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InternLM-Math Base 20B 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

Problem: A rectangle has a length of 21m and a width of 37m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(21 + 37) = <<2(21+37)=2*58=116>>116m.
### Answer: 116

Expected answer: 116

Correct? Yes

🔗 Follow-up

Problem: Now consider that the length of the rectangle has changed to 4m while the width remains at 37m. What is the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(4 + 37) = <<2(4+37)=2*41=82>>82m.
### Answer: 82

Expected answer: 82

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its two lengths and two widths.
The perimeter of the garden is 2(50) + 2(20) = <<2*50+2*20=140>>140 feet.
### Answer: 140

Expected answer: 140

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its two lengths and two widths.
The new perimeter of the garden is 2(15) + 2(20) = <<2*15+2*20=70>>70 feet.
### Answer: 70

Expected answer: 70

Correct? Yes

🔗 Problem

Problem: A rectangle has a width of 7m and a length of 26m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 7*26 = <<7*26=182>>182 m^2.
### Answer: 182

Expected answer: 182

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 7*35 = <<7*35=245>>245 m^2.
### Answer: 245

Expected answer: 245

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 50m and a width of 15m. What is its perimeter?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width).
The perimeter of the rectangle is 2(50+15) = <<2(50+15)=2*65=130>>130m.
### Answer: 130m

Expected answer: 130

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width).
The perimeter of the rectangle is 2(50+9) = <<2(50+9)=2*59=118>>118m.
### Answer: 118m

Expected answer: 118

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of the fence is 5+5+19+19 = <<5+5+19+19=48>>48 meters.
### Answer: 48

Expected answer: 48

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of the fence is 5+5+36+36 = <<5+5+36+36=82>>82 meters.
### Answer: 82

Expected answer: 82

Correct? Yes

🔗 Problem

Problem: A rectangle has sides of length 49m and 6m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(49+6) = <<2*(49+6)=112>>112m.
### Answer: 112

Expected answer: 110

Correct? No

🔗 Problem

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 20B:

 Reasoning: Joshua needs 2(7)+2(26) = <<2*7+2*26=62>>62m of wire to enclose the garden.
### Answer: 62

Expected answer: 66

Correct? No

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 18+11+18+11 = <<18+11+18+11=58>>58 meters.
### Answer: 58

Expected answer: 58

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 18+44+18+44 = <<18+44+18+44=124>>124 meters.
### Answer: 124

Expected answer: 124

Correct? Yes

🔗 Problem

Problem: The dimensions of a rectangle are 22 feet by 39 feet. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 22 feet, and the width of the rectangle is 39 feet. Therefore, the perimeter of the rectangle is 2*(22+39) = <<2*(22+39)=2*61=122>>122 feet.
### Answer: 122

Expected answer: 122

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 26 feet, and the width of the rectangle is 22 feet. Therefore, the perimeter of the rectangle is 2*(26+22) = <<2*(26+22)=2*48=96>>96 feet.
### Answer: 96

Expected answer: 96

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of this rectangle is 45+25+45+25 = <<45+25+45+25=140>>140m.
### Answer: 140

Expected answer: 140

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of this rectangle is 45+25+45+25 = <<45+25+45+25=140>>140m.
### Answer: 140

Expected answer: 134

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 8m and a width of 5m. What is the perimeter of this garden?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this garden is 2*(8+5) = <<2*(8+5)=26>>26m.
### Answer: 26

Expected answer: 26

Correct? Yes

🔗 Follow-up

Problem: The owner of the rectangular garden decided to extend its length to 46m while keeping the width the same, i.e., 5m. What is the new perimeter of the garden now?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this garden is 2*(46+5) = <<2*(46+5)=102>>102m.
### Answer: 102

Expected answer: 102

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 19*17 = <<19*17=323>>323 square units.
### Answer: 323

Expected answer: 323

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The new area of the rectangle is 22*17 = <<22*17=374>>374 square units.
### Answer: 374

Expected answer: 374

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(42+8) = <<2*(42+8)=100>>100 meters.
### Answer: 100

Expected answer: 100

Correct? Yes

🔗 Problem

Problem: A rectangle has a width of 16m and a length of 38m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 16*38 = <<16*38=608>>608 m^2.
### Answer: 608

Expected answer: 608

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and its width.
The new area of the rectangle is 16*22 = <<16*22=352>>352 m^2.
### Answer: 352

Expected answer: 352

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the plot is 28*17 = <<28*17=476>>476m^2.
### Answer: 476

Expected answer: 476

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the plot is 10*17 = <<10*17=170>>170m^2.
### Answer: 170

Expected answer: 170

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 43m and a width of 41m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(43+41) = <<2*(43+41)=2*84=168>>168m.
### Answer: 168

Expected answer: 168

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(18+41) = <<2*(18+41)=2*59=118>>118m.
### Answer: 118

Expected answer: 118

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 44m, and the width is 17m. So, the perimeter is 2*(44+17) = <<2*(44+17)=2*61=122>>122m.
### Answer: 122

Expected answer: 122

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 44m, and the width is 21m. So, the perimeter is 2*(44+21) = <<2*(44+21)=2*65=130>>130m.
### Answer: 130

Expected answer: 130

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 20m, and the width is 2m. So, the perimeter is 2*(20+2) = <<2*(20+2)=44>>44m.
### Answer: 44

Expected answer: 44

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 20m, and the width is 32m. So, the perimeter is 2*(20+32) = <<2*(20+32)=104>>104m.
### Answer: 104

Expected answer: 104

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 4+5+4+5 = <<4+5+4+5=18>>18 meters.
### Answer: 18

Expected answer: 18

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 42+5+42+5 = <<42+5+42+5=94>>94 meters.
### Answer: 94

Expected answer: 94

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 45m and a width of 37m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(45+37) = <<2*(45+37)=2*82=164>>164m.
### Answer: 164

Expected answer: 164

Correct? Yes

🔗 Follow-up

Problem: If the length of the rectangle remains the same at 45m, but the width is increased to 47m, what would be the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(45+47) = <<2*(45+47)=2*92=184>>184m.
### Answer: 184

Expected answer: 184

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 27m and a width of 21m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(27+21) = <<2*(27+21)=102>>102m.
### Answer: 102

Expected answer: 96

Correct? No

🔗 Problem

Problem: A rectangular garden has a length of 30m and a width of 20m. What is the perimeter of the garden?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is the sum of its two lengths and two widths.
The perimeter of the garden is 2(30)+2(20) = <<2*30+2*20=120>>120m.
### Answer: 120

Expected answer: 100

Correct? No

🔗 Problem

Problem: A rectangle has a length of 30m and a width of 27m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(30+27) = <<2*(30+27)=2*57=114>>114m.
### Answer: 114

Expected answer: 114

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 23m and a width of 11m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 23*11 = <<23*11=253>>253m^2.
### Answer: 253

Expected answer: 253

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(3+9) = <<2*(3+9)=30>>30m.
### Answer: 30

Expected answer: 24

Correct? No

🔗 Problem

Problem: A rectangle has a length of 31m and a width of 6m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(31+6) = <<2*(31+6)=76>>76m.
### Answer: 76

Expected answer: 74

Correct? No

🔗 Problem

Problem: A rectangle has a length of 12m and a width of 33m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 12*33 = <<12*33=396>>396m^2.
### Answer: 396

Expected answer: 396

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of this rectangle is 18*33 = <<18*33=594>>594m^2.
### Answer: 594

Expected answer: 594

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its four sides. Since all of the sides are of equal length, the perimeter is 4 times the length of one side. The length of one side is 20m, so the perimeter is 4*20m = <<4*20=80>>80m.
### Answer: 80

Expected answer: 80

Correct? Yes

🔗 Problem

Problem: A rectangular playground has a length of 35m and a width of 9m. What is the perimeter of the playground?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the playground is 2*(35+9) = <<2*(35+9)=2*44=88>>88m.
### Answer: 88

Expected answer: 88

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the playground is 2*(26+9) = <<2*(26+9)=2*35=70>>70m.
### Answer: 70

Expected answer: 70

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 9m and a width of 25m. What is the area of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 9*25 = <<9*25=225>>225 m^2.
### Answer: 225

Expected answer: 225

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 16*9 = <<16*9=144>>144 m^2.
### Answer: 144

Expected answer: 144

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 13m and a width of 43m. What is the area of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 13*43 = <<13*43=572>>572m^2.
### Answer: 572

Expected answer: 559

Correct? No

🔗 Problem

Problem: A rectangle has a width of 27m and a length of 47m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(47+27) = <<2*(47+27)=2*74=148>>148m.
### Answer: 148

Expected answer: 148

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(47+9) = <<2*(47+9)=2*56=112>>112m.
### Answer: 112

Expected answer: 112

Correct? Yes

🔗 Problem

Problem: A rectangular football field has an area of 21m * 42m, calculate the area of the football field.

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the football field is 21m * 42m = <<21*42=882>>882m^2.
### Answer: 882

Expected answer: 882

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is found by multiplying its length by its width.
The area of the rectangle is 27*25 = <<27*25=675>>675m^2.
### Answer: 675

Expected answer: 675

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is found by multiplying its length by its width.
The area of the rectangle is 11*25 = <<11*25=275>>275m^2.
### Answer: 275

Expected answer: 275

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 45m and a width of 48m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(45+48) = <<2*(45+48)=2*93=186>>186m.
### Answer: 186

Expected answer: 186

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(45+34) = <<2*(45+34)=2*79=158>>158m.
### Answer: 158

Expected answer: 158

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 11+7+11+7 = <<11+7+11+7=36>>36 meters.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 10+7+10+7 = <<10+7+10+7=34>>34 meters.
### Answer: 34

Expected answer: 34

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the field is 2*(30+2) = <<2*(30+2)=64>>64m.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the field is 2*(20+2) = <<2*(20+2)=44>>44m.
### Answer: 44

Expected answer: 44

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 2(7)+2(8) = <<2(7)+2(8)=30>>30m.
### Answer: 30

Expected answer: 30

Correct? Yes

🔗 Follow-up

Problem: Consider that the length of the rectangle was originally 19m instead of 7m. How would the new perimeter be calculated then?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the rectangle is 2(19)+2(8) = <<2(19)+2(8)=54>>54m.
### Answer: 54

Expected answer: 54

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 14m and a width of 18m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(14+18) = <<2*(14+18)=76>>76m.
### Answer: 76

Expected answer: 64

Correct? No

🔗 Problem

Problem: A rectangle has a length of 43 units and a width of 3 units. Calculate the perimeter of this rectangle.

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(43+3) = <<2*(43+3)=92>>92 units.
### Answer: 92

Expected answer: 92

Correct? Yes

🔗 Follow-up

Problem: Let's consider the same rectangle from before, but this time, the width has increased to 39 units while the length remained the same at 43 units. What is the new perimeter of this rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(43+39) = <<2*(43+39)=214>>214 units.
### Answer: 214

Expected answer: 164

Correct? No

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 36 meters and the width is 42 meters. Therefore, the perimeter is 2*(36+42) = <<2*(36+42)=2*78=156>>156 meters.
### Answer: 156

Expected answer: 156

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 48 meters and the width is 36 meters. Therefore, the perimeter is 2*(48+36) = <<2*(48+36)=2*84=168>>168 meters.
### Answer: 168

Expected answer: 168

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 1m and a width of 6m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(1+6) = <<2(1+6)=2(7)=14>>14 meters.
### Answer: 14

Expected answer: 14

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(28+6) = <<2(28+6)=2(34)=68>>68 meters.
### Answer: 68

Expected answer: 68

Correct? Yes

🔗 Problem

Problem: A rectangle has two sides measuring 45m and two sides measuring 8m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 45m and the width is 8m. Therefore, the perimeter is 2*(45+8) = <<2*(45+8)=100>>100m.
### Answer: 100

Expected answer: 106

Correct? No

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the garden is 2*(25+37) = <<2*(25+37)=2*62=124>>124 meters.
### Answer: 124

Expected answer: 124

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the garden is 2*(16+37) = <<2*(16+37)=2*53=106>>106 meters.
### Answer: 106

Expected answer: 106

Correct? Yes

🔗 Problem

Problem: A rectangle has length 36m and width 3m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(36+3) = <<2(36+3)=78>>78m.
### Answer: 78

Expected answer: 78

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(1+3) = <<2(1+3)=8>>8m.
### Answer: 8

Expected answer: 8

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the garden is 2*(45+50) = <<2*(45+50)=2*95=190>>190 meters.
### Answer: 190

Expected answer: 190

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 34m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(44+34) = <<2*(44+34)=2*78=156>>156m.
### Answer: 156

Expected answer: 156

Correct? Yes

🔗 Follow-up

Problem: Now suppose the width of the rectangle changes to 36m while the length remains the same at 44m. What would be the new perimeter of the rectangle in this case?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(44+36) = <<2*(44+36)=2*80=160>>160m.
### Answer: 160

Expected answer: 160

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 42 m and a width of 29 m. Calculate the perimeter of the rectangle.

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(42+29) = <<2*(42+29)=200>>200 m.
### Answer: 200

Expected answer: 142

Correct? No

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(44+35) = <<2*(44+35)=2*79=158>>158 meters.
### Answer: 158

Expected answer: 158

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(44+43) = <<2*(44+43)=2*87=174>>174 meters.
### Answer: 174

Expected answer: 174

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 45m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(44+45) = <<2(44+45)=2*89=178>>178m.
### Answer: 178

Expected answer: 178

Correct? Yes

🔗 Follow-up

Problem: The previous rectangle had its width increased from 45m to 48m while keeping the length the same. What is the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(44+48) = <<2(44+48)=2*92=184>>184m.
### Answer: 184

Expected answer: 184

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(3+39) = <<2*(3+39)=84>>84 meters.
### Answer: 84

Expected answer: 84

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(17+39) = <<2*(17+39)=104>>104 meters.
### Answer: 104

Expected answer: 112

Correct? No

🔗 Problem

Problem: A rectangular garden has sides that measure 32m and 12m. What is the perimeter of this garden?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides.
The perimeter of the garden is 2(32+12) = <<2*(32+12)=104>>104m.
### Answer: 104

Expected answer: 88

Correct? No

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the swimming pool is 2*(24+6) = <<2*(24+6)=60>>60 feet.
### Answer: 60

Expected answer: 60

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 27m and a width of 21m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 27*21 = <<27*21=567>>567 m^2.
### Answer: 567

Expected answer: 567

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 27*8 = <<27*8=216>>216 m^2.
### Answer: 216

Expected answer: 216

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The length of the rectangle is 49*2 = <<49*2=98>>98m.
### Answer: 98

Expected answer: 98

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The length of the rectangle is 18*2 = <<18*2=36>>36m.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 50m and a width of 36m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(50+36) = <<2*(50+36)=2*86=172>>172m.
### Answer: 172m

Expected answer: 172

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(50+44) = <<2*(50+44)=2*94=188>>188m.
### Answer: 188m

Expected answer: 188

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 8m and a width of 44m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(8+44) = <<2*(8+44)=100>>100m.
### Answer: 100

Expected answer: 104

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 47*21 = <<47*21=987>>987 square feet.
### Answer: 987

Expected answer: 987

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 21*15 = <<21*15=315>>315 square feet.
### Answer: 315

Expected answer: 315

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 19m and a breadth of 28m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + breadth).
The perimeter of the rectangle is 2(19 + 28) = <<2(19+28)=104>>104m.
### Answer: 104

Expected answer: 94

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width. The length of the rectangle is 39 meters, and the width is 10 meters. The area of the rectangle is 39*10 = <<39*10=390>>390 square meters.
### Answer: 390

Expected answer: 390

Correct? Yes

🔗 Follow-up

Problem: Suppose we discovered an error in the original measurements of the rectangle. The longer side is actually 37 meters instead of 39 meters. With the shorter side still at 10 meters, what is the new area of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The area of a rectangle is the product of its length and width. The length of the rectangle is 37 meters, and the width is 10 meters. The area of the rectangle is 37*10 = <<37*10=370>>370 square meters.
### Answer: 370

Expected answer: 370

Correct? Yes

🔗 Problem

Problem: A rectangle has identical sides, each measuring 19 meters. What is the total perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is the sum of the lengths of its sides. Since the rectangle has identical sides, the perimeter is 2*19 = <<2*19=38>>38 meters.
### Answer: 38

Expected answer: 76

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of the rectangle is 9m * 26m = <<9*26=234>>234m^2.
### Answer: 234m^2

Expected answer: 234

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The new area of the rectangle is 47m * 9m = <<47*9=423>>423m^2.
### Answer: 423m^2

Expected answer: 423

Correct? Yes

🔗 Problem

Problem: A rectangle has one side measuring 19cm and another side measuring 7cm. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 19cm and the width is 7cm. So, the perimeter is 2*(19+7) = <<2*(19+7)=50>>50cm.
### Answer: 50

Expected answer: 52

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The total area of the shutter is 10*6 = <<10*6=60>>60m^2.
### Answer: 60

Expected answer: 60

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of the rectangle is 2(24+15) = <<2(24+15)=2(39)=78>>78 units.
### Answer: 78

Expected answer: 78

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its four sides.
The perimeter of the rectangle is 2(24+10) = <<2(24+10)=2(34)=68>>68 units.
### Answer: 68

Expected answer: 68

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the plot is 32*27 = <<32*27=864>>864 square meters.
### Answer: 864

Expected answer: 864

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the plot is 32*18 = <<32*18=576>>576 square meters.
### Answer: 576

Expected answer: 576

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 15m and the width is 10m. So, the perimeter is 2*(15+10) = <<2*(15+10)=60>>60m.
### Answer: 60

Expected answer: 50

Correct? No

🔗 Problem

Problem: A rectangle has a length of 44m and a width of 37m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(44+37) = <<2(44+37)=2*81=162>>162m.
### Answer: 162

Expected answer: 162

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(25+37) = <<2(25+37)=2*62=124>>124m.
### Answer: 124

Expected answer: 124

Correct? Yes

🔗 Problem

Problem: A rectangle has length of 14 cm and width 8 cm. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 14+8+14+8 = <<14+8+14+8=44>>44 cm.
### Answer: 44

Expected answer: 44

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its length and width.
The perimeter of the rectangle is 14+18+14+18 = <<14+18+14+18=64>>64 cm.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 26m and a width of 28m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(26+28) = <<2*(26+28)=104>>104m.
### Answer: 104

Expected answer: 108

Correct? No

🔗 Problem

Problem: A rectangle has a length of 15m and a width of 37m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(15+37) = <<2*(15+37)=2*52=104>>104m.
### Answer: 104

Expected answer: 104

Correct? Yes

🔗 Follow-up

Problem: The width of the rectangle mentioned in the previous problem is now changed to 19m, while the length is still 15m. What is the new perimeter of this rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(15+19) = <<2*(15+19)=2*34=68>>68m.
### Answer: 68

Expected answer: 68

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 42m and a width of 40m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(42+40) = <<2*(42+40)=2*82=164>>164m.
### Answer: 164

Expected answer: 164

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(42+24) = <<2*(42+24)=2*66=132>>132m.
### Answer: 132

Expected answer: 132

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the garden is 30 feet, and the width of the garden is 21 feet. The perimeter of the garden is 2*(30+21) = <<2*(30+21)=2*51=102>>102 feet.
### Answer: 102

Expected answer: 102

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the garden is 5 feet, and the width of the garden is 21 feet. The perimeter of the garden is 2*(5+21) = <<2*(5+21)=2*26=52>>52 feet.
### Answer: 52

Expected answer: 52

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of the playground is 17*44 = <<17*44=748>>748 square meters.
### Answer: 748

Expected answer: 748

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of the playground is 17*37 = <<17*37=629>>629 square meters.
### Answer: 629

Expected answer: 629

Correct? Yes

🔗 Problem

Problem: Rectangle ABCD has a width of 47m and a length of 50m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is the sum of its 4 sides.
The perimeter of ABCD is 2(47) + 2(50) = <<2*47+2*50=194>>194m.
### Answer: 194

Expected answer: 194

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is the sum of its 4 sides.
The perimeter of ABCD is 2(47) + 2(48) = <<2*47+2*48=196>>196m.
### Answer: 196

Expected answer: 190

Correct? No

🔗 Problem

Problem: A rectangle has a length of 11m and a width of 31m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(11m + 31m) = <<2(11+31)=2*42=84>>84m.
### Answer: 84m

Expected answer: 84

Correct? Yes

🔗 Follow-up

Problem: Suppose we later find out that the actual length of the rectangle is 36m, not 11m as we originally thought, while the width remains the same at 31m. What would then be the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of the rectangle is 2(36m + 31m) = <<2(36+31)=2*67=134>>134m.
### Answer: 134m

Expected answer: 134

Correct? Yes

🔗 Problem

Problem: A rectangular garden has a length of 39m and a width of 19m. What is the area of this garden?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the garden is 39 × 19 = <<39*19=741>>741 m2.
### Answer: 741

Expected answer: 741

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The new area of the garden is 35 × 19 = <<35*19=665>>665 m2.
### Answer: 665

Expected answer: 665

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of one cut wooden plank is 29*11 = <<29*11=319>>319 square centimeters.
### Answer: 319

Expected answer: 319

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of one cut wooden plank is 29*5 = <<29*5=145>>145 square centimeters.
### Answer: 145

Expected answer: 145

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 25m and a width of 11m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(25+11) = <<2*(25+11)=2*36=72>>72m.
### Answer: 72m

Expected answer: 72

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(25+17) = <<2*(25+17)=2*42=84>>84m.
### Answer: 84m

Expected answer: 84

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 49m and a breadth of 9m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + breadth).
The perimeter of the rectangle is 2(49+9) = <<2*(49+9)=2*58=116>>116m.
### Answer: 116

Expected answer: 116

Correct? Yes

🔗 Follow-up

Problem: Suppose that the breadth of the same rectangle was instead 19m, while the length remains the same at 49m. What would be the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + breadth).
The perimeter of the rectangle is 2(49+19) = <<2*(49+19)=2*68=136>>136m.
### Answer: 136

Expected answer: 136

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 16m and a width of 7m. What is the area of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of this rectangle is 16*7 = <<16*7=112>>112 m^2.
### Answer: 112

Expected answer: 112

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of the playground is 2*(14+2) = <<2*(14+2)=2*16=32>>32m.
### Answer: 32

Expected answer: 32

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of the playground is 2*(14+3) = <<2*(14+3)=2*17=34>>34m.
### Answer: 34

Expected answer: 34

Correct? Yes

🔗 Problem

Problem: A square has a side length of 16 units. What is the perimeter of the square?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a square is 4 times the length of one side.
The perimeter of the square is 4*16 = <<4*16=64>>64 units.
### Answer: 64

Expected answer: 64

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2 times the sum of the length and the width.
The perimeter of the rectangle is 2*(22+16) = <<2*(22+16)=76>>76 units.
### Answer: 76

Expected answer: 76

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of this garden is 2*(21+7) = <<2*(21+7)=70>>70 meters.
### Answer: 70

Expected answer: 56

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the rectangle is 20*5 = <<20*5=100>>100 square meters.
### Answer: 100

Expected answer: 100

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and its width.
The area of the new rectangle is 20*17 = <<20*17=340>>340 square meters.
### Answer: 340

Expected answer: 340

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 37m and a width of 43m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(37+43) = <<2*(37+43)=2*80=160>>160m.
### Answer: 160

Expected answer: 160

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of this rectangle is 2*(32+43) = <<2*(32+43)=2*75=150>>150m.
### Answer: 150

Expected answer: 150

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 9m. If the width of the rectangle is 1m, what is the area?

InternLM-Math Base 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 9m * 1m = <<9*1=9>>9m^2.
### Answer: 9

Expected answer: 9

Correct? Yes

🔗 Follow-up

Problem: In the previous problem, we considered a rectangle with a length of 9m and a width of 1m. Now, consider if the width of the rectangle was actually 40m instead of 1m. What would be the new area of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 9m * 40m = <<9*40=360>>360m^2.
### Answer: 360

Expected answer: 360

Correct? Yes

🔗 Problem

Problem: A rectangle's area is the product of 22m and 23m. What is the area of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The area of the rectangle is 22m × 23m = <<22*23=506>>506 m2.
### Answer: 506 m2

Expected answer: 506

Correct? Yes

🔗 Follow-up

Problem: What if we reconsider the area of the rectangle, but this time its width is 6m instead of 23m, while maintaining the length at 22m. What would then be the area of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The area of the rectangle is 22m × 6m = <<22*6=132>>132 m2.
### Answer: 132 m2

Expected answer: 132

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The length of the rectangle is 14 cm and the width is 21 cm. Therefore, the perimeter is 2*(14+21) = <<2*(14+21)=98>>98 cm.
### Answer: 98

Expected answer: 70

Correct? No

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and its width.
The area of the swimming pool is 23*12 = <<23*12=276>>276 square meters.
### Answer: 276

Expected answer: 276

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of this rectangle is 2*(39+8) = <<2*(39+8)=2*47=94>>94 meters.
### Answer: 94

Expected answer: 94

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of this rectangle is 2*(39+23) = <<2*(39+23)=2*62=124>>124 meters.
### Answer: 124

Expected answer: 124

Correct? Yes

🔗 Problem

Problem: A rectangle has a width of 27m and a length of 40m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(40+27) = <<2*(40+27)=2*67=134>>134m.
### Answer: 134

Expected answer: 134

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(44+27) = <<2*(44+27)=2*71=142>>142m.
### Answer: 142

Expected answer: 142

Correct? Yes

🔗 Problem

Problem: A rectangle's length is 35m and its width is 9m. What is the perimeter of the rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(35+9) = <<2*(35+9)=2*44=88>>88m.
### Answer: 88

Expected answer: 88

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of the rectangle is 2*(19+9) = <<2*(19+9)=2*28=56>>56m.
### Answer: 56

Expected answer: 56

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 38m and a width of 11m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(38+11) = <<2*(38+11)=2*49=98>>98m.
### Answer: 98

Expected answer: 98

Correct? Yes

🔗 Follow-up

Problem: In the rectangle mentioned earlier, suppose we now construct another rectangle such that its length remains the same (38m), but the width is now 23m instead of 11m. How would the perimeter of this new rectangle be calculated?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(38+23) = <<2*(38+23)=2*61=122>>122m.
### Answer: 122

Expected answer: 122

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 28*29 = <<28*29=812>>812m^2.
### Answer: 812

Expected answer: 812

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The area of a rectangle is the product of its length and width.
The area of the rectangle is 28*35 = <<28*35=980>>980m^2.
### Answer: 980

Expected answer: 980

Correct? Yes

🔗 Problem

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 20B:

 Reasoning: The perimeter of a square is 4 times the length of one side.
The perimeter of the square is 4*9 = <<4*9=36>>36m.
### Answer: 36

Expected answer: 36

Correct? Yes

🔗 Problem

Problem: A rectangle has a length of 38m and a width of 29m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(38+29) = <<2(38+29)=2(67)=134>>134m.
### Answer: 134m

Expected answer: 134

Correct? Yes

🔗 Follow-up

Problem: Now consider that the rectangle from the previous question actually had a length of 6m instead of 38m. What would be the new perimeter of the rectangle?

InternLM-Math Base 20B:

  Reasoning: The perimeter of a rectangle is 2(length + width). The perimeter of this rectangle is 2(6+29) = <<2(6+29)=2(35)=70>>70m.
### Answer: 70m

Expected answer: 70

Correct? Yes

🔗 Problem

Problem: A rectangle has one side of length 45m and another side of length 47m. Find the perimeter of the rectangle.

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width). The perimeter of the rectangle is 2*(45+47) = <<2*(45+47)=188>>188m.
### Answer: 188

Expected answer: 184

Correct? No

🔗 Problem

Problem: Given a rectangle where the length is 24m and the width is 20m, what is the perimeter of this rectangle?

InternLM-Math Base 20B:

 The perimeter of a rectangle is the sum of the lengths of its sides. The perimeter of this rectangle is 2*24+2*20 = <<2*24+2*20=96>>96m.
### Answer: 96

Expected answer: 88

Correct? No

🔗 Problem

Problem: The length and width of a rectangle are both 32m. What is the perimeter of this rectangle?

InternLM-Math Base 20B:

 Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of this rectangle is 2*(32+32) = <<2*(32+32)=128>>128m.
### Answer: 128

Expected answer: 128

Correct? Yes

🔗 Follow-up

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 20B:

  Reasoning: The perimeter of a rectangle is 2*(length+width).
The perimeter of this rectangle is 2*(32+13) = <<2*(32+13)=90>>90m.
### Answer: 90

Expected answer: 90

Correct? Yes