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प्रश्न
| India is a competitive manufacturing location due to low manpower costs and strong technical and engineering capabilities, which contribute to higher-quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in the 6th year and 22600 sets in the 9th year. |
Based on the above information, answer the following questions.
- In which year will the factory produce 29200 sets?
- Find the production during the 8th year.
- Find the production during the first 3 years.
- Find the difference in production during the 7th year and the 4th year.
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उत्तर
Since production increases uniformly by a fixed number every year, the annual production values form an Arithmetic Progression (AP). Let:
First year production = a
Uniform increase per year = d
From the given data: \[T_6 = a + 5d = 16000 \quad \ldots (1)\]
\[T_9 = a + 8d = 22600 \quad \ldots (2)\]
Subtracting equation (1) from equation (2):
\[(a + 8d) - (a + 5d) = 22600 - 16000\]
\[3d = 6600\]
\[d = 2200\]
Substituting d = 2200 into equation (1):
\[a + 5(2200) = 16000\]
\[a + 11000 = 16000\]
\[a = 5000\]
i. Year in which the factory produces 29200 sets:
Let production in the nth year be \[T_n = 29200\].
\[a + (n - 1)d = 29200\]
\[5000 + (n - 1)2200 = 29200\]
\[(n - 1)2200 = 24200\]
\[n - 1 = \frac{24200}{2200} = 11\]
\[n = 12\]
In the 12th year.
ii. Production during the 8th year: \[T_8 = a + 7d\]
\[T_8 = 5000 + 7(2200)\]
\[T_8 = 5000 + 15400 = 20400\]
The production during the 8th year is 20400 sets.
iii. Production during the first 3 years: \[S_n = \frac{n}{2}[2a + (n - 1)d]\]
For (n = 3): \[S_3 = \frac{3}{2}[2(5000) + (3 - 1)2200]\]
\[S_3 = \frac{3}{2}[10000 + 4400]\]
\[S_3 = \frac{3}{2} \times 14400 = 21600\]
Total production during the first 3 years is 21600 sets.
iv. Difference in production during the 7th year and the 4th year:
\[T_7 - T_4 = (a + 6d) - (a + 3d) = 3d\]
\[T_7 - T_4 = 3 \times 2200 = 6600\]
The difference in production is 6600 sets.
