Top

Download
PDF

FAQ

×

Sorry!, This page is not available for now to bookmark.

The question paper of 10th CBSE Maths will be for 80 marks. There will be four sections of the paper. Section-A a will have all the 1 mark questions and the number of questions in this section is 20. So, the total weightage of section-A is 20 marks. The marks weightage of Section-B is 12 marks where each question will be for 2 marks each. The number of questions in this section is 6. In Section-C, there will be 8 questions asked which carry 3 marks each and the total weightage of this section is 24.

There will be 40 questions in total asked in the question paper, The type of questions is different in each section. There will be objective-type questions from Section-A which will carry 1 mark. Section-B contains short-type questions which are for 2 marks each. The short type questions will be asked in section-C too which will be for 3 marks each. The section-D will have long-type questions which will carry 4 marks each.

These questions will be mostly based on the formulae or describing the part of the formula. The questions will be helpful to find the surface area of the given shape, finding the decimal representation, completing the sequence of Arithmetic progression, the ratio of areas of two similar triangles when the ratio of sides are given, basic probability questions, finding if the pair of straight lines are parallel, nature of roots in a quadratic equation.

The section-C will have short-type questions which will be mostly from the chapter Arithmetic progressions. These will include finding the sum to n terms. The next important topic is from applications of trigonometry like finding the height of the tree if they make a certain angle of elevation or deviation. Questions will also be asked from real numbers (which include finding) if the number is irrational, terminating or non-terminating, etc.

The statistics chapter will have good weightage and the questions can be related to mean median, mode. Coming to the chapter of coordinate geometry, the section formula, distance formula, midpoint of 2 points will be mostly asked in the exam.

There will be an internal choice in this section to answer the paper. In each question, two questions will be asked from which you can answer any one of your choices.

The questions from quadratic equations will be to solve for the value of x. The practical questions from surface areas and volumes will be asked where the real-time examples will be asked and the area or volume should be calculated. The questions from statistics will be related to the mean, median, and mode of the given data.

1. If ΔAB\[\tilde{C}\] ΔPQR and the Ratio of AB and PQ is 4 : 5. Find the Ratio of their Areas and Perimeters.

Given: ΔAB\[\tilde{C}\] ΔPQR

AB : PQ = 4 : 5

\[\frac{AB}{PQ}\] = \[\frac{4}{5}\]

We know that the ratio of areas of 2 similar triangles is equal to the square of the ratio of their corresponding sides.

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{AB}{PQ}\])\[^{2}\]

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{4}{5}\])\[^{2}\]

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = \[\frac{16}{25}\]

Ar(\[\triangle\]ABC) : Ar(\[\triangle\]PQR) = 16 : 25

Hence, the ratio of areas of two triangles is 16 : 25 .

We know that the ratio of areas of 2 similar triangles is equal to the square of their perimeters.

\[\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}\] = (\[\frac{\text{Perimeter of \triangleABC}}{\text{Perimeter of \trianglePQR}}\])\[^{2}\]

\[\sqrt{\frac{Ar(\triangle ABC)}{Ar(\triangle PQR)}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\sqrt{\frac{16}{25}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\sqrt{(\frac{4}{5})^{2}}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

\[\frac{4}{5}\] = (\[\frac{\text{Perimeter of \triangle ABC}}{\text{Perimeter of \triangle PQR}}\])

Perimeter (\[\triangle\]ABC) : Perimeter (\[\triangle\]PQR) = 4 : 5

Hence, the ratio of the perimeters of two similar triangles is 4:5.

Q2. The Point P(x, y) Divides the Line Segment Joining the Points A(1, 4) and B(5, -1). If the x-coordinate of a Point P is 2, Find the Ratio in Which the Point P Divides Point A,B and also Find the y -coordinate of a Point P.

Given: A(1,4) and B(5,-1)

Let Point P(x,y) divides the points A, B in the ratio k : 1

x -coordinate is 2 ⇒ P(2,y)

Section formula: If a point p(x,y) divides (x\[_{1}\],y\[_{1}\]) and (x\[_{2}\],y\[_{2}\]) in the ratio m : n, then

[Image will be Uploaded Soon]

p(x,y) = (\[\frac{mx_{2} + nx_{1}}{m + n}\] , \[\frac{my_{2} + ny_{1}}{m + n}\]

p(2,y) = (\[\frac{5k+1}{k+1}\] , \[\frac{-k+4}{k+1}\])

On comparing the x -coordinate, we get

2 = \[\frac{5k + 1}{k + 1}\]

2(k + 1) = 5k + 1

2k + 2 = 5k + 1

5k - 2k = 2 - 1

3k = 1

k = \[\frac{1}{3}\]

So, the ratio of k : 1 = \[\frac{1}{3}\] : 1

On comparing the y -coordinates, we get

y = \[\frac{-k+4}{k+1}\]

⇒ y = \[\frac{-\frac{1}{3} + 4}{\frac{1}{3} + 1}\]

⇒ y = \[\frac{-\frac{-1+12}{3}}{\frac{1+3}{3}}\]

⇒ y = \[\frac{\frac{11}{3}}{\frac{4}{3}}\]

⇒ y = \[\frac{11}{4}\]

Hence, the ratio in which point P divides points A, B is \[\frac{1}{3}\] : 1 and the value of y -coordinate is \[\frac{11}{4}\].

FAQ (Frequently Asked Questions)

1. Describe the Nature of the Roots of a Quadratic Equation.

Ans: The quadratic polynomial is given by the equation ax² + bx + c = 0, we will get 2 roots for the value of x. The value of x might be a real number, complex number, irrational number. This will decide on the discriminant of the equation. The discriminant of the equation will decide the nature of the roots. The discriminant of a quadratic equation is given by D = b² - 4ac .

So, the values of a, b, c are used to find the nature of the quadratic equation.

Type-1: If D>0 and a perfect square, then the roots are said to be real, rational, and unequal.

Type-2: If D>0 and not a perfect square, then the roots are said to be real, irrational, and unequal

Type-3: If D=0, then the roots are said to be real, rational, and equal

Type-4: If D<0, then the roots are said to be imaginary.

2. Explain Euclid's, Division Lemma.

Ans: Lemma is defined as a statement that is used to prove other statements.

The Euclid’s division Lemma states that if there are any 2 positive integers a, b, then the equation a = bq + r, where 0 ≤ r ≤ b .

Here q is the quotient of the number and r is the remainder of the number and they are whole numbers. The number 'a' is called the dividend and 'b' is called the divisor. Euclid's Division Lemma is generally used to find the Highest Common Factor(HCF) of two positive numbers. HCF can be defined as the largest number which can divide 2 given numbers.