Test Paper # 1 (Mathematics) Test Paper # 2 (Mathematics) Test Paper # 2 (Science and Technology) Test Paper # 2 (Science and Technology) Test Paper # (Social Science) Test Paper # 2 (Social Science) Hints & Solutions # 1(Mathematics) Hints & Solutions # 2(Mathematics) Hints & Solutions # 1 (Science and Technology) 1-2 3-5 6-7 8-9 10 – 12 13 – 15 16 – 21 22 – 28 29 – 35 36 – 39 40 – 45 46 – 52 10.

Hints & Solutions # 2(Science and Technology) 11. Hints & Solutions # 1 (Social Science) 12.Hints & Solutions # 2 (Social Science) Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 1 GENERAL INSTRUCTIONS FOR CBSE TEST PAPERS 1 & 2 SUBJECT : MATHEMATICS CLASS – IX Time : 3 Hr. Max. Marks : 80 GENERAL INSTRUCTION 1. 2.

All questions are compulsory. The questions paper consists of 30 questions divided into the four sections A,B,C and D. Section A contains 10 questions of 1 mark each, section B is of 05 questions of 2 marks each, section C is of 10 questions of 3 marks each and section D is of 05 questions of 6 marks each.

3. 4.Write the serial number of questions before attempting. In questions of construction, the drawing should be neat and exactly as per the given measurements.

5. Use of calculator is not permitted. However, you may ask for mathematical tables. Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 2 TEST PAPER # 1 INSTRUCTIONS : Section A : Q 1 to Q . 10 carries 1 mark each. Section B : Q 11 to Q.

15 carries 2 marks each. Section C : Q 16 to Q. 25 carries 3 marks each. Section D : Q 26 to Q. 30 carries 6 marks each.

SECTION # A 1. 2. 3. Express 0. 05 as a vulager fraction.Find the value of a if (x – a) is a factor of the polynomial x4 – a2x2 + 3x – 6a. The cost of a book is one third the cost of a pen. Write a linear equation in two variables to represent this statement.

4. 5. Find out how many non zero integer solutions can be possible for 3x + 2y = 18. In which quadrant do the following pints lie ? (i) (-3, 2) 6. 7. (ii) (2. 5, 0) The difference of two supplementary angles is 40 0, find the angles.

In the given figure, OP and OQ are opposite rays. Find x. 8.

9. 10. The class marks of distribution are : 6, 10, 14, 18, 22, 26, 30. Find the class size and the class interval.If 7 ? 1 7 +1 ? = a + b 7 , , find the values of a and b. 7 +1 7 ? 1 Find the remainder when x51 + 51 is divided by x+ 1. SECTION # B 11. Plot the points A(2, 0), B(2, 2), C(0, 2) and draw the line segments OA, AB, BC and CO.

What figure do your obtain ? 12. 13. 14. Using factor theorem, show that a – b is the factor of a(b2 – c2) + b(c2 – a2) + c(a2 – b2) Three cubes each of side 6 cm are joined end to end. Find the surface area of the resulting cuboids. The mean of 13 observation is 14. If the mean of the first 7 observation is 12 and that of last 7 observation is 16, find the 7th observation.

15.Find x2 if x = 5 +2 + 5 +1 5 ? 2 . Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 3 SECTION # C 16. 17.

Factorise : p2 + pq + q2 + 1 + 2p + q 4 Let A and B are the remainder when the polynomial y3 + 2y2 – 5ay – 7 and y3 + ay2 – 12y + 6 are divided by y + 1 and y – 2 respectively. If 2A + B = 6 then find the value of a. If the work done by a body on application of a constant force is directly proportional to the distance traveled by the body.

Express this is the form of an equation is two variables and draw the graph of the same by taking the constant forces as 5 units.Also read from the graph the work done when the distance traveled by the body is (i) 2 units (ii) 0 units. Prove that the sum of the three sides of a triangle is greater than the sum of its three median.

Construct a triangle ABC is which BC = 4. 5 cm, ? B = 450 and AB – AC = 2. 5 cm. The perimeter of a triangle is 36 cm and its sides are in the ratio 3 : 4 : 5. Find the area of the triangle. A park in the shape of a quadrilateral ABCD has ? C = 900.

AB = 18 m, BC = 24 m, CD = 10 m and AD = 16m. How much area does it occupy ? 18. 19.

20. 21. 22. 23. 24. A solid cylinder has a total surface area 462 sq.

cm.Its curbed surface area is one-third of the total surface area. Find the volume of the cylinder. The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.. (i) What is the probability that on a given day it was correct ? (ii) What is the probability that it was not correct on a given day ? Plot the points A(0, 5), B(8, 0), (8, 5) and join them. What figure do you obtain ? 25.

SECTION # D 26. 27. Prove that (a + b)3 + (b + c)3 + (c + a)3 – 3(a + b) (b + c) (c + a) = 2 (a3 + b3 + c3 – 3abc) In figure, ABC is a triangle, D is the mid-point of AB, P is any point on BC.Line CQ is drawn parallel to PD to intersect AB at Q. PQ is joined. Show that ar ( ? ABP ) = 1 ar( ? ABC ) . 2 28. O1 and O2 are the centres of two congruent circles intersecting each other at points C and D.

The line joining their centres intersects the circles in points A and B such that AB > O1O2. If CD = 6 cm and AB = 12 cm, determine the radius of either circle. Students of a school stages a rally for cleanliness campaign. They walked through the lanes in two groups.

One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes.If AB = 9m, C = 40 m, CD = 15 m, DA = 28 m and ? B = 900, which group cleaned more area and by how much ? Find the total area cleaned by the students. Draw a Histogram and frequency polygon from the following data : 29.

30. Class Interval Class frequency 21-24 30 26-29 24 31-34 52 36-39 28 41-44 46 46-49 10 Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 4 TEST PAPER # 2 INSTRUCTIONS : Section A : Q. 1 to Q.

10 carry 1 mark each. Section B : Q. 11 to Q. 15 carry 2 marks each. Section C : Q. 16 to Q. 25 carry 3 marks each.

Section D : Q. 26 to Q. 0 carry 6 marks each.

SECTION # A Directions : Answer the questions (1 to 10) 1. 2. Express 0. 418 as a vulager fraction. Find the zero of the polynomial in each of the following cases : p(x) = cx + d, c ? 0, c, d are real numbers.

3. Find a value for a so that each of the following equations may have x = 1, y = 1 as a solution : 3x + ay 6 In supplementary angles one is twice the other. Find the angles.

In the given figure, AOB is a line, determine x. 4. 5. 6. 7.

8. 9. ABCD is a parallelogram. If the two diagonals are equal, find the measure of ? ABC . Find the lateral surface area of a cube of edge 20 m.

The mean of 10, 12, 16, 20, p and 26 is 17. Find the value of p. Which of the following points lie on the x-axis > (i) (1, 1), (ii), (1, 0), (iii), (0, 1) (iv), (0, 0) Calculate the mean of all possible factor of 10. 10.

SECTIONS # B 11. 12. 13. 14. Simplify 7 3 2 5 3 2 ? ? 10 + 3 6+ 5 15 + 3 2 Determine the number of sides of polygon whose exterior and interior angles are in the ratio 1 : 5. Plot the points A(4, 4) and B(-4, 4) and join the lines OA, OB and BA.

What figure do you obtain ? The taxi fare in a city is as follows. For the first kilometre, the fare is Rs. 8 and for the subsequent distance it is Rs. per km. Taking the distance covered as x km. and the total fare as Rs. y, write a linear equation for this information and draw its graph.

15. Free Find the area of triangle two sides of which are 16 cm and 22 cm and perimeter is 64 cm. www. tekoclasses. com Director : SUHAG R.

KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 5 SECTION # C 16. Two regular polygons are such that the ratio between their number of sides is 1 : 2 and the ratio of measures of their interior angles is 3 : 4. Find the number of sides of each polygon. Construct a triangle ABC whose perimeter is 12 cm, ? B = 600 and C = ? 450 17.

18.How many spherical lead shots each 4. 2 cm in diameter can be obtained from a rectangular solid lead with dimensions 66 cm, 42 m and 21 m? 19. Find the mean of each of the following distributions : x: f: 10 4 15 6 20 8 25 18 30 6 35 5 40 3 20. An insurance company selected 2000 drivers at random (i.

e. , without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table : Age of drivers Accidents in one yeaers (in years) 0 1 2 3 Over 3 18 – 29 30 – 50 Above 50 440 505 360 160 125 45 110 60 35 61 22 15 35 18 9Find the probabilities of the following events for a driver chosen at random form the city : (i) being 18 – 29 years of age and having exactly 3 accidents in one year. (ii) being 30 – 50 years of age and having one or more accidents in a year. (iii) having no accident in one year. Factorise : y3 – 7y + 6 21.

22. See the given figure and write the following : Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph.

:(0755) 32 00 000 Page 6 23. (i) The point identified by the coordinates of (-2, -3). (ii) The point identified by the coordinates of (3, -3). iii) The abscissa of point D. (iv) The abscissa of the point H. (v) The coordinates of the point L.

(vi) The coordinates of the point M. Find the area of quadrilateral PQRS whose sides are 9 m, 40 m, 28 m and 15 m respectively and the angle between first two sides is a right angle. The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.

24. 25. Sunita has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops to suffice the needs of their family.She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonal is 160 m, how much area each of them will get > SECTION # D 26.

In the given figure, ? ABC is isosceles with AB = AC. D, E and F are respectively the mid-points of ides BC, CA and AB. Show that the line segment AD is perpendicular to the line segment EF and is bisected by it. 27. ABCD is a rhombus and AB is produced to E and F such that AE = AB = BF. Prove that ED and FC are perpendicular to each other. Prove that the sum of either pair of opposite angles of a cyclic quadrilateral is 1800.Using the above, do the following : In the given figure ABCD is a cyclic quadrilateral.

Side CD is produced on both sides such that ? BCP = 1100 and ? ADQ = 950 Find the value of ? A and ? B . 28. 29. 30. If x3 + mx2 + nx + 6 has x – 2 as a factor and leaves a remainder 3, when divided by x – 3, find the values of m and n. Plot a cumulative frequency diagram for the following distribution : C. l. Frequency 0-9 5 10-19 15 20-29 20 30-39 23 40-49 17 50-59 11 60-69 9 Free www.

tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 7 GENERAL INSTRUCTIONS FOR CBSE TEST PAPER 1 & 2SUBJECT : SCIENCE AND TECHNOLOGY CLASS – IX Time : 21/2 Hr. Max.

Marks : 60 GENERL INSTRUCTION 1. The question paper comprises of two Section A and B. Attempt both the Sections. 2. The candidates are advised to attempt all the questions of Section A and Section B separately. 3. All questions are compulsory. 4.

Marks allocated to each question are indicated against it. 5. Question number 1 to 6 in Section A and 19 to 21 in Section B are very short answer question. These are to be answered in one word or one sentence only.

6. Question number 7 to 12 in Section A and 22 to 24 in Section B are short answer questions.These are to be answered in about 30-40 words each. 7. Question number 13 to 16 in Section A and 25, 26 in Section B are also short answer questions.

These are to be answered in about 40-50 words each. 8. Question number 17 to 18 in Section A and 27 in Section B are long answer questions. These are to be answered in about 70 words each. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph.

:(0755) 32 00 000 Page 8 TEST PAPER # 1 SECTION # A 1. 2. 3. 4. 5. 6. 7.

8. Write the formula for distance covered in nth second. Convert 1 radian (c) in to degree (0) Define Newton’s’ second law of motion.

Sponge is considered as solid yet we are able to compress it. Why ? What is the meaning of 15% solution of NaCl ? Write the names of gases which can be separated from air by fractional distillation. Define scalar and vector quantities and given two examples of each. [1] [1] [1] [1] [1] [1] [2] A body moves from A to B and come back to A and then goes to C point. Find distance and displacement of the body. [2] 9. 10. 11.

Convert 1 KWh into Joule. Write the two reasons of using ultrasonic waves is SONAR (a) Write the conditions to liquefy atmospheric gases (b) What do LPG and CNG stand for ? 2] [2] [2] 12. The mass number of an element is 63. It contains 29 electrons. What is the number of protons and neutrons in its nucleus ? Write the electronic configuration of the element ? [2] Two bodies A and B having mass m and 2m respectively are kept at a distance d apart.

Where should a small particle be placed so that the net gravitational force on it due to the bodies A and B is zero ? [3] 13. Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 9 14. A bullet of mass 10 g is fired from a gun of mass 5. 0 kg.If the velocity of bullet is 250m/s. Find the recoil velocity of the gun.

[3] Calculate the weight of iron which is converted into its oxide by the action of 18g of steam. [3] (a) Define the term mole. (b) If 1 g of carbon contains x atoms, what will be the number of atoms in 1 g of magnesium ? [ C = 12u, Mg = 24u] [3] Prove mathematically that the mechanical energy of a freely falling body remains constant.

[5] (a) On heating, potassium chlorate decomposes to potassium chloride and oxygen. In one experiment 30g of potassium chlorate generates 14. 9g of potassium chloride and 9. 6g of oxygen.What mass of potassium chlorate remains unrecompensed ? (b) Write two characteristic properties of nucleus. Compare these with the properties of an electron.

[5] SECTION # B What do you mean by semiautonomous cell organelles ? [1] Which tissue is responsible for the outer most conversing in animals > What are acute diseases ? Name any three acute diseases. Draw a well labelled diagram of plant cell. [1] [1] [2] 15. 16. 17. 18. 19.

20. 21. 22. 23. it. 24. 25. 6.

27. Explain the main characteristic of phylum thallophytic. Write any two groups included in [2] Write the difference between blood, lymph and serum.Fertilizers which are used in farming are good and bad both, explain why ? Write down four main differences between prokaryotic and eukaryotic cells. Write the difference between striated, non striated and cardiac muscle ? [2] [3] [3] [5] Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 10 TEST PAPER # 2 SECTION # A 1. Can the internal forces acting among the parts of a system change the linear momentum of the system ? [1] In which of the three media-air, water or steel does sound travel the fastest ? A particle is moving with a uniform velocity.What is its acceleration ? State two characteristics of matter demonstrated by : (a) diffusion (b) Brownian motion [1] [1] [1] 2.

3. 4. 5. 6. 7. 8.

9. Define valence shell. What are the fourth and fifth states of mater ? [1] [1] 10. If a bulb of 100 W is lit for 6 hours, how much electric energy would be consumed ? [2] A car moves through 10 km at a speed of 50 km/h and the next 10 km at a speed of 40 km/h.

Calculate its average speed. [2] A dog barks in a park and hear its echo after 2. 5 s. If the sound of its bark got reflected by a nearby building, find the distance between the dog and the building.Take the speed of sound in air as 340 m/s. [2] A man pushes a box of 50 kg with a force of 200 N. What will be the acceleration of the box due to this force ? What would be the acceleration if the mass is halved ? [2] Describe the drawbacks of Rutherford’s atomic model.

[2] 11. 12. 13. Differentiate between solid, liquid and gas on the basis of rigidity and compressibility ? [2] How much momentum will a dumb-bell of mass 15 kg transfer to the floor if it falls from a height of 19. 6 m ? (Take g = 9. 8 m/s2) [3] Derive the first and second equation of motion [3] 14. Free www. tekoclasses.

com Director : SUHAG R.KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 11 15. Write the constituents of gun powder. How can you separate different constituents of gun powder ? [3] Which method is used for the separation of a mixture of alcohol and water.

Draw a well labelled diagram of the apparatus used in above separating technique. [3] Explain the working of human ear with the help of a diagram [5] 16. 17. 18.

(a) The average atomic mass of sample of element X is 16. 2 u. What are the percentage of isotopes 16 8 X and 18 8 X is the sample ? (b) Name an element whose nucleus does not contain any neutrons.

c) Hydrogen has three isotopes written as : 1 2 3 1 H , 1 H , 1H Explain why these isotopes have almost identical chemical properties. [5] SECTION # B 19. 20.

21. 22. 23.

24. 25. What are chronic disease ? Write the three name of chronic diseases. In plant which tissue is responsible for growth ? Write the name of substance. Which is responsible for ozone layer depletion ? Define the term hermaphrodite.

Give two examples. Describe the “Green house effect”. Write the difference between mixed cropping and Inter cropping. [1] [1] [1] [2] [2] [2] Draw an outline of Echiler classification and write down the characteristic features of bryophyte. 3] Write the three main difference between animal cell and plant cell. [3] 26. 27. Write in brief about the five kingdom as explained in ‘Five kingdom classification’.

[5] Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 12 GENERAL INSTRUCTIONS FOR CBSE TEST PAPERS 1 & 2 SUBJECT : SOCIAL SCIENCE CLASS – IX Time : 3 Hr.

Max. Marks : 80 GENERAL INSTRUCTIONS 1. 2.

There are 29 questions in all. All questions are compulsory. Questions from serial number 1 to 10a are 1 mark questions. Answer to these questions may be from one word to one sentence. 3.Questions from serial number 11 to 18 are 3 marks questions Answers to these questions should not exceed 80 words each. 4.

Questions from serial number 19 to 28 are 4 marks questions. Answers to these questions should not exceed 100 words each. 5. Questions No.

29 is map question. Attach the map inside your answer book. Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 13 TEST PAPER # 1 1. Name of the first Indian cricket club. When was it established OR What is Gandhian Cap ? Give the extent of India in latitudes an longitudes What does the ‘breaking of monsoon’ mean ?Why are Himalayan river called Perennial rivers ? Explain the term ‘universal adult franchise’. What is EVM ? Name the tree categories of minister What are the function of FCI ? What is food security ? What is poverty ? Below are given three groups A, B & C o f question s 11 and 12.

Select any one group for answering these two questions. GROUP – A The introduction of railway had an adverse impact on the forest. Justify by giving examples. Who were Kalangs ? Mention any four characteristics of this community ? GROUP – B Explain the major characteristics of pastoral nomodism.The pastoral groups had sustained by a careful consideration of a host of factor”. Explain these factors. GROUPS – C Who was Captain Swing ? What did the name symbolise or represent? Why were threshing machines opposed by the poor in England ? What was Rabindranath Taore’s opinion regarding the national dress ? OR How was the cricket used by the Britishers to spread their policy of racism ? Distinguish between : Evergreen and deciduous forests. What are the Fundamental Rights guaranteed in our Constitution ? In which field do you think India can build the maximum employment opportunity ? Explain ?Explain.

What is the major objectives of Prime Minister Rozgar Yozana ? How is Lok Sabha more powerful than Rajya Sabha ? Who was Napoleon ? Mention any two steps taken by him to modernise France. Explain the racial policy of Hitler or Nazis. www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 14 2. 3.

4. 5. 6. 7. 8.

9. 10. 11. 12. 11. 12. 11. 12.

13. 1 13. 2 14.

15. 16. 17.

18. 19. Free 20. How have advances in technology, especially television technology affected the development of contemporary cricket ? OR 21. 22. 23.

24. 25. 26. 27. 28. 29. 1Give an account of the weather conditions and characteristics of the cold season.

Describe the formation of Himalayas. “It is said elections are the barometer of democracy. ” Elaborate. Enumerate the stages through which a bill moves before it becomes a low. Explain the nature of Fundamental Rights in the Constitution. Why is there need for food security in India ? How did the spread of electricity helps farmers in Palampur ? Discuss the major reasons for poverty in India.

On the given outline political map of the world name and mark : (Any two) (i) A country where Bolsheviks revolution took place (ii) A cotton belt of U. S. A.

iii) A country where Bastar forest movement was launched (iv) An one central power during the first world war. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph.

:(0755) 32 00 000 Page 15 29. 2 In the given outline map of India, locate the following carefully with appropriate symbols and write the name of each item near its location. (i) Simplipal National Park (iii) Sambhar Lake (ii) River Mahanadi (iv) Gulf of Mannar Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 16 TES PAPER # 2 1. What is hyperinflation ? Mention the factor responsible for this.OR Describe the Bloody Sunday. 2. 3.

4. 5. 6. 7. 8. 9. 10.

State in brief the theory of plate tectonics. Which two peninsular rivers flow through troughs ? Distinguish between wind ward side and leeward side. What do you mean by universal adult franchise ? What is a constituency ? What is parliament ? Name the two houses of the Parliament. Define green revolution. Mention some factors on which the quality of population depends. What is fixed capital ? Below are given three groups A, B & C of questions 11 and 12, Select any one group for answering these two questions.

GROUP – A 11. 12. 11. 12. 11. 12. 13. Who are Kalangs ?Why did they attack the Dutch forts at Joana ? Mention any three factors which prompted Samins to revolt against the Dutch.

GROUPS – B Explain the movement of Gujjar Bakarwals of Jammu and Kashmir. Give reasons to explain why the Maasai comunity last their grazing lands. GROUP – C Why were Indian formers reluctant to grow Opium ? Explain the Dust Bowl tragedy. What was the difference between the Amateurs and the professionals ? OR How did the British rule affect the Indian textile industry ? 14. 15.

16. 17. Mention any four characteristics of the Ganga – Brahmaputra delta. Describe four features of democracy as a form of government.What is the role of education in human capital formation ? What is the difference between disguised unemployment and seasonal unemployment ? Free www. tekoclasses.

com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 17 18. 19. 20.

21. 22. 23. 24. 25. 26. 27.

28. 29. 1 Distinguish between a nominal democracy and an ideal democracy.

Why were jews classified as undesirable by the Nazir ? OR Describe the incidents that led to the storming of the Bastille. Discuss the causes of the French Revolution of 1789. OR Why did the Tsarist autocracy collapse in 1917 ? Discuss the mechanism of monsoons.Every species has role to play in the ecosystem.

Elaborate. Explain the manor features of democratic governments. Mention the Fundamental Rights that are given in the Indian constitution. What are the different measures taken to ensure free and fair elections in India. Why is the buffer stock created by the government ? Mention some measures to reduce poverty in India. What are the various activities undertaken in the primary sector secondary sector and the e tertiary sectors ? On the given outline political map of the world name and mark : (Any two) (i) A region of Africa known for pastoralists. ii) A country associated with swing movement. (iii) A country where a revolution took place in 1789. (iv) A test playing country of Asia. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 18 29. 2 In the given outline map of India, locate the following carefully with appropriate symbols and write the name of each item near its location. (ii) River Narmada (iii) Mangrave Forest (iv) Shivalik Range (i) Kaziranga Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 19 HINTS & SOLUTIONS (MATHEMATICS) TEST PAPER # 1 SECTION # A 1. Let x = 0. 5 ? 100x = 5. 05 from eqn (i) & (ii) ? 99x = 5 ? x= 5 99 … (i) ….. (ii) 2. 3. Let p(x) = x4 – a2x2 + 3x – 6+a Since, (x + a) is a factor of p(x), then p(a) = 0 by factor theorem. p(a) = 0 ? (a)4 – a2 (a)2 + 3(a) – 6a = 0 ? a4 – a2 ? a2 + 3a – 6a = 0 ? a4 – a4 – 3a = 0 ? a=0 Let, the cost of pencil = x, and, the cost of a pen = y. Then, ATQ x= 4. 5. 6. 7. 8. 9. ? 3x = y ? 3x – y = 0 Total number of non zero positive integer solutions are 2 for the given equation. An the solutions are x = 4, y = 3 and x = 2 y = 6. (i) In the point (-3, 2) abscissa is negative and ordinate is positive. So it lies in the second quadrant. ii) In the point (2. 5, 0) abscises is positive and ordinate is zero. So it lies on positive X axis. Let first angle be ‘a’ Then second angle will be (a + 40) Now, a + (a + 400) = 1800 ? 2a = 1400 ? a = 700 So, first angle = 700 and second angle = 700 + 400 = 1100 A. T. Q. (x + x + 100 + x + 200) = 1800 ? 3x = 1500 ? x = 500 Class size = 4, 1st class interval = 4 – 8 7 ? 1 7 +1 + = a+b 7 7 +1 7 ? 1 7 ? 1 7 +1 L. H. S. = ? 7 +1 7 ? 1 1 y 3 Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 20 = = ( 7 ? 1)2 ? ( 7 + 1)2 ( 7 + 1)( 7 ? 1) 10. (7 + 1 ? 2 7 ) ? ( 7 + 1 + 2 7 ) ( 7 )2 ? 1)2 8? 2 7 ? 8? 2 7 4 7 2 = = =? 7 7 ? 1 8 3 2 ? ? 7 =a+b 7 3 2 Hence, a = 0 and b = – . 3 Let P(x) = x51 + 51 be the given polynomial If (x + 1) is a factor of P(x) ? P(-1) = 0 but P(-1) = (-1)51 + 51 = -1 + 51 = 50. So, the remainder is 50. SECTION # B 11. 12. 13. 14. On joining OA, AB, BC and CO, we get a square of each side 2 units. By factor theorem, (a – b) will be the factor of the given expression if it vanishes by substituting a = b is it Substituting a = b in the given expression, we have a(b2 – c2) + (c2 – a2) + c(a2 – b2) = b(b2 – c2) + b(c2 – b2) + c(b2 – b2) = b2 – bc2 + bc2 – b3 + c(b2 – b2) = 0 ? a – b) is a factor of a (b2 – c2) + b(c2 – a2) + c(a2 – b. 2) Hence, (a – b) is a factor of the given expression. The dimension of the cuboids so formed are length = 18 cm breath = 6 cm and height = 6 cm. Surface area of cuboids = 2 ( ? b + b ? h + ? h) = 2 ? [18 ? 6 + 6 ? + 18 ? 6] = 504 cm2 Let the 7th observation be p. Now, Sum of 13 observation = 13 ? 14 = 182. Sum of first 7 observation = 12 ? 7 = 84. Sum of last 7 observation = 16 ? 7 = 112. Sum of last 6 observation = 112 – p. ? Sum of 13 observation = Sum of first 7 observation + Sum of last 6 observation = 84 + 112 – p = 196 – p So, 196 – p = 182 ? = 196 – 182 = 14. Thus, the 7th observation is 14. x2 = 5 + 2 + 5 ? 2 + 2 ( 52 ) ? 2 2 5 +1 = 2 5 + 2 2( 5 + 1) = =2 5 +1 5 +1 15. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 21 SECTION # C 16. p2 + pq + q2 + 1 + 2p + q 4 q2 4 = p2 + + 1 + pq + q + 2p 2 17. 18. Let p(y) = y3 + 2y2 – 5ay – 7 and q(y) = y3 + ay2 – 12y + 6 be the given polynomials. Now, A = Remainder when p(y) is divided by (y + 1) ? A = p (-1) ? A = (-1)3 + 2(-1)2 – 5a(-1) – 7 [? p(y) = y3 + 2y2 – 5y – 7] ? A = -1 + 2 + 5a – 7 ? A = 5a – 6 And B = Remainder when q(y) is divided (y – 2) ?B = q(2) ? B = (2)3 + a(2)2 – 12 ? 2 + 6 [? q(y) = y3 + ay2 – 12y + 6] ? B = 8 + 4a – 24 + 6 ? B = 4a – 10 Substituting the values of A and B is 2A + B = 6, we get 2(5a – 6) + (4a – 10) = 6 ? 10a – 12 + 4a – 10 =6 ? 14a – 22 = 6 ? 14a = 28 ? a = 2. y = 5a, where x = distance traveled, y = work done (i) 10 units (ii) 0 unit ?q? ?q? ?q? = ( p)2 + ? ? + ( 1)2 + 2( p)? ? + 2? ?( 1) + 2( 1)(p) ? 2? ?2? ?2? 2 q ? ? = ? p + + 1? [ a 2 + b + c 2 + 2 ab + 2 bc + 2 ca = 2( a + b + c)2 ] 2 ? ? q q ? ? ? ? = ? p + + 1? ? p + + 1? 2 2 ? ? ? ? 19. Given : A ? ABC with AD, Be and CF as medians.To prove : AB + BC + CA > AD + BE + CF Proof : Since, AD is median with D as point of BC ? AB + AC > 2AD …. (i) BE is median with E on AC ? AB + BC > 2BE …. (ii) CF is median on side AB ? AC + BC > 2CF …. (iii) Adding (i), (ii) and (iii), 2(AB + BC + CA) > 2(AD + BE + CF) or AB + BC + CA > AD + BE + CF ? Sum of the three sides of triangle is greater than its three median. www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 22 Free 20. 21. We are given BC = 4. 5 cm, ? B = 450 and AB – AC = 2. 5 cm Steps of Construction : (i) Draw a ray BX and cut off a line segment BC = 4. cm from it. (ii) Construct ? XBY = 450 (iii) Cut off a line segment BC = 2. 5 m from BY. (iv) Join AC. (v) Draw perpendicular bisector of CD cutting BY at a point A. (vi) Join AC. So, ? ABC is the required triangle. Let sides are a, b and c, a = 3x, b = 4x, c = 5x Perimeter = a + b + c ? 36 = 3x + 4x + 5x ? x=3 ? a = 9 cm, b = 12 cm and c = 15 cm ? ? Area = s(s ? a)(s ? b)(s ? c) = 18(18 ? 9)(18 ? 12 )(18 ? 15) = 18 ? 9 ? 6 ? 3 = 54 cm 2 22. A. T. Q. ?C = 900 So, that ? BCD is right angled triangle By using Pythagoras theorem. ? BD2 = BC2 + CD2 ? BD2 = 2a2 + 102 ? BD = 676 ? BD = 26 m Now, ar (?BCD) = 1 ? 24 ? 10 = 120 2 and ar ( ? BCD) = s(s ? a)(s ? b)(s ? c) m2 where a = 18, b = 26, c = 16 and s = 18 + 26 + 16 60 = = 30 2 2 So, ar ( ? ABD ) = 30 ? 14 ? 12 ? 4 = 24 35 23. So, ar (? ABD) = 120 + 35 sq m Let r be the radius of the base and h be the height of the cylinder. Total surface area 2 ? r(r + h ) = 462 cm2 … (i) and curbed surface area ( 2 ? rh ) = ? 462 = 154 cm 2 ? 2 ? rh + 2 ? r 2 = 462 ? 154 + 2? r 2 = 462 ? 2? r 2 = 462 – 154 = 308 ? r2 = 308 ? 7 = 49 2 ? 22 1 3 …. (ii) [From (i)] [From (ii)] ? r = 7 cm ? h= 7 cm. 2 From (ii) we have, the curved surface area, ? rh = 2 ? 22 ? 7 ? h = 154 7 ? Volume of the cylinder = ? r 2 h = 22 7 ? 7 ? 7 ? = 539 cm 3 . 7 2 Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 23 24. The total number of days for which the record is available = 250 (i) P(the forecast was correct on a given day) = (ii) The number of days when the forecast was not correct = 250 – 175 = 75 75 So, P(the forecast was not correct on a given day) = = 0. 3 25. 250 Number of days when the forecast was correct 175 = = 0. 7 Total number of days for which the record is available 250The figure obtained by joining the given three points A, B, & C is right angled triangle. SECTION # D 26. Let a + b = x; b + c = y; c+a=z L. H. S. (a + b)3 + (b + c)3 + (c + a)3 – 3(a + b)(b + c) (c + a) = x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – xz) 1 = (x + y + z) (2×2 – 23y2 + 2z2 – 2xy – 2yz – 2xz) 2 1 = (x + y + z) [(x2 – 2xy + y2) + (y2 + 23yz + z2) + (z2 – 2xz – x2)] 1 = (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2] 2 Now, x + y + z = (a + b) + (b + c) + (c + a) = 2a + 2b + 2c = 2(a + b + c) x – y = a – c , y – z = b – a, z – x = c – b 1 ? L. H. S. = . (a + b + c) [(a – b)2 + (b – a)2 + (c – b)2] 2 = (a + b + c) (a2 – 2ac + c2 + b2 – 2ab + a2 + c2 – 2bc + b2) = 2(a + b + c) (a2 + b2 + c2 – ab – bc – ac) = 2(a2 + b3 + 3 – 3abc) Hence Proved Construction : Join CD. Since, D is the mid-point of AB. So, in ? ABC, CD is the median. 1 ? ar(? BCD) = ar(? ABC) … (i) 2 Since, ? PDQ and ? PDC are on same base PD and between same parallels PD and CQ. ? ar(? PDQ) = ar(? PDC) 1 Now, from (i), ar(? BCD) = ar(? ABC) 2 27. ? ? ? ar(? BDP) + ar(? PDC) ar(? BDP) + ar(? PDQ) ar (? BPQ) 2 1 ar(? ABC) 2 1 = ar(? ABC) 2 1 = ar(? ABC) 2 = Hence proved. Free www. tekoclasses. com Director : SUHAG R.KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 24 28. Let, radius of each circle = r cm AB = 12 cm ? O1O2 = 12 – 2r In quadrilateral O1DO2C, O1D = O2C O1C = O2D ? O1DO2 is a rhombus ? CD ? O1O2 and CD bisect O1O2 ? CP = 1 2 [Radii of congruent circles] ? CD = 3 cm. 1 2 and O1P = (O1O2) = Now in right ? CPO1, (O1C)2 = (O1P02 + (PC)2 ? r2 = (6 – r)2 + (3)2 ? r2 = 36 + r2 – 12r + 9 ? 12r = 45 ? 29. r= Since AB = 9 m and BC = 40 m, ? B = 900, we have : AC = 9 2 + 40 2 m = 81 + 1600 m = 1681 = 41 m 45 12 1 2 (12 – 2r) = (6 – r) cm ? r = 3. 75 cm Therefore, the first group has to clean the area of triangle ABC, which is right angled.Area of ? ABC = = 1 2 1 2 ? base ? height ? 40 ? 9 m2 180 m2. The second group has to cleem the area of triangle ACD, which is scalene has having sides 41 m, 15m and 28m. Here 41 + 15 + 28 m = 42 m 2 Therefore, area of ? ACD = s(s ? a)(s ? b)(s ? c) s= = 42( 42 ? 41)( 42 ? 15)( 42 ? 28) m 2 = 42 ? 1 ? 27 ? 14 m 2 = 126 m 2 So, first group cleaned 180 m2 which is (180 – 126) m2, i. e. , 54m2 more than the area cleaned by the second group. Total area cleaned by all the students = (180 + 126) m2 = 306 m2. 30. C. I. 21-24 26-29 31-34 36-39 41-44 46-49 Frequency 30 24 52 28 46 10 C. I. (adjusted ) 20-25 25-30 30-35 35-40 40-45 45-50 Free ww. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 25 HINTS & SOLUTIONS (MATHEMATICS) TEST PAPER # 2 SECTION # A Directions : Answer the questions (1 to 10) 1. Let x = 0. 418 Then, x = 0. 4181818…. (i) multiplying eq. (i) by 10 10x = 4. 181818 …. (ii) again multiplying eg. (ii) by 100 1000x = 418. 181818 …. (iii) subtracting (ii) from (iii), we get : ? 990x = 414 414 23 . = 990 55 23 Hence, 0. 418 = . 55 ? x= 2. 3. 4. 5. -d/c (i) 3x + ay = 6 If x = 1, y = 1 is a solution, then it must satisfy the equation. ? 3(1) + a(1) = 6 ? a = 6 – 3 =3 600 1200 600 + 400 + 4x = 180 4x = 200 x = 200 200 . 7. 8. Parallelogram whose diagonals are equal is a square hence ? ABC = 900 L. S. A. of a square = 4 ? side2 = 4 ? 202 = 4 ? 400 = 1600 cm2 x= 10 + 12 + 16 + 20 + p + 26 = 17 6 84 + p = 17 6 p 14 + 6 = 17 ? ? ? 9. 10. p = 17 ? 14 = 3 6 i. e. , p = 3 ? 6 = 18. Points of the form (a, 0), i. e. , the points in which ordinate is 0, those points lie on the xaxis. The points in which abscissa is 0, lie on the y-axis. (ii) (1, 0) and (iv) (0, 0). A. T. Q. 1 + 2 + 5 + 10 18 = = 4. 5 4 4 Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 26 SECTION # B 11. 7 3 2 5 3 2 ? ? 10 + 3 6+ 5 15 + 3 2 .. (i) Let, I=A-B-C 7 3 10 + 3 Where A = A= and and and and B= B= B= 2 5 6+ 5 2 5 6? 5 ? 6+ 5 6? 5 2 30 ? 2 ? 5 6? 5 7 3 10 ? 3 ? 10 + 3 10 ? 3 7 3 ( 10 ? 3 ) 10 ? 3 ?A= ? A = 3 ( 10 ? 3 ) B = 2 30 ? 10 ….. (iii) ? A = 30 ? 3 …. (ii) and C= 3 2 15 ? 3 2 3 30 ? 18 3 30 ? 18 ? = = = ? 30 + 6 15 ? 18 ? 3 15 + 3 2 15 ? 3 2 So, I = A – B – C = ( 30 ? 3) ? ( 2 30 ? 10 ) ? ( 30 + 6 ) = 30 ? 3 ? 2 30 + 10 + 30 ? 6 = 2 30 ? 2 30 ? 3 + 10 ? 6 =1 12. 3600 where n is number of sides of polygon. So, 30 ? x = 3600 x = 12 13. Hence proved Let exterior and interior angles be x0 and 5×0 also (x + 5×0) = 1800, (x0 = 300) and x. n =Joining OA, OB and BA, we get a triangle. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 27 14. The given equation is 5x – y + 3 =0 15. Here a = 16 cm, b = 22, let c = x cm Now perimeter = a + b + c ? 64 = 16 + 22 + c c = 26 cm s= Area a + b + c 64 = = 32 2 2 = s(s ? a)(s ? b)(s ? c) = 32( 32 ? 16)( 32 ? 22 )(32 ? 26 ) = 16 ? 16 ? 10 ? 6 = 32 15 cm 2 SECTION # C 16. 2n ? 4 ? Let the number of sides be n and 2n. Then their interior angles are ? ? 90 ? and ? ? n ? 4n – 8 = 3n – 3 N=5 So that the number of sides are 5 and 10. 17. We are given that perimeter of triangle = 12 cm, ?B = 600 and ? C = 450 Steps of Construction (i) Draw a ray PX and cut off a line segment PQ = 12 cm from it. (ii) At P, construct ? YPQ = 300 (1/2 ? 600). (iii) AT Q, construct ? ZOP = 22. 50 (1/2 ? 450) (iv) Let the rays PY and QZ intersect at A. (v) Draw the perpendicular bisector of AP intersecting PQ at a point B. (vi) Draw the perpendicular bisector of AQ intersecting PQ at a point C. (vii) Join AB and AC. So, ? ABC is the required triangle. Free ? 2( 2 n ) ? 4 ? ? 90 ? . ? 2n ? ? 2n ? 4 ? 90 3 n = 2( 2 n ? 4) ? 90 4 2n 2n ? 4 3 = ? 2n ? 2 4 n? 2 3 = n ? 1 4 www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. (0755) 32 00 000 Page 28 18. Dimensions of the rectangular solid are 66 cm, 42 cm 21 cm. Volume of the solid = 66 ? 42 ? 21 cm3 Diameter of a spherical lead shot = 4. 2 cm ? radius = 2. 1 cm 4 22 ? ( 2. 1)3 3 7 Volume of the recangular solid ? Number of lead shots = Volume of one spherical shot 66 ? 42 ? 21 ? 21 = 88 ? ( 2. 1)3 Volume of a spherical lead shot = ? 19. xi 10 15 20 25 30 35 40 Total 20. fi 4 6 8 18 6 5 3 50 fixi 40 90 160 450 180 175 120 1215 So , x = ?f x ? f i i i = 1215 = 24. 30 50 Total number of drivers = 2000. (i) The number of drivers who are 18 – 29 years old and have exactly 3 accidents in one year is 61.So, P(driver is 18 – 29 years old with exactly 3 accidents) = 61 = 0. 0305 = 0. 031 2000 (ii) The number of drivers 30 – 50 years of age and having one or more accidents in one year = 125 + 60 + 22 + 18 = 225 So, P (driver is 30 – 50) years of age and having one or more accidents) = 225 = 2000 0. 1125 ? 0. 113 (iii) The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305 So, P(no accidents) = 21. 1305 = 0. 6525 2000 Let p(y) = y3 – 7y + 6 Constant term f p(y) is 6. Factors of 6 are ±1,±2 ,±6 , p(1) = 1 – 7 + 6 = 0 ? (y – 1) is a factor of p(x). Dividing p(y) by (y – 1), we find the other factor. (y) = (y – 1) (y2 + y – 6) = (y – 1 (y2 + 3y – 2y – 6) = (y – 1) [y(y + 3) – 2(y + 3)] = (y – 1) (y + 3) (y – 2) (i) A (ii) G (iii) 0 22. (iv) 4 (v) (0, -5) (vi) (-4 , 0) Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 29 23. Let PQRS be the given quadrilateral. PQ = 9 m, OR = 40m, SR = 28 m, PS = 15m, ? PQR = 90. 0 ? PR2 = 90 + 42 ? PR = 41 cm. ? Area of ? PQR = In ? PRS, PR = 41, SR = 28, PS = 15 s= 1 1 ? PQ ? QR = ? 9 ? 40 = 180 cm2 2 2 41 + 28 + 15 = 42 2 area of ? PRS = s(s ? a )(s ? b)(s ? c) = 42( 42 ? 41)( 42 ? 28)( 42 ? 15) = 42 ? 1 ? 14 ? 27 = 126 m 2 ? Area of ? PQRS 2 = area ? PQR + area ? PRS = 180 + 126 = 306 m2 24. 25. 4 ? 42 ? ? 28 ? .?? ? = ?? ? . x, where x would be length of wire. 3 ? 2 ? ? 2 ? x = 63 cm Let ABCD be the field. perimeter = 400 m So, each side = 400 m ? 4 = 100 m. i. e. AB = D = 100 Let diagonal BD = 160 m Then semi-perimeter of ? ABD is given by s= Therefore, area of ? ABD 100 + 100 + 160 m = 180 m 2 = 180(180 ? 100)(180 ? 100)(180 ? 160) = 180 ? 80 ? 80 ? 20 m 2 = 4800 m 2 Therefore, each of them will get an area of 4800 m2, and the total area = 2 ? 4800 m2 = 9600 m2. 26. SECTION # D Given : ? ABC is isosceles with AB = AC, E and F are the mid-point of BC, CA and AB.To prove : AD ? EF and is bisected by it Construction : Join D, E and F Proof: DE¦AC and DE = and 1 AB 2 1 DF¦AE and DF = AC 2 [Line segment joining mid-points of two sides of a triangle is parallel to the third side and is half of it. ] DE = DF [? AB = AC] Also AF = AE [? AF = ? DE = AE = AF = DF and also DF¦ AE and DE¦AF ? DEAF is a rhombus. Since diagonals of a rhombus bisect each other at right angles. ? AD ? EF and is bisected by it. Free 1 1 AB, AE = AC] 2 2 www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 30 27. Given : ABCD is a rhombus.AB produced to E and F such that AE = AB = BF Construction : Join ED and CF and produce it to meet at G. To prove : ED ? FC Proof : AB is produced to points E and F such that. AE = AB = BF Also, since ABCD is a rhombus AB = CD = BC = AD … (ii) Now, in ? BCF, BC = BF [From (i) and (ii)] ? ?1 = ? 2 [Exterior angle] ? 3 = ? 1 + ? 2 .. (iii) ? 3 = 2 ? 2 Similarly, AE = AD ? ? 5 = ? 6 ? 4 = ? 5 + ? 6 = 2 ? 5 ? 4 = 2 ? 5 by adding (iii) and (iv) we get ? ? EG ? FC Now in ? EGF 28. .. (iv) [? ?4 and ? 3 are consecutive interior angles] Hence proved. ? 4 + ? 3 = 2 ? 5 + 2 ? 2 800 = 2 (? 5 + ? 2 ) ? 5 + ? 2 + ? EGF = 1800 ? EGF = 900 Given : A cyclic quadrilateral ABCD. To Prove : ? BAD + ? BCD = 1800 and ? ADC + ? CBA = 1800 Construction : Let O be the centre of the circle passing through A, B, C and D. Join OB and OD. Proof. ?BAD = subtended by it at any point on the remaining part of the circle]…. (i) and ? BCD = ? BOD = y , say 2 2 Adding (i) and (ii), we have ? BAD + ? BCD = 1 1 1 reflex ? BOD 2 1 = x , say [The angle subtended by an arc at the centre is double the angle 2 (ab above)…. (ii) Again, as the sum of the angels of a quadrilateral is 3600, therefore, ? ADC + ? CBA = 600 ? ( ? BAD + ? BCD) = 3600 ? 1800 = 1800. 1 ? + y 2 2 1 1 = ( x + y ) = ? 3600 = 180 0 (As X + y = 3600 ) 2 2 IInd Part : ?ADQ + ? ADC = 1800 ? ?ADC = 1800 ? 950 ? ?ADC = 850 ? ?ADC + ? B = 1800 [Linear pair ] ?BCP + ? BCD = 1800 ? BCD = 1800 ? 1100 = 700 Now ? BDC + ? A = 1800 ? ?A = 1800 ? 700 = 1100 [Opp. angles of a cyclic quadrilateral] ? ? = 1800 – 850 ? ?B = 950 29. Let p(x) = x3 + mx2 + nx + 6 be the given polynomial. Then, (x – 2) is a factor of p(x) ? p(2) = 0 [? x – 2 = 0 ? x = 2] ? (2)3 + m(2)2 + n(2) + 6 = 0 ? 8 + 4m + 2n + 6 = 0 ? 4m + 2n = – 14 ? 2m + n = -7 ….. (i) Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. (0755) 32 00 000 Page 31 It is given that p(x) leaves the remainder 3 when it is divided by (x – 3). Therefore. p(3) = 3 ? (3)3 + 3(3)2 + n(3) + 6 = 3 ? 27 + 9m + 3n + 6 = 3 ? 9m + 3n = 3 – 27 – 6 ? 9m + 3n = – 24 – 6 ? 9m + 3n = – 30 ? 3m + n = – 10 …. (ii) Subtract equation (ii) from (i), we get m=-3 Putting m in (i), we get ? 2m + n = – 7 ? 2 (-3) + n = – 7 -6+n=7 ? n=-1 30. C. I. 0-9 10-19 20-29 30-39 40-49 50-59 60-69 True limits -. 5-9. 5 9. 5-19. 5 19. 5-29. 5 29. 5-39. 5 39. 5-49. 5 49. 5-59. 5 59. 5-69. 5 Frequency 5 15 20 23 17 11 9 C. F. 5 20 40 63 80 91 100 Free www. tekoclasses. com Director : SUHAG R.KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 32 HINTS & SOLUTIONS (SCIENCE AND TECHNOLOGY) TEST PAPER # 1 SECTION # A 1. 2. 3. 4. Sn = u + a ( 2 n ? 1) 2 10 = 3600 1800 1800 = = = 57. 30 2? ? 3. 14 Rate of change of momentum or change in momentum per unit time is equal to the applied force. A sponge has minute holes, in which air is trapped. When we press it, the air is expelled out and we are able to compress it. But still it maintain its shape and size as we remove the pressure applied, so it is a solid. 5. 6. 7. 15% solution of Nacl is the solution which contains 15 g of NaCl and 85 g of water.Oxygen, Nitrogen and Argon can be separated by fractional distillation of air. Physical quantities can be divided into two types : (i) Scalar quantity : Any physical quantity, which can be completely specified by its magnitude alone, a scalar quantity or a scalar. Eg : Charge, distance, area, speed, time, temperature, density, volume, work, power, energy, pressure, potential etc. (ii) Vector quantity : Any physical quantity, which required direction in addition to its magnitude it known as a vector quantity. Eg. : Displacement, velocity, acceleration, force, momentum, weight and electric field etc. 8.Distance = AB + BA + AC = a + a + a2 + b2 = 2a + a 2 + b2 = 2a + a 2 + b 2 Displacement = AC = a2 + b2 2 KWh = 1000 Wh = 1000 ? 60 ? 60 Ws = 3600000 J = 3. 6 ? 106 J sea water without being absorbed. (ii) Ultrasonic waves cannot be confused with the noises, such as the voice of engines of ship. It is because the ultrasonic waves are not perceived by human ear. 9. 10. (? Ws = J) (i) Ultrasonic waves have a very high frequency due to which they can penetrate deep in 11. (a) Atmospheric gases can be liquefied by decreasing temperature and increasing pressure. (b) LPG : Liquefied Petroleum Gas. CNG : Compressed Natural Gas. Free ww. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 33 12. In a neutral atom, number of electrons are equal to the number of proton. No. of protons = No. of electrons = 29 Mass number = No. of protons + No. of neutrons So, No. of neutrons (N) = Mass number – No. of protons = 63 – 29 Number of neutrons (N) = 34 Electronic configuration : Electronic configuration is written on the basis of total number of electrons. So, the electronic configuration of the element is 2,8,18,1 It is clear that the particle must be placed on the line AB, suppose it is at a distance x from Let its mass is m’.F1 = Gmm’ towards A x2 G( 2m )m’ towards B. ( d ? x )2 13. A. and that due to B is, F2 = The net force will be zero if F1 = F2 Thus, Gmm’ G(2m )m’ = x2 (d ? x )2 or (d – x)2 = 2×2 d -x= ± 2x d = (1 ± 2 ) x x= d (1 + 2 ) ( d 1+ 2 or (1 ? 2 ) d As x cannot be negative So x = 14. ) Mass of bullet, m = 10 g = 0. 01 kg Mass of Gun, M = 5. 0 kg Velocity of bullet, v = 250 m/s Let the velocity with which Gun recoils is V. By the law of conservation of momentum, Initial momentum of the system = Final momentum of the system. m(0) + M(0) = mv + V V=? 0. 01 ? 250 = ? 0. 5 m / s 5. 0Recoil velocity of the gun is 0. 5 m/s. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 34 15. The reaction between iron and steam occurs as : 3Fe + 4H2O Fe3O4 + 4H2 From chemical equation we can write 4 moles of steam (H2O) are required to convert 3 moles of Fe into its oxide. ? 1 mole of steam (H2O) will convert = 1 mole of steam = 18 g of H2O 1 mole of Fe = 56 g of Fe So, 18 g of steam will convert = 3 4 3 4 mole of Fe into its oxide. ? 56 g of Fe into its oxide = 42 g of FE 18 g of steam will 42g of Fe into its oxide. 16. a) Mole : A mole of a substance is that amount of the substance which contains the same number of particles (atoms, molecules or ions) as there are carbon atoms in 12 g of carbon – 12 element. (b) 1 mole of carbon = Gram Atomic Mass of carbon = 12 g So, 1 of carbon = 1 12 1 12 mole of carbon 1 1 mole of an element = 6. 023 ? 1023 atoms of element mole of carbon = 6. 023 ? 1023 ? 12 atoms = x 1 24 Now, 1 mole of magnesium= Gram Atomic Mass of magnesium = 24 g So, 1 g of Magnesium = mole of magnesium. 1 mole of magnesium contain = 6. 023 ? 1023 atoms 1 1 mole of magnesium contain = ? 0. 023 ? 1023 atoms 24 24According to the questions 1 ? 6. 023 ? 10 23 carbon atoms = x 12 1 x ? ? 6. 023 ? 10 23 magnesium atoms 24 2 1 g of magnesium contains x 2 atoms. x 2 Thus, if 1 gram of carbon contains x atoms, then 1 gram of magnesium will have atoms 17. in it. Let a body of mass m is at rest at a height h from the earth’s surface, when it starts falling, after a distance x (point B) its velocity becomes v and at earth’s surface its velocity is v’. Mechanical energy of the body : At point A : EA= Kinetic energy + Potential energy Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 35 EA = m(0 )2 + mgh 2 EA = mgh 1 mv 2 + mg ( h ? x ) 2 ….. (i) ….. (ii) At point B : EB = From third equation of motion at points A and B v2 = u2 + 2gx ? u=0 On putting the value of v2 in equation (ii) EB = 1 m m(2gx) + 2 mgh – mgx …. (iii) E B = mgx + mgh – mgx EB = mgh At point C : 1 m ( v’ )2 + mg ? 0 . 2 1 E C = m( v’ )2 2 EC = …. (iv) From third equation of motion at points A and C. (v’)2 = u2 + 2gh ? u=0 So, (v’)2 = 2gh On putting the value of (v’)2 in equation (iv) EC = 1 m( 2gh ) 2 EC = mgh or …. (iv) From equation (i), (iii) and (v) EA = EB = EC Hence, the mechanical energy of a freely body will be constant. i. e.Total energy of the body during free fall, remains constant at all positions. The form of energy, however keeps on changing. At point A, energy is entirely potential energy and at point C, it is entirely kinetic energy In between A and C, energy is partially potential and partially kinetic. This variation of energy is shown in figure. Total mechanical energy stays constant (mgh) throughout. Thus is an isolated system, where only conservative forces cause energy changes, the kinetic energy and potential energy can change, but the mechanical energy of the system (which is sum of kinetic energy and potential energy) cannot change.We can, therefore, equate the sum of kinetic energy and potential energy at one instant to the sum of kinetic energy and potential energy at another instant without considering intermediate states. This law has been found to be valid in every situation. No violation, whatsoever, of this law has ever been observed. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 36 18. (a) 2KCIO3 ? 2KCI + 3O2 ? > 30g 14. 9g 9. 6g Suppose mass of under composed KCIO3 = xg According to law of conservation of mass the total mass before and after the reaction remains constant. 0. 0 = x + 14. 9 + 9. 6 x = 5. 5 g (b) (i) Nucleus is present in the centre of atom and it contains protons and neutrons. (ii) Nucleus is positively charged centre having appreciable mass. Electron is present in extra nuclear resion of atom. Electron is negatively charged particle with negligible mass. SECTION – B 19. 20. 21. 22. Mitochondria, plastid and centrioles have their own DNA molecules so they are called as semi – autonomous cell organelles. Epithelium Diseases which last for short periods and are severe are called acute diseases. Examples of Acute diseases : Cough, cold, cholera, typhoid. 23. i) Division Thallophyta : • Thallus : Undifferentiated plant body i. e. absence of root, stem & leaves. • Their is no vascular system. • Reproductive organs are single -celled and there is no embryo formation after fertilization. • Dominant gametophyte. Algae and Fungi. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 37 24. Blood : It is reddish coloured vascular tissue that flows inside blood vessels by means of pumping activity of heart, pH is 7. 4. Colour is bright red in oxygenated form and purple in deoxygenerated form Blood has two parts, fluid plasma (55%) and blood cells (45%).There are three types of blood cells 1. Red blood Cells. Eosinophils Granulocytes 2. White Blood Cell Agranulocytes Lymphocyt 3. Blood Platelets Lymph : It is light yellow fluid connectivees tissue which is formed from tissue fluid and filtered out blood. Lymph is devoid of red blood corpuscles and blood platelets. Proteins are fewer, so are white blood corpuscles. However, lymphocytes are most abundant. Lymph flows lymph capillaries and lymph vessels. All places they pass through lymph nodes and lymph organs where lymphocytes multiply and mature. They are also sites for entrapping microbes. Lymph’s is ultimately passed into blood.Most of the organs and tissues pour their secretions and excretions into lymph instead of blood. Lymph’s is middle man between tissues and blood. Serum. Serum is plasma from which fibrinogen is removed. Advantages of Fertilizers. • Fertilizers are nutrient specific, so these supply specific elements to the soil. • Fertilizers are compact and in concentrated form, So they are easy to store and transport. They are even required in very small amount. • Fertilizers are readily absorbed by the plants because they are soluble in water. • Fertilizers are available throughout the year at the places. Fertilizers are a factor in higher yields of high cost farming. Disadvantages of Fertilizers. Fertilizers are non-biodegradable. The excessive use of fertilizer causes. • Water pollution in lakes and rives due to eutrophication which makes the water unfit for human consumption and even kills the aquatic animals. • Change in the nature of soil (or soil chemistry) making it either too alkaline or too acidic. For example, when we use ammonium sulphate fertilizer in the same soil again and again, the soil becomes acidic whereas when we use sodium fertilizer in the same soil again and again, the soil becomes alkaline. Fertilizers are quite expensive. They push the cost of crop production. • Fertilizers give only short term benefits. • They harm soil microorganisms. • Nonreplenishment of organic matter destroys the crumb structure of soil affecting both hydration. www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 38 Neutrophils Basophils Monocytes 25. Free 26. Characters 1. Nuclear body Prokaryotic Cells Incipient nucleus, No nuclear membrane Nucleolus absent No mitosis Single closed loop, (histones absent) In internal membranes, (chloroplasts absent) Absent 2.Photosynthetic apparatus 3. Golgi bodies, Chloroplast, Endoplasmic reticulum, Mitochondria, Lysosomes 4. Ribosomes Eukaryotic Cells True nucleus, Nuclear membrane, present Nucleolus present Mitosis found Multiple chromosomes, (histones present in chromosome) In chloroplasts Present 70 S type 80 S type 27. (i) Striated muscles : They are also called as voluntary muscles because these are under the control of one’s will. Muscle fibers or cells are multinucleated and unbranched. Each fiber enclosed by thin membrane which is called as sarcolemma. Cytoplasm is called as sarcoplasm. These muscles get tired and need rest. ii) Non striated muscles : They are involuntary muscles also called as smooth muscles. These muscle fibers are uninucleated & spindle shaped. They are not enclosed by membrane but many fibers are joined together in bundles. Such muscles are found in the walls of stomach, intestine, urinary bladder, bronchi, iris of eye etc. peristaltic movements i n alimentary canal are brought about by smooth muscles. (iii) Cardiac muscle fibers : They are also involuntary muscles. Only found in the walls of heart. Their structure is in between the striated & non-striated muscles. They are uninucleated & branched. Branches are united by intercalated disc.In these muscles rhythmic contraction & relaxation occurs throughout the life. Free www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 39 HINTS & SLUTIONS (SCIENCE AND TECHNOLOGY) TEST PAPER # 2 SECTION # 1. 4. No 2. Steel 3. Zero The characteristics of matter which can be demonstrated by diffusion and Brownian motion are (i) matter is made up of tiny particles. (ii) the particles of matter are constantly moving. The outer most shell of an atom is known at the valence shell. Fourth and fifth states of matter are Plasma state and Bose-Einstein condensate respectively.Given : Power = 100 watt Time 6 hours = 21600 s Energy consumed = Power ? Time = 100 ? 21600 = 216 ? 104 J s1 = 10 km, v1 = 50 km/h, s2 = 10 km, v2 = 40 km/h Average speed = t1 = s 1 10 1 = = h, v 1 50 5 Total distnce Total time taken t2 = s2 10 1 = = h v 2 40 4 5. 6. 7. 8. Average speed = 9. s 1 + s 2 10 + 10 = = 44. 4 km / h 1 1 t1 + t2 + 5 4 Let the distance of dog from the building be d. Here, v = 340 m/s, t = 2. 5 s We know, d = vt 340 ? 2. 5 = = 425 m 2 2 10. Given : Mass, m = 50 kg Force, F = 200 N Acceleration, a = F 200 = = 4 ms ? 2 m 50 The acceleration of the box will be 4 m/s2 11.Now, if the mass is halved, the acceleration would be doubled, it would be 8 ,/s2. Defects of Rutherford’ model : (1) Rutherford did not specify the number of electrons in each orbit. (2) According to electromagnetic theory, if a charged particle (electron) is accelerated around another charged particle (nucleus) then there would be continuous irradiation of energy. This loss of energy would slow down the speed of electron and eventually the electron would fall into the nucleus. But such a collapse does not occur. Rutherford’s model could not explain this theory. www. tekoclasses. com Director : SUHAG R. KARIYA (SRK Sir), Bhopal Ph. (0755) 32 00 000 Page 40 Free 12. State Solid Liquid Gas Rigidity Compressibility Solids are rigid. They have definite Solids are almost incompressible. shape and definite volume. Liquids have fixed volume but no Liquids are relatively more fixed shape. compressile than solid. Gas has neither fixed volume nor Gas has high compressibility. fixed shape. 13. 14. Here, u = 0, s = 19. 6 m, a = g = 9. 8 m/s2 Using equation, v2 = u2 + 2as ? v2 = 0 + 2 ? 9. 8 ? 19. 6 ? v = 19. 6 m/s Momentum of the dumb-bell just before touching the ground is given by, P1 = mv = 15 ? 19. 6 = 294 m/s Finally, the dumb-bell comes to rest, i. . its final momentum is zero. ? Momentum transferred to the floor = 294 – 0 = 294 kg m/s 1st Equation of Motion : Consider a body having initial velocity u. Suppose it is subjected to a uniform acceleration ‘a’ so that after time ‘t’ its final velocity becomes v. Now we known, Acceleration = a= v? u t change in velocity Time or v = u + at or v = at + u ….. (i) 2nd Equation of Motion : Suppose a body has an initial velocity ‘u’ and uniform acceleration ‘a’ for time ‘t’ so that its final velocity becomes ‘v’. The distance traveled by moving body in time t is s, then the average velocity = (u + v)/2.Distance traveled = Average velocity ? time ? u+v? ? u + u + at ? s=? ?t ? s = ? ?t 2 ? 2 ? ? ? ? 2 ut + at 2 ? 2 u + at ? ? ? s=? ?t ? s = ? ? ? 2 ? 2 ? ? ? 1 s = ut + at 2 2 (As , v = u + at ) ……(ii) 15. 16. Gun powder is a mixture of potassium nitrate, sulphur and charcoal. The mixture of gun powder can be separated by using suitable solvent in following steps (a) When water is added to the mixture then potassium nitrate is dissolved in water while charcoal and sulphur remain insoluble. Insoluble sulphur and charcoal are then separated by filtrating. (b) This residue is dissolved in carbon disulphide.Sulhpur is soluble in CS2 while charcoal is insoluble in it. These two constituents can then be separated by filtration. To separate a mixture of two or more miscible liquids for which the difference in boiling points is less than 25K, fractional distillation process is used. Alcohol (b. p. 780C) and water (b. p. 1000C) are miscible liquids having boiling point difference 220C or 22K. So, mixture of alcohol and water can be separated by fractional distillation. Apparatus used in separation of mixture of alcohol and water is given below- Free www. tekoclasses. com Director : SUHAG R.KARIYA (SRK Sir), Bhopal Ph. :(0755) 32 00 000 Page 41 17. The sound waves (coming from a sound producing body) are collected by the pinna of outer ear. These sound waves pass through the ear canal and fall on the ear-drum. Sound waves consist of compression (high pressure regions) and rarefactions (low pressure regions). When the compression of sound wave strikes the ear-drum, the pressure on the outside of ear-drum increases and pushes the ear-drum inwards and when the rarefaction of sound wave falls on the ear-drum, the pressure on the outside of ear-drum decreases and it moves outward.Thus, when the sound waves fall on the ear-drum the ear-drum starts vibrating back and forth rapidly. The vibrating ear-drum causes a small bone hammer to vibrate. From hammer, vibrations are passed on to the second bone anvil and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval windows and passes its vibrations to the liquid in the cochlea. Due to this, the liquid in the cochlea beings to vibrate. The vibrating liquid of cochlea sets up electrical impulses in the nerve cells present in it.These electrical impulses are carried by auditory nerve to the brain. The brain interprets these electrical impulses as sound and we get the sensation of hearing. 18. (a) Let the percentage of Average atomic mass 16 8 X be (a) then percentage of will be(100 – a) 16 ? a + 18( 100 ? a) = = 16. 2 100 a = 90 Thus, percentage of 16 X = 90 8 Percentage of 18 X = (100 ? 90) = 10 8 (b) Hydrogen (1 H ) does not contain any neutron. It has only 1 electron and 1 proton. 1 (c) Three isotope of hydrogen 1H1, 1H2 and 1H3 have one proton but number of neutrons are different in different isotopes.Chemical properties depend on number of electrons and number of electrons are similar in all three isotopes. So, these three isotopes have identical chemical properties. SECTION # B 19. The diseases which are long lasting are called chronic diseases. Chronic diseases have drastic long term effects on people’s health. Examples of Chronic diseases : Tuberculosis, arthritis, diabetes, cancer, cardiovascular diseases, elephantiasis. Maristematic tissue. Chlorofluorocarbons (CFC) 20. 21.