0 deepaksoniGuru Asked: March 17, 20212021-03-17T19:44:46+05:30 2021-03-17T19:44:46+05:30In: ICSE Find the sub-triplicate ratio of (i) 1: 216 (ii) 1/8: 1/125 (iii) 27a3: 64b3 0 Question no6. From ML aggarwal book, class10, chapter 7, ratio and proportion We have been given the ratio in the form of a:b and we have to find sub triplicate ratio of them icse boardml aggarwal solutionratio and proportion Share Facebook 1 Answer Voted Oldest Recent AAREKH Guru 2021-03-24T14:15:35+05:30Added an answer on March 24, 2021 at 2:15 pm Solution: (i) 1: 216 We know that Sub-triplicate ratio of 1: 216 = âˆ›1: âˆ›216 By further calculation = (1^{3})^{1/3}: (6^{3})^{1/3} = 1: 6 (ii) 1/8: 1/125 We know that Sub-triplicate ratio of 1/8: 1/125 = (1/8)^{1/3}: (1/125)^{1/3} It can be written as = [(1/2)^{3}]^{1/3}: [(1/5)^{3}]^{1/3} So we get = Â½: 1/5 = 5: 2 (iii) 27a^{3}: 64b^{3} We know that Sub-triplicate ratio of 27a^{3}: 64b^{3}Â = [(3a)^{3}]^{1/3}: [(4b)^{3}]^{1/3} So we get = 3a: 4b 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Question 21. (a) In the figure (i) given below, PQ is a tangent to the circle at A, ... Question 13. (a) In the figure (i) given below, AB = 8 cm and M is mid-point of ... Question 3. The tangent to a circle of radius 6 cm from an external point P, is of ... Question 16. In the given figure, chords AB and CD of the circle are produced to meet at ... Question 15. (a) Prove that a cyclic parallelogram is a rectangle. (b) Prove that a cyclic rhombus is ...

Solution:(i) 1: 216

We know that

Sub-triplicate ratio of 1: 216 = âˆ›1: âˆ›216

By further calculation

= (1

^{3})^{1/3}: (6^{3})^{1/3}= 1: 6

(ii) 1/8: 1/125

We know that

Sub-triplicate ratio of 1/8: 1/125 = (1/8)

^{1/3}: (1/125)^{1/3}It can be written as

= [(1/2)

^{3}]^{1/3}: [(1/5)^{3}]^{1/3}So we get

= Â½: 1/5

= 5: 2

(iii) 27a

^{3}: 64b^{3}We know that

Sub-triplicate ratio of 27a

^{3}: 64b^{3}Â = [(3a)^{3}]^{1/3}: [(4b)^{3}]^{1/3}So we get

= 3a: 4b