OSSSC / OSSC MATH AND REASONING

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𝐀𝐧𝐲 PROMOTION RELATED CONTACT OWNER @BUNNYTAPAS Join our YouTube Channel: https://shorturl.at/Ulwjw 🔴 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐟𝐟𝐚𝐢𝐫𝐬 @CURRENT_AFFAIRS_REGULARLY 🔴𝐄𝐧𝐠𝐥𝐢𝐬𝐡- @BLACKBOOK_ENGLISH

Network:ଓଡ଼ିଆ ବ୍ୟାକରଣ India33 457 Education17 102...

📈 Analytical overview of Telegram channel OSSSC / OSSC MATH AND REASONING

Channel OSSSC / OSSC MATH AND REASONING (@pinnacle_math_reasoning) in the English language segment is an active participant. Currently, the community unites 11 531 subscribers, ranking 17 102 in the Education category and 33 457 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 11 531 subscribers.

According to the latest data from 02 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 147 over the last 30 days and by 0 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 20.61%. Within the first 24 hours after publication, content typically collects 4.23% reactions from the total number of subscribers.
Post reach: On average, each post receives 2 377 views. Within the first day, a publication typically gains 488 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
Thematic interests: Content is focused on key topics such as ʜᴇʀᴇ, cgle, cgl, mcqs, mcq.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“𝐀𝐧𝐲 PROMOTION RELATED CONTACT OWNER @BUNNYTAPAS Join our YouTube Channel: https://shorturl.at/Ulwjw 🔴 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐟𝐟𝐚𝐢𝐫𝐬 @CURRENT_AFFAIRS_REGULARLY 🔴𝐄𝐧𝐠𝐥𝐢𝐬𝐡- @BLACKBOOK_ENGLISH”

Thanks to the high frequency of updates (latest data received on 03 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

11 531

Subscribers

No data24 hours

+227 days

+14730 days

2 377

Post views

~ 48824 hours

~ 70648 hours

20.61%

Engagement rate

~ 1

Posts per day

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Posts Archive

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients are 1, 11 and 2, respectively. What is sum of the two numbers ?

Anonymous voting

11 535

Ans. (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

Ans. (c) 156 – 6 = 150 181 – 6 = 175 331 – 6 = 325 Required number = HCF of 150, 175 and 325 = 25 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 What is the greatest number by which when 156, 181 and 331 are divided, the remainder is 6 in each case?

Anonymous voting

11 535

Ans. (a) : ∵Length = 15 m 17 cm = 1517 cm. Breadth = 9m 43 cm = 943 cm. On taking HCF of 1517 cm and 943 cm H.C.F = 41 Then the least number of tiles placed ( 1517 × 943 ) / (41 × 41 ) = 37 × 23 = 851 ✅

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

Ans. (d): LCM of the number 6, 7 and 8 = 168 The largest number of three digits · 999 On dividing 999/ 168 the Remainder is 159. ∴ 999 – 159 = 840 The number 840 is completely divisible by 168. ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The largest three digit number that is exactly divisible by 6, 7 and 8 is:

Anonymous voting

11 535

Ans. (b) : ∵ The LCM of 56, 57 and 58 is K. ∴ 59 is a prime number. L.C.M of 56, 57, 58 and 59 = 59 × K = 59K ✅✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 If the Least Common Multiple of 56, 57 and 58 is K, then what will be the Least Common multiple of 56, 57, 58 and 59?

Anonymous voting

11 535

Ans. (b) : Let the three numbers be 3x, 4x and 5x respectively. LCM of all three numbers = 3×4×5×x = 60 x LCM = 1800 ∴60 x = 1800 ⇒ x = 30 ∴ Second number = 4x= 4×30= 120 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 The proportion among three numbers is 3:4:5 and their LCM is 1800. Then the second number is:

Anonymous voting

11 535

Ans. (b) Given, 5 digit number · 538xy This number is divisibe by 3, 7 and 11 Hence, the number will be a multiple of (L.C.M) 3,7 and 11. ∴ L.C.M of 3, 7 and 11 = 231 538xy = 53823 ∴ x = 2 y = 3 ∴ x²+ y²= 2² + 3² => x2 + y2 = 13✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 If the 5-digit number 538xy is divisible by 3, 7 and 11, then the value of (x² + y²) is:

Anonymous voting

11 535

Ans. (d) L.C.M of 2, 3, 4, 5, 6 and 7 =420 Hence x = 420 k + 1 According to the condition The value of x lies between 2000 and 2500 On putting k = 5 ∴ x = 420 × 5 + 1 x = 2101 ∴ The sum of the digit of x = 2 + 1 + 0 + 1 = 4 ✅

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 Let x be the least number of 4 digits that when divided by 2, 3, 4, 5, 6 and 7 leave a remainder of 1 in each case. If x lies between 2000 and 2500, then what is the sum of the digits of x?

Anonymous voting

11 535

Ans. (a) : L.C.M of 15, 18 and 36 15 = 3 × 5 18 = 2 × 3 × 3 36 = 2 × 2 × 3 × 3 LCM = 180 Required number = 180 k + 9 => 180 × 6 + 9 = 1089 Hence, on putting k = 6, the number is divisible by 11

11 535

𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘 What is the least number which when divided by 15, 18 and 36 leaves the same remainder 9 in each case and is divisible by 11 ?

Anonymous voting