CBSE Class 10 Question Paper 2023 Solution Maths Standard

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2023
MATHEMATICS PAPER CODE 30/1/1

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
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3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done
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These are in the nature of Guidelines only and do not constitute the complete answer. The students
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Scheme. If there is any variation, the same should be zero after deliberation and discussion. 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.
6 Evaluators will mark ( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be marked.
Evaluators will not put right (✓) while evaluating which gives an impression that answer is correct
and no marks are awarded. This is most common mistake which evaluators are committing.
7 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 totaled up and written in the left-hand margin and
encircled. This may be followed strictly.
8 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.
9 In Q1-Q20, if a candidate attempts the question more than once (without canceling the previous
attempt), marks shall be awarded for the first attempt only and the other answer scored out
with a note “Extra Question”.
10 In Q21-Q38, if a student has attempted an extra question, answer of the question deserving
more marks should be retained and the other answer scored out with a note “Extra Question”.
11 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
12 A full scale of marks __________(example 0 to 80/70/60/50/40/30 marks as given in Question Paper)
has to be used. Please do not hesitate to award full marks if the answer deserves it.

1

13 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day
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). This is in view of the reduced syllabus and number of
questions in question paper.
14 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 totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● 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 X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
15 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked
as cross (X) and awarded zero (0)Marks.
16 Any un assessed portion, non-carrying over of marks to the title page, or totaling error detected by
the candidate 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.
17 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot
Evaluation” before starting the actual evaluation.
18 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totaled and written in figures and words.
19 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once
again reminded that they must ensure that evaluation is carried out strictly as per value points for
each answer as given in the Marking Scheme.

2

MARKING SCHEME
MATHEMATICS (Subject Code–041)
(PAPER CODE: 30/1/1)

Q. No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION A
Questions no. 1 to 18 are multiple choice questions (MCQs) and questions
number 19 and 20 are Assertion-Reason based questions of 1 mark each

The graph of y = p ( x ) is given, for a polynomial p ( x) . The number of
1.
zeroes of p(x) from the graph is

(A) 3 (B) 1 (C) 2 (D) 0

Sol. (B) 1 1
The value of k for which the pair of equations kx = y + 2 and 6 x = 2 y + 3
2.
has infinitely many solutions,

(A) is k = 3 (B) does not exist (C) is k = −3 (D) is k = 4

Sol. (B) does not exist 1
3. If p − 1, p + 1 and 2p + 3 are in A.P., then the value of p is
(A) − 2 (B) 4 (C) 0 (D) 2

Sol. (C) 0 1
In what ratio, does x-axis divide the line segment joining the points
4.
A(3, 6) and b(−12, −3) ?
(A) 1: 2 (B) 1: 4 (C) 4 :1 (D) 2 :1

3

Sol. (D) 2 : 1 1
In the given figure, PQ is tangent to the circle centred at O.
5.
If AOB = 95 , then the measure of ABQ will be

(A) 47.5 (B) 42.5 (C) 85 (D) 95

Sol. (A) 47.5 1
4sin A + 3cos A
6. If 2 tan A = 3 , then the value of is
4sin A − 3cos A
7 1
(A) (B) (C) 3 (D) does not exist
13 13

Sol. (C) 3 1
1 1
7. If  ,  are the zeroes of a polynomial p( x) = x 2 + x − 1 , then 𝛼 +𝛽 equals to

−1
(A) 1 (B) 2 (C) − 1 (D)
2

Sol. (A) 1 1
The least positive value of k , for which the quadratic equation
8.
2 x 2 + kx − 4 = 0 has rational roots, is
(A) 2 2 (B) 2 (C) 2 (D) 2
Sol. (B) 2 1
3 2 0 0
 4 tan 30 − sec 45 + sin 60  is equal to
9. 2 0 2

5 −3 1
(A) −1 (B) (C) (D)
6 2 6
Sol. (A) – 1 1

4

Curved surface area of a cylinder of height 5 cm is 94.2 cm 2 . Radius of the
10.
cylinder is (Take  = 3.14)
(A) 2 cm (B) 3 cm (C) 2.9 cm (D) 6 cm

Sol. (B) 3 1
The distribution below gives the marks obtained by 80 students on a test :
11.
Marks Less Less Less Less Less Less
than 10 than than than 40 than than
20 30 50 60
Number of 3 12 27 57 75 80
Students
The modal class of this distribution is :
(A) 10-20 (B) 20-30
(C) 30-40 (D) 50-60
Sol. (C) 30 – 40 1
The curved surface area of a cone having height 24 cm and radius 7 cm, is
12.
(A) 528 cm 2 (B) 1056 cm 2 (C) 550 cm 2 (D) 500 cm 2

Sol. (C)550cm2 1

13. The distance between the points (0,2√5) and (−2√5, 0) is
(A) 2√10 units (B) 4√10units (C) 2√20 units (D) 0
Sol. (A) 2√10 units 1
−2 2
14. Which of the following is a quadratic polynomial having zeroes and ?
3 3
4 9
(A) 4 x 2 − 9 (D) 5(9 x − 4)
2
(B) (9 x 2 + 4) (C) x 2 +
9 4
Sol. (D) 5 (9x2 – 4) 1
If the value of each observation of a statistical data is increased by 3, then the
15.
mean of the data
(A) remains unchanged (B) increases by 3
(C) increases by 6 (D) increases by 3n

Sol. (B) increases by 3 1
5

Probability of happening of an event is denoted by p and probability of non-
16.
happening of the event is denoted by q . Relation between p and q is
(A) p + q = 1 (B) p = 1, q = 1
(C) p = q − 1 (D) p + q + 1 = 0

Sol. (A) p + q = 1 1
A girl calculates that the probability of her winning the first prize in a lottery
17.
is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40 (B) 240 (C) 480 (D) 750
Sol. (C) 480 1

In a group of 20 people, 5 can't swim. If one person is selected at random,
18.
then the probability that he/she can swim, is
3 1 1
(A) (B) (C) 1 (D)
4 3 4
Sol. (A) 3/4 1

19. Assertion (A): Point P ( 0, 2 ) is the point of intersection of y-axis with the
line 3 x + 2 y = 4 .
Reason (R): The distance of point P ( 0, 2 ) from x-axis is 2 units.

(B) Both Assertion (A) and Reason ( R) are correct but Reason ( R) is
Sol. 1
not the correct explanation of Assertion (A)

6

Assertion (A): The perimeter of ABC is a rational number.
20.
Reason (R): The sum of the squares of two rational numbers is always
rational.

Sol. (D) Assertion (A) is false but Reason (R) is true 1

SECTION B
This section comprises of Very Short Answer (VSA) type questions of 2
marks each.
Solve the pair of equations x = 3 and y = −4 graphically.
21(a).
Sol. Correct graph of both the equations. 1
Solution of equation is x = 3, y = – 4 1

OR
Using graphical method, find whether following system of linear equations
21(b).
is consistent or not:
x = 0 and y = −7

Sol. Correct graph of y = – 7 and x = 0 1
As y = – 7 is intersecting x = 0 at (0, – 7)
So, system of equations is consistent 1
In the given figure, XZ is parallel to BC. AZ = 3 cm, ZC = 2 cm,
22.
BM =3 cm and MC = 5 cm. Find the length of XY.

7

AX 3 AZ ½
Sol. As XZ || BC Therefore = = – (i)
XB 2 ZC
 AXY   ABM ½

𝐴𝑋 𝑋𝑌
⇒ 𝐴𝐵 = 𝐵𝑀 or 5 = 3
3 𝑋𝑌 ½

9
½
 𝑋𝑌 = 5 or 1·8 cm

23(a). If sin + cos = 3 , then find the value of sin  cos .

Sol. sin  + cos  = 3
squaring both sides
sin2  + cos2  + 2 sin cos  = 3 1
 1 + 2 sin  cos  = 3 ½
½
 sin  cos  = 1

OR
1
23(b). If sin  = and cot  = 3 , then find the value of cosecα+ cosecβ
2

8

Sol. 1 ½
cosec  = = 2
sin 
1
cosec  = 1 + cot 2  = 1+3 =2

½
 cosec  + cosec  = 2 + 2 or 2 ( 2 + 1)

Find the greatest number which divides 85 and 72 leaving remainders 1 and
24.
2 respectively.

We have to find HCF of 85 – 1 = 84 and 72 – 2 = 70.
Sol. 1
HCF of 84 and 70 = 14 1

A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random
25.
from the bag. Find the probability that drawn ball is
(i) red (ii) yellow.
Total No of Balls=9
Sol.
4 1
(i) P(drawn ball is red) =
9
2 1
(ii) P(drawn ball is yellow) =
9

SECTION C
This section comprises of Short Answer (SA) type questions of 3 marks
each.
Half of the difference between two numbers is 2. The sum of the greater
26.
number and twice the smaller number is 13. Find the numbers.
Let the numbers be x and y, x > y
Sol.
1
Therefore (x – y) = 2 — (i) 1
2
and 2y + x = 13 — (ii) 1
Solving equations (i) and (ii)
x = 7, y = 3 1

9

27. Prove that 5 is an irrational number.
Sol. Let √𝟓 be a rational number.
𝐩 ½
∴ √𝟓 = 𝐪 , where q≠0 and let p & q be co-primes.

5q2 = p2 ⟹ p2 is divisible by 5 ⟹ p is divisible by 5
1
⟹ p = 5a, where ‘a’ is some integer ----- (i)
25a2 = 5q2 ⟹ q2 = 5a2 ⟹q2 is divisible by 5 ⟹ q is divisible by 5
½
⟹ q = 5b, where ‘b’ is some integer ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are co-primes.
1
∴ √𝟓 is an irrational number.
If ( −5,3) and (5,3) are two vertices of an equilateral triangle, then find
28.
coordinates of the third vertex, given that origin lies inside the triangle. (Take
3 = 1.7)

Let the third vertex be (x,y)
Sol.
A (-5,3) B(5,3) C(x,y)
AB=10=AC 1
AC2=100
(-5-x)2+(3-y)2 = (5-x)2+(3-y)2
1
20x =0

x=0 ½

(3-y)2=75
3-y = ±5√3
y=3-5√3
y= -5.5
½
The coordinates of the third vertex are (0,-5.5)

Two tangents TP and TQ are drawn to a circle with centre O from an external
29(a).
point T. Prove that PTQ = 2OPQ .

10

TP = TQ
Sol.
⇒  TPQ =  TQP 1

Let  PTQ be 
180° – 𝜃 𝜃
⇒  TPQ =  TQP = = 90 – 2 1
2

Now  OPT = 90
𝜃 𝜃
⇒ OPQ = 90 – (90 – 2 ) = 2
1
 PTQ = 2  OPQ

OR
In the given figure, a circle is inscribed in a quadrilateral ABCD in which
29(b).
B = 900 . If AD 17 cm, AB = 20 cm and DS = 3 cm, then find the radius of
the circle.

11

Sol.

DR = DS = 3 cm ½

 AR = AD – DR = 17 – 3 = 14 cm 1
 AQ = AR = 14 cm ½

 QB = AB – AQ = 20 – 14 = 6 cm ½

Since QB = OP = r  radius = 6 cm ½

tan  + sec  − 1 1 + sin 
30. Prove that: =
tan  − sec  +  cos 

(𝑡𝑎𝑛 𝜃 + 𝑠𝑒𝑐 𝜃) – (𝑠𝑒𝑐 2 𝜃 – 𝑡𝑎𝑛 2 𝜃)
Sol. LHS= 1
𝑡𝑎𝑛 𝜃 – 𝑠𝑒𝑐 𝜃 + 1
(𝑡𝑎𝑛 𝜃 + 𝑠𝑒𝑐 𝜃) (1 – 𝑠𝑒𝑐 𝜃 + 𝑡𝑎𝑛 𝜃) 1
= 𝑡𝑎𝑛 𝜃 – 𝑠𝑒𝑐 𝜃 + 1

= 𝑡𝑎𝑛𝜃 + 𝑠𝑒𝑐𝜃 ½
1 + 𝑠𝑖𝑛 𝜃
= = RHS ½
𝑐𝑜𝑠 𝜃

A room is in the form of cylinder surmounted by a hemi-spherical dome.
31(a).
The base radius of hemisphere is one-half the height of cylindrical part.
 1408  3
Find total height of the room if it contains   m of air. Take
 21 
 22 
 = 
 7 

12

Sol. Let h be height of cylindrical part and r be radius of hemisphere ½
2 1408 1
Volume of room = 2 r 3 +  r 3 =
3 21

r = 2 ½
Therefore, h=4 ½
Height of the room is = 6m ½

OR
An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled
31(b).
th
1
in it so that lower part of the cone which is   of the volume of the
6
cone is unfilled but hemisphere is formed on the top. Find volume of the
ice-cream. (Take  = 3.14)

Sol. 1 1
Volume of the cone = =    9 12 = 36 cm3
3
5
Volume of ice-cream in the cone = 6 × 36 × 𝜋 = 30𝜋𝑐𝑚3 ½
2 1
Volume of ice-cream on top = 3 × 27 × 𝜋 = 18𝜋𝑐𝑚3

Total volume of the ice-cream = (30 + 18 ) = 48 cm3

= 48  3.14 = 150.72cm3 ½

SECTION D
This section comprises of Long Answer (LA) type questions of 5 marks
each.

13

If a line is drawn parallel to one side of a triangle to intersect the other
32.
two sides at distinct points, prove that the other two sides are divided in
the same ratio.

Sol. Correct Given, to prove, figure, construction 2
Correct proof 3
33(a). The angle of elevation of the top of a tower 24 m high from the foot of
another tower in the same plane is 600. The angle of elevation of the top
of second tower from the foot of the first tower is 300. Find the distance
between two towers and the height of the other tower. Also, find the
length of the wire attached to the tops of both the towers.

Sol.

1 mark
for
correct
figure

Let AB and CD be the given towers.
1 h
tan 30 = =  x = h 3 __ (i) 1
3 x
24 24
tan 60 = 3 = x= or 8√3 __ (ii) 1
x 3
using (i) and (ii)
1 1
x = 8 3 and h = 8 +
2 2

length of wire = BE 2 + x 2 = 256 + 192 = 448 m = 8 7 m 1

14

OR
A spherical balloon of radius r subtends an angle of 600 at the eye of an
33(b).
observer. If the angle of elevation of its centre is 450 from the same point,
then prove that height of the centre of the balloon is 2 times its radius.

Sol.

1 mark
for
correct
figure

Let Point B represents observer.
  QBP = 60;  ABO = 45
1
Using geometry  PBO =  60 = 30 1
2
r 1
Now, = sin 30 =  OB = 2r — (i)
OB 2 1

OA 1
Also = sin 45 =  OB = OA 2 (ii) 1
OB 2

Using (i) and (ii) OA = 2 r 1
or height of center of balloon = 2 r units

A chord of a circle of radius 14 cm subtends an angle of 600 at the centre.
34.
Find the area of the corresponding minor segment of the circle. Also find
the area of the major segment of the circle.

Sol. 22 60 1 3 1+1
Area of minor segment = 14 14  − 14 14 
7 360 2 2
 308 
= − 49 3  cm2 or 17.9cm2 1
 3 

15

22 308
Area of major segment = 7 × 14 × 14 − ( 3 − 49√3)
308 1
= 616 − + 49 3
3
 1540 
= + 49 3  cm2 or 598.1cm2 1
 3 
th th
The ratio of the 11 term to 17 term of an A.P. is 3:4. Find the ratio of
35(a).
5th term to 21st term of the same A.P. Also, find the ratio of the sum of
first 5 terms to that of first 21 terms.

Sol. a + 10 d 3 1
Given =
a + 16 d 4
 4a + 40d = 3a + 48d
1
 a = 8d (i)
a a + 4d 3
therefore 5 = =7 using( i)
a 21 a + 20 d 1

a5 : a21 = 3 : 7
5
s5 (2a + 4 d ) 5  20 d 25
= 2 = = 2
s 21 21
(2a + 20 d ) 21  36 d 189
2
Therefore, S5:S21=25:189
OR
250 logs are stacked in the following manner:
35(b).
22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so
on (as shown by an example). In how many rows, are the 250 logs placed and
how many logs are there in the top row?

Sol.
Let the number of rows be n.
A.P. formed is 22, 21, 20, 19, .........
16

Here a = 22, d = – 1 Sn = 250 1
n
 250 = [44 + (n – 1) (– 1)] 1
2
1
 n2 – 45n + 500 = 0
 (n – 25) (n – 20) = 0
n  25  n = 20 1
logs in top row = a20 = 22 + 19 (– 1) = 3 1

SECTION E
This section comprises of 3 case-study based questions of 4 marks each.
While designing the school year book, a teacher asked the student that the
36.
length and width of a particular photo is increased by x units each to double
the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based on the above information, answer the following questions:
(I) Write an algebraic equation depicting the above information.
(II) Write the corresponding quadratic equation in standard form.
(III) What should be the new dimensions of the enlarged photo?

OR
Can any rational value of x make the new area equal to 220cm 2

Sol. (i) (18 + x) (12 + x) = 2(18 ×12) 1
(ii) x2 + 30x – 216 = 0
1
(iii) Solving : x2 + 30x – 216 = 0
 (x + 36) (x – 6) = 0
x  -36   x = 6. 1
1
new dimensions are 24 cm  18 cm

17

OR
(iii) If (18 + x) (12 + x) = 220
1
then x2 + 30x – 4 = 0
1
Here D = 900 + 16 = 916 which is not a perfect square.

Thus we can’t have any such rational value of x.

India meteorological department observes seasonal and annual rainfall
37.
every year in different sub-divisions of our country.

It helps them to compare and analyse the results. The table given below shows
sub-division wise seasonal (monsoon) rainfall (mm) in 2018:
Rainfall (mm) Number of Sub-divisions
200-400 2
400-600 4
600-800 7
800-1000 4
1000-1200 2
1200-1400 3
1400 -1600 1
1600-1800 1
Based on the above information, answer the following questions:
(I) Write the modal class.
(II) Find the median of the given data.
18

OR
(II) Find the mean rainfall in this season.
(III) If sub-division having at least 1000 mm rainfall during monsoon season,
is considered good rainfall sub-division, then how many sub-divisions
had good rainfall?

Sol. (i) Modal Class is 600-800 1

𝑁
(ii) 2 = 12, median class is 600 – 800 ½

Rainfall xi fi cf.
200 – 400 300 2 2
400 – 600 500 4 6
600 – 800 700 7 13
800 – 1000 900 4 17
1000 – 1200 1100 2 19
1200 – 1400 1300 3 22
½ for
1400 – 1600 1500 1 23 correct
table
1600 – 1800 1700 1 24
24

200
Median = 600 + (12 – 6)
7
5400
= 7 or 771·4 1

OR
(ii)
Rainfall xi fi f ixi
200 – 400 300 2 600
400 – 600 500 4 2000

19

600 – 800 700 7 4900
800 – 1000 900 4 3600
1000 – 1200 1100 2 2200 1 for
correct
1200 – 1400 1300 3 3900 table
1400 – 1600 1500 1 1500
1600 – 1800 1700 1 1700
24 20400

20400 1
Mean = = 850
24

1
(iii) Sub-divisions having good rainfall = 2 + 3 + 1 + 1 = 7.

The discus throw is an event in which an athlete attempts to
38.
throw a discus. The athlete spins anti-clockwise around one and
a half times through a circle, then releases the throw. When released,
the discus travels along tangent to the circular spin orbit.

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point
O is centre of the circle and ABO = 300 . PQ is parallel to OA.

20

Based on above information:
(a) find the length of AB.
(b) find the length of OB.
(c) find the length of AP.
OR
Find the length of PQ

Sol. 1 75 1
(i)tan 30 = =
3 AB 2
 AB = 75 3 cm 1
2
1 75 1
(ii)sin 30 = 2 = 𝑂𝐵
2
1
 OB = 150 cm
2

(iii) QB = 150 – 75 = 75 cm 1
 Q is mid point. of OB

Since PQ ll AO therefore P is mid point of AB
75 3 1
Hence AP = cm.
2
OR
1
(iii) QB = 150 – 75 = 75 cm
2
Now,  BQP   BOA

𝑄𝐵 𝑃𝑄
 𝑂𝐵 = 𝑂𝐴
1 𝑃𝑄
 2 = 75 1
75 1
 PQ = 2 cm
2

21

22