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 About the author Maya
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 Reply
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 Reply
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
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