---
title: "CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2022-basic-430-1-1
dateModified: 2026-10-07T15:26:52+00:00
---

# CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions

Basic · Code 430/1/1 · 40 marks · 120 minutes

Solved CBSE Class 10 Maths Basic board paper from 2022, set 430/1/1. Every question has step-by-step working written to the CBSE marking scheme, with the marks for each answer.

Official question paper: https://www.cbse.gov.in/cbsenew/question-paper/2022/X/Math_B.zip. Solutions written by SwaVid.

Free PDF (11 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2022-basic-430-1-1-4eddca0d4d.pdf

## SECTION A

Question numbers 1 to 6 carry 2 marks each.

### Question 1 (a)

*2 marks · Short answer · Arithmetic Progressions*

In an AP, if $a = 50$, $d = -4$ and $S_n = 0$, then find the value of $n$.

**Solution**

1. Use the formula for the sum of $n$ terms of an AP: $S_n = \frac{n}{2}[2a + (n-1)d]$.
2. Substitute the given values $a = 50$, $d = -4$, and $S_n = 0$: $0 = \frac{n}{2}[2(50) + (n-1)(-4)]$.
3. Simplify the equation: $100 - 4n + 4 = 0$, which gives $104 - 4n = 0$.
4. Solve for $n$: $4n = 104$, so $n = 26$.

**Answer:** $n = 26$

> Common mistake: Forgetting that $n$ must be a positive integer or making sign errors while expanding the bracket.

### Question 1 (OR)

*2 marks · Short answer · Arithmetic Progressions*

Find the sum of the first twelve 2-digit multiples of $7$, using an AP.

**Solution**

1. The first twelve 2-digit multiples of $7$ form an AP: $14, 21, 28, \dots$ up to $12$ terms.
2. Here, the first term $a = 14$, the common difference $d = 7$, and the number of terms $n = 12$.
3. Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ and substitute the values: $S_{12} = \frac{12}{2}[2(14) + (12-1)7]$.
4. Calculate the value: $S_{12} = 6[28 + 77] = 6[105] = 630$.

**Answer:** $630$

> Common mistake: Starting the AP from $7$ instead of the first 2-digit multiple which is $14$.

### Question 2

*2 marks · Short answer · Surface Areas and Volumes*

A solid metallic sphere of radius $3\text{ cm}$ is melted and recast into the shape of a solid cylinder of radius $2\text{ cm}$. Find the height of the cylinder.

**Solution**

1. Volume of the solid sphere = Volume of the solid cylinder.
2. Write the formula for the volume of a sphere: $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = 36\pi$.
3. Write the formula for the volume of a cylinder: $\pi R^2 h = \pi (2)^2 h = 4\pi h$.
4. Equate both volumes: $4\pi h = 36\pi$, which gives $h = 9\text{ cm}$.

**Answer:** $9\text{ cm}$

> Common mistake: Calculating surface area instead of volume when shapes are melted and recast.

### Question 3 (a)

*2 marks · Short answer · Quadratic Equations*

Find the nature of the roots of the quadratic equation $x^2 - 5x + 9 = 0$.

**Solution**

1. Compare the given quadratic equation $x^2 - 5x + 9 = 0$ with the standard form $ax^2 + bx + c = 0$ to get $a = 1$, $b = -5$, and $c = 9$.
2. Find the discriminant $D = b^2 - 4ac$.
3. Substitute the values: $D = (-5)^2 - 4(1)(9) = 25 - 36 = -11$.
4. Since $D < 0$, the given quadratic equation has no real roots.

**Answer:** No real roots

> Common mistake: Making sign errors while calculating the square of negative numbers in the discriminant.

### Question 3 (OR)

*2 marks · Short answer · Quadratic Equations*

Write a quadratic equation with roots $-3$ and $5$.

**Solution**

1. Let the roots be $\alpha = -3$ and $\beta = 5$.
2. Find the sum of the roots: $\alpha + \beta = -3 + 5 = 2$.
3. Find the product of the roots: $\alpha \beta = (-3)(5) = -15$.
4. Use the standard quadratic equation form $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ to get $x^2 - 2x - 15 = 0$.

**Answer:** $x^2 - 2x - 15 = 0$

> Common mistake: Reversing the sign of the sum of roots coefficient in the quadratic equation.

### Question 4

*2 marks · Short answer · Statistics*

Find the mode of the following frequency distribution:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Class & $0 - 20$ & $20 - 40$ & $40 - 60$ & $60 - 80$ & $80 - 100$ \\
\hline
Frequency & $8$ & $7$ & $12$ & $5$ & $3$ \\
\hline
\end{tabular}

**Solution**

1. Identify the modal class, which is the class with the maximum frequency ($12$), so the modal class is $40 - 60$.
2. Write down the values: lower limit $l = 40$, class size $h = 20$, modal frequency $f_1 = 12$, preceding frequency $f_0 = 7$, and succeeding frequency $f_2 = 5$.
3. Use the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Substitute the values: $\text{Mode} = 40 + \left(\frac{12 - 7}{2(12) - 7 - 5}\right) \times 20 = 40 + \left(\frac{5}{24 - 12}\right) \times 20 = 40 + \frac{100}{12} = 40 + 8.33 = 48.33$.

**Answer:** $48.33$

> Common mistake: Confusing $f_0$ (frequency of the preceding class) with $f_2$ (frequency of the succeeding class).

### Question 5

*2 marks · Short answer · Quadratic Equations*

Solve the quadratic equation $2x^2 - 5x - 1 = 0$ for $x$.

**Solution**

1. Compare the given equation $2x^2 - 5x - 1 = 0$ with the standard form $ax^2 + bx + c = 0$ to get $a = 2$, $b = -5$, and $c = -1$.
2. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.
3. Substitute the values of $a$, $b$, and $c$ into the formula to get $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-1)}}{2(2)}$.
4. Simplify the expression to obtain $x = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4}$.

**Answer:** $x = \frac{5 \pm \sqrt{33}}{4}$

> Common mistake: Making sign errors while substituting negative values of $b$ and $c$ into the quadratic formula.

### Question 6

*2 marks · Short answer · Circles*

In Figure 1, if tangents $PA$ and $PB$ drawn from a point $P$ to a circle with centre $O$, are inclined to each other at an angle of $70^\circ$, then find the measure of $\angle POA$.

**Solution**

1. Note that the radius is perpendicular to the tangent at the point of contact, so $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
2. Consider the quadrilateral $OAPB$; the sum of all interior angles is $360^\circ$, so $\angle AOB + \angle APB = 180^\circ$.
3. Substitute the given angle $\angle APB = 70^\circ$ to find $\angle AOB = 180^\circ - 70^\circ = 110^\circ$.
4. In triangles $\triangle OPA$ and $\triangle OPB$, $OP$ is common, $OA = OB$, and $PA = PB$, making $\triangle OPA \cong \triangle OPB$ by SSS congruence.
5. Therefore, $\angle POA = \frac{1}{2} \angle AOB = \frac{1}{2} \times 110^\circ = 55^\circ$.

**Answer:** $\angle POA = 55^\circ$

> Common mistake: Confusing angle $\angle AOB$ with angle $\angle POA$ and forgetting to divide by 2.

## SECTION B

Question numbers 7 to 10 carry 3 marks each.

### Question 7

*3 marks · Short answer · Statistics*

The frequency distribution given below shows the weight of $40$ students of a class. Find the median weight of the students.
\begin{tabular}{|c|c|}
\hline
Weight (in kg) & Number of Students \\
\hline
$40 - 45$ & $9$ \\
\hline
$45 - 50$ & $5$ \\
\hline
$50 - 55$ & $8$ \\
\hline
$55 - 60$ & $9$ \\
\hline
$60 - 65$ & $6$ \\
\hline
$65 - 70$ & $3$ \\
\hline
\end{tabular}

**Solution**

1. Construct the cumulative frequency table: Classes $40-45$ (f: $9$, cf: $9$), $45-50$ (f: $5$, cf: $14$), $50-55$ (f: $8$, cf: $22$), $55-60$ (f: $9$, cf: $31$), $60-65$ (f: $6$, cf: $37$), $65-70$ (f: $3$, cf: $40$).
2. Here total number of observations $N = 40$, so $\frac{N}{2} = 20$.
3. The cumulative frequency just greater than $20$ is $22$, corresponding to the median class $50 - 55$.
4. Using the median formula $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$, where $l = 50$, $cf = 14$, $f = 8$, and $h = 5$.
5. Substitute the values: $\text{Median} = 50 + \left( \frac{20 - 14}{8} \right) \times 5 = 50 + \frac{6 \times 5}{8} = 50 + 3.75 = 53.75\text{ kg}$.

**Answer:** $53.75\text{ kg}$

> Common mistake: Taking the wrong cumulative frequency or incorrect lower limit of the median class.

### Question 8 (a)

*3 marks · Short answer · Circles*

Draw a circle of radius $4\text{ cm}$. Construct a pair of tangents to the circle from a point $6\text{ cm}$ away from its centre.

**Solution**

1. Draw a circle of radius $4\text{ cm}$ with centre $O$ and mark a point $P$ at a distance of $6\text{ cm}$ from $O$.
2. Bisect the line segment $OP$ to obtain its midpoint $M$.
3. Taking $M$ as centre and $PM$ as radius, draw a circle to intersect the given circle at two points, say $T_1$ and $T_2$.
4. Join $PT_1$ and $PT_2$ to get the required pair of tangents.

**Answer:** Required tangents constructed successfully.

> Common mistake: Not drawing the perpendicular bisector accurately or measuring the wrong radius.

### Question 8 (OR)

*3 marks · Short answer · Triangles*

Draw a line segment $PQ = 7.5\text{ cm}$. Divide it in the ratio $3 : 1$.

**Solution**

1. Draw a line segment $PQ = 7.5\text{ cm}$.
2. Draw a ray $PX$ making an acute angle with $PQ$.
3. Locate $3 + 1 = 4$ points ($A_1, A_2, A_3, A_4$) on ray $PX$ at equal intervals.
4. Join $A_4$ to $Q$, and draw a line through $A_3$ parallel to $A_4Q$ intersecting $PQ$ at point $R$ to divide it in the ratio $3 : 1$.

**Answer:** Line segment divided in the ratio $3 : 1$.

> Common mistake: Making incorrect number of divisions on the ray or drawing parallel lines inaccurately.

### Question 9

*3 marks · Proof · Some Applications of Trigonometry*

In Figure 2, the angles of elevation of the top of a tower $AB$ of height '$h$' m, from two points $P$ and $Q$ at a distance of $x\text{ m}$ and $y\text{ m}$ from the base of the tower respectively and in the same straight line with it, are $60^\circ$ and $30^\circ$, respectively. Prove that $h^2 = xy$.

**Solution**

1. Given: Height of tower $AB = h$, distance $AP = x$, distance $AQ = y$, $\angle APB = 60^\circ$, and $\angle AQB = 30^\circ$.
2. In right-angled triangle $ABP$, $\tan 60^\circ = \frac{AB}{AP} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$.
3. In right-angled triangle $ABQ$, $\tan 30^\circ = \frac{AB}{AQ} \implies \frac{1}{\sqrt{3}} = \frac{h}{y} \implies h \sqrt{3} = y$.
4. Multiply the two equations for $h$: $h \times h = (x\sqrt{3}) \left(\frac{y}{\sqrt{3}}\right)$.
5. Simplify the product: $h^2 = xy$. Hence proved.

**Answer:** $h^2 = xy$

> Common mistake: Interchanging the distances $x$ and $y$ with the wrong angles of elevation.

### Question 10

*3 marks · Short answer · Statistics*

The following table shows the age of patients admitted in a hospital during a particular week:
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
Age (in years) & $5 - 15$ & $15 - 25$ & $25 - 35$ & $35 - 45$ & $45 - 55$ & $55 - 65$ \\
\hline
Number of Patients & $5$ & $12$ & $20$ & $24$ & $15$ & $4$ \\
\hline
\end{tabular}
Find the mean age of the patients.

**Solution**

1. Set up a frequency distribution table with class intervals, frequencies ($f_i$), class marks ($x_i$), and the product $f_i x_i$ for each class.
2. For class $5 - 15$, class mark $x_1 = \frac{5 + 15}{2} = 10$, and $f_1 x_1 = 5 \times 10 = 50$.
3. For class $15 - 25$, class mark $x_2 = 20$, and $f_2 x_2 = 12 \times 20 = 240$.
4. For class $25 - 35$, class mark $x_3 = 30$, and $f_3 x_3 = 20 \times 30 = 600$.
5. For class $35 - 45$, class mark $x_4 = 40$, and $f_4 x_4 = 24 \times 40 = 960$.
6. For class $45 - 55$, class mark $x_5 = 50$, and $f_5 x_5 = 15 \times 50 = 750$.
7. For class $55 - 65$, class mark $x_6 = 60$, and $f_6 x_6 = 4 \times 60 = 240$.
8. Find the sum of frequencies: $\sum f_i = 5 + 12 + 20 + 24 + 15 + 4 = 80$.
9. Find the sum of products: $\sum f_i x_i = 50 + 240 + 600 + 960 + 750 + 240 = 2840$.
10. Apply the direct method formula for the mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2840}{80} = 35.5$.

**Answer:** $35.5$ years

> Common mistake: Calculating incorrect class marks by subtracting instead of adding the lower and upper limits, or making arithmetic errors in $\sum f_i x_i$.

## SECTION C

Question numbers 11 to 14 carry 4 marks each.

### Question 11 (a)

*4 marks · Long answer · Surface Areas and Volumes*

A spherical glass vessel has a cylindrical neck $8\text{ cm}$ long and $1\text{ cm}$ in radius. The radius of the spherical part is $9\text{ cm}$. Find the amount of water (in litres) it can hold, when filled completely.

**Part (i)**

1. Radius of the cylindrical neck ($r$) = $1\text{ cm}$, height ($h$) = $8\text{ cm}$.
2. Volume of the cylindrical neck = $\pi r^2 h = \pi (1)^2 (8) = 8\pi\text{ cm}^3$.
3. Radius of the spherical part ($R$) = $9\text{ cm}$.
4. Volume of the spherical part = $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (9)^3 = 972\pi\text{ cm}^3$.
5. Total volume of the vessel = $8\pi + 972\pi = 980\pi\text{ cm}^3$.
6. Total volume = $980 \times \frac{22}{7} = 3080\text{ cm}^3$.
7. Amount of water in litres = $\frac{3080}{1000} = 3.08\text{ litres}$ (or using $3.14$ as $\pi$, $321.39\text{ cm}^3$ giving $3.21\text{ litres}$). Using $\pi = \frac{22}{7}$, volume is $3.08\text{ litres}$.

Answer (i): $3.08\text{ litres}$

**Answer:** $3.21\text{ litres}$

> Common mistake: Forgetting to add the volume of both parts or using incorrect radii for the cylinder and sphere.

### Question 11 (OR)

*4 marks · Long answer · Surface Areas and Volumes*

From a solid cylinder, whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.

**Part (i)**

1. Given height of the cylinder ($h$) = $2.4\text{ cm}$ and diameter ($d$) = $1.4\text{ cm}$, so radius ($r$) = $0.7\text{ cm}$.
2. Slant height of the conical cavity ($l$) = $\sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$.
3. Total surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of cone + Area of the base of the cylinder.
4. CSA of cylinder = $2\pi r h = 2 \times \frac{22}{7} \times 0.7 \times 2.4 = 10.56\text{ cm}^2$.
5. CSA of cone = $\pi r l = \frac{22}{7} \times 0.7 \times 2.5 = 5.50\text{ cm}^2$.
6. Area of the base = $\pi r^2 = \frac{22}{7} \times (0.7)^2 = 1.54\text{ cm}^2$.
7. Total surface area = $10.56 + 5.50 + 1.54 = 17.60\text{ cm}^2$.

Answer (i): $17.6\text{ cm}^2$

**Answer:** $11.99\text{ cm}^2$

> Common mistake: Omitting the area of the circular base of the cylinder in the total surface area.

### Question 12

*4 marks · Proof · Circles*

In Figure 3, the tangent $l$ is parallel to the tangent $m$ drawn at points $A$ and $B$ respectively to a circle centred at $O$. $PQ$ is a tangent to the circle at $R$. Prove that $\angle POQ = 90^\circ$.

**Solution**

1. Given: Tangents $l$ and $m$ at $A$ and $B$ to a circle with centre $O$ are parallel. $PQ$ is a tangent at $R$ intersecting $l$ at $Q$ and $m$ at $P$.
2. To prove: $\angle POQ = 90^\circ$.
3. Construction: Join $OR$.
4. Consider $\triangle OAQ$ and $\triangle ORQ$: $OA = OR$ (radii of the same circle), $AQ = RQ$ (lengths of tangents drawn from an external point $Q$ are equal), and $OQ = OQ$ (common side).
5. Therefore, $\triangle OAQ \cong \triangle ORQ$ by SSS congruency criterion, which implies $\angle AOQ = \angle ROQ$ (CPCT). Let $\angle AOQ = \angle ROQ = \angle 1$.
6. Similarly, consider $\triangle OBP$ and $\triangle ORP$: $OB = OR$ (radii), $BP = RP$ (tangents from $P$), and $OP = OP$ (common).
7. Therefore, $\triangle OBP \cong \triangle ORP$, which implies $\angle BOP = \angle ROP$ (CPCT). Let $\angle BOP = \angle ROP = \angle 2$.
8. Since $AB$ is a line segment passing through the centre $O$ representing the diameter, $AOB$ is a straight line, so the sum of angles on straight line $AB$ is $180^\circ$.
9. Thus, $2\angle 1 + 2\angle 2 = 180^\circ$, which gives $\angle 1 + \angle 2 = 90^\circ$.
10. Since $\angle POQ = \angle ROQ + \angle ROP = \angle 1 + \angle 2$, we get $\angle POQ = 90^\circ$. Hence proved.

**Answer:** Hence proved that $\angle POQ = 90^\circ$.

> Common mistake: Forgetting to state that tangents from an external point are equal or misidentifying the straight line angles.

### Question 13 (a)

*2 marks · Case-based · Arithmetic Progressions*

Based on the given information about old clothes and a footmat (rug) where stitches in circular rows make a pattern: $6, 12, 18, 24, \dots$
Check whether the given pattern forms an AP. If yes, find the common difference and the next term of the AP.

**Part (i)**

1. Given sequence of stitches is $6, 12, 18, 24, \dots$
2. Check the difference between consecutive terms: $a_2 - a_1 = 12 - 6 = 6$, $a_3 - a_2 = 18 - 12 = 6$, $a_4 - a_3 = 24 - 18 = 6$.
3. Since the difference between each term and its preceding term is constant ($d = 6$), the given pattern forms an AP.

Answer (i): Yes, it forms an AP with common difference $d = 6$.

**Part (ii)**

1. The next term is the fifth term of the AP ($a_5$).
2. $a_5 = a_4 + d = 24 + 6 = 30$.

Answer (ii): The next term of the AP is 30.

**Answer:** The pattern forms an AP with common difference 6 and next term 30.

> Common mistake: Checking only one pair of consecutive terms instead of verifying multiple pairs.

### Question 13 (b)

*2 marks · Case-based · Arithmetic Progressions*

Write the $n^{\text{th}}$ term of the AP. Hence, find the number of stitches in the $10^{\text{th}}$ circular row.

**Part (i)**

1. The given AP has first term $a = 6$ and common difference $d = 6$.
2. The $n^{\text{th}}$ term of an AP is given by $a_n = a + (n - 1)d$.
3. Substitute $a = 6$ and $d = 6$: $a_n = 6 + (n - 1)6 = 6 + 6n - 6 = 6n$.

Answer (i): The $n^{\text{th}}$ term is $6n$.

**Part (ii)**

1. To find the number of stitches in the $10^{\text{th}}$ row, substitute $n = 10$ in the $n^{\text{th}}$ term formula.
2. $a_{10} = 6(10) = 60$.

Answer (ii): The number of stitches in the $10^{\text{th}}$ row is 60.

**Answer:** The nth term is $6n$ and the 10th term is 60.

> Common mistake: Multiplying incorrectly or confusing $n$ with the common difference.

### Question 14 (a)

*2 marks · Case-based · Some Applications of Trigonometry*

The TV Tower in Pitampura, Delhi stands vertically on the ground. From a point '$A$' on the ground, the angle of elevation of top of the tower (point '$B$') is $60^\circ$. There is a point '$C$' on the tower which is $78\text{ m}$ (approx.) above the ground. The angle of elevation of the point $C$ from point $A$ is found to be $30^\circ$.
Draw a well-labelled figure, based on the information given above.

**Part (i)**

1. Diagram: Draw a vertical line segment $BD$ representing the TV Tower of total height $h$ standing on ground level $AD$.
2. Mark point $A$ on the ground at a distance from the base $D$. Join $AB$, showing the angle of elevation of the top $B$ from $A$ as $60^\circ$.
3. Mark point $C$ on the tower $BD$ such that $CD = 78\text{ m}$. Join $AC$, showing the angle of elevation of point $C$ from $A$ as $30^\circ$.

Answer (i): Diagram drawn and labelled as per the given data.

**Answer:** A well-labelled figure showing the TV tower, point A, point C, and the respective angles of elevation $60^\circ$ and $30^\circ$.

> Common mistake: Interchanging the angles of elevation for the top of the tower and point C.

### Question 14 (b)

*2 marks · Case-based · Some Applications of Trigonometry*

Find the height of the tower and the distance of the tower from point $A$.

**Part (i)**

1. Let the height of the tower be $h$ and the distance from point $A$ to the foot of the tower be $x$.
2. In the right-angled triangle, $\tan 60^\circ = \frac{h}{x}$.
3. Since $\tan 60^\circ = \sqrt{3}$, we get $h = x\sqrt{3}$.

Answer (i): Distance of the tower from point $A$ is $20\text{ m}$ and height of the tower is $20\sqrt{3}\text{ m}$.

**Answer:** Height of the tower is $20\sqrt{3}\text{ m}$ and distance from point A is $20\text{ m}$.

> Common mistake: Taking the incorrect trigonometric ratio or interchanging sine and tangent values.

## Frequently asked questions

### What is the paper pattern and section breakdown for the CBSE Class 10 Maths Basic 2022 Set 430/1/1 question paper?

The question paper is divided into three sections totaling 40 marks with a time duration of 120 minutes. Section A contains 6 questions for 12 marks, Section B has 4 questions for 12 marks, and Section C includes 6 questions for 16 marks.

### How are the marks distributed across the different sections in this paper?

Section A carries 12 marks across 6 questions, Section B carries 12 marks across 4 questions, and Section C carries 16 marks across 6 questions. Together, the paper totals 40 marks to be completed in 120 minutes.

### Which chapters carry the most marks in the CBSE Class 10 Maths Basic 2022 Set 430/1/1 paper?

Circles carries the highest weightage with 9 marks, followed by Statistics with 8 marks and Some Applications of Trigonometry with 7 marks. Arithmetic Progressions and Surface Areas and Volumes carry 6 marks each, while Quadratic Equations carries 4 marks.

### How should students write answers to secure full marks in this Class 10 Maths Basic paper?

Students should write step-by-step solutions with proper mathematical formulas and clean diagrams wherever necessary. Showing clear working helps examiners award step marks even if the final calculation has a minor error.

### Is the solutions PDF for this CBSE Class 10 question paper free to access?

Yes, the complete solutions PDF for the CBSE Class 10 Maths Basic 2022 Set 430/1/1 paper is available for free on SwaVid. Students can download it to check their answers and understand the correct method of solving each question.

## Related pages

- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

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