CTET 2 Math and Science Jul 2019 Paper

© examsiri.com

Question : 51 of 120

Marks: +1, -0

If

x = 2^3 × 3^2 × 5^3 × 7^3

,

y = 2^2 × 3^3 × 5^4 × 7^3

, and

z = 2^4 × 3^4 × 5^2 × 7^5

,
then H.C.F. of x, y and z is:

$(15)^3 × 7^4$
$(30)^3 × 7^3$
$30 × 7^5$
$(30)^2 × 7^3$

Solution:

HCF (x, y, z) =

2^2 × 3^2 × 5^2 × 7^3

= 900 × 7^3

= (30)^2 × 7^3

© examsiri.com

Go to Question:

Prev Question

Next Question