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Q.

Suppose a consumer has total budget ₹ 20 and cost of both the goods (x1, x2) is ₹ 5, then arrange the following available bundle sets on suitable place in the table below:

(4, 3) (2, 2) (3, 1) (4, 0) (3, 3) (4, 5) (2, 3) (1, 3)

Within budget limit
Beyond budget limit

OR

In table given below write the appropriate answer in the blank space a, b, c, d, e, f, g and h respectively.

Sl. No.% change in the price of item% change in demand for the itemImpact on expenditureThe nature of (ed) elasticity of demand
1+10− 08(a)e < 1
2+10− 12Decrease(b)
3+10− 10(c)e = 1
4− 10+ 15(d)e > 1
5− 10+ 07Increase(e)
6− 10+ 10Unchanged(f)
7+10− 06Increase(g)
8− 10+ 18(h)e > 1
Rajasthan RbseRBSE Rajasthan Senior Secondary (Class-12) Commerce Board 2026Subjective· 4mImportance★★★★★
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Budget line: 5x1 + 5x2 = 20 ⇒ x1 + x2 = 4. Affordable bundles have x1 + x2 ≤ 4. OR: complete the elasticity table a=Increase, b=e>1, c=Unchanged, d=Increase, e=e<1, f=e=1, g=e<1, h=Increase.

Main question — sorting the bundles:

Total budget = ₹20 and the price of each good is ₹5, so the budget constraint is 5x1 + 5x2 ≤ 20, i.e. x1 + x2 ≤ 4. A bundle is affordable (within the budget) if its two quantities add up to 4 or less, and it lies beyond the budget if they add up to more than 4.

Bundle (x1, x2)x1 + x2Cost (₹5 each)Position
(4, 3)735Beyond budget
(2, 2)420Within budget (on the line)
(3, 1)420Within budget (on the line)
(4, 0)420Within budget (on the line)
(3, 3)630Beyond budget
(4, 5)945Beyond budget
(2, 3)525Beyond budget
(1, 3)420Within budget (on the line)

Result:

Within budget limit(2, 2)(3, 1)(4, 0)(1, 3)
Beyond budget limit(4, 3)(3, 3)(4, 5)(2, 3)

OR alternative — completing the elasticity table:

Rule used: (i) elasticity ed = |% change in demand| ÷ |% change in price|; (ii) effect on consumer expenditure (P × Q): when price rises, expenditure increases if demand is inelastic (e<1), is unchanged if unitary (e=1), and decreases if elastic (e>1) — and the reverse when price falls.

Sl.%ΔP%ΔQed valueImpact on expenditureNature
1+10−80.8(a) Increasee < 1
2+10−121.2Decrease(b) e > 1
3+10−101.0(c) Unchangede = 1
4−10+151.5(d) Increasee > 1
5−10+70.7Increase (given)(e) e < 1
6−10+101.0Unchanged(f) e = 1
7+10−60.6Increase(g) e < 1
8−10+181.8(h) Increasee > 1
…

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