NCERT Solutions Class 7 Maths Part 2 Chapter 7: Finding the Unknown (Hinglish)

Ganita Prakash Grade 7 (2026-27) ke Part 2 Chapter 7, Finding the Unknown, ke har sawaal ke jawab: wazan waali paheliyan, text ke saare sawaal, har Figure it Out, Math Talk aur Try This, Mind the Mistake, itihaas ke sawaal aur A Magic Trick. Page numbers kitaab ke hisaab se hain (pages 164–190). Har jawab ko waapas uski equation mein rakhkar check kiya gaya hai. Chhote notes: Part 2 Chapter 7 ke notes.

Anjaan wazan (pages 164–166)

Latakne waala taraazu kaise kaam karta hai. Upar likhi sankhya dandi se latke kul wazan ki hai. Seedhi dandi ka matlab hai dono taraf barabar wazan, to har taraf kul ka aadha hai.

Fig. 7.1. Kul 16, to har taraf 8. Left: 3 + kachhua + 3 = 8, to kachhua 2 hai. Right: phool 8 hai.

Fig. 7.2. Kul 24, to har taraf 12. Left: 2 + machhli + 2 = 12, to machhli 8 hai. Right: 8 + submarine = 12, to submarine 4 hai.

Fig. 7.3. Kul 8, to har taraf 4. Kitaab 4 hai. Do pencil box ek chhoti dandi pe seedhe latke hain, to har pencil box 2 hai.

Fig. 7.4. Kul 18, to har taraf 9. Left: 5 + 4 baadal = 9, to har baadal 1 hai. Right: 3 bijli = 9, to har bijli 3 hai.

Fig. 7.5. Kul 40, to har taraf 20. Right: 4 mukut = 20, to har mukut 5 hai. Left: 5 + 5 ratna = 20, to har ratna 3 hai.

Fig. 7.6. Bread ke teen slice 2 + 2 + 2 = 6 hain, to do ande 6 ke hain aur har anda 3 hai.

Fig. 7.7. Chaar cross 16 ke hain. Doosri taraf 4 + do chhalle hain, to do chhalle 12 ke aur har chhalla 6 hai.

Fig. 7.8. Tarbooz 10 hai. Doosri taraf 4 + do kele hain, to har kela 3 hai.

Math Talk: apne jawabon pe charcha karo. Har jawab ka kaaran ek hi tarah diya ja sakta hai: dandi seedhi hai, to dono taraf barabar hain; jaane hue wazan tay kar dete hain ki unknown kya hona chahiye; aur value waapas rakhne pe dono taraf barabar aate hain.

Fig. 7.9. Dono palde mein 2 kg ka baat hai. Dono se hata do: bori 10 kg ki hai.

Fig. 7.10. Har palde se ek bori hatao. Bachi ek bori 10 kg + 4 kg ke barabar hai, to bori 14 kg ki hai.

Fig. 7.11. Paanch boriyan, do boriyon aur 10 kg, 10 kg aur 1 kg ke barabar hain. Har palde se do boriyan hatao: teen boriyan 21 kg ke barabar, to bori 7 kg ki hai.

Fig. 7.12. 90 boriyan aur 50 kg, 60 boriyon aur 500 kg ke barabar hain. Dono palde se 60 boriyan aur 50 kg hatao: 30 boriyan 450 kg ke barabar, to bori 15 kg ki hai.

Matchstick pattern (pages 166–167)

Theek 99 tiliyan kis position mein lagti hain? Position n pe 2n + 1 tiliyan lagti hain, to 2n + 1 = 99 chahiye. Position 49.

Math Talk: 2n + 1 = 99 se n nikaalne ke tareeke. Steps ulto: 1 ghataao to 98, phir aadha karo to 49. Ya values aazmaate jaao jab tak 2n + 1, 99 na ho jaaye. Ya gino: position 1 pe 3 tiliyan hain aur har agli position 2 jodti hai, to 3 + 2(n − 1) = 99 se bhi n = 49 aata hai.

Kya theek 200 tiliyon se koi arrangement ban sakta hai? Nahi. 2n hamesha even hai, to 2n + 1 hamesha odd hai aur kabhi 200 nahi ho sakta (2n + 1 = 200 ke liye n = 99.5 chahiye).

Equations banaana (page 168)

Figs. 7.6 se 7.11 ke liye equations banao aur solve karo.

• Fig. 7.6 (anda e): 2e = 6, to e = 3.
• Fig. 7.7 (chhalla y): 4 + 2y = 16, to y = 6.
• Fig. 7.8 (kela b): 2b + 4 = 10, to b = 3.
• Fig. 7.9 (bori s): s + 2 = 10 + 2, to s = 10.
• Fig. 7.10: 2s = s + 10 + 4, to s = 14.
• Fig. 7.11: 5s = 2s + 10 + 10 + 1, to 3s = 21 aur s = 7.

Har value wahi hai jo taraazu se mili thi.

Math Talk: 5 equations banao aur unhe solve karne ke tareeke dhoondho. Jaise x + 9 = 15 (x = 6), 4y = 28 (y = 7), 3z − 2 = 13 (z = 5), p ÷ 4 = 5 (p = 20) aur 2q + 3 = q + 8 (q = 5). Har ek trial and error se ya, jaldi, dono taraf ek hi step se operations ulat kar solve ho sakti hai.

Equations ko tareeke se solve karna (pages 168–171)

Kya 2n + 1 = 99 ka koi aur solution ho sakta hai? Nahi. n mein har 1 badhne pe LHS 2 badhta hai, to sirf n = 49 se theek 99 aata hai.

5x − 4 = 7 trial and error se solve karo. x = 2 se 6 aata hai aur x = 3 se 11, to x 2 aur 3 ke beech hai. Decimal aazmaao: x = 2.2 se 11 − 4 = 7. To x = 2.2, yaani 11/5. Isse dikhta hai ki andaaze mein kitna samay lagta hai.

15 + 8 = 23 ke dono taraf 10 jodna. 15 + 8 + 10 = 33 aur 23 + 10 = 33, to barabari bani rehti hai. Dono taraf ek hi sankhya ghataane, guna karne ya bhaag dene pe bhi yahi hota hai (zero se bhaag chhodkar).

Example 1. 14593 − 1459 + 145 − 14 = 13353 − 88 = 13265, lamba expression hal kiye bina. Ye isliye chalta hai kyunki ghataana jod ko ulat deta hai: dono taraf se 88 ghataane se + 88 hat jaata hai.

Example 2. 23 × 41 × 11 × 8 = 5,80,888 ÷ 7 = 82,984. Haan, ye dono taraf 7 se bhaag dene jaisa hi hai, jisse LHS se factor 7 hat jaata hai.

Example 3. − (−67) hataane ke liye dono taraf (−67) jodo: 12345 − 5432 + 135 − 24 = 7091 + (−67) = 7024. (Check: 12345 − 5432 = 6913, + 135 = 7048, − 24 = 7024.) Page 170 pe kitaab + 135 ki jagah + 132 chhaapti hai; + 132 se jawab 7021 aata, to 7024 page 169 ke sawaal ka hi jawab hai.

Example 4. Dono taraf 8/9 se bhaag do, yaani 9/8 se guna karo: (35/113) × 24 × 14 = (94080/1017) × (9/8) = 11760/113. Check: 35 × 24 × 14 = 11760.

5x − 4 = 7 solve karo aur check karo. Dono taraf 4 jodo: 5x = 11. Dono taraf 5 se bhaag do: x = 11/5. Check: 5 × 11/5 − 4 = 11 − 4 = 7, jo RHS hai.

Example 5. 11y + (−5) = 61 se 11y = 66 aur y = 6. Check: 11 × 6 + (−5) = 66 − 5 = 61.

Example 6. 6y + 7 = 4y + 21: 4y ghataao to 2y + 7 = 21, 7 ghataao to 2y = 14, to y = 7. Check: 49 = 49.

Figure it Out (page 172)

1. Solve karo aur check karo.

• (a) 3x − 10 = 35: 3x = 45, to x = 15. Check: 45 − 10 = 35.
• (b) 5s = 3s: dono taraf se 3s ghataao, 2s = 0, to s = 0. Check: 0 = 0.
• (c) 3u − 7 = 2u + 3: dono taraf se 2u ghataao, u − 7 = 3, to u = 10. Check: 23 = 23.
• (d) 4(m + 6) − 8 = 2m − 4: 4m + 16 = 2m − 4, to 2m = −20 aur m = −10. Check: 4 × (−4) − 8 = −24 aur −20 − 4 = −24.
• (e) u ÷ 15 = 6: dono taraf 15 se guna karo, u = 90. Check: 90 ÷ 15 = 6.

2. Aisi equation banao jiska koi solution na ho. x + 4 = x + 5. Kisi sankhya se 4 zyaada kabhi usi sankhya se 5 zyaada ke barabar nahi hota; dono taraf se x ghataao to 4 = 5 bachta hai, jo galat hai. Ek aur: 2x + 1 = 2(x + 3).

Problems hal karna (pages 174–179)

Example 7. Kya Ranjana theek 100 tiles laga sakti hai? Step k mein 3k + 1 tiles lagti hain, to 3k + 1 = 100, 3k = 99 aur k = 33. Haan, Step 33 mein.

Example 8. Madhubanti kitne dost bula sakti hai? Kul p log hon to 25p + 50 = 500, to 25p = 450 aur p = 18. Uske 5 logon ka parivaar hata do, to wo 13 dost bula sakti hai. Srikanth ka tareeka bhi yahi deta hai: 25(f + 5) = 450, to f + 5 = 18 aur f = 13.

Example 9. Jawab m = 7 check karo. Jahnavi: 4000 + 650 × 7 = 4000 + 4550 = ₹8,550. Sunita: 5050 + 500 × 7 = 5050 + 3500 = ₹8,550. Barabar, to jawab sahi hai.

Example 10. Teeno tareekon se x = 21. Check: 28 × (21 + 4) + 300 = 28 × 25 + 300 = 700 + 300 = 1000.

Example 11. Riyaz ki trick. 4x − 4 = 24 se 4(x − 1) = 24, x − 1 = 6 aur x = 7. Doosri sankhyayein aazmaao: 5 se 2, 8 aur 16 aata hai; 10 se 7, 28 aur 36. Aakhri jawab hamesha 4 × (shuru ki sankhya) − 4 hota hai. Aasaan niyam: aakhri jawab ko 4 se bhaag do aur 1 jodo, shuru ki sankhya mil jaayegi (24 ÷ 4 + 1 = 7).

Example 12. Kanche. y + (y + 30) = 60 se 2y + 30 = 60, 2y = 30 aur y = 15. Suresh ke paas 15 kanche aur Ramesh ke paas 45. Check: 15 + 45 = 60 aur 45 − 15 = 30.

Equations banaana (pages 179–180)

Math Talk: aisi equations jinka solution y = 5 ho. Jaise 2y − 1 = 9, y ÷ 5 = 1, 10 − y = 5 aur 4y = 3y + 5. Har ek y = 5 se shuru karke dono taraf ek hi operation karke bani hai, ya y = 5 rakhkar check ho sakti hai.

Chains ko neeche se upar banao. Pehli chain: 5 = y; 6 ghataao to −1 = −6 + y; y ghataao to −1 − y = −6, yaani −y − 1 = −6; −1 se guna karo to y + 1 = 6. Doosri chain: y = 5; 2 jodo to y + 2 = 7; 3 se guna karo to 3y + 6 = 21; 6 ghataao to 3y = 15. Tulna: upar jaane mein neeche ke har step ka ulta operation, ulte kram mein lagta hai.

Bina hisaab kiye chains ki har equation ka unknown. Har ek mein y = 5, kyunki har equation pichhli equation se dono taraf ek hi step karke bani hai, jo solution nahi badalta.

Example 13. 100x + 75 = 250 ke liye ek asli sthiti. Ek school ₹x ki 100 pencil khareedta hai aur ₹75 delivery deta hai; bill ₹250 hai. Tab 100x = 175 aur x = 1.75, to ek pencil ₹1.75 ki hai.

Figure it Out (page 181)

1. Paanch equations jinka solution x = −2 ho. x + 5 = 3; 3x = −6; 2x + 7 = 3; 10 − x = 12; x ÷ 2 = −1. x = −2 rakhne pe har ek sahi hoti hai.

2. Unknown nikaalo.

• (a) 2y = 60: y = 30.
• (b) −8 = 5x − 3: 5x = −5, to x = −1.
• (c) −53w = −15: w = 15/53.
• (d) 13 − z = 8: z = 5.
• (e) k + 8 = 12 − k: 2k = 4, to k = 2.
• (f) 7m = m − 3: 6m = −3, to m = −1/2.
• (g) 3n = 10 + n: 2n = 10, to n = 5.

3. Teen ankon ki sankhya. Ikaai ka ank u maano. Tab dahaai ka ank u − 3 aur saikda ka ank u − 6 hai, to u + (u − 3) + (u − 6) = 15, 3u = 24 aur u = 8. Sankhya hai 258. Check: 2 + 5 + 8 = 15.

4. Eent. Wazan w kg maano. Tab w = w/2 + 1, to w/2 = 1 aur w = 2 kg. Check: 2 ka aadha 1 hai, aur 1 + 1 = 2.

5. Sankhya. n/4 + 9 = n, to 9 = 3n/4 aur n = 12. Check: 12 ÷ 4 + 9 = 3 + 9 = 12.

6. Diya hai 4k + 1 = 13, to 4k = 12 aur k = 3:

• (a) 8k + 2 = 2(4k + 1) = 26
• (b) 4k = 12
• (c) k = 3
• (d) 4k − 1 = 11
• (e) −k − 2 = −5

Mind the Mistake, Mend the Mistake (pages 181–182)

1. 4x + 6 = 10. Galat step: 4x = 10 + 6. + 6 doosri taraf jaakar − 6 banta hai: 4x = 10 − 6 = 4, to x = 1.
2. 7 − 8z = 5. Step 8z = 7 − 5 = 2 sahi hai, par z = 4 galat hai: 2 ko 8 se bhaag dene ki jagah 8 ko 2 se bhaag diya. z = 2/8 = 1/4.
3. 2v − 4 = 6. Galat step: v − 4 = 6 − 2 (factor 2 sirf ek term se hataakar ghata diya). Pehle 4 ko le jaao: 2v = 10, to v = 5.
4. 5z + 2 = 3z − 4. Galat step: 5z + 3z = −4 + 2. Dono le jaayi gayi terms ka sign badalna chahiye: 5z − 3z = −4 − 2, to 2z = −6 aur z = −3.
5. 15w − 4w = 26. Step 15w = 26 + 4w sahi hai, par 26 + 4w, 30 nahi hota. Seedha tareeka: 11w = 26, to w = 26/11.
6. 3x + 1 = −12. Galat step: x + 1 = −12/3 (sirf 3x aur −12 ko bhaag diya). Pehle 1 ghataao: 3x = −13, to x = −13/3.
7. 4(4q + 2) = 50. Galat step: 4(4q) = 50 − 2 (bracket ke andar ka 2 asal mein 4 × 2 = 8 hai). 16q + 8 = 50, to 16q = 42 aur q = 21/8.
8. −2(3 − 4x) = 14. Galat step: −6v − 8x = 14. “v” chhapaai ki chook hai, aur −2 × (−4x) = + 8x hota hai. To −6 + 8x = 14, 8x = 20 aur x = 5/2.
9. 3(7y + 4) = 9 + 5y. Do galat steps: 3 se bhaag dene mein 5y ko bhi bhaag dena tha, aur 7y − 5y + 4 = 3 se 2y = 3 − 4 aata hai, 4 − 3 nahi. Sahi: 21y + 12 = 9 + 5y, to 16y = −3 aur y = −3/16.

A Pinch of History (pages 182–184)

Example 16: Bhaskaracharya ka ghode waala sawaal. 300 + 6x = 10x − 100 se 400 = 4x, to ghoda ₹100 ka hai. Check: 300 + 600 = 900 aur 1000 − 100 = 900. (Kitaab ise Example 16 kehti hai, haalaanki Examples 14 aur 15 kahin nahi hain.)

Ax + B = Cx + D ka formula. 5x + 4 = 3x + 8 ke liye: x = (8 − 4) ÷ (5 − 3) = 2. 3x − 6 = 2x + 4 ke liye 3, −6, 2 aur 4 lo: x = (4 − (−6)) ÷ (3 − 2) = 10. Brahmagupta ka niyam x = (D − B) ÷ (A − C) bachat waale sawaal ke liye bhi m = 1050 ÷ 150 = 7 deta hai.

Formula se 2x + 3 = 4x + 5 solve karo. x = (5 − 3) ÷ (2 − 4) = 2 ÷ (−2) = −1. Check: 2(−1) + 3 = 1 aur 4(−1) + 5 = 1.

Figure it Out (pages 185–189)

1. Khaali jagah integers se bharo.

• (a) 5 × ___ − 8 = 37: 5 × ___ = 45, to khaali jagah mein 9.
• (b) 37 − (33 − ___) = 35: 33 − ___ = 2, to khaali jagah mein 31.
• (c) −3 × (−11 + ___) = 45: −11 + ___ = −15, to khaali jagah mein −4.

2. Ranju ke note. Har tarah ke n note maano. 50n + 100n = 750, to 150n = 750 aur n = 5. ₹50 ke 5 note aur ₹100 ke 5 note.

3. Dhabbe. 3 dhabbe aur 4 dikhte bindu hain. Agar har dhabba b bindu chhupaata hai, to 3b + 4 = 25, to 3b = 21 aur b = 7.

4. Machines. (a) Machine 3 jodti hai, 4 se guna karti hai aur 5 ghataati hai. Aakhir se ulto: 43 + 5 = 48, 48 ÷ 4 = 12, 12 − 3 = 9; aur 75 + 5 = 80, 80 ÷ 4 = 20, 20 − 3 = 17. (b) Input do raaston se jaata hai, × 3 aur + 3, aur nateeje ghataaye jaate hain: 3x − (x + 3) = 2x − 3 (12 se: 36 − 15 = 21). To 2x − 3 = 63 se x = 33, aur 2x − 3 = 227 se x = 115.

5. Inputs. Do baar 3 se bhaag dene pe 5 aata hai, to input hai 5 × 3 × 3 = 45. Do baar 4 ghataane pe −11 aata hai, to input hai −11 + 4 + 4 = −3.

6. Taxi. 800 + 20k = 2200, to 20k = 1400 aur k = 70 km.

7. Do sankhyayein. x + 3x = 76, to 4x = 76 aur x = 19. Sankhyayein hain 19 aur 57.

8. Khidki ki grill. Upar se neeche: 3 cm ki patti, phir g cm ke chhe gap jinke beech 2 cm ki paanch chhadein, aur neeche 3 cm ki patti (neeche ki patti ko upar jaisi 3 cm maana). To 3 + 6g + 5 × 2 + 3 = 34, 6g + 16 = 34, 6g = 18 aur gap 3 cm hai.

9. Juice aur milkshake. Milkshake ₹m ka maano; juice ₹(m − 15) ka. 4(m − 15) + 7m = 600, to 11m − 60 = 600, 11m = 660 aur m = 60. Milkshake ₹60, fruit juice ₹45. Check: 4 × 45 + 7 × 60 = 180 + 420 = 600.

10. Diya hai 28p − 36 = 98. To 28p − 38 = 98 − 2 = 96, aur 14p − 19, 28p − 38 ka aadha hai, to 14p − 19 = 48.

11. Galtiyan dhoondho aur theek karo.

• (a) 6x + 9 = 66: 6 aur 66 ko 6 se bhaag diya par 9 ko nahi. Sahi: 6x = 57, to x = 57/6 = 9.5. Check: 57 + 9 = 66.
• (b) 14y + 24 = 36: koi galti nahi. Aadha karne pe 7y + 12 = 18, to 7y = 6 aur y = 6/7.
• (c) 4x − 5 = 9x + 8: − 5 doosri taraf jaakar + 5 banna chahiye tha, aur aakhir mein 3 ÷ (−5) ulta likh diya. Sahi: 4x = 9x + 13, to −5x = 13 aur x = −13/5.

12. Triangles ke angles. Triangle ke angles ka jod 180° hota hai. Pehla triangle (do bhujaayein barabar, to neeche ke angles barabar): y + (y + 15) + (y + 15) = 180, 3y + 30 = 180, y = 50, to angles 50°, 65° aur 65° hain. Doosra: x + (x − 10) + (x + 10) = 180, 3x = 180, x = 60, to angles 60°, 50° aur 70° hain.

13. Chaar equations jinme u = 6. u + 4 = 10; 3u = 18; u ÷ 2 = 3; 2u − 5 = 7.

14. Bakhshali ka sawaal. Pehle ko a mile, to doosre ko 2a, teesre ko 3 × 2a = 6a aur chauthe ko 4 × 6a = 24a. a + 2a + 6a + 24a = 33a = 132, to pehle vyakti ko 4 milta hai (aur baakiyon ko 8, 24 aur 96).

15. Giraffe. h = h/2 + 2.5, to h/2 = 2.5 aur giraffe 5 m ka hai.

16. Tiliyon ke pattern. Pehli figure: har position mein ek square judta hai, jo 3 nayi tiliyon se banta hai, aur position 1 pe 6 tiliyan hain, to position n pe n squares aur 3n + 3 tiliyan. Doosri figure: har position position 1 ke 4 squares aur 13 tiliyon mein 3 squares aur 9 tiliyan jodti hai, to position n pe 3n + 1 squares aur 9n + 4 tiliyan.

• (a) Position 11 pe squares: pehli figure mein 11, doosri mein 34.
• (b) Position 11 pe tiliyan: pehli mein 36, doosri mein 103.
• (c) 85 tiliyan: pehli figure mein 3n + 3 = 85 se 3n = 82, aur 82, 3 ka multiple nahi, to nahi; doosri figure mein 9n + 4 = 85 se n = 9, to haan, position 9.
• (d) 150 tiliyan: pehli figure mein 3n + 3 = 150 se n = 49, to haan, position 49; doosri figure mein 9n + 4 = 150 se 9n = 146, aur 146, 9 ka multiple nahi, to nahi.

17. Sankhya. x + 36 = 10x, to 9x = 36 aur x = 4. Check: 4 + 36 = 40 = 10 × 4.

18. Solve karo.

• (a) 5(r + 2) = 10: r + 2 = 2, to r = 0.
• (b) −3(u + 2) = 2(u − 1): −3u − 6 = 2u − 2, to −5u = 4 aur u = −4/5.
• (c) 2(7 − 2n) = −6: 7 − 2n = −3, to 2n = 10 aur n = 5.
• (d) 2(x − 4) = −16: x − 4 = −8, to x = −4.
• (e) 6(x − 1) = 2(x − 1) − 4: 4(x − 1) = −4, x − 1 = −1, to x = 0.
• (f) 3 − 7s = 7 − 3s: −4s = 4, to s = −1.
• (g) 2x + 1 = 6 − (2x − 3): 2x + 1 = 9 − 2x, to 4x = 8 aur x = 2.
• (h) 10 − 5x = 3(x − 4) − 2(x − 7): RHS hai 3x − 12 − 2x + 14 = x + 2, to 10 − 5x = x + 2, 6x = 8 aur x = 4/3.

19. Start se End tak ka raasta. Har box solve karo; uska jawab us teer pe likha hai jis pe aage jaana hai. Start: 8x = 20 + 3x, x = 4 → 2x − 9 = −3, x = 3 → −2x = −42, x = 21 → 15 = 19 − 4x, x = 1 → 2x + 3 = x + 5, x = 2 → 2x + 5 = 3(x − 1), x = 8 → 2(x + 1) − 10 = 18, x = 13 → 8m + 8 = −72, m = −10 → −4 = 16 − 5k, k = 4 → 2x − 9 = 3 − x, x = 4 → End: 30 = 4 − 50n, n = −13/25. Box −7 = 11 − 3x (x = 6) raaste pe nahi hai.

20. Try This: bachche aur gadhe. d gadhe maano; tab 28 − d bachche hain. Pair: 4d + 2(28 − d) = 80, to 2d + 56 = 80, 2d = 24 aur d = 12. 12 gadhe aur 16 bachche. Check: 48 + 32 = 80 pair.

A Magic Trick (page 190)

Ye hamesha 8 kyun deta hai? Sankhya ko x kaho. 2 se guna karne pe 2x, 10 jodne pe 2x + 10, 2 se bhaag dene pe x + 5, sochi hui sankhya ghataane pe 5, aur 3 jodne pe 8. x kat jaata hai, isliye har sankhya 8 pe pahunchti hai.

Apni trick banao. Jaise: koi sankhya socho, 4 jodo, 3 se guna karo, 12 ghataao, 3 se bhaag do. Tumhe hamesha wahi sankhya waapas milegi jisse shuru kiya: x + 4, 3x + 12, 3x, x.