InternLM-Math Base 7B on 4.MD.A.3
Common Core Standard 4.MD.A.3
This standard covers the following:
Apply the area and perimeter formulas for rectangles in real world and mathematical problems.
Problem: A rectangle has a length of 21m and a width of 37m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
| |assistant| Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. The perimeter of the rectangle is 2*21+2*37 = <<2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*37=2*21+2*
Expected answer: 116
Correct? No
Problem: A rectangular garden has a length of 50 feet and a width of 20 feet. What is the perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(50+20) = <<2x(50+20)=2x(70)=140>>140 ### Answer: 140
Expected answer: 140
Correct? Yes
Problem: A landscaper plans to reduce the length of the rectangular garden to 15 feet while keeping the width the same. What will the new perimeter of the garden be?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the garden is 2x(15+20) = <<2x(15+20)=2x(35)=70>>70 ### Answer: 70
Expected answer: 70
Correct? Yes
Problem: A rectangle has a width of 7m and a length of 26m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the width and the length. The product of 7m and 26m is 182m2. ### Answer: 182m2
Expected answer: 182
Correct? Yes
Problem: The width of the rectangle stays the same at 7m, but the length extends by 9m to become 35m. What is the area of the rectangle now?
InternLM-Math Base 7B:
Reasoning: The area of the rectangle is 7m × 35m = 245m2. ### Answer: 245m2
Expected answer: 245
Correct? Yes
Problem: A rectangle has a length of 50m and a width of 15m. What is its perimeter?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is <<50m>>50m. The width of the rectangle is <<15m>>15m. The perimeter of the rectangle is 2 times the sum of the length and the width. 2 times the sum of 50m and 15m is 2 times 65m, or <<2*65=130>>130m. ### Answer: 130
Expected answer: 130
Correct? Yes
Problem: The rectangle mentioned in the previous problem had its width reduced to 9m while the length remained the same. What is the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is <<50m>>50m. The width of the rectangle is <<9m>>9m. The perimeter of the rectangle is 2 times the sum of the length and the width. 2 times the sum of 50m and 9m is 2 times 59m, or <<2*59=118>>118m. ### Answer: 118
Expected answer: 118
Correct? Yes
Problem: A rectangular fence has a side length of 5m and another side of 19m. What is the total distance, in meters, around the fence (perimeter)?
InternLM-Math Base 7B:
Reasoning: The total distance, in meters, around the fence (perimeter) is 5+5+19+19 = <<5+5+19+19=48>>48m. ### Answer: 48
Expected answer: 48
Correct? Yes
Problem: The owner of the fence decided to extend the side that originally measured 19m to a new length of 36m, while the other side remains the same. After the extension, what will be the new total distance, in meters, around the fence (perimeter)?
InternLM-Math Base 7B:
Reasoning: The new total distance, in meters, around the fence (perimeter) is 5+5+19+36 = <<5+5+19+36=60>>60m. ### Answer: 60
Expected answer: 82
Correct? No
Problem: A rectangle has sides of length 49m and 6m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(6m+49m) = <<2x(6m+49m)=2x(55m) = 2x(55m) = 110m>>110m ### Answer: 110
Expected answer: 110
Correct? Yes
Problem: Suppose we were mistaken in the original measurement, and one side of the rectangle was actually 17m instead of 49m. What would the new perimeter of the rectangle be?
InternLM-Math Base 7B:
Reasoning: 2x(6m+17m) = <<2x(6m+17m)=2x(23m) = 2x(23m) = 46m>>46m ### Answer: 46
Expected answer: 46
Correct? Yes
Problem: Joshua plans to lay a wire around his rectangular garden. If the garden is 7m wide and 26m long, how much wire will Joshua need to enclose the garden completely?
InternLM-Math Base 7B:
Reasoning: Joshua's garden is 7m by 26m. The perimeter of a rectangle is 2 times the width plus 2 times the length. 2*7+2*26= <<2*7=14, 2*26=52. 14+52=66>>66m. ### Answer: 66
Expected answer: 66
Correct? Yes
Problem: A rectangle has a length of 18 meters and a width of 11 meters. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is 18 and the width of the rectangle is 11. The formula for the perimeter of a rectangle is 2(l+w). 2(18+11) = 2(29) = <<2*29=58>>58. The perimeter of the rectangle is 58 meters.
Expected answer: 58
Correct? Yes
Problem: Suppose the width of the rectangle changed to 44 meters, while the length remained at 18 meters. What would be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is 18 and the width of the rectangle is 44. The formula for the perimeter of a rectangle is 2(l+w). 2(18+44) = 2(62) = <<2*62=124>>124. The perimeter of the rectangle is 124 meters.
Expected answer: 124
Correct? Yes
Problem: The dimensions of a rectangle are 22 feet by 39 feet. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The formula for the perimeter of a rectangle is 2 × (width + height). 2 × (22+39) = 2 × 61 = 122 ### Answer: 122
Expected answer: 122
Correct? Yes
Problem: If the length of the rectangle is now 26 feet instead of 39 feet, while the width remains the same at 22 feet, what would then be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new width is 2 × 22 = 44. The new perimeter is 2 × (22+44) = 2 × 66 = 132 ### Answer: 132
Expected answer: 96
Correct? No
Problem: A rectangle has two sides measuring 45m and the other two sides measuring 25m each. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the two longer sides, 2x45m, and 2 times the sum of the two shorter sides, 2x25m. 2x45m+2x25m = 2x(45m+25m) = 2x(70m) = 2x70m = 140m. The perimeter of the rectangle is 140m. ### Answer: 140m
Expected answer: 140
Correct? Yes
Problem: In the previous rectangle question, suppose one of the sides measuring 45m was incorrectly assessed. The actual measurement is found to be 42m. With this new information, what would be the adjusted perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the two longer sides, 2x45m, and 2 times the sum of the two shorter sides, 2x25m. 2x45m+2x25m = 2x(45m+25m) = 2x(70m) = 2x70m = 140m. The perimeter of the rectangle is 140m. ### Answer: 140m
Expected answer: 134
Correct? No
Problem: A rectangular garden has a length of 8m and a width of 5m. What is the perimeter of this garden?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*8+2*5= <<2*8+2*5=32>>32 ### Answer: 32
Expected answer: 26
Correct? No
Problem: In a rectangle, the length is 19 units and the width is 17 units. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the length and the width. The area of the given rectangle is 19 ⋅ 17 = <<19⋅17=323>>323. ### Answer: 323
Expected answer: 323
Correct? Yes
Problem: In the previous rectangle problem, what if the length was 22 units instead of 19 units while the width remained the same at 17 units? What would be the new area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new area of the rectangle is 22 ⋅ 17 = <<22⋅17=374>>374. ### Answer: 374
Expected answer: 374
Correct? Yes
Problem: A rectangle has a length of 42 meters and a width of 8 meters. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 12 centimeters and a width of 8 centimeters. What is the area of the rectangle?
Expected answer: 100
Correct? No
Problem: A rectangle has a width of 16m and a length of 38m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of its width and length. The area of the rectangle is 16*38 = <<16*38=608>>608m^2. ### Answer: 608
Expected answer: 608
Correct? Yes
Problem: In the rectangle from the previous problem, suppose we replace the length of 38m with a new length of 22m. What would the new area of the rectangle be?
InternLM-Math Base 7B:
Reasoning: The new area of the rectangle is 16*22 = <<16*22=352>>352m^2. ### Answer: 352
Expected answer: 352
Correct? Yes
Problem: Consider a rectangular plot of land. The plot has a length of 28m and a width of 17m. What is the area of the plot?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. The area of the plot is 28 × 17 = <<28 × 17= 476>>476m2. ### Answer: 476
Expected answer: 476
Correct? Yes
Problem: Suppose instead of the original 28m, the length of the rectangular plot of land were only 10m while keeping the width the same at 17m. What would then be the area of the rectangular plot of land?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. The area of the plot is 10 × 17 = <<10 × 17= 170>>170m2. ### Answer: 170
Expected answer: 170
Correct? Yes
Problem: A rectangle has a length of 43m and a width of 41m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. The perimeter of the rectangle is 2*<<43+41=84>>43+2*<<43+41=84>>41 = 2*<<43+41=84>>84 = <<2*43+2*41=2*84=168>>168m. ### Answer: 168
Expected answer: 168
Correct? Yes
Problem: Suppose the rectangle from the previous problem had its length reduced to 18m while keeping the width the same. What would be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new length of the rectangle is 18m. The new width of the rectangle is 41m. The new perimeter of the rectangle is 2*18+2*<<2*18+2*41=2*59=118>>41 = 2*<<2*18+2*41=2*59=118>>59 = 2*<<2*18+2*41=2*59=118>>118m. ### Answer: 236
Expected answer: 118
Correct? No
Problem: A rectangle has one side that measures 44m, and the other side that measures 17m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length of the sides. 2x(44+17) = 2x(61) = <<2x(61)=122>>122m ### Answer: 122m
Expected answer: 122
Correct? Yes
Problem: In the previous problem, we considered a rectangle with one side that measures 44m, and the other side that measures 17m. What would be the perimeter of the rectangle if instead the length of the second side were 21m?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length of the sides. 2x(44+21) = 2x(65) = <<2x(65)=130>>130m ### Answer: 130m
Expected answer: 130
Correct? Yes
Problem: A rectangle's perimeter is twice the sum of its width and length. If the rectangle's width is 2m and its length is 20m, what is its perimeter?
InternLM-Math Base 7B:
Reasoning: The rectangle's width is 2m and its length is 20m. The sum of the width and length is 2m+20m = <<2m+20m=22m>>22m. The rectangle's perimeter is 2(22m) = <<2(22m)=44m>>44m. ### Answer: 44m
Expected answer: 44
Correct? Yes
Problem: If the width of the rectangle is increased from 2m to 32m, while keeping the length the same at 20m, what is the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new width of the rectangle is 32m. The new sum of the width and length is 32m+20m = <<32m+20m=52m>>52m. The new rectangle's perimeter is 2(52m) = <<2(52m)=104m>>104m. ### Answer: 104m
Expected answer: 104
Correct? Yes
Problem: A rectangle has a length of 4 meters and a width of 5 meters. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. The perimeter of the rectangle is 2*5+2*4 = <<2*5+2*4=18>>18 meters. ### Answer: 18
Expected answer: 18
Correct? Yes
Problem: Consider the same rectangle from before, but now suppose its length is extended to 42 meters while the width remains 5 meters. What would then be the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of the rectangle is 2*5+2*42 = <<2*5+2*42=110>>110 meters. ### Answer: 110
Expected answer: 94
Correct? No
Problem: A rectangle has a length of 45m and a width of 37m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*45m + 2*37m = <<2*45m+2*37m=162m>>162m.
Expected answer: 164
Correct? No
Problem: A rectangle has a length of 27m and a width of 21m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
|
Expected answer: 96
Correct? No
Problem: A rectangular garden has a length of 30m and a width of 20m. What is the perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The garden has a width of 20m and a length of 30m. The perimeter of the garden is 2*20+2*30 = <<2*20+2*30=100>>100m. ### Answer: 100
Expected answer: 100
Correct? Yes
Problem: Suppose the width of the rectangular garden is increased to 48m, while the length remains the same at 30m. What would be the new perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The new width of the garden is 48m. The new length of the garden is 30m. The new perimeter of the garden is 2*48+2*30 = <<2*48+2*30=136>>136m. ### Answer: 136
Expected answer: 156
Correct? No
Problem: A rectangle has a length of 30m and a width of 27m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. The perimeter of this rectangle is 2*30+2*27 = <<2*30+2*27=114>>114m. ### Answer: 114
Expected answer: 114
Correct? Yes
Problem: A rectangle has a length of 23m and a width of 11m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 23 × 11 = <<23*11=253>>253m2.
Expected answer: 253
Correct? Yes
Problem: A rectangle has a length of 3m and a width of 9m. What would be the total perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length of the rectangle, plus 2 times the width of the rectangle. The width of the rectangle is 9m, and the length of the rectangle is 3m. The total perimeter of the rectangle is 2(9m) + 2(3m) = <<2(9m)+2(3m)=30m>>30m. ### Answer: 30m
Expected answer: 24
Correct? No
Problem: A rectangle has a length of 31m and a width of 6m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Expected answer: 74
Correct? No
Problem: A rectangle has a length of 12m and a width of 33m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 12 × 33 = <<12*33=396>>396m2. ### Answer: 396m2
Expected answer: 396
Correct? Yes
Problem: Now, consider that the length of the rectangle was initially 18m instead of 12m, while the width remains the same at 33m. What would be the area of the rectangle in this scenario?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 18 × 33 = <<18*33=594>>594m2. ### Answer: 594m2
Expected answer: 594
Correct? Yes
Problem: A rectangle has a length of 20m and all of its sides are of equal length. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: A rectangle with a length of 20m and all of its sides of equal length is a square. The perimeter of a square is 4 times the length of a side. 4*20 = <<4*20=80>>80m ### Answer: 80
Expected answer: 80
Correct? Yes
Problem: A rectangular playground has a length of 35m and a width of 9m. What is the perimeter of the playground?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2(35+9) = 2(44) = <<2*44=88>>88 ### Answer: 88
Expected answer: 88
Correct? Yes
Problem: Suppose the length of the rectangular playground was initially stated incorrectly and it's actually 26m, while the width remains 9m. What would be the correct perimeter of the playground in this case?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2(26+9) = 2(35) = <<2*35=70>>70 ### Answer: 70
Expected answer: 70
Correct? Yes
Problem: A rectangle has a length of 9m and a width of 25m. What is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 9 × 25 = <<9*25=225>>225. The area of the rectangle is 225m2.
Expected answer: 225
Correct? Yes
Problem: Suppose the length of the rectangle was actually 16m and the width remained the same at 9m. What would be the new area of the rectangle in this case?
InternLM-Math Base 7B:
Reasoning: The new area of the rectangle is 16 × 9 = <<16*9=144>>144m2.
Expected answer: 144
Correct? Yes
Problem: A rectangle has a length of 13m and a width of 43m. What is the area of the rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 13m and a width of 43m. What is the area of the rectangle?
Expected answer: 559
Correct? No
Problem: A rectangle has a width of 27m and a length of 47m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the width plus 2 times the length. 2(27) + 2(47) = 54+94 = 148m.
Expected answer: 148
Correct? Yes
Problem: Suppose the width of the rectangle is actually 9m instead of 27m. What would be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2(9) + 2(47) = 18+94 = 112m.
Expected answer: 112
Correct? Yes
Problem: A rectangular football field has an area of 21m * 42m, calculate the area of the football field.
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the two sides. 21m * 42m = 21m * 3m * 14m = 3m * 21m * 14m = 3m * 14m * 21m = 42m * 21m = 21m * 42m = <<21m*42m=882m2>>882m2 ### Answer: 882m2
Expected answer: 882
Correct? Yes
Problem: A rectangle's area is found by multiplying its length by its width. If the width of a certain rectangle is 27m and the length is 25m, what is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is 27*25 = <<27*25=675>>675m2. ### Answer: 675
Expected answer: 675
Correct? Yes
Problem: Suppose the width of the rectangle in the previous problem was not 27m, but was instead 11m, with the length remaining at 25m. What would then be the new area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new area of the rectangle is 11*25 = <<11*25=275>>275m2. ### Answer: 275
Expected answer: 275
Correct? Yes
Problem: A rectangle has a length of 45m and a width of 48m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2 ⋅ 45 + 2 ⋅ 48 = 90 + 96 = 186m.
Expected answer: 186
Correct? Yes
Problem: In the previous example, consider if the width of the rectangle was actually 34m instead of 48m. Now, what would be the perimeter of that rectangle?
InternLM-Math Base 7B:
Reasoning: 2 ⋅ 45 + 2 ⋅ 34 = 90 + 68 = 158m.
Expected answer: 158
Correct? Yes
Problem: A rectangle has a length of 11 meters and a width of 7 meters. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. P = 2(7) + 2(11) = 14 + 22 = 36 ### Answer: 36
Expected answer: 36
Correct? Yes
Problem: Suppose the length of the rectangle was instead 10 meters while the width remained 7 meters. What would be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. P = 2(7) + 2(10) = 14 + 20 = 34 ### Answer: 34
Expected answer: 34
Correct? Yes
Problem: A rectangular field has a length of 30m and a width of 2m. Can you calculate the perimeter of the field?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length of the long side, and 2 times the width of the short side. The perimeter of the field is 2 times 30m + 2 times 2m = 2*30m+2*2m= <<2*30m+2*2m= 2m*30+2m*2m= 60m+4m= 64m>> 64m. ### Answer: 64m
Expected answer: 64
Correct? Yes
Problem: As a follow-up to the previous problem, let's say that the length of the rectangular field is reduced to 20m instead of 30m, while the width remains the same. What would the new perimeter of the field be?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length of the long side, and 2 times the width of the short side. The perimeter of the field is 2 times 20m + 2 times 2m = 2*20m+2*2m= 2m*20+2m*2m= 40m+4m= 44m. ### Answer: 44m
Expected answer: 44
Correct? Yes
Problem: A rectangle has one side that is 7m long and another side that is 8m long. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
|
Expected answer: 30
Correct? No
Problem: A rectangle has a length of 14m and a width of 18m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 8m and a width of 5m. What is the perimeter of this rectangle?
Expected answer: 64
Correct? No
Problem: A rectangle has a length of 43 units and a width of 3 units. Calculate the perimeter of this rectangle.
InternLM-Math Base 7B:
|
Expected answer: 92
Correct? No
Problem: A rectangle has one side that measures 36 meters and another side that measures 42 meters. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2(36+42) = <<2(36+42)=2(78)=156>>156 ### Answer: 156
Expected answer: 156
Correct? Yes
Problem: Now suppose that the longer side of the rectangle measures 48 meters instead of 42. What would be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2(36+48) = <<2(36+48)=2(84)=168>>168 ### Answer: 168
Expected answer: 168
Correct? Yes
Problem: A rectangle has a length of 1m and a width of 6m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(1+6) = <<2x(1+6)=2x7=14>>14m ### Answer: 14m
Expected answer: 14
Correct? Yes
Problem: Now consider that the length of the rectangle has changed to 28m, while the width remains at 6m. What is the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(28+6) = <<2x(28+6)=2x34=68>>68m ### Answer: 68m
Expected answer: 68
Correct? Yes
Problem: A rectangle has two sides measuring 45m and two sides measuring 8m. What is the perimeter of this rectangle?
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: 106
Correct? No
Problem: A rectangular garden has a length of 25 meters and a width of 37 meters. What is the perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the width + 2 times the length. 2*25 + 2*37 = 50 + 74 = 124.
Expected answer: 124
Correct? Yes
Problem: Suppose we reduce the length of the garden from 25 meters to 16 meters, while the width remains unchanged at 37 meters. What would be the new perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the garden is 2*16 + 2*37 = 32 + 74 = 106.
Expected answer: 106
Correct? Yes
Problem: A rectangle has length 36m and width 3m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(3+36) = <<2x(3+36)=2x39=78>>78m.
Expected answer: 78
Correct? Yes
Problem: Suppose we made a change to the rectangle from the previous problem. Instead of a length of 36m, we shortened it to only 1m. What would be the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2x(3+1) = <<2x(3+1)=2x4=8>>8m.
Expected answer: 8
Correct? Yes
Problem: A rectangular garden has a length of 45 meters and a width of 50 meters. Calculate the perimeter of the garden.
InternLM-Math Base 7B:
Reasoning: The formula for the perimeter of a rectangle is <<P=2L+2W>>. P=2(45)+2(50) = 2(45+50) = 2(95) = 190 ### Answer: 190
Expected answer: 190
Correct? Yes
Problem: A rectangle has a length of 44m and a width of 34m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*44+2*34 = <<2*44+2*34=176>>176.
Expected answer: 156
Correct? No
Problem: A rectangle has a length of 42 m and a width of 29 m. Calculate the perimeter of the rectangle.
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2 · 42 m + 2 · 29 m = 2 · 71 m = <<2 · 71 m=142>>142 m. The perimeter of the rectangle is 142 m. ### Answer: 142
Expected answer: 142
Correct? Yes
Problem: Suppose the width of the rectangle was actually 38 m instead of 29 m. What would then be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2 · 42 m + 2 · 38 m = 2 · 80 m = <<2 · 80 m=160>>160 m. The new perimeter of the rectangle is 160 m. ### Answer: 160
Expected answer: 160
Correct? Yes
Problem: A rectangle has a length of 44 meters and a width of 35 meters. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
| |assistant| Reasoning: The length of the rectangle is <<44>>44 meters and the width of the rectangle is <<35>>35 meters. The formula for the perimeter of a rectangle is 2 x (l+w). 2 x (l+w) = 2 x (<<44>>44+<<35>>35) = 2 x (<<44+35=79>>79) = <<2x79=158>>158. The perimeter of the rectangle is <<158>>158 meters.
Expected answer: 158
Correct? Yes
Problem: Consider the same rectangle from the previous problem, but now the width has changed to 43 meters while the length remains the same at 44 meters. What is the new perimeter of the rectangle?
InternLM-Math Base 7B:
|assistant| Reasoning: The length of the rectangle is <<44>>44 meters and the width of the rectangle is <<43>>43 meters. The formula for the perimeter of a rectangle is 2 x (l+w). 2 x (l+w) = 2 x (<<44>>44+<<43>>43) = 2 x (<<44+43=87>>87) = <<2x87=174>>174. The perimeter of the rectangle is <<174>>174 meters.
Expected answer: 174
Correct? Yes
Problem: A rectangle has a length of 44m and a width of 45m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*44+2*45 = <<2*44+2*45=168>>168 ### Answer: 168
Expected answer: 178
Correct? No
Problem: A rectangle has a length of 3 meters and a width of 39 meters. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2 times 3 = <<2*3=6>>6. 2 times 39 = <<2*39=78>>78. 6+78 = <<6+78=84>>84. The perimeter of the rectangle is 84 meters.
Expected answer: 84
Correct? Yes
Problem: Suppose the length of the rectangle was actually 17 meters instead of 3 meters, keeping the width the same. What would be the new perimeter of the rectangle now?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2 times 17, which is <<2*17=34>>34, plus 2 times 39, which is 78. 34+78 = <<34+78=112>>112. The new perimeter of the rectangle is 112 meters.
Expected answer: 112
Correct? Yes
Problem: A rectangular garden has sides that measure 32m and 12m. What is the perimeter of this garden?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is the sum of the length of the sides. The perimeter of the garden is 12m+12m+32m+32m = <<12m+12m+32m+32m=76m>>76m. ### Answer: 76m
Expected answer: 88
Correct? No
Problem: A rectangular swimming pool has a length of 24 feet and a width of 6 feet. What is the perimeter of the swimming pool?
InternLM-Math Base 7B:
Expected answer: 60
Correct? No
Problem: A rectangle has a length of 27m and a width of 21m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. The area of the rectangle is 27 × 21 = <<27*21=567>>567m2. ### Answer: 567
Expected answer: 567
Correct? Yes
Problem: In the previous problem, the rectangle had a length of 27m and a width of 21m. Now, consider that the width of the rectangle is instead 8m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. The area of the rectangle is 27 × 8 = <<27*8=216>>216m2. ### Answer: 216
Expected answer: 216
Correct? Yes
Problem: A rectangle has twice the length of a square. If the square's side is 49m, what is the length of the rectangle?
InternLM-Math Base 7B:
Reasoning: The length of the square is 49m. The length of the rectangle is 2 ⋅ 49m = <<2⋅49=98>>98m. ### Answer: 98
Expected answer: 98
Correct? Yes
Problem: In the previous problem, we found the length of the rectangle when the square's side was 49m. Now, consider a square with a side length of 18m. What would be the length of the rectangle if it's still twice the length of the square?
InternLM-Math Base 7B:
Reasoning: The length of the square is 18m. The length of the rectangle is 2 ⋅ 18m = <<2⋅18=36>>36m. ### Answer: 36
Expected answer: 36
Correct? Yes
Problem: A rectangle has a length of 50m and a width of 36m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*50m + 2*36m = 2*<<2*50m+2*36m=2*86m=172m>>86m = 172m.
Expected answer: 172
Correct? Yes
Problem: In our previous rectangle, if we now increase the width from 36m to 44m, while keeping the length the same, what will be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2*50m+2*44m = 2*<<2*50m+2*44m=2*94m=188m>>94m = 188m.
Expected answer: 188
Correct? Yes
Problem: A rectangle has a length of 8m and a width of 44m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*8+2*44= <<2*8+2*44=176>>176 ### Answer: 176
Expected answer: 104
Correct? No
Problem: If a rectangle has a length of 47 feet and a width of 21 feet, what is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the length and the width. The product of 47 and 21 is <<47*21=987>>987.
Expected answer: 987
Correct? Yes
Problem: Now, suppose we adjusted the measurements of the previous rectangle. If the length of the rectangle remained the same at 21 feet, but the width of the rectangle was reduced to 15 feet, what would the new area of the rectangle be?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the length and the width. The product of 21 and 15 is <<21*15=315>>315.
Expected answer: 315
Correct? Yes
Problem: A rectangle has a length of 19m and a breadth of 28m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the breadth.
Expected answer: 94
Correct? No
Problem: One side of a rectangle is four times the length of the other side. If the longer side measures 39 meters, what is the area of the rectangle when the shorter side is 10 meters?
InternLM-Math Base 7B:
Reasoning: The other side of the rectangle is 1/4 the length of the longer side. The other side is 1/4(39) = <<1/4(39)=9.75>>9.75 meters. The area of the rectangle is 9.75*10 = <<9.75*10=97.5>>97.5 m2. ### Answer: 97.5
Expected answer: 390
Correct? No
Problem: A rectangle has identical sides, each measuring 19 meters. What is the total perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and width. The length and width of the rectangle are 19 meters. The total perimeter of the rectangle is 2(19+19) = 2(38) = <<2*38=76>>76.
Expected answer: 76
Correct? Yes
Problem: The rectangle mentioned in the previous problem has been resized. The new rectangle now has a length of 19 meters but a width of 6 meters. What is the new total perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of a rectangle is 2 times the sum of the length and width. The new length and width of the rectangle are 19 meters and 6 meters. The new total perimeter of the rectangle is 2(19+6) = 2(25) = <<2*25=50>>50.
Expected answer: 50
Correct? Yes
Problem: The area of a rectangle is given by the equation length * width. If the length is 9m and the width is 26m, what is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is given by the equation length * width. The area of the given rectangle is 9m * 26m = <<9m*26m=234m2>>234m2. ### Answer: 234m2
Expected answer: 234
Correct? Yes
Problem: The rectangle now has a new dimension with its length increased to 47m while the width is still 9m. What is the new area of the rectangle?
InternLM-Math Base 7B:
Expected answer: 423
Correct? No
Problem: A rectangle has one side measuring 19cm and another side measuring 7cm. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is the sum of the four sides. The two sides of 7cm and 19cm are the same. The other two sides are 7cm and 19cm. 7+7+19+19 = 7*2+19*2 = 7*2+7*2+19*2-7*2 = 7*2+19*2-7*2 = 2*7+2*19 = 2*(7+19) = 2*26 = 2*2*13 = 2*2*2*13/2 = 2*2*2*13/2*2/2 = 2*2*2*13/2*2 = 2*2*2*13/4 = 2*2*2*13/2*2/2 = 2*2*2*13/2*2*2/2 = 2*2*2*13/2*2*2*2/2 = 2*2*2*13/2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2/2 = 2*2*2*13/2*2*2*2*2*2*2*2*2*2*2*2*2*2*
Expected answer: 52
Correct? Yes
Problem: A shutter is in the shape of a rectangle. Its length is 10m and its width is 6m. What is the total area of the shutter?
InternLM-Math Base 7B:
Reasoning: The total area of the shutter is 6 · 10 = <<6 · 10=60>>60m2. ### Answer: 60
Expected answer: 60
Correct? Yes
Problem: A rectangle has a length of 24 units and a width of 15 units. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is <<24>>24 units. The width of the rectangle is <<15>>15 units. The formula for the perimeter of a rectangle is P = 2l + 2w. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2 ⋅ <<15>>15. P = 2 ⋅ <<24>>24 + 2
Expected answer: 78
Correct? Yes
Problem: Now, imagine the rectangle's width was reduced to 10 units, while the length remained 24 units. What would be the new perimeter of the rectangle under these conditions?
InternLM-Math Base 7B:
Reasoning: The length of the rectangle is <<24>>24 units. The new width of the rectangle is <<10>>10 units. The formula for the perimeter of a rectangle is P = 2l + 2w. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2 ⋅ <<10>>10. P = 2 ⋅ <<24>>24 + 2
Expected answer: 68
Correct? Yes
Problem: If a rectangular plot of land has a length of 32m and a width of 27m, what is the area of the plot in square meters?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of its length and its width. The area of the plot is 32*27= <<32*27=864>>864 square meters. ### Answer: 864
Expected answer: 864
Correct? Yes
Problem: In the previous scenario, suppose the width of the rectangular plot of land was actually 18m rather than 27m. What would then be the area of the plot in square meters?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of its length and its width. The area of the plot is 32*18= <<32*18=576>>576 square meters. ### Answer: 576
Expected answer: 576
Correct? Yes
Problem: A rectangle has one side that measures 15m and another side that measures 10m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is the sum of the four sides. The two sides that are 10m long and the two sides that are 15m long. 10+10+15+15 = 50m. ### Answer: 50m
Expected answer: 50
Correct? Yes
Problem: Now consider the rectangle has one side that measures 33m instead of 15m while the other side still measures 10m. What is the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is the sum of the four sides. The two sides that are 10m long and the two sides that are 33m long. 10+10+33+33 = 86m. ### Answer: 86m
Expected answer: 86
Correct? Yes
Problem: A rectangle has a length of 44m and a width of 37m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2(44m) + 2(37m) = <<2(44m)+2(37m)= 2(44m+37m) = 2(81m) = 162m>>162m ### Answer: 162m
Expected answer: 162
Correct? Yes
Problem: What would the perimeter of the rectangle be if the length was 25m instead of 44m but the width remained the same at 37m?
InternLM-Math Base 7B:
Reasoning: 2(25m) + 2(37m) = <<2(25m)+2(37m)= 2(25m+37m) = 2(62m) = 124m>>124m ### Answer: 124m
Expected answer: 124
Correct? Yes
Problem: A rectangle has length of 14 cm and width 8 cm. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width. 2x(8+14) = <<2x(8+14)=2x(22)=44>>44.
Expected answer: 44
Correct? Yes
Problem: If the length of the rectangle remains 14 cm, but the width is increased to 18 cm, what would be the new perimeter of the rectangle now?
InternLM-Math Base 7B:
Reasoning: 2x(18+14) = <<2x(18+14)=2x(32)=64>>64.
Expected answer: 64
Correct? Yes
Problem: A rectangle has a length of 26m and a width of 28m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*28+2*26=2*28+2*26=<<2*28=56, 2*26=52. 56+52=108>>108.
Expected answer: 108
Correct? Yes
Problem: In the previous scenario, if the rectangle was actually a square with all sides measuring 26m, what would the new perimeter be?
InternLM-Math Base 7B:
Reasoning: The new perimeter is 4*26=4*26=<<4*26=104>>104m.
Expected answer: 104
Correct? Yes
Problem: A rectangle has a length of 15m and a width of 37m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 15m and a width of 37m. What is the perimeter of this rectangle?
Expected answer: 104
Correct? No
Problem: A rectangle has a length of 42m and a width of 40m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2(42) + 2(40) = 2(42+40) = 2(82) = <<2(82)=164>>164 ### Answer: 164
Expected answer: 164
Correct? Yes
Problem: The rectangle you were previously given now has a length of 42m and the width has been adjusted to 24m. What is the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2(42) + 2(24) = 2(42+24) = 2(66) = <<2(66)=132>>132 ### Answer: 132
Expected answer: 132
Correct? Yes
Problem: A rectangular garden has one side that measures 21 feet and another side that measures 30 feet. What is the perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is the sum of the lengths of the sides. The two sides of the garden that are 21 feet long are the same, and the two sides of the garden that are 30 feet long are the same. The perimeter of the garden is 2(21) + 2(30) = 42 + 60 = <<42+60=102>>102. ### Answer: 102
Expected answer: 102
Correct? Yes
Problem: Now, imagine that the side of the garden that initially measured 30 feet was actually only 5 feet long. What would be the new perimeter of the garden?
InternLM-Math Base 7B:
Reasoning: The new side of the garden is 5 feet. The new perimeter of the garden is 2(21) + 2(5) = 42 + 10 = <<42+10=52>>52. ### Answer: 52
Expected answer: 52
Correct? Yes
Problem: A certain playground is in the shape of a rectangle. The area of the playground is given as the product of 17 and 44. How many square meters is the playground?
InternLM-Math Base 7B:
Reasoning: 17*44 = <<17*44=748>>748 The playground is 748 square meters.
Expected answer: 748
Correct? Yes
Problem: Let's assume that the length of the playground was mistakenly recorded. Instead of 44m, the correct measurement is 37m. Given this information, how many square meters is the playground now?
InternLM-Math Base 7B:
Reasoning: 17*37 = <<17*37=629>>629 The playground is 629 square meters.
Expected answer: 629
Correct? Yes
Problem: Rectangle ABCD has a width of 47m and a length of 50m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The formula for the perimeter of a rectangle is 2*(length+width). 2*50+2*47 = 2*<<2*50+2*47=194>>194 ### Answer: 194
Expected answer: 194
Correct? Yes
Problem: Now consider that the rectangle ABCD's dimensions have changed slightly. The width is still 47m, but the length has adjusted to 48m. What is the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The formula for the perimeter of a rectangle is 2*(length+width). 2*48+2*47 = 2*<<2*48+2*47=194>>194 ### Answer: 194
Expected answer: 190
Correct? No
Problem: A rectangle has a length of 11m and a width of 31m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the width plus 2 times the length. The perimeter of the rectangle is 2*11+2*31= <<2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11+2*31=2*11
Expected answer: 84
Correct? No
Problem: A rectangular garden has a length of 39m and a width of 19m. What is the area of this garden?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 39 × 19 = <<39 × 19=741>>741m2.
Expected answer: 741
Correct? Yes
Problem: If the length of the rectangular garden was actually 35m instead of 39m, keeping the width the same, what would be the new area of the garden?
InternLM-Math Base 7B:
Reasoning: 35 × 19 = <<35 × 19=665>>665m2.
Expected answer: 665
Correct? Yes
Problem: A carpenter wants to install a wooden floor in a rectangular hallway. The wooden plank has a width of 29 cm and the carpenter has cut it to a length of 11 cm. How many square centimeters of area does one cut wooden plank cover?
InternLM-Math Base 7B:
Reasoning: 1 plank = 1*11*29 = 1*11* <<1*11*29=319>>319 ### Answer: 319
Expected answer: 319
Correct? Yes
Problem: If instead the carpenter decided to cut the same 29 cm wide wooden plank into shorter lengths of 5 cm each, how many square centimeters of area does one of these shorter cut wooden planks cover?
InternLM-Math Base 7B:
Reasoning: 1 plank = 1*5*29 = 1*5* <<1*5*29=145>>145 ### Answer: 145
Expected answer: 145
Correct? Yes
Problem: A rectangle has a length of 25m and a width of 11m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*11=<<2*11=22>>22, 2*25=<<2*25=50>>50. 22+50=<<22+50=72>>72.
Expected answer: 72
Correct? Yes
Problem: Now, suppose the rectangle's width is changed from 11m to 17m. What will be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: 2*11=<<2*11=22>>22, 2*25=<<2*25=50>>50. 22+50=<<22+50=72>>72.
Expected answer: 84
Correct? No
Problem: A rectangle has a length of 49m and a breadth of 9m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the breadth.
Expected answer: 116
Correct? No
Problem: A rectangle has a length of 16m and a width of 7m. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. 16 × 7 = <<16*7=112>>112m2.
Expected answer: 112
Correct? Yes
Problem: The perimeter of a rectangular playground is found by the formula 2*(Length + Breadth). If the length of the playground is 14m and the breadth is 2m, what is the perimeter of the playground?
InternLM-Math Base 7B:
Reasoning: 2*(14m+2m) = <<2*(14m+2m) = 2*16m = 32m>>32m ### Answer: 32m
Expected answer: 32
Correct? Yes
Problem: Now, consider that the breadth of the rectangular playground is changed to 3m, keeping the length the same as before (14m). What is the new perimeter of the playground?
InternLM-Math Base 7B:
Reasoning: 2*(14m+3m) = <<2*(14m+3m) = 2*17m = 34m>>34m ### Answer: 34m
Expected answer: 34
Correct? Yes
Problem: A square has a side length of 16 units. What is the perimeter of the square?
InternLM-Math Base 7B:
Reasoning: A square has 4 equal side lengths. 4x16 = <<4x16=64>>64. The perimeter of the square is 64 units.
Expected answer: 64
Correct? Yes
Problem: In the previous problem, the square's side length was 16 units. Now consider a rectangle with a length of 22 units and a width of 16 units. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: 22+16+22+16 = <<22+16+22+16=76>>76. The perimeter of the rectangle is 76 units.
Expected answer: 76
Correct? Yes
Problem: A rectangular garden has a width of 7 meters and a length of 21 meters. What is the perimeter of this garden?
InternLM-Math Base 7B:
Reasoning: The width of the garden is 7 meters, and the length of the garden is 21 meters. The perimeter of the garden is 2*7+2*21 = 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*21= 2*7+2*
Expected answer: 56
Correct? Yes
Problem: A rectangle has a length of 20 meters and a width of 5 meters. What is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the length and the width. The area of the rectangle is <<20*5=100>>100 square meters. ### Answer: 100
Expected answer: 100
Correct? Yes
Problem: In the previous problem, we calculated the area of a rectangle with a length of 20 meters and a width of 5 meters. This time, instead of having the width as 5 meters, we have a new rectangle where the width is 17 meters. With the length still being 20 meters, what is the area of this new rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the length and the width. The area of the rectangle is <<20*17=340>>340 square meters. ### Answer: 340
Expected answer: 340
Correct? Yes
Problem: A rectangle has a length of 37m and a width of 43m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. P = 2(37m) + 2(43m) = 74m + 86m = 160m ### Answer: 160m
Expected answer: 160
Correct? Yes
Problem: Suppose the length of the rectangle is now changed to 32m while the width remains the same at 43m. What will be the new perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2(32m) + 2(43m) = 64m + 86m = 150m ### Answer: 150m
Expected answer: 150
Correct? Yes
Problem: A rectangle has a length of 9m. If the width of the rectangle is 1m, what is the area?
InternLM-Math Base 7B:
Expected answer: 9
Correct? No
Problem: A rectangle's area is the product of 22m and 23m. What is the area of the rectangle?
InternLM-Math Base 7B:
Reasoning: The product of 22m and 23m is 2*3m*2*3m= <<2*3m*2*3m=12*9m*2=36m*2=72m2>>72m2. ### Answer: 72m2
Expected answer: 506
Correct? No
Problem: A rectangle has one side that measures 14 cm and another side that measures 21 cm. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the length and width. The sum of the length and width is 14+21= <<14+21=35>>35. 2 times 35 is <<2*35=70>>70. ### Answer: 70
Expected answer: 70
Correct? Yes
Problem: Consider the previous rectangle and suppose its side that previously measured 21 cm now measures 47 cm. What is the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2 times the sum of the length and width. The sum of the length and width is 14+47= <<14+47=61>>61. 2 times 61 is <<2*61=122>>122. ### Answer: 122
Expected answer: 122
Correct? Yes
Problem: In Clara's backyard, there is a rectangular swimming pool. The length of the pool is 23 meters, while the width is 12 meters. What is the area of the swimming pool in Clara's backyard?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is length × width. The area of the pool is 23 × 12 = <<23 × 12=276>>276 square meters. ### Answer: 276
Expected answer: 276
Correct? Yes
Problem: A rectangle has a length of 39 meters and a width of 8 meters. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
| |assistant| Reasoning: The length of the rectangle is 39 meters, and the width is 8 meters. The perimeter of a rectangle is 2 times the length plus 2 times the width. 2 times 39 = <<2*39=78>>78. 2 times 8 = <<2*8=16>>16. 78+16 = <<78+16=94>>94. The perimeter of the rectangle is 94 meters.
Expected answer: 94
Correct? Yes
Problem: Suppose the width of the rectangle was actually 23 meters instead of 8 meters. What would the perimeter of the rectangle be in this case?
InternLM-Math Base 7B:
Reasoning: The width of the rectangle is 23 meters. 2 times 23 = <<2*23=46>>46. 2 times 39 = <<2*39=78>>78. 78+46 = <<78+46=124>>124. The perimeter of the rectangle is 124 meters.
Expected answer: 124
Correct? Yes
Problem: A rectangle has a width of 27m and a length of 40m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the width + 2 times the length. 2*27 + 2*40 = 2* <<2*27+2*40=2*67>>67 = 134m ### Answer: 134m
Expected answer: 134
Correct? Yes
Problem: What if the length of the rectangle was 44m instead of 40m while keeping the width still at 27m? What would the new perimeter of the rectangle be?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the width + 2 times the length. 2*27 + 2*44 = 2* <<2*27+2*44=2*91>>91 = 182m ### Answer: 182m
Expected answer: 142
Correct? No
Problem: A rectangle's length is 35m and its width is 9m. What is the perimeter of the rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the length plus 2 times the width. 2*35+2*9 = <<2*35+2*9= 2*44=88>>88m.
Expected answer: 88
Correct? Yes
Problem: Suppose we resized the rectangle and shortened its length from 35m to 19m, while keeping the width the same. What would the new perimeter of the rectangle be?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2*19+2*9 = <<2*19+2*9= 2*28=56>>56m.
Expected answer: 56
Correct? Yes
Problem: A rectangle has a length of 38m and a width of 11m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 38m and a width of 11m. What is the perimeter of this rectangle?
Expected answer: 98
Correct? No
Problem: One side of a rectangle is 28m long, and the other side is 29m long. What is the area of this rectangle?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the two side lengths. The area of the given rectangle is 28*29= <<28*29=812>>812m2. ### Answer: 812
Expected answer: 812
Correct? Yes
Problem: Suppose the side of the rectangle previously mentioned gets extended to 35m instead of 29m, while the other side remains 28m. What would be the new area of the rectangle in this case?
InternLM-Math Base 7B:
Reasoning: The area of a rectangle is the product of the two side lengths. The area of the new rectangle is 28*35= <<28*35=980>>980m2. ### Answer: 980
Expected answer: 980
Correct? Yes
Problem: A square has a side length of 9m. What is the total length of fencing needed to enclose the square's perimeter?
InternLM-Math Base 7B:
Reasoning: A square has 4 equal sides. 9m x 4 = <<9m*4=36m>>36m ### Answer: 36m
Expected answer: 36
Correct? Yes
Problem: A rectangle has a length of 38m and a width of 29m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
| |user| A rectangle has a length of 38m and a width of 29m. What is the perimeter of this rectangle?
Expected answer: 134
Correct? No
Problem: A rectangle has one side of length 45m and another side of length 47m. Find the perimeter of the rectangle.
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2 times the sum of the two side lengths. The perimeter of the given rectangle is 2*<<45+47=92>>92 = <<2*92=184>>184m. ### Answer: 184
Expected answer: 184
Correct? Yes
Problem: Suppose the length of one side of the rectangle is altered from 47m to 46m. What would be the new perimeter of the rectangle now?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the rectangle is 2*<<45+46=91>>91 = <<2*91=182>>182m. ### Answer: 182
Expected answer: 182
Correct? Yes
Problem: Given a rectangle where the length is 24m and the width is 20m, what is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2*length + 2*width. The perimeter of the given rectangle is 2*24 + 2*20 = <<2*24+2*20=88>>88m. ### Answer: 88
Expected answer: 88
Correct? Yes
Problem: Continuing from the previous scenario, imagine the length of the rectangle stayed the same at 24m, but the width was reduced to 17m. Now, what would be the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The new perimeter of the given rectangle is 2*24 + 2*17 = <<2*24+2*17=78>>78m. ### Answer: 78
Expected answer: 82
Correct? No
Problem: The length and width of a rectangle are both 32m. What is the perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2*32+2*32 = <<2*32+2*32=128>>128m. ### Answer: 128
Expected answer: 128
Correct? Yes
Problem: The length of the rectangle from the previous problem remains the same, but the width of the rectangle is now adjusted to 13m. What is the new perimeter of this rectangle?
InternLM-Math Base 7B:
Reasoning: The perimeter of a rectangle is 2*32+2*13 = <<2*32+2*13=100>>100m. ### Answer: 100
Expected answer: 90
Correct? No