Playing with Numbers unit ke topics pe poori practice, ek hi post mein: number ka general form (place value se kholna), digits reverse karke add-subtract ke game, digit puzzles (cryptarithm), 2, 3, 4, 5, 6, 8, 9, 10 aur 11 ki divisibility tests aur unka “kyun”, remainder, digital root, even-odd (parity) aur lagaataar numbers ke sum. Pehle chhota recap (saare naye word wahin samjhaaye hain), phir 10 MCQs, fill in the blanks, true/false, short aur long answer, HOTS, aur aakhir mein answer key. Har sawaal ka hal Hinglish mein step by step hai. Yeh post apne current chapter Book 1 Ch 5: Number Play ke saath padhiye: chapter se samjho, yahan se practice karo. 🔢
🧠 Concept Recap: Pehle Yeh Yaad Kar Lo
📖 Pehle Book 1 Ch 5 padho: Number Play (hub). Place value aur count ki story ke liye: Book 1 Ch 3: A Story of Numbers (hub).
📏 Pehle yeh word aur sign samjho (har word yahin samjhaaya hai, aage yahi use honge):
- Digit (ank) = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 mein se koi ek sign. Number in digits se banta hai: 47 mein do digits hain, 4 aur 7.
- Place value (sthaaniya maan) = digit ki place ke hisaab se uski price. 47 mein 4 dahai (tens) ki place pe hai, to uski price 40; 7 ikai (ones) ki place pe hai, price 7. Teen-digit number mein sabse baayi place saikda (hundreds) hoti hai.
- Letters jaise a, b, c, n, A, B = kisi anjaan digit (ya number) ki place rakha “empty box”.
- ⚠️ Pakka rule: letter number ke saath ho to matlab kya? Is post mein hamesha yahi do rule chalenge:
(1) Multiply hamesha × ke saath likha hai. “10 × a” matlab 10 times a, “11 × A” matlab 11 times A, “9 × A” matlab 9 times A. Is post mein multiply kabhi bhi × ke bina (number aur letter ko chipka kar) nahi likha gaya.
(2) Digits/letters bina × ke saath-saath likhe hon, to woh ek number hai, aur hum use “do-digit number 3A” ya “teen-digit number abc” likh kar bataenge. Do-digit number 3A matlab dahai ki place 3 aur ikai ki place A (A = 8 ho to number 38). Isliye do-digit number 3A = 10 × 3 + A = 30 + A; yeh 3 × A nahi hai. Isi tarah do-digit number ab matlab a dahai aur b ikai wala number (a × b nahi).
Pehchaan: × dikhe to multiply; × na dikhe to number. - General form (vistrit roop) = number ko place value se khol kar likhna. Do-digit number ab = 10 × a + b (jaise 47 = 10 × 4 + 7). Teen-digit number abc = 100 × a + 10 × b + c (jaise 365 = 100 × 3 + 10 × 6 + 5).
- Reverse (ulta number) = digits ka order ulat dena: 47 ka reverse 74, 365 ka reverse 563.
- Multiple (gunaj) = kisi number ko 1, 2, 3 … se multiply karke milne wale number. 11 ke multiples: 11, 22, 33 … Factor (gunankhand) = jo number poora-poora divide kar de: 12 ke factors 1, 2, 3, 4, 6, 12. Common factor (saajha factor) = jo number dono numbers ka factor ho, yaani dono ko poora divide kar de: 4 aur 6 ka common factor 2 hai (4 = 2 × 2, 6 = 2 × 3). 1 to har number ka factor hota hai, isliye “1 ke alawa koi common factor nahi” ka matlab hai dono mein kuch bhi saajha nahi.
- Divisible (poora bantna) = jab division dene pe kuch na bache. 12, 3 se divisible hai kyunki 12 = 3 × 4 (kuch nahi bacha). 13, 3 se divisible nahi.
- ÷ (bhaag) = equal parts mein divide karna. Remainder (shesh, baaki) = division ke baad jo bach jaaye: 47 ko 5 se divide karo to 47 = 5 × 9 + 2, yaani quotient 9 aur remainder 2.
- Divisibility test (bhaajyata ki pehchaan) = bina poora divide kiye jaldi pata karna ki number divisible hai ya nahi (neeche table mein saare rule hain).
- Digit sum (ankon ka jod) = number ke saare digits ka sum: 2357 ka digit sum = 2 + 3 + 5 + 7 = 17. Digital root = digit sum ka bhi digit sum baar-baar lo jab tak ek hi digit na bache: 17 → 1 + 7 = 8.
- Even (sam) aur odd (visham) = even number 2 se divisible (0, 2, 4, 6, 8 pe khatam); odd number 2 se divisible nahi (1, 3, 5, 7, 9 pe khatam). Parity = number even hai ya odd, yahi uski parity.
- Consecutive numbers (lagaataar numbers) = ek ke baad ek aane wale numbers, jaise 23, 24, 25. Middle (beech wala) number = beech mein aane wala, jaise 23, 24, 25 mein 24.
- Cryptarithm (ank paheli) = aisa sum ya multiply jisme kuch digits ki place letters hon. Rule: ek letter = hamesha ek hi digit; alag letters = alag digits; number ka pehla digit 0 nahi hota.
- > (bada), < (chhota) = 9 > 4 padho “9, 4 se bada”; 4 < 9 padho “4, 9 se chhota”.
- Minus wale (negative) numbers = zero se neeche ke numbers, jaise thermometer pe 0 se neeche ka temperature. −99 padho “minus 99”: yeh 99 ka reverse hai, kyunki dono ko add karo to 0 bachta hai (99 − 99 = 0). Isi tarah −2 aur +2 milkar 0, −1 aur +1 milkar 0. Aur “c − 100 × c” mein ek c mein se 100 times c hataya, to 99 times c kam pad gaya, yaani −99 × c.
- → (arrow) = “ban jaata hai” ya “iska matlab”. “17 → 1 + 7 = 8” padho “17 se 1 + 7 = 8 ban jaata hai”. Chitra mein ⇄ (do-tarfa arrow) = digits ki place aapas mein ulat dena.
- Algebra wala reasoning = kisi baat ko letters se likh kar dikhaana ki woh har number ke liye sach hai, sirf ek example ke liye nahi.
Worked example (pehle ek aasaan sawaal): 58 ka general form: 58 = 10 × 5 + 8. Iska reverse 85 = 10 × 8 + 5. Dono ka sum 58 + 85 = 143 = 11 × 13. Dekho, 11 ka multiple aaya! Aage hum dikhaenge ki aisa hamesha kyun hota hai.
1. General form aur digits reverse karne ke game
| Game | Letters se (har number ke liye) | Example |
|---|---|---|
| Do-digit number + uska reverse | (10 × a + b) + (10 × b + a) = 11 × a + 11 × b = 11 × (a + b) → hamesha 11 ka multiple | 36 + 63 = 99 = 11 × 9 |
| Do-digit number − uska reverse (bada − chhota) | (10 × a + b) − (10 × b + a) = 9 × a − 9 × b = 9 × (a − b) → hamesha 9 ka multiple | 82 − 28 = 54 = 9 × 6 |
| Teen-digit number − uska reverse | (100 × a + 10 × b + c) − (100 × c + 10 × b + a) = 99 × a − 99 × c = 99 × (a − c) → hamesha 99 ka multiple (beech ka digit cancel ho jaata hai) | 642 − 246 = 396 = 99 × 4 |
| Do-digit number − uska digit sum | (10 × a + b) − (a + b) = 9 × a → hamesha 9 ka multiple | 57 − 12 = 45 = 9 × 5 |
2. Divisibility tests (bina bhaag kiye pehchaan)
| Kisse divisible? | Rule | Example |
|---|---|---|
| 2 | Last digit 0, 2, 4, 6 ya 8 (even) | 3458 ✅ (last 8) |
| 5 | Last digit 0 ya 5 | 735 ✅ |
| 10 | Last digit 0 | 4170 ✅ |
| 3 | Digit sum 3 se divisible | 411: 4 + 1 + 1 = 6 ✅ |
| 9 | Digit sum 9 se divisible | 738: 7 + 3 + 8 = 18 ✅ |
| 4 | Last do digits ka number 4 se divisible (kyunki 100 = 4 × 25) | 5316: 16 = 4 × 4 ✅ |
| 8 | Last teen digits ka number 8 se divisible (kyunki 1000 = 8 × 125) | 7120: 120 = 8 × 15 ✅ |
| 6 | 2 se bhi aur 3 se bhi divisible (dono test pass) | 114: even ✅ aur 1 + 1 + 4 = 6 ✅ |
| 11 | Daayein se count karo: odd places (1st, 3rd, 5th …) ke digits ka sum aur even places (2nd, 4th …) ke digits ka sum nikaalo. Dono ka difference (bada − chhota) 0 ho ya 11 ka multiple ho | 5918: odd place 8 + 9 = 17, even place 1 + 5 = 6, difference 17 − 6 = 11 ✅ |
3 aur 9 wala rule kyun chalta hai? 100 = 99 + 1 aur 10 = 9 + 1. To 100 × a + 10 × b + c = 99 × a + 9 × b + (a + b + c). 99 × a aur 9 × b to hamesha 9 (aur 3) ke multiple hain, isliye whole number tabhi 9 (ya 3) se bantega jab bacha hua part, yaani digit sum (a + b + c), bante. Isi se remainder bhi milta hai: number ko 9 se divide karne pe wahi remainder bachta hai jo uske digit sum ko 9 se divide karne pe.
3. Even-odd (parity) ke rule
| Add / multiply | Result | Example |
|---|---|---|
| even + even | even | 4 + 6 = 10 |
| odd + odd | even | 3 + 5 = 8 |
| even + odd | odd | 4 + 5 = 9 |
| odd numbers ki count odd ho (jaise 3, 5, 7 odd numbers add karo) | odd | 1 + 3 + 5 = 9 |
| koi bhi number × even | even | 7 × 4 = 28 |
| odd × odd | odd | 3 × 7 = 21 |
4. Lagaataar numbers ka sum
3 lagaataar numbers ka sum = 3 × beech wala number (23 + 24 + 25 = 72 = 3 × 24). 5 lagaataar numbers ka sum = 5 × beech wala (6 + 7 + 8 + 9 + 10 = 40 = 5 × 8). Kyun? Beech wale se ek chhota aur ek bada, dono milkar beech wale ke do times ke equal: 23 + 25 = 48 = 2 × 24.
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📝 Section A: 10 MCQs
Har sawaal ke 4 options mein se ek sahi chuno. Pehle khud socho, phir “Hal dekho” kholo.
Q1. Ek do-digit number ka dahai (tens) digit 7 aur ikai (ones) digit 3 hai. Iska general form kya hai?
Hal (step by step) dekho
Sahi jawab: (b) 10 × 7 + 3
Step 1: dahai ki place wale digit ki price 10 times hoti hai: 7 dahai = 10 × 7 = 70.
Step 2: ikai wala digit jaisa hai waisa: 3.
Step 3: general form = 10 × 7 + 3 = 70 + 3 = 73 ✅
Dhyaan: 7 + 3 = 10 sirf digit sum hai, number nahi. 100 × 7 + 3 = 703 to teen-digit number ban gaya.
Q2. Koi bhi do-digit number lo aur usme uska reverse number (digits reverse karke) add karo. Sum hamesha kis number se divisible hoga?
Hal (step by step) dekho
Sahi jawab: (c) 11
Step 1: do-digit number ab = 10 × a + b, uska reverse do-digit number ba = 10 × b + a.
Step 2: sum = 10 × a + b + 10 × b + a = 11 × a + 11 × b = 11 × (a + b).
Step 3: sum 11 ke multiply wala hai, to hamesha 11 se divisible. Example: 47 + 74 = 121 = 11 × 11 ✅
Q3. Ek do-digit number mein se uska reverse number subtract karo (bade mein se chhota). Difference hamesha kis number se divisible hoga?
Hal (step by step) dekho
Sahi jawab: (a) 9
Step 1: (10 × a + b) − (10 × b + a) = 10 × a − a + b − 10 × b = 9 × a − 9 × b = 9 × (a − b).
Step 2: isliye difference hamesha 9 se divisible. Example: 82 − 28 = 54 = 9 × 6 (aur 6 = 8 − 2, yaani digits ka difference) ✅
Q4. Teen-digit number abc mein se uska reverse cba subtract karo (a > c). Difference hamesha kis number se divisible hoga?
Hal (step by step) dekho
Sahi jawab: (b) 99
Step 1: abc = 100 × a + 10 × b + c aur cba = 100 × c + 10 × b + a.
Step 2: subtract karo: 100 × a − a + 10 × b − 10 × b + c − 100 × c = 99 × a − 99 × c = 99 × (a − c). Beech wala digit b cancel ho gaya.
Step 3: difference hamesha 99 se divisible. Example: 642 − 246 = 396 = 99 × 4 ✅
Q5. Inme se kaunsa number 9 se divisible hai?
Hal (step by step) dekho
Sahi jawab: (a) 3456
Rule: digit sum 9 se divisible ho to number bhi 9 se divisible.
3456: 3 + 4 + 5 + 6 = 18, aur 18 = 9 × 2 ✅. 2345: sum 14 ❌. 1234: sum 10 ❌. 5678: sum 26 ❌.
Jawab: 3456. Check: 9 × 384 = 3456 ✅
Q6. Inme se kaunsa number 11 se divisible hai?
Hal (step by step) dekho
Sahi jawab: (c) 5918
Rule: daayein se odd places ke digits ka sum aur even places ke digits ka sum; difference 0 ya 11 ka multiple ho.
5918: daayein se digits 8 (1st), 1 (2nd), 9 (3rd), 5 (4th). Odd place: 8 + 9 = 17. Even place: 1 + 5 = 6. Difference 17 − 6 = 11 ✅
5917: 7 + 9 = 16 aur 1 + 5 = 6, difference 10 ❌. 1234: 4 + 2 = 6 aur 3 + 1 = 4, difference 2 ❌. 4320: 0 + 3 = 3 aur 2 + 4 = 6, difference 3 ❌.
Check: 11 × 538 = 5918 ✅
Q7. Inme se kaunsa number 8 se divisible hai?
Hal (step by step) dekho
Sahi jawab: (b) 7120
Rule: last teen digits ka number 8 se divisible ho (kyunki 1000 = 8 × 125, to hazaar wala part hamesha 8 se bantta hai).
7120: last teen digits 120 = 8 × 15 ✅. 124 = 8 × 15 + 4 ❌. 130 = 8 × 16 + 2 ❌. 140 = 8 × 17 + 4 ❌.
Jawab: 7120 (7120 = 8 × 890 ✅).
Q8. Kinhi bhi 5 lagaataar (consecutive) numbers ka sum hamesha kis number se divisible hota hai?
Hal (step by step) dekho
Sahi jawab: (c) 5
Step 1: beech wala number m maano. Paanch numbers: m − 2, m − 1, m, m + 1, m + 2.
Step 2: add karo: −2, −1, +1, +2 aapas mein cancel ho jaate hain (−2 − 1 + 1 + 2 = 0), bachta hai 5 × m.
Step 3: isliye sum hamesha 5 se divisible. Example: 6 + 7 + 8 + 9 + 10 = 40 = 5 × 8 ✅. 10 hamesha nahi: 1 + 2 + 3 + 4 + 5 = 15, jo 10 se divisible nahi.
Q9. 7 odd numbers ka sum kaisa hoga?
Hal (step by step) dekho
Sahi jawab: (b) Hamesha odd
Step 1: do odd numbers ka sum even hota hai (3 + 5 = 8). To 7 odd numbers mein 3 pairs bana lo: teen even sum, aur ek odd number akela bacha.
Step 2: even + even + even = even, aur even + odd = odd.
Jawab: hamesha odd. Example: 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 (odd) ✅
Q10. 9876 ka digital root kya hai?
Hal (step by step) dekho
Sahi jawab: (b) 3
Step 1: digit sum: 9 + 8 + 7 + 6 = 30.
Step 2: 30 ek digit nahi, to phir se add karo: 3 + 0 = 3.
Jawab: digital root 3. (Matlab 9876 ko 9 se divide karo to remainder 3: 9876 = 9 × 1097 + 3 ✅)
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✍️ Section B: Fill in the Blanks
Blank (_____) mein sahi word ya number bharo.
Q1. Teen-digit number abc (a saikda, b dahai, c ikai) ka general form _____ hai.
Jawab aur reason
Jawab: 100 × a + 10 × b + c
Saikda ki place wala digit 100 times, dahai wala 10 times, ikai wala jaisa hai. Jaise 365 = 100 × 3 + 10 × 6 + 5.
Q2. Koi number 5 se tabhi divisible hai jab uska last digit _____ ho.
Jawab aur reason
Jawab: 0 ya 5
5 ke multiples 5, 10, 15, 20 … hamesha 5 ya 0 pe khatam hote hain. Jaise 735 ✅, 742 ❌.
Q3. Number 4□7 (beech ka digit gaayab) 9 se divisible hai. Empty boxes □ ka digit _____ hai.
Jawab aur reason
Jawab: 7
Step 1: digit sum = 4 + 7 + □ = 11 + □.
Step 2: yeh 9 ka multiple hona chahiye. □ ek digit hai (0 se 9), to 11 + □ sirf 18 ho sakta hai, yaani □ = 18 − 11 = 7.
Check: 477 = 9 × 53 ✅
Q4. Even + odd = _____
Jawab aur reason
Jawab: odd
Jaise 4 + 5 = 9 aur 10 + 7 = 17. Even ke saath ek odd judne se ek “akela” bach jaata hai, isliye result odd.
Q5. 75 ko 8 se divide karne pe remainder _____ bachta hai.
Jawab aur reason
Jawab: 3
8 ke multiples dekho: 8 × 9 = 72 aur 8 × 10 = 80 (75 se bada). To 75 = 8 × 9 + 3, yaani remainder 3.
Q6. Koi number 6 se tabhi divisible hai jab woh 2 se bhi aur _____ se bhi divisible ho.
Jawab aur reason
Jawab: 3
6 = 2 × 3, aur 2 aur 3 ka koi common factor (1 ke alawa) nahi. Isliye dono test pass hon to number 6 se divisible. Jaise 114: even ✅ aur digit sum 6 ✅, to 114 = 6 × 19 ✅.
Q7. Ek do-digit number mein se uska digit sum subtract karne pe jawab hamesha _____ ka multiple hota hai.
Jawab aur reason
Jawab: 9
(10 × a + b) − (a + b) = 9 × a. Jaise 57 − (5 + 7) = 57 − 12 = 45 = 9 × 5.
Q8. 1001 = 7 × 11 × _____
Jawab aur reason
Jawab: 13
7 × 11 = 77, aur 77 × 13 = 1001. Isi wajah se abcabc jaise number (254254) 7, 11 aur 13 teeno se divisible hote hain (HOTS Q2 dekho).
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✅❌ Section C: True / False
Batao statement sahi (True) hai ya galat (False). Galat ho to use sahi karke likho.
Q1. Jo number 9 se divisible hai, woh 3 se bhi divisible hai.
Jawab aur reason
✅ True
Sahi. 9 = 3 × 3, to 9 ka har multiple 3 ka bhi multiple hai. Jaise 63 = 9 × 7 = 3 × 21.
Q2. Jo number 3 se divisible hai, woh 9 se bhi divisible hai.
Jawab aur reason
❌ False
Galat. Jaise 12 = 3 × 4, par 12 ko 9 se divide karo to 12 = 9 × 1 + 3 (3 bacha). Sahi: “9 se divisible number 3 se bhi divisible hota hai, par reverse zaroori nahi.”
Q3. Do odd numbers ka sum odd hota hai.
Jawab aur reason
❌ False
Galat. Do odd numbers ka sum even hota hai: 3 + 5 = 8, 7 + 9 = 16. Kyunki dono ke “akele” bache parts milkar ek pair ban jaate hain.
Q4. Jo number 4 se bhi aur 6 se bhi divisible ho, woh 24 se zaroor divisible hoga.
Jawab aur reason
❌ False
Galat. 12 ko dekho: 12 = 4 × 3 aur 12 = 6 × 2, par 12 to 24 se chhota hai, divisible nahi. Wajah: 4 aur 6 mein 2 common factor hai, isliye rule sirf aise numbers pe chalta hai jinka koi common factor na ho (jaise 2 aur 3 → 6). Sahi: “4 aur 6 dono se divisible number 12 se divisible hota hai.”
Q5. Kisi bhi do-digit number aur uske reverse number ka sum 11 se divisible hota hai.
Jawab aur reason
✅ True
Sahi. (10 × a + b) + (10 × b + a) = 11 × (a + b). Jaise 58 + 85 = 143 = 11 × 13.
Q6. Jis number ka last digit 6 ho, woh 6 se divisible hota hai.
Jawab aur reason
❌ False
Galat. 16 ka last digit 6 hai, par 16 = 6 × 2 + 4 (4 bacha). 6 ka test: 2 aur 3 dono se divisible. 16 ka digit sum 7, jo 3 se divisible nahi.
Q7. Do lagaataar numbers ka multiply (product) hamesha even hota hai.
Jawab aur reason
✅ True
Sahi. Do lagaataar numbers mein ek even aur ek odd hota hai (jaise 7 aur 8). Even se multiply karne pe result hamesha even: 7 × 8 = 56.
Q8. 1001, 11 se divisible hai.
Jawab aur reason
✅ True
Sahi. Daayein se digits 1, 0, 0, 1. Odd place: 1 + 0 = 1, even place: 0 + 1 = 1. Difference 0 ✅. Check: 11 × 91 = 1001.
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🟢 Section D: Short Answer Questions
2 se 3 marks wale sawaal. Har step likhne ki habit daalo.
Q1. Digit puzzle (cryptarithm) hal karo: 3A + A3 = 121. (Yahan 3A aur A3 dono do-digit numbers hain, multiply nahi: do-digit number 3A matlab dahai 3 aur ikai A; do-digit number A3 matlab dahai A aur ikai 3.)
Step-by-step hal
Step 1 (general form, place value se): do-digit number 3A = 10 × 3 + A = 30 + A. Do-digit number A3 = 10 × A + 3.
Step 2 (add karo): 30 + A + 10 × A + 3. Akele numbers alag add karo: 30 + 3 = 33. A wale parts alag add karo: ek A aur 10 times A milkar 11 times A, yaani A + 10 × A = 11 × A. To sum = 33 + 11 × A. (Dhyaan: 11 × A matlab 11 times A; yeh multiply hai, do-digit number nahi.) Kyunki 33 = 11 × 3, yeh sum 11 × (3 + A) bhi hai (number + reverse = 11 × digits ka sum).
Step 3: 11 × (3 + A) = 121, aur 121 = 11 × 11. To 3 + A = 11, yaani A = 11 − 3 = 8.
Check: 38 + 83 = 121 ✅
Q2. Dikhao ki kisi bhi do-digit number mein se uske digits ka sum subtract karo to jawab 9 ka multiple aata hai. 73 se check karo.
Step-by-step hal
Step 1 (letters se): do-digit number ab = 10 × a + b, digit sum = a + b.
Step 2: subtract karo: 10 × a + b − a − b. b aur −b kat gaye, aur 10 × a − a = 9 × a.
Step 3: 9 × a matlab 9 times a, to jawab hamesha 9 ka multiple (a koi bhi digit ho).
Check (73): 73 − (7 + 3) = 73 − 10 = 63 = 9 × 7 ✅ (aur 7 = dahai ka digit, jaisa 9 × a kehta hai).
Q3. Number 52□6 (ek digit gaayab) 11 se divisible hai. □ ka digit nikaalo.
Step-by-step hal
Step 1 (place count karo, daayein se): 6 = 1st, □ = 2nd, 2 = 3rd, 5 = 4th.
Step 2: odd places ka sum = 6 + 2 = 8. Even places ka sum = □ + 5.
Step 3: difference 0 ya 11 ka multiple hona chahiye. □ + 5 kam se kam 5 aur zyada se zyada 14 hai. 8 aur (□ + 5) equal hon to □ = 8 − 5 = 3. (Difference 11 ke liye □ + 5 = 19 chahiye, jo mumkin nahi.)
Check: 5236 = 11 × 476 ✅
Q4. Bina long division kiye batao: 2357 ko 9 se divide karne pe remainder kya bachega?
Step-by-step hal
Step 1: number ko 9 se divide karne pe wahi remainder bachta hai jo uske digit sum ko 9 se divide karne pe (kyunki 1000, 100, 10 har ek 9 ke multiple + 1 hain).
Step 2: digit sum = 2 + 3 + 5 + 7 = 17.
Step 3: 17 = 9 × 1 + 8, to remainder 8. (Digital root bhi 1 + 7 = 8.)
Check: 9 × 261 + 8 = 2349 + 8 = 2357 ✅
Q5. (a) Teen lagaataar numbers ka sum 72 hai. Numbers nikaalo. (b) Kya teen lagaataar numbers ka sum 100 ho sakta hai? Reason do.
Step-by-step hal
(a) Step 1: teen lagaataar numbers ka sum = 3 × beech wala number.
Step 2: 3 × beech wala = 72, to beech wala = 72 ÷ 3 = 24.
Step 3: numbers: 23, 24, 25. Check: 23 + 24 + 25 = 72 ✅
(b) sum hamesha 3 ka multiple hota hai. 100 ka digit sum 1 hai, jo 3 se divisible nahi (100 = 3 × 33 + 1). Isliye nahi, sum 100 nahi ho sakta.
Q6. 7128 ki divisibility 2, 3, 4, 8, 9 aur 11 se check karo (har ek ka reason likho).
Step-by-step hal
2: last digit 8 (even) ✅
3 aur 9: digit sum 7 + 1 + 2 + 8 = 18. 18 = 3 × 6 ✅ aur 18 = 9 × 2 ✅
4: last do digits 28 = 4 × 7 ✅
8: last teen digits 128 = 8 × 16 ✅
11: daayein se 8, 2, 1, 7. Odd place 8 + 1 = 9, even place 2 + 7 = 9, difference 0 ✅
Jawab: 7128 in saare numbers se divisible hai. Check: 7128 = 8 × 891 = 9 × 792 = 11 × 648 ✅
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📘 Section E: Long Answer Questions
4 se 5 marks wale sawaal. Hal mein har step ka reason bhi likha hai.
Q1. Letters (general form) se dikhao ki teen-digit number abc mein se uska reverse cba subtract karne pe (a > c) jawab hamesha 99 ka multiple aata hai, aur 99 se divide karne pe (a − c) milta hai. Phir 754 se check karo.
Step-by-step hal
Step 1 (general form): abc = 100 × a + 10 × b + c. Reverse cba = 100 × c + 10 × b + a.
Step 2 (subtract karo, ek-ek place): saikda: 100 × a − a = 99 × a. Dahai: 10 × b − 10 × b = 0 (beech ka digit cancel ho gaya). Ikai: c − 100 × c = −99 × c.
Step 3: difference = 99 × a − 99 × c = 99 × (a − c). To yeh 99 ka multiple hai, aur 99 se divide karne pe (a − c) bachta hai.
Step 4 (754 se check): a = 7, b = 5, c = 4. Reverse = 457. Difference 754 − 457 = 297.
Step 5: 99 × (7 − 4) = 99 × 3 = 297 ✅. Aur 297 ÷ 99 = 3 = 7 − 4 ✅
Q2. Do digit puzzles hal karo (har letter ek digit hai; alag letters = alag digits). Yahan 1A, 9A aur AB teeno do-digit numbers hain (inke beech × nahi, isliye number): (a) 1A × A = 9A, yaani do-digit number 1A ko digit A se multiply karo to do-digit number 9A milta hai. (b) AB + 37 = 9A, yaani do-digit number AB mein 37 add karo to do-digit number 9A milta hai.
Step-by-step hal
(a) Step 1: do-digit number 1A ko A se multiply kiya, aur jawab do-digit number 9A hai, jiska ikai digit A hai. To A × A ka last digit bhi A hona chahiye. Aise digits: 0 (0 × 0 = 0), 1 (1 × 1 = 1), 5 (5 × 5 = 25), 6 (6 × 6 = 36).
Step 2 (har ek try karo): A = 0: 10 × 0 = 0, 90 nahi ❌. A = 1: 11 × 1 = 11, 91 nahi ❌. A = 5: 15 × 5 = 75, 95 nahi ❌. A = 6: 16 × 6 = 96 ✅
Jawab (a): A = 6.
(b) Step 1 (general form, place value se): do-digit number AB = 10 × A + B (dahai A, ikai B). Do-digit number 9A = 10 × 9 + A = 90 + A (dahai 9, ikai A).
Step 2 (sawaal ko multiply ke saath likho): 10 × A + B + 37 = 90 + A.
Step 3: dono taraf se ek A subtract karo (weighing scale ke dono pans se same cheez remove karo to balance bani rehti hai). Baayein: 10 times A mein se ek A gaya, to 9 times A bacha, yaani 10 × A − A = 9 × A. Daayein: 90 + A − A = 90. Ab: 9 × A + B + 37 = 90.
Step 4: dono taraf se 37 subtract karo: 9 × A + B = 90 − 37 = 53. (Dhyaan: yahan 9 × A matlab 9 times A; yeh multiply hai, do-digit number 9A nahi.)
Step 5 (A dhundho): B ek digit hai (0 se 9). Har A try karo: 9 × 4 = 36, to B = 53 − 36 = 17, jo digit nahi ❌. 9 × 5 = 45, to B = 53 − 45 = 8 ✅. 9 × 6 = 54, jo 53 se bada hai, to B ke liye kuch nahi bachta ❌ (A aur bada lo to aur bhi bada). 4 se chhote A pe B aur bhi bada (17 se zyada) aayega ❌. To sirf A = 5, B = 8 (dono alag digits ✅).
Check: do-digit number AB = 58, do-digit number 9A = 95 (dahai 9, ikai A = 5). 58 + 37 = 95 ✅
Q3. Teen-digit number abc = 100 × a + 10 × b + c ko tod kar samjhao ki 9 ki divisibility test (digit sum) kyun chalti hai. Phir 738 se dikhao.
Step-by-step hal
Step 1: 100 = 99 + 1 aur 10 = 9 + 1. To 100 × a = 99 × a + a aur 10 × b = 9 × b + b.
Step 2: number = 99 × a + a + 9 × b + b + c = (99 × a + 9 × b) + (a + b + c).
Step 3: 99 × a + 9 × b = 9 × (11 × a + b), jo hamesha 9 ka multiple hai.
Step 4: to whole number 9 se tabhi bantega jab bacha part (a + b + c), yaani digit sum, 9 se bante. Yahi test hai! (3 ke liye bhi same, kyunki 99 aur 9 dono 3 ke multiple hain.)
Step 5 (738): 99 × 7 + 9 × 3 + (7 + 3 + 8) = 693 + 27 + 18 = 738 ✅. Digit sum 18 = 9 × 2, to 738 bhi 9 se divisible: 738 = 9 × 82 ✅
Q4. Number ka magic: “Koi bhi number socho. Use 2 se multiply karo, 10 add karo, phir 2 se divide kar do, aur jo number socha tha woh subtract kar do. Jawab hamesha 5 aayega!” (a) 13 se karke dikhao. (b) Letter n se samjhao ki yeh hamesha kyun hota hai.
Step-by-step hal
(a) 13 × 2 = 26. 26 + 10 = 36. 36 ÷ 2 = 18. 18 − 13 = 5 ✅
(b) Step 1: socha hua number n. 2 se multiply: 2 × n (2 times n).
Step 2: 10 add karo: 2 × n + 10.
Step 3: 2 se divide: (2 × n + 10) ÷ 2 = n + 5 (dono parts ko half kiya: 2 × n ka half n, 10 ka half 5).
Step 4: socha hua n subtract karo: n + 5 − n = 5. n cancel ho gaya!
Jawab: n kuch bhi ho, jawab hamesha 5. Yahi “algebra wala reasoning” hai: ek example se nahi, letters se har number ke liye saabit kiya.
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🧩 Section F: HOTS (Soch Wale Sawaal)
HOTS matlab Higher Order Thinking Skills: aise sawaal jinme formula se aage thoda mind lagana padta hai.
Q1. Kya 5 odd numbers ka sum 50 ho sakta hai? Reason do.
Soch aur hal dekho
Soch: parity dekho. Odd numbers ki count odd (5) hai, to sum hamesha odd hoga (MCQ Q9 jaisa reasoning: do-do ki pair even, ek akela odd bachta hai).
50 even hai. Isliye nahi, kabhi nahi ho sakta, chaahe koi bhi odd numbers lo.
Q2. Koi teen-digit number do baar likh kar chhe-digit number banao (jaise 254 se 254254). Dikhao ki aisa number hamesha 7, 11 aur 13 teeno se divisible hota hai.
Soch aur hal dekho
Step 1: 254254 = 254000 + 254 = 254 × 1000 + 254 × 1 = 254 × 1001.
Step 2: yahi har teen-digit number abc ke liye: abcabc = abc × 1001.
Step 3: 1001 = 7 × 11 × 13 (7 × 11 = 77, 77 × 13 = 1001).
Step 4: to abcabc = abc × 7 × 11 × 13, isliye 7, 11 aur 13 teeno se divisible ✅
Q3. Aaj Saturday (Shanivaar) hai. Aaj se theek 100 days baad kaunsa day hoga?
Soch aur hal dekho
Soch: day har 7 days mein dohraate hain, to sirf 100 ko 7 se divide karne ka remainder kaam ka hai.
Step 1: 100 = 7 × 14 + 2 (14 poore week = 98 days, phir 2 days bache).
Step 2: 98 days baad phir Saturday. Uske 2 days aage: Sunday, phir Monday.
Q4. Number 3□5□ (do digits gaayab) 9 se bhi aur 10 se bhi divisible hai. Dono empty digits nikaalo. Kya ek se zyada jawab ho sakte hain?
Soch aur hal dekho
Step 1 (10 wala test): last digit 0 hona chahiye. To number 3□50.
Step 2 (9 wala test): digit sum = 3 + □ + 5 + 0 = 8 + □. Yeh 9 ka multiple ho. □ ek digit (0 se 9), to 8 + □ ya to 9 ya 18 ho sakta hai.
Step 3: 8 + □ = 9 to □ = 1 ✅. 8 + □ = 18 to □ = 10 ❌ (digit nahi).
Jawab: sirf ek: 3150. Check: 3150 = 9 × 350 ✅ aur 3150 = 10 × 315 ✅
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🔑 Answer Key (Ek Nazar Mein)
Pehle saare sawaal khud karo, phir yahan milaao. Jahan mistake ho, upar us sawaal ka step-by-step hal dobara padho.
| Section | Answers |
|---|---|
| A. MCQ | 1: (b) · 2: (c) · 3: (a) · 4: (b) · 5: (a) · 6: (c) · 7: (b) · 8: (c) · 9: (b) · 10: (b) |
| B. Fill in the Blanks | 1: 100 × a + 10 × b + c · 2: 0 ya 5 · 3: 7 · 4: odd · 5: 3 · 6: 3 · 7: 9 · 8: 13 |
| C. True / False | 1: T · 2: F · 3: F · 4: F · 5: T · 6: F · 7: T · 8: T |
| D. Short Answer | 1: A = 8 (38 + 83 = 121) · 2: 10 × a + b − (a + b) = 9 × a; 73 − 10 = 63 = 9 × 7 · 3: □ = 3 (5236) · 4: Remainder 8 · 5: (a) 23, 24, 25 (b) Nahi, 100 teen ka multiple nahi · 6: Saare (2, 3, 4, 8, 9, 11) se divisible |
| E. Long Answer | 1: 99 × (a − c); 754 − 457 = 297 = 99 × 3 · 2: (a) A = 6 (16 × 6 = 96) (b) A = 5, B = 8 (58 + 37 = 95) · 3: Number = 9 ka multiple + digit sum; 738 = 9 × 82 · 4: (a) 5 (b) (2 × n + 10) ÷ 2 − n = 5 |
| F. HOTS | 1: Nahi (5 odd ka sum hamesha odd) · 2: abcabc = abc × 1001 = abc × 7 × 11 × 13 · 3: Monday (remainder 2) · 4: 3150 (sirf ek jawab) |
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👩🏫 CharuMam Ka Tip
Number ke sawaal mein pehle general form likho (10 × a + b ya 100 × a + 10 × b + c): bahut si “magic” wali baatein letters likhte hi saaf dikh jaati hain. Divisibility ke liye table yaad rakho: 2, 5, 10 → last digit; 4 → last do; 8 → last teen; 3, 9 → digit sum; 11 → odd-even places ka difference; 6 → 2 aur 3 dono. Aur aakhir mein check line zaroor likho (multiply karke wapas number banao). 🔢
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👉 Aage Kya?
Pehle chapter Number Play ke posts padh lo (link neeche): wahan divisibility, remainder, parity aur cryptarithm har ek ko alag post mein reasoning ke saath samjhaaya hai. Place value ki story ke liye Book 1 Ch 3: A Story of Numbers bhi dekho. Phir yeh practice dobara karo aur score milao. Agla practice unit: Algebraic Expressions, Identities and Factorisation, jald aa raha hai!
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