Solution to 3)Let x=n+ε, where n is an integer and 0≤ε<1. If ⌊55x/7⌋=8n, then which of the … - Sikademy
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3)Let x=n+ε, where n is an integer and 0≤ε<1. If ⌊55x/7⌋=8n, then which of the following cannot be a value of x: a) 14.3 b) 37.7 c) 46.9 d) 51.6 4) Let x=n+ε, where n is an integer and 0≤ε<1. If ⌈33x/8⌉=5n, then which of the following is a valid value of x: a) 5.8 b) 4.7 c) 6.1 d) 2.5

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Here's the Solution to this Question

(i) As x=n+ε, 0≤ε<1


Given relation is- \dfrac{55x}{7}=8n

(a) Let x=14.3, then n=14 as x=n+ε


Taking LHS=\dfrac{55\times 14.3}{7}=112.35


Taking RHS=8(14)=112

(b) Let x=37.7, then n=37


Taking LHS=\dfrac{55\times 37.7}{7}=296.4


Taking RHS=8(37)=296


(c) Let x=46.9, then n=46


Taking LHS=\dfrac{55\times 46.9}{7}=368.5


Taking RHS=8(46)=368

(d) Let x=51.6, then n=51


Taking LHS=\dfrac{55\times 51.6}{7}=405.42


Taking RHS=8(51)=408


The difference between the LHS and RHS is more in part D,

Hence The value of x can not be 51.6


(ii) Given Relation is \dfrac{33x}{8}=5n


(a) Let x=5.8 , then n=5


Taking LHS =\dfrac{33\times 5.8}{8}=23.925


Taking RHS=5(5)=25


(b) Let x=4.7 , then n=4


Taking LHS =\dfrac{33\times 4.7}{8}=19.3875

Taking RHS=5(4)=20


(c) Let x=6.1 , then n=6

Taking LHS =\dfrac{33\times 6.1}{8}=25.1625


Taking RHS=5(6)=30


(d) Let x=2.5 , then n=2


Taking LHS =\dfrac{33\times 2.5}{8}=10.3125


Taking RHS=5(2)=10


The difference between the RHS and LHS is significant in part (a) and (C)

Therefore x can not have the values 5.8 and 6.1



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Question ID: mtid-5-stid-8-sqid-3073-qpid-1772