The weighted aggregative index number for the following table is given as 153.61 considering Laspeyres method. Find the approximate value of k.

$\begin{matrix} P_{0} & Q_{0} & P_{1} & Q_{1} H & 0.5 & 20 & 2.5 & 25 I & 10.5 & 10 & 7 & 11 J & 3.5 & 8 & 4 & 12 K & 7 & 25 & 14 & 30 L & 15 & 13 & 22 & 18 \end{matrix}$

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Solution

The correct option is C
20

Look at the table below:

$\begin{matrix} P_{0} & Q_{0} & P_{1} & Q_{1} & P_{1} Q_{0} & P_{0} Q_{0} H & 0.5 & k & 2.5 & 25 & 2.5 k & 0.5 k I & 10.5 & 10 & 7 & 11 & 70 & 105 J & 3.5 & 8 & 4 & 12 & 32 & 28 K & 7 & 25 & 14 & 30 & 350 & 175 L & 15 & 13 & 22 & 18 & 286 & 195 Total & 738 + 2.5 k & 503 + 0.5 k \end{matrix}$

$P 01 = \frac{\sum P_{1} Q_{0}}{\sum P_{0} Q_{0}} \times 100 153.61 = \frac{738 + 2.5 k}{503 + 0.5 k} \times 100 \Rightarrow 1.5361 \times (503 + 0.5 k) = 738 + 2.5 k \Rightarrow 772.66 + 0.77 k = 738 + 2.5 k \Rightarrow 34.66 = 1.73 k \Rightarrow k = 20$

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$\begin{matrix} Class interval & 1 - 10 & 11 - 20 & 21 - 30 & 31 - 40 Frequency & 7 & 12 & 15 & 13 \end{matrix}$

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Marks in Maths	10	11	10	11	11	14	12	12	13	10
Marks in Statistics	20	21	22	21	23	23	22	21	24	23
Marks in Maths	13	12	11	12	10	14	14	12	13	10
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15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10

15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10

The weighted aggregative index number for the following table is given as 153.61 considering Laspeyres method. Find the approximate value of k. P0Q0P1Q1H0.5202.525I10.510711J3.58412K7251430L15132218

The correct option is C 20 Look at the table below: P0Q0P1Q1P1Q0P0Q0H0.5k2.5252.5k0.5kI10.51071170105J3.584123228K7251430350175L15132218286195Total738+2.5k503+0.5k P01=∑P1Q0∑P0Q0×100153.61=738+2.5k503+0.5k×100⇒1.5361×(503+0.5k)=738+2.5k⇒772.66+0.77k=738+2.5k⇒34.66=1.73k⇒k=20

15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10