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The frequency distribution of the age of students in a class of 40 students is given below. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-1wig{font-weight:bold;text-align:left;vertical-align:top} .tg .tg-baqh{text-align:center;vertical-align:top} Age 15 16 17 18 19 20 No of Students 5 8 5 12 $$x$$ $$y$$ If the mean deviation about the median is 1.25, then $$4x+5y$$ is equal to :

JEE · Math · previous-year question

  1. A.43
  2. B.46
  3. C.44correct
  4. D.47

Answer

C. 44

Explanation

.tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} Age No. of Students CF 15 5 5 16 8 13 17 5 18 18 12 30 19 $$x$$ $$30+x$$ 20 $$y$$ $$30+x+y$$ $$\begin{aligned} & 30+x+y=40 \\ & x+y=10 \end{aligned}$$ Median $$=\left(\frac{n+1}{2}\right)^{\text {th }}$$ observation $$=\frac{40+1}{2}=\frac{41}{2}$$ Median $$=18$$ Mean deviation about median $$\begin{aligned} & 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\ & 15+16+5+x+2 y=50 \\ & x+2 y=14 \\ & \quad x+y=10 \\ & \Rightarrow \quad x=4 \\ & \Rightarrow \quad y=6 \\ & \begin{aligned} 4 x+5 y & =24+20 \\ & =44 \end{aligned} \end{aligned}$$

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