[4 marks sum]

Question 14 Marks

The following data has been arranged in ascending order.$0, 1, 2, 3, x + 1, x + 5, 20, 21, 26, 29.$Find the value of $x,$ if the median is $5.$

Answer

Data in ascending order:
$0,1,2,3, x+1, x+5,20,21,26,29$
Here, total number of observation $=n=10($even$)$
Median$=5 $
$\Rightarrow \frac{\left(\frac{ n }{2}\right)^{\text {th }} \text { term }+\left(\frac{ n }{2}+1\right)^{\text {th }} \text { term }}{2}=5 $
$\Rightarrow \frac{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }}{2}=5$
$\Rightarrow \frac{(x+1)+(x+5)}{2}=5$
$\Rightarrow 2 x +6=10 $
$\Rightarrow 2 x =4 $
$\Rightarrow x=2 . $

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Question 24 Marks

$3, 8, 10, x, 14, 16, 18, 20$ are in the ascending order and their median is $13.$ Calculate the numerical value of $x.$

Answer

$3,8,10, x, 14,16,18,20$ is already in ascending order
Here $N =8$
$\therefore\left(\frac{ N }{2}\right)^{\text {th }}$term
$=\left(\frac{8}{2}\right) $
$=4^{\text {th }}$term
$=x $
$\left(\frac{ N }{2}+1\right)^{\text {th }}$term
$=4+1 $
$=5^{\text {th }}$term
$=14$
Given, median $=13$
$\therefore \frac{x+14}{2}=13$
$\therefore x+14$
$=13 \times 2 $
$=26$
$\Rightarrow x$
$=26-14$
$=12 .$

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Question 34 Marks

Calculate the median of the following sets of number:$1, 9, 10, 8, 2, 4, 4, 3, 9, 1, 5, 6, 2$ and $4.$

Answer

Arranging these numbers in ascending order,
$1,1,2,2,3,4,4,4,5,6,8,9,9,10 $
Here $N =14 $
$\therefore\left(\frac{ N }{2}\right)^{\text {th }}$term
$=\left(\frac{14}{2}\right) $
$=7^{\text {th }}$term
$=4 $
$\left(\frac{ N }{2}+1\right)^{\text {th }}$term
$=7+1 $
$=8^{\text {th }}$term
$=4 $
$\therefore$ Median
$=\frac{4+4}{2} $
$=\frac{8}{2} $
$=4$
Or
Median$=4 .$

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Question 44 Marks

Find the median of the following sets of numbers.$25, 11, 15, 10, 17, 6, 5, 12.$

Answer

$25,11,15,10,17,6,5,12$
Arranging the data in ascending order,
$5,6,10,11,12,15,17,25$
Here $N =8$
$\therefore\left(\frac{ N }{2}\right)^{\text {th }}$term
$=\left(\frac{8}{2}\right) $
$=4^{\text {th }}$term
$=11 $
$\left(\frac{ N }{2}+1\right)^{\text {th }}$term
$=4+1 $
$=5^{\text {th }}$term
$=12$
$\therefore$ Median
$=\frac{11+12}{2} $
$=\frac{23}{2} $
$=11.5$
Or
Median$=11.5$

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Question 54 Marks

The runs scored by a cricket player in the last $30$ innings are:$75, 125, 36, 89, 154, 56, 12, 28, 96, 142, 78, 54, 30, 88, 116, $$104, 55, 84, 10, 29, 31, 08, 24, 136, 117, 22, 99, 80, 112, 35.$Arrange these scores in an ascending order and answer the following:

Find the highest score.
Find the number of centuries scored by him.
Find the number of times he scored over $50.$
Find the number of times he failed to score a $50.$

Answer

$ \operatorname{ar}(\triangle APD )=\frac{\sqrt{3} s ^2}{4}$
$\operatorname{ar}(\triangle APD )=\frac{\sqrt{3} \times 8^2}{4}$
$\operatorname{ar}(\triangle APD )=\frac{\sqrt{3} \times 64}{4}$
$\operatorname{ar}(\triangle APD )=\sqrt{3} \times 16=16 \sqrt{3} \ cm ^2$
$\operatorname{ar}(\triangle APD )=\frac{1}{2} \times \operatorname{ar}($parallelogram $\text{ABCD}) $
$($The area of a triangle is half that of a parallelogram on the same base and between the same parallels$)$
$\Rightarrow \operatorname{ar}($ parallelogram $A B C D)=2 \times \operatorname{ar}(\triangle A P D)$
$\Rightarrow \operatorname{ar}($ parallelogram $A B C D)=2 \times 16 \sqrt{3} \ cm ^2$
$\Rightarrow \operatorname{ar}($ parallelogram $ABCD )=32 \sqrt{3} \ cm ^2$

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MATHEMATICS [4 marks sum]

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