MathsCBSE Class 10 · Basic · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic sample paper 2023-24: question 36

From the official CBSE sample question paper for 2023-24, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

CASE STUDY 1

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing 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 6th year and 22600 in 9th year.

1)
In which year, the production is 29,200 sets?
1 mark
2)
Find the production in the 8th year.
1 mark
2)
Find the production in first 3 years.
1 mark
3)
Find the difference of the production in 7th year and 4th year.
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Equal yearly increases mean the productions form an AP; write the 6th and 9th terms using $a_{n} = a + (n - 1) d$ .

Hint 2

Subtract the two equations to find $d$ , then find $a$ .

Hint 3

For the first 3 years use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ ; for the difference of the 7th and 4th terms, remember it equals $3 d$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Since production increases by a fixed number every year, the yearly productions form an AP with first term $a$ and common difference $d$ . Given $a_{6} = 16000$ and $a_{9} = 22600$ :
$a + 5 d = 16000$ ...(1)
$a + 8 d = 22600$ ...(2)
Subtracting (1) from (2): $3 d = 6600 \Rightarrow d = 2200$ . Then $a = 16000 - 5 (2200) = 16000 - 11000 = 5000$ .
1) Let the production be 29200 in the $n$ th year. Using $a_{n} = a + (n - 1) d$ :
$29200 = 5000 + (n - 1) (2200)$
½ mark
$(n - 1) (2200) = 24200 \Rightarrow n - 1 = 11 \Rightarrow n = 12$ .
The production is 29,200 sets in the 12th year.
½ mark
2) Production in the 8th year: $a_{8} = a + 7 d = 5000 + 7 (2200)$
½ mark
$= 5000 + 15400 = 20400$ . The production in the 8th year is 20,400 sets.
½ mark
2) (OR) Production in the first 3 years: $S_{3} = \frac{n}{2} [2 a + (n - 1) d] = \frac{3}{2} [2 (5000) + 2 (2200)]$
½ mark
$= \frac{3}{2} (10000 + 4400) = \frac{3}{2} (14400) = 21600$ . The production in the first 3 years is 21,600 sets.
½ mark
3) $a_{4} = a + 3 d = 5000 + 3 (2200) = 5000 + 6600 = 11600$
½ mark
$a_{7} = a + 6 d = 5000 + 6 (2200) = 5000 + 13200 = 18200$
½ mark
Difference: $a_{7} - a_{4} = 18200 - 11600 = 6600$ sets. (Check: $a_{7} - a_{4} = 3 d = 3 \times 2200 = 6600$ .)
1 mark

Answer: 1) 12th year; 2) 20,400 sets (OR first 3 years: 21,600 sets); 3) 6,600 sets

Where the marks are

Sets up $a + 5 d = 16000$ , $a + 8 d = 22600$ and gets $d = 2200$ , $a = 5000$ (needed to start)
Writes $29200 = 5000 + (n - 1) 2200$ (½ mark)
Gets $n = 12$ , i.e. the 12th year (½ mark)
Writes $a_{8} = 5000 + 7 (2200)$ (½ mark)
Gets 20400 sets (½ mark)
OR: uses $S_{3} = \frac{3}{2} [10000 + 4400]$ (½ mark) and gets 21600 sets (½ mark)
Finds $a_{4} = 11600$ (½ mark) and $a_{7} = 18200$ (½ mark)
States the difference as 6600 sets (1 mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Finding nth Term from the Sum of n Terms

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic sample paper 2023-24: question 36

Hints

Step-by-step solution

Revise the chapter

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