CBSE Class 10

NCERT Solutions for Class 10 Hindi Kshitij Chapter 17 संस्कृति

These Solutions are part of NCERT Solutions for Class 10 Hindi. Here we have given NCERT Solutions for Class 10 Hindi Kshitij Chapter 17 संस्कृति.

प्रश्न-अभ्यास

(पाठ्यपुस्तक से)

प्रश्न 1.
लेखक की दृष्टि में ‘सभ्यता’ और ‘संस्कृति की सही समझ अब तक क्यों नहीं बन पाई है?
उत्तर
‘सभ्यता और संस्कृति की सही समझ नहीं बनने के मुख्य कारण निम्नलिखित हैं

1. लेखक के अनुसार हम सभ्यता और संस्कृति का अर्थ स्पष्ट किए बिना मनमाने ढंग से दोनों शब्दों का भरपूर प्रयोग करते हैं।
2. इन शब्दों के आगे अनेक विशेषण भी लगा देते हैं; जैसे-भौतिक सभ्यता, आध्यात्मिक सभ्यता आदि। इस प्रकार के गलत-सलत प्रयोग करने की वजह से हम सभ्यता और संस्कृति के अंतर को समझने में हम आज भी अक्षम हैं।
3. लोग अपने अलग-अलग विचार प्रस्तुत करते रहते हैं, अलग-अलग ढंग से परिभाषित करते हैं। अतः इन दोनों शब्दों में अर्थ की दृष्टि से सही समझ नहीं।
   बन पाई है।

प्रश्न 2.
आग की खोज एक बहुत बड़ी खोज क्यों मानी जाती है? इस खोज के पीछे रही प्रेरणा के मुख्य स्रोत क्या रहे होंगे?
उत्तर
आवश्यकता ही आविष्कार की जननी है जब आग का आविष्कार नहीं हुआ था तब आग की खोज मनुष्य के लिए सबसे बड़ी प्रसन्नता रही होगी। आग का महत्त्व और उपयोग सबसे अधिक है। अन्य बहुत से कार्यों में आग की सबसे अधिक उपयोगिता भोजन पकाने में है। ठंड से बचने के उपाय, अंधकार के भय को प्रकाश से दूर करना आदि-आदि अनेक कारण खोज के पीछे रहे होंगे।

प्रश्न 3.
वास्तविक अर्थों में ‘संस्कृत व्यक्ति,’ किसे कहा जा सकता है?
उत्तर

1. ऐसा व्यक्ति जो अपनी योग्यता के आधार पर नए तथ्य की खोज करता है वह व्यक्ति वास्तविक अर्थों में ‘संस्कृत व्यक्ति’ कहा जा सकता है।
2. न्यूटन ने अपनी योग्यता के आधार पर गुरुत्वाकर्षण के सिद्धांत का आविष्कार किया। यह सिद्धांत नया था, इसलिए उसे संस्कृत-व्यक्ति कहना उचित है।
3. जिन्होंने भी अपनी योग्यता से आग या सुई-धागे का आविष्कार किया होगा, वे ‘संस्कृत व्यक्ति’ रहे होंगे।
4. इसी प्रकार जनकल्याण की भावना से निहित व्यक्ति जन-कल्याण के सूत्र स्थापित करता है तो वह संस्कृत व्यक्ति कहलाता है।

प्रश्न 4.
न्यूटन को संस्कृत मानव कहने के पीछे कौन से तर्क दिए गए हैं? न्यूटन द्वारा प्रतिपादित सिद्धांतों एवं ज्ञान की कई दूसरी बारीकियों को जानने वाले लोग भी न्यूटन की तरह संस्कृत नहीं कहला सकते, क्यों?
उत्तर
लेखक के अनुसार संस्कृत व्यक्ति वह है जो अपनी योग्यता के आधार पर नए तथ्यों का आविष्कार कर नए तथ्य के दर्शन किए हों। न्यूटन ने भी सर्वप्रथम गुरुत्वाकर्षण के सिद्धांत का आविष्कार किया। इसलिए वह संस्कृत मानव था। उसने अपने सिद्धांत से सुशिक्षितों को परिचित कराया।

दूसरी ओर न्यूटन के सिद्धांत से परिचित होने के बाद न्यूटन से भी अधिक ज्ञान रखने वाले उसी प्रकार संस्कृत व्यक्ति नहीं कहला सकते, जिस प्रकार पूर्वजों से प्राप्त वस्तु संतान को अनायास ही मिल जाती है तो संतान संस्कृत नहीं कहला सकती है।

अतः आविष्कर्ता, आविष्कार का जनक संस्कृत व्यक्ति होती है, अन्य नहीं। अतः लेखक की परिभाषा के अनुसार न्यूटन से भी अधिक बारीकियों का ज्ञान रखने वाले न्यूटन से अधिक सभ्य कहे जा सकते हैं, संस्कृत व्यक्ति नहीं ।

प्रश्न 5.
किन महत्त्वपूर्ण आवश्यकताओं की पूर्ति के लिए सुई-धागे का आविष्कार हुआ होगा?
उत्तर
जिन महत्त्वपूर्ण आवश्यकताओं की पूर्ति के लिए मनुष्य ने सुई-धागे का आविष्कार किया वे निम्नलिखित रही होंगी-

1. जब वह सर्दी-गर्मी को सहन करने में असमर्थ हो गया होगा । फिर शीतोष्ण से बचने के लिए उपाय ढूँढ़ते हुए सुई-धागे का आविष्कार किया होगा।
2. आवश्यकतानुसार शरीर को सजाने की प्रवृत्ति ने जन्म लिया होगा और किस तरह कपड़ों के दो टुकड़े जोड़े जा सकते हैं। इस आवश्यकता से सुई-धागे का आविष्कार किया होगा।
3. मनुष्य के मन में यह भी आया होगा कि किस तरह शरीर को अच्छी तरह ढका जा सकता है और सुई-धागे का आविष्कार कर लिया होगा।

प्रश्न 6.
“मानव संस्कृति एक अविभाज्य वस्तु है” किन्ही दो प्रसंगों का उल्लेख कीजिए जब
(क) मानव संस्कृति को विभाजित करने की चेष्टाएँ की गई।
(ख) जब मानव संस्कृति ने अपने एक होने का प्रमाण दिया।
उत्तर

1. भारत अनेक संस्कृतियों और संप्रदायों का देश है। यहाँ रहे रही मुख्य रूप से दो संस्कृतियों हिंदू और मुसलमानों के मध्य विद्वेष फैलाकर अंग्रेजों ने मानवे संस्कृति को विभाजित करने का प्रयास किया। हालाँकि फलस्वरूप हिंदू और मुसलमान के नाम पर भारत और पाकिस्तान का निर्माण तो हुआ किंतु दोनों देशों के जनमानस के सोचने-समझने का तरीका आज भी एक है। वे आज भी अपने-अपने सरहदों में मानव संस्कृति के वाहक हैं क्योंकि मानव संस्कृति के बीच में कोई लकीर नहीं खींची जा सकती।।
2. जब-जब अमानवीय कृत्य से मनुष्य संत्रस्त हुए या दानवता ने अपने पैर पसारे हैं तो सभी भेद भुलाकर मानव-जाति मानवता के नाम पर उठ खड़ी हुई है। ऐतिहासिक दृष्टि से जापान में गिराए गए परमाणु बम के विनाश से संपूर्ण धरती काँप गई तो विश्व के लोगों ने एक होकर विरोध किया।

प्रश्न 7.
आशय स्पष्ट कीजिए-
मानव की जो योग्यता उससे आत्म-विनाश के साधनों का आविष्कार कराती है, हम उसे उसकी संस्कृति कहें या असंस्कृति?
उत्तर
मानव-संस्कृति अपनी सुरक्षा के प्रति सदैव से चिंतित रही है। मानव जब अपनी योग्यता से मानव-हित की दृष्टि से और आत्महित की दृष्टि से आविष्कार करता है। तभी दूसरी ओर आत्मरक्षा का चिंतन मनुष्य को विनाश की ओर प्रेरित करता है। फिर मानव उस आविष्कार का कल्याण से अलग दुरुपयोग करने लगता है और विनाश के साधनों को जुटाने लगता है। यह मनुष्य की असंस्कृति ही है। रचना और अभिव्यक्ति

प्रश्न 8.
लेखक ने अपने दृष्टिकोण से सभ्यता और संस्कृति की एक परिभाषा दी है। आप सभ्यता और संस्कृति के बारे में क्या सोचते हैं, लिखिए।
उत्तर
सभ्यता और संस्कृति एक ओर तो एक-दूसरे के पूरक हैं। एक के अभाव में दूसरे को स्पष्ट करना कठिन है। सभ्यता दृश्य और स्थूल है तथा संस्कृति अदृश्य और सूक्ष्म है। संस्कृति एक विचार है। हमारे आदर्श पुरुषों ने, मनीषियों ने जो सूत्र अपने अनुभवों से स्थापित किए हैं, जो मानव-हित में हैं वे अनुकरणीय सूत्र संस्कृति हैं।

सभ्यता दृष्ट है, वह जीवन जीने की कला है। सभ्यती बदलती रहती है। एक ही पुरुष अलग-अलग स्थानों पर अलग-अलग सभ्यता को अपनाता है। जैसे एक पुरुष अपने घर जूते उतार कर कुश के आसन पर बैठकर पवित्रता से भोजन करता है वहीं दूसरे के घर उत्सव में बिना जूते उतारे, खड़े होकर भोजन करता है। ये दोनों उसकी सभ्यता हैं। अतः सभ्यता एक रीति है, रिवाज है। अलग-अलग स्थानों पर, अलग-अलग ऋतुओं में अलग-अलग वस्त्र पहनना उसकी सभ्यता है।

भाषा-अध्ययन

प्रश्न 9.
निम्नलिखित सामासिक पदों का विग्रह करके समास का भेद भी लिखिए-

उत्तर

Hope given NCERT Solutions for Class 10 Hindi Kshitij Chapter 17 are helpful to complete your homework.

CBSE Sample Papers for Class 10 Maths Paper 6

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 6

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Use prime factorization method to find the HCF of 300 and 1000.

Question 2.
Find the value of k for which the quadratic equation 2x² + 5x + k = 0 has equal roots.

Question 3.
Find the 101th term of the A.P. – 4, – 2, 0, 2, 4, …

Question 4.
∆ABC is similar to ∆DEF. If AB = 10 cm, DE = 12 cm, perimeter of ∆ABC = 25 cm, find the perimeter of ∆DEF.

Question 5.
Find the distance of A(9, – 40) from origin.

Question 6.
Given that tan A = \(\frac { 1 }{ \sqrt { 5 } } \) what is the value of \(\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+2{ sin }^{ 2 }A } \) ?

SECTION-B

Question 7.
Use Euclid’s algorithm to find the HCF of 612 and 408. Also find their LCM using their HCF.

Question 8.
Find the 10th term from the end of the A.P. 60, 130, 200, …, 2160.

Question 9.
Solve for x and y : ax + by = a² + b²; bx + ay = 2ab.

Question 10.
Find a point on the Y-axis which is equidistant from the points A( – 4, 3) and B(6, 5).

Question 11.
Mean and median of a slightly asymmetric data are 52 and 54.5 respectively. Find the mode of the data.

Question 12.
A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting :
(i) A black king or a red queen
(ii) A non-face card.

SECTION-C

Question 13.
Show √7 is irrational.

OR

Show n(n + 1) (n – 1) is divisible by 3 for any positive integer ‘n’.

Question 14.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).

Question 15.
Solve graphically : 3x – 2y = 5, 2x + 3y = 12. Also find area bounded by these lines with X-axis.

Question 16.
ABCD is a quadrilateral. Diagonals AC and BD meet at point O. Also AO x DO = BO x CO. Show ABCD is a trapezium.

OR

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.

Question 17.
Find the area of the quadrilateral whose vertices taken in order are P( – 5, – 3), Q( – 4, – 6), R(2, – 1) and S(1, 2).

Question 18.
Prove that : \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } \)

OR

Prove that : \(\frac { 1+cosA }{ sinA } +\frac { sinA }{ 1+cosA } =2cosecA\)

Question 19.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.

Question 20.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Question 21.
A solid sphere made of copper has a diameter of 6 cm. It is melted and recast into small spherical balls of diameter 2 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls obtained from the given sphere.

OR

A square field and an equilateral triangular park have equal perimeters. Cost of tilling the field at the rate of ₹ 5 per m² is ₹ 720. What is the cost of grassing the park at the rate of ₹ 10 per m² correct to nearest rupee ?

Question 22.
The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution.

SECTION-D

Question 23.
In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.

Question 24.
180 logs are stacked in the following manner : 19 logs in the bottom row, 18 in the next row, 17 in the row next to it and so on. In how many rows are the 180 logs placed and how many logs are in top row ?

Question 25.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

OR

State and prove the Basic Proportionality Theorem.

Question 26.
If tan A = n tan B and sin A = m sin B, prove that \({ cos }^{ 2 }A=\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)

Question 27.
The angle of elevation of the top of a building at a point on the level ground is 30°. After walking a distance of 100 m towards the foot of the building along the horizontal line through the foot of the building on the same level ground, the angle of the elevation of the top of the building is 60°. Find the height of the building.

OR

From the top of a building 100 m high, the angles of depression of the top and bottom of a tree are observed to be 45° and 60° respectively. Find the height of the tree.

Question 28.
Draw ∆ABC with AB = 5 cm, ∠B = 60° and BC = 4 cm. Now construct a triangle whose sides are \(\\ \frac { 4 }{ 5 } \) times the corresponding sides of ∆ABC. Give the steps of construction.

Question 29.
A metallic right circular cone 20 cm high and whose semi-vertical angle is 30° is cut into two parts from the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

OR

A solid consisting of a right circular cone of height 120 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. How many litres of water is left in the cylinder, if the radius of cylinder is 60 cm and its height 180 cm ? (Use π = \(\\ \frac { 22 }{ 7 } \))

Question 30.
The following table gives the yield per hectare of wheat of 60 farms in a village.

Draw a more than ogive for the above data.

SOLUTIONS

SECTION-A

Solution 1:
Given numbers are 300 and 1000.
∵ 300 = 3 x 2² x 5²
1000 = 2³ x 5³
∴ HCF (300, 1000) = 2² x 5² = 100. Ans.

Solution 2:
Given quadratic equation is,
2x² + 5x + k = 0
Here, A = 2, B = 5, C = k
Equation has equal roots, if
D = 0
=> B² – 4AC = 0
=> 25 – 4 x 2 x k = 0
=> 8k = 25
=> k = \(\\ \frac { 25 }{ 8 } \)

Solution 3:
Given A.P. is – 4, – 2, 0, 2, 4,…
Here, a = – 4, d = 2, n = 101
a₁₀₁ = a + 100d = – 4 + 100 x 2
= 196.

Solution 4:
Given, AB = 10 cm, DE = 12 cm, pr. (∆ABC) = 25 cm
It is given that ∆ABC ~ ∆DEF
=> \(\frac { pr.\left( \triangle ABC \right) }{ pr.\left( \triangle DEF \right) } =\frac { AB }{ DE } \)
=> \(\frac { 25 }{ pr.\left( \triangle DEF \right) } =\frac { 10 }{ 12 } \)
=> pr.(∆DEF) = 30 cm

Solution 5:
Given points are A(9, – 40) and O(0, 0).
Using distance formula,
OA = \(\sqrt { { (9-0) }^{ 2 }+{ (-40-0) }^{ 2 } } \)
= \(\sqrt { { 9 }^{ 2 }+{ (-40) }^{ 2 } } \)
= \(\sqrt { 81+1600 } \)
= \(\sqrt { 1681 } \)
= 41 units.

Solution 6:
Given
tan A = \(\frac { 1 }{ \sqrt { 5 } } \)
We have,
\(\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+{ sin }^{ 2 }A } \)
Dividing N_r and D_r by cos² A, we get

SECTION-B

Solution 7:
Given numbers are 612 and 408.
Now, using Euclid’s division algorithm
612 = 408 x 1 + 204
408 = 204 x 2 + 0
HCF = 204
We know that HCF x LCM = a x b
=> 204 x LCM = 612 x 408
=> LCM = \(\\ \frac { 612X408 }{ 204 } \)
= 1224

Solution 8:
Given AP is 60, 130, 200,…., 2160.
Here, l = 2160, d = 130 – 60 = 70
∴ nth term from end = l – (n – l)d
10th term from end = 2160 – (10 – 1) x 70
= 2160 – 630 = 1530. Ans.

Solution 9:
Given equations are,
ax + by = a² + b² …(i)
and bx + ay = 2ab …(ii)
On multiplying equation (i) by a and equation (ii) by b, we get

Solution 10:
Given points are A( – 4, 3) and B(6, 5).
Let the point on Y-axis be P(0, b). (Given)
Now, PA = PB
=> PA² = PB²
=> (0 + 4)² + (b – 3)² = (0 – 6)² + (b – 5)²
=> 16 + b² – 6b + 9 = 36 + b² – 10b + 25
=> b² – 6b + 25 = b² – 10b + 61
=> 4b = 36
=> b = 9
The required point is (0, 9). Ans.

Solution 11:
Given : Mean = 52, Median = 54.5.
We know, Mode = 3 Median – 2 Mean
= 3 x 54.5 – 2 x 52
= 163.5 – 104
=> Mode = 59.5. Ans.

Solution 12:
Total number of cards = 52
No. of black king cards = 2
No. of red queen cards = 2
No. of face cards = 12
(i) P(black king or red queen) = P(black king) + P(red queen)
\(=\frac { 2 }{ 52 } +\frac { 2 }{ 52 } =\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
(ii) P(non-face card) = 1 – P(face card)
\(=1-\frac { 12 }{ 52 } =\frac { 40 }{ 52 } =\frac { 10 }{ 13 } \)

SECTION-C

Solution 13:
Let √7 is a rational number.
Then, √7 can be written in the form
√7 = \(\\ \frac { p }{ q } \)
such that q ≠ 0
and p and q are co prime integers
On squaring both sides,

Solution 14:
We have,
Dividend = x³ – 3x² + x + 2
Quotient = x – 2
Remainder = – 2x + 4
We know,

Solution 15:
Given equations are,
3x – 2y = 5…(i)
and 2x + 3y = 12…(ii)
On solving, we get
l₁ : 3x – 2y = 5

Solution 16:
Given : AO x DO = BO x CO
To prove : ABCD is a trapezium.
Proof : We have,
AO x DO = BO x CO

=> \(\\ \frac { AO }{ OC } \) = \(\\ \frac { BO }{ DO } \)
Also ∠1 = ∠2
∴ By SAS similarity axiom,
∆AOB ~ ∆COD
∠3 = ∠4
∠3 and ∠4 from a pair of alterate angles
AB || CD
Hence, ABCD is a trapezium.
Hence Proved.

OR

Given, ∆ABC and ∆PQR in which AD and PM are medians drawn on sides BC and QR respectively. It is also given that
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AD }{ PM } \)

Solution 17:
Given, vertices of quadrilateral are P(- 5, – 3), Q(- 4, – 6), R(2, – 1) and S(1, 2).
Construction : Join R.

Now, quadrilateral PQRS is divided into two triangles, ∆PSR and ∆PQR.
∴ar(\(\Box \) PQRS) = ar (∆PQR) + ar (∆PSR)
= \(\\ \frac { 1 }{ 2 } \) |- 5( – 6 + 1) – 4( – 1 + 3) + 2( – 3 + 6) | + \(\\ \frac { 1 }{ 2 } \) | – 5(2 + 1) + 1( – 1 + 3) + 2( – 3 – 2) |
= \(\\ \frac { 1 }{ 2 } \) | 25 – 8 + 6 | + \(\\ \frac { 1 }{ 2 } \) | – 15 + 2 – 10 |
= \(\\ \frac { 1 }{ 2 } \) | 23 | + \(\\ \frac { 1 }{ 2 } \) | – 23 |
= 23 sq. unit.

Solution 18:
Given
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)

Solution 19:
Given : l and m are two parallel tangents, n is the intercept of tangents which intersect circle at point C.
To prove : ∠AOB = 90°.
Construction : Join OP and OC.
Proof : In ∆OPA and ∆OAC, we have

OP = OC [Radii]
AP = AC
[Tangents to a circle from an external point are equal in length]
AO = AO (common)

Solution 20:
Given ar(∆ABC) = 17320.5 cm²
let radius of circle be r
Then,AB = BC = CA = 2r

Solution 21:
Given,
Diameter of sphere of copper = 6 cm
Radius, R = \(\\ \frac { 1 }{ 2 } \) = 3 cm
Diameter of small spherical balls = 2 cm
Radius, r = \(\\ \frac { 2 }{ 2 } \) = 1 cm
Let number of small spheres be n.
Then,
n x Volume of small sphere = Volume of large sphere

Solution 22:

SECTION-D

Solution 23:
Let marks obtained in mathematics be x and marks obtained in science be y.
According to the question,
x + y = 28
=> y = 28 – x…(i)
Also (x + 3).(y – 4) = 180 ,
=> (x + 3) (28 – x – 4) = 180
=> (x + 3) (24 – x) = 180
=> 24x – x² + 72 – 3x = 180
=> x² – 21x + 108 = 0
=> x² – 12x – 9x + 108 = 0
=> x(x – 12) – 9(x – 12) = 0
=> (x – 12) (x – 9) = 0
=> x = 12 or x = 9.
∴Marks obtained in mathematics = 12 or 9.
If marks obtained in mathematics = 12 then marks obtained in science = 28 – 12 = 16
If marks obtained in mathematics = 9 then marks obtained in science = 28 – 9 = 19

Solution 24:
Let number of rows be ‘n’.
No. of logs in bottom row = 19
In next row = 18 and so on Also, total logs = 180
=> 19 + 18 + 17 + … (n terms) = 180
This is an A.P.
Here, a = 19, d = – 1, n = ?, S_n = 180
We know,

Solution 25:
Given : A right-angled triangle ABC in which ∠B = 90°.
To prove : (Hypotenuse)² = (Base)² + (Perpendicular)²
i.e„ AC² = AB² + BC²
Construction : From B, draw BD ⊥ AC.
Proof : In triangles ADB and ABC, we have
∠ADB = ∠ABC
and, ∠A = ∠A

Solution 26:
Given
sin A = m sin B,
tan A = n tan B
=> \(m=\frac { sinA }{ sinB } ,n=\frac { tanA }{ tanB } \)

Solution 27:
Let AB be the building, C and D be the points of observation.
Given : DC = 100 m
Let BC = x m and AB = h m

Solution 28:
Steps of construction :
(1) Draw a line segment BC = 6 cm.
(2) Construct ∠YBC = 60°.
(3) Taking B as centre and radius 5 cm, draw on arc, intersecting BY at A.
(4) Join AC. Thus, ∆ABC is obtained.
(5) Draw a ray BX such that ∠CBX is an acute angle and BX lies in exterior of ∆ABC.
(6) Mark B₁ B₂, B₃ and B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
(7) Join B₃ to C.
(8) From B₄, draw a line parallel to B₃C to meet BC produced at C’.
(9) From C’ draw a line parallel to CA to meet AB produced at A’.
Thus, ∆A’BC’ is the required triangle whose sides are \(\\ \frac { 4 }{ 3 } \) of corresponding sides of ∆ABC.

Solution 29:
Let VAD be the right circular cone of height AC = 20 cm. Suppose the cone is cut by a plane parallel to its base at a point B such that AB = BC

Solution 30:

We hope the CBSE Sample Papers for Class 10 Maths Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 6, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 10

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 10.

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

❖ All questions are compulsory.
❖ The question paper consists of 30 questions divided into four sections – A, B, C and D.
❖ Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
❖ There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
❖ Use of calculator is not permitted.

SECTION – A

Question 1.
Find the distance of the point P(-6, 8) from the origin.

Question 2.
∆ABC is similar to ∆DEF. Area of ∆ABC is nine-sixteenth the area of ∆DEF. If perimeter of ∆DEF is 24 cm then find the perimeter of ∆ABC.

Question 3.
Write whether \(\frac { 2\surd 45+3\surd 20 }{ 2\surd 5 }\) on simplification gives a rational or an irrational number.

Question 4.
Find the sum and product of zeroes of the polynomial 7x² + 5x – 2.

Question 5.
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.

Question 6.
∆ABC is right angled at A and ∆PQR is right angled at P. If cos B = cos Q, show ∠B = ∠Q.

SECTION – B

Question 7.
Find the fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3).

Question 8.
Show that 5 – √3 is irrational.

Question 9.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to drawn. Find the probability that the card drawn is :
(i) A face card
(ii) A red face card
(iii) ‘2’ of spades

Question 10.
A jar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marble at random from the jar is \(\frac { 1 }{ 3 }\), and the probability of selecting a green marble at random is \(\frac { 4 }{ 9 }\) . How many white marbles does the jar contain ?

Question 11.
If the product of zeroes of the polynomial ax² – 6x – 6 is 4, find the value of ‘a’.

Question 12.
Solve for x and y : \(\frac { x }{ 2 }\) + \(\frac { y }{ 4 }\) = 3, 2x – y = 4.

SECTION – C

Question 13.
In what ratio does the X-axis divide the line segment joining the points (-4, -6) and (-1, 7) ? Also find the coordinates of the point of division.

Question 14.
Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the point P, Q and R.
OR
If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4. Find the value of b and c.

Question 15.
In the given figure, a circle is inscribed in ∆PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the length QM, RN and PL.

Question 16.
Find the largest number which can divide 1002 and 1288 leaving remainder 1 in each case.

Question 17.
Find the area of the circle in which a square of area 64 cm² is inscribed. (Use π = 3.14)
OR
A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

Question 18.
How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.
OR
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.

Question 19.
Solve for x and y : \(\frac { 44 }{ x + y }\) + \(\frac { 30 }{ x – y }\) = 10 and \(\frac { 55 }{ x + y }\) +\(\frac { 40 }{ x – y }\) = 13, where x ≠ y and x ≠ -y.
OR
Solve for x and y : 1001x + 999y = 2; 999x + 1001y = -2.

Question 20.
Determine the values of a and b for which the given system of equations has infinitely many solutions : (2a – 1) x + 3y – 5 = 0 and 3x + (b – 1) y – 2 = 0.

Question 21.
If cosec 3A = sec (A – 36°), where 3A is an acute angle, find the value of A.

Question 22.
Find the mode of the following data:

SECTION – D

Question 23.
Draw a circle of radius 3 cm. Take a point P at a distance of 5.5 cm from the centre of the circle. From point P, draw tangents to the circle. Write steps of construction.

Question 24.
Show that the lengths of tangents drawn from an external point to a circle are equal.

Question 25.
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Question 26.
A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
OR
A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Use π = 3.14)

Question 27.
An express train covers a distance of 240 km at a particular speed. Another train whose speed is 12 km/hr less takes one hour more to cover the same distance. Find the speed of the express train.
OR
A shopkeeper buys a number of books for ₹ 80. If he had bought 4 more books for the same amount, each book would have cost him ₹ 1 less. How many books did he buy ?

Question 28.
Mr. Singh repays his total loan of ₹ 118000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month, what amount will be paid by him in the 30th instalment ? What amount of loan does he still have to pay after the 30th instalment ?
OR
A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief ?

Question 29.
The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 m, find the height of the chimney.

Question 30.
The mean of the following frequency distribution is 62.8 and the sum of all frequency is 50. Compute the missing frequencies f₁ and f₂:

SOLUTIONS

SECTION – A

Solution 1.
Given, point P(-6, 8) and origin O(0, 0)
Using distance formula

Solution 2.

Solution 3.

Solution 4.
Given polynomial is 7x² + 5x – 2 = 0
On comparing with ax² + bx + c = 0,
we get,
a = 7, b = 5, c = -2
Sum of zeroes = \(\frac { -b }{ a }\) = \(\frac { -5}{ 7 }\)
Product of zeroes = \(\frac { c }{ a }\) = \(\frac { -2}{ 7 }\)

Solution 5.
Given A.P. is 2x, x + 10, 3x + 2.
Common difference between two consecutive terms of A.P. is same.
b – a = c – b
⇒ 2b = a + c
⇒ 2(x + 10) = 2x + 3x + 2
⇒ 2x + 20 = 5x + 2
⇒ 20 – 2 = 5x – 2x
⇒ 18 = 3x
⇒ x = 6.

Solution 6.
Given : ∠A = ∠P = 90° ……(i)
and
cos B = cos Q
⇒ \(\frac { AB }{ BC }\) = \(\frac { PQ }{ QR }\) ……(ii)
From equation (i) and (ii),
∆ABC ~ ∆DEF [S.A.S.]
∠B = ∠Q.
Hence Proved.

SECTION – B

Solution 7.
Given, coordinates of parallelogram ABCD are A(-2, 3), B(6, 7) and C(8, 3)
Let the coordinates of D be (α, β).
We know that diagonals of parallelogram bisect each other.
So, mid-point of AC = mid-point of BD


Coordinates of vertex D = (0, -1).

Solution 8.
Let us assume on the contrary that 5 – √3 is rational number. Then there exist co-prime positive integers x and y such that
5 – √3 = \(\frac { x }{ y }\)
5 – \(\frac { x }{ y }\) = √3
\(\frac { 5y – x }{ y }\) = √3
Since x, y are integers, so \(\frac { 5y – x }{ y }\) is a rational number
√3 is rational.
This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 5 – √3 is an irrational number.
Hence Proved.

Solution 9.
Given :
Total cards = 52
(i) We know that there are 12 face cards
P(a face cards) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(ii) We know that there are 6 red face cards
P(Red face cards) = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)
(iii) There is only one card of ‘2’ of spades
P(2 of spade) = \(\frac { 1 }{ 52 }\)

Solution 10.
We have,
Total marbles = 54
P(blue marble) = \(\frac { 1 }{ 3 }\)
P(green marble) = \(\frac { 4 }{ 9 }\)
Let number of white marbles be x.

Solution 11.
Given polynomial is,
ax² – 6x – 6
On comparing with Ax² + Bx + C = 0, we get
A = a, B = -6, C = -6
Product of roots = \(\frac { C }{ A }\)
4 = \(\frac { -6 }{ a }\)
a = \(\frac { -6 }{ 4 }\) = \(\frac { -3 }{ 2 }\)
Value of a = \(\frac { -3 }{ 2 }\)

Solution 12.
We have,

SECTION – C

Solution 13.

Solution 14.
We have, points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) into five equal parts.

Solution 15.
Given, PQ = 10 cm, PR = 12 cm and QR = 8 cm.
Let the length of QM be x cm
Thus,
RM = (8 – x) cm
and QL = x cm [LQ and QM are tangents from Q]
RN = (8 – x) cm [RN and RM are tangents from R]
Now, PN = PR – RN = [12 – (8 – x)] cm = (4 + x) cm
and PN = PL [PL and PN are tangents from P]
Again, QL = PQ – PL = [10 – (4 + x)] cm = (6 – x) cm
But QL = x cm [From above]
x = 6 – x
⇒ 2x = 6
⇒ x = 3 cm
QM = 3 cm
RN = (8 – 3) cm = 5 cm
and PL = (4 + 3) cm = 7 cm.

Solution 16.
Let x be the required number which divides 1002 and 1288 leaving remainder 1 each case.
This means x perfectly divide (1002 – 1) and (1288 – 1) i.e., 1001 and 1287.
So, x is the HCF of 1001, 1287
Using Euclid’s division alogrithm,
1287 = 1001 x 2 + 286
1001 = 286 x 3 + 143
286 = 143 x 2 + 0
Hence, HCF of 1001 and 1287 is 143.

Solution 17.
Given : Area of square = 64 cm²
Let side of square and radius of circle be a cm and r cm, respectively.
Now, we know

Solution 18.
Given:
External length of the box = 36 cm
External breadth of the box = 25 cm
External height of the box = 16.5 cm
Width of the box = 1.5 cm
Internal length of the box = 36 – 1.5 = 34.5 cm [Internal length = External length – Width]
Internal breadth of the box = 25 – 1.5 = 23.5 cm
Internal height of the box = 16.5 – 1.5 = 15 cm
Volume of iron in the box = External volume of the box – Internal volume of the box
= 36 x 25 x 16-5 – 34.5 x 23.5 x 15
= 14850 – 12161.25
= 2688.75 cm³
1 cm³ iron = 7.5 gram
2688.75 cm³ iron = 7.5 x 2688.75 gram
= 20165.625 gram.
OR
Let the radii of circular ends of frustum be R cm and r cm and its slant height be 1 cm.
Then, 2πR = 18 cm ⇒ πR = 9 cm
and 2πr = 6 cm ⇒ πr = 3 cm
Also, l = 4 cm
Curved surface area of the frustum = πl(R + r) = l(πR + πr) = 4(9 + 3) = 48 cm².

Solution 19.
Given equations are,

Solution 20.
Given system of equation is,
(2a – 1)x + 3y – 5 = 0 …..(i)
and 3x + (b – 1)y – 2 = 0 …..(ii)

Solution 21.
We have,
cosec 3A = sec (A – 36°)
sec (90° – 3A) = sec (A – 36°) [cosec A = sec (90° – A)]
90° – 3A = A – 36°
90° + 36° = A + 3A
4A = 126°
A = 31.5°.

Solution 22.

SECTION – D

Solution 23.
Steps of construction :
1. Draw a circle of radius 3 cm taking O as centre.
2. Draw a point P at a distance of 5.5 cm from the centre and join OP.
3. Draw the perpendicular bisector of OP, intersecting OP at point M.
4. Taking MP as radius and M as centre, draw a circle which cut the given circle at two points A and B.
5. Join A to P and B to P.
Thus, AP and BP are required tangents.

Solution 24.
Let P be a point outside a circle from which two tangents PA and PB are drawn.
To prove : PA = PB.
Proof : In ∆PBO and ∆AOP,
OP = OP [Common]
OB = OA [Radius of the circle]
∠OBP = ∠OAP
[Both 90°, any point joined from centre to point of contact of tangent is perpendicular]
By R.H.S. congruency rule,
∆PBO = ∆AOP
PA = PB [cpct]
So, tangents are equal when drawn from an exterior point to the circle.
Hence Proved.

Solution 25.
Let AB be the tower of height ‘h’ m, BC be the shadow of tower when sun’s altitude is 60° and BD be the shadow of tower when sun’s altitude is 30°.

Solution 26.
Given:
Height of cylindrical bucket, H = 32 cm
Radius of cylindrical bucket, R = 18 cm
and Height of cone, h = 24 cm
Let r be the radius of cone.
According to the question,
Volume of cone = Volume of cylinder
\(\frac { 1 }{ 3 }\) πr²h = πR²H

Solution 27.
Given:
Let Total distance to be covered = 240 km
Speed of Express train = x km/h
Speed of another train = (x – 12) km/h

Solution 28.
Let total installment paid be ‘n’.
Since, the amount of each installment increases by ₹ 100 every month, so installments are paid in A.P.
Given : Total loan, S_n = 118000
Amount of first installment, a = 1000
Increase in amount of each installment, d = 100
Amount to be paid in 30th installment = a₃₀
We know that
a_n = a + (n – 1 )d
a₃₀ = 1000 + (30 – 1)100 = 1000 + 2900
a₃₀ = 3900
So, amount to be paid in 30th installment is ₹ 3900.
Total amount paid in 30th installment = \(\frac { 30 }{ 2 }\) [2 x 1000 + 29 x 100]
= 15[2000 + 2900] = 15 x 4900 = 73500
Amount left to be paid = 118000 – 73500
= ₹ 44500.
OR
Let total time be n minutes.
Since, policeman runs after 1 minutes, so he will catch the thief in (n – 1) minutes.
Total distance covered by thief = 100 m/minute x n minute =(100n) m
Now, total distance covered by the policeman = (100) m + (100 + 10) m + (100 + 10 + 10) m +…+ (n – 1) terms
= 100 + 110 + 120 +…+ (n – 1) terms

⇒ n² – 6n + 3n – 18 = 0
⇒ n(n – 6) + 3(n – 6) = 0
⇒ (n + 3) (n – 6) = 0
n = 6 or n = – 3 (Neglect)
Hence, policeman will catch the thief in (6 – 1) i.e., 5 minutes.

Solution 29.
Let AB be the tower and CD be the chimney of height h m.
AB = 40 m CD = h m
Let the distance between the foot of the tower and the foot of the chimney be x m.

Solution 30.
Let the assumed mean be A = 50 and class size (h) = 20.

We hope the CBSE Sample Papers for Class 10 Maths Paper 10 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 10, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 9

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 9

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Write down the decimal expansion of rational number \(\\ \frac { 14588 }{ 625 } \)

Question 2.
Check whether the following equation is a quadratic equation : (x – 2) (x + 1) = (x – 1) (x + 5).

Question 3.
In figure, O is the centre of a circle, AB is a chord and AP is the tangent at A. If ∠AOB = 120°, then find ∠BAP.

Question 4.
If 2 cosec² θ (1 – cos θ) (1 + cos θ) = k + 2, then find the value of k.

Question 5.
Find the area of a sector of a circle with radius 4 cm, if angle of the sector is 60°. (Use π = 3.14)

Question 6.
A bag contains cards numbered from 1 to 50. A card is drawn from the bag. Find the probability that the number on this card is divisible by 3 and 5.

SECTION-B

Question 7.
Show that 7√5 is irrational.

Question 8.
How many two-digit number are divisible by 5 ?

Question 9.
Find the point on the X-axis which is equidistant from (2, – 5) and (- 2, 9).

Question 10.
If cos A = \(\\ \frac { 3 }{ 5 } \) then calculate sin A and cot A.

Question 11.
Prove the identity : \(\frac { cosA }{ 1+sinA } +\frac { 1+sinA }{ cosA } =2secA\)

Question 12.
Find the area of a quadrant of a circle whose circumference is 44 cm. (Use \(\pi =\frac { 22 }{ 7 } \))

SECTION-C

Question 13.
Show that any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5) where q is some integer.

Question 14.
Solve for x and y : \(\frac { 10 }{ x+y } +\frac { 2 }{ x-y } =4\quad and\quad \frac { 15 }{ x+y } -\frac { 5 }{ x-y } =-2\)

OR

Solve for x : \(\frac { 1 }{ x+1 } +\frac { 3 }{ 5x+1 } =\frac { 5 }{ x+4 } ;x\neq -1,-\frac { 1 }{ 5 } ,-4\)

Question 15.
Find the roots of the equation 2x² + x – 4 = 0 by the method of completing the square.

OR

If ad ≠ bc, then prove that the equation (a² + b²) x² + 2(ac + bd)x + (c² + d²) = 0 has no real roots.

Question 16.
For what value of n are the nth terms of two A.Ps : 63, 65, 67, … and 3,10,17, … equal ?

OR

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.

Question 17.
Prove the identity : (cosec A – sin A) (sec A – cos A) = \(\frac { 1 }{ tanA+cotA } \)

Question 18.
In figure, PQ = 12 cm, PR = 9 cm and O is the radius of the circle. Find the area of the shaded region. (Use π = 3.14)

OR

In figure, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm show that area of shaded region \(25\left( \sqrt { 3 } -\frac { \pi }{ 6 } \right)\) cm²

Question 19.
Prove that the parallelogram circumscribing a circle is a rhombus.

Question 20.
In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Question 21.
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

1. a king of black colour,
2. a face card,
3. queen of hearts.

Question 22.
The given distribution shows the marks, out of 50 obtained by 100 students in a test:

Find the mean and median

SECTION-D

Question 23.
Find the 40th term of the A.P. 8,10,12, …, if it has a total of 40 terms and hence find the sum of its last 10 terms.

Question 24.
The total cost of a certain length of a piece of cloth is Rs 100. If the piece was 5 m longer and each metre of cloth costs Rs 1 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre ?

OR

A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it ?

Question 25.
Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (- 2, – 3).

OR

In what ratio does the point \(\left( \frac { 24 }{ 11 } ,y \right) \) divide the line segment joining the points P(2, – 2) and Q(3, 7) ? Also find the value of y.

Question 26.
Prove that the lengths of the tangents drawn from external point to a circle are equal.

Question 27.
Construct a triangle ABC in which AB = 8 cm, BC = 10 cm and AC = 6 cm. Then construct another triangle whose sides are \(\\ \frac { 3 }{ 5 } \) of the corresponding sides of ∆ABC.

Question 28.
The angles of depression of the top and bottom at a 12 m tall building from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multi-storey building.

OR

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. (Use √3 = 1.732)

Question 29.
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of base of the cone is 21 cm and its volume is \(\\ \frac { 2 }{ 3 } \) of the volume of the hemisphere. Calculate the height of the cone and the surface area of the toy. (Use \(\pi =\frac { 22 }{ 7 } \))

Question 30.
A box contains cards bearing numbers from 5 to 69. If one card is drawn at random from the box, find the probability that it bears :
(i) a one-digit number,
(ii) a number divisible by 5,
(iii) an odd number less than 30,
(iv) a prime number between 50 to 69.

SOLUTIONS

SECTION-A

Solution 1.
Let
\(x=\frac { 14588 }{ 625 } =\frac { 14588 }{ { 5 }^{ 4 } } \)
= \(\frac { 14588\times { 2 }^{ 4 } }{ { 2 }^{ 4 }\times { 5 }^{ 4 } } =\frac { 14588\times { 2 }^{ 4 } }{ 10^{ 4 } } \)
= \(\frac { 233408 }{ { 10 }^{ 4 } } \)
= 23.3408
So, decimal expression is 23.3408

Solution 2.
We have,
(x – 2)(x + 1) = (x -1) (x + 5)
x² – 2x + x – 2 = x² + 5x – x – 5
x² – x – 2 = x² + 4x – 5
– 5x + 3 = 0
Since, the above equation is not of the form ax² + bx + c = 0
Therefore, the given equation is not a quadratic equation.

Solution 3.
Given
∠AOB = 120° [Radii]
Since AO = OB
∴ ∆AOB is an isosceles triangle.
=> ∠OAB = ∠OBA [cpct]
∴ In ∆ABOB
∠OAB + ∠AOB + ∠OBA =180° [Angle sum property]
=> 2∠OAB + 120° = 180°
=> ∠OAB = ∠OBA = 30°
Now, ∠OAP = 90° [∴ Tangent AP ⊥ radius OA]
Thus, ∠BAP = ∠OAP – ∠OAB
= 90° – 30° = 60°
Hence, ∠BAP = 60°. Ans.

Solution 4.
We have,
2 cosec² θ(1- cos θ) (1 + cos θ) = k + 2
=> 2 cosec² θ (1 – cos² θ) = k + 2 [∵ (a + b)(a – b) = a²-b²)]
=> 2 cosec² θ sin² θ = k + 2 [∵ 1 – cos² θ = sin² θ]
=> \(2\times \frac { 1 }{ { sin }^{ 2 }\theta } \times { sin }^{ 2 }\theta =k+2\)
=> 2 = k + 2
=> k = 0 Ans.

Solution 5.
Given sector is OAPB.
Radius of circle, r = 4 cm
Angle of sector, θ = 60°
Now,

Hence, the area of the sector is 8.37 cm².

Solution 6.
Total number of possible outcomes = 50 = n(S)
Let E be the event that the number on the card drawn is divisible by both 3 and 5.
∴ E = {15, 30,45}
∴ n(E) = 3
∴ \(P(E)=\frac { n(E) }{ n(S) } =\frac { 3 }{ 50 } \)

SECTION-B

Solution 7:
Let us assume that 7√5 is rational.
Therefore, we can find co-prime a and b(≠ 0) such that
7√5 = \(\\ \frac { a }{ b } \)
or
√5 = \(\\ \frac { a }{ 7b } \)
Now, since a, b and 7 are integers, therefore \(\\ \frac { a }{ 7b } \) is rational and so √5 is rational, which is not true as √5 is irrational.
Hence, we conclude that 7√5 is irrational.

Solution 8:
Two-digit numbers that are divisible by 5 are
10,15,20,…, 90,95.
It is an A.P. with a = 10, d = 15 – 10 = 5 and a_n = 95
We know a_n = a + (n – 1 )d
=> 95 = 10 + (n – 1)5
=> 85 = (n – 1)5
=> 17 = n – 1
=> n =18
So, there are 18 two-digit numbers that are divisible by 5.

Solution 9:
Let the required point on the X-axis be (a, 0).
Then, according to question,
Distance of (a, 0) and (2, – 5) = Distance of (a, 0) and ( – 2, 9)
=> \(\sqrt { { (2-a) }^{ 2 }+{ (-5-0) }^{ 2 } } =\sqrt { { (-2-a) }^{ 2 }+{ (9-0) }^{ 2 } } \)
Squaring on both sides, we get
(2 – a)² + 5² = (- 2 – a)² + 9²
=> 4 + a² – 4a + 25 = 4 + a² + 4a + 81
=> – 8a = + 56
=> a = – 7
∴ ( – 7, 0) is the point which is equidistant from (2, – 5) and (- 2, 9).

Solution 10:
Let ABC be a triangle, then
\(cosA=\frac { AB }{ AC } =\frac { 3 }{ 5 } \)
Let AB = 3k and AC = 5k

Solution 11:
L.H.S = \(\frac { cosA }{ 1+sinA } +\frac { 1+sinA }{ cosA } \)
\(\frac { { cos }^{ 2 }A+{ (1+sinA) }^{ 2 } }{ (1+sinA)cosA } \)

Solution 12:
Let r be the radius of the circle,
Given
circumference = 2πr = 44 cm

SECTION-C

Solution 13:
Let k be any integer.
Using Euclid’s division lemma with k and 6, we get
k = 6q + r, where 0 ≤ r ≤ 6
For r = 0
k = 6q i.e., even number
For r = 1
k = 6q + 1 i.e., odd number
For r = 2
k = 6q + 2 i.e., even number
For r = 3
k = 6q + 3 i.e., odd number
For r = 4
k = 6q + 4 i.e., even number
For r = 5
k = 6q + 5 i.e., odd number
Therefore, any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5).

Solution 14:
Given equations are,
\(\frac { 10 }{ x+y } +\frac { 2 }{ x-y } =4 \)
and \(\frac { 15 }{ x+y } -\frac { 5 }{ x-y } =-2\)

Solution 15:
Given equation is,
2x² + x – 4 = 0
On multiplying the equation by 1/2 ,we get

Solution 16:
First A.P. is 63, 65, 67, …
Here, a = 63
d = 65 – 63 = 2
.’. nth term, a_n = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
= 61 + 2n

Solution 17:
L.H.S = (cosec A – sin A) (sec A – cos A)
= \(\left( \frac { 1 }{ sinA } -sinA \right) \left( \frac { 1 }{ cosA } -cosA \right) \)

Solution 18:
Since, the angle in a semi-circle is a right angle
∠RPQ = 90°
Thus, PQR represents a right angled triangle
where PR = 9 cm, PQ = 12 cm
In ∆PQR,using Pythagoras theorm

Solution 19:
Consider a parallelogram ABCD circumscribing a circle. Sides AB, BC, CD and AD touch the circle at points P, Q, R and S respectively.

Since the lengths of the tangents drawn from an external point to a circle are equal.
AP =AS …(i)
PB =BQ …(ii)
DR =DS …(ii)
CR = CQ …(iv)
On adding all equations, we get
(AP + PB) + (DR + CR) = (AS + DS) + (BQ + CQ)
=> AB + CD = AD + CB
=> 2AB = 2BC [∵ in a parallelogram, opposite sides are equal]
=> AB =BC
Since, adjacent sides of a parallelogram are equal
∴ ABCD is a rhombus.

Solution 20:
Given : XY and X’Y’ are parallel.
Tangent AB is another tangent which touches the circle at C.

Solution 21:
Total number of outcomes = 52 = n(S).
(i) Let E denote the event of getting a king of black colour.
There are two cards in this category i.e., king of club and king of spade.

Solution 22:

SECTION-D

Solution 23:
The given A.P. is 8,10,12,…
So, a = 8, d = 10 – 8 = 2.
We know that an = a + (n – 1 )d
So, 40th term = a₄₀ = 8 + 39 x 2
= 8 + 78
= 86
Hence, the 40th term of the given A.P. is 86.
Now, since the A.P. has a total of 40 terms. So, to find the sum of last n terms, we take

Hence, the sum of last 10 terms of the given A.P. is 770.

Solution 24:
Let the length of the piece of the cloth be x m.
Total cost of the piece of cloth = Rs 100
Then cost per metre = Rs \(\\ \frac { 100 }{ x } \)
New length = (x + 5) m
Since the cost of the piece of cloth remains unchanged

Solution 25:
We have, A(4, -1) and B(- 2, – 3)
Let P and Q be the points of trisection of A and B.
i.e., AP = PQ = QB

Solution 26:
Given, PA and PB are two tangents drawn from an external point P to the circle C(0, r).
Construction: Join OA and OB.
To prove : PA = PB
Proof : Since, tangent at any point to a circle is perpendicular to the
radius through the point of contact
PA ⊥ OA and PB ⊥ OB

Solution 27:
Steps of the construction :
(1) Draw a line segment BC = 10 cm.
(2) Taking B point as centre and radius 8 cm, draw an arc.
(3) Similarly, taking point C as centre and radius equal to 6 cm, draw an arc. Let the two arcs intersect at point A.
(4) Join AB, AC.
Thus, ∆ABC is obtained.
(5) Draw a ray BX making an acute angle with line BC on the opposite side of vertex A.
(6) Mark 5 points B₁ ,B₂ ,B₃, B₄ and B₅ on ray BX such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
(7) Join CB₅.
(8) Through B₃, draw a line parallel to CB₅ intersecting BC at point C.
(9) Draw a line parallel to line AC through C’ intersecting AB at A’.
Thus, ∆A’BC’ is obtained whose sides are \(\\ \frac { 3 }{ 5 } \) of corresponding sides of ∆ABC.

Solution 28:
Let AB be the tall building of height 12 m and CD be the multi-storey building.
Let CD = h m
Given: ∠DBE = 30° and ∠DAC = 60°
Here, AB = CE = 12 m
DE = CD – CE = (h – 12)m

Solution 29:
Let
h = height of the cone
Radius of the cone = Radius of hemisphere
= 21 cm

Solution 30:
Numbers of possible outcomes = 69 – 5 + 1 = 65
n(S) = 65
(i) Let A be the event of getting one digit number.
Favourable outcomes = {5,6,7,8,9}
n(A) = 5

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These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 1.

CBSE Sample Papers for Class 10 Maths Paper 1

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time Allowed: 3 hours
Max. Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
After how many decimal places will the decimal expansion of the number 47/2^{3 .} 5²
terminate ? [1]

Question 2.
If two zeroes of polynomial ƒ(x) = x³ – 4x² – 3x + 12 are √3 and – √3, then find third zero. [l]

Question 3.
Find the smallest positive value of k for which the equation x² + kx + 9 = 0 has real roots. [1]

Question 4.
The coordinates of the points P and Q are (4, – 3) and (- 1, 7), respectively. Find the abscissa of a point R on the line segment PQ such that \(\frac { PR }{ PQ } \) = 3/5. [1]

Question 5.
In the adjoining figure, PA and PB are tangents from a point P to a circle with centre O. Show that quadrilateral that OAPB is cyclic. [1]

Question 6.
If for some angle θ, cot 2θ = 1/√3, then find the value of sin 30, where 2θ < 90°. [1]

SECTION-B

Question 7.
Is there any natural number n for which 6ⁿ ends with the digit 0 ? Give reasons in support of your answer. [2]

Question 8.
Which term of the A.P. : 5, 17, 29, 41, … will be 120 more than its 15th term ? [2]

Question 9.
The length of a line segment is 10 units. If one end is (2, – 3) and the abscissa of the other end is 10, then its ordinate is either 3 or – 9. Give justification for the two answers. [2]

Question 10.
Find the value of c for which the pair of equations cx – y = 2 and 6x – 2y = 4 will have infinitely many solutions. [2]

Question 11.
More than ogive and less than ogive for a given data with mean 23 meet at (20, 22). Find its mode. [2]

Question 12.
A letter of English alphabets is chosen at random. Find the probability that it is a letter of the word ‘MATHEMATICS’. [2]

SECTION-C

Question 13.
Find the zeroes of the polynomial p(x) = 4√3x² – 2√3x – 2√3 and verify the relationship between the zeroes and the coefficients. [3]
OR
On dividing the polynomial ƒ(x) = x³ – 5x² + 6x – 4 by a polynomial g(x), the quotient q(x) and remainder r(x) are x – 3 and – 3x + 5, respectively. Find the polynomial g(x).

Question 14. 
If the sum of the first n terms of an A.P. is 4n – n² , what is the 10^th term and the n^th term ? [3]
OR
Flow many terms of the A.P. : 9, 17, 25, … must be taken to give a sum 636 ?

Question 15.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find the value of x and y. [3]

Question 16.
The sides AB, BC and median AD of a ∆ABC are respectively proportional to the sides PQ, QR and the median PM of ∆PQR. Show that ∆ABC ~ ∆PQR. [3]
OR
ABC is an isosceles triangle right angled at B. Two equilateral triangles ACD and ABE are constructed on the sides AC and AB, respectively. Find the ratio of the area of ∆ABE and ∆ACD.

Question 17.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm, respectively. Find the sides AB and AC. [3]
OR
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Question 18.
Evaluate:

Question 19.
In the figure, ABC is a triangle right angled at A. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. [3]

Question 20.
A bag contains white, black and red balls only. A ball is drawn at random from the bag. The probability of getting a white ball is 3/10 and that of a black ball is 2/5. Find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag. [3]

Question 21.
Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer. [3]

Question 22.
A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank. [3]

SECTION-D

Question 23.
Construct an isosceles triangle whose base is 6 cm and altitude 5 cm and then construct another triangle whose sides are \(\frac { 7 }{ 5 } \) of the corresponding sides of the isosceles triangle. [4]

Question 24.
Prove that :

Question 25.
If the price of a book is reduced by ₹5, a person can buy 5 more books for ? 300. Find the original list price of the book. [4]
OR
Solve for x :

Question 26.
State and prove the basic proportionality theorem. [4]
OR
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Question 27.
A building is made by keeping the lower window of a building at a height of 2 m above the ground and its upper window 6 m vertically above the lower window in view to have proper sunlight. At certain instant, the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground. [4]

Question 28.
A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [4]

Question 29.
Solve the equations 5x – y = 5 and 3x – y = 3 graphically. Find area of the triangle formed between the lines and Y-axis. [4]
OR
A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Question 30.
The following table shows the ages of the patients admitted in a hospital during a month :

Find the mean and the mode of the data given above. [4]

SOLUTIONS
SECTION-A

Answer 1.
Let

x will terminate after 3 decimal places.

Answer 2.
Let α = √3, β = √3 be the given zeroes and γ be the third zero, then
α + β + γ = \(-\frac { -4 }{ 1 } \)
√3 – √3 + γ = 4
γ = 4
Hence, third zero is 4.

Answer 3.
Given equation is,
Here A = 1, B = k, C = 9
∵ Roots are real
∵ D ≥ 0
⇒ B² – 4AC ≥ 0
⇒ k² -4 × 1 × 9 ≥ 0
⇒ k²- 36 ≥ 0
⇒ k² ≥ 36
⇒ k² ≥ 6²
⇒ k ≥ 6 or k ≤ -6
So,the least positive value for which the given equation has real roots is k = 6.

Answer 4.
The coordinates of P and Q are (4,-3) and (-1,7) respectively.
Let coordinate of R be (x,y).

Now,

By section formula,

Thus, abscissa of point R is 1.

Answer 5.
In a circle, radius is perpendicular to the tangent at the point of contact.
∵ ∠OAP = 90°
and ∠OBP = 90°
=> ∠OAP + ∠OBP = 90° + 90° = 180°
∵ Opposite angles are supplementary .
∵ Quadrilateral OAPB is cyclic.

Answer 6.
Here cot 2θ = \(\frac { 1 }{ \sqrt { 3 } }\) = cot 60°
⇒  2θ = 60° ⇒ θ = 30°
∵ sin 3θ = sin 90° = 1.

SECTION-B

Answer 7.
6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
6ⁿ cannot end in zero for any natural number ‘n’.
To a number end in zero, ‘2′ and ‘5’ both must be factor of the number.
Factor ‘5’ is not present in 6ⁿ so it will not end in ‘0’ for any natural number ‘n’

Answer 8.
Given A.P. is 5, 17, 29, 41..
Here,
a = 5, d = 17 – 5 = 12
Let n^th term of the A.P. is 120 more than its 15th term.
Then, according to the question,
a_n = 120 + a₁₅
a + (n-1)d = 120 + a + 14d
(n-1) × 12 = 120 + 14 × 12
(n-1) × 12 = 288
n-1 = 24
n = 25
∵ Required term is 25^th term.

Answer 9.
Let AB be the line segment of length 10 cm having coordinates A(2, – 3) and B(10, b).
Here


For x = 10, there will be two points which are at a distance of 10 units from A.

Answer 10.
Given pair of equations are,
cx – y = 2
6x – 2y =4
On comparing the equations with a₁x + b₁y + c₁ = 0’and a₂x + b₂y + c₂, respectively, we get
a₁= c : b₁ = – 1, c₁ = 2
a₂ = 6 : b₂ = – 2, c₂ = 4
System has infinitely many solutions, if
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } }\)
⇒ \(\frac { c }{ 6 } =\frac { 1 }{ 2 } =\frac { 2 }{ 4 } \)
⇒ c = 3.

Answer 11.
Given : Mean, \(\bar { x }\) = 23, Median, m₀ = 20 (x-coordinate of point of intersection)
We know,
Mode, M₀ = 3m₀ – 2\(\bar { x }\)
=3 × 20 – 2 × 23 = 60 – 46 = 14

Answer 12.
Favourable cases : M, A, T, H, E, I, C, S
=> No. of favourable cases = 8
=> n(E) = 8
Total cases = 26
=> n(S) = 26
Probability = \(\frac { n(E) }{ n(S) } =\frac { 8 }{ 26 } =\frac { 4 }{ 13 } \)

SECTION-C

Answer 13.
Given polynomial is,
p(x) = 4√3x² – 2√3x – 2√3 = 0
=> 2√3 (2x² – x -1) =0
=> 2√3 (2x²-2x + x-1) =0
2√3 [2x(x – 1) + 1(x – 1)] =0
2√3 (2x + 1)(x – 1) =0
⇒  2x + 1 = 0 ⇒ x = \(-\frac { 1 }{ 2 }\)
and x – 1 = 0 ⇒ x = 1
∵ x = \(-\frac { 1 }{ 2 }\), 1
Now
Sum of zeroes = \(-\frac { 1 }{ 2 } +1=\frac { 1 }{ 2 } =\frac { -Coefficient\quad of\quad x }{ { Coefficient\quad of\quad x }^{ 2 } } \)
Product of zeroes = \(-\frac { 1 }{ 2 } \times 1=\frac { 1 }{ 2 } =-\frac { 1 }{ 2 } =\frac { Constant\quad term }{ { Coefficient\quad of\quad x }^{ 2 } } \)
Hence, relationship between zeroes and coefficient is verified.
OR
We have,
Dividend : ƒ(x) = x³ – 5x² + 6x – 4
Divisor : g(x)
Quotient : q(x) = x – 3
Remainder : r(x) = – 3x + 5
We know,
Dividend = Divisor x Quotient + Remainde
OR
Divisor = \(\frac { Dividend-Remainder }{ Quotient } \)

Answer 14.

Answer 15.
15. Let the vertices of parallelogram be A( 1, 2), B(4, y), C(x, 6) and D(3, 5). We know, diagonals of a parallelogram bisect each other.
Mid-point of AC = Mid-point of BD
By Mid-point formula,

Answer 16.

Answer 17.
Given :
BD = 8 cm
DC = 6 cm
Radius of circle, r = 4 cm
AF = x
Let
We know, tangents to a circle from a same external point are equal
∵ BD = BF = 8 cm
AF = AE = x

DC = CE = 6 cm
AB = (x+8) cm
BC = (8+6) cm
AC = (6+x) cm

Now in ∆ABC
Semi-perimeter, s = \(\frac { AB+BC+CA }{ 2 } \)
\(\frac { X+8+14+x+6 }{ 2 } =\frac { 2x+28 }{ 2 } =x+14\)
We know,


OR
Given :
Circle with centre 0 with exterior point T having TP and TQ as tangents.
To prove : ∠PTQ = 2∠OPQ
Construction : Join PQ.
Proof : TP = TQ

[Tangents to a circle from an external point are equal]
⇒ ∆TPQ is isosceles.
⇒ ∠1 = ∠2
[Angles opposite to eqaul sides of a triangle are equal in length]
Let
∠PTQ = θ
In ∆PTQ,
∠1 + ∠2 + θ = 180°
∠1 + ∠1 = 180° – θ
2∠1 = 180° – θ
∠1 = \(\frac { { 180 }^{ 0 }-\theta }{ 2 } =({ 90 }^{ 0 }-\frac { \theta }{ 2 } )\)
∠TPO = 90°
[Radius is perpendicular to the tangent at the point of contact]
∵ ∠OPQ = 90° – ∠1 = 90° – \(-({ 90 }^{ 0 }-\frac { \theta }{ 2 } )=\frac { \theta }{ 2 }\)

From equation (i) and (ii),
∠OPQ = \(\frac { 1 }{ 2 }\) ∠PTQ
OR
∠PTQ = 2∠OPQ

Answer 18.
We have,

Answer 19.
We have, AB = 3 cm, AC = 4 cm
in right ∆BAC, by Pythagoras theorem
BC² = AB² + BC²
= (3²) + (4²) = 9+16=25
BC = √25 = 5 cm

Answer 20.

Answer 21.

Answer 22.
Given :
Rectangular tank :
Length = 11 m
Breadth = 6 m
Level of water, H = 5 m
Cylindrical tank :
Radius = 3.5 m
Let the level of water in cylinderical tank be h.
Then,
Vol. of water in cylindrical tank = Vol. of water in rectangular tank
πr²h = L × B × H
\(\frac { 22 }{ 7 } \) × 3.5 × h = 1 × 6 × 5
22 × 0.5 × 3.5 × h = 11 × 6 × 5
h = \(\frac { 11\times 6\times 5 }{ 22\times 0.5\times 3.5 } \)
h = 8.57 m.

SECTION-D

Answer 23.
Steps of construction :

• Draw a line segment AB = 6 cm.
• Draw the perpendicular bisector XY of AB intersecting AB at D.
• With D as centre and radius 5 cm, draw an arc cutting XY at C.
• Join AC and BC. Thus, ∆ABC is obtained.
• Draw ray AX such that ∠BAX is acute.
• Mark A₁ A₂,…A₇ on AX such that AA₁ = A₁A₂ =…….= A₆A₇.
• Join A₅ to B.
• From A₇, draw a line parallel to A₅B to meet AB produced at B’.
• From B’, draw a line parallel to BC to meet AC produced at C’.

Thus, ∆AB’C’ is the required triangle, whose sides are \(\frac { 7 }{ 5 } \) of the corresponding sides of ∆ABC.

Answer 24.

Answer 25.
Let price of a book be ₹ x and number of books be n.
Then nx = 300
Also (x – 5) (n + 5) = 300
=> nx + 5x -5n-25 = 300
5x – 5n – 25 =0
x – n – 5 = 0
We have,
n = \(\frac { 300 }{ x } \)
Equation (ii) becomes,
x – \(\frac { 300 }{ x } \) – 5 = 0
x² – 5x – 300 = 0
x² – 20x + 15x – 300 = 0
x(x-20) + 15(x-20) = 0
(x-20) (x+15) = 0
x = 20, -15
x ≠ – 15, as price of a book cannot be negative.
∵ price of each book is ₹20

OR

We have,
\(\frac { 1 }{ (x-1)(x-2) } +\frac { 1 }{ (x-2)(x-3) } \frac { 2 }{ 3 }\) x ≠ 1,2,3
\(\frac { (x+3)+(x-1) }{ (x-1)(x-2)(x-3) } =\frac { 2 }{ 3 }\)
3(x-3) + 3(x- 1) = 2(x – 1) (x – 2) (x-3)
3x – 9 + 3x – 3 = 2(x – 1) (x – 2) (x-3)
6x – 12 = 2(x – 1) (x – 2) (x-3)
6(x – 2) = 2(x – 1) (x – 2) (x-3)
3 =(x-1) (x-3)
3 = x²– 3x – x + 3
x² – 4x = 0
x(x – 4)= 0
x = 0 or 4

Answer 26.
Statement : The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.
Given : A ∆ABC is which D is the mid-point of side AB and the line DE is drawn parallel to BC,
meeting AC at E.
To prove : E is the mid-point of Ac.
Proof : In ∆ABC,
DE || BC
∴ \(\frac { AD }{ BD } =\frac { AE }{ EC } \)
Since, D is the mid-point of AB
∴ AD = BD
Thus, from equation (i),
\(\frac { AE }{ EC } =1\)
⇒ AE = EC
Hence, E bisects AC.

OR

Given : Two triangles ABC and DEF such that ΔABC ~ ΔDEF
To prove :

\(\frac { Area(\triangle ABC) }{ Area(\triangle DEF) } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } \)

Construction : Draw AL ⊥ BC and DM ⊥ EF
Proof : Since, similar triangles are equiangular and their corresponding sides proportional,
Therefore, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
and \(\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF } \)
Thus, in ∆ALB and ∆DME, we have
∠ALB = ∠DME = 90°
and ∠B = ∠E
So, by AA-criterion of similarity, we have
∆ALB ~ ∆DME
⇒ \(\frac { AL }{ DM } =\frac { AB }{ DE } \)
From (i) and (ii), we get
\(\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF } =\frac { AL }{ DM } \)

Answer 27.

Answer 28.
Let r be the radius of well, h be the height of the well, x be the width of embankment and H be the height of embankment.
Then, 2r = 3
=> r = \(\frac { 3 }{ 2 } \) m;
x = 4 m and h = 14 m
R = r + x = (4 + \(\frac { 3 }{ 2 } \) ) m = \(\frac { 11 }{ 2 } \) m
Volume of embankment (hollow cylinder) = Volume of well cylinder

Answer 29.
Given system of equation is
5x – y = 5
and 3x – y = 3


Triangle formed by these lines and Y-axis is ABC.
∴ Area of ∆ ABC = \(\frac { 1 }{ 2 } \) × BC × OA
= \(\frac { 1 }{ 2 } \) × 2 × 1 = 1 sq.unit
OR
Let the two digit number be l0x + y.
Then, according to the question,
Case 1:
l0x + y = 8(x + y) – 5
l0x + y = 8x + 8y – 5
2x – 7y = – 5
Case II:
16(x – y) + 3 = 10x + y
16x – 16y + 3 = 10x + y
6x – 17y = – 3
Multiplying equation (i) by 3 and then subtracting it with equation (ii), we get

Substituting the value of y in equation (i),
2x – 7 × 3 = – 5
⇒  2x – 21 = – 5
2x = – 5 + 21
⇒  2x =16
⇒  x = 8
Therefore, required number is 83.

Answer 30.

Mean, \(\bar { x }\) =\(\frac { \Sigma fx }{ \Sigma f } =\frac { 2830 }{ 80 }\) = 35.375 yrs.
Model class = group with highest frequency
= 35 – 45
So, Lower limit of the modal class, l = 35
Frequency of modal class, ƒ₁ = 23
Frequency of class preceding modal class, ƒ₀ = 21
Frequency of class successing model class, ƒ₂ = 14

We hope the CBSE Sample Papers for Class 10 Maths Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 1, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 4

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 4

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
If P(a/3, 4) is the mid-point of the line segment joining the points Q(- 6, 5) and R(- 2, 3), then find the value of ‘a’.

Question 2.
It is given that ∆ABC ~ ∆PQR, with = \(\frac { BC }{ QR } =\frac { 1 }{ 4 } \) Then, find \(\frac { ar(PRQ) }{ ar(BCA) } \).

Question 3.
HCF and LCM of a and b are 19 and 152 respectively. If a = 38 then find the value of b.

Question 4.
Find a quadratic polynomial, the sum and product of whose zeroes are 6 and 5, respectively.

Question 5.
If x = 2 and x = 3 are roots of 3x² – 2kx + 2m = 0, find the value of k and m.

Question 6.
Evaluate : 2 tan² 45° + 4 sin² 30° – 8 cos² 60°.

SECTION-B

Question 7.
Determine the least number that is divisible by all the numbers from 2 to 10 (both inclusive).

Question 8.
Find the coordinates of the point which is equidistant from the three vertices of the ∆AOB as shown in the given figure.

Question 9.
1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Saket has purchased a lottery ticket, what is the probability of winning a prize ?

Question 10.
A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

Question 11.
Determine the value of k for which the system : 2x – 3y = 7; (k + 2)x – (2k + 1 )y = 3(2k – 1), represents coincident lines on the graph.

Question 12.
Determine the sum of first 99 even natural numbers.

SECTION-C

Question 13.
Prove that exactly one out of any three consecutive positive integers is divisible by 3.

Question 14.
Find the area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b).

Question 15.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

OR

O is an interior point of a rectangle ABCD. Prove that OB² + OD² = QA² + OC².

Question 16.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Question 17.
Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.

OR

A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate Rs 25 per m².

Question 18.
A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1 m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

OR

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

Question 19.
If the remainder on division of x³ + 2x² + kx + 3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx – 18.

Question 20.
Solve graphically : 3x – y = 3; 2x – y = – 2.

OR

2 audio and 3 video cassettes together cost 426, whereas 3 audio and 2 video cassettes cost Rs 350. What are the prices of an audio and a video cassettes ?

Question 21.
If cos (A – B) = \(\frac { \sqrt { 3 } }{ 2 } \), cos (A + B) = \(\frac { 1 }{ 2 } \), 0°<A+B≤90° find cot (A+3B)

Question 22.
The following table gives the daily wage distribution of 50 casual workers in a locality. Find the modal daily wage of a worker.

SECTION-D

Question 23.
A tower subtends an angle a at a point A in the plane of its base and the angle of depression of the foot of the tower from a point which is b metres just above A is p. Prove that the height of the tower is b tan α cot β .

OR

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.

Question 24.
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.

Question 25.
In figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.

Question 26.
The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the height of rainfall in cm.

OR

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 Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm². (Take π = 3.14)

Question 27.
The sum of the squares of the two consecutive natural numbers is 313. Find the numbers.

OR

Sum of two numbers is 16 and the sum of their reciprocals is 16/63. Find the numbers.

Question 28.
There are 25 trees at equal distance of 5 metre in a line with a well, the distance of the well from the nearest tree being 10 metre. A gardener waters all the trees separately starting from the well ’ and he returns to the well after watering each tree to get water for the next. Find the total distance, the gardener will cover in order to water all the trees.

Question 29.
If x = tan A + sin A and y = tan A – sin A, then prove that: x² – y² = √xy.

Question 30.
The students of Class X of a school decided to donate their pocket money to purchase mineral water bottles for the people using contaminated water in a nearby village. They packed the mineral water bottles in different boxes. These boxes contained varying number of mineral water bottles. The following table shows the distribution of mineral water bottles according to the number of boxes :

Find the mean number of mineral water bottles kept in a packing box.

SOLUTIONS

SECTION-A

Solution 1:
Given : P is the mid-point of QR.
So by mid-point formula,

Solution 2:
Given
∆ABC ~ ∆PQR
and \(\frac { BC }{ QR } =\frac { 1 }{ 4 } \)

Solution 3:
We know,
Product of two numbers = H.C.F. × L.C.M.
⇒ a x b = H.C.F. x L.C.M.
⇒ 38 x b = 19 x 152
⇒ b = \(\frac { 19\times 152 }{ 38 } \) = 76.
∴ b = 76.

Solution 4:
Given : Sum of zeroes = 6
and product of zeroes = 5
We know, quadratic polynomial is given as,
= k[x² – (Sum of zeroes)x + Product of zeroes]
= k[x² – 6x + 5] where k ≠ 0.

Solution 5:
Given quadratic equation is,
x² – 2kx + 2m = 0
Since, 2 and 3 are roots of the given quadratic equation.
So, putting x = 2, 3 in given equation, we get
3(2)² – 2k(2) + 2m = 0
⇒ 12 – 4k + 2m = 0
and 3(3)² – 2k(3) + 2m=0
⇒ 27 – 6k + 2m = 0
Subtracting equation (i) and (ii), we get

k = \(\frac { 15 }{ 2 } \)
Substituting the value of k in equation (i), we get
12 – 4 x \(\frac { 15 }{ 2 } \) + 2m = 0
12 – 30 + 2m = 0
2m = 18 ⇒ m = 9
∴ k = \(\frac { 15 }{ 2 } \) and m = 9.

Solution 6:
We have,

SECTION-B

Solution 7:
Given numbers are 2, 3, 4, 5, 6, 7, 8, 9, 10
L.C.M. of 2, 4, 8 = 8
L.C.M. of 3, 9 = 9
L.C.M. of 5, 10 = 10
L.C.M. of 6, 2, 4, 8, 3, 9 = 8 x 9 = 72
L.C.M. of 2, 3, 4, 6, 8, 9, 5, 10 = 72 x 5 = 360
L.C.M. 2, 3, 4, 5, 6, 8, 9, 10, 7 = 360 x 7 = 2520
∴ 2520 is least number which is divisible by all the numbers from 2 to 10.

Solution 8:
Since ∠AOB = 90°
So, ∆AOB is right angle triangle.
We know that in a right angled triangle, mid-point of hypotenuse is equidistant from all three vertices of right angle triangle.
Mid-point of hypotenuse AB = \(\frac { 2x+0 }{ 2 } ,\frac { 0+2y }{ 2 } \) = (x, y)
Hence, C(x, y) is the equidistant from the three vertices of the ∆AOB.

Solution 9:
Total lottery tickets = 1000
∴ n(S) = 1000
Total winner tickets = 5
∴ n(E) = 5
Probability of winning the prize
P(E) = \(\frac { n(E) }{ n(S) } =\frac { 5 }{ 1000 } \) = 0.005.

Solution 10:
Let E be the event of B’s chance of throwing a higher number than 9.
⇒ E= {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
∴ Total chances, n(E) = 6
Hence, probability of throwing a higher number than 9
P(E) = \(\frac { n(E) }{ n(S) } =\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)

Solution 11:
2x – 3y – 7 = 0
and (k+2)x – (2k+1)y – 3(2k-1)=0

Solution 12:
Sum of first 99 even natural numbers
i.e., 2 + 4 + 6 + 8 …
These are in arithmetic progression with
a = 2, d = 4 – 2 = 2, n = 99
Sum of K terms of A.P. is given as

Hence, sum of first 99 even natural number = 9900.

SECTION-C

Solution 13:
Let the number be a = 3q + r where r = 0,1, 2
Case I: When r = 0,
Then a = 3q
It is divisible by 3.
Case II: When r = 1,
Then a = 3q + 1
It is not divisible by 3.
Case III: When r = 2,
Then a = 3q + 2
It is not divisible by 3
These are also three consecutive numbers hence exactly one out of any three consecutive number is divisible by 3.

Solution 14:
Given vertices of triangles are
A(a, b + c), B(b, c + a) and C(c, a + b)
We know, area of a triangle is given as

Solution 15:
Given
∆ABC and ∆DEF are two similar triangles, AP and DQ are medians of triangles respectively

Solution 16:
Given : ABCD is a quadrilateral circumscribing a circle.
To prove : ∠AOD + ∠BOC = 180°
∠AOB + ∠COD = 180°
Construction : Join OP, OQ, OR and OS.

Solution 17:
Let the radius of third circle be r cm.
Area enclosed by two concentric circles I and II of radii 3.5 cm and 7 cm, respectively

Solution 18:
Given : Dimensions of cuboidal block = 4.4 m x 2.6 m x 1 m
Internal radius of pipe, r = 30 cm = 0.30 m
Thickness of cylindrical pipe = 5 cm
External radius of pipe, R = 30 + 5 cm = 35 cm = 0.35 m
Now, volume of cuboidal block = 4.4 x 2.6 x 1 m³ = 11.44 m³
Volume of the hollow cylindrical pipe = π(R² – r²)h
= π(0.35² – 0.30²)h

Solution 19:
x³ + 2x² -9x-18
it is divided by x – 3
x – 3 = 0 x = 3
From the given data, we get
f(3) = 21
⇒ (3)³ + 2(3)² + k(3) = 21
⇒ 27 + 18 + 3k + 3 = 21
⇒ 48 + 3k = 21
⇒ 3k = – 27
⇒ k = – 9
∴ Polynomial becomes
x³ + 2x² + (-9) x + 3
x³ + 2x² – 9x + 3
We know,
Dividend = Quotient × Divisor + Remainder
⇒ x³ + 2x² – 9x + 3 = Quotient × (x – 3) + 21
⇒ x³ + 2x² – 9x – 18 = Quotient × (x – 3)

Solution 20:
Given equations are
3x – y = 3
and 2x – y = – 2
From Equations (i),
3x – y = 3


The point of intersection of these lines is (5, 12).
x = 5 and y = 12.
OR
Let the price of audio cassette be Rs x
and the price of video cassette be Rs y
According to the question,
2x + 3y = 426 …(i)
and 3x + 2y = 350 ….(ii)
Adding equations (i) and (ii), we get
5x + 5y = 776
⇒ x + y = 155.20 ….(iii)
Subtracting equation (ii) from equation (i), we get
– x + y =76 …(iv)
Adding equations (iii) and (iv), we get
2y = 231.20
⇒ y = \(\frac { 231.20 }{ 2 } \) = 115.60
Putting the value of y in equation (iii), we get
x + 115.60 = 125.20
⇒ x = 155.20 – 115.60
⇒ x = 39.60
Thus, Price of audio cassette = Rs 115.60
Price of video cassette = Rs 39.60.

Solution 21:
We have,
\(cos(A-B)=\frac { \sqrt { 3 } }{ 2 } \) …(i)
and \(cos(A+B)=\frac { 1 }{ 2 } \) ….(ii)
From equation (i)

Solution 22:

Modal class = Group with highest frequency
= 15000 – 20000
Now,

= 15000 + 3000
Mode = 18000
Modal daily wages = Rs 18000.

SECTION-D

Solution 23:
Let OP be the tower of height h m
Given AB = b m
Let distance between tower and point A(AO) = x m
In ∆AOP,
∠O = 90°

Solution 24:
Steps of Construction :
(1) Draw a circle with centre O and radius 3 cm.
(2) Draw a diameter of the circle. Extend it on both sides upto P and Q such that OP = OQ = 7 cm.
(3) Draw perpendicular bisectors of OP and OQ which meet them at M and N respectively.
(4) With M as centre and MP as radius and with N as centre and NQ as radius, draw circles which intersect the given circle at A, B and C, D respectively.
(5) Join PA, PB, QC, QD which are the required tangents.

Solution 25:
Given : A circle with centre of radius 5 cm and OT = 13 cm
Since, PT is a tangent at P and OP is a radius through P
OP ⊥ PT
In ∆OPT

Solution 26:
Given
Dimesions of roof = 22m x 20m
Diameter of cylinder = 2 m
Radius , r = \(\\ \frac { 2 }{ 2 }\) = 1m

Solution 27:
Let the two consecutive number be x and (x + 1).
Then, according to the question,
x² + (x + 1)² = 313
=> x² + x² + 2x + 1 – 313 = 0
=> 2x² + 2x – 312 = 0
=> x² + x – 156 = 0
=> x² + 13x – 12x – 156 = 0
=> x(x + 13) – 12(x + 13) = 0
=> (x + 13) (x – 12) = 0
Either x + 13 = 0 => x = – 13, rejected as it is not natural number.
or x – 12 = 0 => x = 12
∴ Consecutive numbers are 12 and 13.
OR

Solution 28:
Distance covered in watering the nearest tree = 2 x 10 = 20 m
Distance covered in watering the second tree = 15 x 2 = 30 m
Distance covered in watering third tree = 20 x 2 = 40 m
Sum of the distance covered = 20 + 30 + 40 + …

Solution 29:
Given
x = tanA + sin A
y = tanA – sinA
We know

Solution 30:
As this series is an inclusive one, we can make it exclusive by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit of each class interval. Thus, we have

We hope the CBSE Sample Papers for Class 10 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 4, drop a comment below and we will get back to you at the earliest.