Question

let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46
please answer fast

Answer 1

Given :- Let a and b are positive integers such that 90 < a + b < 99 and 0.9 < a/b < 0.91 . Find (ab/46) ?

Answer :-

→ 90 < a + b < 99 --------- Eqn.(1)

→ 0.9 < a/b < 0.91 -------- Eqn.(2)

taking Eqn.(2),

→ 0.9 < a/b < 0.91

→ 0.9b < a < 0.91b

adding b ,

→ 0.9b + b < a + b < 0.91b + b

using Eqn.(1) now , we get,

→ 0.9b + b < 99 => 1.9b < 99 => b < 52.1

→ 0.91b + b > 90 => 1.91 b > 99 => b > 51.8

since it is given that, b is a positive integer ,

→ b = 52 .

putting this value we get,

→ 0.9 * 52 < a < 0.91 * 52

→ 46.8 < a < 47.3

→ a = 47 .

but,

→ a + b < 99

→ 52 + 47 < 99

→ 99 < 99

therefore, b must be 51 and a must be 46 .

hence,

→ (ab)/46

→ (51 * 46)/46

→ 51 (Ans.)

Note :- if first condition was 90 ≤ a + b ≤ 99 , a will be 52 and b will be 47 .

Learn more :-

if the positive square root of (√190 +√ 80) i multiplied by (√2-1) and the

product is raised to the power of four the re...

https://brainly.in/question/26618255

Answer 2

Given : a and b be positive integers such that 90<a+b<99 and 0.9<a/b<0.91

To Find : ab/46​

Solution:

90<a+b<99

0.9<a/b<0.91

0.9b < a

a < 0.91b

0.9b < a

a+b<99

=> a < 99-b

=> 0.9b <  99-b

=> 1.9b < 99

=> b < 52.1

a < 0.91b

90<a+b

=> 90 - b < a

=> 90 - b < 0.91b

=> 90 < 1.91b

=> 47.12 < b

47.12 < b  < 52.1

=> b = 48 , 49 , 50 , 51 , 52

0.9b < a  < 0.91b

b = 48  

=> 43.2 < a < 43.68    no integer values of a

b = 49  

=> 44.1 < a < 44.59    no integer values of a

b = 50  

=> 45 < a < 45.5   no integer values of a

b = 51

=> 45.9 < a < 46.41     so a = 46

b = 52

=> 46.8 < a < 47.32     so a = 47

but 52 + 47 not less than 99.

b = 51  and a = 46   only satisfy

ab /46  = 46 * 51/46  = 51

Learn More

Find ab/46​

https://brainly.in/question/40052768