Class X - Math Standard - Paper - 2 - Practice for Everyone

Kindly share you feedback about the website – Click here

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. A letter of English alphabet is chosen at random, what is the probability that the letter so chosen is a consonant?

(a) 5/26

(b) 21/26

(d) 7/13

View Answer

Ans. (b) 21/26

Since total number in English alphabet is 26, in which 5 vowels and 21 consonants

In this case total possible outcome

n(S) = 26

Favourable outcome is a consonant. Thus

n(E₂) = 21

P(E₂) = n(E₂)/n(S) = 21/26

2. What is the HCF of smallest primer number and the smallest composite number?

(a) 2

(b) 4

(d) 8

View Answer

Ans. (a) 2

Smallest prime number is 2 and smallest composite number is 4. HCF of 2 and 4 is 2.

3. If A (5, 2), B (2, -2) and C (-2, t) are the vertices of a right angled triangle with ∠B = 90º, then the value of t will be

(a) 1

(b) 2

(d) 4

View Answer

Ans. (a) 1

Now AB² = (2 – 5)² + (-2-2)² = 9 + 16 = 25

BC² = (-2 – 2)² + (t + 2)² = 16 + (t + 2)²

AC² = (5 + 2)² + (2 – t)² = 49 + (2 – t)²

Since ΔABC is a right angled triangle

AC² = AB² + BC²

49 + (2 – t)² = 25 + 16 + (t + 2)²

49 + 4 – 4t + t² = 41 + t² + 4t + 4

t = 1

4. The sum and product of zeroes of a quadratic polynomial are 6 and 9 respectively. The quadratic polynomial will be

(a) x² + 9x – 6

(b) x² + 6x + 9

(d) x² + 6x – 9

View Answer

Ans. (c) x² – 6x + 9

Sum of zeroes = α + β = 6

Product of zeroes = α β = 9

Now p(x) = x² – (α + β)x + α β

Thus x² – 6x + 9

5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. The dimensions of garden will be

(a) 20 m by 16 m

(b) 36 m by 10 m

(d) 20 m by 16 m

View Answer

Ans. Let the length of the garden be x and the width be y.

Perimeter of rectangular garden

p = 2(x + y)

Since half perimeter is given as 36 m,

x + y = 36 ….. (1)

Also, x = y + 4

Or x – y = 4 ….. (2)

Adding (1) and (2) we have

2x = 40 => x = 20

Subtracting eq(2) from (1), we have

2y = 32 => y = 16

Hence, length is 20 m and width is 16 m.

6. The quadratic equation x² + x – 5 = 0 has

(a) two distinct real roots

(b) two equal real roots

(d) more than 2 real roots

View Answer

Ans. (a) two distinct real roots

We have x² + x – 5 = 0

Here, a = 0, b = 1, c = -5

Now, D² – 4ac

= (1)² – 4 x 1 x (-5)

= 21 > 0

7. Which of the following equations has 2 as a root?

(a) x² – 4x + 5 = 0

(b) x² + 3x – 12 = 0

(d) 3x² – 6x – 2 = 0

View Answer

Ans. (c) 2x² – 7x + 6 = 0

8. What happens to value of cos θ when θ increases from 0º to 90º.

(a) cos θ decreases from 1 to 0.

(b) cos θ increases from 0 to 1.

(d) cos θ decreases from 1 to ½

View Answer

Ans. (a) cos θ decreases from 1 to 0.

9. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below

The number of athletes who completed the race in less than 14.6 second is :

(a) 11

(b) 71

(d) 130

View Answer

Ans. (c) 82

The number of athletes who completed the race in less than 14.6

= 2 + 4 + 5 + 71 = 82

10. For what value of k, the pair of linear equations kx – 4y = 3, 6x – 12y = 9 has an infinite number of solutions ?

(a) k = 2

(b) k ≠ 2

(d) k = 4

View Answer

Ans. (a) k = 2

We have kx – 4y = 3

and 6x – 12y = 9

where, a₁ = k, b₁ = 4, c₁ = -3

a₂ = 6, b₂ = -12, c₂ = -9

Condition for infinite solutions:

a₁/a₂ = b₁/b₂ = c₁/c₂

k/6 = -4/-12 = 3/9

Hence k = 2

11. 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

(d) 6 m

View Answer

Ans. (a) 12 m

Ans. Height pf big pole CD = 20 m

Height of small pole AB = 14 m

DE = CD – CE

= CD – AB [AB = CE]

= 20 – 14 = 5 m

In ΔBDE, sin 30° = DE/BD

½ = 6/BD => BD = 12 m

Thus length of wire is 12 m

12. Which term of an AP, 21, 42, 63, 84, … is 210?

(a) 9th

(b) 10th

(d) 12th

View Answer

Ans. (b) 10th

Let nth term of given AP be 210,

First term a = 21

Common difference d = 42 – 21 = 21

And a_n = 210

In an AP, a_n = a + (n – 1)d

210 = 21 + (n – 1)21

n = 10

13. The perimeters of two similar triangles ΔABC and ΔPQR are 35 cm and 45 cm respectively, then the ratio of the areas of the two triangles is ………

(a) 2/9

(b) 7/9

(d) 3/4

View Answer

Ans. (c) 49/81

We know that

1. Ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

2. Ratio of the perimeter of two similar triangles is the same as the ratio of their corresponding sides.

Therefore, in result ratio of areas of two similar triangles is equal to the ration of the squares of their perimeters.

ar (ΔABC)/ar (ΔPQR) = 35²/45² = 7²/9² =49/91

14.

(a) 1

(b) 2 tan² θ

(d) 2 sec² θ

View Answer

Ans. (a) 1

15. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(a) 10 m

(b) 15 m

(d) 24 m

View Answer

Ans. (a) 10 m

16. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

(a) 4 π r²

(b) 6 π r²

(d) 8 π r²

View Answer

Ans. (a) 4 π r²

Because curved surface area of a hemisphere is (a) 2 π r² and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.

Hence, the curved surface area of new solid = 2 π r²+ 2 π r² = 4 π r²

17. If zeroes of the polynomial x² + 4x + 2a are a and 2/a then the value of a is

(a) 1

(b) 2

(d) 4

View Answer

Ans. (a) 1

Product of (zeroes) roots,

c/a = 2a/1 = α x 2/ α = 2

or 2a = 2

Thus a = 1

18. If radii of two concentric circles are 4 cm and 5 cm, then the length of each of one circle which is tangent to the other circle, is

(a) 3 cm

(b) 6 cm

(d) 1 cm

View Answer

Ans. (b) 6 cm

AB is tangent at P and CP is radius at P. Tangent at any point of circle is perpendicular to the radius through the point of contact.

Thus CP ⊥ AB

Now, in right triangle PAC

By Pythagoras theorem we have

AP² = AC² – PC² = 5² – 4² = 25 – 16 = 9

AP = 3 cm

So, length of chord,

AB = 2AP = 2 x 3 = 6 cm

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: Pair of linear equations: 9x + 3y +12 = 0, 8x + 6y + 24 = 0 have infinitely many solutions.

Reason: Pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have infinitely many solutions, if a a₁/a₂ = b₁/b₂ = c₁/c₂

(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).

(d) Assertion (A) is false but reason (R) is true.

View Answer

Ans. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

From the given equations, we have

9/18 = 3/6 = 12/24

½ = ½ = ½ i.e., a₁/a₂ = b₁/b₂ = c₁/c₂

20. Assertion : If nth term of an AP is 7 – 4n, then its common differences is -4.

Reason : Common difference of an AP is given by d = a_n+1 − a_n .

(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).

(d) Assertion (A) is false but reason (R) is true.

View Answer

Ans. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Assertion, an = 7 – 4n

d = a_n+1– a_n

= 7 – 4(n + 1) – (7 – 4n)

= 7 – 4n – 4 – 7 + 4n = -4

Section – B

Section B consists of 5 questions of 2 marks each.

21. If two positive integers p and q are written as p = a²b³ and q = a³ b, where a and b are prime numbers than verify LCM (p, q) x HCF (q, q) = pq

View Answer

Ans. We have p = a²b³ = a x a x b x b x b

And q = a³b = a x a x a x b

Now LCM (p, q) = a x a x a x b x b x b

= a³b³

And HCF (p, q) = a x a x b = a²b

LCM (p, q) x HCF (p, q) = a³b³ x a²b

= a⁵b⁴

= a²b³ x a³b

= pq

22. If the nth term of an AP -1 4, , 9 1, , 4 ….. is 129. Find the value of n.

View Answer

Ans. Let the first term be a, common difference be d and nth term be a_n

We have a = – 1 and d = 4 – (-1) = 5

-1 +(n – 1) x 5 = a_n

-1 + 5n – 5 = 129

5n = 135

n = 27

Hence 27^th term is 129

Write the nth term of the AP

View Answer

Ans. Let the first term be a, common difference be d and nth term be a_n.

We have a = 1/m

d = (1+ m)/m – 1/m = 1

a_n 1/m + (n – 1)1

Hence a_n 1/m + n – 1

23. If the mid-point of the line segment joining the points A(3, 4) and B (k, 6) is P (x , y) and x + y − 10 = 0, find the value of k.

View Answer

Ans. If P (x, y) is midpoint of A (3, 4) and B (k, 6), then we have

(3 + k)/2 = x and y = (4 + 6)/2 = 10/2 = 5

Substituting above value in x + y – 10 = 0 we have

(3 + k) /2 + 5 – 10 = 0

(3 + k)/2 = 5

3 + k = 10 => 10 – 3 = 7

24. In figure, AP, AQ and BC are tangents of the circle with centre O. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then what is the length of AP?

View Answer

Ans. Due to tangents from external points,

BP = BR, CR = CQ and AP = AQ

Perimeter of ΔABC,

AB + BC + AC = AB + BR + RC + AC

5 + 4 + 6 = AB + BP +CQ + AC

15 = AP + AQ

15 = 2AP

AP = 15/2 = 7.5 cm

Two chords AB and CD of a circle intersect at E such that AE = 2.4 cm, BE = 3.2 cm and CE = 1.6 cm. What is the length of DE?

View Answer

Ans.

Applying the rule, AE x EB = CE x ED

2.4 x 3.2 = 1.6 x ED

ED = 4.8 cm

25. Two coins are tossed together. Find the probability of getting both heads or both tails.

View Answer

Ans. Possibilities are HH, HT, TH, TT out of which HH and TT are favourable.

n(S) = 4

n(E) = 2

P(HH or TT), P(E) = n(E)/n(S) = 2/4 = 1/2

Section – C

Section C consists of 6 questions of 3 marks each.

26. Find HCF and LCM of 378, 180 and 420 by prime factorization method. Is HCF × LCM of these numbers equal to the product of the given three numbers?

View Answer

Ans. Finding prime factor of given number we have,

378 = 2 x 3³ x 7

180 = 2² x 3² x 5

420 = 2² x 3 x 7 x 5

HCF (378, 180, 420) = 2 x 3 = 6

LCM (378, 180, 420) = 2² x 3³ x 5 x 7

= 3780

HCF x LCM = 6 x 3780 = 22680

Product of given numbers

378 x 180 x 420

= 28576800

Hence, HCF x LCM ≠ Product of three numbers.

27. In Figure, in ΔABC, DE || BC such that AD = 2.4 cm, AB = 3.2 cm and AC = 8 cm, then what is the length of AE?

View Answer

Ans.

28. Prove that:

View Answer

Ans.

29. From a point P, which is at a distant of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR are drawn to the circle, then the area of the quadrilateral PQOR (in cm²).

View Answer

Ans. As per the given question we draw the figure as below:

Hence OQ is radius and QP is tangent at Q, since radius is always perpendicular to tangent at point of contact, ΔOQP is right angle triangle.

Now, PQ = √(OP² – OR²)

= √(13² – 5²) = √(169 – 25)

= √144 = 12 cm

Area of triangle ΔOQP,

Δ = ½(OQ)(QP) = ½ x 12 x 5 = 30

Area of quadrilateral PQOR

2 x ΔPOQ = 2 x 30 = 50 cm²

30. In the given figure, a chord AB of the circle with centre O and radius 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Hence find the area of major segment ALBQA. (Use π = 3.14)

View Answer

Ans. Area of sector OAPB,

90/360 π(10)² = 25 π cm²

Area of ΔAOB, = ½ x 10 x 10 = 50 cm²

Area of minor segment AQBP,

= (25 π – 50) cm²

= 25 x 3.14 – 50

= 78.5 – 50 = 28.5 cm²

Also area of circle = π(10)²

= 3.14 x 100 = 314 cm²

Area of major segment = ALBQA = 314 – 28.5

= 285.5 cm²

Find the area of shaded region shown in the given figure where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

View Answer

Ans. Since OAB is an equilateral triangle, we have

∠AOB = 60

Area of shaded region = Area of major sector + (Area of ΔAOB – Area of minor sector)

31. A die is thrown once. Find the probability of getting a number which

(i) is a prime number

(ii) lies between 2 and 6.

View Answer

Ans. Total outcomes n(S) = 6

(i) is a prime number

Prime numbers are 2, 3 and 5

n(E₁) = 3

P(prime no.)

P(E₁) = n(E₁)/n(S) = 3/6 = ½

(ii) lies between 2 and 6

n(E2) = 3

P(lies between 2 and 6)

P(E₂) = n(E₂)/n(S) = 3/6 = 1/2

A die is thrown twice. Find the probability that

(i) 5 will come up at least once.

(ii) 5 will not come up either time.

View Answer

Ans. There are 6 x 6 = 36 possible outcome. Thus sample space for two die is

n(S) = 36

(i) 5 will come up at least once

Favourable case are (5, 1), (5, 2), (5, 3), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5) and (6,5) thus 11 case.

Number of favourable outcome

n(E₁) = 11

Probability that 5 will come up at least once

P(E₁) = n(E₁)/n(S) = 11/36

(ii) 5 will not come up either time

Probability that 5 will come up either time

P(not E) = 1 – P(E)

= 1 – 11/36 = 36 – 11/36

= 25/36

Section – D

Section D consists of 4 questions of 5 marks each.

32. Solve the following pair of linear equations graphically:

X + 3y = 12, 2x – 3y = 12

Also shade the region bounded by the line 2x – 3y = 2 and both the co-ordinate axes.

View Answer

Ans. We have x + 3y = 6 => y = (6 – x)/3 …..(1)

And 2x – 3y = 12 => y = (2x – 12)/3

Plotting the above points and drawing lines joining them, we get the following graph.

The two lines intersect each other at point B(6, 0).

Hence, x = 6 and y = 0

Again ΔOAB is the region bounded by the line 2x – 3y = 12 and both the co-ordinate axes.

For what values of a and b does the following pair of linear equations have infinite number of solution?

2x + 3y = 7, a(x + y) – b(x – y) = 3a+ b – 2.

View Answer

Ans. We have 2x + 3y – 7 = 0

Here a1 = 2, b1 = 3, c1 = -7

And a(x + y) – b(x – y) = 3a + b – 2

ax + ay – bx + by = 3a + b – 2

(a – b) x + (a+ b)y – (3a + b – 2) = 0

Here a₂ = a – b, b₂ = a + b, c₂ = -(3a + b – 2)

For infinite many solutions

a₁/a₂ = b₁/b₂ = c₁/c₂

33. Show that A(-1, 0), B(3, 1), C(2, 2) and D(-2, 1) are the vertices of a parallelogram ABCD.

View Answer

Ans.

Here Mid-point of AC = Mid-point of BD

Since diagonals of a quadrilateral bisect each other, ABCD is a parallelogram.

34. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR

View Answer

Ans. It is given that in ΔABC and ΔPQR, AD and PM are their medians,

such that AB/PQ = AD/PM = AC/PR

We have produce AD to E such that AD = DE and produce PM to N such that PM = MN. We join CE and RN. As per given condition we have drawn the figure below.

In ΔABD and ΔEDC,

AD = DE ( By construction)

∠ADB = ∠EDC (VOA)

BD = DC (AD is a median)

By SAS congruency

ΔABD ≅ ΔEDC

AB = CE (By CPCT)

Similarly PQ = RN and ∠A = ∠2

AB/PQ = AD/PM = AC/PR (Given)

CE/RN = 2AD/2PM = AC/PR

CE/RN = AE/PN = AC/PR

By SSS similarlity, we have

ΔAEC ~ ΔPNR

∠3 = ∠4

∠1 = ∠2

∠1 + ∠3 = ∠2 + ∠4

By SAS similarity, we have

ΔABC ~ ΔPQR (Hence proved)

Find the length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm.

View Answer

Ans. As per given condition we have drawn the below

We have AB = BC = CD = AD = 5 cm and AC = 6 cm

Since AO = OC, AO = 3 cm

Here ΔAOB is right angled triangle as diagonals of rhombus intersect at right angle.

By Pythagoras theorem,

OB = 4 cm

Since DO = OB, BD = 8 cm, length of the other diagonal = 2(BO) where BO = 4 cm

Hence BD = 2 x BO = 2 x 4 = 8 cm.

35. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pipe, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)

View Answer

Ans. As per question the figure is shown below:

Radius of lower cylinder, R = 12 cm

Height of lower cylinder, H = 220 cm

Radius of upper cylinder, r = 8 cm

Height of upper cylinder, h = 60 cm

Volume of solid iron pole,

Section – E

Case study based questions are compulsory.

36. Auditorium, the part of a public building where an audience sits, as distinct from the stage, the area on which the performance or other object of the audience’s attention is presented. In a large theatre an auditorium includes a number of floor levels frequently designed as stalls, private boxes, dress circle, balcony or upper circle, and gallery. A sloping floor allows the seats to be arranged to give a clear view of the stage. The walls and ceiling usually contain concealed light and sound equipment and air extracts or inlets and may be highly decorated.

In an auditorium, seats are arranged in rows and columns. The number of rows are equal to the number of seats in each row. When the number of rows are doubled and the number of seats in each row is reduced by 10, the total number of seats increases by 300.

(i) If x is taken as number of row in original arrangement, write the quadratic equation that describes the situation?

View Answer

Ans. Since number of rows are equal to the number of seats in each row in original arrangement, total seats are x²

In new arrangement row are 2x and seats in each row are x – 10. Hence total 2x (x – 10) seats are there.

Total seats are 300 more than previous seats so total number of seats are x² + 300

Thus 2x (x – 10) = x² + 300

2x² – 20x = x2 + 300

x² – 20x – 300 = 0

(ii) How many number of rows are there in the original arrangement?

View Answer

Ans. We have x² – 20x – 300 = 0

x²– 30x + 10x – 300 = 0

x(x – 30) + 10 (x – 30) = 0

(x – 30)(x + 10) = 0 => x = 30, -10

(iii) How many number of seats are there in the auditorium in original arrangement? How many number of seats are there in the auditorium after re-arrangement?

View Answer

Ans. Number of seats in original arrangement,

x² = 30² = 900

Total seats in rearrangement = 30² + 300

= 900 + 300 = 1200

(iv) How many number of columns are there in the auditorium after re-arrangement?

View Answer

Ans. Number of row are 30 in original arrangement. In rearrangement number of rows are 2 x 30 = 60

Number of column after rearrangement,

= Total seats/Row = 1200/60 = 20 column.

37. Drawbridge: A drawbridge is a bridge that can be moved in order to stop or allow passage across it. Modern drawbridges are often built across large, busy waterways. They can be lifted to allow large ships to pass or lowered to allow land vehicles or pedestrians to cross.

A drawbridge is 60-metre-long when stretched across a river. As shown in the figure, the two sections of the bridge can be rotated upward through an angle of 30°.

(i) If the water level is 5 metre below the closed bridge, find the height h between the end of a section and the water level when the bridge is fully open.

(ii) How far apart are the ends of the two sections when the bridge is fully opened, as shown in the figure?

View Answer

Ans. It may be easily seen that length of each section of bridge is 60/2 = 30 m. We draw a diagram of the situation as shown below. Let h be height between the end of a section and the water level, when bridge is fully opened. Let d be distance between the end of a section, when bridge is fully opened.

(i) When the bridge is fully open, height is 20 meter between the end of a section and the water level.

(ii) When the bridge is fully opened the ends of the two section are 8.04 metre apart.

38. Life insurance is a contract between an insurance policy holder and an insurer or assurer, where the insurer promises to pay a designated beneficiary a sum of money upon the death of an insured person (often the policy holder). Depending on the contract, other events such as terminal illness or critical illness can also trigger payment. The policy holder typically pays a premium, either regularly or as one lump sum.

SBI life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

(i) What is the median value of age?

(ii) What will be the upper limit of the modal class?

(iii) What is the mode value of age?

(iv) Find the mean value of age using empirical relation.

View Answer

Ans. The given table is cumulative frequency distribution.

We write the frequency distribution as given below:

Class X – Math Standard – Paper – 2