Top 5 Questions on Real Numbers in 5 Mins 🔥 | Class 10 Maths Session 2022-23 | Ashish Gupta

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[Abbas_]

1. Ans. LCM(a, b) - x³y³
2. Ans. No
3. Ans. Product of prime number is called composite number
4. Ans. √2 is an irrational number
5. Ans. LCM(12, 15, 18) - 180
Therefore three bells will toll after
180
🙃🙃🙃🙃🙃

[pranshusengar]

(1) x³y³
(5)180minute
Yes, I have done all the questions

[vaishnavipandey]

Ans 1 ~ x raise to the power 3, y raise to the power 3
Ans 2 ~ Since prime factorization of 9 does not contain 2 and 5 as its prime factors. So, by the uniqueness of fundamental theorem of arithmetic it cannot end with the digit 0
Ans 3 ~ Since it has more than 2 factors, hence it is a composite number
Ans 4 ~ It is an irrational number (By contradiction method)
Ans 5 ~ LCM of 12, 15, 18 = 180
Thank you sir

[Shreyanshhu]

It is really amazing series. Hope it would be very much helpful for board❤❤❤❤❤

[sketchworld]

1q=lcm=x3y3, hcf=xy
2q= it is not end with digit 0 if no. End ls with digit 0 prime factor must contain pair or 2x5
3q=3473, it is composite or prime no.
5q=120 minutes

[masteryoda]

Lcm is 180 who all got 360 at first attempt😂😂😂

[VIRATKHOLI-n]

Sir I had found your channel some times ago on regular scrolling on homepage but it helps me to score 100

[youtubebabagaming]

I will definitely score 95 +in maths✨💥💥💥

[FnF_United]

Sir aise hi sabhii chapters ke laa dijiye Thoda thoda krke ho jata hai Parr pura NCERT sochne se hi darr lagata hai

[Shreyanshhu]

Answers of This test By Piyush

Ans 1. X³Y³

Ans 2 . 9ñ cannot end with digit 0 because in prime factorization of 9 , 2 and 5 doesn't occur so by fundamental theorem of arithmetic 9ñ cannot end with digit 0


Ans 3. (3×5×13×46)+23
(3×5×13×2×23)+23
23(3×5×13×2)+1)
Since the number is divisble by 23 and (3×5×13×2)+1)
therefore (3×5×13×46)+23 is a composite vnumber

Ans 4. Let us assume that √2 is a rational number with p and q as co-prime integers and q ≠ 0

√2 = p/q

On squaring both sides

2q2 = p2
(2 divides q2. Therefo minutre, 2 also divides q.)

Let p = 2x
2q2 = p2

2q2 = (2m)2

2q2 = 4m2

q2 = 2m2

(2 divides m2. Therfore, 2 also divides m.)

Since p and q both divisible with 2 as a common mltiple which means that p and q are not co-prime nvumbers as their HCF is 2.
So there is an contradictionv.

Ans 5. 180 minutes

[AshutoshKumar-wyyb]

sir plz plz plz reply me,
i startes learning from you since last 4 months and really OBSSESSED by your teaching style

[Shriyanshi-oenp]

It's really helpful...thank u sir 😊

[movies.cop]

sir ye questions aate hai aapne kraye the sir ji

[syahi]

9-3², since it has no 2&5 so it will not end with 0

[punyanshpatial]

Hatts off to u and your efforts sir ❤️

[lazylife]

Love you sir are you best teacher ❤️❤️❤️ according to me

[sunitharai]

Sir is this video usefull for basic students 💥🙂💖

[rishabh]

Thank you so much sir 💖💖 sir please ye bata do ki agar mai ye series + 10 hours wala session kr lu tho mere 80/80 aajayega?? maine aaj ka expected wala live session bhi attend Kiya tha ❤❤.

[luckylalvishwakarma]

in question no. 3 the given no. is composite no.

[s.pidey_x]

Sir aap padhaye huye h, sare ke sare aate h ..
Love from muzzafarpur sir❤