Answer these questions based on the following data.
In a standard 5 class, the students are arranged according to the increasing order of their marks in an examination. The marks obtained by every student, starting from the third student, are the sum of the marks obtained by the earlier two students.

Marks scored by students are in A.P. If the $12^{t h}$ student got 809 marks and the $2^{n d}$ student got 6 marks, then what is the sum of the marks obtained by the first ten students?

815

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803

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800

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Cannot be determined

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Solution

The correct option is B
803

By observation, we get
$T_{12} = S_{(n - 2)} + t_{2}$
It is given that
$T_{12}$ = 809 and $t_{2}$ =6
$T_{12} = S_{(12 - 2)} + t_{2}$
$\Rightarrow t_{12} = S_{10} + t_{2}$
$\Rightarrow S_{10} = 809 - 6 = 803$

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\begin{matrix} 8 & 7 & 12 & 17 & 20 & 11 & 10 & 10 & 16 \\ 21 & 28 & 25 & 13 & 9 & 27 & 18 & 20 & 19 \\ 7 & 29 & 16 & 15 & 20 & 22 & 27 \end{matrix}

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Q. Study the line graph and answer the given questions. The given graph indicates the marks scored in two subjects namely physics and chemistry by 5 students in an examination. Use the information to answer the question.

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Q. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions:

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The correct option is B 803 By observation, we get T12=S(n−2)+t2 It is given that T12 = 809 and t2 =6 T12=S(12−2)+t2 ⇒t12=S10+t2 ⇒S10=809−6=803

The correct option is B
803

By observation, we get
$T_{12} = S_{(n - 2)} + t_{2}$
It is given that
$T_{12}$ = 809 and $t_{2}$ =6
$T_{12} = S_{(12 - 2)} + t_{2}$
$\Rightarrow t_{12} = S_{10} + t_{2}$
$\Rightarrow S_{10} = 809 - 6 = 803$