9TH SLIP TEST 1 PRINCIPLE OF EVALUATION

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PRINCIPLE OF EVALUATION / VALUATION KEY

Class: IX
Subject: Mathematics
Slip Test - 1
Max Marks: 20

General Valuation Guidelines:

  • Award step-marks strictly according to the breakdown indicated below.
  • Alternative mathematically correct methods should be given full credit.
  • Do not deduct marks for minor calculation mistakes if the logical procedure is entirely correct (deduct max ½ mark for computational error).

SECTION I (Max Marks: 6)

Q1. Express 0.36 in
p
q
form [ Marks: 6 ]
Step Description / Marking Points Marks Breakdown
Let x = 0.3666... — (Eq 1) 1 Mark
Multiply Eq (1) by 10 (since 1 non-repeating digit): 10x = 3.666... — (Eq 2) 1 Mark
Multiply Eq (2) by 10 (periodicity = 1): 100x = 36.666... — (Eq 3) 1.5 Marks
Subtract Eq (2) from Eq (3): 100x − 10x = 36.666... − 3.666... ⇨ 90x = 33 1.5 Marks
Simplifying to
x =
33
90
 = 
11
30
(Final Answer) 		1 Mark
Q2 (OR). Factorise 4x2 + 9y2 + 25z2 − 12xy − 30yz + 20zx [ Marks: 6 ]
Step Description / Marking Points Marks Breakdown
Identifying identity: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 1.5 Marks
Rewriting terms with proper signs: (−2x)2 + (3y)2 + (5z)2 + 2(−2x)(3y) + 2(3y)(5z) + 2(5z)(−2x)
(OR equivalent sign combination: (2x)2 + (-3y)2 + (-5z)2)		 2.5 Marks
Expressing as square: (−2x + 3y + 5z)2 or (2x − 3y − 5z)2 1 Mark
Final factorised form: (−2x + 3y + 5z)(−2x + 3y + 5z) 1 Mark

SECTION II (Max Marks: 8)

Q3. Visualise 6.27 (6.277) on number line up to 3 decimal places [ Marks: 4 ]
Step Description / Marking Points Marks Breakdown
Locating 6.277 between 6 and 7 & drawing Step 1 (magnifying 6 to 7) 1 Mark
Locating 6.277 between 6.2 and 6.3 & drawing Step 2 (magnifying 6.2 to 6.3) 1 Mark
Locating 6.277 between 6.27 and 6.28 & drawing Step 3 (magnifying 6.27 to 6.28) 1 Mark
Accurately pointing and labeling 6.277 on the final magnified number line 1 Mark
Q4. Find a and b in
√3 + √2
√3 − √2
 =  a + b√6
[ Marks: 4 ]
Step Description / Marking Points Marks Breakdown
Multiplying numerator and denominator by rationalising factor (√3 + √2) 1 Mark
Expanding numerator: (√3 + √2)2 = 3 + 2 + 2√6 = 5 + 2√6 1 Mark
Expanding denominator: (√3)2 − (√2)2 = 3 − 2 = 1 1 Mark
Comparing 5 + 2√6 with a + b√6 ⇨ a = 5, b = 2 1 Mark

SECTION III (Max Marks: 6)

Q5. If p(x) = x2 − 3x + 2, find p(−1), p(0), p(1) and p(2) [ Marks: 2 ]
Step Description / Marking Points Marks Breakdown
p(−1) = (−1)2 − 3(−1) + 2 = 1 + 3 + 2 = 6 ½ Mark
p(0) = 02 − 3(0) + 2 = 2 ½ Mark
p(1) = (1)2 − 3(1) + 2 = 1 − 3 + 2 = 0 ½ Mark
p(2) = (2)2 − 3(2) + 2 = 4 − 6 + 2 = 0 ½ Mark
Q6. Locate √10 on the number line [ Marks: 2 ]
Step Description / Marking Points Marks Breakdown
Splitting using Pythagoras theorem: 10 = 32 + 12 ⇨ Base = 3 units, Height = 1 unit 1 Mark
Constructing right-angled triangle on number line & drawing arc of radius √10 to mark point on line 1 Mark
Q7. Multiply (3 − √2) with (√2 + 3) [ Marks: 2 ]
Step Description / Marking Points Marks Breakdown
Rearranging terms into (3 − √2)(3 + √2) and applying identity (a − b)(a + b) = a2 − b2 1 Mark
Calculation: 32 − (√2)2 = 9 − 2 = 7 1 Mark
Source: eTeacherDesk

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