 7B:
Reasoning: 9.8+20= <<9.8+20=29.8>>29.8 ### Answer: 29.8
Expected answer: 29.8000000000000
Correct? Yes
Problem: A water bottle holds 7.55 liters of water initially. If you add 17.4 more liters of water, how many liters are in the water bottle?
InternLM-Math Base 7B:
Reasoning: 7.55+17.4 = <<7.55+17.4=25.95>>25.95.
Expected answer: 24.9500000000000
Correct? No
Problem: If the eastern direction on a compass is represented by 25.348 degrees, and the southeastern direction is represented by 15.313 degrees. What is the difference between these two directions in degrees?
InternLM-Math Base 7B:
Reasoning: The difference between 25.348 and 15.313 is <<25.348-15.313=10.035>>10.035 degrees. ### Answer: 10.035
Expected answer: 10.0350000000000
Correct? Yes
Problem: Now consider a new scenario, where the eastern direction on a compass is represented by 28 degrees instead of 25.348. What is the difference between this new eastern direction and the previous southeastern direction in degrees?
InternLM-Math Base 7B:
Reasoning: The difference between 28 and 15.313 is <<28-15.313=12.687>>12.687 degrees. ### Answer: 12.687
Expected answer: 12.6870000000000
Correct? Yes
Problem: A specific computer model costs $20.7. The store offers an upgrade that costs an additional $6.18. What will be the total cost for the upgraded computer model?
InternLM-Math Base 7B:
Reasoning: 20.7+6.18 = <<20.7+6.18=27.88>>27.88 ### Answer: 27.88
Expected answer: 26.8800000000000
Correct? No
Problem: The weight of a teddy bear is 29.2 grams. It loses 9.1 grams of its stuffing. Together with it, a small ribbon weighing 6.62 grams is attached to it. How much does the teddy bear with missing stuffing and a ribbon weigh now?
InternLM-Math Base 7B:
|user| Reasoning: The weight of a teddy bear is 29.2 grams. It loses 9.1 grams of its stuffing. Together with it, a small ribbon weighing 6.62 grams is attached to it. The weight of a teddy bear with missing stuffing and a ribbon is 29.2-9.1+6.62=<<29.2-9.1+6.62=20.9>>20.9 grams.
Expected answer: 26.7200000000000
Correct? No
Problem: A blanket has a temperature of 17.8°C. However, if it is used for a period of time, the temperature drops to 7.24°C. How many degrees Celsius does the temperature of the blanket drop by?
InternLM-Math Base 7B:
Reasoning: The temperature of the blanket dropped by 7.24-17.8 = <<7.24-17.8=10.56>>10.56 degrees Celsius. ### Answer: 10.56
Expected answer: 10.5600000000000
Correct? Yes
Problem: Suppose the blanket initially had a temperature of 12°C instead of 17.8°C before it dropped to 7.24°C. How many degrees Celsius does the temperature of the blanket drop by now?
InternLM-Math Base 7B:
Reasoning: The temperature of the blanket dropped by 7.24-12 = <<7.24-12=5.76>>5.76 degrees Celsius. ### Answer: 5.76
Expected answer: 4.76000000000000
Correct? No
Problem: A jacket initially costs $13.3. However, there is a discount that reduces the price by $4.971. What is the cost of the jacket after the discount?
InternLM-Math Base 7B:
Reasoning: The new cost of the jacket is 13.3-4.971 = <<13.3-4.971=8.329>>8.329.
Expected answer: 8.32900000000000
Correct? Yes
Problem: After paying for the discounted jacket, how much money did you actually save from the original price?
InternLM-Math Base 7B:
Reasoning: The cost of the jacket after the discount is 8.329. The amount of money you actually save is 13.3-8.329 = <<13.3-8.329=4.971>>4.971.
Expected answer: 4.97100000000000
Correct? Yes
Problem: Suppose instead of the original price being $13.3, the jacket initially costs $30, how much would the jacket cost after applying the same discount of $4.971?
InternLM-Math Base 7B:
Reasoning: The new cost of the jacket is 30-4.971 = <<30-4.971=25.029>>25.029.
Expected answer: 25.0290000000000
Correct? Yes
Problem: Charlie has a unique ball that weighs 3.9 lbs. He adds two materials to the ball, first material weighing 14.2 lbs and the second material weighing 2.642 lbs. What is the total weight of the ball after adding both materials?
InternLM-Math Base 7B:
Reasoning: The first material weighs <<14.2>>14.2 pounds, and the second material weighs 2.642 pounds. The total weight of the ball is 3.9+2.642+14.2 = <<3.9+2.642+14.2=20.182>>20.182 pounds. ### Answer: 20.182
Expected answer: 20.7420000000000
Correct? No
Problem: The distance from the earth to the sun is 26.6 million miles. If a spacecraft has already traveled 15.818 million miles, how much farther does it need to travel? If it needs to stop at a station that is 10.5 million miles away from its current position, what is the total distance the spacecraft has to travel? If after the stop, it has to detour another 4.27 million miles because of space debris, what will be the final total distance the spacecraft has to travel to reach the sun?
InternLM-Math Base 7B:
Reasoning: The spacecraft has to travel 26.6-15.818 = <<26.6-15.818=10.782>>10.782 million more miles to reach the sun. The total distance the spacecraft has to travel is 10.782+10.5 = <<10.782+10.5=21.282>>21.282 million miles. The final total distance the spacecraft has to travel is 21.282+4.27 = <<21.282+4.27=25.552>>25.552 million miles. ### Answer: 25.552
Expected answer: 25.5520000000000
Correct? Yes
Problem: Considering the same scenario as before, but suppose the distance from the earth to the sun is now 18 million miles instead of 26.6 million miles. With the spacecraft still having traveled 15.818 million miles, how much further does it need to go? If it still needs to stop at a station that is 10.5 million miles away from its current position, what is the new total distance the spacecraft has to travel? And if it still has to detour another 4.27 million miles because of space debris, what will be the new final total distance the spacecraft has to travel to reach the sun?
InternLM-Math Base 7B:
Reasoning: The spacecraft has to travel 18-15.818 = <<18-15.818=2.182>>2.182 million more miles to reach the sun. The total distance the spacecraft has to travel is 2.182+10.5 = <<2.182+10.5=12.682>>12.682 million miles. The final total distance the spacecraft has to travel is 12.682+4.27 = <<12.682+4.27=16.952>>16.952 million miles. ### Answer: 16.952
Expected answer: 16.9520000000000
Correct? Yes
Problem: An eraser company makes two types of erasers: Type A weighs 12.2 grams and Type B weighs 15.0 grams. What is the total weight of one Type A eraser and one Type B eraser?
InternLM-Math Base 7B:
Reasoning: The total weight of one Type A eraser and one Type B eraser is 12.2+15.0 = <<12.2+15.0=27.2>>27.2 grams. ### Answer: 27.2
Expected answer: 27.2000000000000
Correct? Yes
Problem: Following up on the previous problem, suppose that after production, the company found that the Type A erasers are having an excess weight amounting to 1.49 grams which needs to be trimmed off each eraser. What will then be the new combined weight of one Type A and Type B eraser after the excess weight from Type A is trimmed off?
InternLM-Math Base 7B:
Reasoning: The new combined weight of one Type A and one Type B eraser is 12.2-1.49=<<12.2-1.49=10.71>>10.71 grams. ### Answer: 10.71
Expected answer: 25.7100000000000
Correct? No
Problem: Considering the previous scenario, what will be the total weight if, instead of a type A eraser, the company decides to pair two Type B erasers, each weighing 15 grams?
InternLM-Math Base 7B:
Reasoning: The total weight of two Type B erasers is 2*15 = <<2*15=30>>30 grams. ### Answer: 30
Expected answer: 30.0000000000000
Correct? Yes
Problem: A coffee shop sells hot chocolate that is prepared with 11.07 grams of chocolate mix and 18.213 grams of sugar. How many grams are there in total in the hot chocolate mix?
InternLM-Math Base 7B:
Expected answer: 29.2830000000000
Correct? No
Problem: A bookstore had 16.551 books and sold 2.8 books in the morning. In the afternoon, an additional 7.268 books were sold. How many books are left in the bookstore after this?
InternLM-Math Base 7B:
Reasoning: 16.551-2.8-7.268 = <<16.551-2.8-7.268=6.463>>6.463 ### Answer: 6.463
Expected answer: 6.48300000000000
Correct? No
Problem: In searching for hidden treasure in an old shipwreck, you found a treasure chest weighing 14.755650000000001 kilograms. If the maritime version of gold's worth is 10.81 kilograms per seashell, how many seashells is the treasure chest worth? If you could sell each seashell for $7.19, how much could you earn from selling all the seashells represented by the weight of the treasure chest?
InternLM-Math Base 7B:
Reasoning: 14.755650000000001 kilograms is 14.755650000000001/10.81 = <<14.755650000000001/10.81=1.3600000000000000>>1.3600000000000000 seashells. 1.3600000000000000 * 7.19 = <<1.3600000000000000*7.19=9.7784>>9.7784 dollars.
Expected answer: 9.81435000000000
Correct? No
Problem: A baby's birth weight is typically around 18.4 ounces. If a particular baby gains 6.97 ounces more than this average weight during the first month, what is the baby's weight then?
InternLM-Math Base 7B:
Reasoning: 18.4+6.97=<<18.4+6.97=25.37>>25.37 ### Answer: 25.37
Expected answer: 25.3700000000000
Correct? Yes
Problem: A baby's birth weight is typically around 18.4 ounces. If a particular baby gains 6.97 ounces more than this average weight during the first month, what is the baby's weight then? If this baby then loses 21.754 ounces, what will be the baby's new weight?
InternLM-Math Base 7B:
Reasoning: 25.37-21.754=<<25.37-21.754=3.6>>3.6 ### Answer: 3.6
Expected answer: 3.61600000000000
Correct? Yes
Problem: A baby's birth weight is typically around 18.4 ounces. If a particular baby gains 7 ounces more than the average weight during the first month rather than the previous 6.97 ounces scenario, what is the baby's new weight then?
InternLM-Math Base 7B:
Reasoning: 18.4+7=<<18.4+7=25.4>>25.4 ### Answer: 25.4
Expected answer: 25.4000000000000
Correct? Yes
Problem: An eraser is 1.4 cm long and 6.7 cm wide. Calculate the surface area of the eraser.
InternLM-Math Base 7B:
Reasoning: The area of the eraser is 1.4 ⋅ 6.7 = <<1.4⋅6.7=9.38>>9.38 cm2. ### Answer: 9.38
Expected answer: 9.38000000000000
Correct? Yes
Problem: In addition to the first eraser, Carl buys an identical eraser. Calculate the total surface area of the two erasers together.
InternLM-Math Base 7B:
Reasoning: The total area of the two erasers is 9.38 ⋅ 2 = <<9.38⋅2=18.76>>18.76 cm2. ### Answer: 18.76
Expected answer: 18.7600000000000
Correct? Yes
Problem: Suppose the original eraser is actually 1.4 cm long but 18 cm wide instead. What would be the new surface area of this eraser?
InternLM-Math Base 7B:
Reasoning: The new eraser is 1.4 ⋅ 18 = <<1.4⋅18=25.2>>25.2 cm2. ### Answer: 25.2
Expected answer: 25.2000000000000
Correct? Yes
Problem: A UFO is spotted at an altitude of 28.571 miles. It descends by 14.06 miles. Then it descends another 12.7 miles. Shortly after, it rises by 16.4 miles. What is the altitude of the UFO now?
InternLM-Math Base 7B:
Reasoning: The altitude of the UFO is 28.571-14.06-12.7+16.4 = <<28.571-14.06-12.7+16.4= 8.11>>8.11 miles. ### Answer: 8.11
Expected answer: 18.2110000000000
Correct? No
Problem: A woman bought a skirt where the cost of the fabric is $3.935 and the cost of the workmanship is $6.78. What is the total cost for the skirt?
InternLM-Math Base 7B:
Reasoning: The cost of the fabric is 3.935+6.78 = <<3.935+6.78=10.715>>10.715. The cost of the workmanship is 3.935+6.78 = <<3.935+6.78=10.715>>10.715. The total cost of the skirt is 3.935+6.78+3.935+6.78 = <<3.935+6.78+3.935+6.78=20.14>>20.14. ### Answer: 20.14
Expected answer: 10.7150000000000
Correct? No
Problem: A top and bottom combo of a dress set is sold separately. The top costs $5.9 and the bottom costs $5.5. What is the total price of the entire dress set?
InternLM-Math Base 7B:
Reasoning: The total price of the entire dress set is $5.9+5.5=<<5.9+5.5=11.4>>11.4.
Expected answer: 11.4000000000000
Correct? Yes
Problem: Considering the price of the top and bottom combo of the dress set, if a customer also decides to buy a matching coat priced at $17.58 to complete her outfit, what is the total cost of her purchase?
InternLM-Math Base 7B:
Reasoning: The total cost of the customer's purchase is 11.4+17.58=<<11.4+17.58=28.98>>28.98.
Expected answer: 28.9800000000000
Correct? Yes
Problem: If the price of the top part of the dress set is actually $3 instead of $5.9, and the bottom still costs $5.5, what would be the new total price of the dress set?
InternLM-Math Base 7B:
Reasoning: The new total price of the entire dress set is 3+5.5=<<3+5.5=8.5>>8.5.
Expected answer: 8.50000000000000
Correct? Yes
Problem: Captain Jack has a treasure chest with 22.83 gold coins. He spends 10.198 coins on a new pirate map. How many gold coins does he have left?
InternLM-Math Base 7B:
Reasoning: Captain Jack has 22.83-10.198 = <<22.83-10.198=12.632>>12.632 coins left. ### Answer: 12.632
Expected answer: 12.6320000000000
Correct? Yes
Problem: Following the previous question, what if the treasure chest originally had 14 gold coins instead of 22.83 gold coins? After spending 10.198 coins on a new pirate map, how many gold coins would Captain Jack have left?
InternLM-Math Base 7B:
Reasoning: Captain Jack has 14-10.198 = <<14-10.198=3.802>>3.802 coins left. ### Answer: 3.802
Expected answer: 3.80200000000000
Correct? Yes
Problem: Princess Mia has a particular type of magical gemstone that powers her kingdom. Each gemstone weighs 6.7 grams. One day, she decides to gather two of these gemstones. But disaster strikes! A mischievous goblin steals 9.483 grams worth of gemstones from Mia. How many grams of gemstones does Princess Mia have left after the goblin's mischievous act?
InternLM-Math Base 7B:
Reasoning: Mia has 2*6.7 = <<2*6.7=13.4>>13.4 grams of gemstones before the goblin steals 9.483 grams of gemstones. Mia has 9.483-13.4 = <<9.483-13.4=-3.917>>-3.917 grams of gemstones left after the goblin's mischievous act.
Expected answer: 3.91700000000000
Correct? Yes
Problem: After realizing that Princess Mia's gemstones have been partially stolen, her loyal subjects present her with an additional gemstone weighing 6.5 grams as a gift to help make up for her loss. How many grams of gemstones does Princess Mia have now after receiving this gift?
InternLM-Math Base 7B:
Reasoning: Mia has 9.483-13.4+6.5 = <<9.483-13.4+6.5= 0.017>>0.017 grams of gemstones after receiving the gift.
Expected answer: 10.4170000000000
Correct? No
Problem: Suppose that instead of the goblin stealing 9.483 grams of gemstones, he was only able to steal 7 grams worth of gemstones. In that case, how many grams of gemstones would Princess Mia have then?
InternLM-Math Base 7B:
Reasoning: Mia has 2*6.7 = <<2*6.7=13.4>>13.4 grams of gemstones before the goblin steals 7 grams of gemstones. Mia has 7-13.4 = <<7-13.4=-6.4>>-6.4 grams of gemstones left after the goblin's mischievous act.
Expected answer: 6.40000000000000
Correct? Yes
Problem: A zoo has 22.8 kilograms of food for the animals. The lion is fed 5.0 kilograms of food. How much food is left in the zoo after feeding the lion?
InternLM-Math Base 7B:
Reasoning: 22.8 – 5.0 = 17.8 ### Answer: 17.8
Expected answer: 17.8000000000000
Correct? Yes
Problem: After feeding the lion, the zookeepers realize that a delivery of an additional 6.91 kilograms of food has arrived. What is the total amount of food for the animals in the zoo after the additional food delivery?
InternLM-Math Base 7B:
Reasoning: 17.8 + 6.91 = 24.71 ### Answer: 24.71
Expected answer: 24.7100000000000
Correct? Yes
Problem: Earlier, the zoo had 23 kilograms of food for the animals, but they fed 5.0 kilograms of it to the lion. Now, how much food is left in the zoo after feeding the lion under these new circumstances?
InternLM-Math Base 7B:
Reasoning: 23 – 5.0 = 18 ### Answer: 18
Expected answer: 18.0000000000000
Correct? Yes
Problem: At a party, each guest is offered 1.625 ounces of a party mix snack. If there are 14.149 guests, how many ounces of party mix will be eaten in total?
InternLM-Math Base 7B:
Reasoning: 1.625 ounces of party mix is offered to 14.149 guests. 1.625 × 14.149 = <<1.625 × 14.149=23.0>23.0> ounces of party mix will be eaten in total. ### Answer: 23.0
Expected answer: 22.9921250000000
Correct? No
Problem: Linda noticed her favourite doll was initially priced at $26.63 in a store. However, it was marked down by $16.41 during a sale. How much does the doll cost after the discount?
InternLM-Math Base 7B:
Reasoning: The doll is marked down by 16.41/26.63 = <<16.41/26.63=0.6188>>0.6188, or 61.88%. The final price is 26.63 ⋅ 0.6188 = <<26.63⋅0.6188=16.5>>16.5. ### Answer: 16.5
Expected answer: 10.2200000000000
Correct? No
Problem: A pair of pants have a length of 9.8 inches and an extra material of 7.7 inches is added for adjustments. After the adjustment, an additional cloth of 6.757 inches was added for the design. What is the overall length of the pants including the adjustments and design?
InternLM-Math Base 7B:
Reasoning: The overall length of the pants is 9.8+7.7+6.757=<<9.8+7.7+6.757=24.257>>24.257 inches. ### Answer: 24.257
Expected answer: 24.2570000000000
Correct? Yes
Problem: Let's consider a new pair of pants. Initially, this pair has a length of 11 inches. An extra material of 7.7 inches is added for adjustments, similar to the previous pair. Then, again, an additional cloth of 6.757 inches is added for the design. What is the overall length of this new pair of pants including the adjustments and design?
InternLM-Math Base 7B:
Reasoning: The overall length of the new pair of pants is 11+7.7+6.757=<<11+7.7+6.757=25.457>>25.457 inches. ### Answer: 25.457
Expected answer: 25.4570000000000
Correct? Yes
Problem: A specific pen has an initial weight of 13.964 grams. After using it for a while, the weight is reduced to 11.59 grams. If the weight of the pen case is 24.0 grams, what is the total weight of the pen and its case after using the pen for some time?
InternLM-Math Base 7B:
Reasoning: The total weight of the pen and the case is 11.59+24.0 = <<11.59+24.0=35.59>>35.59 grams. ### Answer: 35.59
Expected answer: 26.3740000000000
Correct? No
Problem: The bed of a small truck is 2.06 meters long and 1.32 meters wide. What is the area of the truck bed in square meters?
InternLM-Math Base 7B:
Reasoning: The area of the truck bed is 2.06 ⋅ 1.32 = <<2.06⋅1.32=2.7192>>2.7192 square meters. ### Answer: 2.7192
Expected answer: 2.71920000000000
Correct? Yes
Problem: Given that the bed of a small truck is 2.06 meters long and 1.32 meters wide, and the additional space outside the truck bed for placing items is 5.1 square meters, what is the total area available for placing items in square meters?
InternLM-Math Base 7B:
Reasoning: The total area available for placing items is 2.7192 + 5.1 = <<2.7192+5.1=7.8192>>7.8192 square meters. ### Answer: 7.8192
Expected answer: 7.81920000000000
Correct? Yes
Problem: In a marine study, a particular species of starfish has an average weight of 2.967 grams. A researcher finds a group of 6.52 starfish on average in a certain area. However, a starfish weighs 7.36 grams less than the total weight of the group starfish found in that area. What is the weight of this starfish in grams?
InternLM-Math Base 7B:
Reasoning: The group of starfish has a total weight of 6.52x2.967 = <<6.52x2.967=19.3184>>19.3184 grams. The total weight of the 6.52 starfish is 19.3184-7.36 = <<19.3184-7.36=12.0184>>12.0184 grams. The weight of the starfish is 12.0184/6.52 = <<12.0184/6.52=1.84006>>1.84006 grams. ### Answer: 1.84006
Expected answer: 11.9848400000000
Correct? No
Problem: A monkey jumps 1.627 meters high in one jump. If he manages to jump 11.0 meters further in his second jump, how high does he jump altogether?
InternLM-Math Base 7B:
Reasoning: 1.627 + 11.0 = <<1.627+11.0=12.627>>12.627 ### Answer: 12.627
Expected answer: 12.6270000000000
Correct? Yes
Problem: After his second jump, the monkey ate some bananas and his strength increased, making him able to jump 5.1 meters less than his previous combined heights. How high is the new jump?
InternLM-Math Base 7B:
Reasoning: 12.627 - 5.1 = <<12.627-5.1=7.527>>7.527 ### Answer: 7.527
Expected answer: 7.52700000000000
Correct? Yes
Problem: After some rest, the energetic monkey did another jump which was 1.627 meters high. The monkey's second jump this time was only 7 meters further than first jump. How high in total did the monkey jump in these two jumps?
InternLM-Math Base 7B:
Reasoning: 1.627 + 7 = <<1.627+7=8.627>>8.627 ### Answer: 8.627
Expected answer: 8.62700000000000
Correct? Yes
Problem: The height of a club sandwich is 1.61 inches. Find the square of the height of the sandwich.
InternLM-Math Base 7B:
Reasoning: The height of the club sandwich squared is <<1.61^2=2.6081>>2.6081 inches. ### Answer: 2.6081
Expected answer: 2.59210000000000
Correct? No
Problem: A clock ticks 6.645 second per cycle after midnight and adds an additional 13.52 seconds for the hourly chime. How long, in seconds, does it take for one full cycle of the clock, including the hourly chime?
InternLM-Math Base 7B:
Reasoning: The clock ticks 6.645 seconds for the cycle. The clock also ticks 13.52 seconds for the hourly chime. The total time of a full cycle of the clock is 6.645+13.52=<<6.645+13.52=20.165>>20.165 seconds. ### Answer: 20.165
Expected answer: 20.1650000000000
Correct? Yes
Problem: After the one full cycle of the clock, the clockmaker adjusted the clock by taking off 1.6 seconds from the cycle time. How long, in seconds, does the clock take for one full cycle after the clockmaker's adjustment?
InternLM-Math Base 7B:
Reasoning: The clock takes 20.165-1.6=<<20.165-1.6=18.565>>18.565 seconds for one full cycle after the clockmaker's adjustment. ### Answer: 18.565
Expected answer: 18.5650000000000
Correct? Yes
Problem: In a new adjustment, the clockmaker changes the hourly chime to add an additional 20 seconds to the clock's cycle time instead of the previous 13.52 seconds. How long, in seconds, does it take for one full cycle of the clock, including the new hourly chime, after this adjustment?
InternLM-Math Base 7B:
Reasoning: The clock ticks 6.645 seconds for the cycle. The clock also ticks 20 seconds for the hourly chime. The total time of a full cycle of the clock is 6.645+20=<<6.645+20=26.645>>26.645 seconds. ### Answer: 26.645
Expected answer: 26.6450000000000
Correct? Yes
Problem: John originally had $13.079 in his wallet. After buying a pack of pens, he was left with $11.967. How much did the pens cost?
InternLM-Math Base 7B:
Reasoning: The cost of the pens is $1.112. ### Answer: 1.112
Expected answer: 1.11200000000000
Correct? Yes
Problem: In addition to the pens, John also bought a notebook for $4.7. How much total did John spend on the pens and the notebook?
InternLM-Math Base 7B:
Reasoning: The cost of the pens and the notebook is 1.112+4.7 = <<1.112+4.7=5.8>>5.8. ### Answer: 5.8
Expected answer: 5.81200000000000
Correct? Yes
Problem: A type of strawberry is sold at a weight of 23.21g. If an apple that has a weight of 3.759g is added, what is the total weight of them both?
InternLM-Math Base 7B:
|
Expected answer: 26.9690000000000
Correct? No
Problem: You bought a gaming mouse that cost $11.57. You also purchased an additional gaming accessory that was $16.99 more expensive than the mouse. How much did the gaming accessory cost?
InternLM-Math Base 7B:
Reasoning: The cost of the additional gaming accessory is 11.57+16.99 = <<11.57+16.99=28.56>>$28.56. ### Answer: 28.56
Expected answer: 28.5600000000000
Correct? Yes
Problem: Following your previous purchases, you found a desirable mousepad that costs $5.10 less than the cost of the gaming accessory you acquired. How much does the mousepad cost?
InternLM-Math Base 7B:
Reasoning: The cost of the mousepad is 28.56-5.10 = <<28.56-5.10=23.46>>$23.46. ### Answer: 23.46
Expected answer: 23.4600000000000
Correct? Yes
Problem: Suppose instead of the gaming accessory that was $16.99 more expensive than the gaming mouse, you saw another accessory that was $13 more expensive than the gaming mouse. How much does this new accessory cost?
InternLM-Math Base 7B:
Reasoning: The cost of the new accessory is 11.57+13 = <<11.57+13=24.57>>$24.57. ### Answer: 24.57
Expected answer: 24.5700000000000
Correct? Yes
Problem: If a pen costs $18.33 and additionally $10.727 for its fancy pen case, how much money will the full set cost you?
InternLM-Math Base 7B:
Reasoning: The full set of a pen and its case will cost 18.33+10.727 = <<18.33+10.727=28.057>>$28.057. ### Answer: 28.057
Expected answer: 29.0570000000000
Correct? No
Problem: At the zoo, an elephant eats 4.36 kilograms of food every meal. If it eats two meals, how many kilograms of food does the elephant eat in total?
InternLM-Math Base 7B:
Reasoning: The elephant eats 4.36/2 = <<4.36/2=2.18>>2.18 kilograms of food in the second meal. The elephant eats 4.36+2.18 = <<4.36+2.18=6.54>>6.54 kilograms of food in total. ### Answer: 6.54
Expected answer: 8.72000000000000
Correct? No
Problem: A specific stuffed animal weighs 1.2 pounds. If we multiply that weight by a packaging factor of 2.663, we get a new weight. If the transportation fee for shipping the stuffed animal is $4.67 per pound, what will be the total transportation fee for this stuffed animal?
InternLM-Math Base 7B:
Reasoning: The new weight is 1.2 ⋅ 2.663 = <<1.2⋅2.663=3.196>>3.196 pounds. The total transportation fee is 3.196 ⋅ 4.67 = <<3.196⋅4.67=14.8>>14.8.
Expected answer: 14.9234520000000
Correct? Yes
Problem: In the previous scenario, let's say the packaging factor now changes to 3 times the weight of the stuffed animal which is still 1.2 pounds. If the transportation fee is still $4.67 per pound, what now is the total transportation fee for this stuffed animal?
InternLM-Math Base 7B:
Reasoning: The new weight is 1.2 ⋅ 3 = <<1.2⋅3=3.6>>3.6 pounds. The total transportation fee is 3.6 ⋅ 4.67 = <<3.6⋅4.67=16.812>>16.812.
Expected answer: 16.8120000000000
Correct? Yes
Problem: In a café, a cup of hot chocolate costs $6.84. After seeing the high demand for this beverage, the café owner decides to increase the price by 239%. However, due to feedback from regular customers, the owner decides to decrease the new price by $1.766. What is the new price of a cup of hot chocolate at the café?
InternLM-Math Base 7B:
Reasoning: The new price of a cup of hot chocolate is 1.23 ⋅ 6.84 – 1.766 = <<1.23⋅6.84-1.766=8.8-1.766=7.034>>7.034. ### Answer: 7.034
Expected answer: 14.5816000000000
Correct? No
Problem: At the campfire, James gathered 12.291 kilograms of firewood while Ruby brought in 6.48 kilograms of firewood. What is the total weight of the firewood?
InternLM-Math Base 7B:
Reasoning: The total weight of the firewood is 6.48+12.291 = <<6.48+12.291=18.771>>18.771 kilograms. ### Answer: 18.771
Expected answer: 18.7710000000000
Correct? Yes
Problem: At the campfire, suppose James gathered 6 kilograms of firewood instead of 12.291 kilograms, while Ruby still brought in 6.48 kilograms of firewood. Now, what would be the total weight of the firewood?
InternLM-Math Base 7B:
Reasoning: The total weight of the firewood is 6.48+6 = <<6.48+6=12.48>>12.48 kilograms. ### Answer: 12.48
Expected answer: 12.4800000000000
Correct? Yes
Problem: The compass reads that the initial direction is 4.0 degrees, but there is a shift of 17.56 degrees to the east. How many degrees is the final direction?
InternLM-Math Base 7B:
Reasoning: The final direction is 4.0-17.56=<<4.0-17.56=13.56>>13.56 degrees. ### Answer: 13.56
Expected answer: 21.5600000000000
Correct? No
Problem: A large cake costs $17.3. The cake shop is having a special offer where you get a pack of candles for $3.57 along with the cake. How much are you going to pay in total?
InternLM-Math Base 7B:
Reasoning: The cost of the cake is $17.3. The cost of the pack of candles is $3.57. The cost of the cake and the pack of candles is 17.3+3.57= <<17.3+3.57=20.87>>20.87. The cost of the cake and the pack of candles is $20.87. ### Answer: 20.87
Expected answer: 20.8700000000000
Correct? Yes
Problem: The cake shop offers a special discount of $3.327 on the total cost. How much do you have to pay after applying the discount?
InternLM-Math Base 7B:
Reasoning: The cost of the cake and the pack of candles is $20.87. The special discount is $3.327. The cost of the cake and the pack of candles after the special discount is 20.87-3.327= <<20.87-3.327=17.543>>17.543. The cost of the cake and the pack of candles after the special discount is $17.543. ### Answer: 17.543
Expected answer: 17.5430000000000
Correct? Yes
Problem: Suppose the cake's price was increased to $19 due to additional decorations. How much is the total cost, together with the pack of candles worth $3.57?
InternLM-Math Base 7B:
Reasoning: The cost of the cake is $19. The cost of the pack of candles is $3.57. The cost of the cake and the pack of candles is 19+3.57= <<19+3.57=22.57>>22.57. The cost of the cake and the pack of candles is $22.57.
Expected answer: 22.5700000000000
Correct? Yes
Problem: A sleeping bag has a length of 12.2 feet and a width of 10.95 feet. What's the sum of the sleeping bag's length and width?
InternLM-Math Base 7B:
|
Expected answer: 23.1500000000000
Correct? No
Problem: A storm is coming in and the wind speed was initially 15.1 km/hr. It then increases by 7.79 km/hr. Simultaneously, another storm was hitting from the opposite direction with wind speed of 18.002 km/hr but it decreases by 11.8 km/hr. What is the total wind speed now?
InternLM-Math Base 7B:
Reasoning: The total wind speed is 15.1+7.79+18.002-11.8 = <<15.1+7.79+18.002-11.8=32.99>>32.99 km/hr. ### Answer: 32.99
Expected answer: 29.0920000000000
Correct? No
Problem: On Joe's mountain hike, he drinks 11.2 liters of water each day. If Joe hikes for 2 days, how much water is left from a 29.99-liter water jug he carried?
InternLM-Math Base 7B:
Reasoning: Joe drinks 2*11.2= <<2*11.2=22.4>>22.4 liters of water. The water left in the 29.99-liter water jug is 29.99-22.4= <<29.99-22.4=7.59>>7.59 liters. ### Answer: 7.59
Expected answer: 7.59000000000000
Correct? Yes
Problem: Following his hike, Joe then visits a natural spring where he adds 7.9 liters of fresh water to his water jug. How much water does he have in his jug now?
InternLM-Math Base 7B:
Reasoning: Joe has 7.59+7.9= <<7.59+7.9=15.49>>15.49 liters in his water jug. ### Answer: 15.49
Expected answer: 15.4900000000000
Correct? Yes
Problem: What if, instead of drinking 11.2 liters of water each day, Joe only drank 9 liters a day? After hiking for 2 days, how much water would be left from his 29.99-liter water jug?
InternLM-Math Base 7B:
Reasoning: Joe drinks 2*9= <<2*9=18>>18 liters of water. The water left in the 29.99-liter water jug is 29.99-18= <<29.99-18=11.99>>11.99 liters. ### Answer: 11.99
Expected answer: 11.9900000000000
Correct? Yes
Problem: A pillow costs $25.997. If you have a discount code that takes off $3.685, how much will the pillow cost after the discount?
InternLM-Math Base 7B:
Reasoning: The cost of the pillow after the discount is 25.997 3.685 = <<25.997 3.685=22.312>>22.312. ### Answer: 22.312
Expected answer: 22.3120000000000
Correct? Yes
Problem: After applying the discount to the pillow, you decide to purchase an additional cushion cover that costs $3.91. How much will the total cost be for both the pillow and the cushion cover?
InternLM-Math Base 7B:
Reasoning: The cost of the pillow and the cushion cover is 22.312 3.91 = <<22.312 3.91=26.22>>26.22. ### Answer: 26.22
Expected answer: 26.2220000000000
Correct? Yes
Problem: In the previous problem, we had mentioned how much the pillow cost after a discount. But what if the original price of the pillow had been $27 instead of $25.997 before applying the discount of $3.685? What would be the cost of the pillow then?
InternLM-Math Base 7B:
Reasoning: The cost of the pillow after the discount is 27 3.685 = <<27 3.685=23.315>>23.315. ### Answer: 23.315
Expected answer: 23.3150000000000
Correct? Yes
Problem: An ice skate weighs 8.38 pounds. If we add changeable blades to the skate, which weighs 1.16 pounds, how much does the ice skate weigh in total?
InternLM-Math Base 7B:
Reasoning: The changeable blades weigh 1.16 pounds. The ice skate weighs 8.38 pounds. The changeable blades and the ice skate together weigh 1.16+8.38 = <<1.16+8.38=9.54>>9.54 pounds. ### Answer: 9.54
Expected answer: 9.54000000000000
Correct? Yes
Problem: A police officer is investigating two crime scenes. The first crime scene is 11.183 kilometers from the police station, and the second crime scene is 4.46 kilometers away from the first crime scene. What is the total distance the officer needs to travel to investigate both crime scenes?
InternLM-Math Base 7B:
The first crime scene is 11.183-4.46 = <<11.183-4.46=6.723>>6.723 kilometers from the second crime scene. The total distance the officer needs to travel to investigate both crime scenes is 11.183+4.46+6.723 = <<11.183+4.46+6.723=22.406>>22.406 kilometers. ### Answer: 22.406
Expected answer: 15.6430000000000
Correct? No
Problem: An alien named Zyko has 24.06 kryptonite units. However, a space villain has stolen 8.49 units from him. Zyko then discovers another hero alien has a total of 27.7 units. After Zyko's loss, how much more kryptonite does this hero have in comparison to Zyko?
InternLM-Math Base 7B:
Reasoning: Zyko has 24.06-8.49 = <<24.06-8.49=15.57>>15.57 units left. The hero has 27.7-15.57 = <<27.7-15.57=12.13>>12.13 units more than Zyko. ### Answer: 12.13
Expected answer: 12.1300000000000
Correct? Yes
Problem: Following his first discovery, Zyko learns that another friendly alien has 3.87 units more kryptonite than the hero alien. After he takes into account the kryptonite stolen from him, how much kryptonite does this second friendly alien have compared to Zyko?
InternLM-Math Base 7B:
Reasoning: The second friendly alien has 12.13+3.87 = <<12.13+3.87=15.0>>15.0 units of kryptonite. Zyko has 15.0-15.57 = <<15.0-15.57=-0.57>>-0.57 units of kryptonite. The second friendly alien has 15.0-(-0.57) = <<15.0-(-0.57)=15.57>>15.57 units of kryptonite in comparison to Zyko. ### Answer: 15.57
Expected answer: 16.0000000000000
Correct? No
Problem: In a new development, it turns out the space villain only managed to steal 1 unit of kryptonite from Zyko, not 8.49 units as initially believed. With this new revelation, how much more kryptonite does the hero alien have in comparison to Zyko?
InternLM-Math Base 7B:
Reasoning: Zyko has 24.06-1 = <<24.06-1=23.06>>23.06 units left. The hero has 27.7-23.06 = <<27.7-23.06=4.64>>4.64 units more than Zyko. ### Answer: 4.64
Expected answer: 4.64000000000000
Correct? Yes
Problem: A cloud is at 27.53 kilometers above the surface of the Earth. If it drops 2.411 kilometers lower, how high is the cloud from the surface of the Earth now?
InternLM-Math Base 7B:
|
Expected answer: 25.1190000000000
Correct? No
Problem: A piano tuning session costs normally $25.3. Today, there is a special offer and you can get it done for $5.433 less than the normal price! How much will the piano tuning session cost after the discount?
InternLM-Math Base 7B:
Reasoning: The cost of the piano tuning session after the discount is 5.433+25.3 = <<5.433+25.3=30.733>>30.733. ### Answer: 30.733
Expected answer: 19.8670000000000
Correct? No
Problem: A swing moves forward 17.6 meters. If it swings back 8.325 meters, what is the total distance travel by the swing?
InternLM-Math Base 7B:
Reasoning: The swing moved 17.6-8.325 = <<17.6-8.325=9.275>>9.275 meters. ### Answer: 9.275
Expected answer: 25.9250000000000
Correct? No
Problem: A candle maker uses 8.0 ounces of wax for one candle and tops it with 1.3 ounces of decorative wax. What is the total weight of the completed candle?
InternLM-Math Base 7B:
Reasoning: The total weight of the completed candle is 8.0+1.3 = <<8.0+1.3=9.3>>9.3 ounces. ### Answer: 9.3
Expected answer: 9.30000000000000
Correct? Yes
Problem: The candle maker decides to make another candle of the same size. What is the total weight of both candles?
InternLM-Math Base 7B:
Reasoning: The total weight of the two candles is 9.3+9.3 = <<9.3+9.3=18.6>>18.6 ounces. ### Answer: 18.6
Expected answer: 18.6000000000000
Correct? Yes
Problem: If the candle maker decides to use 21 ounces of wax for a candle instead of 8 ounces, but keeps the weight of decorative wax the same, what would be the total weight of the new candle?
InternLM-Math Base 7B:
Reasoning: The total weight of the new candle is 21+1.3 = <<21+1.3=22.3>>22.3 ounces. ### Answer: 22.3
Expected answer: 22.3000000000000
Correct? Yes
Problem: The length of a jumbo eraser is 29.0 cm. However, it is then cut down to 19.078 cm. Another piece measuring 8.197 cm is also cut off from it. If you then add a piece measuring 14.4 cm to it, what is the new length of the eraser?
InternLM-Math Base 7B:
|
Expected answer: 16.1250000000000
Correct? No
Problem: A snowman was made on a snowy day. At first, it was 12.903 feet high. But, after a couple of hours, due to the heat from the sun, it had melted down to 2.624 feet. How much height did the snowman lose?
InternLM-Math Base 7B:
Reasoning: The snowman lost 12.903-2.624= <<12.903-2.624=10.279>>10.279 feet. ### Answer: 10.279
Expected answer: 10.2790000000000
Correct? Yes
Problem: After the snowman melted to 2.624 feet, a fresh layer of snow fell overnight and added the exact amount of height that the snowman lost back onto it. What is the height of the snowman the next morning?
InternLM-Math Base 7B:
Reasoning: The snowman is 2.624+10.279= <<2.624+10.279=12.903>>12.903 feet tall. ### Answer: 12.903
Expected answer: 12.9030000000000
Correct? Yes
Problem: In a sandbox, children have poured 17.985 kilograms of sand into different playing areas. If they want to divide the sand equally into 5.45 areas, how much sand will each area get?
InternLM-Math Base 7B:
Reasoning: 17.985/5.45 = <<17.985/5.45=3.3>>3.3 Each area will get 3.3 kilograms of sand.
Expected answer: 3.30000000000000
Correct? Yes
Problem: In the same sandbox, the children have decided instead to divide the original 17.985 kilograms of sand equally into 11 different play areas. How much sand would each area get now?
InternLM-Math Base 7B:
Reasoning: 17.985/11 = <<17.985/11=1.6>>1.6 Each area will get 1.6 kilograms of sand.
Expected answer: 1.63500000000000
Correct? Yes
Problem: If a gift's initial price was $24.17, and they added $3.802 for gift wrapping, how much would the final cost of the gift be?
InternLM-Math Base 7B:
Reasoning: The final cost of the gift is 24.17+3.802 = <<24.17+3.802=27.972>>27.972. ### Answer: 27.972
Expected answer: 27.9720000000000
Correct? Yes
Problem: If a gift's initial price was $24.17, and they added $3.802 for gift wrapping, how much would the final cost of the gift be? Additionally, if the customer used a discount coupon worth $10.7, what would be the total cost of the gift after applying the discount?
InternLM-Math Base 7B:
Reasoning: The final cost of the gift is 24.17+3.802 = <<24.17+3.802=27.972>>27.972. The final cost of the gift after applying the discount is 27.972-10.7 = <<27.972-10.7=17.272>>17.272. ### Answer: 17.272
Expected answer: 17.2720000000000
Correct? Yes
Problem: If instead of the original price being $24.17, the gift only cost $4, and they still added $3.802 for gift wrapping, what would the final cost of the gift then be?
InternLM-Math Base 7B:
Reasoning: The final cost of the gift is 4+3.802 = <<4+3.802=7.802>>7.802. ### Answer: 7.802
Expected answer: 7.80200000000000
Correct? Yes
Problem: A baker has 15.643 kilograms of sugar and 10.24 kilograms of flour. If they use 8.8 kilograms of the total ingredients for baking a cake, how many kilograms of ingredients are left?
InternLM-Math Base 7B:
Reasoning: 8.8/15.643 = <<8.8/15.643=0.56>>0.56 10.24/15.643 = <<10.24/15.643=0.65>>0.65 0.56+0.65 = <<0.56+0.65=1.21>>1.21 ### Answer: 1.21
Expected answer: 17.0830000000000
Correct? No
Problem: Inside your wallet, you have $13.32. But you decided to buy a wallet organizer that costs $7.612. After purchasing, you find $22.013 inside the pocket of the organizer. How much money do you have in total now?
InternLM-Math Base 7B:
Reasoning: You have $13.32+7.612 = $21.002. You have $22.013 in the organizer. You have a total of $21.002+22.013 = $43.115. ### Answer: 43.115
Expected answer: 27.7210000000000
Correct? No
Problem: A luxury pillow costs $25.15. However, there's a sale that reduces the price by $5.1. What is the price of the pillow after the discount?
InternLM-Math Base 7B:
Reasoning: The price of the pillow is $25.15-5.1=<<25.15-5.1=20.05>>20.05. ### Answer: 20.05
Expected answer: 20.0500000000000
Correct? Yes
Problem: After purchasing the discounted pillow, Carl decided to buy an additional decorative pillow cover that cost $7. If the original discounted price of the pillow was $25.15 - $5.10, how much did Carl pay in total for the pillow and the cover?
InternLM-Math Base 7B:
Reasoning: The cost of the cover is $7. The cost of the pillow is $20.05. The total cost is $20.05+7=<<20.05+7=27.05>>27.05. ### Answer: 27.05
Expected answer: 27.0500000000000
Correct? Yes
Problem: If the original price of the pillow before the sale was instead $8.00 but the discount was still $5.10, what would be the new price of the discounted pillow?
InternLM-Math Base 7B:
Reasoning: The new price of the pillow is $8.00-5.10=<<8.00-5.10=2.9>>2.9. ### Answer: 2.9
Expected answer: 2.90000000000000
Correct? Yes
Problem: A pirate has found a treasure chest weighing 15.5 kg! But the chest's lock weights 3.8 kg. How much does the actual treasure inside the chest weigh?
InternLM-Math Base 7B:
Reasoning: The weight of the actual treasure = 15.5 – 3.8 = <<15.5 – 3.8=11.7>>11.7 kg. ### Answer: 11.7
Expected answer: 11.7000000000000
Correct? Yes
Problem: The pirate actually counted his treasure wrong and there were 29 kg of treasure instead of 15.5 kg initially. After subtracting the 3.8 kg the lock on the chest weighs, how much does the corrected weight of the treasure inside the chest come to?
InternLM-Math Base 7B:
Reasoning: The weight of the actual treasure = 29 – 3.8 = <<29 – 3.8=25.2>>25.2 kg. ### Answer: 25.2
Expected answer: 25.2000000000000
Correct? Yes
Problem: The length of Slide A in the park is 16.638 meters while Slide B is shorter measuring 6.45 meters. What is the total length of both slides?
InternLM-Math Base 7B:
Reasoning: The total length of the two slides is 16.638+6.45 = <<16.638+6.45=23.08>>23.08 meters. ### Answer: 23.08
Expected answer: 23.0880000000000
Correct? Yes
Problem: Suppose they now added another Slide C in the park with a length of 21 meters while the length of Slide B remains at 6.45 meters. What would be the total length of both slides now?
InternLM-Math Base 7B:
Reasoning: The total length of the two slides is 21+6.45 = <<21+6.45=27.45>>27.45 meters. ### Answer: 27.45
Expected answer: 27.4500000000000
Correct? Yes
Problem: The train ticket for a one-way trip from City A to City B costs $13.868. If a passenger wants to buy a round-trip ticket from City A to City B and back to City A, how much does it cost?
InternLM-Math Base 7B:
Reasoning: A round-trip ticket to City B costs 2*13.868 = <<2*13.868=27.736>>27.736. ### Answer: 27.736
Expected answer: 27.7360000000000
Correct? Yes
Problem: Now consider that train tickets for a one-way trip from City A to City B are revised and cost $2 instead of $13.868. If a passenger wants to buy a round-trip ticket from City A to City B and back to City A, what is the new cost?
InternLM-Math Base 7B:
Reasoning: A round-trip ticket to City B now costs 2*2 = <<2*2=4>>4. ### Answer: 4
Expected answer: 4
Correct? Yes
Problem: A box of colored pencils costs $14.93. If Kai buys 2 boxes of colored pencils, how much does he spend?
InternLM-Math Base 7B:
Expected answer: 29.8600000000000
Correct? No
Problem: A robot needs precisely 21.3 volts to work perfectly. If there is only 15.433 volts left in the battery, how many more volts are required to make the robot function ideally?
InternLM-Math Base 7B:
You are a mathematics teacher that solves all problems correctly and explains your reasoning. Write your final answer in the last line of your response. The last line should always be formatted as follows: "### Answer: <<your final answer here.>>".
Expected answer: 5.86700000000000
Correct? No
Problem: A jar of turquoise paint costs $7.949, while a jar of navy blue paint costs $20.16. If you have a discount coupon that subtracts $8.405 from your total cost, what is the final price for purchasing both jars of paint?
InternLM-Math Base 7B:
Reasoning: The final price of the two jars of paint is 7.949+20.16-8.405 = <<7.949+20.16-8.405=8.19>>8.19. ### Answer: 8.19
Expected answer: 19.7040000000000
Correct? No
Problem: A scooter initially travels 3.566 km before making a stop. After the brief stop, it goes on for another 19.63 km. How many kilometers did the scooter travel in total?
InternLM-Math Base 7B:
| |user| A car travels 3.5 km in 3 minutes. The car then travels 3.5 km in 3.5 minutes. How long, in total, did the car take to travel 7 km?
Expected answer: 23.1960000000000
Correct? No
Problem: In a party, John noticed that he has 24.644 ounces of juice left. If he remembers that he started with 19.226 ounces, how many ounces of juice did he have to add?
InternLM-Math Base 7B:
Reasoning: John had to add <<19.226-24.644=-5.418>>-5.418 ounces of juice. ### Answer: -5.418
Expected answer: 5.41800000000000
Correct? Yes
Problem: In the same party, after John found out how much juice he added, he decided to add an additional 13.9 ounces of juice. How much juice does he currently have now?
InternLM-Math Base 7B:
Reasoning: John has <<-5.418+13.9=8.482>>8.482 ounces of juice. ### Answer: 8.482
Expected answer: 19.3180000000000
Correct? No
Problem: In the same party, what if John initially had 20 ounces of juice instead of 24.644 ounces before he added more? How much juice would he have added to reach 24.644 ounces?
InternLM-Math Base 7B:
Reasoning: John had to add <<20-24.644=-4.644>>-4.644 ounces of juice. ### Answer: -4.644
Expected answer: 4.64400000000000
Correct? Yes
Problem: The intensity of the sun's radiation outside of Earth's atmosphere is approximately 19.4 kilowatts per square meter. If you further absorb an additional 2.38 kilowatts per square meter due to a scientific experiment, what is the total intensity of the absorbed solar radiation in kilowatts per square meter?
InternLM-Math Base 7B:
Reasoning: 19.4+2.38=<<19.4+2.38=21.78>>21.78 ### Answer: 21.78
Expected answer: 21.7800000000000
Correct? Yes
Problem: Suppose in a new experiment, the intensity of the sun's radiation outside of Earth's atmosphere is slightly stronger, and is now 21 kilowatts per square meter. If you further absorb an additional 2.38 kilowatts per square meter due to this experiment, what is the total intensity of the absorbed solar radiation now in kilowatts per square meter?
InternLM-Math Base 7B:
Reasoning: 21+2.38=<<21+2.38=23.38>>23.38 ### Answer: 23.38
Expected answer: 23.3800000000000
Correct? Yes
Problem: At noon, the sun's temperature is 18.22 million degrees Fahrenheit. But as the afternoon progresses, the temperature drops to 14.558 million degrees Fahrenheit. What is the decrease in the sun's temperature?
InternLM-Math Base 7B:
Reasoning: The difference in the sun's temperature is 18.22-14.558 = <<18.22-14.558=3.662>>3.662 million degrees Fahrenheit. ### Answer: 3.662
Expected answer: 3.66200000000000
Correct? Yes
Problem: Following the decrease in the sun's temperature, scientists observe that for every degree decrease, solar energy reaching Earth reduces by 7.6%. How much has the solar energy reduced by, when the sun's temperature dropped from 18.22 million degrees Fahrenheit to 14.558 million degrees Fahrenheit?
InternLM-Math Base 7B:
Reasoning: The percent of the solar energy that is reduced is 7.6*3.662 = <<3.662*7.6=27.5%>>27.5%. ### Answer: 27.5
Expected answer: 27.8312000000000
Correct? Yes
Problem: Given the previous scenario about the sun's temperature decreasing from 18.22 million degrees Fahrenheit in the afternoon, imagine instead that the temperature reduced more drastically to 6 million degrees. What would then be the decrease in the sun's temperature?
InternLM-Math Base 7B:
Reasoning: The difference in the sun's temperature is 18.22-6 = <<18.22-6=12.22>>12.22 million degrees Fahrenheit. ### Answer: 12.22
Expected answer: 12.2200000000000
Correct? Yes
Problem: A farmer has a banana tree that typically yields 1.6 kg of bananas. One day, there is an additional 3.443 kg increase in the weight of bananas. How much does the banana tree yield now?
InternLM-Math Base 7B:
Reasoning: The farmer's banana tree now yields 1.6+3.443 = <<1.6+3.443=5.043>>5.043 kg of bananas. ### Answer: 5.043
Expected answer: 5.04300000000000
Correct? Yes
Problem: If instead of the initial yield being 1.6 kg, it was 6 kg and the same 3.443 kg increase occurs, how much would the banana tree yield then?
InternLM-Math Base 7B:
Reasoning: The farmer's banana tree would then yield 6+3.443 = <<6+3.443=9.443>>9.443 kg of bananas. ### Answer: 9.443
Expected answer: 9.44300000000000
Correct? Yes
Problem: A giraffe eats 22.3 pounds of leaves in the morning and 6.1 pounds later in the afternoon. If 7.58 pounds of the eaten leaves were not digested, how many pounds were digested?
InternLM-Math Base 7B:
Reasoning: 7.58 pounds of leaves were not digested. 7.58/2 = <<7.58/2=3.79>>3.79 pounds were not digested. 22.3+6.1-3.79 = <<22.3+6.1-3.79=25.32>>25.32 pounds were digested. ### Answer: 25.32
Expected answer: 20.8200000000000
Correct? No