Verified Solution • 2025–26 Session moderate

QJEEM24Ma7395: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7395 Solution: Let and . Let the mean, the mean deviation about the mean and…

Let a,b,c∈N\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{N} and \mathrm{a}< \mathrm{b}< \mathrm{c} . Let the mean, the mean deviation about the mean and the variance of the 5 observations 9,25,a,b,c9,25, a, b, c be 18, 4 and 1365\frac{136}{5} , respectively. Then 2a+b−c2 a+b-c is equal to ________

Step-by-Step Verified Solution

amp;a,b,c∈Namp;alt;blt;camp; Mean =18amp;9+25+a+b+c5=18amp;34+a+b+c=90amp;a+b+c=56\begin{aligned}& a, b, c \in N \\& a< b < c \\& \text { Mean }=18 \\& \frac{9+25+a+b+c}{5}=18 \\& 34+a+b+c=90 \\& a+b+c=56\end{aligned}

amp;∣9−18∣+∣25−18∣+∣a−18∣+∣b−18∣+∣c−18∣5=4amp;9+7+∣a−18∣+∣b−18∣+∣c−18∣=20amp;∣a−18∣+∣b−18∣+∣c−18∣=4amp;1365=706+a2+b2+c25−(18)2amp;⇒136=706+a2+b2+c2−1620amp;⇒a2+b2+c2=1050amp; Consider alt;19lt;blt;camp; Solving a=17,b=19,c=20amp;2a+b−camp;34+19−20amp;=33\begin{aligned}& \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\& 9+7+|a-18|+|b-18|+|c-18|=20 \\& |a-18|+|b-18|+|c-18|=4 \\& \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\& \Rightarrow 136=706+a^2+b^2+c^2-1620 \\& \Rightarrow a^2+b^2+c^2=1050 \\& \text { Consider } a<19 < b< c \\& \text { Solving } a=17, b=19, c=20 \\& 2 a+b-c \\& 34+19-20 \\& =33\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring