10TH SLIP TEST 1 PRINCIPLE OF EVALUATION

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PRINCIPLE OF EVALUATION (KEY PAPER)

Class: X
Subject: Mathematics
Max Marks: 20
Q1. Prove √2 − √5 is irrational 6 Marks (AS 2: Reasoning & Proof)
Step / Expected Solution Marks Allocation
Let us assume to the contrary that √2 − √5 is rational.
∴ √2 − √5 = a/b, where a, b ∈ ℤ, b ≠ 0, and gcd(a, b) = 1.		 1 M
Rearranging terms: √2 − (a/b) = √5 1 M
Squaring on both sides:
(√2 − a/b)2 = (√5)2
2 + (a2/b2) − 2√2(a/b) = 5		 2 M
Simplifying for √2:
2√2(a/b) = (a2/b2) − 3 ⇒ √2 = (a2 − 3b2) / (2ab)		 1 M
Since a and b are integers, (a2 − 3b2) / (2ab) is rational, which implies √2 is rational.
This contradicts the fact that √2 is irrational. Hence √2 − √5 is irrational.		 1 M
Total for Q1 (Option A) 6 M
Q2. [OR] Verify n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 6 Marks (AS 2: Reasoning & Proof)
Step / Expected Solution Marks Allocation
Writing given sets in roster form:
A = {2, 3, 5, 7, 11, 13, 17, 19} ⇒ n(A) = 8
B = {1, 2, 3, 6} ⇒ n(B) = 4		 2 M
Finding A ∪ B = {1, 2, 3, 5, 6, 7, 11, 13, 17, 19} ⇒ n(A ∪ B) = 10 1.5 M
Finding A ∩ B = {2, 3} ⇒ n(A ∩ B) = 2 1 M
Substituting values into RHS:
RHS = n(A) + n(B) − n(A ∩ B) = 8 + 4 − 2 = 10
LHS = n(A ∪ B) = 10
LHS = RHS. Hence verified.		 1.5 M
Total for Q2 (Option B) 6 M
Q3. Set Operations from Venn Diagram 4 Marks (AS 5: Visualization & Representation)
Step / Expected Solution Marks Allocation
i) X ∪ Y = {C, I, L, P, E, N, A, U, T} 1 M
ii) X ∩ Y = {P, E, N} 1 M
iii) X − Y = {C, I, L} 1 M
iv) Y − X = {A, U, T} 1 M
Total for Q3 4 M
Q4. Median Formula & Explanation of Terms 4 Marks (AS 3: Communication)
Step / Expected Solution Marks Allocation
Formula:

Median  =  l  +  [
n
2
 −  cf
f
]  ×  h
1.5 M
Explanation of Terms:
• l = lower limit of median class
• n = number of observations (∑f)
• cf = cumulative frequency of class preceding the median class
• f = frequency of median class
• h = class size / width of median class 		2.5 M
Total for Q4 4 M
Q5. HCF of 24 and 33 using Euclid Division Lemma 2 Marks (AS 1: Problem Solving)
Step / Expected Solution Marks Allocation
Applying Euclid Division Lemma (a = bq + r):
33 = 24 × 1 + 9
24 = 9 × 2 + 6
9 = 6 × 1 + 3
6 = 3 × 2 + 0		 1.5 M
Remainder is 0. Divisor at this stage is 3.
∴ HCF(24, 33) = 3		 0.5 M
Total for Q5 2 M
Q6. Roster Form of Set A 2 Marks (AS 3: Communication)
Step / Expected Solution Marks Allocation
Identifying multiples of 3 less than 15: 3, 6, 9, 12 1 M
Writing in Roster Form: A = {3, 6, 9, 12} 1 M
Total for Q6 2 M
Q7. Prime Factorization of 480 2 Marks (AS 1: Problem Solving)
Step / Expected Solution Marks Allocation
Prime factorizing 480:
480 = 2 × 2 × 2 × 2 × 2 × 3 × 5		 1 M
Expressing in exponential form: 480 = 25 × 31 × 51 1 M
Total for Q7 2 M
Source: eTeacherDesk

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