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Class- X Exam – 2023-24
Mathematics – Standard
Time Allowed: 3 Hours Maximum Marks : 80
General Instructions :
1. This Question Paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each
3. Section B has 5 questions carrying 02 marks each.
4. Section C has 6 questions carrying 03 marks each.
5. Section D has 4 questions carrying 05 marks each.
6. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided.
8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated.
Section – A
Section A consists of 20 questions of 1 mark each.
1. Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is
(a) -1
(b) -8
(c) 7
(d) -9
View AnswerAns. (c) 7
4th term of first AP
a4 = -1 + (4 – 1)d = -1 + 3d
4th term of second AP
a4’ = -8 + (4 – 1)d = -8 + 3d
Now the difference between their 4th terms
a4’ – a4 = (-1 + 3d) – (-8 + 3d)
= -1 + 3d + 8 – 3d = 7
2. For the following distribution:
The modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60
View AnswerAns. (c) 30 – 40
Class 30 – 40 has the maximum frequency 30, therefore this is modal class.
3. If one zero of the polynomial (3x2 + 8x + k) is the reciprocal of the other, then value of k is
(a) 3
(b) -3
(c) 1/3
(d) 3
View AnswerAns. (a) 3
Let the zeroes be α and 1/ α
Product of zeroes, α x 1/ α = constant/coefficient of x2
1 = k/3 => k = 3
4. (x2 + 1)2 – x2 = 0 has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root
View AnswerAns. (c) no real roots
(x2 + 1)2 – x2 = 0
X4 + 1 + 2x2 – x2 = 0
X4 + x2 + 1 = 0
(x2)2 + x2 + 1 = 0
Let x2 = y
y2 + y + 1 = 0
Comparing with ax2 + by + c = 0, we get a = 1, b = 1 and c = 1
Discriminant D = b2 – 4ac
= (1)2 – 4(1)(1) = -3
Since D < 0, equation has no real roots
5. The LCM of smallest two-digit composite number and smallest composite number is
(a) 12
(b) 4
(c) 20
(d) 44
View AnswerAns. (c) 20
6. Which of the following statement is false?
(a) All isosceles triangles are similar.
(b) All quadrilateral is similar.
(c) All circles are similar.
(d) None of the above
View AnswerAns. (a) All isosceles triangles are similar
Isosceles triangle is a triangle in which two side of equal length. Thus two isosceles triangles may not be similar. Hence statement given in option (a) is false
Thus (a) is correct option.
7. If a regular hexagon is inscribed in a circle of radius r, then its perimeter is
(a) 3r
(b) 6r
(c) 9r
(d) 12r
View AnswerAns. (b) 6r
Side of the regular hexagon inscribed in a circle of radius r is also r, the perimeter is 6r
8. If 4 tan θ = 3, then (4 sin θ – cos θ) /(4 sin θ + cos θ) is equal to
(a) 2/3
(b) 1/3
(c) 1/2
(d) 3/4
View AnswerAns. (c) ½
Given 4 tan θ= 3
tan θ=3/4
(4sin θ – cos θ)/(4sin θ + cos θ) =[4(sin θ/cos θ) – 1]/ [4(sin θ/cos θ) + 1]= (4 tan θ – 1)/(4 tan θ + 1)
[4(3/4) – 1]/[4(3/4) + 1] = (3 – 1)/(3 + 1) = 1/2
9. The top of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30c with the horizontal, then the length of the wire is
(a) 12 m
(b) 10 m
(c) 8 m
(d) 6 m
View AnswerAns. (a) 12 m
Ans. Height of big pole CD = 20 m
Height of small pole AB = 14 m
DE = CD – CE
= CD – AB [AB = CE]
= 20 – 14 = 6 m
In ΔBDE, sin 30 = DE/BD
½ = 6/BD => BD = 12 m
10. A ladder, leaning against a wall, makes an angle of 60c with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.
(a) 5 m
(b) 5.5 m
(c) 2.5 m
(d) 4.5 m
View AnswerAns. (a) 5 m
As per given in question we have drawn figure below
In ΔACB with ∠C = 60, we get
Cos 60 = 2.5/AC
½ = 2.5/AC
AC = 2 x 2.5 = 5 m
11. In the figure, PQ is parallel to MN. If KP/PM = 4/13 = and KN = 20.4 cm then find KQ
(a) 4.8 cm
(b) 4.6 cm
(c) 4.4 cm
(d) 4.2 cm
View AnswerAns. (a) 4.8 cm
In the given figure PQ || MN, thus
KP/PM = KQ/QN (By BPT)
KP/PM = KQ/(KN – KQ)
4/13 = KQ/(20.4 – KQ)
4 x 20.4 – 4KQ = 13KQ
17KQ = 4 x 20.4
KQ = 20.4 x 4/17 = 4.8 cm
12. Ratio of lateral surface areas of two cylinders with equal height is
(a) 1 : 2
(b) H : h
(c) R : r
(d) None of these
View AnswerAns. (c) R : r
2 πRh : 2 πrh = R : r
13. In the formula
for finding the mean of grouped data di ’s are deviation from a of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the class marks
View AnswerAns. (c) mid points of the classes
Mid point of classes (xi – a)
Where, xi = (upper limit + lower limit)/2
14. If the zeroes of the quadratic polynomial ax2 + bx + c, where c ≠ 0, are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have same sign
(d) c and b have the same sign
View AnswerAns. (c) c and a have same sign
Let f(x) = ax2 + bx + c
Let α and β are zeroes of this polynomial
Then, α + β = -b/a
And α β = c/a
Since α = β, then α and β must be of same sign, i.e., either both are positive or both are negative. In both case
α β > 0
c/a > 0
Both c and a are of same sign
15. Two coins are tossed simultaneously. The probability of getting at most one head is
(a) 1/4
(b) 1/2
(c) 2/3
(d) 3/4
View AnswerAns. (d) ¾
All possible outcomes are {HH, HT, TH, TT}
Thus n(S) = 4
Favourable outcomes are {HT, TH, TT}
n(E) = 3
Probability of getting at most one head,
P(E) = n(E)/n(S) = 3/4
16. Someone is asked to take a number from 1 to 100. The probability that it is a prime, is
(a) 8/25
(b) 1/4
(c) 3/4
(d) 13/50
View AnswerAns. (b) 1/4
Prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 57, 61, 67, 71, 73, 79, 83, 89 and 97 i.e., 25 outcome
n(S) = 100
n(E) = 25
P(E) = n(E)/n(S) = 25/100 = 1/4
17. For which value(s) of p, will the lines represented by the following pair of linear equations be parallel 3x – y – 5 = 0
6x – 2y – p = 0
(a) all real values except 10
(b) 10
(c) 5/2
(d) 1/2
View AnswerAns. (a) all real values except 10
We have, 3x – y – 5 = 0
6x – 2y – p = 0
Here a1 = 3, b1 = -1, c1 = -1
And a2= 6, b2 = -2, c2 = -p
Since given lines are parallel,
a1/a2=b1/b2 ≠c1/c2
-1/-2 ≠ -5/-p
P ≠ 5 x 2 => p ≠ 10
18. The coordinates of a point A on y -axis, at a distance of 4 units from x -axis and below it are
(a) (4, 0)
(b) (0, 4)
(c) (-4, 0)
(d) (0, -4)
View AnswerAns. (d) (0, -4)
Because the point is 4 units down the x-axis i.e., coordinate is -4 and on y axis abscissa is 0. So, the coordinates of point A is (0, -4)
In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correction option.
19. Assertion : If a wire of length 22 cm is bent in the shape of a circle, then area of the circle so formed is 40 cm2.
Reason : Circumference of the circle = length of the wire.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
View AnswerAns. (d) Assertion (A) is false but reason (R) is true
We have 2 πr = 22
R = 3.5 cm
Area of the circle = 22/7 x 3.5 x 3.5
= 38.5 cm2
20. Assertion : The slant height of the frustum of a cone is 5 cm and the difference between the radii of its two circular ends is 4 cm. Then the height of the frustum is 3 cm.
Reason : Slant height of the frustum of the cone is given by
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
View AnswerAns. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
l = 5 cm, R – r = 4 cm
5 = √((4)2 + h2)
16 + h2 = 25
h2 = 25 – 16 = 9
h = 3 cm
Section – B
Section B consists of 5 questions of 2 marks each.
21. In the figure, PQRS is a trapezium in which PQ || RS. On PQ and RS, there are points E and F respectively such that EF intersects SQ at G. Prove that EQ x GS = GQ x FS.
Ans. In ΔGEQ and Δ GFS,
Due to vertical opposite angle,
∠EGQ = ∠FGS
Due to alternate angle,
∠EQG = ∠FSG
Thus by AA similarity we have
ΔGEQ ~ ΔGFS
EQ/FS = GQ/GS
EQ x GS = GQ x FS
22. In figure, O is the centre of the circle and LN is a diameter. If PQ is a tangent to the circle at K and ∠KLN = 30º, find ∠PKL.
Ans. Since OK and OL are radius of circle, thus
OK = OL
Angles opposite to equal sides are equal,
∠OKL = ∠OLK = 30°
Tangent is perpendicular to the end point of radius,
∠OKP = 90°
Now ∠PKL = ∠OKP – ∠L
= 90° – 30° = 60°
23. Evaluate:
Ans.
24. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting :
(i) a non face card,
(ii) a black king.
View AnswerAns. Total cards n(S) = 52
(i) There are 12 face cards and thus 40 non-face cards
n(E1) = 40
P(non-faces), P(E1) = n(E1)/n(S) = 40/52 = 10/13
(ii) there are 2 black king
n(E2) = 2
P(black king), P(E2) = P(E2) = n(E2)/n(S) = 2/52 = 1/26
OR
In a family of two children find the probability of having at least one girl.
View AnswerAns. All possible outcomes,
S = {GG, GB, BG, BB}
Total number of possible outcomes,
n(S) = 4
Favourable outcomes are GG, GB, BG
Thus n(E) = 3
P(at least one girl),
P(E) = n(E)/n(S) = ¾
25. In an equilateral triangle of side 24 cm, find the length of the altitude.
View AnswerAns. Let ΔABC be an equilateral triangle of side 24 cm and AD is altitude which is also a perpendicular bisector of side BC. This is shown in figure given below
BD = BC/2 = 24/2 = 12 cm
AB = 24 cm
AD = √(AB2 – BD2) =√(242 – 122)
= √(576 – 144) = √432 = 12√3 cm
OR
In the given figure, PQR is a triangle right angled at Q and XY || QR. If PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the length of PR and QR.
Ans. Since XY || OR, by BPT we have
PX/XQ = PY/YR
½ = PY /(PR – PY)
= 4 / (PR – 4)
PR – 4 = 8 => PR = 12 cm
In right ΔPQR, we have
QR2 = PR2 – PQ2
= 122 – 62
= 144 – 36 = 108
Thus QR = 6√3 cm
Section – C
Section C consists of 6 questions of 3 marks each.
26. Verify whether 2, 3 and 1/2 are the zeroes of the polynomial p(x) = 2x3 – 11x2 +17x – 6.
View AnswerAns. If 2, 3 and ½ are the zeroes of the polynomial p(x), then these must satisfy p(x) = 0
(1) 2 p(x) = 2x3 – 11x2 +17x – 6
P(2) = 0
(2) p(3) = 2x3 – 11x2 +17x – 6
P(3) = 0
(3) p(1/2) = 0
Hence 2, 3, and ½ are the zeroes of p(x)
27. Solve for x and y:
Ans.
28. In the given figure, if ∠ACB = ∠CDA, AC = 6 cm and AD = 3 cm, then find the length of AB.
Ans. In ΔABC and ΔACD we have
∠ACB = ∠CDA (given)
∠CAB = ∠CAD (common)
By AA similarity criterion we get
ΔABC ~ ΔACD
Thus AB/AC = BC/CD = AC/AD
Now AB/AC=AC/AD
AC2 AB x AD
62 = AB x 3
AB = 36/3 = 12 cm
OR
In the given figure, BC || PQ and BC = 8 cm, PQ = 4 cm, BA = 6.5 cm AP = 2.8 cm. Find CA and AQ.
Ans.
In ΔABC and ΔAPQ, AB = 6.5 cm, BC = 8 cm, PQ = 4 cm and AP = 2.8 cm
We have BC || PQ
Due to alternate angles
∠CBA = ∠AQP
Due to vertically opposite angles,
∠BAC = ∠PAQ
Due to AA similarity,
ΔABC ~ ΔAQP
AB/AQ = BC/QP = AC/AP
6.5/AQ = 8/4 = AC/AP
AQ = 6.5/2 = 3.25 cm
AC = 2 x 2.5 = 5.6 cm
29. Prove that
Ans.
30. A toy is in the form of a cone radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy. Use π =22/7
View AnswerAns. As per question the figure is shown below. Here total surface area of the toy is equal to the sum of surface area of hemisphere and curved surface area of cone.
Radius r = 7/2 = 3.5 cm
And height h = 12 cm
Slant height of cone,
l = √(r2 + h2) = √(3.52 + 122) = 12.5
Total surface area of the toy
= Surface area of hemisphere + Curved surface area of cone
= 2 πr2 + πrl
= πr(2r + l)
= 22/7 x 3.5 x (2 x 3.5 + 12.5)
= 11 x 19.5
= 214.5 cm2
OR
A well of diameter 4 m dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
View AnswerAns. Radius of earth dug out r = 4/2 = 2 m
Depth of the earth d = 21
Volume of earth πr2d = 22/7 x (2)2 x 21
=22 x 4 x 3 = 265 m2
Width of embankment = 3 m
Outer radius of ring = 2 + 3 = 5 m
Let the height of embankment be h
Volume of embankment,
π (R2 – r2) h = 264
22/7 x (52 – 22) x h = 264
22/7 x (25 – 4) x h = 264
22/7 x 21 x h = 264
22 x 3 x h = 264
H = (264 x 7)/(22 x 21) = 4
Height of embankment is 4 m
31. Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.
View AnswerAns. Assume that (5 + 3√2) is a rational number. Therefore, we can write it in the form of p/q where p and q are co-prime integers and q ≠ 0.
Now 5 + 3√2 = p/q
Where q ≠ 0 and p and q are integers.
Rewriting the above expression as,
3√2 = p/q – 5
√2 = (p – 5q)/3q
Here (p – 5q)/3q is rational because p and q are co-prime integers, thus √2 should be a rational number. But √2 is irrational. This contradicts the given fact that √2 is irrational. Hence (5 + 3√2) is an irrational number.
Section – D
Section D consists of 4 questions of 5 marks each.
32. Solve for x:
Ans.
OR
Solve for x: 4x2 + 4bx – (a2 – b2) = 0
View AnswerAns.
33. A right triangle ABC, right angled at A is circumscribing a circle. If AB = 6 cm and BC = 10 cm, find the radius r of the circle.
View AnswerAns. As per question we draw figure shown below.
In triangle ΔABC,
AC = √102 – 62) = 8
Area of triangle ΔABC,
ΔABC = ½ x AB x AC
= ½ x 6 x 8 = 24 cm2
Here we have joined AO, BO and CO
For area of triangle we have
ΔABC = ΔOBC + ΔOCA + ΔOAB
24 = ½ rBC + ½ rAC + ½rAB
= ½r(BC + AC + AB)
= ½r (6 + 10 + 8) = 12 r
Or 12r = 24
r = 2cm
34. Daily wages of 110 workers, obtained in a survey, are tabulated below:
Compute the mean daily wages a modal daily wages of these workers.
View AnswerAns. Let a = 170 be assumed mean.
OR
The mean of the following distribution is 18. Find the frequency of the class 19-21.
Ans.
35. Two friends Seema and Aditya work in the same office at Delhi. In the Christmas vacations, both decided to go to their hometown represented by Town A and Town B respectively in the figure given below. Town A and Town B are connected by trains from the same station C (in the given figure) in Delhi. Based on the given situation answer the following questions:
(i) Who will travel more distance, Seema or Aditya, to reach to their hometown?
(ii) Seema and Aditya planned to meet at a location D situated at a point D represented by the mid-point of the line joining the points represented by Town A and Town B. Find the coordinates of the point represented by the point D.
View AnswerAns. (i) Distance travelled by Seema
CA = √[(-4 – 1)2 + (4 – 7)2]
= √(-5)2 + (-3)2
= √25 + 9 = √34 units
Thus distance travelled by Seema is √34 units.
Similarly, distance travelled by Aditya
CB = √[(4 + 4)2 + (4 – 2)2]
=√(82 + 22)=√(64 + 4) = √68 units
Distance travelled by Aditya is √68 units and Aditya travels more distance.
(ii) Since, D is mid-point of town A and town B
D = [(1 + 4)/2, (7 + 2)/2] = (5/2, 9/2)
Section – E
Case study based questions are compulsory.
36. Cubic Coating: Frozen specimens are stored in a cubic metal box that is x inches on each side. The box is surrounded by a 2-inch-thick layer of foam insulation.
(i) Find a polynomial function V(x) that gives the total volume in cubic inches for the box and insulation.
(ii) Find the total volume if x is 10 inches.
(iii) Use the remainder theorem to find the total volume when x is 10 inches.
View AnswerAns. Since 2 inches of foam is added all around the box, the sides are now x + 4 inches each.
Volume of a cube of side x + 4,
V(x) = (x + 4)3 = (x + 4) (x2 + 8x + 16)
= x(x2 + 8x + 16) + 4(x2 + 8x + 16)
V(x) = x3 + 12x2 + 48x + 64
(ii) The total volume if x is 10 inches,
V(10) = 103 + 12 x 102 + 48 x 10 + 64
= 2744 in2
37. Wilt Chamberlain : Wilton Norman “Wilt” Chamberlain was an American basketball player, and played in the NBA during the 1960s. At 7 feet 1 inch, he was the tallest and heaviest player in the league for most of his career, and he was one of the most famous people in the game for many years. He is the first and only basketball player to score 100 points in an NBA game.
In the 1961–1962 NBA basketball season, Wilt Chamberlain of the Philadelphia Warriors made 30 baskets. Some of the baskets were free throws (worth 1 point each) and some were field goals (worth 2 points each). The number of field goals was 10 more than the number of free throws.
(i) How many field goals did he make?
(ii) How many free throws did he make?
(iii) What was the total number of points scored?
(iv) If Wilt Chamberlain played 5 games during this season, what was the average number of points per game?
View AnswerAns. Let x be the free throw and y be the fixed goal.
As per question
x + y = 30
y = x + 10
Solving x = 10, y = 20
Thus he made 20 fixed goal.
(ii) Free throw x = 10
(iii) Point scored = 10 + 2 x 20 = 50
(iv) Average point 50/5 = 10
38. Underground water tank is popular in India. It is usually used for large water tank storage and can be built cheaply using cement-like materials. Underground water tanks are typically chosen by people who want to save space. The water in the underground tank is not affected by extreme weather conditions. The underground tanks maintain cool temperatures in both winter and summer. Electric pump is used to move water from the underground tank to overhead tank.
Ramesh has build recently his house and installed a underground tank and overhead tank. Dimensions of tanks are as follows:
Underground Tank: Base 2 m x 2 m and Height 1.1 m.
Overhead tank: Radius 50 cm and Height 175 cm
(i) What is the capacity of the underground tank?
(ii) What is the ratio of the capacity of the underground tank to the capacity of the overhead tank?
(iii) If curved part of overhead tank need to be painted to save it from corrosion, how much area need to be painted? If water is filled in the overhead tank at the rate of 11 litre per minute, the tank will be completely filled in how much time?
View AnswerAns. (i) Volume of underground tank,
lbh = 2 x 2 x 1.1 = 4.4 m3
Since 1 m3 is equal to 100 litres,
4.4 m3 = 4.4 x 1000 = 4400 litres
(ii) Radius of overhead is 50 cm i.e., ½ meter and height is 175 cm i.e., 1.75 = 7/4 m
Thus volume of overhead tank,
π r2h = 22/7 x ½ x ½ x 7/4 = 11/8 m3
Capacity of sump/Capacity of overhead tank = lbh/ πr2h= 4.4/11/8 = 3.2
(iii) C.S.A of cylindrical tank
2 πrh = 2 x 22/7 x ½ x 7/4 = 5.5 m2
Volume of water in cylindrical tank is 11/8 m3
11/8 m3 = 11/8 x 1000 litres
Thus time taken to fill tank,
11/8 x 1000 x 1/11 = 125 minutes
OR
(iv) If the amount of water in the underground tank, at an instant, is 2400 litres, find then the water level in the underground tank at that instant.
View AnswerAns. volume of water in underground tank
= 2400 litres = 2.4 m3
Then, V = lbh
2 x 2 x h = 2.4
H = 2.4/(2 x 2) = 0.6 m = 60 cm