CBSE Class 10 Question Paper 2022 Solution Maths Basic

Strictly Confidential : (For Internal and Restricted use only)
Secondary School Examination
Term–II, 2022

Marking Scheme : MATHEMATICS (Basic) (Subject Code : 241)
[ Paper Code : 430/4/2 ]

General Instructions :
1. You are aware that evaluation is the most important process in the actual and correct assessment of the
candidates. A small mistake in evaluation may lead to serious problems which may affect the future of
the candidates, education system and teaching profession. To avoid mistakes, it is requested that before
starting evaluation, you must read and understand the spot evaluation guidelines carefully.

2. ‘‘Evaluation policy is a confidential policy as it is related to the confidentiality of the examinations
conducted, evaluation done and several other aspects. Its leakage to public in any manner could
lead to derailment of the examination system and affect the life and future of millions of
candidates. Sharing this policy/document to anyone, publishing in any magazine and printing in
Newspaper/ Website, etc., may invite action under IPC.’’

3. Evaluation is to be done as per instruction provided in the Marking Scheme. It should not be done
according to one’s own interpretation or any other consideration. Marking Scheme should be strictly
adhered to and religiously followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed for their correctness
otherwise and marks be awarded to them. In Class-X, while evaluating two competency based
questions, please try to understand given answer and even if reply is not from marking scheme
but correct competency is enumerated by the candidate, marks should be awarded.

4. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first
day, to ensure that evaluation has been carried out as per the instructions given in the Marking Scheme.
The remaining answer books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.

5. Evaluators will mark () wherever answer is correct. For wrong answer ‘’ be marked. Evaluators will
not put right kind of mark while evaluating which gives an impression that answer is correct and no
marks are awarded. This is most common mistake which evaluators are committing.

6. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for
different parts of the question should then be totalled up and written in the left-hand margin and
encircled. This may be followed strictly.

7. If a question does not have any parts, marks must be awarded in the left-hand margin and encircled.
This may also be followed strictly.

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8. If a student has attempted an extra question, answer of the question deserving more marks should be
retained and the other answer scored out.

9. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

10. A full scale of marks _____ (example 0–40 marks as given in Question Paper) has to be used. Please do
not hesitate to award full marks if the answer deserves it.

11. Every examiner has to necessarily do evaluation work for full working hours, i.e., 8 hours everyday and
evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects
(Details are given in Spot Guidelines).

12. Ensure that you do not make the following common types of errors committed by the Examiner in the
past :
 Leaving answer or part thereof unassessed in an answer book
 Giving more marks for an answer than assigned to it
 Wrong totalling of marks awarded on a reply
 Wrong transfer of marks from the inside pages of the answer book to the title page
 Wrong questionwise totalling on the title page
 Wrong totalling of marks of the two columns on the title page
 Wrong grand total
 Marks in words and figures not tallying
 Wrong transfer of marks from the answer book to online award list
 Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly
and clearly indicated. It should merely be a line. Same is with the  for incorrect answer).
 Half or a part of answer marked correct and the rest as wrong, but no marks awarded.

13. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as
() and awarded zero (0) Mark.

14. Any unassessed portion, non-carrying over of marks to the title page, or totalling error detected by the
candidates shall damage the prestige of all the personnel engaged in the evaluation work as also of the
Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the instructions
be followed meticulously and judiciously.

15. The examiners should acquaint themselves with the guidelines given in the guidelines for spot
evaluation before starting the actual evaluation.

16. Every examiner shall also ensure that all the answers are evaluated, marks carried over to the title page,
correctly totalled and written in figures and words.

17. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application
and also separately as a part of the re-evaluation process on payment of the processing charges.

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MARKING SCHEME
Secondary School Examination TERM–II, 2022
MATHEMATICS (Basic) (Subject Code–241)
[Paper Code : 430/4/2 ]
Q. No. EXPECTED ANSWER / VALUE POINTS Marks
SECTION—A
1(a) a1  3, a2  4, d  a2  a1  4  3  7 ½
1
a36  a  35d  3  35(7)
½
 242
OR
1(b) 27  3  3  3  3 3
48  4  4  3  4 3
75  5  5  3  5 3
d  4 3 3 3  3 1
 Next two terms are 6 3, 7 3 or 108, 147 .
½+ ½
2. Mode = 3 Median – 2 Mean
1 1
 Median  [Mode + 2 Mean]  [175  338]  171 1+1
3 3
3. Modal class is 20–40 1
Median class is 40–60 1
4(a) Volume of solid = volume of cone + volume of hemisphere
 1½
1 2 
  (1) (1)   (1)
2 3

3 3
  cm3 ½
OR
4(b)
22 2 ½
3  r  462
7
½
r 7 m
2 22 3
V  7 ½
3 7
2156 3
m or 718.67m 3 ½
3

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5. P A Q
AOB is a diameter 1
 Distance between parallel tangents = Diameter = 13 cm O ½
13
 Radius   6  5 cm ½
2 R B S
6. The given equation x2  ax  1  0 will have two equal roots, if
(a)2  4(1)(1)  0 1

 a2  4
 a  2 1

SECTION—B
7. * Drawing a line segment AB of length 6 cm and making acute angle(s) 1
4
* Locating the point X on the line segment AB such that AX= AB 2
5
8. a  (n 1)d  252
12  (n  1)6  252 ½
(n 1)6  240  n 1  40
n  41
1
 Middle term = 21st term
½
a21  a  20d
 12  20(6)  132 1
9. D Correct
10 m Figure:
B 30° 1
E

25 m 25 m

A C
DE
In right BDE, sin 30 
BD
1 10
   BD  20 1
2 BD
 Length of wire = 20 m
10 1
 tan 30   BE  10 3 m
BE 3 1

Distance between the poles = 10 3m

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10(a) x  3 is one root of the quadratic equation
 2(3)2  p (3)  30  0  18  3 p  30  0 1

p  16 ½

 2x2  16x  30  0
1
 x 2  8x  15  0  (x  5)(x  3)  0
x  5, 3  Other Root = 5 ½

OR
10(b) Let Breadth of the park = x m

 Length = (2x + 5) m ½
Area  250  (2x  5)x  250 ½
2x2  5x  250  2x 2  5x  250  0
1
(x 10)(2x  25)  0
 x  10  0 , 2x  25  0
25
x  10 x (Not possible)
2
 Length  2x  5  25 m and Breadth  x  10 m 1

SECTION—C
D R C Correct
11(a)
5 4 figure
6 3 1
S Q
7 O 2
8 1

A P B
 OBP   OBQ (sss)
 1  2 cpct
1
Similarly 3  4, 5  6 and 7  8
As 1  2  3  4  5  6  7  8  360
 28  21  24  25  360  8  1  4  5  180 1
 AOB  COD  180 similarly BOC  AOD  180 1
OR

11(b)

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We know that tangents drawn from the external point to the circle are equal
 BP  BQ 
 1½
CP  CR 
AQ  AR 
2AQ  AQ  AR
 (AB  BQ)  (AC  CR) 1
½
 AB  BP  AC  CP
 AB  AC  BP  CP
 AB  AC  BC
1 1
AQ  ( AB  AC  BC)
2
12. X Correct
B
30° Figure:
45° 1
100 m
30° 45°
D C A
100
In right ABC , tan 45 
AC
1
AC  100 m
100 1
In right ABD , tan 30   AD  100 3 m
AD
 CD = Distance between ships  AD  AC

CD  100 3 100  73.2m 1
Note: If 3 = 1.732 is not used in calculation of the distance CD, ½ mark to be
deducted.

Daily Distance x Number of cf fx
13. (in km) Buses
(f )
100–120 110 12 12 1320

120–140 130 14 26 1820

140–160 150 8 34 1200

160–180 170 6 40 1020

180–200 190 10 50 1900

50 7260

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N 50
(a)   25
2 2
 Median Class  120  140 ½
N 
 C
Median  l  
2   h  120  25  12  20 1
f 14
130
Median  120   120  18  57  138  57 km
7 ½
 fx 7260
(b) Mean    145  2 km
f 50 1 for
 fx &
1 for Mean

14. r  21 cm
h  28 cm
28 cm

(a) Volume of 1 toy = volume of hemisphere + volume of cone ½
21 cm
2 3 1 2
 r  r h
3 3
1
 r 2 [2r  h]
3
1 22 1
   21 21[2  21  28]
3 7
= 22  21 70  32340 cm3
Volume of 50 toys  50  32340
½
 1617000 cm3
2 3
r
Volume of hemishere 3
(b)  1
Volume of cone 1 2
r h
3
3
 1
2
Required ratio is 3:2

***

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