𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
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𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹 (@gate2027updates) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 365 obunachidan iborat bo'lib, Taʼlim toifasida 13 992-o'rinni va Hindiston mintaqasida 28 875-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 14 365 obunachiga ega bo‘ldi.
26 Iyul, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 565 ga, so‘nggi 24 soatda esa 20 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 16.74% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 3.52% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 2 401 marta ko‘riladi; birinchi sutkada odatda 505 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 3 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent ace, wallah, notes, lectures, pdfs kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi ma’lumot 27 Iyul, 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.
14 365
Obunachilar
+2024 soatlar
+1597 kunlar
+56530 kunlar
Postlar arxiv
Q) The average of 5 consecutive odd numbers is 27. The largest number is:
Repost from 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
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Q) A's salary is 25% more than B's salary. By what percent is B's salary less than A's salary?
IITM GATE 2027 Website is Live !!https://gate.iitm.ac.in/ https://gate.iitm.ac.in/ You can check the updated syllabus directly from the website
Q) Let A be a singular Hermitian matrix with an eigenvalue λ. The minimum eigenvalue of the matrix A² is:
Q) In how many ways can 5 people be seated around a circular table?
Q) In how many ways can 5 people be seated around a circular table?
Q) Statement: "The eigenvalues of a unitary matrix can take only 1 or -1."
Q) Let X be an eigenvector of matrix A. If B = P⁻¹AP, then the eigenvector Y of matrix B corresponding to the same eigenvalue is:
Q) If A is a singular square matrix of order n (where n >= 2), then Adj(A) is:
Q) If A is a 3x3 matrix with eigenvalues 1, 2, and 3, then what is the value of det(A^2 + A + I)?
Detailed Explanation1. How do we get a zero? A trailing zero is created every time you multiply a 2 and a 5 (because 2 × 5 = 10). In 100! (which is 100 × 99 × 98 × ... × 1), there are plenty of even numbers, meaning we have more than enough 2s. Therefore, the number of trailing zeros is determined entirely by how many 5s are hiding inside the numbers from 1 to 100. 2. Counting the 5s (Step 1): First, we count every number that is a multiple of 5 (5, 10, 15, 20... up to 100). Formula: 100 ÷ 5 = 20 So, there are 20 numbers that give us at least one 5. 3. The Trap (Why the answer isn't 20): Some numbers contain more than one 5. Take the number 25, for example. 25 is 5 × 5 (it has two 5s). The numbers 25, 50, 75, and 100 all have an "extra" 5 hiding in them. Our first step only counted one of them! 4. Counting the extra 5s (Step 2): To count those extra 5s, we divide 100 by 25 (which is 5 squared). Formula: 100 ÷ 25 = 4 So, there are 4 extra 5s. (Note: We don't need to check for 125 (5 cubed) because 125 is bigger than 100). 5. Final Total: Add them together: 20 (from the multiples of 5) + 4 (from the multiples of 25) = 24 total 5s. Because there are exactly 24 fives (and plenty of twos to pair them with), there will be exactly 24 trailing zeros at the end of 100!.
Q) How many zeros are there at the end of 100 factorial?
Q) How many zeros are there at the end of 100 factorial?
Q) If it takes 1 hour to boil 1 egg in a pot, how long to boil 4 eggs together?
