**Solutions of Class 10 Unit Test -1(2022-23) Mathematics-G.D Lancer’s Public School Delhi**

Solutions of Class 10 Unit Test -1(2022-23) Mathematics-G.D Lancer’s Public School Delhi is created here for helping the class 10 students in clearing their doubts about incoming exams and tests. All students of class 10 are needed special attention in mathematics because it is the subject that helps you to increase your aggregate percentage marks in the exam. The solutions of the class 10 Unit Test will give you an idea about the technics of solving the question papers. The unit test of class 10 is presented here is taken from a reputed public school of Delhi G.D Lancer Public School. The solution to the test is created by an expert in maths in a step-by-step way so every student can understand the solution easily. The total questions in this test are 11 and divided into three sections A, B, and C. Section A contains 6 questions and each question is of one mark, section B contains 3 questions and each question is of 2 marks and section C contains 2 questions and each question is of 4 marks. The maximum mark for this maths test is 20.

**Solutions of Class 10 Unit Test -1(2022-23) Mathematics-G.D Lancer’s Public School Delhi**

Q1.The number √(5+3)/√(5-3) is

(a)Rational number (b) Irrational number

(c)An Integer (d)A natural number

Solution.(a)Rational number

The given number is

√(5+3)/√(5-3)

=√8/√2 = 2√2/√2 =2

2 is a rational number.therefore given number is rational

Q2.If a,b are both positive rational number then is (√a + √b)(√a – √b)

(a)Rational number (b) Irrational number

(c)An Integer (d)both rational as well as irrational

Solution.(a)Rational number

The given number is

(√a + √b)(√a – √b)

Applying the identity (x+y)(x -y) = x² -y²

√a² – √b² = a – b

Since it is given that a and c are positive rational number, according to closure property of rational number (a -b) must be a rational number,therefore the given number is a rational number.

Q3.If am ≠bl then system of equations ax +by = c,lx +my = n has

(a)Unique solution (b) May or may not have solution

(c)Infinitely many the solution (d)no solution

Solution.(a)Unique solution

Since it is given to us

am ≠bl

a/l ≠b/m

⇒a_{1}/a_{1 }≠b_{1}/b_{2 }

Comparing the coefficients of the given pair of the linear equations with the standard pair of the linear equation a_{1}x+b_{1}y +c_{1}=0 and a_{2}x+b_{2}y +c_{2}=0

Q4.Form a quadratic polynomial whose zeroes are 5 +√3 and 5 -√3

Solution.The given zeroes ,α= 5 +√3 and β=5 -√3

The required polynomial is given as

x² -(α +β)x + αβ

x² -( 5 +√3 +5 -√3)x + (5 +√3 )(5 -√3)

x² -10x + (5² -√3²)

x² -10x +(25 -3)

x² -10x +22

Hence the given quadratic polynomial is x² -10x +22

Q5.What number is to be added to the polynomial x² -5x +4, so that 3 becomes the zero of the polynomial

Solution. Let k is added to the given polynomial so that 3 is the zero of the given polynomial x² -5x +4

The polynomial will become

p(x) =x² -5x +4 +k

Then it should satisfy 3 for p(x) =0

3² -5×3 +4 +k =0

9 -15 +4 +k =0

-2 +k =0

k =2

Therefore 2 is added to the given polynomial so that 3 is its zero

Q6.If sum of the zeroes of the polynomial p(x) =(k² -14)x² -2x -12 is 1,then k takes the value

(a)14 (b)-14 (c)2 (d)±4

Solution. (d)±4

The sum of the zeroes is given as

α +β = -b/a, where a is the coefficient of the quadratic term and b is the coefficient of linear term

In the given quadratic equation p(x) =(k² -14)x² -2x -12

a = k² -14, b = -2 and c =-12

α +β = -(-2)/(k²-14)

Since it is given to us α +β =1

1 =2/(k²-14)

k²-14 = 2

k² = 16

k =≠4

**Solutions of Class 10 Unit Test -1(2022-23) Mathematics-G.D Lancer’s Public School Delhi**

**Section-B**

Q7.Prove that √5 is irrational.

Ans. Let √5 is a rational number

Where a and b are co-prime number ( study **Number system**)

b√5 = a

Squaring both sides

5 b² = a²………..(i)

5 is one of the factors of a²

∴ 5 will also be one of the factors of a

Therefore a can be written as a multiple of 5

a = 5c ( where c is another positive integer)

Putting the value of ‘a’ in equation number (i)

5 b² = (5c)² = 25c²

b² = 5c²…………(ii)

5 is one of the factors of b²

∴ 5 will also be one of the factors of b

It is evident from eq.(i) and (ii) that 5 is a common factor in ‘a’ and ‘b’, so it is contrary to the fact we have supposed that √5 is a rational number

Therefore √5 is an irrational number.

**Solutions of Class 10 Unit Test -1(2022-23) Mathematics-G.D Lancer’s Public School Delhi**

### Section-C

Q10.Solve the following graphically

2x -3y -4 =0

x – y -1 = 0

Shade the triangle between x-axis and given lines.

Solution. The given pair of the linear equation is

2x -3y -4 =0

x – y -1 = 0

The solutions of the equations 2x -3y -4 =0 and x – y -1 = 0 are given as

Now drawing the graph of both equations

The graph of the both equations intersect at (-1,-2),therefore the solutions of the equation is x=-1 and y=-2

Q11.Show that 15^{n}

can not end with the digits 0,2,4,6 and 8 for any natural number

Solution.

The number given is

15^{n}

= (3×5)^{n}

= 3^{n}×5^{n}

3^{n} ×5^{n}can not end at at 0,2,4,6 and 8 for any value n =1,2,3…. because 3 and 5 both are odd numbers

As an example

If n = 1

then 3^{1}×5^{1}=15

For n =2

3^{2}×5^{2}=225

And so on

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