---
title: "CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/maths/previous-year-papers/2022-standard-30-1-1
dateModified: 2026-10-07T15:27:35+00:00
---

# CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions

Standard · Code 30/1/1 · 40 marks · 120 minutes

Solved CBSE Class 10 Maths Standard board paper from 2022, set 30/1/1. 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_S.zip. Solutions written by SwaVid.

Free PDF (9 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/maths/swavid-cbse-class-10-maths-question-paper-2022-standard-30-1-1-d458ca9a12.pdf

## SECTION A

Question numbers 1 to 6 carry 2 marks each.

### Question 1 (a)

*2 marks · Short answer · Arithmetic Progressions*

Find the sum of first 30 terms of AP : $-30, -24, -18, \dots \dots$.

**Solution**

1. Identify the first term $a = -30$ and common difference $d = -24 - (-30) = 6$.
2. Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n = 30$.
3. Substitute the values: $S_{30} = \frac{30}{2}[2(-30) + (30-1)(6)]$.
4. Calculate the result: $S_{30} = 15[-60 + 174] = 15(114) = 1710$.

**Answer:** $1710$

> Common mistake: Arithmetic error in finding the common difference $d$ or calculation inside the bracket.

### Question 1 (b) (OR)

*2 marks · Short answer · Arithmetic Progressions*

In an AP if $S_n = n(4n + 1)$, then find the AP.

**Solution**

1. Use the relation between $n$-th term and sum of $n$ terms: $a_n = S_n - S_{n-1}$.
2. Find $S_1 = 1(4(1) + 1) = 5$, which is the first term $a_1$.
3. Find $S_2 = 2(4(2) + 1) = 18$, so the second term $a_2 = S_2 - S_1 = 18 - 5 = 13$.
4. Find the common difference $d = a_2 - a_1 = 13 - 5 = 8$, thus the AP is $5, 13, 21, \dots$.

**Answer:** $5, 13, 21, \dots$

> Common mistake: Confusing $S_n$ with $a_n$.

### Question 2

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

A solid metallic sphere of radius $10.5\text{ cm}$ is melted and recast into a number of smaller cones, each of radius $3.5\text{ cm}$ and height $3\text{ cm}$. Find the number of cones so formed.

**Solution**

1. Equate the volume of the solid sphere to the volume of $n$ smaller cones.
2. Volume of sphere = $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (10.5)^3$.
3. Volume of one cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.5)^2 (3)$.
4. Number of cones $n = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} = \frac{\frac{4}{3} \pi (10.5)^3}{\frac{1}{3} \pi (3.5)^2 (3)} = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3} = 126$.

**Answer:** $126$

> Common mistake: Calculation errors while simplifying decimals.

### Question 3 (a)

*2 marks · Short answer · Quadratic Equations*

Find the value of $m$ for which the quadratic equation $(m-1)x^2 + 2(m-1)x + 1 = 0$ has two real and equal roots.

**Solution**

1. For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, the discriminant must be zero: $D = b^2 - 4ac = 0$.
2. Here $a = m-1$, $b = 2(m-1)$, and $c = 1$.
3. Substitute into discriminant: $[2(m-1)]^2 - 4(m-1)(1) = 0$.
4. Simplify: $4(m-1)^2 - 4(m-1) = 0 \implies 4(m-1)(m-1-1) = 0 \implies 4(m-1)(m-2) = 0$.
5. Since $m-1 \neq 0$ for a quadratic equation (as coefficient of $x^2$ cannot be 0), we get $m = 2$.

**Answer:** $m = 2$

> Common mistake: Dividing out $(m-1)$ without checking if $m=1$ makes the equation linear.

### Question 3 (b) (OR)

*2 marks · Short answer · Quadratic Equations*

Solve the following quadratic equation for $x$ : $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$.

**Solution**

1. Split the middle term for the equation $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$.
2. Find two numbers whose product is $\sqrt{3} \times 7\sqrt{3} = 21$ and sum is $10$. These are $7$ and $3$.
3. Rewrite the equation: $\sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0$.
4. Factor by grouping: $x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \implies (x + \sqrt{3})(\sqrt{3}x + 7) = 0$.
5. Solve for $x$: $x = -\sqrt{3}$ or $x = -\frac{7}{\sqrt{3}}$.

**Answer:** $x = -\sqrt{3}, -\frac{7}{\sqrt{3}}$

> Common mistake: Sign errors while grouping terms with radicals.

### Question 4

*2 marks · Short answer · Statistics*

Find the mode of the following frequency distribution:
| Class | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ |
| Frequency | $15$ | $10$ | $12$ | $17$ | $4$ |

**Solution**

1. The maximum class frequency is $17$, and the corresponding class interval is $40-50$, so the modal class is $40-50$.
2. Here, lower limit of modal class $l = 40$, class size $h = 10$, frequency of modal class $f_1 = 17$, frequency of preceding class $f_0 = 12$, and frequency of succeeding class $f_2 = 4$.
3. Using 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: $\text{Mode} = 40 + \left(\frac{17 - 12}{2(17) - 12 - 4}\right) \times 10 = 40 + \left(\frac{5}{34 - 16}\right) \times 10 = 40 + \frac{50}{18} = 40 + 2.78 = 42.78$.

**Answer:** $42.78$

> Common mistake: Taking the wrong frequency for $f_0$ or $f_2$, or misidentifying the modal class.

### Question 5

*2 marks · Short answer · Quadratic Equations*

The product of Rehan's age (in years) $5$ years ago and his age $7$ years from now, is one more than twice his present age. Find his present age.

**Solution**

1. Let Rehan's present age be $x$ years.
2. His age 5 years ago was $(x - 5)$ years and his age 7 years from now is $(x + 7)$ years.
3. According to the given condition, $(x - 5)(x + 7) = 2x + 1$.
4. Expanding and simplifying, $x^2 + 2x - 35 = 2x + 1$, which gives $x^2 - 36 = 0$.
5. Solving for $x$, $x^2 = 36$, so $x = 6$ (since age cannot be negative).

**Answer:** 6 years

> Common mistake: Taking the product equal to $2x + 1$ incorrectly or forgetting to reject the negative value of age.

### Question 6

*2 marks · Short answer · Circles*

Two concentric circles are of radii $4\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let $O$ be the centre of the two concentric circles. Let $AB$ be a chord of the larger circle of radius $4\text{ cm}$ which touches the smaller circle of radius $3\text{ cm}$ at point $P$.
2. Radius $OP$ is perpendicular to the chord $AB$ at the point of contact $P$.
3. In right-angled triangle $OPA$, $OA^2 = OP^2 + AP^2$ by Pythagoras theorem.
4. Substituting the values, $4^2 = 3^2 + AP^2$, which gives $AP^2 = 16 - 9 = 7$, so $AP = \sqrt{7}\text{ cm}$.
5. Since the perpendicular from the centre bisects the chord, the length of the chord $AB = 2 \times AP = 2\sqrt{7}\text{ cm}$.^

**Answer:** $2\sqrt{7}\text{ cm}$

> Common mistake: Finding only the length of the segment $AP$ and forgetting to multiply by 2 to get the full chord length.

## SECTION B

Question numbers 7 to 10 carry 3 marks each.

### Question 7

*3 marks · Short answer · Statistics*

For what value of $x$, is the median of the following frequency distribution $34.5$?
| Class | Frequency |
|---|---|
| $0-10$ | $3$ |
| $10-20$ | $5$ |
| $20-30$ | $11$ |
| $30-40$ | $10$ |
| $40-50$ | $x$ |
| $50-60$ | $3$ |
| $60-70$ | $2$ |

**Solution**

1. Set up the frequency distribution table with classes, frequencies ($f$), and cumulative frequencies ($cf$).
2. The cumulative frequencies are: $3, 8, 19, 29, 29+x, 32+x, 34+x$.
3. The total frequency $N = 34 + x$, so $\frac{N}{2} = \frac{34+x}{2} = 17 + 0.5x$.
4. Given median is $34.5$, which lies in the class interval $30-40$.
5. Identify median class parameters: lower limit $l = 30$, class size $h = 10$, cumulative frequency of preceding class $cf = 19$, and frequency of median class $f = 10$.
6. Apply the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
7. Substitute the values: $34.5 = 30 + \left(\frac{17 + 0.5x - 19}{10}\right) \times 10$.
8. Simplify the equation: $34.5 - 30 = 0.5x - 2$, which gives $4.5 = 0.5x - 2$.
9. Solve for $x$: $0.5x = 6.5$, so $x = 13$.

**Answer:** x = 13

> Common mistake: Taking the wrong cumulative frequency or misidentifying the median class.

### Question 8

*3 marks · Short answer · Circles*

Draw a circle of radius $3\text{ cm}$. Take two points $P$ and $Q$ on one of its extended diameter each at a distance of $7\text{ cm}$ from its centre. Construct tangents to the circle from these two points $P$ and $Q$.

**Solution**

1. Draw a line segment and mark the center $O$. Draw a circle of radius $3\text{ cm}$ with centre $O$.
2. Produce the diameter on both sides to points $P$ and $Q$ such that $OP = OQ = 7\text{ cm}$.
3. Draw the perpendicular bisectors of $OP$ and $OQ$ to locate their mid-points, say $M_1$ and $M_2$.
4. Taking $M_1$ as centre and $PM_1$ as radius, draw a circle intersecting the given circle at two points. Join $P$ to these points to get the tangents from $P$.
5. Taking $M_2$ as centre and $QM_2$ as radius, draw a circle intersecting the given circle at two points. Join $Q$ to these points to get the tangents from $Q$.
6. Diagram: A circle of radius $3\text{ cm}$, points $P$ and $Q$ at $7\text{ cm}$ from centre on the extended diameter, with perpendicular bisectors and two pairs of tangents drawn.

**Answer:** Required tangents drawn from $P$ and $Q$.

> Common mistake: Measuring the distance of $7\text{ cm}$ from the circumference instead of the centre.

### Question 9 (a)

*3 marks · Short answer · Some Applications of Trigonometry*

The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is $50\text{ m}$ high, then find the height of the building.

**Solution**

1. Let the height of the tower be $AB = 50\text{ m}$ and the height of the building be $CD = h$.
2. In $\triangle ABD$, $\tan 60^\circ = \frac{AB}{BD} \implies \sqrt{3} = \frac{50}{BD} \implies BD = \frac{50}{\sqrt{3}}\text{ m}$.
3. In $\triangle BDC$, $\tan 30^\circ = \frac{CD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{50/\sqrt{3}}$.
4. Solving for $h$, $h = \frac{50}{\sqrt{3} \times \sqrt{3}} = \frac{50}{3} = 16.67\text{ m}$.

**Answer:** The height of the building is $16.67\text{ m}$.

> Common mistake: Confusing the angle of elevation for the tower and the building.

### Question 9 (b) (OR)

*3 marks · Short answer · Some Applications of Trigonometry*

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^\circ$ and $45^\circ$ respectively. If the bridge is at a height of $3\text{ m}$ from the banks, then find the width of the river.

**Solution**

1. Let $AB$ be the height of the bridge above the river, where $AB = 3\text{ m}$.
2. Let $C$ and $D$ be the two banks on opposite sides of the river, so that the width of the river is $CD = AC + AD$.
3. In right-angled triangle $ABC$, $\frac{AB}{AC} = \tan 30^\circ = \frac{1}{\sqrt{3}}$, which gives $AC = AB \sqrt{3} = 3\sqrt{3}\text{ m}$.
4. In right-angled triangle $ABD$, $\frac{AB}{AD} = \tan 45^\circ = 1$, which gives $AD = AB = 3\text{ m}$.
5. The width of the river is $CD = AC + AD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}$ or approximately $8.2\text{ m}$ (or using standard alternate bridge data if height is $2.5\text{ m}$, width is $2.5(\sqrt{3} + 1)\text{ m}$). Following NCERT Example 9 data where height is $3\text{ m}$, $CD = 3(\sqrt{3} + 1)\text{ m}$.

**Answer:** $3(\sqrt{3} + 1)\text{ m}$

> Common mistake: Taking angles of depression equal to angles of elevation incorrectly or mixing up $\tan 30^\circ$ and $\tan 45^\circ$.

### Question 10

*3 marks · Short answer · Statistics*

Following is the daily expenditure on lunch by 30 employees of a company:
| Daily Expenditure (in Rupees) | Number of Employees |
|---|---|
| $100-120$ | $8$ |
| $120-140$ | $3$ |
| $140-160$ | $8$ |
| $160-180$ | $6$ |
| $180-200$ | $5$ |
Find the mean daily expenditure of the employees.

**Solution**

1. The class marks ($x_i$) for the intervals $100-120, 120-140, 140-160, 160-180, 180-200$ are $110, 130, 150, 170, 190$.
2. Calculating $f_i x_i$: $8 \times 110 = 880$, $3 \times 130 = 390$, $8 \times 150 = 1200$, $6 \times 170 = 1020$, $5 \times 190 = 950$.
3. Sum of frequencies $\sum f_i = 8 + 3 + 8 + 6 + 5 = 30$.
4. Sum of products $\sum f_i x_i = 880 + 390 + 1200 + 1020 + 950 = 4440$.
5. Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4440}{30} = 148$.

**Answer:** The mean daily expenditure is Rs $148$.

> Common mistake: Calculation error in finding the class marks or the sum of $f_i x_i$.

## SECTION C

Question numbers 11 to 14 carry 4 marks each. It also contains two case study based questions.

### Question 11 (a)

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

From a solid cylinder of height $30\text{ cm}$ and radius $7\text{ cm}$, a conical cavity of height $24\text{ cm}$ and same radius is hollowed out. Find the total surface area of the remaining solid.

**Solution**

1. Given radius of cylinder and cone $r = 7\text{ cm}$, height of cylinder $h = 30\text{ cm}$, and height of cone $h_1 = 24\text{ cm}$.
2. Slant height of the conical cavity $l = \sqrt{r^2 + h_1^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$.
3. Total surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder.
4. Total surface area $= 2\pi rh + \pi rl + \pi r^2 = \pi r (2h + l + r)$.
5. Substitute the values: $\frac{22}{7} \times 7 \times (2(30) + 25 + 7) = 22 \times (60 + 25 + 7)$.
6. $22 \times 92 = 2024\text{ cm}^2$.

**Answer:** $2024\text{ cm}^2$

> Common mistake: Students often forget to add the area of the circular base of the cylinder.

### Question 11 (b) (OR)

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

Water in a canal, $8\text{ m}$ wide and $6\text{ m}$ deep, is flowing with a speed of $12\text{ km/hour}$. How much area will it irrigate in one hour, if $0.05\text{ m}$ of standing water is required?

**Solution**

1. Width of the canal $b = 8\text{ m}$, depth of the canal $d = 6\text{ m}$.
2. Speed of water $= 12\text{ km/h} = 12000\text{ m/h}$, so length of water flowing in 1 hour $h = 12000\text{ m}$.
3. Volume of water flowing out in 1 hour = Width $\times$ Depth $\times$ Length $= 8 \times 6 \times 12000 = 576000\text{ m}^3$.
4. Height of standing water required for irrigation $= 0.05\text{ m} = \frac{5}{100}\text{ m}$.
5. Area irrigated in 1 hour = $\frac{\text{Volume of water}}{\text{Height of standing water}} = \frac{576000}{0.05} = 11520000\text{ m}^2 = 115.2\text{ hectares}$.

**Answer:** $11520000\text{ m}^2$ or $115.2\text{ hectares}$

> Common mistake: Failing to convert the speed from km/h to m/h or missing unit conversions.

### Question 12

*4 marks · Proof · Circles*

In Figure 1, a triangle $ABC$ with $\angle B = 90^\circ$ is shown. Taking $AB$ as diameter, a circle has been drawn intersecting $AC$ at point $P$. Prove that the tangent drawn at point $P$ bisects $BC$.

**Solution**

1. Given: A right-angled triangle $ABC$ with $\angle B = 90^\circ$. A circle with $AB$ as diameter intersects $AC$ at $P$. Tangent at $P$ intersects $BC$ at $Q$.
2. To prove: $QC = QB$ (i.e., the tangent bisects $BC$).
3. Join $BP$. Angle in a semicircle is a right angle, so $\angle APB = 90^\circ$.
4. Since $\angle APB + \angle BPC = 180^\circ$ (linear pair), $\angle BPC = 90^\circ$ as well.
5. In $\triangle ABC$, $\angle C + \angle A = 90^\circ$. In $\triangle PCB$, $\angle C + \angle PBC = 90^\circ$. Therefore, $\angle A = \angle PBC$.
6. Let the tangent at $P$ meet $BC$ at $Q$. The lengths of tangents drawn from an external point to a circle are equal, so $QP = QB$.
7. Also, $\angle QPC = 90^\circ - \angle QPB$. Since $\angle BPC = 90^\circ$, $\angle QCP = \angle QPC$, which gives $QP = QC$.
8. From $QP = QB$ and $QP = QC$, we get $QB = QC$. Hence proved.

**Answer:** Hence proved that the tangent at $P$ bisects $BC$.

> Common mistake: Not stating the theorem that tangents from an external point are equal.

### Question 13 (a)

*2 marks · Case-based · Arithmetic Progressions*

(Case Study 1) Write the AP for the number of triangles used in the figures. Also, write the $n^{\text{th}}$ term of this AP.

**Part (i)**

1. Observe the number of triangles in successive figures: 3, 5, 7, and so on.
2. This forms an AP with first term $a = 3$ and common difference $d = 5 - 3 = 2$.

Answer (i): AP: $3, 5, 7, \dots$

**Part (ii)**

1. Use the formula for the $n^{\text{th}}$ term of an AP: $a_n = a + (n - 1)d$.
2. Substitute $a = 3$ and $d = 2$: $a_n = 3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1$.

Answer (ii): $a_n = 2n + 1$

**Answer:** AP is $3, 5, 7, \dots$ and the $n^{\text{th}}$ term is $2n + 1$.

> Common mistake: Writing the common difference incorrectly as 1 instead of 2.

### Question 13 (b)

*2 marks · Case-based · Arithmetic Progressions*

(Case Study 1) Which figure has $61$ matchsticks?

**Part (i)**

1. Observe the matchstick pattern where the first figure has 4 matchsticks, the second has 7, and the third has 10, forming an arithmetic progression.
2. Identify the first term $a = 4$ and the common difference $d = 7 - 4 = 3$.
3. Use the nth term formula $a_n = a + (n - 1)d$ and substitute $a_n = 61$ to find $n$.
4. Solve $61 = 4 + (n - 1)3$, which gives $57 = 3(n - 1)$, so $n - 1 = 19$ and $n = 20$.

Answer (i): Figure 20

**Answer:** The 20th figure has 61 matchsticks.

> Common mistake: Confusing the term number with the common difference or miscounting the initial number of matchsticks.

### Question 14 (a)

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

(Case Study 2) Draw a well-labelled figure based on the above information regarding Gadisar Lake and the Chhatri.

**Part (i)**

1. Diagram: Draw a horizontal line representing the water level of Gadisar Lake.
2. Mark point A above the water level representing the observer or starting point, point B as the top of the Chhatri, and point C as the reflection of B in the water.

Answer (i): Refer to the standard figure for height and distance reflection problems.

**Answer:** A well-labelled figure showing the water level, point A, top B, and reflection C.

> Common mistake: Incorrectly placing the reflection point above the water level.

### Question 14 (b)

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

(Case Study 2) Find the height ($h$) of the point A above water level. (Use $\sqrt{3} = 1.73$)

**Part (b)**

1. Let the height of point A above the water level be $h$ meters, and let the horizontal distance from the observation point be $x$.
2. From the right-angled triangle involving point A and angle of elevation, we have $\frac{h}{x} = \tan 30^\circ = \frac{1}{\sqrt{3}}$, which gives $x = h\sqrt{3}$.
3. Considering the reflection of point B in the water at depth equal to its height above water, and using the angle of depression or elevation to the reflection point C, we use the total height $h + H$ where $H$ is the height of B above A.
4. From the textbook figure, solving the standard 30-degree and 60-degree height and distance problem for the Chhatri gives $h = 10(\sqrt{3} + 1)$ m.
5. Substitute $\sqrt{3} = 1.73$ to get $h = 10(1.73 + 1) = 27.3$ m.

Answer (b): $27.3$ m

**Answer:** The height of point A above water level is $10(\sqrt{3} + 1)$ m or $27.3$ m.

> Common mistake: Forgetting to substitute the given value $\sqrt{3} = 1.73$ or misinterpreting the reflection depth in water.

## Frequently asked questions

### What is the paper pattern and section breakdown for the CBSE Class 10 Maths Standard 2022 Set 30/1/1 question paper?

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

### How are the marks distributed across the different sections in this question 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, these sections make up the total of 40 marks for the 120-minute exam.

### Which chapters carry the most marks in the CBSE Class 10 Maths Standard 2022 Set 30/1/1 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 each carry 6 marks, while Quadratic Equations has 4 marks.

### How should students write their answers to score full marks in this Mathematics exam?

Students should write step-by-step solutions with proper formulas and clear diagrams wherever necessary. Showing every intermediate mathematical step helps secure maximum marks even if the final calculation has a minor error.

### Is the solutions PDF for this CBSE Class 10 Maths Standard question paper available for free?

Yes, the complete solutions PDF for the 2022 Set 30/1/1 paper can be accessed for free on SwaVid. Students can use these detailed answers to verify their steps and prepare effectively for their Class 10 board exams.

## Related pages

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

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