# Maths Semester-1 ICSE Specimen Paper Solved Class-10

Maths Semester-1 ICSE Specimen Paper Solved Class-10 for practice.  Step by step solutions of ICSE Class-10 specimen model sample paper .  During solutions of semester-1 Maths specimen paper we explain with figure , graph, table whenever necessary so that student can achieve their goal in next upcoming exam of council .

## Maths Semester-1 ICSE Specimen Paper Solved Class-10 for practice

 Board ICSE Class 10th (x) Subject Maths Topic Semester-1 ICSE Specimen Paper Solved Syllabus on bifurcated syllabus (after reduction) session 2021-22 Question Type MCQ/ Objective (as prescribe by council) Total question section-A=16 Section-B=6 Section-C=3 Total=25 Max mark section-A=1 mark x16 Que =16 Section-B=2 mark x 6 Que=12 Section-C=4 mark x 3  Que= 12 Total         =25

Click Below  for Answer Key of Board Paper

Answer Key of Sem-1 ICSE Maths Board Paper (6th December)

### ICSE Maths Specimen Paper Solution Class-10 for Semester-1 (below)

Section-A (12 mark)

Warning – please view Question Paper first before viewing solution

Question-1

If matrix A is of order 3 x 2 and matrix B is of order 2 x 2 then the matrix AB is of order

(a) 3 x 2 (b) 3 x 1 (c) 2 x 3 (d) 1 x 3

Solution

Correct option is a)   3×2

Explanation

Order of A:3×2

Order of B:2×2

Multiplication of matrices is possible if and only if the number of columns of first matrix is equal to the number of rows of second matrix

In AB

No. of columns in A is 2

No. of rows in B is 2

∴AB exists

Order of AB is ( number of rows of A x number of columns of B)

∴ Order of AB is (3×2)

#### Question-2

The percentage share of SGST of total GST for an Intra-State sale of an article is
(a) 25% (b) 50% (c) 75% (d) 100%

Solution

option (b) 50%

Explanation

if there is intrastate selling then total gst earn distribute equally to state SGST and central CGST

#### Question-3

𝐴𝐵𝐶𝐷 𝑖𝑠 𝑎 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑤𝑖𝑡ℎ 𝐴𝐵 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐷𝐶. 𝑇ℎ𝑒𝑛 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 ∆𝐴𝑂𝐵

(a) ∆𝐴𝐷𝐵 (b) ∆𝐴𝐶𝐵 (c) ∆𝐶𝑂𝐷 (d) ∆𝐶𝑂D

Solutions

option (c) ∆𝐶𝑂D is correct

Explanation

in ∆ AOB and in ∆ COD

 Statement reason angle BAC=  angle OCD alternate angle ABO=  angle ODC alternate So ∆ AOB similar ∆ COD by A.A.

#### Question-4

The mean proportion between 9 and 16 is
(a) 25 (b) 144 (c)7 (d) 12

Solution

option (d) 12 is correct

explanation

when A,B,C are in continued proportion then B is said to be the mean proportional.

let us take A=9 and C=16 and B as X. then

A:X : : X:C

9:X : : X:16

Therefore, x = 12.

So, the mean proportional of 9 and 16 is 12.

#### Question-5

A man deposited ₹ 500 per month for 6 months and received ₹3300 as the
maturity value. The interest received by him is: –
(a) 1950 (b) 300 (c) 2800 (d) none of these

Solution

option (b) is correct

explanation

total money deposited in 6 month= 500 x 6= 3000

so The interest received by him is= 3300-3000=300

#### Question-6

The solution set representing the following number line is….
(a) {x: x ∈ R, -3 ≤ x < 2}
(b) {x: x ∈ R, -3 < x < 2}
(c) {x: x ∈ R, -3 < x ≤ 2}
(d) {x: x ∈ R, -3 ≤ x ≤2}

Solution

option (a) {x: x ∈ R, -3 ≤ x < 2} is correct

explanation

shaded dot over 3 show that 3 also include while hollow dot over 2 show that 2 is not included but less than 2

#### Question-7

The first three terms of an arithmetic progression (A. P.) are 1, 9, 17, then the next two terms are
(a) 25 and 35 (b) 27 and 37 (c) 25 and 33 (d) none of these

Solution

option  (c) 25 and 33 correct

explanation

in AP common deference should be same

Here deference is 9-1=8

So next term obtain by just adding 8 to previous term

#### Question-8

If ∆𝐴𝐵𝐶 ~ ∆𝑄𝑅𝑃 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒
(a) 𝐴𝐵/𝑄𝑅=𝐵𝐶/𝑅𝑃
(b) 𝐴𝐶/𝑄𝑅=𝐵𝐶/𝑅𝑃
(c) 𝐴𝐵/𝑄𝑅=𝐵𝐶/𝑄𝑃
(d) 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑅P

Solution

Option (a) 𝐴𝐵/𝑄𝑅=𝐵𝐶/𝑅𝑃 is correct

explanation

in ∆𝐴𝐵𝐶 ~ ∆𝑄𝑅𝑃

A𝐵/𝑄𝑅     (taking first 2 letter of similar ∆)

𝐵𝐶/𝑅𝑃    (taking last 2 letter of similar ∆)

#### Question-9

If x∈ 𝑊 , then the solution set of the inequation -x > -7, is
(a) {8,9,10…} (b) {0,1,2,3,4,5,6} (c) {0,1,2,3 …} (d) { -8, -9, -10…}

Solution

option (b) {0,1,2,3,4,5,6} is correct

explanation

-x > -7

-x by -1 < -7 by -1

sign of inequality change if multiply by negative number

x < 7

So vale of x is less than 7 which is possible in option (b) {0,1,2,3,4,5,6}

0 also included as it is Whole number

#### Question-10

The roots of the quadratic equation 4𝑥2- 7x + 2 = 0 are 1.390, 0. 359.The roots correct to 2 significant figures are
(a) 1.39 and 0.36

(b) 1.3 and 0.35

(c) 1.4 and 0.36

(d) 1.390 and 0.360

Solution

option (c) 1.4 and 0.36 is correct

explanation

in 1.390 taking 2 figure is 1.39 =1.4

and 0. 359 taking 2 figure is 0.36

#### Question-11

1.5, 3, x and 8 are in proportion, then x is equal to

(a) 6 (b) 4 (c) 4.5 (d) 16

Solution

option (b) 4 is correct

explanation

1.5/ 3,=  x / 8 are in proportion

1.5 x 8 = 3 x x (cross multiplication)

on solving

x=4

#### Question-12

If a polynomial 2𝑥2 – 7x – 1 is divided by (x + 3), then the remainder is
(a) – 4 (b) 38 (c)-3 (d) 2

solution

option (b)38  is correct

explanation

when x+3=0

then x= -3

putting x=-3 in  2𝑥2 – 7x – 1

= 2(-3)²-7(-3)-1

= 2×9 + 21-1

=18+21-1

= 39-1

=38

#### Question-13

If 73 is the nth term of the arithmetic progression 3, 8, 13, 18…, then ‘n’ is
(a) 13 (b) 14 (c) 15 (d) 16

Solution

option (c) 15 is correct

explanation

here

a=3

d=8-3

=5

an = 73

an = a + (n − 1)d.

73= 3 +  (n − 1) 5

73-3 = (n − 1) 5

70 = (n − 1) 5

70 / 5 = n-1

14= n-1

14+1 = n

15= n

#### Question-14

The roots of the quadratic equation x2+2x+1 = 0 are
(a) Real and distinct
(b) Real and equal

(c) Distinct

(d) Not real / imaginary

Solution

option (b) Real and equal  is correct

explanation

if  b24ac = 0 then root are Real and equal

#### Question-15

Which of the following statement is not true?
(a) All identity matrices are square matrix
(b) All null matrices are square matrix
(c) For a square matrix number of rows is equal to the number of columns
(d) A square matrix all of whose elements except those in the leading diagonal are zero is the diagonal matrix

Solution

option ((b) All null matrices are square matrix

explanation

A square matrix all of whose elements except those in the leading diagonal are zero is the diagonal identity or unit matrix

#### Question-16

If (x – 2) is a factor of the polynomial x3 + 2×2 – 13 x + k, then ‘k’ is equal to
(a) -10 (b)26 (c)-26 (d) 10

Solution

option (d) 10 is correct

explanation

p(x) = x^3 + 2x^2 – 13x + k

For (x – 2) to be the factor of p(x) = x^3 + 2x^2 – 13x + k

Thus (2)3 + 2(2)2 – 13(2) + k = 0

⇒ 8 + 8 – 26 + k = 0

⇒ 16-26 +k=0

⇒ -10 = – k

⇒ k = 10

### Section-B (12 marks)

Maths Semester-1 ICSE Specimen Paper Solved Class-10

#### Question-17

A man deposited ₹1200 in a recurring deposit account for 1 year at 5% per annum simple interest. The interest earned by him on maturity is
(a) 14790 (b) 390 (c) 4680 (d) 780

Solution

option (b) 390 is correct

explanation

Deposit per month = Rs. 1200
Period = 1 years = 12 months
∴ Total principal for 1 month = (P × n(n +1))/2

= ₹ (1200 × 12 × 13)/2

= ₹ 93600

∴ Interest = PRT/100

=(93600 x 5 x 1 )/ (100 x 12)

= ₹ 390

#### Question-18

If x2 – 4 is a factor of polynomial x3+ x2- 4x- 4, then its factors are
(a) (x-2) (x+2) (x+1)
(b) (x-2) (x+2) (x-1)
(c) (x-2) (x-2) (x+1)
(d) (x-2) (x-2) (x-1)

Solution

option (a) (x-2) (x+2) (x+1) is correct

explanation

x3 + x2 – 4x – 4
Let x + 1 = 0
∴ x = -1
On substituting value of x in the expression
∴ f(-1) = (-1)+ (-1)– 4(-1) -4 = 0
Clearly x + 1 is a factor of
f(x) = x3 + x2 – 4x – 4
∴ f(x) = (x + 1) (x2 – 4)   …(By actual division)
= (x + 1) (x – 2) (x + 2)

#### Question-19

The following bill shows the GST rates and the marked price of articles A and B:

BILL: GENERAL STORE

Articles  MP   Rate
A       ₹300   12%
B      ₹1200    5%
The total amount to be paid for the above bill is: –
(a) 1548 (b) 1596 (c) 1560 (d) 1536

Solution

option (b) 1596 is correct

Explanation

Amount for A = 300 + 300 x 12/100

= 300+36=336

Amount for B = 1200 + 1200 x 5/100

= 1200 + 60 = 1260

total amount = Amount for A + Amount for B

= 336 + 1260

=1596

#### Question-20

The solution set for the linear inequation -8 ≤ x – 7 < – 4, x∈ 𝐼 is
(a) {x: x ∈ R, -1≤ x < 3}
(b) {0, 1, 2, 3}
(c) {-1, 0, 1, 2, 3}
(d) { -1, 0, 1, 2}

Solution

option (d) { -1, 0, 1, 2} is correct

Explanation

-8 ≤ x – 7 and x – 7 < -4

-8 +7 ≤ x  and x  < -4 +7

-1 ≤ x and x  < 3

-1 ≤ x   < 3

#### Question-21

If 5𝑎/7𝑏 =4𝑐/3𝑑 , then by Componendo and dividendo
(a) 5𝑎+7𝑏/5𝑎−7𝑏=4𝑐−3𝑑/4𝑐+ 3𝑑
(b) 5𝑎−7𝑏/5𝑎+7𝑏=4𝑐+3𝑑/4𝑐−3𝑑
(c) 5𝑎+7𝑏/5𝑎−7𝑏=4𝑐+3𝑑/4𝑐−3𝑑
(d) 5𝑎+7𝑏/5𝑎+7𝑏=4𝑐−3𝑑/4𝑐−3d

Solution

option (c) 5𝑎+7𝑏/5𝑎−7𝑏=4𝑐+3𝑑/4𝑐−3𝑑 is correct

Explanation

5𝑎/7𝑏 =4𝑐/3𝑑

5𝑎+7𝑏/5𝑎−7𝑏=4𝑐+3𝑑/4𝑐−3𝑑  applying Componendo and dividendo

#### Question-22

Solutions

option (b) [4 0−9 49]  is correct

explanation

A² =A x A =

### Section -C

Maths Semester-1 ICSE Specimen Paper Solved Class-10

Question-23

The distance between station A and B by road is 240 km and by train it is 300 km. A car starts from station A with a speed x km/hr whereas a train starts from station B with a speed 20km/hr more than the speed of the car

(i) The time taken by car to reach station B is

Solutions

option (a) 240/x  is correct

explanation

speed of car = x

Time = ?

Time = dis/ speed

= 240 / x

(ii) The time taken by train to reach station A

Solutions

option (d) 300 /𝑥+20  is correct

explanation

Dis by rail = 300

speed of train = x+20

Time = ?

Time = dis/ speed

= 300 / x+20

(iii) If the time taken by train is 1 hour less than that taken by the car, then the quadratic equation formed is

Solutions

option (b) x2 + 80x –4800=0  is correct

explanation

According condition

Time by car – Time by train = 1

(240 / x )  – (300 / x+20) =1

on simplification

x2 + 80x –4800=0

(iv) The speed of the car is
(a) 60km/hr (b) 120km/hr (c) 40km/hr (d) 80km/hr

Solutions

option (c) 40km/hr  is correct

explanation

x2 + 80x –4800=0

x2 + 120x -40x –4800=0

x (x+120)-40(x+120)=0

(x+120) (x-40)= 0

applying zero product rule

either x+120 = 0

then x = -120    but speed can be negative hence ignore this value

or x-40=0

then x= 40

#### Question-24

In the given triangle PQR, AB || QR, QP || CB and AR intersects CB at O.

(i) The triangle similar to ΔARQ is

(a) ΔORC (b) ΔARP (c) ΔOBR (d) ΔQRP

Solutions

option (a) ΔORC  is correct

explanation

ΔARQ  and ΔORC

∠ARQ= ∠ORC common

∠AQR= ∠OCR  alternate

So ΔARQ  similar to ΔORC  by AA axiom

(ii) ΔPQR ~ΔBCR by axiom

Solutions

option (b)AAA  is correct

explanation

ΔPQR  and  ΔBCR

∠QPR= ∠CBR  Alternate

∠PRQ= ∠BRC  common

hence ΔPQR ~ΔBCR by AA axiom

(iii) If QC =6 cm, CR = 4 cm, BR = 3 cm. The length of RP is
(a) 4.5 cm (b) 8cm (c) 7.5cm (d) 5cm

Solutions

option  (c) 7.5 cm  is correct

explanation

in ΔPQR and similar ΔBCR

PQ/BC = QR/CR = PR/BR

or     QR/CR = PR/BR

or     10/4 = PR/3

on simplify PR= 15/2 = 7.5

(iv) The ratio PQ: BC is
(a) 2 : 3 (b) 3 : 2 (c) 5 : 2 (d) 2 : 5

Solutions

option  (c) 5 : 2  is correct

explanation

PQ/BC = QR/CR

PQ/BC = 10/4

PQ/BC = 5/2 = 5:2

Question-25

The nth term of an arithmetic progression (A.P.) is (3n + 1)
(i) The first three terms of this A. P. are
(a) 5, 6, 7 (b) 3, 6, 9 (c) 1, 4, 7 (d) 4, 7, 10

Solutions

option  (d) 4, 7, 10  is correct

explanation

An= (3n + 1)

A1 = 3 x 1 +1 = 4

A2 = 3 x 2 +1= 7

A3 = 3 x 3 +1 = 10

Hence term are 4, 7, 10

(ii) The common difference of the A.P. is
(a) 3 (b)1 (c) -3 (d) 2

Solutions

option  (a) 3   is correct

explanation

common difference of the A.P = term – proceeding term

= 7-4=3

(iii) Which of the following is not a term of this A.P.
(a) 25 (b) 27 (c) 28 (d) 31

Solutions

option  (b) 27   is correct

explanation

term can be obtain by adding common difference to proceeding term

hence

4, 7, 10, 13,16,19,22,25,28,31

It is clear that 27 is not a term of this AP

(iv) Sum of the first 10 terms of this A.P. is
(a) 350 (b) 175 (c) -95 (d) 70

Solutions

option  (b) 175   is correct

explanation

Sn = (n/2)(2a + (n – 1)d)

S10= (10/2) ((2 × 4) + ((10 – 1) × (3))

= (5) (8 + (27))

= 5 (35)

=175

— End of Semester-1 ICSE Maths Specimen Paper Solutions

thanks

### 20 thoughts on “Maths Semester-1 ICSE Specimen Paper Solved Class-10”

1. In no. 15 it is said thai all null matrix are square matrix how can this statement be true?

• now focus on only sem -2 please

⭐⭐⭐⭐⭐

• thanks keep in touch

• Are all solutions correct or can we consider these as correct answers to Maths Semester 1 Specimen paper

• yes

3. Thank You

• ok keep in touch

4. Thank a lot for this ….but I need more questions to practice for my boards!

Answer 1 is (a) 3 x 2
Because in question A is 3×2 and B is 2×2.

Answer 15 is (b) All null matrices are square matrix.
This is wrong because a null matrix can be non-square also.
For e.g.
[0 0 0 0] is also null matrix but is not a square matrix.

• during solutions
mistakely put the value of order b 2×1 in place of 2×2
hence ans become wrong
but now corrected

6. is the 1st and 15th one correct?

• please now focus on sem-2

7. Thank you so much …..it would be even more better if u try to share in pdf form

• thanks for suggestion but pdf can not be update time to time as syllabus/pattern change very soon in icse/isc due to cowid 19

8. thnks pandey tutoria