AmbifulWorksheets
Board exam practice paper
CBSE · Class 10 · Mathematics (Basic) (241)

Mathematics (Basic): Board Exam Practice Paper

Time allowed: 3 hours · Maximum marks: 80 · AW-10MB-BP-70241

Name: Date: Score: ____ / 80
  1. This paper has 38 questions. All questions are compulsory.
  2. It is in five sections: A, B, C, D and E.
  3. Section A has 20 multiple-choice questions (18 MCQs and 2 assertion–reason questions) of 1 mark each.
  4. Section B has 5 very short answer questions of 2 marks each.
  5. Section C has 6 short answer questions of 3 marks each.
  6. Section D has 4 long answer questions of 5 marks each.
  7. Section E has 3 case-based questions of 4 marks each, with sub-parts of 1, 1 and 2 marks.
  8. Some questions have an internal choice (marked OR): answer only one of them.
  9. Draw neat figures wherever needed. Take π = 22/7 wherever needed, unless the question says otherwise.
  10. Use of a calculator is not allowed.
Section A: 20 questions of 1 mark each
1.
[1]The LCM of 100 and 140 is:
  1. 20
  2. 240
  3. 14000
  4. 700
2.
[1]The product of the zeroes of the polynomial x2 + 4x − 45 is:
  1. 45
  2. −4
  3. −46
  4. −45
3.
[1]The solution of 4x − 6y = 8 and 3x + 6y = −36 is:
  1. (4, −4)
  2. (−3, −4)
  3. (−4, −4)
  4. (−4, 4)
4.
[1]If x + y = 27 and x − y = −13, then (x, y) is:
  1. (27, −13)
  2. (7, 20)
  3. (20, 7)
  4. (7, −20)
5.
[1]Which of these is a quadratic equation?
  1. 9x + 2 = 0
  2. (x − 3)2 = x2 + 2
  3. x3 + 9x + 2 = 0
  4. 5x2 + 9x + 2 = 0
6.
[1]The sum of the first 14 multiples of 3 is:
  1. 315
  2. 588
  3. 42
  4. 105
7.
[1]The sum of the first 29 terms of 16, 14, 12, … is:
  1. −348
  2. 406
  3. −390
  4. −696
8.
[1]In △ABC, DE ∥ BC with D on AB and E on AC. If AD = 20 cm, DB = 28 cm and AE = 5 cm, then EC is:
  1. 12 cm
  2. 7 cm
  3. 8 cm
  4. 25/7 cm
9.
[1]△ABC ~ △DEF. AB = 3 cm, DE = 12 cm and the perimeter of △ABC is 11 cm. The perimeter of △DEF is:
  1. 46 cm
  2. 11 cm
  3. 44 cm
  4. 55 cm
10.
[1]The midpoint of (4, 6) and (6, 14) is:
  1. (2, 8)
  2. (5, 10)
  3. (10, 5)
  4. (10, 20)
11.
[1]cosec A equals:
  1. cos A / sin A
  2. 1 / cos A
  3. 1 / tan A
  4. 1 / sin A
12.
[1]If cosec A = 13/5, then tan A equals:
  1. 12/5
  2. 5/12
  3. 12/13
  4. 5/13
13.
[1]The tangent from a point P to a circle of radius 8 cm is 6 cm long. The distance of P from the centre is:
  1. 2 cm
  2. 11 cm
  3. 14 cm
  4. 10 cm
14.
[1]A line that meets a circle at two points is called a:
  1. diameter
  2. tangent
  3. secant
  4. radius
15.
[1]A solid is a cylinder of radius 7 cm and height 19 cm with a cone of the same radius and height 18 cm on top. Its volume (π = 22/7) is:
  1. 4,850 cm3
  2. 2,926 cm3
  3. 5,698 cm3
  4. 3,850 cm3
16.
[1]For the data below, the modal class is:
Class: 80–100, 100–120, 120–140, 140–160, 160–180, 180–200
Frequency: 19, 17, 10, 17, 12, 12
  1. 100–120
  2. 120–140
  3. 80–100
  4. 140–160
17.
[1]The class mark of the class 175–200 is:
  1. 25
  2. 200
  3. 212.5
  4. 187.5
18.
[1]If P(E) = 0.1, then P(not E) is:
  1. 1
  2. 0.8
  3. 0.9
  4. 0.1
19.
[1]Assertion (A): When a die is thrown, the probability of getting a number greater than 4 is 1/3.
Reason (R): The probability of an impossible event is 0.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
20.
[1]Assertion (A): In △ABC, DE ∥ BC with AD = 15 cm, DB = 9 cm and AE = 15 cm, so EC = 10 cm.
Reason (R): A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B: 5 questions of 2 marks each
21.
[2]Find the HCF and LCM of 6 and 9, and check that HCF × LCM = 6 × 9.
OR
[2]Express 66 as a product of its prime factors.
22.
[2]One zero of 2x2 + 2x − 12 is 2. Find the other zero.
23.
[2]Evaluate: (3 tan2 30° + 4 cos2 45°) ÷ (3 tan2 45°)
24.
[2]Tangents PA and PB are drawn from an external point P to a circle with centre O. If ∠AOB = 100°, find ∠APB.
25.
[2]Find y if the distance between (5, 8) and (−4, y) is 15 units.
OR
[2]Find the perimeter of the triangle with vertices (4, 3), (9, 3) and (4, 15).
Section C: 6 questions of 3 marks each
26.
[3]Prove that √3 is irrational. Hence show that 7 − 8√3 is irrational.
27.
[3]The sum of the digits of a two-digit number is 15. The number obtained by reversing the order of its digits is 27 less than the original number. Find the number.
28.
[3]PA and PB are tangents drawn from an external point P to a circle with centre O and radius 120 cm. If PA = 27 cm, find OP and the perimeter of the quadrilateral OAPB.
OR
[3]TG is a diameter of a circle with centre O. Prove that the tangents drawn at T and G are parallel. A third tangent touches the circle at D and meets the tangents at T and G at W and B respectively. If TW = 15 cm and GB = 16 cm, find WB.
29.
[3]Show that the points P(−7, −3), Q(−4, −6) and R(−7, −9) are the vertices of a right isosceles triangle.
30.
[3]If cot θ = 5/12, where θ is an acute angle, find the value of (sec θ + tan θ)/(sec θ − tan θ).
31.
[3]The following table gives the marks obtained by the students of a class in a test. The mean of the data is 25. Find the missing frequency p.
Class interval (marks) and frequency (number of students): 0–10: 7, 10–20: 13, 20–30: 5, 30–40: p, 40–50: 12.
OR
[3]The following table gives the weights (in kg) of the students of a class. Find the mean using the direct method.
Class interval (weight in kg) and frequency (number of students): 30–35: 10, 35–40: 4, 40–45: 3, 45–50: 18, 50–55: 5.
Section D: 4 questions of 5 marks each
32.
[5]Kabir, who is 0.9 m tall, walks away from the base of a lamp-post in a straight line at a speed of 1.2 m/s. The lamp is 4.5 m above the ground. (a) Find the length of Kabir's shadow after 8 seconds. (b) How far from the base of the lamp-post will Kabir be when the shadow is 1.9 m long?
33.
[5]Find two consecutive even positive integers, the sum of whose squares is 3700.
OR
[5]A father is 23 years older than his son. 3 years ago, the product of their ages (in years) was 78. Find their present ages.
34.
[5]Isha makes a solid paperweight in the shape of a cone standing on a hemisphere of the same radius. The radius is 28 cm, the height of the cone is 21 cm and its slant height is 35 cm. Find (i) the total surface area of the paperweight and (ii) its volume. (Use π = 22/7.)
35.
[5]From a point on a bridge across a river, the angles of depression of the two banks, on opposite sides of the river, are 45° and 30°. The point is 48 m above the water. Find the width of the river, and how far the point is horizontally from each bank. (Take √3 = 1.73.)
OR
[5]The angles of elevation of the top of a chimney from two points on the ground, on the same side of it and in a straight line with its foot, are 30° and 60°. The two points are 46 m apart. Find the height of the chimney and the distance of the nearer point from its foot. (Take √3 = 1.73.)
Section E: 3 case-based questions of 4 marks each
36.
Sharing a round treat
At Ananya's birthday party, a round cake of diameter 21 cm is cut from the centre into 5 equal slices. Each slice is a sector of a circle, and its curved edge is the icing border. Asha wants to work out the size of each slice. (Use π = 22/7.)
(i)
[1]What is the angle at the centre of one slice?
(ii)
[1]Find the area of the top of one slice.
(iii)
[2]Find the perimeter of the top of one slice.
OR
[2]The guests eat 3 slices. Find the area of the top of the round cake that is left.
37.
Seats in the school hall
The school assembly hall has 25 rows of seats. The first row has 12 seats, and each row has 2 more seats than the row in front of it. The school is getting the assembly hall ready for its annual day. The numbers of seats in the rows form an arithmetic progression.
(i)
[1]How many seats are there in row 19?
(ii)
[1]Which row has 32 seats?
(iii)
[2]Find the total number of seats in the assembly hall.
OR
[2]The first 8 rows are kept for parents. How many seats are kept for parents?
38.
Checking for defects
A workshop moulds 240 water bottles in a day. The supervisor checks them and finds that 200 are good, 37 have minor defects and the rest have major defects. Kabir, a customer, will buy only a good one, while a shopkeeper will buy one that has no major defect. One of the day's water bottles is picked at random.
(i)
[1]Find the probability that Kabir will buy it.
(ii)
[1]Find the probability that the shopkeeper will buy it.
(iii)
[2]Find the probability that it is not good.
OR
[2]Find the probability that neither Kabir nor the shopkeeper will buy it.
AmbifulWorksheets
Marking scheme
CBSE · Class 10 · Mathematics (Basic) (241)

Marking scheme

Mathematics (Basic): Board Exam Practice Paper · AW-10MB-BP-70241 · For the parent or teacher: award the marks shown for each step, even when the final answer is wrong.

QAnswer and marks for each stepMarks
Section A: 20 questions of 1 mark each
1(d) 700
LCM = 20 × 5 × 7 = 700
1
2(d) −45
product of zeroes = c/a = −45
1
3(c) (−4, −4)
solve: x = −4, y = −4
1
4(b) (7, 20)
adding: 2x = 14, x = 7; y = 20
1
5(d) 5x2 + 9x + 2 = 0
the highest power of x is 2, and the x2 term does not cancel
1
6(a) 315
3(1 + 2 + … + 14) = 3 × 105
1
7(a) −348
29/2 × (2 × 16 + 28 × (−2)) = −348
1
8(b) 7 cm
AD/DB = AE/EC: 20/28 = 5/EC, so EC = 7
1
9(c) 44 cm
perimeters are in the ratio of corresponding sides: 11 × 4/1 = 44
1
10(b) (5, 10)
((4 + 6)/2, (6 + 14)/2)
1
11(d) 1 / sin A1
12(b) 5/12
cosec A = 13/5 gives sides 5 : 12 : 13; tan A = 5/12
1
13(d) 10 cm
OP = √(82 + 62) = 10
1
14(c) secant1
15(d) 3,850 cm3
πr2h1 + (1/3)πr2h2 = 2,926 + 924
1
16(c) 80–100
the highest frequency is 19
1
17(d) 187.5
(175 + 200) ÷ 2 = 187.5
1
18(c) 0.9
1 − 0.1 = 0.9
1
19(b) Both A and R are true, but R is not the correct explanation of A.
A is true: 2 of the 6 outcomes are favourable. R is true but does not explain A.
1
20(d) A is false, but R is true.
A is false: EC = AE × DB/AD = 9, not 10. R is true.
1
Section B: 5 questions of 2 marks each
21HCF = 3, LCM = 18; 3 × 18 = 54 = 6 × 9
  1. 6 = 2 × 3, 9 = 32 (2)
2
21 OR2 × 3 × 11
  1. divide by primes in turn: 2 × 3 × 11 (2)
2
22−3
  1. sum of zeroes = −b/a = −1, so the other zero = −1 − 2 = −3 (2)
2
231
  1. 3 ÷ 3 = 1 (2)
2
2480°
  1. OAPB has two right angles, so ∠APB = 180° − 100° = 80° (2)
2
25y = 20 or y = −4
  1. (−9)2 + (y − 8)2 = 225, so y − 8 = ±12 (2)
2
25 OR30 units
  1. sides 5, 12 and √(52 + 122) = 13 (2)
2
Section C: 6 questions of 3 marks each
26Suppose √3 = r/s, where r and s are integers with no common factor other than 1, and s ≠ 0. Then 3s2 = r2, so 3 divides r2, and since 3 is prime, 3 divides r. Write r = 3c: then 3s2 = 9c2, so s2 = 3c2, and 3 divides s as well. So 3 is a common factor of r and s, a contradiction. Hence √3 is irrational. Now suppose 7 − 8√3 = x, a rational number. Then √3 = (7 − x)/8, which is rational, since x, 7 and 8 are rational. This contradicts the irrationality of √3, so 7 − 8√3 is irrational.
  1. Assuming √3 = r/s (r, s coprime) and getting 3s2 = r2, so 3 divides r (1)
  2. Showing 3 also divides s, a contradiction, so √3 is irrational (1)
  3. Writing √3 = (7 − x)/8 and concluding that 7 − 8√3 is irrational (1)
3
27The number is 96.
  1. Let the tens digit be x and the units digit be y: x + y = 15 and (10x + y) − (10y + x) = 27, that is, x − y = 3 (1)
  2. Adding and subtracting the equations: x = 9, y = 6 (1)
  3. The number is 96 (1)
3
28OP = 123 cm; perimeter of OAPB = 294 cm
  1. OA ⟂ PA (radius ⟂ tangent), so △OAP is right-angled at A (1)
  2. OP = 123 cm (1)
  3. PB = PA, perimeter = 294 cm (1)
3
28 ORThe tangent at T is ⟂ OT and the tangent at G is ⟂ OG (radius ⟂ tangent). OT and OG lie along the same line TG, so both tangents are ⟂ TG, and two lines perpendicular to the same line are parallel. WD = WT and BD = BG (tangents from an external point), so WB = 15 + 16 = 31 cm
  1. Both tangents are ⟂ TG (radius ⟂ tangent) (1)
  2. Lines perpendicular to the same line are parallel (1)
  3. WB = 31 cm using equal tangents (1)
3
29PQ = QR = 3√2 and PQ2 + QR2 = PR2, so it is a right isosceles triangle, right-angled at Q
  1. PQ2 = 18, QR2 = 18, PR2 = 36 (1)
  2. PQ = QR = 3√2, so the triangle is isosceles (1)
  3. PQ2 + QR2 = 36 = PR2, so by the converse of Pythagoras' theorem it is right-angled at Q (1)
3
30(sec θ + tan θ)/(sec θ − tan θ) = 25
  1. cot θ = 5/12: take the adjacent side = 5k and the opposite side = 12k; then the hypotenuse = √(52 + 122)k = 13k, so sin θ = 12/13, cos θ = 5/13, tan θ = 12/5 (1)
  2. Substituting: (13/5 + 12/5)/(13/5 − 12/5) (1)
  3. = 25 (1)
3
31p = 3
  1. Class marks xi: 5, 15, 25, 35, 45; Σfi = 37 + p, Σfixi = 895 + 35p (1)
  2. (895 + 35p)/(37 + p) = 25 (1)
  3. Solving, p = 3 (1)
3
31 ORMean = 43
  1. Class marks xi: 32.5, 37.5, 42.5, 47.5, 52.5; fixi: 325, 150, 127.5, 855, 262.5 (1)
  2. Σfi = 40, Σfixi = 1,720 (1)
  3. Mean = Σfixi/Σfi = 1,720/40 = 43 (1)
3
Section D: 4 questions of 5 marks each
32(a) 2.4 m (b) 7.6 m
  1. Similar triangles identified (AA), with a correct ratio (1)
  2. Distance walked = 9.6 m and the proportion set up (1)
  3. Shadow = 2.4 m (1)
  4. Proportion for (b) (1)
  5. Distance = 7.6 m (1)
5
33Let the integers be x and x + 2: x2 + (x + 2)2 = 3700, that is, 2x2 + 4x − 3696 = 0, so x2 + 2x − 1848 = 0, so (x − 42)(x + 44) = 0, giving x = 42 or x = −44. The integers are positive, so x = −44 is rejected. The integers are 42 and 44.
  1. Forming the equation: Let the integers be x and x + 2: x2 + (x + 2)2 = 3700, that is, 2x2 + 4x − 3696 = 0 (2)
  2. Solving: x = 42 or x = −44 (2)
  3. Rejecting x = −44: the integers are 42 and 44 (1)
5
33 ORLet the son's present age be x years; the father's is x + 23. Then (x − 3)(x + 20) = 78, so x2 + 17x − 138 = 0, so (x − 6)(x + 23) = 0, giving x = 6 or x = −23. Age cannot be negative, so x = −23 is rejected. The son is 6 years old and the father is 29 years old.
  1. Forming the equation: Let the son's present age be x years; the father's is x + 23. Then (x − 3)(x + 20) = 78, so x2 + 17x − 138 = 0 (2)
  2. Solving: x = 6 or x = −23 (2)
  3. Rejecting x = −23: son 6 years, father 29 years (1)
5
34(i) 8,008 cm2; (ii) 63,242 2/3 ≈ 63,242.67 cm3
  1. Curved surface area of the hemisphere = 2πr2 = 2 × 22/7 × 282 = 4,928 cm2 (1)
  2. Curved surface area of the cone = πrl = 22/7 × 28 × 35 = 3,080 cm2 (1)
  3. (i) Total surface area = 4,928 + 3,080 = 8,008 cm2 (1)
  4. Volume = (1/3)πr2h + (2/3)πr3 = 17,248 + 45,994 2/3 (1)
  5. (ii) = 63,242 2/3 ≈ 63,242.67 cm3 (1)
5
35Width = 48(√3 + 1) m ≈ 131.0 m; the point is 48 m from the first bank and 48√3 m ≈ 83.0 m from the second
  1. Angles of depression equal the angles of elevation from the banks (alternate angles) (1)
  2. First distance = 48 m (1)
  3. Second distance = 48√3 m (1)
  4. Width = 48(√3 + 1) m (1)
  5. ≈ 131.0 m (√3 = 1.73) (1)
5
35 ORHeight = 23√3 m ≈ 39.8 m; the nearer point is 23 m from the foot
  1. Correct figure described and the two tan equations formed (1)
  2. Equations correct with tan 30° = 1/√3, tan 45° = 1, tan 60° = √3 (1)
  3. Solving for the unknown distance (1)
  4. Height = 23√3 m (1)
  5. ≈ 39.8 m (√3 = 1.73) (1)
5
Section E: 3 case-based questions of 4 marks each
36 (i)72°
  1. 360° ÷ 5 — 72° (1)
1
36 (ii)69.3 cm2
  1. Area = θ/360 × πr2 = 72/360 × 22/7 × 10.5 × 10.5 — 69.3 cm2 (1)
1
36 (iii)34.2 cm
  1. Arc = 72/360 × 2 × 22/7 × 10.5 = 13.2 cm (1)
  2. Perimeter = 13.2 + 2 × 10.5 = 34.2 cm (1)
2
36 (iii) OR138.6 cm2
  1. Slices left = 5 − 3 = 2 (1)
  2. Area left = 2 × 69.3 = 138.6 cm2 (1)
2
37 (i)48 seats
  1. a19 = a + (n − 1)d = 12 + 18 × 2 — 48 seats (1)
1
37 (ii)Row 11
  1. 12 + (n − 1) × 2 = 32 gives n − 1 = 10 — Row 11 (1)
1
37 (iii)900 seats
  1. S = n/2 [2a + (n − 1)d] = 25/2 [2 × 12 + 24 × 2] (1)
  2. = 900 seats (1)
2
37 (iii) OR152 seats
  1. S = 8/2 [2 × 12 + 7 × 2] (1)
  2. = 152 seats (1)
2
38 (i)5/6
  1. P = 200/240 = 5/6 — 5/6 (1)
1
38 (ii)79/80
  1. Good or minor defect = 200 + 37 = 237; P = 237/240 = 79/80 — 79/80 (1)
1
38 (iii)1/6
  1. Not good = 37 + 3 = 40 (1)
  2. P = 40/240 = 1/6 (1)
2
38 (iii) OR1/80
  1. Neither buys only when it has a major defect: 240 − 200 − 37 = 3 (1)
  2. P = 3/240 = 1/80 (1)
2