ICSE 2021 Maths Question Paper Solved Semester 1 for Class 10

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ICSE Class 10 Maths Question Paper 2021 Solved Semester 1

Maximum Marks: 40
Time Allowed: One and a half hours

• You will not be allowed to write during the first 10 minutes.
• This time is to be spent in reading the question paper.
• ALL QUESTIONS ARE COMPULSORY.
• The marks intended for questions are given in brackets [ ]

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Question 1.
If (x + 2) is a factor of polynomial x³ – kx² – 5x + 6, then the value ofk is : [1]
(a) 1
(b) 2
(c) 3
(d) -2
Solution:
(b) 2
[Here, (- 2)³ – k(- 2)² – 5(- 2) + 6 = 0
-8 – 4k + 10 + 6 = 0
4k = 8
k = 2]

Question 2.
The solution set of the inequation x – 3 ≥ – 5, x ∈ R is : [1]
(a) {x : x > – 2, x ∈ R}
(b) {x : x ≤ – 2, x ∈ R}
(c) {x : x ≥ -2, x ∈ R}
(d) {-2, -1, 0, 1, 2}
Solution:
(c) {x : x ≥ – 2, x ∈ R}
[Here, x – 3 ≥ -5, x ∈ R
x ≥ – 5 + 3
x ≥ -2]

Question 3.
The product AB of two matrices A and B is possible if : [1]
(a) A and B have the same number of rows.
(b) the number of columns of A is equal to the number of rows of B.
(c) the number of rows of A is equal to the number of columns of B.
(d) A and B have the same number of columns.
Solution:
(b) the number of columns of A is equal to the number of rows of B.

Question 4.
If 70, 75, 80, 85 are the first four terms of an Arithmetic Progression, then the 10^th term is : [1]
(a) 35
(b) 25
(c) 115
(d) 105
Solution:
(c) 115
[Here, a = 70, d = 75 – 70 = 5 and n = 10
∴ a₁₀ = a + 9 d
= 70 + 9 × 5
= 70 + 45
= 115]

Question 5.
The selling price of a shirt excluding GST is ₹ 800. If the rate of GST is 12%, then the total price of the shirt is : [1]
(a) ₹ 704
(b) ₹ 96
(c) ₹ 896
(d) ₹ 1848
Solution:
(c) ₹ 896
[Total price of the shirt = ₹ 800 + ₹ \(\frac{12}{100}\) × 800 = ₹ 896]

Question 6.
Which of the following quadratic equations has 2 and 3 as its roots ? [1]
(a) x² – 5x + 6 = 0
(b) x² + 5x + 6 = 0
(c) x² – 5x – 6 = 0
(d) x² + 5x – 6 = 0
Solution:
(a) x² – 5x + 6 = 0
[Required quadratic equation
= (x – 2) (x – 3)
= x² – 5x + 6]

Question 7.
If x, 5.4, 5, 9 are in proportion, then x is : [1]
(a) 3
(b) 9.72
(c) 25
(d) \(\frac{25}{3}\)
Solution:
(a) 3
[Here, x : 5.4 = 5 : 9
⇒ x = \(\frac{5 \times 5.4}{9}\) = 3]

Question 8.
Mohit opened a Recurring deposit account in a bank for 2 years. He deposits ₹ 1000 every month and receives ₹ 25500 on maturity. The interest he earned in 2 years is: [1]
(a) ₹ 13500
(b) ₹ 3000
(c) ₹ 24000
(d) ₹ 1500
Solution:
(d) ₹1500 ;
[Here, Interest earned in 2 years
= ₹ 25500 – ₹ 2 × 12 × 1000
= ₹ 25500 – ₹ 24000
= ₹ 1500]

Question 9.
In the given figure AB = 24 cm, AC = 18 cm, DE = 12 cm, DF = 9 cm and ∠BAC = ∠EDF. Then ∆ABC ~ ∆DEF by the condition : [1]

(a) AAA
(b) SAS
(c) SSS
(d) AAS
Solution:
(b) SAS
[Here, ∠BAC = ∠EDF, \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{24}{12}\) = 2, \(\frac{\mathrm{AC}}{\mathrm{DF}}\) = \(\frac{18}{9}\) = 2]

Question 10.
If A = \(\left[\begin{array}{ll}
5 & 10 \\
3 & -4
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), then AI is equal to : [1]
(a) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
5 & 10 \\
-3 & 4
\end{array}\right]\)
(c) \(\left[\begin{array}{rr}
5 & 10 \\
3 & -4
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
15 & 15 \\
-1 & -1
\end{array}\right]\)
Solution:
(c) \(\left[\begin{array}{rr}
5 & 10 \\
3 & -4
\end{array}\right]\)

Question 11.
The polynomial x³ – 2x² + ax + 12 when divided by (x + 1) leaves a remainder 20, then ‘a’ is equal to : [1]
(a) -31
(b) 9
(c) 11
(d) -11
Solution:
(b) 9
[Here, (- 1)³ – 2(- 1)² + a(- 1) + 12 = 20
-1 – 2 – a + 12 = 0 ⇒ a = 9]

Question 12.
In an Arithmetic Progression (A.P.) if, first term is 5, common difference is -3 and the n^th term is 7, then n is equal to . [1]
(a) 5
(b) 17
(c) -13
(d) 7
Solution:
(a) 5
[Here, a_n = – 7, a = 5, d = – 3
∴ a + (n – 1)d = -7
⇒ 5 + (n – 1) (-3) = – 7
⇒ n – 1 = \(\frac{-7-5}{-3}\) = 4
⇒ n = 1 + 4 = 5]

Question 13.
In the given figure PQ is parallel to TR, then by using condition of similarity: [1]

Solution:
(b) \(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O Q}{O T}\)
[Here, ∆PQO ~ ∆RTO by AA similarity .
axiom \(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O Q}{O T}\)].

Question 14.
If a, b, c and d are proportional, then \(\frac{a+b}{a-b}\) is equal to : [1]
(a) \(\frac{c}{d}\)
(b) \(\frac{d}{c}\)
(c) \(\frac{c-d}{c+d}\)
(d) \(\frac{c+d}{c-d}\)
Solution:
(d) \(\frac{c+d}{c-d}\)
[Here, a, b, c and d are in proportion
∴\(\frac{a}{b}\) = \(\frac{c}{d}\)
By using componendo and dividendo, we have
\(\frac{a+b}{a-b}\) = \(\frac{c+d}{c-d}\)]

Question 15.
The first four terms, of an Arithmetic Progression (A.P.), whose first term is 4 and common difference is-6 are: [1]
(a) 4, -10, -16, -22
(b) 4, 10, 16, 22
(c) 4, – 2, -8, -14
(d) 4, 2, 8, 14
Solution:
(c) 4, -2, -8, -14
[Here, a = 4 and d = – 6
∴ a₁ = 4, a₂ = 4 – 6 = – 2, a₃ = – 2 – 6 = – 8 and a₄ = – 8 – 6 = – 14]

Question 16.
One of the roots of the quadratic equation x² – 8x + 5 = 0 is 7.3166. The root of the equation correct to 4 significant figures is: [1]
(a) 7.3166
(b) 7.317
(c) 7.316
(d) 7.32
Solution:
(c) 7.316

Question 17.
(x + 2) and (x + 3) are two factors of the polynomial x³ + 6x² + 11x + 6. If this polynomial is completely factorised the result is: [2]
(a) (x – 2) (x + 3) (x + 1)
(b) (x + 2) (x – 3) (x – 1)
(c) (x + 2) (x + 3) (x – 1)
(d) (x + 2) (x + 3) (x + 1)
Solution:
(d) (x + 2)(x + 3)(x + 1)
[Here, (x + 2)(x + 3) = x² + 5x + 6

Question 18.
The sum of the first 20 terms of the Arithmetic Progression 2, 4, 6, 8, … is :
(a) 400
(b) 840
(c) 420
(d) 800
Solution:
(c) 420
[Here, a = 2, d = 4 – 2 = 2, n = 20
∴ S₂₀ = \(\frac{20}{2}\)[2 × 2 + (20 – 1)²]
= 10[4 + 38]
= 420]

Question 19.
The solution set on the number line of the linear inequation : 2y – 6 < y + 2 ≤ 2y, y ∈ N is:  [2]

Solution:

[Here, 2y – 6 < y + 2 ; y + 2 ≤ 2y, y ∈ N y < 8 ; 2 ≤ y ⇒ 2 ≤ y < 8; y ∈ N]

Question 20.
If x, y, z are in continued proportion, then (y² + z²) : (x² + y²) is equal to: [2]
(a) z : x
(b) x : z
(c) zx
(d) (y + z) : (x+y)
Solution:
(a) z : x
[Given that x, y, z are in continued proportion
∴ \(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\frac{\mathrm{y}}{\mathrm{z}}\) ⇒ y² = xz
Now \(\frac{y^2+z^2}{x^2+y^2}=\frac{x z+z^2}{x^2+x z}=\frac{z(x+z)}{x(x+z)}=\frac{z}{x}\) or z : x]

Question 21.
The marked price of an article is ₹ 5000. The shopkeeper gives a discount of 10%. If the rate of GST is 12%, then the amount paid by the customer including GST is: [2]
(a) ₹ 5040
(b) ₹ 6100
(c) ₹ 6272
(d) ₹ 6160
Solution:
Solution :
(a) ₹ 5040
[Amount paid by the customer
= (5000 – 10% of 5000) + 12% of (5000 – 10% of 5000)
= 4500 + 540 = ₹ 5040]

Question 22.
If A = \(\left[\begin{array}{ll}
3 & 5 \\
I & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 4 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
I & -I \\
2 & I
\end{array}\right]\), then 5A – BC is equal to :

Solution:

Question 23.
In the given figure ABCD is a trapezium in which DC is parallel to AB.
AB = 16 cm and DC = 8 cm, OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x – 1) cm.
Using the given information answer the following questions.

(i) From the given figure name the pair of similar triangles: [1]
(a) ∆OAB, ∆OBC
(b) ∆COD, ∆AOB
(c) ∆ADB, ∆ACB
(d) ∆COD, ∆COB
Solution:
(b) ∆COD, ∆AOB
[∵ ∆COD ~ ∆AOB by AA similarity axiom]

(ii) The corresponding proportional sides with respect to the pair of similar triangles obtained in (i) : [ 1 ]
(a) \(\frac{C D}{A B}=\frac{O C}{O A}=\frac{O D}{O B}\)
(b) \(\frac{A D}{B C}=\frac{O C}{O A}=\frac{O D}{O B}\)
(c) \(\frac{A D}{B C}=\frac{B D}{A C}=\frac{A B}{D C}\)
(d) \(\frac{O D}{O B}=\frac{C D}{C B}=\frac{O C}{O A}\)
Solution:
(a) \(\frac{C D}{A B}=\frac{O C}{O A}=\frac{O D}{O B}\)

(iii) The ratio of the sides of the pair of similar triangles is: [1]
(a) 1 : 3
(b) 1 : 2
(c) 2 : 3
(d) 3 : 1
Solution:
(b) 1 : 2
[Here, \(\frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{OD}}{\mathrm{OB}}=\frac{8 \mathrm{~cm}}{16 \mathrm{~cm}}=\frac{1}{2}\)]

(iv) Using the ratio of sides of the pair of similar triangles the values of x and y are respectively: [1]
(a) x = 4.6, y = 7
(b) x = 7, y = 7
(c) x = 6.5, y = 7
(d) x = 6.5, y = 2
Solution:
(c) x = 6.5, y = 7
[Now, \(\frac{x-1}{11}=\frac{5}{y+3}=\frac{1}{2}\)
2x – 2 = 11 and y + 3 = 10
x = \(\frac{13}{2}\) = 6.5 and y = 7]

Question 24.
7vvo cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then:

(i) the number of litres of diesel used by car X is : [ 1 ]
(a) \(\frac{72}{x-3}\) litres
(b) \(\frac{72}{x+3}\) litres
(c) \(\frac{72}{x}\) litres
(d) \(\frac{12}{x}\) litres
Solution:
(c) \(\frac{72}{x}\) litres

(ii) the number of litres of diesel used by car Y is : [ 1 ]
(a) \(\frac{72}{x-3}\) litres
(b) \(\frac{72}{x+3}\) litres
(c) \(\frac{72}{x}\) litres
(d) \(\frac{12}{x+3}\) litres
Solution:
(b) \(\frac{72}{x+3}\) litres

(iii) If car X used 4 litres of diesel more than car Y in the journey. then: [1]

Solution:
(c) \(\frac{72}{x}-\frac{72}{x+3}\) = 4

(iv) The amount of diesel used by the car X is : [1]
(a) 6 litres
(b) 12 litres
(c) 18 litres
(d) 24 litres
Solution:
(b) 12 litres
[Here, 72 \(\left(\frac{x+3-x}{x^2+3 x}\right)\) = 4
⇒ x² + 3x – 54 = 0
⇒ (x – 6) (x + 9) = 0
⇒ x = 6 as x ≠ -9]
Required number of litres = \(\frac{72}{6}\) = 12]

Question 25.
Joseph has a recurring deposit account in a bank for two years at the rate of 8% per annum simple interest.

(i) If at the time of maturity Joseph receives ₹ 2000 as interest, then the monthly instalment is: [1]
(a) ₹ 1200
(b) ₹ 600
(c) ₹ 1000
(d) ₹ 1600
Solution:
(c) ₹ 1000
[Here, P × \(\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\) = 1
⇒ P × \(\frac{24 \times 25}{24} \times \frac{8}{100}\) = 2000
⇒ P = 2000 ÷ 2 = 1000]

(ii) The total amount deposited in the bank: [1]
(a) ₹ 25000
(b) ₹ 24000
(c) ₹ 26000
(d) ₹ 23000
Solution:
(b) ₹ 24000
[Amount deposited = ₹ 1000 × 2 × 12
= ₹ 24000]

(iii) The amount Joseph receives on maturity is : [ 1 ]
(a) ₹ 27000
(b) ₹ 25000
(c) ₹ 26000
(d) ₹ 28000
Solution:
(c) ₹ 26000
[Amount on maturity = ₹ 24000 + ₹ 2000 = ₹ 26000]

(iv) If the monthly instalment is ₹ 100 and the rate of interest is 8%, in how many months Joseph will receive ₹ 52 as interest ? [1]
(a) 18
(b) 30
(c) 12
(d) 6
Solution:
(c) 12
[Here, P = ₹ 100, r = 8%, Interest = ₹ 52
Now, 100 × \(\frac{n(n+1)}{2 \times 12}\) \(\frac{8}{100}\) = 52
n(n + 1) = \(\frac{52 \times 2 \times 12}{8}\) = 156 = 12(13)
= 12(12 + 1)
Hence, n = 12