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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-14-ex-14-1/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 14
Chapter Name	Statistics
Exercise	Ex 14.1
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:

Since the values of x_i and f_i are small, so we have used direct method to find the mean.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Here h = 20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Solution:

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method

Solution:

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Here h = 3

Step deviation method.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:
Here h = 0.04

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Here h = 10

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-14-ex-14-3/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 14
Chapter Name	Statistics
Exercise	Ex 14.3
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Solution:

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution:
Given, Median = 28.5 which lies in the class (20 – 30).
Here, l = 20, f = 20, cf = 5 + x, h = 10, n = 60

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Solution:

Question 4.
The lengths of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to 117.5 – 126.5,126.5 -135.5,…, 171.5 -180.5.)
Solution:

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.
Solution:

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:

For mean

Question 7.
The distribution below gives the weight of 30 students of a class. Find the median weight of the students.

Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-13-ex-13-4/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.4
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular .ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Given: upper diameter = 4 cm ⇒ upper radius = \(\frac { 1 }{ 2 }\) = 2 cm = R
lower diameter = 2 cm ⇒ lower radius = \(\frac { 2 }{ 2 }\) = 1 cm = r
height of glass = 14 cm

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Given: upper circumference of the frustum = 18 cm

Slant height (l) = 4 cm
We have C.S.A of the frustum = π (r₁ + r₂)l
Putting values from equation (i) and (ii), we get
Curved surface area = (πr₁ + πr₂)l = (9 + 3) x 4 = 12 x 4 = 48 cm²

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:
Radius of open side (r₁) = 10 cm
Radius of upper base (r₂) = 4 cm
Slant height (l) = 15 cm

Total surface area of the cap = C.S.A. of the frustum + Area of upper base
= 660 cm² + 50.28 cm² = 710.28 cm²

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm2. (Take π = 3.14)
Solution:
Radius of the lower end (r₁) = 8 cm
Radius of the upper end (r₂) = 20 cm
Height of the frustum (h) = 16 cm

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac { 1 }{ 16 }\) cm, find the length of the wire.
Solution:
Let ADC is a cone with vertical angle 600.
Now, cone is cut into two parts, parallel to its base at height 10 cm.
Radius of larger end of the frustum = R₁

A wire be formed having diameter \(\frac { 1 }{ 16 }\) cm and length be H cm
Volume of wire so obtained = πr²H

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-11-ex-11-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 11
Chapter Name	Constructions
Exercise	Ex 11.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2

In each of the following, give the justification of the construction also:

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction:
1. Draw a circle of radius 6 cm.
2. Mark a point P at a distance 10 cm from the centre O.
3. Here OP = 10 cm, draw perpendicular bisector of OP, which intersects OP at O’.
4. Take O’ as centre, draw a circle of radius O’O, which passes through O and P and intersects the previous circle at points T and Q.
5. Join PT and PQ, measure them, these are the required tangents.

Justification:
Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PT = PQ.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction:
1. Draw concentric circles of radius OA = 4 cm and OP = 6 cm having same centre O.
2. Mark these circles as C and C’.
3. Points O, A and P lie on the same line.
4. Draw perpendicular bisector of OP, which intersects OP at O’.
5. Take O’ as centre, draw a circle of radius OO’ which intersects the circle C at points T and Q.
6. Join PT and PQ, these are the required tangents.
7. Length of these tangents are approx. 4.5 cm.

Justification:
Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PT = PQ.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction:

1. Draw a circle of radius 3 cm, having centre O. Mark this circle as C.
2. Mark points P and Q along its extended diameter such that OP = 7 cm and OQ = 7 cm.
3. Draw perpendicular bisector of OQ, intersecting OQ at O’.
4. Draw perpendicular bisector of OP intersecting OP at O”.
5. Take O’ as centre and draw a circle of radius OO’ which passes through points O and Q, intersecting circle C at points R and T.
6. Take O” as centre and draw a circle of radius O”P which passes through points O and P, intersecting the, circle C at points S and U.
7. Join QR and QT; PS and PU, these are the required tangents.
Justification:
Join OR.
In ∆OQR,
OR ⊥ QR [Radius ± to tangent]
OQ² = OR² + QR²
⇒ (7)² = (3)² + QR² ⇒ 49 = 9 + OR²
⇒ 40 = QR² ⇒ QR = 2 \(\sqrt{10}\) cm
Similary,
QT = 2 \(\sqrt{10}\) cm
PS = 2 \(\sqrt{10}\) cm
PU = 2 \(\sqrt{10}\) cm
A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ QR = QT and PS = PU

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction:

1. Draw a circle of radius 5 cm.
2. As tangents are inclined to each other at an angle of 60°.
∴ Angle between the radii of circle is 120°. (Use quadrilateral property)
3. Draw radii OA and OB inclined to each other at an angle 120°.
4. At points A and B, draw 90° angles. The arms of these angles intersect at point P.
5. PA and PB are the required tangents.
Justification:
In quadrilateral AOBP,
AP and BP are the tangents to the circle.
Join OP.
In right angled AOAP, OA ⊥ PA [Radius is ⊥ to tangent]

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ PA = PB

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:

1. Draw a line segment AB = 8 cm.
2. Taking A as centre draw a circle C of radius 4 cm and taking B as centre draw a circle C’ of radius 3 cm.
3. Draw perpendicular bisector of AB, which intersects AB at point O.
4. Taking point O as centre draw a circle of radius 4 cm passing through points A and B which intersect circle C at P and S and circle C’ at points R and Q.
5. Join AQ, AR, BP and BS. These are the required tangents.
Justification:
Join AP.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ AQ = AR and BP = BS.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ZB = 90°. BD is the perpendicular Burn B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction:
1. Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and ZB = 90°.
2. From B, draw BD perpendicular to AC.
3. Draw perpendicular bisector of BC which intersect BC at point O’.
4. Take O’ as centre and O’B as radius, draw a circle C’ passes through points B, C and D.

5. Join O’A and draw perpendicular bisector of O’A which intersect O’A at point K.
6. Take K as centre, draw an arc of radius KO’ intersect the previous circle C’ at T.
7. Join AT, AT is required tangent.
Justification:
∠BDC = 90°
∴ BC acts as diameter.
AB is tangent to circle having centre O’

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.
∴ AB = AT.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction:
1. Draw a circle C’ with the help of a bangle, for finding the centre, take three non-collinear points A, B and C, lying on the circle. Join AB and BC and draw perpendicular bisector of AB and BC, both intersect at a point O,
‘O’ is centre of the circle.

2. Take a point P outside the circle. Join OP.
3. Draw perpendicular bisector of OP, which intersects OP at point O’.
4. Take O’ as the centre with OO’ as radius draw a circle which passes through O and P, intersecting previous circle C’ at points R and Q.
5. Join PQ and PR.
6. PQ and PR are the required pair of tangents.
Justification:
Join OQ and OR.
In AOQP and AOPR, OQ = OR [Radii of the circle]
OP = OP [Common]
∠Q = ∠R = 90° [Radius is ⊥ to tangent]
∆OQP ≅ ∆ORP [by RHS]
∴ PQ = PR [By C.P.C.T]
A pair of tangents can be drawn to a circle from an external point lying outside the circle. These two tangents are equal in lengths.
∴ PQ = PR

We hope the NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2, drop a comment below and we will get back to you at the earliest.

Check the below NCERT MCQ Questions for Class 6 Sanskrit Chapter 14 अहह आः च with Answers Pdf free download. MCQ Questions for Class 6 Sanskrit with Answers were prepared based on the latest exam pattern. We have provided अहह आः च Class 6 Sanskrit MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-6-sanskrit-chapter-14/

Students can also read NCERT Solutions for Class 6 Sanskrit Chapter 14 Questions and Answers at LearnInsta. Here all questions are solved with a detailed explanation, It will help to score more marks in your examinations.

मञ्जूषायाः सहायतया गद्यांश पूरयत। (मञ्जूषा की सहायता से गद्यांश पूरा कीजिए।)
Complete the extract with help from the box.

अवकाशम्, अवकाशस्य, आनय, इव, एवम्, सेवायाम्, दास्यामि, परिश्रमी, चतुरः
अजीजः सरलः …………….. च आसीत्। सः स्वामिनः एव …………………. लीनः आसीत्। एकदा सः गृहं गंतुम् ………………. वाञ्छति। स्वामी ……………. आसीत्। सः चिंतयति-‘अजीजः ……………… न कोऽपि अन्यः कार्यकुशलः। एष …………. अपि वेतनं ग्रहीष्यति।’ ……………… चिंतयित्वा स्वामी कथयति-‘अहं तुभ्यम् अवकाशस्य वेतनस्य च सर्वं धनं ………….. । परम् एतदर्थं त्वं वस्तुद्वयम् …………………

Answer

Answer:
परिश्रमी, सेवायाम्, अवकाशम्, चतुरः, इव, अवकाशस्य, एवम्, दास्यामि, आनय।

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गद्यांश पठित्वा अधोदत्तान् प्रश्नान् उत्तरत। (गद्यांश पढ़कर निम्नलिखित प्रश्नों के उत्तर दीजिए।)
Read the extract and answer the following questions.

अजीजं दृष्ट्वा स्वामी चकितः भवति। स्वामी शनैः शनैः पेटिकाम् उद्घाटयति। पेटिकायां लघुपात्रद्वयम् आसीत्। प्रथमं सः एकं लघुपात्रम् उद्घाटयति। सहसा एका मधुमक्षिका निर्गच्छति। तस्य च हस्तं दशति। स्वामी उच्चै वदति-“अहह!।” द्वितीय लघुपात्रम् उद्घाटयति। एका अन्या मक्षिका निर्गच्छति। सः ललाटे दशति। पीडितः सः अत्युच्चैः चीत्करोति-“आः” इति। अजीजः सफलः आसीत्। स्वामी तस्मै अवकाशस्य वेतनस्य च पूर्णं धनं ददाति।

Question 1.
(क) अजीजं दृष्ट्वा कः चकित:? ………………….
(ख) स्वामी काम् उद्घाटयति? ………………..
(ग) लघुपात्रद्यम् कस्याम् आसीत्? ………………
(घ) पात्रात् का निर्गच्छति? …………………..

Answer

Answer:
(क) स्वामी
(ख) पेटिकाम्
(ग) पेटिकायाम्
(घ) मधुमक्षिका

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Question 2.
(क) अन्या मधुमक्षिका किं करोति?
(ख) तदा स्वामी किं करोति?

Answer

Answer:
(क) अन्या मधुमक्षिका ललाटे दशति।
(ख) तदा स्वामी अत्युच्यैः (अति + उच्चैः) चीत्करोति।

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Question 3.
(क) ‘नीचैः’ इति पदस्य विलोमं लिखत।
(ख) ‘विस्मितः’ इति पदस्य पर्यायं लिखत।

Answer

Answer:
(क) उच्चैः।
(ख) चकितः।

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Question 4.
यथानिर्देशं रिक्तस्थानि पूरयत। यथा
एकं लघुपात्रम्।
(i) …………. पेटिका।
(ii) …………… सेवकः

Answer

Answer:
(i) एका।
(ii) एकः।

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मधुमक्षिका निर्गच्छति। मधुमक्षिकाः निर्गच्छन्ति।
(i) सा दशति। ……………… …………..
(ii) सः उद्घाटयति। ……………… ……………

Answer

Answer:
(i) ताः दशन्ति।
(ii) ते उद्घाटयन्ति।

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Question 5.
लङ्लकारे परिवर्तयत-
(i) स्वामी चकितः भवति।
(ii) सः चकितः भवति।

Answer

Answer:
(i) स्वामी चकितः अभवत्।
(ii) सः उच्चैः अवदत्।

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लुट्लकारे परिवर्तयत
(i) स्वामी चीत्कारं करोति।
(ii) स: उच्चैः वदति।
(iii) सः तस्मै पूर्णं धनं ददाति।

Answer

Answer:
(i) स्वामी चीत्कारं करिष्यति।
(ii) सः चकितः भविष्यति।
(iii) सः तस्मै पूर्ण धनं दास्यति।

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मञ्जूषातः उचितं क्रियापदं चित्वा वाक्यपूर्तिं कुरुत- (मञ्जूषा की सहायता से उचित क्रियापद चुनकर वाक्य पूरे कीजिए)
Pick out the correct options and complete the sentences.

प्रार्थयति, परिभ्रमति, पश्यति, प्राप्नोति, प्राप्स्यति, मिलति, निर्गच्छति, पृच्छति।
एतत् श्रुत्वा अजीजः वस्तुद्वयम् आनेतुं …………….. । सः इतस्ततः ………………..। जनान् ……………….| आकाशं ……………… धरां ……………| परं सफलतां नैव ………………. । चिंतयति,
परिश्रमस्य धनं सः नैव। कुत्रचित् एका वृद्धा ………………

Answer

Answer:
निर्गच्छति, परिभ्रमति, पृच्छति, पश्यति, प्रार्थयति, प्राप्नोति, प्राप्स्यति, मिलति।

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यथानिर्देशम् लकारपरिवर्तन कृत्वा वाक्यानि पुनः लिखत। (निर्देशानुसार लकार बदलकर वाक्य पुनः लिखिए।)
Change tense as per directions and rewrite the sentences.

(क) अहं तुभ्यं सर्वं धनं दास्यामि। – (लट्लकारे)
(ख) सः आकाशं पश्यति। – (लङ्लकारे)
(ग) अजीज! त्वम् वस्तुद्यम् आनयति। – (लोटलकारे)
(घ) अजीजः सफलः अभवत्। – (लुटलकारे)
(ङ) अजीजः कार्यकुशलः आसीत्। – (लट्लकारे)

Answer

Answer:
(क) अहं तुभ्यं सर्वं धनं ददामि।
(ख) सः आकाशं अपश्यत्।
(ग) अजीज! त्वम् वस्तुद्वयम् आनय।
(घ) अजीजः सफलः भविष्यति।
(ङ) अजीजः कार्यकुशलः अस्ति।

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उचितेन विकल्पेन रिक्तस्थानानि पूरयत- (उचित विकल्प द्वारा रिक्त स्थान भरिए)
Fill in the blanks with the correct option.

1. (i) सहसा एका …………….. निर्गच्छति। (वृद्धा, पेटिका, मधुमक्षिका)
(ii) अजीजः ……………. आनेतुं निर्गच्छति। (लघुपात्रम्, वस्तुद्वयम्, धनम्)
(iii) सः ताम् सर्वां ……………. श्रावयति। (कथाम्, वृद्धाम्, व्यथाम्)
(iv) कुत्रचित् एका वृद्धा …………….. (मिलति, कथयति, ददाति)
(v) स्वामी …………….. पेटिकाम् उद्घाटयति। (उच्चैः, शनैः शनैः, एकदा)

Answer

Answer:
(i) मधुमक्षिका
(ii) वस्तुद्वयम्
(iii) व्यथाम्
(iv) मिलति
(v) शनैः शनैः।।

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2. (i) अजीजः स्वामिनः ………….. लीनः आसीत्। (सेवा, सेवाम्, सेवायाम्)
(ii) सः चिंतयति ………. धनं सः नैव प्राप्स्यति। (परिश्रमी, परिश्रमस्य, परिश्रमः)
(iii) सहसा ……………. मधुमक्षिका निर्गच्छति। (एकः, एका, एकम्)
(iv) अजीजं …………. स्वामी चकितः भवति। (दृष्ट्वा, द्रष्टुम्, पश्यति)
(v) मधुमक्षिका ……………. दशति। (हस्तः, हस्तम्, हस्तेन)

Answer

Answer:
(i) सेवायाम्
(ii) परिश्रमस्य
(iii) एका
(iv) दृष्ट्वा
(v) हस्तम्।

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3. (i) अजीजः गृहं गन्तुम् अवकाशं …………….. (पृच्छति, वाञ्छति, मिलति)
(ii) सः सफलतां नैव …………….. (पश्यति, परिभ्रमति, प्राप्नोति)
(iii) प्रथमं सः एकं लघुपात्रम् ………………… (निर्गच्छति, ददाति, उद्घाटयति)
(iv) अन्या मक्षिका ललाटे ……………….. (ददाति, दशति, चीत्करोति)
(v) कुत्रचित् एका वृद्धा …………… (श्रावयति, चिन्तयति, मिलति)

Answer

Answer:
(i) वाञ्छति
(ii) प्राप्नोति
(iii) उद्घाटयति
(iv) दशति
(v) मिलति।

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