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Ex 2.5
Ex 2.5, 1 (ii)
Ex 2.5, 1 (iii) Important
Ex 2.5, 1 (iv)
Ex 2.5, 1 (v)
Ex 2.5, 2 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 2 (ii) Important
Ex 2.5, 2 (iii)
Ex 2.5, 3 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 3 (ii) Important
Ex 2.5, 3 (iii)
Ex 2.5, 4 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 4 (ii)
Ex 2.5, 4 (iii) Important
Ex 2.5, 4 (iv)
Ex 2.5, 4 (v)
Ex 2.5, 4 (vi)
Ex 2.5, 5 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 5 (ii) Important
Ex 2.5, 6 (i)
Ex 2.5, 6 (ii)
Ex 2.5, 6 (iii)
Ex 2.5, 6 (iv) Important
Ex 2.5, 7 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 7 (ii)
Ex 2.5, 7 (iii) Important
Ex 2.5, 8 (i)
Ex 2.5, 8 (ii)
Ex 2.5, 8 (iii) Important
Ex 2.5, 8 (iv) Important
Ex 2.5, 8 (v)
Ex 2.5, 9 (i)
Ex 2.5, 9 (ii)
Ex 2.5, 10 (i) Important Deleted for CBSE Board 2023 Exams
Ex 2.5, 10 (ii)
Ex 2.5, 11 Deleted for CBSE Board 2023 Exams
Ex 2.5,12 Important
Ex 2.5,13
Ex 2.5, 14 (i)
Ex 2.5, 14 (ii) Important
Ex 2.5, 15 (i)
Ex 2.5, 15 (ii) Important
Ex 2.5, 16 (i) Deleted for CBSE Board 2023 Exams
Ex 2.5, 16 (ii) Important You are here
Last updated at Aug. 25, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 2.5, 16 What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : 12ky2 + 8ky – 20k First we factorize 12ky2 + 8ky – 20k = 4k (3y2 + 2y – 5) We factorize (3y2 + 2y – 5) separately 3y2 + 2y – 5 = 3y2 + 5y – 3y – 5 = y (3y + 5) – 1 (3y + 5) = (3y + 5) (y – 1) Thus, 12ky2 + 8ky – 20k = 4k (3y + 5) (y – 1) So, Volume = 4k (3y + 5) (y – 1) We know that Volume of cuboid = Length × Breadth × Height Possible dimensions can be Length = 4k, Breadth = 3y + 5 , Height = y – 1