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CBSE Class 10 Maths Basic 2022 Question Paper 430/1/1 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/1/1 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 430/1/1 Set 430/1/2 Set 430/1/3 Set 430/2/1 Set 430/2/2 Set 430/2/3 Set 430/3/1 Set 430/3/2 Set 430/3/3 Set 430/4/1 Set 430/4/2 Set 430/4/3

Q12 marksVery Short AnswerArithmetic Progressions

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

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Answer:

n = 26

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 0$
$\frac{n}{2} [100 + (n - 1) (- 4)] = 0$
Since $n \neq = 0$ , $100 - 4 n + 4 = 0$
$4 n = 104$ , so $n = 26$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

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

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Answer: 630

The 2-digit multiples of 7 are 14, 21, 28, ... (an AP with $a = 14$ , $d = 7$ )
For $n = 12$ : $S_{12} = \frac{12}{2} [2 (14) + 11 (7)]$
$= 6 (28 + 77) = 6 \times 105 = 630$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q22 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

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

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Answer: 9 cm

Volume of sphere $= \frac{4}{3} π (3)^{3} = 36 π$ cm $^{3}$
Volume of cylinder $= π (2)^{2} h = 4 π h$
Volumes are equal: $4 π h = 36 π$
$h = 9$ cm

Also asked in: 2022 Basic 430/1/2

Q32 marksVery Short AnswerQuadratic Equations

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

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Answer: No real roots (discriminant

= - 11 < 0

)

Here $a = 1$ , $b = - 5$ , $c = 9$
$D = b^{2} - 4 a c = 25 - 36 = - 11$
Since $D < 0$ , the equation has no real roots.

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q3 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

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

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Answer:

x^{2} - 2 x - 15 = 0

Sum of roots $= - 3 + 5 = 2$
Product of roots $= (- 3) (5) = - 15$
Required equation: $x^{2} - (sum) x + product = 0$
$x^{2} - 2 x - 15 = 0$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q42 marksVery Short AnswerStatistics

Find the mode of the following frequency distribution :
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 8, 7, 12, 5, 3

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Answer:

\frac{145}{3} \approx 48.33

Highest frequency is 12, so the modal class is 40–60
$l = 40$ , $h = 20$ , $f_{1} = 12$ , $f_{0} = 7$ , $f_{2} = 5$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{12 - 7}{24 - 7 - 5} \times 20$
$= 40 + \frac{5}{12} \times 20 = 40 + 8.33 = 48.33$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q52 marksVery Short AnswerQuadratic Equations

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

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Answer:

x = \frac{5 + 33}{4}

x = \frac{5 - 33}{4}

$a = 2$ , $b = - 5$ , $c = - 1$
$D = b^{2} - 4 a c = 25 + 8 = 33$
$x = \frac{- b \pm D}{2 a} = \frac{5 \pm 33}{4}$

Also asked in: 2022 Basic 430/1/3

Q62 marksVery Short AnswerCircles

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 $7 0^{\circ}$ , then find the measure of $∠ P O A$ .

Diagram for CBSE 2022 Class 10 Maths question 6

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Answer:

∠ P O A = 5 5^{\circ}

$O A ⊥ P A$ , so $∠ O A P = 9 0^{\circ}$
OP bisects $∠ A P B$ (tangents from an external point), so $∠ O P A = 3 5^{\circ}$
In $△ O A P$ : $∠ P O A = 18 0^{\circ} - 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q73 marksShort AnswerStatistics

The frequency distribution given below shows the weight of 40 students of a class. Find the median weight of the students.
Weight (in kg): 40–45, 45–50, 50–55, 55–60, 60–65, 65–70
Number of Students: 9, 5, 8, 9, 6, 3

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Answer: 53.75 kg

Cumulative frequencies: 9, 14, 22, 31, 37, 40
$N = 40$ , $\frac{N}{2} = 20$ , so the median class is 50–55
$l = 50$ , $c f = 14$ , $f = 8$ , $h = 5$
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 50 + \frac{20 - 14}{8} \times 5$
$= 50 + 3.75 = 53.75$ kg

Q83 marksShort AnswerCirclesNot in current syllabus

Draw a circle of radius 4 cm. Construct a pair of tangents to the circle from a point 6 cm away from its centre.

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Answer: Construction (see steps).

Draw a circle with centre O and radius 4 cm; mark P with OP = 6 cm
Bisect OP; let M be its midpoint
With M as centre and radius MO, draw a circle cutting the given circle at A and B
Join PA and PB; these are the required tangents (each about 4.5 cm long)

Also asked in: 2022 Basic 430/1/2

Q8 (OR) (OR)3 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment PQ = 7.5 cm. Divide it in the ratio 3 : 1.

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Answer: Construction (see steps).

Draw PQ = 7.5 cm and a ray PX making an acute angle with PQ
Mark 4 equal points $A_{1}, A_{2}, A_{3}, A_{4}$ on PX
Join $A_{4} Q$ ; through $A_{3}$ draw a line parallel to $A_{4} Q$ meeting PQ at R
Then PR : RQ = 3 : 1 (by BPT), i.e. PR = 5.625 cm, RQ = 1.875 cm

Also asked in: 2022 Basic 430/1/2

Q93 marksShort AnswerSome 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 m and y m from the base of the tower respectively and in the same straight line with it, are $6 0^{\circ}$ and $3 0^{\circ}$ , respectively. Prove that $h^{2} = x y$ .

Diagram for CBSE 2022 Class 10 Maths question 9

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Answer: Proved.

In right $△ B A P$ : $tan 6 0^{\circ} = \frac{A B}{A P} = \frac{h}{x}$ , so $h = 3 x$ ... (1)
In right $△ B A Q$ : $tan 3 0^{\circ} = \frac{A B}{A Q} = \frac{h}{y}$ , so $h = \frac{y}{3}$ ... (2)
Multiplying (1) and (2): $h^{2} = 3 x \times \frac{y}{3} = x y$
Hence $h^{2} = x y$ .

Also asked in: 2022 Basic 430/1/3

Q103 marksShort AnswerStatistics

The following table shows the age of patients admitted in a hospital during a particular week :
Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65
Number of Patients: 5, 12, 20, 24, 15, 4
Find the mean age of the patients.

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Answer: 35.5 years

Class marks $x_{i}$ : 10, 20, 30, 40, 50, 60
$f_{i} x_{i}$ : 50, 240, 600, 960, 750, 240
$\sum f_{i} = 80$ , $\sum f_{i} x_{i} = 2840$
Mean $= \frac{2840}{80} = 35.5$ years

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q114 marksLong AnswerSurface Areas and Volumes

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

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Answer: 3.08 litres

Volume of spherical part $= \frac{4}{3} π (9)^{3} = 972 π$ cm $^{3}$
Volume of neck $= π (1)^{2} (8) = 8 π$ cm $^{3}$
Total volume $= 980 π = 980 \times \frac{22}{7} = 3080$ cm $^{3}$
$= \frac{3080}{1000} = 3.08$ litres

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q11 (OR) (OR)4 marksLong AnswerSurface Areas and Volumes

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

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Answer: 17.6 cm

^{2}

$r = 0.7$ cm, $h = 2.4$ cm
Slant height $l = 2. 4^{2} + 0. 7^{2} = 6.25 = 2.5$ cm
TSA = curved surface of cylinder + top base + curved surface of cone
$= 2 π r h + π r^{2} + π r l = π r (2 h + r + l)$
$= \frac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6$ cm $^{2}$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q124 marksLong AnswerCircles

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 $∠ P O Q = 9 0^{\circ}$ .

Diagram for CBSE 2022 Class 10 Maths question 12

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Answer: Proved.

Join OR. $O A ⊥ l$ and $O B ⊥ m$ ; since $l ∥ m$ , A, O, B are collinear (AB is a diameter).
In $△ O A Q$ and $△ O R Q$ : $O A = O R$ (radii), $Q A = QR$ (tangents from Q), $O Q$ common
So $△ O A Q ≅ △ O R Q$ (SSS), giving $∠ A O Q = ∠ R O Q$
Similarly $△ O B P ≅ △ O R P$ , giving $∠ B O P = ∠ R O P$
$∠ A O Q + ∠ R O Q + ∠ R O P + ∠ B O P = 18 0^{\circ}$ (AOB is a straight line)
$2∠ R O Q + 2∠ R O P = 18 0^{\circ}$ , so $∠ R O Q + ∠ R O P = 9 0^{\circ}$
Hence $∠ P O Q = 9 0^{\circ}$ .

Q134 marksCase StudyArithmetic Progressions

Do you know old clothes which are thrown as waste not only fill the landfill site but also produce very harmful greenhouse gas. So, it is very important that we reuse old clothes in whatever way we can.
The picture given below on the right, shows a footmat (rug) made out of old t-shirts yarn. Observing the picture, you will notice that a number of stitches in circular rows are making a pattern : 6, 12, 18, 24, ...
Based on the above information, answer the following questions :
(a) Check whether the given pattern forms an AP. If yes, find the common difference and the next term of the AP. (2)
(b) Write the $n^{t h}$ term of the AP. Hence, find the number of stitches in the $1 0^{t h}$ circular row. (2)

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Answer: (a) Yes, it is an AP; common difference 6, next term 30 (b)

a_{n} = 6 n

; 60 stitches

(a) $12 - 6 = 18 - 12 = 24 - 18 = 6$ , so the pattern is an AP with $d = 6$
Next term $= 24 + 6 = 30$
(b) $a_{n} = a + (n - 1) d = 6 + 6 (n - 1) = 6 n$
$a_{10} = 6 \times 10 = 60$ stitches

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q144 marksCase StudySome Applications of Trigonometry

The following TV Tower was built in 1988 and is located in Pitampura, Delhi. It has an observation deck. Observe the picture given below :
The TV Tower stands vertically on the ground. From a point ‘A’ on the ground, the angle of elevation of top of the tower (point ‘B’) is $6 0^{\circ}$ . There is a point ‘C’ on the tower which is 78 m (approx.) above the ground.
The angle of elevation of the point C from point A is found to be $3 0^{\circ}$ .
(a) Draw a well-labelled figure, based on the information given above. (2)
(b) Find the height of the tower and the distance of the tower from point A. (2)

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Answer: (a) Right triangle with tower BD vertical, C on BD with CD = 78 m,

∠ C A D = 3 0^{\circ}

∠ B A D = 6 0^{\circ}

(b) Height = 234 m; distance =

783 \approx 135.1

(a) Let D be the foot of the tower. Draw vertical BD, mark C on BD with CD = 78 m, A on the ground; join AC and AB with $∠ C A D = 3 0^{\circ}$ and $∠ B A D = 6 0^{\circ}$
(b) In right $△ A D C$ : $tan 3 0^{\circ} = \frac{C D}{A D}$ , so $A D = 783$ m $\approx 135.1$ m
In right $△ A D B$ : $tan 6 0^{\circ} = \frac{B D}{A D}$ , so $B D = 783 \times 3 = 234$ m
Height of tower = 234 m; distance from A = $783$ m

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

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