Candidates of four schools appear in a mathematics test- The data were as follows- - School No-of-Candidates Average-Score- I 60 75- II 48 80- III Not-available 55- IV 40 50- - If the average score of the candidates of all the four schools is 66- find the number of candidates that appeared from school III-

Let the students from 3rd school be x Average score = ( 60×75 ) + ( 80×48 ) + ( 55× x) + ( 50 × 40 )/ 60 + 48 + x + 40 66 = 4500+3840+55x + 2000/148 +x 66(148 + ...

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Standard X

Mathematics

Finding Mean for Grouped Data When Class Intervals Are Not Given

Candidates of...

Question

Candidates of four schools appear in a mathematics test. The data were as follows:
$\begin{matrix} S c h o o l & N o . o f C a n d i d a t e s & A v e r a g e S c o r e I & 60 & 75 I I & 48 & 80 I I I & N o t a v a i l a b l e & 55 I V & 40 & 50 \end{matrix}$
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

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Solution

Let the students from 3rd school be x

Average score

$= \frac{(60 \times 75) + (80 \times 48) + (55 \times x) + (50 \times 40)}{60 + 48 + x + 40}$

$66 = \frac{4500 + 3840 + 55 x + 2000}{148 + x}$

$66 (148 + x) = 10340 + 55 x$

$9768 + 66 x = 10340 + 55 x$

$11 x = 572; x = 52$

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Candidates of four schools appear in a mathematics test. The data were as follows: SchoolNo. of CandidatesAverage ScoreI6075II4880IIINot available55IV4050 If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

Let the students from 3rd school be x Average score =(60×75)+(80×48)+(55×x)+(50×40)60+48+x+40 66=4500+3840+55x+2000148+x 66(148+x)=10340+55x 9768+66x=10340+55x 11x=572 ; x=52

Let the students from 3rd school be x

Average score

$= \frac{(60 \times 75) + (80 \times 48) + (55 \times x) + (50 \times 40)}{60 + 48 + x + 40}$

$66 = \frac{4500 + 3840 + 55 x + 2000}{148 + x}$

$66 (148 + x) = 10340 + 55 x$

$9768 + 66 x = 10340 + 55 x$

$11 x = 572; x = 52$