ICSE Specimen Class 10 Maths Sample Paper 2024 Solution
ICSE Board Class 10 Maths Sample Paper 2024 Solution www.cisce.org. New Sample Paper Solution ICSE Board 2024 Maths all Question Answers by Expert Teachers. Students of Class X who are preparing for Board Exam 2024 can solved here Maths Sample Paper.
SECTION A
Choose the correct answers to the questions from the given options.
(Do not copy the question, write the correct answers only.)
(i) = [−1, 2] = [1 −2; 0 3]
Which of the following operation is possible?
(a) A – B
(b) A + B
(c) AB
(d) BA
Solution :
(c) AB
(ii) If 2 + + 6 = ( − 2 )( − 3) for all values of x, then the value of k is:
(a) – 5
(b) – 3
(c) – 2
(d) 5
Solution :
(a) – 5
x2 + 5x + 6
x2 – 3x – 2x + 6
(iii) A retailer purchased an item for ₹1500 from a wholesaler and sells it to a customer at 10% profit. The sales are intra-state and the rate of GST is 10%. The amount of GST paid by the customer:
(a) ₹15
(b) ₹30
(c) ₹150
(d) ₹165
Solution :
(d) ₹165
Profit for the retailer
= 1500 × 10/100
= 150
Selling = 1500 + 150
= 1650
GST = 1650 × 10/100
= 165 Rs
(iv) If the roots of equation 2 − 6 + = 0 are real and distinct, then value of k is:
(a) > –9
(b) > –6
(c) < 6
(d) < 9
Solution :
(d) < 9
D > 0, b2 – 4ac > 0
(-6)2 – 4 × 1 × k > 0
36 – 4k > 0
36 > 4k
9 > k
(v) Which of the following is/are an Arithmetic Progression (A.P.)?
- 1, 4, 9, 16,……….
- √3, 2√3, 3√3, 4√3,………
- 8, 6, 4, 2,………
(a) only 1.
(b) only 2.
(c) only 2. and 3.
(d) all 1., 2. and 3.
Solution :
(c) only 2. and 3.
2√3 – √3 = √3
3√3 – 2√3 = √3
-2, -2, -2
(vi) The table shows the values of x and y, where x is proportional to y
What are the values of M and N?
x | 6 | 12 | N |
y | M | 18 | 6 |
(a) M = 4, N = 9
(b) M = 9, N = 3
(c) M = 9, N = 4
(d) M = 12, N = 0
Solution :
(c) M = 9, N = 4
2/3 = 6/M = N/6
2M = 18
M = 9
2/3 = N/6
3N = 12
N = 4
(vii) In the given diagram, ∆ABC ~ ∆PQR and AD/PS = 3/8. The value of AB : PQ is:
(a) 8 : 3
(b) 3 : 5
(c) 3 : 8
(d) 5 : 8
Solution :
(c) 3 : 8
(viii) A right angle triangle shaped piece of hard board is rotated completely about its
hypotenuse, as shown in the diagram. The solid so formed is always:
1.) a single cone
2.) a double cone
Which of the statement is valid?
(a) only 1.
(b) only 2.
(c) both 1. and 2.
(d) neither 1. nor 2.
Solution :
(b) only 2.
(ix) Event A: The sun will rise from east tomorrow.
Event B: It will rain on Monday.
Event C: February month has 29 days in a leap year.
Which of the above event(s) has probability equal to 1?
(a) all events A, B and C
(b) both events A and B
(c) both events B and C
(d) both events A and C
Solution :
(d) both events A and C
(x) The three vertices of a scalene triangle are always equidistant from a fixed point.
The point is:
(a) Orthocentre of the triangle.
(b) Incentre of the triangle.
(c) Circumcentre of the triangle.
(d) Centroid of the triangle.
Solution :
(c) Circumcentre of the triangle.
(xi) In a circle with radius R, the shortest distance between two parallel tangents is equalto∶
(a) R
(b) 2R
(c) 2πR
(d) πR
Solution :
(b) 2R
(xii) An observer at point E, which is at a certain distance from the lamp post AB, findsthe angle of elevation of top of lamp post from positions C, D and E as α, β and γ.It is given that B, C, D and E are along a straight line.Which of the following condition is satisfied?
(a) tanα> tan β
(b) tan β < tan γ
(c) tan γ > tan α
(d) tan α < tan β
(xiii) 1. Shares of company A, paying 12%, ₹100 shares are at ₹80.
2.) Shares of company B, paying 12%, ₹100 shares at ₹100.
3.) Shares of company C, paying 12%, ₹100 shares are at ₹120.
Shares of which company are at premium?
(a) Company A
(b) Company B
(c) Company C
(d) Company A and C
Solution :
(c) Company C
∵ M.V > F.V
(xiv) Which of the following equation represent a line passing through origin?
(a) 3x – 2y + 5 = 0
(b) 2x – 3y = 0
(c) x = 5
(d) y = –6
Solution :
(b) 2x – 3y = 0
an + by – c = 0
an + by = 0
(xv) For the given 25 variables: , , … … … … … .
Assertion (A): To find median of the given data, the variate needs to be arrangedin ascending or descending order.
Reason (R): The median is the central most term of the arranged data.
(a) A is true, R is false
(b) A is false, R is true
(c) both A and R are true
(d) both A and R are false
Solution :
(c) both A and R are true
Question 2
(i) Shown below is a horizontal water tank composed of a cylinder and two hemispheres. The tank is filled up to a height of 7 m. Find the surface area of the tank in contact with water.
Solution :
C.S.A of cylinder + 2 C.S.A of hemisphere
= 1/2 (2πrh + 2 × +2πr2)
= 1/2 2 × 22/7 × 20 + 2 × 2 × 22/7 × 7)
= 1/2 (2 × 22 × 20 + 4 × 22 × 7
= 440 + 44 × 7 = 1
= 440 + 308
= 748 m2
(ii) In a recurring deposit account for 2 years, the total amount deposited by a person is ₹ 9600. If the interest earned by him is one-twelfth of his total deposit, then find:
(a) the interest he earns.
(b) his monthly deposit.
(c) the rate of interest
Solution :
(a) Total = 9600,
Interest = 1/12 × 9600
= ₹ 800
(b) let the mothlydeposite is P
Then, 2y = 24 month
= 24 × P = 960
= P = 9600/24
= P = ₹ 400
(c) 9 = p × n (n + 1)/2 × 12 × r/100
= 800 = 400 × 24 × 25/2 × 12 × r/100
= 800 × 100r
= r = 800/100
= r = 8%
(iii) Find:
(a) (sin + )2
(b) ( + )2
Using the above results prove the following trigonometry identity.
(sin + )2 + (cos + sec )2 = 7 + 2 + 2θ
Solution :
(a) (sinθ + cosecθ)2 (a + b)2 = a2 + 2ab + b2
= sin2θ + cosec2θ + 2/sin2θ + 1 + cot2θ +2)
= sin2θ + cos2θ + 3
(b) (cosθ + secθ)2
= cos2θ + sec2θ + 2cosθ. secθ
= cos2θ + sec2θ + 2
= cos2θ + 1 + tan2θ + 2/cos2θ + tan2θ + 3
(sin + )2 + (cos + sec )2 = 7 + 2 + 2θ
= siin2θ + cot2θ + 3 + cos2θ tan2θ + 3
= 6 + 1 + tan2θ + cot2θ
= 7 + tan2θ + cot2θ
Question 3
(i) If a, b and c are in continued proportion, then prove that:
3a2 + 5ab + 7b2/3b2 + 5 bc + 7c2 = a/c
Solution :
3a2 + 5ab + 7b2/3b2 + 5 bc + 7c2 = a/c
a, b and c are in continued property
a/b = b/c = k
b = ck
b2 = ac
Now,
3a2 + 5ab + 7(ac)/3(ac) + 5bc + 7c2
= a(3a + 5b + 7c)/c (3a + 5b + 7c)
= 9/c
(ii) In the given diagram, O is the centre of circle circumscribing the ∆ABC. CD is perpendicular to chord AB. ∠OAC=32°. Find each of the unknown angles x, y and z.
Solution :
ΔAOC
= x + 32 + 32 = 180
= x + 64 = 180
= x = 180 – 64
= x = 116
y = 1/2x
y = 1/2 × 116
y = 58 ∵ (angle subistands by an arc at the cenrtre is double the angle subistands by it at the corresponding segment)
In ΔBOC
= y + z + 90 = 180
= 58 + z + 90 = 180
z = 180 – 90 – 58
z = 90 – 658
z = 32
(iii) Study the graph and answer each of the following:
(a) Name the curve plotted
(b) Total number of students
(c) The median marks
(d) Number of students scoring between 50 and 80 marks
Solution :
(a) Ogive
(b) 40
(c) 56
(d) 25
SECTION B
Questions 4
(i) If A = [4 -4; -4 4], find A2. If A2 = P A, then find the value of p.
Solution:
A = [4 -4; -4 4] find A2
A×A = [4 -4; -4 4] [4 -4; -4 4]
[16 -16; +16 -16 -16 16; -16 +16]
A2 = [32 -32; -32 32]
A2 = PA => [32 -32; -32 32] = P [4 -4; -4 4]
= [32 -32; -32 32] = [4p -4p; -4p 4p]
4p = 32
P = 8
(ii) Solve the given equation x2 – 4x – z = 0 and express your answer correct to two places of decimal.
(You may use mathematical tables for this question)
Solution:
x2 – 4x – z = 0
a = 1, b = -4, c = -2
x = -b ± √b2-4ac/2a
x = – (-4) ± √(-4)2 – 4×1×(-2)/2×1
x = 4±√16+8/2×1
x = 4±√24/2 => x = 4±√2×2×2×3/3
= 4±2√6/2
= 4±2√6/2
= 2 (2±√6)/2
x = 2 + √6, x = 2 – √6
= 2+2.45, x = 2-2.45
= 4.45 or – 0.45
(iii) In the given diagram, △ABC is right angled at B. BDFE is a rectangle.
AD = 6cm, CE = 4 cm and BC = 12cm
(a) Prove that △ADF ~△FEC
(b) Prove that △ADF ~ △ABC
(c) Find the length of FE
(d) Find area △ADF : area △ABC
Solution:
Let, ∠A = θ
Then,∠C = 90 – θ
In △CEF
∠C + E + ∠F = 180
90 – θ + 90 + ∠F = 180
∠F + 180 – θ = 180
∠F = θ
(i) In △ADF and △FEC
∠D = ∠E (each 90°)
∠A = ∠F
△ADF ~ △FEC (A.A)
(ii) In △ADF and △ABC
∠D = ∠B (Given)
∠A = ∠A (common)
△ADF ~ △ABC (A.A)
(iii) Same △ADF ~ △FEC
AD/FE = DF/EC = AF/FC
6/FE = 8/4
6/FE = 2
2FE = 6
FE = 3 cm
(d) In △ADF ~ △ABC
or△ADF/or △ABC = (DF/BC)2
= (8/12)2
= (2/3)2 = 4/9 => 4:9
Question 5
(i) Shown below is a table illustrating the monthly income distribution of a company with 100 employees.
Monthly Income (in ₹10, 000) | 0 – 4 | 4 – 8 | 8 – 12 | 12 – 16 | 16 – 20 | 20 – 24 |
Number of employees | 55 | 15 | 06 | 08 | 12 | 4 |
Using step- deviation method, find the mean monthly income of an employee.
Solution:
Monthly | Number | xi | di | fi di |
0 – 4 | 55 | 2 | -2 | -110 |
4 – 8 | 15 | 6 | -1 | -15 |
8 – 12 | 06 | 10 | 0 | 0 |
12 – 16 | 08 | 14 | 1 | 8 |
16 – 20 | 12 | 18 | 2 | 24 |
20 – 24 | 4 | 22 | 3 | 12 |
100 | -81 |
Mean = A + Σfi ui/Σfi× h
= 10 + -81/100 × 4
= 10 + -324/100
=> 10 – 3.24
= 6.76
(ii) The following bill shows the GST rate and the marked price of articles:
Vidhyut Electronics | ||||
S. No. | Item | Marked Price | Quantity | Rate of GST |
(a) | LED TV set | ₹ 12000 | 01 | 28% |
(b) | MP4 player | ₹ 5000 | 01 | 18% |
Find the total amount to be paid (including GST) for the above bill.
Solution :
LED Tv = 12000
Tax = 28%
Amount = 12000 × 28/100
= 3360
(a) 12000 + 3360 = 15360
(b) 5000 + 900 = 5900
So, 15360 + 5900
= 21260
(iii) In the given figure, O is the centre of the circle and AB is a tangent to the circle at B.
If ∠PQB=55°
(a) find the value of the angles x, y and z.
(b) prove that RB is parallel to PQ.
Solution :
ΔPOB ∠B = 90°, ∠O = 55°, ∠P = ?
∠P + ∠O + ∠B = 180°
∠P + 55 + 90 = 180 °
∠P = 180 – 145
∠P = 35°
Z = ∠P (angle in the alternate segment)
Z = 55°
∠ SOB = 2∠SPB (∵ angle subtends by chord at the center is double angle subtends by it at the remaining circle)
Y = 2 × 35 = 70°
X = ∠P (angle in the same segment)
X = 35°
Question 6
(i) There are three positive numbers in a Geometric Progression (G.P.) such that:
(a) their product is 3375
(b) the result of the product of first and second number added to the product of second andthird number is 750.Find the numbers.
Solution :
a/r, a, ar
Product of three number is = 3375
a/r × a × ar = 3375
= a3 = 3375
= a = ∛3375
= a = 3 × 5 = 15
(b) a/r . a + a. ar = 750
= 15/r . 15 k+ 15. 15 .r = 750
= 225/r + 225 .r = 750
= 225(1/r + r) = 750/225
= 750/225
= 30/9
= 10/3
3(1 + r2) = 10r
= 3r2 – 9r – r + 3 = 0
= 3r(r – 3) – 1 (r – 3) = 0
= r = 3 or r = 1/3
(i) a = 15, r = 3
15/3, 15, 15 × 3
= 5, 15, 45
(ii) a = 15, r = 1/3
= 15/1/3, 15, 15 × 1/3
= 45, 15, 5
(ii) The table given below shows the ages of members of a society.
Age (in years) | Number of Members of the Society |
25 – 35 | 05 |
35 – 45 | 32 |
45 – 55 | 69 |
55 – 65 | 80 |
65 – 75 | 61 |
75 – 85 | 13 |
(iii) A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius ofthe cylindrical part are 7 m each and the total height of the tent is 14 m. Find the:
(a) quantity of air contained inside the tent.
(b) radius of a sphere whose volume is equal to the quantity of air inside the tent.
Solution :
Volume of tent = volume of cylinder + volume of ore
= πr²h + 1/3 (πr²h)
= π (7)² × h + 1/3 π 7²h
= 22/7 × 7 × 7 + 1/3 × 22/7 × 7 × 7
= 22 × 49 + 1/3 × 22 × 49
= 22 × 49 (1 + 1/3)
= 22 × 49 (4/3)
= 4/3 × 22 × 49 cm3
Volume of sphere, 4/3 × 22 × 49
= 4/3 πr3 = 4/5 × 22 × 49
= 22/7 × r3 = 22 × 49
= r3 = 22 × 49 × 7/22
= r = 3√49 × 7
= 7 cm
Question 7
(i) The line segment joining A(2,-3) and B(-3, 2) is intercepted by the -axis at the point Mand the y axis at the point N. PQ is perpendicular to AB produced at R and meets the y- axisat a distance of 6 units from the origin O, as shown in the diagram, at S. Find the:
(a) coordinates of M and N
(b) coordinates of S
(c) slope of AB.
(d) equation of line PQ
Solution :
(a) M = (-1, 0) and N = 0,-1)
(b) S = 0, 6)
(c) Slop of AB,
Slop = y2 – y1/x2 – x1
Slop of line AB = -3 – 2/2 – (-3)
= -5/2 + 3
= -5/5
= – 1
(d) Let the slop of PO be M,
Slop of PO × Slop of AB = – 1
M × -1 = -1
M, = -1/-1
M = 1
line PO passes through point P (0, 6)
(y – y1) m(x – x)
(y – 6) = 1 (x – 0)
Y – 6 = x
x – y + 6 = 0
(ii) The angle of depression of two ships A and B on opposite sides of a light house of height100m are respectively 42oand 54o. The line joining the two ships passes through the footof the lighthouse.
(a) Find the distance between the two ships A and B.
(b) Give your final answer correct to the nearest whole number.
(Use mathematical tables for this question)
Solution :
In ΔADC
tan 54 = CD/AD
1.3764 = 100/AD
1.3764 × AD = 100
AD = 10/1.3764
AD = 100/1.37
AD = 10000/134
tan 42 = CD/BD
= 0.9004 = 100/BD
= BD = 100/0.9004
= 1000/0.90 = 1000/90 = 1000/9
AB = AD + BD
Question 8 :
(i) Solve the following inequation write the solution set and represent it on the real number line.
3 – 2x ≥ x + 1 – x/3 > 2x/5, x ∈ R
Solution :
3 – 2x ≥ x + 1 – x/3 > 2x/5, x ∈ R
= 3 – 2x ≥ x + 1 – x/3
= 3 – 2x ≥ 3x + 1 – x/3
= 3 – 2x ≥, 2x + 1/3
= 9 – 6x ≥, 2x + 1
= 9 – 1 ≥, 2x – 6x
= 8 ≥ 8x
= 8/8 ≥ x
= 1 ≥ x
Now,
x + 1 – x/3 > 2x/5
= 3x + 1 – x/3 > 2x/5
= 2x + 1/3 > 2x/5
= 10x + 5 > 6x
= 4x > – 5
x > -5/4
∴x∈ (-5/4 < x ≤ 1)
(ii) ABCD is a cyclic quadrilateral in which BC = CD and EF is a tangent at A.
∠CBD = 43° and ∠ADB = 62°. Find:
(a) ∠ADC
(b) ∠ABD
(c) ∠FAD
Solution :
(i) ∠ADC = 62 + 43 = 105
(ii) ∠ABC + ∠ABD = 180° (opposite angle of
∠ABC + 105 = 180
∠ABC + 108 – 105
∠ABC = 75°
∠ABD + DBC = 75
∠ABD + 43 = 75
∠ABD = 75 – 43
∠ABD = 32°
∠FAD = ∠ABD (Only in the alternate segment)
∠FAD = 32°
(iii) A (a, b), B(-4, 3) and C(8,-6)are the vertices of a ∆ABC. Point D is on BC such that BD : DC is 2 : 1 and M (6, 0) is mid point of AD. Find:
(a) coordinates of point D.
(b) coordinates of point A.
(c) equation of a line passing through M and parallel to line BC.
Solution :
(i)Co ordinate of D = (2 × 3 – 4/2 + 1, -12 + 3/2 + 1)
= 12/3, -9/3 = 4, -3)_
(ii) (0, 6) (6, 0) 4, -3)
mid point = (x1 + x2/2, y1 + y2/2)
(6, 0) = (a + 4/2, b – 3/7)
= a + 4/2 = 6
= a + 4 = 12
a = 8
and b – 3/2 = 0
b – 3/2 = 0
b – 3 = 0
b = 1
Therefore, Co – ordinate of AB = (8, 3)
(iii) Slope of BC = y2 – y1/x2 – x1
= -6 – 3/8 – (-4)
= -9/-2
= -3/4
Since the lines parallel to BC therefore y
slope of line = -3/4
Square of line having slop = -3/4 and planethrough M (6, 0)
(y – y1) = (x – x1)
(y – 0) = -3/4 (x – 6)
y = -3/4 (x – 6)
4y = -3/x + 18
∴3x + 4y – 18 = 0
Question 9
(i) Using componendo and dividend, find the value of x2 when:
x3 + 3x/3x2 + 1 = 14/13
Solution :
x3 + 3x/3×2 + 1 = 14/13
= x3 + 3x + 3×2 + 1/x3 + 3x -3×2 – 1 = 14 + 13/14 – 13
= (x + 1)3/(x – 1)3 = 27/1
= ∛((x + 1)3/(x – 1)3
= ∛27/1
= x + /x – 1 = 3/1
= x + 1 + x – 1/x + 1 – (x – 1) = 3 + 1/3 – 1
= 2x/x + 1 – x + 1 = 4/2 = 2
2x /2 = 2
∴x = 2
(ii) The total expense of a trip for certain number of people is ₹18000. If three more people join them, then the share of each reduces by ₹3000. Taking x to be the original number of people, form a quadratic equation in x and solve it to find the value of x.
Solution :
₹ 18000
Number of people = x
(i) share of each people = 18000/x
(ii) Then number of people = x + 3
then share of each people = 18000/ x + 3
18000/x + 18000/x + 3 = 3000
18000(1/x – 1/x + 3) = 3000
x + 3 – x/x(x + 1) = 3000/18000
= 3/x2 + 3x = 1/6
= 18 = x2 + 3x
= x2 + 3x – 18 = 0
= x2 + 6x – 3x – 18 = 0
= x(x + 6) – 3 (x + 6) = 0
= (x – 3) (x + 6) = 0
= x = 3 or x = 6
Question 10
(i) Using remainder and factor theorem, factorize completely, the given polynomial :
2 3 − 9 2 + 7 + 6
Solution :
2x3 − 9x 2 + 7x + 6
= 2x3 – 9x2 + 7x + 6
= 2 × 8 – 9 × 4 + 14 + 6
= 16 – 36 – + 14 + 6
= 36 – 36 = 0
f(2) = 0
(x – 2) is the factor
f(x)
= 0
Other factory, (fn)
2x2 – 5x – 3 = 2x2 + 6x + x – 3
= 2x(x – 3) + 1 (x – 3)
= (2x + 1) (x – 3)
(ii) Each of the letter of the word “HOUSEWARMING”” is written on cards and put in a bag.
If a card is drawn at random from the bag after shuffling, what is the probability that theletter on the card is∶
(a) a vowel
(b) one of the letters of the word SEWING.
(c) not a letter from the word WEAR.
Solution :
(a) 5/12
(b) 6/12 = 1/2
(c) 8/12 = 2/3
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