MathsCBSE Class 10 · Standard · Board exam 2023, set 30/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2023: question 36

Set in the CBSE Class 10 board exam of 2023, paper 30/4/1, 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-basedPair of Linear Equations in Two Variables

Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey ₹ x per student and Cricket ₹ y per student. School 'P' decided to award a total of ₹ 9,500 for the two games to 5 and 4 students respectively; while school 'Q' decided to award ₹ 7,370 for the two games to 4 and 3 students respectively.

Based on the above information, answer the following questions:

(i)
Represent the following information algebraically (in terms of x and y).
1 mark
(ii)(a)
What is the prize amount for hockey?
2 marks
OR
(ii)(b)
Prize amount on which game is more and by how much?
2 marks
(iii)
What will be the total prize amount if there are 2 students each from two games?
1 mark

Figure in the paper: Photograph of two silver-coloured trophies placed side by side, shown beside the case study text. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/4/1.

Hints

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

Hint 1

Each school's total equals (number of hockey students × x) + (number of cricket students × y).

Hint 2

Eliminate y by making its coefficients equal: multiply the first equation by 3 and the second by 4.

Hint 3

After finding x, substitute it back to get y, then compare or add as asked.

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.

(i) School P gives 5 hockey prizes and 4 cricket prizes worth ₹9500 in all:
$5 x + 4 y = 9500$ ...(1)
½ mark
(i) School Q gives 4 hockey prizes and 3 cricket prizes worth ₹7370 in all:
$4 x + 3 y = 7370$ ...(2)
½ mark
(ii)(a) Multiply (1) by 3 and (2) by 4:
$15 x + 12 y = 28500$
$16 x + 12 y = 29480$
Subtracting the first from the second gives $x = 980$ .
So the prize amount for hockey is ₹980 per student.
2 marks
(ii)(b) Solving as in (a) gives $x = 980$ .
Putting this in (1): $4 y = 9500 - 4900 = 4600$ , so $y = 1150$ .
1 mark
(ii)(b) Since $1150 > 980$ , the cricket prize is more, by $₹ (1150 - 980) = ₹170$ .
1 mark
(iii) For 2 students in each game:
Total $= 2 x + 2 y = 2 (980 + 1150) = 2 \times 2130 = ₹4260$
1 mark

Answer: (i) 5x + 4y = 9500, 4x + 3y = 7370; (ii)(a) ₹980; (ii)(b) Cricket, by ₹170; (iii) ₹4260

Where the marks are

Forms $5 x + 4 y = 9500$ (½ mark)
Forms $4 x + 3 y = 7370$ (½ mark)
(ii)(a) Solves to get hockey prize ₹980 (2 marks)
(ii)(b) Finds $x = 980$ , $y = 1150$ (1 mark)
(ii)(b) States the cricket prize is more, by ₹170 (1 mark)
(iii) Total $2 (x + y) = ₹4260$ (1 mark)

Revise the chapter

Chapter: Pair of Linear Equations in Two Variables (notes, formulas, where marks are lost and every past question)
Topic: Word Problems on Linear Equations

Practise this chapterBoard-style questions from Pair of Linear Equations in Two Variables, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/4/1 (Standard): 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 Standard 2023: question 36

Hints

Step-by-step solution

Revise the chapter

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