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Let $S=\{6,10,7,13,5,12,8,11,9\},$ and $a=\sum_{x\in S}(x-9)^{2}\:\&\: b=\sum_{x\in S}(x-10)^{2}.$ Then

  1. $a<b$
  2. $a>b$
  3. $a=b$
  4. None of these
in Numerical Ability by Boss (17.5k points)
recategorized by | 53 views
0
@jothee
0
$D$?
0

@ankitgupta.1729

this as well?

+2

@`JEET

here, $a= \sum_{x=5}^{13} (x-9)^2$

and $b= \sum_{x=5}^{13} (x-10)^2 = \sum_{x=5}^{13} ((x-9) - 1)^2 = \sum_{x=5}^{13}[(x-9)^2 + 1 - 2(x-9)] = \sum_{x=5}^{13}(x-9)^2 + \sum_{x=5}^{13}1 - 2\sum_{x=5}^{13}(x-9) = a + \sum_{x=5}^{13}1 - 2\sum_{x=5}^{13}(x-9)$

$b-a = \sum_{x=5}^{13}1 - 2\sum_{x=5}^{13}(x-9)$

now, check $b-a = 0$ (or) $b-a > 0$ (or) $b-a < 0$

 

+1
Thanks

2 Answers

+3 votes

Here $a=\displaystyle \sum_{x\in S} (x-9)^2$ and $b=\displaystyle \sum_{x\in S} (x-10)^2$

$\begin{align}\therefore a-b &=\sum_{x\in S} \left\{ (x-9)^2-(x-10)^2 \right\}; ~~[\text{Distributive law of sum}] \\&=\sum_{x\in S} \left( 2x-19 \right); ~~[\text{using formula }a^2-b^2=(a+b)(a-b)]\\&=2\sum_{x\in S}x-19\sum_{x\in S}1;~~[\text{using the distributive law of sum and }\sum cf(x)=c \sum f(x)]\\&=2 \sum_{x\in S}x-19\times|S| \\&=2(6+10+7+13+5+12+8+11+9) -19\times 9 \\&=-9 \\ \Rightarrow a-b &<0;~~[\because-9<0]\\ \therefore a&<b\end{align}$

 

So the correct answer is A.

by Active (3.6k points)
+2 votes
$\underline{\textbf{Answer:}\Rightarrow}\;\;\mathbf{A.}$

$$\begin {align}\textbf{Given:}\;\;\;\;\;\;\;\mathrm a &= \sum_{x=5}^{13}(x-9)^2\\ \text b &= \sum_{x=5}^{13}(x-10)^2 \\&=\sum_{x = 5}^{13}((x-9)-1)^2\\&=\sum_{x = 5}^{13}[(x-9)^2 + 1 -2(x-9)] \\&=\sum_{x = 5}^{13}(x-9)^2 + \sum_{x=5}^{13}1-2\sum _{x = 5}^{13}(x-9)\\&=a + \sum_{x = 5}^{13}1-2 \underbrace {\sum_{x=5}^{x = 13}(x-9)}_\text {= 0 }\\&=a + 9 -2(-4-3-2-1-0+1+2+3+4)\\&=a + 9 \end {align}$$

Now, we get:

$$\begin{align} &b = a + 9\\ &\Rightarrow b-a > 0\\ &\Rightarrow b\gt a \\ &\text{or,} ~a \lt b \end {align}$$

$\therefore \textbf A$ is the correct answer.
by Boss (19.2k points)
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