Case Study II
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 in the 9th year.
Based on this information, answer the following questions:
The production of the TV sets during the first year was :
(a) 4000
(b) 4500
(c) 5000
(d) 5500
What was the uniform increase in the production of TV sets every year?
(a) 1800
(b) 2400
(c) 1600
(d) 2200
The production of the TV sets during the 8th year was:
(a) 20000
(b) 20400
(c) 21200
(d) 22800
The total production of the TV sets during first 6 years was:
(a) 56000
(b) 72000
(c) 66000
(d) 63000
The average production of the TV sets during first 6 years was:
(a) 10500
(b) 11000
(c) 11500
(d) 12000

Question

Case Study II

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 in the 9th year.

Based on this information, answer the following questions:

The production of the TV sets during the first year was :
(a) 4000
(b) 4500
(c) 5000
(d) 5500
What was the uniform increase in the production of TV sets every year?
(a) 1800
(b) 2400
(c) 1600
(d) 2200
The production of the TV sets during the 8th year was:
(a) 20000
(b) 20400
(c) 21200
(d) 22800
The total production of the TV sets during first 6 years was:
(a) 56000
(b) 72000
(c) 66000
(d) 63000
The average production of the TV sets during first 6 years was:
(a) 10500
(b) 11000
(c) 11500
(d) 12000

KnowledgeBoat · Accepted Answer

1\. Since, production of TV sets in a factory increases uniformly by a fixed number every year. The production in each year forms an A.P. Let production of the TV sets during the first year be a and the common difference in production each year be d. We know that, an = a + (n - 1)d Given, Production in the 6th year = 16000. ⇒ a6 = a + (6 - 1)d ⇒ 16000 = a + 5d ⇒ a + 5d = 16000 ....(1) Given, Production in the 9th year = 22600. ⇒ a9 = a + (9 - 1)d ⇒ 22600 = a + 8d ⇒ a + 8d = 22600 ....(2) Subtracting Eq. (1) from Eq. (2), we get : ⇒ a + 8d - (a + 5d) = 22600 - 16000 ⇒ a + 8d - a - 5d = 6600 ⇒ 3d = 6600 ⇒ d = ⇒ d = 2200. Substituting d in equation (1), we get : ⇒ a + 5(2200) = 16000 ⇒ a + 11000 = 16000 ⇒ a = 16000 - 11000 ⇒ a = 5000. Hence, option (c) is the correct option. 2\. From question (1), The uniform increase is the common difference, d. d = 2200. Hence, option (d) is the correct option. 3\. Given, The production of TV sets in each year forms an A.P.: 5000, 7200, 9400, ..... a = 5000 n = 8 d = 2200 We know that, an = a + (n - 1)d ⇒ a8 = 5000 + (8 - 1)2200 = 5000 + 7(2200) = 5000 + 15400 = 20400. Production of the TV sets during the 8th year = 20400. Hence, option (b) is the correct option. 4\. Given, The production of TV sets in each year forms an A.P.: 5000, 7200, 9400..... a = 5000 l = 16000 n = 6 We know that, Sn = (a + l) ⇒ S6 = (5000 + 16000) = 3 × (21000) = 63000. Total production of the TV sets during first 6 years = 63000 Hence, option (d) is the correct option. 5\. From part 4 we have, S6 = 63000 The average production is the total production divided by the number of years: Average = = = 10500. Hence, option (a) is the correct option.

Mathematics

AP

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