(o i 1') [...] ... [...] ... [...] the procedure.

(o i 4') [...] what? [...] ... [...] you will see. [...] Square 2. [...] Multily [3 by] 2. The height is 6. [That is] the procedure.

(o i 10') [...] what? [...]

(o i 18') h1 Problem (i') [...]. What are [the length and width]?

(o i 19') [...] You will see 3. Break 3 in half. You will see 1;30.

(o i 20') [... Release the rešiprošal of 1;30.] You will see 0;40. The ratio of the width. Release the rešiprošal of 12, the ratio of the depth.

(o i 21') [You will see 0;05.] Multiply 0;05 by 1. You will see 0;05. Multiply by 0;40. You will see 0;03 20.

(o i 22') Multiply 0;03 20 by 0;05. You will see 0;00 16 40. Find the reciprocal of 0;00 16 40. You will see 3 36. Multiply 3 36 by 1;10. You will see 4 12. The square-side is 6. Multiply 6 by 0;05. The length is 0;30. Multipy 6 by 0;03 20.

(o i 24') The width is 0;20. Multiply 6 by 1. You will see 6, the depth. That is the procedure.

(o i 26') h1 Problem (ii') An excavation. The length is equal to the depth. I removed 1, the volume. I summed my ground-area and the volume: 1;10. The (sum of the) length and width is 0;50. What are the length and width?

(o i 27') You: Multiply 0;50 by 1, the ratio. You will see 0;50. Multiply 0;50 by 12. You will see 10.

(o i 28') Square 0;50. You will see 0;41 40. Multiply by 10. You will see 6;56 40. Find its reciprocal. You will see 0;08 38 44.

(o i 29') Multiply by 1;10. You will see 0;10 04 48. The square-sides are 0;36, 0;24, 0;42.

(o i 30') Multiply 0;36 by 0;50. The length is 0;30. Multiply 0;24 by 0;50. The width is 0;20. <Multiply> 0;36 by 10. The depth is 6. The procedure.

(o i 32') h1 Problem (iii') An excavation. The depth is as much as the length. I removed 1, the volume. I summed my ground-area and volume: 1;10. The length exceeded the width by 0;10.

(o i 33') You: Put down 1 and 12, the ratios. Multiply 0;10, the excess, by 1. You will see 0;10. Multiply by 12. You will see 2.

(o i 34') Square 0;10. You will see 0;01 40. Multiply by 2. You will see 0;03 20. Find the reciprocal of 0;03 20. You will see 18.

(o i 35') Multiply by 1;10. You will see 21. The square-sides are 3, 2, 21. Multiply [0;10 by 3]. The length is 0;30.

(o i 36') Multiply 0;10 by 2. The width is 0;20. Multiply 3 by 2. You will see 6. The depth is [6]. The procedure.

(o i 38') h1 Problem (iv') An excavation. The length is as much as the depth. I removed a volume.I summed my ground-area and the volume: 1;10. The length is 0;30. What is the width?

(o i 39') You: multiply 0;30, the length, by 12. You will see 6, the depth. Add 1 to 6. You will see 7.

(o i 40') The reciprocal of 7 cannot be found. What should I put to 7 that will give me 1;10? Put 0;10. Solve the reciprocal of 0;30, the length.

(o i 41') You will see 2. Multiply 0;10 by 2. You will see 0;20, the width. The procedure.

(o i 43') h1 Problem (v') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume: 1;10. The width is 0;20. <What is> the length?

(o i 44') You: multiply 0;20 by 12. You will see 4. Multiply 4 by 1;10. You will see 4;40.

(o i 45') Break in 1/2 0;20, the width. You will see 0;10. Square 0;10. You will see 0;01 40. Add to 4;40.

(o i 46') You will see 4;41 40. The square-side is 2;10. Subtract 0;10 that you squared and you will see 2. Find the reciprocal of 4. you will 0;15. Multiply by 2.

(o i 48') You will see 30, the length. The procedure.

(o ii 1') h1 Problem (vi') [...] add [...] 4;30, 53;20, 1;45 ... [...]

(o ii 3') The length is 0;30. Multiply 0;22 30 by 0;53 20. The width is 0;20. [...] The procedure.

(o ii 5') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume.

(o ii 6') I took a 7th. I added (it) to my ground-area and it was 0;20. [The length is] 0;30. [What is the width?]

(o ii 7') You: multiply 0;30 by 12. You will see 6, the depth. [Add] 1 to 6.

(o ii 8') You will see 7. Take a 7th. You will see 1. Sum 1 and 1.

(o ii 9') You will see 2. Find the reciprocal of 2. You will see 0;30. Multiply 0;30 by 0;20, the sum.

(o ii 10') You will see 0;10. Find the reciprocal of 0;30, the length. You will see 2. Multiply 2 by 0;10. [The width is 0;20.] The procedure.

(o ii 12') h1 Problem (vii') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume: 1;10 I took a 7th of it. I added (it) to my ground-area: 0;20. The width was 20.

(o ii 14') You: multiply 0;20 by 7. You will see 2;20. Multiply 0;20, the width, by 12.

(o ii 15') You will see 4. Multiply 4 by 2;20. You will see 9;20. Add 1 to 7.

(o ii 16') You will see 8. Multiply 8 by 0;20. You will see 2;40. Break 1/2 of 2;40. [Square (it).]

(o ii 17') You will see 1;46 10. Add to 9;20. You will see 11;06 40.

(o ii 18') The square-side is 3;20. Subtract the 1;20 that you squared. You will see 2.

(o ii 19') Find the riciprocal of 4. You will see 0;15. Multiply 0;15 by 2. [The length is] 0;20. The procedure.

(o ii 21') h1 Problem (viii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as that by which the the reciprocal exceeds its reciprocal pair. I removed [a volume] of 16. What are the length, width, and depth?

(o ii 23') You: find the reciprocal of 12. You will see [0;05]. Multiply 0;05 by 16. You will see [1];20.

(o ii 24') The reciprocal is 1;20. [Find] the reciprocal of 1;20. You will see [0;45]. Its recipocal pair is 0;45. The depth is [16]. The procedure.

(o ii 26') h1 Problem (ix') An excavation. The length is as much as a reciprocal. The width is as much as [its reciprocal pair]. <The depth is> as much as that by which the reciprocal exceeds its reciprocal pair.

(o ii 27') I removed a volume of 26. What are the reciprocal, [its reciprocal pair, and the depth]?

(o ii 28') You: solve the reciprocal of 12. You will see 0;05. Multiply 36 by 0;05 <<by 36>>.

(o ii 29') You will see 3. Break 3 in 1/2. [You will see 1;30.] The recriprocal is 1;30. Its reciprocal pair is 0;40. The depth is 36. The procedure.

(o ii 31') h1 Problem (x') An excavation. The length is as much as a reciprocal. [The width is as much as its reciprocal pair]. The depth is as much as the total of the reciprocal and its reciprocal pair.

(o ii 32') I removed a volume of 26. What are the reciprocal, its reciprocal pair, and the depth?

(o ii 33') You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 26.

(o ii 34') You will see 2;10. Break 2;10 in 1/2. Square (it). You wil see 1;10 25. <Subtract 1 from 1;10 25. You will see 0;10 25.>

(o ii 35') The square-side is 0;25. Add and subtract (it) to <1>;05. You will see 1;30 and 0;40.

(o ii 36') The reciprocal is 1;30. Its reciprocal pair is 0;40. The depth is 26. The procedure.

(o ii 38') h1 Problem (xi') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as that by which the reciprocal exceeds its reciprocal pair, subtracted from the reciprocal. I removed a volume of 6. [What are] the reciprocal and its reciprocal pair?

(o ii 40') You: find the reciprocal of 12. You will see 0;05. Multiply by 6. You will see 0;30.

(o ii 41') Find the reciprocal of 0;30. You will see 2. The reciprocal is 0;30, its reciprocal pair 2, the depth 6. The procedure.

(o ii 43') h1 Problem (xii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The [depth] is as much as the total of the reciprocal and its recriproal pair. [I removed] a volume of 30.

(o ii 44') You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 30, the volume.

(o ii 45') You will see 2;30. Break 2;30 in 1/2. Square (it). You will see [1;33] 45.

(o ii 46') Subtract 1 from 1;33 45. You will see 0;33 45. The square-side is 0;45.

(o ii 47') Add and subtract from 1;45. You will see 2 and 0;30. The procedure.

(o ii 49') h1 Problem (xiii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as the reciprocal pair.

(o ii 50') I removed a volume of 20. What are the reciprocal, its reciprocal pair, and the depth?

(o ii 51') You: find the reciprocal of 12. Multiply by 20. You will see 1;40.

(o ii 52') The reciprocal is 1;40. Its reciprocal pair is 0;36. The depth is 20. The procedure.

(r i 1) h1 Problem (xiv') An excavation. The depth is as much as I squared and 7 cubits. I removed 3 20, the volume.

(r i 2) What are the length, width, and depth?

(r i 3) You: take a 7th of 7. You will see 1. Find the reciprocal of 12. You will see 0;05.

(r i 4) Multiply 0;05 by 1. You will see 0;05. Multiply 0;05 by 12. You will see 1.

(r i 5) Square 0;05. Multiply 0;00 25 by 1. You will see 0;00 25. Find the reciprocal of 0;00 25. You will see 2 24. Multiply 2 24 by 3 20, the volume. You will see 8 00 00. (With)what does it square itself?

(r i 7) It squares itself (with sides of) 1 00, 1 00, 8. Multiply 0;05 by 1 00. You will see 5. The length is 5 cubits.

(r i 8) Multiply 8 by 1. The depth is 8 cubits. The procedure.

(r i 10) h1 Problem (xv') An excavation. The depth is as much as I squared and 7 cubits. I removed a volume of <3 15> <<13>>.

(r i 11) What are the length, width, and depth?

(r i 12) You: do as before. (W)ith what does 7 48 square itself?

(r i 13) Its square-sides are 6, 6, and 13. It squares itself (with a side of) 6. The depth is 13. The procedure.

(r i 15) h1 Problem (xvi') An excavation. The depth is as much as I made square. I removed a volume of 1;30. <What are> the length, width, and depth?

(r i 16) You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 1;30. <You will see> 0;07 30.

(r i 17) The square-side is 0;30. Multiply 0;30 by 1. It squares itself (with a side of) 0;30. Multilpy 0;30 by 12. The depth is 6. The procedure.

(r i 19) h1 Problem (xvii') An excavation. The depth is as much as I made square and 1 cubit excess. I removed a volume of 1;45.

(r i 20) You: Multiply 0;05, the excess by 1, the ratio. You will see 0;05. Muliply by 12. You will see 1.

(r i 21) Square 0;05. You will se 0;00 25. Multiply 0;00 25 by 1. You will see 0;00 25. [Find] the reciprocal of 0;00 25.

(r i 22) You will see 2 24. Multiply 2 24 by 1;45. [You will see] 4 12.

(r i 23) In the square-side, 1 added, the square-sides are 6 (and) 1. Multiply 6 by 0;05. You will see 0;30. It squares itself (with a side of) <0;30>. The depth is 7!(6). The procedure.

(r i 25) h1 Problem (xviii') An excavation. The depth is 3;20. I removed a volume of 27;46 40. The length exceeds the width by 0;50.

(r i 26) You: find the reciprocal of 3;20, the depth. You will see 0;18. Multiply by 27;46 40, the volume.

(r i 27) You will see 8;20. Break 50 in 1/2. Square (it). You will see 0;10 25.

(r i 28) Add to 8;20. You will see 8;30 25.

(r i 29) The square-side is 2;55. [Put it down] twice. Add to 1, subtract from 1.

(r i 30) You will see 3;20, the length, (and) 2;30, the width. The procedure.

(r i 32) h1 Problem (xix') An excavation. The depth is 3;20. [I removed a volume of] 27;46 40. [I summed the length and width:] 5;50.

(r i 33) You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by [27;46 40].

(r i 34) You will see 8;20. Break 1/2 of 5;50. Square (it). [You will see] 8;[30 25].

(r i 35) Subtract 8;20 from inside it. [You will see] 0;10 25. [The square-side is 0;25.]

(r i 36) Add and subtract from 2;55. [The length is] 3;20. [The width is 2;30.] The procedure.

(r i 38) h1 Problem (xx') An excavation. The depth is 3;20. [I removed] a volume of 27;46 40. [The width exceeds the depth by as much as 2/3 of the length.]

(r i 39) You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by 27;[46 40].

(r i 40) You will see 8;20. Multiply 8;20 by 0;40. [You will see] 5;33 [20. Multiply 3;20, the depth, by 0;05: 0;16 40.]

(r i 41) Start again. Break 1/2 of 0;16 40. You will see 0;08 20. Square (it). [Add 0;01 09 26 40 to 5;33 20.]

(r i 42) (With) what does it square itself? Put down 2;21 40 twice. [Add and subtract] 0;08 20.