OSSSC / OSSC MATH AND REASONING

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Tarmoq:ଓଡ଼ିଆ ବ୍ୟାକରଣ Hindiston33 457 Taʼlim17 102...

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OSSSC / OSSC MATH AND REASONING (@pinnacle_math_reasoning) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 11 535 obunachidan iborat bo'lib, Taʼlim toifasida 17 102-o'rinni va Hindiston mintaqasida 33 457-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 11 535 obunachiga ega bo‘ldi.

02 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 147 ga, so‘nggi 24 soatda esa 0 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 20.61% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 4.23% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 2 377 marta ko‘riladi; birinchi sutkada odatda 488 ta ko‘rish yig‘iladi.
Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 4 ta reaksiya keladi.
Tematik yo‘nalishlar: Kontent ʜᴇʀᴇ, cgle, cgl, mcqs, mcq kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“𝐀𝐧𝐲 PROMOTION RELATED CONTACT OWNER @BUNNYTAPAS Join our YouTube Channel: https://shorturl.at/Ulwjw 🔴 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐟𝐟𝐚𝐢𝐫𝐬 @CURRENT_AFFAIRS_REGULARLY 🔴𝐄𝐧𝐠𝐥𝐢𝐬𝐡- @BLACKBOOK_ENGLISH”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 03 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

11 535

Obunachilar

Ma'lumot yo'q24 soatlar

+227 kun

+14730 kun

2 377

Post ko'rishlar

~ 48824 soatlar

~ 70648 soatlar

20.61%

Muloqot nisbati

~ 1

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11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The average of 39 numbers is zero. How many of those numbers can be greater than zero?

Anonymous voting

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The average of nine 2 digit numbers is decreased by 6 when the digits of one of the 2 digit numbers is interchanged. Find the difference between the digits of that number.

Anonymous voting

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The average of five numbers, given in a particular order, is 32. The average of the first three numbers is 28, while that of the last three numbers is 34. what is the average of the first two numbers?

Anonymous voting

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 What is the least number of square tiles required to pave the floor of a room 15m 17cm long and 9m 43cm broad?

Anonymous voting

11 535

a) : 1149 – 827 = 322 1310 – 1149 = 161 1310 – 827 = 483 Largest desired number = H.C.F of 322, 161 and 483 = 161 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 Which is the largest number that divides 827, 1149 and 1310 to leave the same remainder in each case?

Anonymous voting

11 535

On taking L.C.M of 12, 18, 24 and 28 LCM = 2×2×2×3×3×7 LCM = 8×9×7 = 72×7 = 504 ∴ The required number = 504 + 5 = 509 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 What is the least number which when divided by 12, 18, 24 and 28 we obtain 5 as the remainder in each case?

Anonymous voting

11 535

Ans. (d) : The H.C.F of both the numbers is 23 So, both the numbers will be divisible by 23. The greatest 3-digits number which is divisible by 23 = 989 The smallest 4-digits number which is divisible by 23 = 1012 Sum of both numbers = 989 + 1012 = 2001 ✅✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 What is the sum of the greatest three digit number and the smallest four digit number such that their HCF is 23?

Anonymous voting

11 535

Ans. (a) : Let the numbers be 15x and 15y. According to the question, 15x + 15y = 240 =>x + y = 16 Hence, the possible pairs of both the numbers = (1, 15), (3, 13), (5, 11), (7, 9) ∴ Possible number of pairs = 4 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The sum of two positive number is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.

Anonymous voting

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 Let x be the number divisible by 16,24,30,36 and 45, and x is also a perfect square. What is the remainder when x is divided by 123?

Anonymous voting

11 535

Ans. (a) : Product of two numbers = LCM × HCF ∴ LCM × HCF = 120 × 6 = 720 ∴ 720 = Product of two numbers From option (a) 24 × 30 = 720 Hence, option (a) is correct. ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 If the HCF of two numbers is 6 and their LCM is 120, one such pair of numbers is:

Anonymous voting

11 535

Ans. (c): LCM of two number = 60 HCF of two numbers = 3 Difference of both numbers =3 Let the numbers be 3x and 3x + 3. 3x (3x+3) = 60×3 => x = 4 Sum of the numbers = 3x+3x+3 = 6x+3 = 24 +3 = 27 ✅