Students can refer to the following Sample Paper ICSE Class 10 Mathematics Set I with Answers provided below based on the latest syllabus and examination guidelines issued for ICSE Mathematics. All specimen papers have been prepared covering all chapters given in ICSE Mathematics book for Class 10. You should also refer to ICSE Class 10 Mathematics Solutions.
Sample Paper ICSE Class 10 Mathematics Set I with Answers
Class–X
Mathematics
(Maximum Marks : 80)
(Time allowed : Two and half hours)
Instructions :
(i) Attempt all questions from Section A and any four questions from Section B.
(ii) All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.
(iii) Omission of essential working will result in loss of marks.
(iv) The intended marks for questions or parts or questions are given in brackets [].
(v) Mathematical tables are provided.
SECTION A
Question 1
(a) Find the value of ‘k’ if 4x3 − 2x2 + kx + 5 leaves
remainder – 10 when divided by 2x + 1. [3]
(b) Amit deposits 1600 per month in a bank for 18 months in a recurring deposit account. If he gets 31,080 at the time of maturity, what is the rate of interest per annum? [3]
(c) The price of an article is ` 9350 which includes VAT at 10%. Find how much less a customer pays for the article, if the VAT on the article decreases by 3%. [4]
Sol:
(a) f(x) = 4x3 – 2x2 + kx + 5
2x + 1 = 0
2x = 0 – 1
2x = – 1
x=-(1/2)
As given in question when 4x3 – 2x2 + kx + 5 divided by 2x + 1 leaves remainder – 10.
f(-1/2) =-10
4(-1/2)3-2(-1/2)2+k(-1/2)+5 =-10
4x(-1/8×)-2x(1/4)-(k/2)+5 =-10
-(1/2)-(1/2)-(k/2)+5 =-10
-1-(k/5)+ =– 10
-(k/2) = – 10 – 14
-(k/2) = – 14
k = – 14 × – 2 Þ k = 28
(b) P = ` 1600, x = 18 months, M.V. = 31080, r = ?
M.P. = P X n+PXx(x+1)r/2400
31080 =1600×18(1600x18x19xr/2400)
2280 = 2 × 6 × 19r
r =2280/12×19
r =10%
(c) Let S.P. =₹x
X+Xx10/100 =9350
X = 9350×100/110
X =₹8500
Customer pays 8500X3/100 =255 less if VAT decreases by 3%
Question 2
(a) Solve the following inequalities and represent your solution on the real number line :
(b) Find the 16th term of the A.P. 7, 11, 15, 19…. Find the sum of the first 6 terms. [3]
(c) In the given figure CE is a tangent to the circle at point C. ABCD is a cyclic quadrilateral. If
ÐABC = 93° and ÐDCE = 35°.
Find : (i) ∠ADC, (ii) ∠CAD, (iii) ∠ACD
Sol:
(b) a = 7, d = 11 – 7 = 4, n = 16
Tn = a + (n – 1)d
T16 = 7 + 15 × 4
T16 = 7 + 60 = 67
Sn = N/2[2 + ( − 1) ]
S6 = 6/2(2 × 7 + 5 × 4)
= 3(34) = 102
(c) In given fig. EC is tangent and touches circle at C and ABCD is cyclic quadrilateral
∠B + ∠D = 180° (Sum D opp. of cyclic quadrilateral is 180°)
93° + ∠D = 180°
∠D = 180° – 93° = 87°
∠ADC = 87°
Now, OC is radius and CF is tangent
∠ACD + ∠DCE = 90°
∠ACD = 90° – ∠DCE
∠ACD = 90° – 35° = 55°
Now, In ΔACD
∠ACD + ∠CDA + ∠DAC = 180°
55° + 87° + ∠DAC = 180°
∠DAC = 180° – 142° = 38°
Question 3 (a) Prove the following identity :
(c) For what value of ‘k’ will the following quadratic equation :
(k + 1)x2 − 4kx + 9 = 0 have real and equal roots?
Solve the equations.
Soll:
– 18 – y = – 6
– y = – 6 + 18
– y = 12
y = – 12
Now, 3x – 6 = 0
3x = 6
X= 6/3
x = 2
(c) (k + 1)x2 – 4kx + 9 = 0 for real and equal roots
D = 0
b2 – 4ac = 0
(– 4k)2 – 4×(k + 1)(9) = 0
16k2 – 36(k + 1) = 0
16k2 – 36k – 36 = 0
8k2 – 18k – 18 = 0
4k2 – 9k – 9 = 0
4k2 – 12k + 3k – 9 = 0
4k(k – 3) + 3(k – 3) = 0
(4k + 3)(k – 3) = 0
k = 3 or k =-(3/4)
4x2 – 12x + 9 = 0
4x2 – 6x – 6x + 9 = 0
2x(2x – 3) – 3(2x – 3) = 0
(2x – 3)(2x – 3) = 0
x =3/4
Question 4
(a) A box consists of 4 red, 5 black and 6 white balls. One ball is drawn out at random. Find the probability
that the ball drawn is :
(i) black (ii) red or white [3]
(b) Calculate the median and mode for the following distribution :
(c) A solid cylinder of radius 7 cm and height 14 cm is melted and recast into solid spheres each of radius 3.5 cm. Find the number of spheres formed.
Sol:
(a) Total balls = 4 red + 5 black + 6 white = 15
(i) P(black balls) = 5/15=1/3
(ii) P(red or white) = 10/15=2/3
Median = (1+n/2)th observation
= (17+1/2)th=(18/2)th
= 9th = 52
(ii) highest frequency is 5, hence mode = 52.
(c) Cylinder (r’) = Radius of cylinder
= 7 cm
Height = 14 cm
Circle r = radius = 3·5
Vol. of cylinder/Vol. of sphere = No. of sphere formed
πr2h/(4/3πr3) =7 x 7 x 14 x 3 x 1000/4 x 35 x 35 x 35
= 12
SECTION B
Question 5
(a) The 2nd and 45th term of an arithmetic progression are 10 and 96 respectively. Find the first term and
(b) If A =
find matrix B such that A2 – 2B =3A + 5I where I is a 2 x 2 identity matrix. [3]
(c) With the help of a graph paper, taking 1 cm=1 unit along both x and y axis :
(i) Plot points A (0, 3), B (2, 3), C (3, 0), D (2, – 3), E (0, -3).
(ii) Reflect points B, C and D on the y axis and name them as B’, C’ and D’ respectively.
(iii) Write the co-ordinates of B’, C’ and D’.
(iv) Write the equation of line B’ D’.
(v) Name the figure BCDD’C’B’B.
Sol:
(a) Given, T2 = 10
a + (2 – 1)d = 10
a + d = 10 …(i)
and T45 = 96
a + 44d = 96 …(ii)
on subtracting eqn. (i) from (ii),
a + 44d – a – d = 96 – 10
43d = 86
d = 2
by putting d = 2 in eqn. (i)
a + 2 = 10
a = 8
Now, S15 =15/2(2 × 8 + 14 × 2)
=15/2(16 + 28)
=15/2(44)
= 15 × 22 = 330
Hence, first term, a = 8
Common difference, d = 2
Sum of first 15 terms S15 = 330
(b)
(ii) See the graph as reflection of points B, C and D are marked as B’, C’ and D’ on the y-axis.
(iii) B'(– 2, 3)
C'(– 3, 0)
D'(– 2, – 3)
(iv) x = – 2
(v) Hexagon
Question 6
(a) In ΔABC and ΔEDC, AB is parallel to ED. BD =1/3 BCand AB = 12.3 cm.
(i) Prove that DABC ~ DEDC
(ii) Find DE
(iii) Find : area of EDC / area of ABC
(b) Find the ratio in which the line joining (– 2, 5) and (– 5, – 6) is divided by the line y = – 3. Hence find the point of intersection. [3]
(c) The given solid figure is a cylinder surmounted by a cone. The diameter of the base of the cylinder is 6 cm. The height of the cone is 4 cm and the total height of the solid is 25 cm. Take π =22/7
Find the :
(i) Volume of the solid
(ii) Curved surface area of the solid
Give your answers correct to the nearest whole number
Sol: (a) (i) Given : In fig. AB || ED and ED·BD =1/3 BC and AB = 12·3 cm.
To prove : ∠ABC ~ ∠EDC
Proof : In ΔABC & ΔEDC
∠C = ∠C
∠CAB = ∠CED
ΔABC ~ ΔEDC
AB/ED=BC/DC=AC/EC
(ii) In ΔABC and ΔEDC CD/BC=DE/AB
BC-BD/BC=DE/AB
3BD- BD/BC=DE/AB
2BD/3BD=DE/AB
2/3=DE/AB
2/3=DE/12.3
DE=2 x12.3/3
DE = 8·2 cm
(iii) area of EDC/area ofABC=ED2/AB2
=(8.2 )2/(12.3)2=67.24/151.29=0.44
Let the line y = – 3 divide the line joining A and B in k : 1
Now, y =-6k+5/k+1
and also, y = – 3
-6k+5/k+1= – 3
– 6k + 5 = – 3k – 3
– 3k = – 8
The line y = – 3 divides the coordinate in 8:3 ratio.
x =(8x-6)+(2×1)/8+3
=-48+2/11=− 46/11
and y = – 3
coordinates of point of intersection
=(-46/11,-3)
(c) (i) Cone h = 4 cm
r = 3 cm
Cylinder h’ = 25 – 4 = 21 cm
r = 3 cm
Volume of solid = Volume of cone + volume of cylinder
=1/3πr2h+πr2h
=4422/7
= 631·71 = 632 cm3
C.S.A of the solid = C.S.A. of cone + C.S.A of cylinder
= πrl + 2πrh’
= 47·14 + 396
= 443·14 cm
Question 7
(a) In the given figure, PAB is a secant and PT a tangent to the circle with centre O.
If ∠ATP = 40°, PA = 9 cm and AB = 7 cm
Find :
(i) ÐAPT
(ii) length of PT [3]
(b) The 1st and the 8th term of a GP are 4 and 512 respectively. Find :
(i) the common ratio
(ii) the sum of its first 5 terms. [3]
(c) The mean of the following distribution is 49. Find the missing frequency ‘a’.
Sol:
(a) (i) ∠BAT = 90° [Angle in semi-circle]
∠PAT = 90°
In ΔPAT,
∠PAT + ∠PTA + ∠APT = 180° [Angle sum property of D]
90° + 40° + ∠APT = 180°
130°+ ∠APT = 180°
∠APT = 50°
(ii) PT2 = PA × PB
= 9 × 16
PT2 = 144
PT = 12 cm
a = 4
T8 = 512
ar8–1 = 512
ar7 = 512
r7 = 512/4
r7 = 128
r7 = (2)7
r = 2
(i) Common ratio (r) = 2
(ii) S5 = ?, a = 4, r = 2, n = 5
Sn = a(rn-1)/(r-1)
= 4(25-1)/(2-1)
= 4(32-1)/(1)
= 4(31)=124
(75 + a) – 1 = (– 30 + a)h
(– 30 + a)20 = – 75 – a
⇒ – 600 + 20a = – 75 – a
⇒ 21a = – 75 + 600
⇒ a = 525/21
a = 25
Question 8
(a) Prove the following identity :
(sinA + cosecA)2 + (cosA + secA)2 = 5 + sec2A· cosec2A. [3]
(b) Find the equation of the perpendicular bisector of line segment joining A(4, 2) and B(– 3, – 5). [3]
(c) Using properties of proportion, find x : y if : x3+12x/6x2+8 = 1f3+12y/9y2+27
Sol:
(a) sin2A + cosec2A + 2sinAcosecA + cos2A + sec2A + 2cosA secA
sin2A + cos2A + 2sinA ×1/sinA+ 2cosA×1/cosA+ cosec2A + sec2A
1 + 2 + 2 + cosec2A + sec2A
5+1/sin2 A+1/cos2 A
5_1/sin Acos A
5 + cosec2A·sec2A
sin Acos A
5 + cosec2A·sec2A
=-5-2/-3-4
=-7/-7=1
Slope of line perpendicular to line AB
m2 = – 1
Equation of line passing through point P which is mid-point of AB and perpendicular to AB :y=(-3/2)=(-1)(X-1/2)
y+3/2=-x+1/2
y + x =1/2-3/2
y + x =-2/2
y + x = – 1
y + x + 1 =0
(C) x3+12 x / 6x2+8 = y3+27y / 9y2+27
Applying componendo and dividendo
x3+12x+6x2+8/x3+12 x -6x2-8 = y3+27y+9y2+27/y3+27y-9y2-27
(x)3 + 3 (2)2 (x) + 3(2)(x)2 + (2)3 / (x)3 – 3(2)2(x) – 3(2)(x)2 – (2)3 = y3 + 3(3)2y + 3(3)y2 + (3)3 / y3 + 3(3)2y – 3(3)(y)2 – (3)3
(x+2)3/(x-2)3 = (y+3)3/(y-3)3
On taking cube root both sides, we get
(x+2)/(x-2) = (y+3)/(y-3)
Again applying componendo and dividendo
(x+2)+(x-2) / (x+2)-(x-2) = (y+3)+(y-3) /(y+3)-(y-3)
x+2+x-2 / x+2-x-2 = y+3+y-3 / y+3 -y+3
2x/4 = 2y/6
x/2 = y/3
3x = 2y
x/y = 2/3
x:y = 2:3
Question 9
(a) The difference of the squares of two natural numbers is 84. The square of the larger number is 25 times the smaller number. Find the numbers. [4]
(b) The following table shows the distribution of marks in Mathematics :
With the help of a graph paper, taking 2 cm = 10 units along one axis and 2 cm = 20 units along the other axis, plot an ogive for the above distribution and use it to find the :
(i) median.
(ii) number of students who scored distinction marks (75% and above)
(iii) number of students, who passed the examination if pass marks is 35%.
Sol:
(a) Let the two number be x and y, x > y
x2 – y2 = 84 …(i)
x2 = 25y …(ii)
Putting eqn. (ii) in (i)
25y – y2= 84
y2 – 25y + 84 = 0
y2 – 21y – 4y + 84 = 0
y(y – 21) – 4(y – 21) = 0
(y – 21)(y – 4) = 0
or y = 21, y = 4
Taking y = 4
x = 10
On the graph paper, we plot the following points A(10, 7), B(20, 28), C(30, 54), D(40, 71), E(50, 84), F(60, 105), G(70, 147) and H(80, 180). Join all these point by free hand.
Question 10
(a) Prove that two tangents drawn from an external point to a circle are of equal length. [3]
(b) From the given figure find the :
(i) Coordinates of points P, Q, R.
(ii) Equation of the line through P and parallel to QR.
(c) A manufacturer sells an article to a wholesaler with marked price ` 2000 at a discount of 20% on the marked price. The wholesaler sells it to a retailer at a discount of 12% on the marked price. The retailer sells the article at the marked price. If the VAT paid by the wholesaler is ` 11.20, find the :
(i) Rate of VAT
(ii) VAT paid by the retailer.
(a) Given : In fig. AP and PB are the two tangents.
To prove : PA = PB
Proof : In fig. OA ^ AP
∠PAO = 90°
(The angle between radius and tangent at the point of contact is 90°)
Similarly ∠PBO = 90°
In ΔPAO & ΔPBO OP = OP (Common)
OA = OB (Radius)
∠PAO = ∠PBO (each 90°)
ΔPAO ∠PBO [by R.H.S.]
∴ PA = PB (by c.p.c.t)
(b) (i) Coordinates of P, Q, R
P ⇒ (4, – 3)
Q ⇒ (– 2, – 2)
R ⇒ (2, 0)
(ii) Slope of line QR, m1 = 0 + 2 / 2+ 2 = 2/4 = 1/2
Slope of line, passing through P and parallel to line QR, m2 = m1 = 1/2
Equation of line
y-(-3) = 1/2(x-4)
y+3 = 1/2(x-4)
2y + 6 = x – 4
2y – x = – 12
2y – x + 12 = 0
(c) Marked price of an article = ₹2000
Manufacturer gives a discount to the wholesaler = 20% on marked price
∴ Selling price of manufacturer = Cost-price for wholesaler
= 2000 – 20/100×2000
= 2000 – 400 = ₹1600
Selling price of wholesaler = cost price for retailer
= 2000 – 20/100×2000
[Wholesaler gives a discount of 12% to retailer]
= 2000 – 240
= ₹ 1760
∴ Selling price of retailer = cost price of customer
= 2000
11·20 =R/100(1760-1600)
R=7%
(ii) Government received a VAT from the retailer
=7/100x(2000-1760)
=7/100×240
= ₹16·8
Question 11
(a) Mr. Sharma receives an annual income of ` 900 in buying ` 50 shares selling at ` 80. If the dividend declared is 20%, find the :
(i) Amount invested by Mr. Sharma.
(ii) Percentage return on his investment. [3]
(b) Two poles AB and PQ are standing opposite each other on either side of a road 200 m wide. From a point R between them on the road, the angles of elevation of the top of the poles AB and PQ are 45° and 40° respectively. If height of AB = 80 m, find the height of PQ correct to the nearest metre. [3]
(c) Construct a triangle PQR, given RQ = 10 cm,
ÐPRQ = 75° and base RP = 8 cm.
Find by construction :
(i) The locus of points which are equidistant from QR and QP.
(ii) The locus of points which are equidistant from P and Q.
(iii) Mark the point O which satisfies conditions (i) and (ii).
Sol:
(a) Income = 900, F.V. = 50, M.V. = 80 dividend = 20%
Income = No. of shares × F.V. ×dividend/100
900 = No of shares × 50 ×20/100
No. of shares =900×100/50×20= 90 shares
(i) Total Investment by Mr. Sharma = No. of Shares × M.V.
= 90 × 80
= ₹7200
(ii) Percentage return =(Income/Investment)× 100
= (900/7200)× 100 = 12·5%
In fig. AB and PQ are two poles standing opposite to each other and distance between them is 200 m.
Now, In ΔABR
tan 45° =AB/BR
1 =AB/BR
BR = AB = 80
In ΔPQR, tan 40° =x/120
x = 120 × tan 40°
= 120 × 0·83909
= 100·68 m
x = 100 m
1. Draw a line segment PR = 8 cm.
2. From the point R, draw ∠QRP = 75°.
3. Taking radius 10 cm, with R as centre cut RQ = 10 cm.
4. Join PQ, Here ΔPQR is required triangle.
(i) Draw angle bisector of ∠PQR
(ii) Perpendicular bisector of line PQ.
(ii) Point O