Guess Papers SCIENCE SCIE NCE & MA MATHS Class-9
5 2
1
P 4 9 A 2x 2 1 1 4 + 4 ) C 3 4 6 2
2
P
Table of Contents Science (Set(Set-1) 1) Questions
Page 4-10
Science (Set(Set-1) 1) Solutions
Page 12-21
Science (Set(Set-2) 2) Questions
Page 24-29
Science (Set(Set-2) 2) Solutions
Page 31-39
Mathematics (Set(Set-1) 1) Questions
Page 42-48
Mathematics (Set(Set-1) 1) Solutions
Page 50-66
Mathematics (Set (Set-2) -2) Questions
Page 69-75
Mathematics Mathe matics (Set-2) Solutio Solutions ns
Page 77-95
Follow Us on
www.meritnation.com
Table of Contents Science (Set(Set-1) 1) Questions
Page 4-10
Science (Set(Set-1) 1) Solutions
Page 12-21
Science (Set(Set-2) 2) Questions
Page 24-29
Science (Set(Set-2) 2) Solutions
Page 31-39
Mathematics (Set(Set-1) 1) Questions
Page 42-48
Mathematics (Set(Set-1) 1) Solutions
Page 50-66
Mathematics (Set (Set-2) -2) Questions
Page 69-75
Mathematics Mathe matics (Set-2) Solutio Solutions ns
Page 77-95
Follow Us on
www.meritnation.com
SCIENCE (Set-1) Time allowed: 3 hours
Maximum marks: 90
General instructions: (i) The question paper comprises of two sections, sec tions, A and B. You You are to attempt both the sections. (ii) All quesstions are compulsory. (iii) All questions of Section A and all questions of Section B are to be attempted separately. (iv) Question numbers 1 to 3 in Section A are one-mark questions. These are to be answered in one word or one sentence. (v) Quetion numbers 4 to 7 in Section A are two-mark questions. These are to be answered in about 30 words each. (vi) Question numbers 8 to 19 in Section A are three-mark questions. These are to be answered in about 50 words each. (vii) Question numbers 20 to 24 in Section A are five-mark questions. These are to be answered in about 70 words each. (viii) Question numbers 25 to 42 in Section B are multiple choice questions based on practical skills. Each question is a one-mark question. que stion. You You are to select selec t one most appropriate response out of the four provided to you.
Follow Us on
www.meritnation.com
Section A 1. Name two processes by which oxygen and carbon dioxide are cycled in the atmosphere. 2. What is the work done by the force of gravity on a satellite moving around the earth in a circular orbit? 3. Number of moles What are I and II? 4. How does water pollution lead to a decrease in fish population? 5. Around 50 people in a locality consumed contaminated water. Out of these 50 people, only 20 people suffered from diarrhoea whereas others remained disease-free. Why? 6. A substance of volume 40 cm3 and mass 80 g is immersed in water. The density of water is 1 g/cm3 . Will the substance float or sink in water? 7. Is there any relationship between
?
8. Give the characteristic features of the division to which Marsilea and Equisetum belong. 9. What role is played by water in soil formation? 10. What is the law of conservation of energy? A ball of mass 500 g is dropped from a tower of height 80 m. Determine the velocity with which the ball strikes the ground. 11.Complete the given table.
Follow Us on
www.meritnation.com
12. A student observed a warm-blooded, vertebrate specimen with hollow bones. Which class does this specimen belongs to? Write the characteristic features of this class. 13. (i) Distinguish between the physical notations ‘g’ and ‘G’. (ii) A body having a mass 20 kg on earth is placed on the surface of moon. Calculate the change in the weight of the body. 14. If the time period of ripples in water is 0.2 s, then what is the speed of the wave? (Given that the wavelength of the ripples is 10 cm) 15. A force of 15 N acts on an object for a distance of 3 m, as shown in the given figure.
What is the work done on the box? 16. Why the use of antibiotics is not effective against common cold? 17. State some of the distinguishing features of mammals that differentiates them from other animals. 18. (a)What is the number of moles in 30.110 × 1024 molecules of nitrogen dioxide? (b) Of one mole of baking soda or one mole of washing soda, which contains greater number of molecules? 19. (a) Calculate the number of electrons and neutrons present in
.
(b) Write the electronic configuration and number of neutrons present in
.
20. Answer the following questions. (a) What is the nitrogen-cycle? (b) How are CFCs responsible for the depletion of the ozone layer? (c) Explain why the atmospheric temperature is increasing gradually. 21. Answer the following questions. (a) Describe the sub-phylum that lies at the border line between chordates and non-chordates. (b) Classification throws light on the evolutionary history of an organism. Justify this statement. 22. (a) State Archimedes’ principle. (b) Steel sinks in water, but a ship made of iron and steel floats. Why? (c) Give two examples of the application of Archimedes’ principle.
Follow Us on
www.meritnation.com
23. (a) What are isotopes? Name the three isotopes of hydrogen. (b) An element X is available in nature in the form of two isotopes: Calculate the average atomic mass of atom X. (c) What is the relationship between the following two atoms? Explain.
(75 %) and
(25 %).
and
24. A body having mass m starts, from rest, falling freely from a height h. At a height x above the ground, its velocity becomes v . Derive the expression for its total energy at i) height h ii) height x iii) ground
Section B 25. I- Needle like leaves II- Presence of Capsule III- Presence of rementia IV- Presence of heterospores Which of the above features belong to Pinus? (a) II and II I (b) II and IV (c) I and II I (d) I and IV 26. The given picture represents two different plants.
The presence of which of the following features is similar in the given plants? (a) Adventitious roots (b) Circinate leaves (c) Capsule (d) Spores
Follow Us on
www.meritnation.com
27. Male and female cockroaches can be distinguished by the (a) number of segmented parts (b) number of pairs of legs (c) presence of the anal cerci (d) presence of the anal style 28. A student observed cycloid scales on the specimen of a fish. Which of the following fish specimens he might have observed? (a) Scoliodon (b) Torpedo (c) Trygon (d) Labeo 29. Rohan was observing the external features of a rose plant. Which of the following features was not observed by him in the plant? (a) Presence of leaves in alternate manner (b) Leaves showing parallel venation (c) Presence of two cotyledons (d) Petals in the multiple of five 30. The given image represents one of the stages in the life cycle of a mosquito.
Which of the following labelled parts serve as the respiratory organ in the given stage? (a) P (b) Q (c) R (d) S
Follow Us on
www.meritnation.com
31. Which of the following relations between the angle of incidence of sound and the angle of reflection of sound is correct for a sound wave reflecting on a hard surface?
(a)
(b)
(c)
(d)
32. In an experiment on studying the laws of reflection of sound, the positions of the clock and ear is shown in the figure below. At what angle from the normal line, the second tube is placed so that the listener gets the best reflected sound?
(a) 30 (b) 45 (c) 55 (d) 60
ₒ
ₒ
ₒ
ₒ
33. Naked seeds are produced by the members of (a) gymnosperms (b) pteridophytes (c) angiosperms (d) bryophytes
Follow Us on
www.meritnation.com
34. What would we mean if we say that the density of wood is 25 kg/m3 ? (a) Volume of 1 kg of wood is 25 cm3 . (b) Volume of 1 kg of wood is 25 m3 . (c) Volume of 25 kg of wood is 1 m3 . (d) Volume of 25 kg of wood is 1 cm3 . 35. A stone is suspended from a spring balance and its weight is recorded. The stone is then immersed completely in brine solution, water, alcohol and petrol one by one. The loss in the weight of the stone is maximum in (a) brine (b) water (c) petrol (d) alcohol 36. When a body is immersed in a liquid it undergoes an apparent loss in its weight due to (a) upward thrust (b) decrease in its mass (c) decrease in its volume (d) decrease in the density of the body 37. If a 10 kg body with base dimensions 2 m the body on the ground is (a) 2 N.m-2 (b) 7 N.m-2 (c) 9 N.m-2 (d) 14 N.m-2
7 m is resting on ground, then the pressure exerted by
38. Which of the following statements regarding pressure is incorrect? (a) School bags have wide straps to reduce the pressure. (b) Pressure decreases with the increase in area of contact of the body. (c) Pressure on the ground is more when a person is walking than standing. (d) Pressure increases with increase in area of contact of the body. 39. In an experiment conducted to determine the velocity of a pulse propagated through a long stretched spring, the jerk given to create a pulse is (a) along the length of the string (b) at an angle of 60 with the string (c) at an angle of 45 with the string (d) perpendicular to the length of the string ₒ
ₒ
Follow Us on
www.meritnation.com
40. Consider the given image. If a pulse is created at position A and it takes 5 seconds to reach the fixed end of the spring, then what is the velocity of the pulse?
(a) 2 ms-1 (b) 4 ms-1 (c) 8 ms-1 (d) 12 ms-1 41. A student performs an experiment to study the rusting of iron nails and notes that a rusty nail is heavier in mass than a non-rusty nail. Which of the following inferences is correct on the basis of the given observation? (a) Rusting of iron nail does not follow the law of conservation of mass. (b) Mass of oxygen and water was added to the mass of the rusty nail. (c) There was an error in weighing the rusty nail. (d) The process of rusting was incomplete. 42. In an experiment, 115 g of copper was allowed to react with 48 g of oxygen gas to form copper (I I) oxide. After the reaction, no residue was left. What amount of copper (I I) oxide was formed after the experiment? (a) 48 g (b) 67 g (c) 115 g (d) 163 g
Follow Us on
www.meritnation.com
SCIENCE (Set-1) SOLUTIONS
Follow Us on
www.meritnation.com
SC IE NCE (Set-1) SOLU TI ON S Answers:
Section A
( 1-mark)
1. Carbon dioxide as well as oxygen is cycled through the environment by the processes of photosynthesis and respiration. 2. The satellite moves around the earth in a circular path. Earth’s gravitational force acts on the satellite at right angles to the direction of motion of the satellite. So, the work done by the earth on the satellite moving around it in a circular orbit is zero.
3.
Thus, I and I I corresponds to Given Mass and Avogadro number. 4. When chemicals (excess of fertilisers and pesticides) enter water bodies, they cause an increase in the growth of algae (this process is called eutrophication). This excess growth covers the surface of water and reduces the amount of oxygen available to the aquatic organisms. Consequently, the fishes die and hence their population decreases. 5. Around 50 people in a locality consumed contaminated water. Out of these 50 people only 20 people suffered from diarrhoea whereas others remained disease-free. This happened because of the following factors: • A healthy and well-nourished body is less likely to catch a disease when exposed to diseasecausing agents. On the other hand, a poorly nourished body will easily become diseased. • A person can have a genetic predisposition to catch diarrhoea on immediate exposure. The twenty people who caught diarrhoea might be genetically more prone for catching the disease. 6. It is known that density of substance = 80 ÷ 40 = 2 g/cm3 Since the density of the substance is more than that of water, the substance will sink in water. 7. Both carbon and nitrogen have the same mass number i.e., 13, but different atomic numbers i.e., 6 and 7 respectively. Thus, they are isobars.
Follow Us on
www.meritnation.com
(3-mark) 8. Equisetum and Marsilea are pteridophytes. Some of the general features of pteridophytes are:
i. The plant body is differentiated into root, shoot, and leaves. ii. They have vascular tissue for conducting food and water. iii. They produce naked embryos called spores. 9. Water helps in soil formation in the following ways: (i) Water enters the rocks through cracks in them. When temperature falls, water freezes. As ice occupies more space than water, the crack expands when water freezes. This expansion eventually leads to the breaking down of rocks. (ii) When water flow along the rocks, it creates a friction between water and rocks, which results in the weathering of rocks. This leads to the formation of soil. 10. According to the law of conservation of energy, the total energy of a system always remains constant, i.e., the sum of potential and kinetic energies of a system remains the same. Mass of the ball, m = 500 g = 0.5 kg Height of the tower, h = 80 m At the top of the tower: Potential energy of the ball, E P = mgh = 0.5 × 9.8 × 80 = 392 J Kinetic energy of the ball, E K = 0 (Since velocity = 0) ∴Total energy of the ball at the top, ET = E P + E K = 392 J On the ground: Potential energy of the ball, E P = 0 (Since height = 0) Kinetic energy of the ball, E K = Where v is the velocity with which the ball strikes the ground ∴Total energy of the
ball on the ground,
As per the law of conservation of energy, the total energy of the system remains constant. ∴E T = E G
Hence, the ball strikes the ground with a velocity of 39.6 m/s.
Follow Us on
www.meritnation.com
11. The atomic number of an element is the number of protons present in an atom. For carbon: Number of protons = 6 Thus, atomic number of carbon = 6 For argon: It is given that atomic number of argon is 18. Thus, the number of electrons present in the nucleus of argon atom is 18. For sodium: Mass number of an element is the total number of protons and neutrons present in the nucleus of an atom. Thus, mass number of sodium = Number of protons + Number of neutrons = 11 + 12 Therefore, mass number of sodium = 23 12. A warm-blooded, vertebrate specimen with hollow bones belongs to class Aves. Some of the characteristic features of this class are: • Most of them have feathers. • They possess a beak. • Forelimbs are modified into wings for flight. • Hind limbs are modified for walking. • Heart is four-chambered. • Respiration is through lungs. • They are warm-blooded animals. • They are oviparous. 13. (i) The given table lists some of the differences between acceleration due to gravity (g) and universal gravitational constant (G).
(ii) Weight (w) is defined as the magnitude of gravitational force exerted on a body. It is given by: w = mg Where m = Mass of the body = 20 kg g = Acceleration due to gravity On the surface of earth, g = 9.8 m/s2 Weight of the body on the surface of earth, we = 20 × 9.8 = 196 N On the surface of moon:
Follow Us on
www.meritnation.com
14. Speed, wavelength, and time period of sound waves are related as:
Speed It is given that: Wavelength = 10 cm = 10 × 10−2 m Time period = 0.2 s Speed = = 0.5 m/s Therefore, the speed of the ripple is 0.5 m/s.
15. Work done (W) by force (F) on the box is equal to the product of applied force and the displacement caused in the box (s) by that force. It can be expressed as Work done = Force × Displacement Applied force = 15 N Displacement in the box = 3 m Work done = 15 × 3 = 45 J 16. Antibiotics block a particular biochemical mechanism in the target organisms. Penicillin, for example, blocks the synthesis of the cell wall in bacteria. Therefore, antibiotics can be used effectively against a large number of bacteria. However, it cannot be used against viral infections such as common cold. Viruses do not have cellular structures or biochemical pathways of their own. They survive on the cellular apparatus of their host. Hence, it is difficult to find virus specific targets. Thus, the use of antibiotics is not effective against common cold. 17. Some of the distinguishing features of mammals which are not present in other group of animals are as follows: • They have two pairs of limbs for walking, running, or flying. • Their skin has hair as well as sweat glands. Hair on their body protects them in winters and sweat glands keep their body cool in summers since they are warm-blooded animals. • The heart is four-chambered. • They give birth to young ones i.e., they are viviparous with some exceptions. 18. (a) 6.022 × 1023 molecules are present in 1 mole of nitrogen dioxide. 30.110 × 1024 molecules are present in
=
= 5 × 101 moles. Hence, 50 moles of nitrogen dioxide contains 30.110 × 1024 molecules. (b)The number of molecules present in 1 mole of any substance, whether it is baking soda or washing soda, is fixed. One mole of a substance contains 6.022 × 1023 particles.
Follow Us on
www.meritnation.com
19. (a)Number of electrons in
= 15 + 3 = 18
Number of protons present = 15 Number of neutrons = Mass number − Number of protons = 31 − 15 = 16 (b) For chlorine, Atomic number = 17 Electronic configuration = 2, 8, 7 No. of neutrons = Atomic mass − Atomic number = 35 − 17 = 18 (5-mark)
20. (a) Nitrogen-cycle is a biogeochemical cycle. It describes the transformation of atmospheric nitrogen to simpler molecules in soil and water that get converted to more complex molecules and eventually are converted back to the simple nitrogen molecule in the atmosphere. (b) CFCs or chloro fluoro carbons are carbon compounds having both fluorine and chlorine. On reaching the ozone layer, CFCs react with the ozone molecules. This results in a reduction of the ozone layer, thereby causing its depletion. (c) The rise in atmospheric temperature is due to the release of large amount of greenhouse gases such as carbon dioxide and methane. The greenhouse gases trap the infrared radiations coming from sun in the atmosphere. This effect is known as greenhouse effect. With the increase in the amount of greenhouses gases in the atmosphere, the temperature of earth also increases. This increase in average earth’s atmosphere is known as global warming. 21. (a) Sub-phylum Protochordata lies at the border line between chordates and non-chordates. These animals are primitive chordates and have given rise to modern day chordates. General features of Protochordata are as follows: • They are exclusively marine animals. They often live in burrows. • They show an organ system level of organisation and are triploblastic. • Their body is not segmented and is bilaterally symmetrical. • Notochord is present in some stages during their lives. • Body cavity is enterocoelom. (b) Most of the life forms seen today have arisen from the accumulation of changes in the body design of ancient organisms. This slow change in the body design of an organism over a long period of time is termed as evolution. Classification allows things to be identified and categorised on the basis of structure and function of an organism and accordingly, they can be referred as primitive or advanced organisms. This helps in predicting the line of evolution. 22. (a) When an object is wholly or partially immersed in a liquid, it experiences a buoyant force (or upthrust) which is equal to the weight of the liquid displaced by the object. (b) A ship is not a solid block of iron and steel. In fact, it is a hollow object containing a lot of air in it. Air has a very low density. Due to the presence of a lot of air, the average density of a ship becomes less than that of water. Due to this, it floats in water. (c) Two examples of the application of Archimedes’ principle: (i) The hydrometers used for determining the density of liquids are based on Archimedes’ principle. (ii) The lactometers used for determining the purity of milk are based on Archimedes’ principle.
Follow Us on
www.meritnation.com
23. (a) Isotopes are defined as the atoms of the same element, having the same atomic number, but different mass numbers. The three isotopes of hydrogen are − (i) Protium (ii) Deuterium (iii) Tritium (b)The average atomic mass of X can be calculated as:
Thus, the average atomic mass of X is 35.5 u. (c)The two atoms are isotones that have the same number of neutrons. Number of neutrons present in
= 14 − 6 = 8
Number of neutrons present in
= 16 − 8 = 8
24. (a) i) Since at height h, the body is at rest, kinetic energy, EK = 0 Potential energy, E P = mgh (Where g = acceleration due to gravity) ∴ Total energy, E = E P = mgh ii) At height x from the ground, E P = mg x
From third equation of motion: 2 2 v = u + 2 g (h − x ) [where u = initial velocity = 0] Or, v 2 = 2 g (h − x ) ∴ Total energy = mgx + = mgx + = mgh
Follow Us on
www.meritnation.com
24.
Follow Us on
www.meritnation.com
Section B
25. Needle-like leaves and presence of heterospores (microspores and megaspores) are the characteristic features of Pinus. Hence, the correct option is (d). 26. I and II corresponds to moss (bryophyte) and fern (pteridophyte). Both the plants produce spores for reproduction. In moss, spores are enclosed in capsule whereas in fern, spores are present on the underside of the leaves. Hence, the correct option is (d). 27. Male and female cockroaches can be distinguished by the presence of the anal style. Male cockroaches possess anal style in its 9th segment, which is found to be absent in female cockroaches. Hence, the correct option is (d). 28. Presence of cycloid scales is a feature of bony fish, to which Labeo belongs. Hence, the correct option is (d). 29. Rose is a dicot plant, so its leaves show reticulate venation. Hence, the correct option is (b). 30. The given image represents the larva stage of the life cycle of a mosquito. The labelled parts of the larva of a mosquito are as follows: P- Tracheal gills Q- Abdomen R- Respiratory siphon S- Thorax The respiratory siphon of the larva acts as the respiratory organ. Hence, the correct option is (c). 31. According to the law of reflection of sound, the angle of incidence of sound wave is equal to the angle of reflection. ∴
Hence, the correct option is (c). 32. According to the law of reflection, the angle of incidence should be equal to the angle of reflection. Therefore, the listener will get the best reflected sound if angle θ is equal to 30 . Hence, the correct option is (a). ₒ
33. Naked seed is a characteristic feature of gymnosperms. Hence, the correct option is (a). 34. Density of the body is given by
If the density of the substance is 25 kg/m3 , it means that the volume of 25 kg of wood is 1 m3 . Hence, the correct option is (c). Follow Us on
www.meritnation.com
35. If the body is fully immersed in a liquid, the apparent loss the weight of the body is equal to the weight of the liquid displaced. And weight of the liquid displaced by the body is directly proportional to the density of the liquid. As the density of brine solution is highest among the given liquids, the loss in the weight of the stone is maximum in brine solution. Hence, the correct option is (a). 36. If the body is completely immersed in a liquid, then the apparent loss in the weight of the body is equal to the weight of the liquid displaced by the body. Therefore, it experience an upward thrust due to this displaced liquid. Hence, the correct option is (a). 37. We know
38. Pressure decreases with an increase in surface area. Hence, the correct option is (d). 39. In the experiment conducted to determine the velocity of a pulse propagated through a long stretched spring, the jerk should be perpendicular to the length of the spring to produce the pulse. Hence, the correct option is (d). 40. We know
Follow Us on
www.meritnation.com
41. Iron rusts as it reacts with oxygen and water. So, the mass of a rusty iron nail increases due to the addition of oxygen and water. 6Fe + 3O2 + xH2 O → 2Fe2O3. x H2O Hence, the correct option is (c). 42. Reaction between copper and oxygen: 2Cu + O2 → 2CuO According to the law of conservation of mass, the mass of reactants is equal to the mass of products. ∴ Mass of copper + Mass of oxygen = Mass of copper (II) oxide (115 + 48) g = Mass of copper (I I) oxide Thus, 163 g of copper (I I) oxide was formed after the experiment. Hence, the correct option is (d).
Follow Us on
www.meritnation.com
SCIENCE (Set-2) QUESTIONS
Follow Us on
www.meritnation.com
SCIENCE (Set-2) Time allowed: 3 hours
Maximum marks: 90
General instructions: (i) The question paper comprises of two sections, A and B. You are to attempt both the sections. (ii) All quesstions are compulsory. (iii) All questions of Section A and all questions of Section B are to be attempted separately. (iv) Question numbers 1 to 3 in Section A are one-mark questions. These are to be answered in one word or one sentence. (v) Quetion numbers 4 to 7 in Section A are two-mark questions. These are to be answered in about 30 words each. (vi) Question numbers 8 to 19 in Section A are three-mark questions. These are to be answered in about 50 words each. (vii) Question numbers 20 to 24 in Section A are five-mark questions. These are to be answered in about 70 words each. (viii) Question numbers 25 to 42 in Section B are multiple choice questions based on practical skills. Each question is a one-mark question. You are to select one most appropriate response out of the four provided to you.
Follow Us on
www.meritnation.com
Section A 1. What is meant by the term ‘pollutant’? 2. Give an example each of bulk phenomenon and surface phenomenon. 3. Calculate the number of moles present in 220 g of Co2 . 4. The people living in large towns and cities use a large amount of water obtained from tube wells. What are the problems that they might face in future? 5. Describe the categories of class Pisces. 6. Steel sinks in water, but a ship made of iron and steel floats. Why? 7. If K and L shells of an atom are filled and there are 4 electrons in the M shell, then what is the atomic number and valency of this atom? 8. Write the salient features of the only phylum consisting of pseudocoelomates. 9. Explain the factors affecting the movement of air. 10. Which of the two objects has a larger capacity of doing work, a ball weighing 15 g thrown horizontally at 90 km/hr or a bullet weighing 0.5 g shot horizontally at 500 km/hr? 11. Calculate the molecular weight of the following. (i) Hydrated copper sulphate (ii) Potassium dichromate (iii) Ammonium nitrate [O = 16 amu; H = 1 amu; N = 14 amu; Cu = 64 amu; S = 32 amu; Cr = 52 amu, K=39] 12. What are the differences between pteridophytes and gymnosperms? 13. Answer the following questions. (i) While drawing water from a well, why is it easier to lift the bucket of water when it is still inside water than when it is in the air? (ii) An object of weight 98 N is immersed in water. How much upthrust is required for the object to float? 14. Draw a labelled diagram of the human ear and explain its working. 15.(i) How much energy is possessed by an object of mass 10 kg when it is at a height of 2 m above the ground? (ii) A certain household device consumes 280 units of energy during a month. How much is its monthly consumption of energy in joules? Follow Us on
www.meritnation.com
16. ‘Prevention is better than cure’. Justify this statement. 17. (a) How are aves different from amphibians? Write at least four points of difference. (b) Can mammals lay eggs? Justify with help of examples. 18. (i)What is the molecular formula of a compound containing potassium and bromine? (ii) Explain why the formula of the phosphate of sodium is Na3 PO4 , while that of the phosphate of calcium is Ca3(PO4)2. 19.(i)What are the drawbacks of Rutherford’s atomic model? Write any 4 points. (ii)Why do isotopes of elements have similar chemical properties but different physical properties? 20. Answer the following questions. (i) Name the factors responsible for the formation of soil. (ii) Explain the importance of ozone. (iii) Explain the importance of nitrogen-fixing bacteria. Where are they found? 21. (a) Explain how animals can be classified based on the type of body cavity. (b) Which animal is used in the process of vermicomposting? Write some of the salient features of this animal. 22. (i) An object of mass 40 g has a volume of 20 cm3 . Calculate the density of the object. If the density of water is 1 g/cm3, state whether the object will float or sink in water. Explain why. (ii) An object floats in water with half of its volume immersed in water. What will be its relative density? 23. (i) How could Bohr’s model explain the stability of an atom? (ii) Which of the following is the arrangement of electrons for Mg2+ ? Why?
(iii) Draw the electronic structure of calcium atom. Will it form cation or anion? 24. (i) Name the conditions to define work scientifically. (ii) What is the work done in pushing a box with a force of 4 N through a distance of 12 m? (iii) A weightlifter lifts a weight with a mass 120 kg to a height of 194 cm in 4 s. Calculate the power applied by the weightlifter.
Follow Us on
www.meritnation.com
Section B 25. Which of the following features does not belongs to Spirogyra? (a) Double layered cell wall (b) Ribbon-like chloroplast (c) Presence of rhizoids (d) Filamentous body 26. Pileus, gills and annulus are the characteristic features of (a) Funaria (b) Agaricus (c) Spirogyra (d) Dryopteris 27. Compound eyes and jointed appendages are the characteristic features of (a) Labeo rohita (b) Ascaris lumbricoides (c) Pheretima posthuma (d) Periplaneta americana 28. Which of the following chemicals is used for preserving specimens? (a) Dilute HCl (b) Formalin (c) Acetone (d) Ether 29. While observing the external features of a particular plant, a student noticed parallel venation in its leaves. Which of the following plants the student might have observed? (a) Onion (b) Mango (c) Tomato (d) Sweet Pea 30. Which stage of the mosquito life cycle is also known as the wriggler? (a) Egg (b) Pupa (c) Adult (d) Larva
Follow Us on
www.meritnation.com
31. A student is asked to perform an experiment to verify the law of reflection of so und. For this, he measured the angle of reflection for different values of angle of incidence and plotted them graphically. Which of following graphs represents the correct relationship between the angle of incidence and the angle of reflection?
32. A sound wave travels in a medium in time T. If at any given instant (t = 0), a given region has minimum density, then when will the density of this region be minimum again? (a) t = T (b) t = T/2 (c) t = T/3 (d) t = T/4 33. What is the effect on the density of a solid body if its volume is reduced to half without affecting its mass? (a) Becomes half (b) Remains same (c) Becomes twice (d) Becomes thrice 34. For determining the density of a brass block using a spring balance and a measuring cylinder, the following steps are suggested. However, they are not in the correct sequence. (I) Immerse the brass block in water. (I I) Note the level of water in the measuring cylinder. (I II) Note the water level with the brass block completely immersed in the measuring cylinder. (I V) Note the weight of the brass block using a spring balance. What would be the correct sequence? (a) IV, II, II I and I (b) IV, II, I and III (c) IV, II, II I and I (d) I, IV, II and III 35. Mohan took a measuring cylinder and filled it with 45 mL of water. He then immersed the stone completely in it and noted the level of water in the measuring cylinder, which was 60 mL. He found that the stone appears lighter when placed in water. The loss in weight of the stone is (a) 5 g (b) 15 g (c) 5000 g (d) 15000 g
Follow Us on
www.meritnation.com
36. A air balloon has a volume of 200 cm3 . What would be the minimum force applied by a child to immerse it completely in the water? (a) 50 gf (b) 100 gf (c) 200 gf (d) 400 gf 37. Rohit used a cuboid of dimensions 7 cm × 9 cm × 11 cm and calculated the pressure exerted by all the faces on sand. Which of the following observations made by him is correct? (a) Pressure is minimum for face 7 cm × 9 cm. (b) Pressure is minimum for face 7 cm × 11 cm. (c) Pressure is maximum for face 9 cm × 11 cm. (d) Pressure is maximum for face 7 cm × 9 cm. 38. When we push a stone using our thumb on a wood, it does not go into the wood. But when we push a drawing pin into the wood using our thumb with the same force, it goes into the wood. What is correct reason for this? (a) Tip of pin has large surface area so pressure applied is larger (b) Tip of pin has small surface area so pressure applied is large. (c) Pin has large density so pressure applied is larger. (d) Pin has smaller density so pressure applied is larger. 39. In an experiment to find out the speed of a pulse propagated through a stretched spring, the spring used should be (a) long, hard and rigid (b) short, hard and rigid (c) long, soft and flexible (d) short, soft and flexible 40. A strong transverse horizontal pulse is created at one end of a string. It is observed that the wave completes its journey along its length before fading out. The initial and final readings on a stopwatch are shown below.
If the length of the string is L m, then the speed of the pulse through the string is
(a)
Follow Us on
(b)
(c)
(d)
www.meritnation.com
41. In an experiment, 7.2 g of magnesium was allowed to react with 16 g of oxygen gas to form magnesium oxide. After the reaction was complete, the residue was weighed. What was the weight of the residue? (a) 46.4 g (b) 39.2 g (c) 30.4 g (d) 23.2 g 42. A student dropped zinc discs in copper sulphate solution to carry out a reaction. On weighing the discs and solution separately after some time, he found that the mass of zinc discs had increased, but the mass of the solution had decreased. Which of the following inferences is correct regarding the given experiment? (a) Reaction between zinc discs and copper sulphate solution was incomplete. (b) Atomic mass of zinc is greater than that of copper. (c) There was an error in weighing the contents. (d) Law of conservation of mass was obeyed.
Follow Us on
www.meritnation.com
SCIENCE (Set-2) SOLUTIONS
Follow Us on
www.meritnation.com
SC IE NCE (Set-2) SOLU TI ON S Answers:
Section A
(1-mark)
1. Waste substances which are harmful to nature are known as pollutants. 2. Bulk phenomenon − Boiling Surface phenomenon − Evaporation 3. Molecular weight of CO2 =12 + 32 = 44 g ∴ 44 g of CO 2 = 1 mole of Co2 1 g of CO 2 = 220 g of CO 2 =
mole of Co2 moles of Co2
= 5 moles of Co2 (2-mark)
4. Tube wells take out the water from the ground. If a large amount of ground water is used for a long time, then ground water level or the water table will drop. This will eventually lead to death of plants and trees. Also, the arranging for the water for day-to-day activities will also become very difficult. 5. The class Pisces consists of exclusively marine-dwelling animals such as fishes.
(i) Cartilaginous fishes: The skeleton of these fishes is made up entirely of cartilage. Scoliodon, torpedo, and sharks are examples of cartilaginous fishes. (ii) Bony fishes: The skeleton of these fishes is partly or wholly composed of bones. Rohu, sea horse, and Anabas are examples of bony fishes.
6. A ship is not a solid block of iron and steel. In fact, it is a hollow object containing a lot of air in it. Air has a very low density. Due to the presence of a lot of air, the average density of a ship becomes less than that of water that causes it to float in water. 7. The electronic configuration of the given element is:
Hence, the atomic number of the element is 14 and the valency of the element is 4. Follow Us on
www.meritnation.com
(3-mark)
8. Phylum Nematoda consist of pseudocoelomate organisms. General features of phylum Nematoda: • They are bilaterally symmetrical and triploblastic. • They have a cylindrical body with tapering ends. • They have an organ system level of organisation. • They are parasitic animals. 9. The movement of air results in diverse atmospheric phenomena. It is affected by the uneven heating of the atmosphere in different regions of the earth, which in turn is due to the heating of air and the formation of water vapour. The rotation of the earth and the presence of mountains in the paths of wind also affect the movement of air. 10. Energy = Capacity of doing work.
11. (i) 250 amu (ii) 294 amu (iii) 80 amu 12. Pteridophytes have roots, stems, leaves, and specialized vascular tissues for the conduction of water and food. Flowers and seeds are absent and they reproduce through spores. Pteridophytes are known as cryptogams as they have hidden or inconspicuous reproductive organs. Gymnosperms have well-differentiated roots, stems, and leaves. They have a xylem and a phloem for conduction of water and food. The gymnosperms are called phanerogams as they bear seeds and have well-differentiated sex organs.
Follow Us on
www.meritnation.com
13. (i) Water in the well provides upthrust to the bucket of water when it is still inside the water. This makes the apparent weight of the bucket lesser than its weight in the air such that the net force required to lift it is reduced. This makes it easier to lift the bucket when it is still inside the water than when it is in the air. (ii) For an object to float, the upthrust it experiences must balance out the weight of the object. Therefore, when weight of an object is equal to the upthrust it experiences, the object floats on the liquid. Hence, upthrust required in this case is 98 N. 14.
Working of the human ear:
When a sound wave falls on the eardrum, it starts vibrating back and forth rapidly. The vibrations pass on from the eardrum to the hammer, the anvil and finally to the stirrup. The vibrating stirrup strikes the membrane of the oval window and passes its vibration on to the liquid in the cochlea. This produces electrical impulses in the nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are then interpreted by the brain as sound, and we get the sensation of hearing. 15. (i) It is given that:
Follow Us on
www.meritnation.com
16. Even though most infectious diseases can be cured with the help of medicines, it is still said that prevention is better than cure. Even after cure, diseases leave a long lasting effect on the overall health of a person. The long lasting drawbacks related with the diseases are enumerated as follows: • During any disease, functions of the body are damaged and take time to recover completely. • Due to a prolonged treatment, the patient remains bedridden for a long time. It affects the mental status of a person. • A person suffering from an infectious disease can spread it to others. Hence, it is said that prevention is better than cure. 17. (a)
(b) Mammals can lay eggs. Egg-laying mammals are called oviparous mammals. They include Platypus, Echidna, etc. 18. (i)Potassium as well as bromine form monovalent ions i.e., K+ and Br- respectively. Hence, they combine in the ratio 1:1 to form the neutral compound KBr. (ii)This can be answered on the basis of the combining capacities of atoms.
The formula of this compound is Na3 Po4 .
The formula of this compound is Ca3 (PO 4 )2 .
Follow Us on
www.meritnation.com
19. (i) Drawbacks of Rutherford’s atomic model: (a) It cannot explain the stability of an atom on the basis of classical mechanics and electromagnetic theory. (b) If the electrons were stationary, then the strong electrostatic force of attraction between the dense nucleus and the electrons would pull the electrons towards the nucleus. Thus, it cannot explain the stability of an atom. (c) Rutherford’s model does not give any idea about the distribution of electrons around the nucleus (i.e. the electronic structure of the atom) and about their energy. (d) It cannot explain the atomic spectra. (ii) Isotopes of elements have same atomic numbers but different atomic masses. The chemical properties of elements depend on the number of electrons present in an atom and not neutrons. Hence, isotopes have similar chemical properties as the number of electrons present in them are equal. (5-mark)
20. (i) The factors that influence the formation of soil are the sun, water, wind and living organisms such as lichens and mosses. (ii) The ozone layer of the atmosphere absorbs the harmful ultraviolet radiations of the sun, and prevents them from reaching the earth’s surface. (iii) Plants cannot utilise atmospheric nitrogen as such. Nitrogen-fixing bacteria convert atmospheric nitrogen molecules into nitrates and nitrites. These are then utilised by plants. Nitrogen-fixing bacteria can be found in the roots nodules of leguminous plants. These nitrogenfixing bacteria may also be free living. 21. (a) Coelom is the body cavity that is used by mesoderm. Based on the type of coelom development, animals can be differentiated into the following: (i) Acoelomates: The animals which do not possess a body cavity are known as acoelomates. (ii) Coelomates: Animals possessing a body cavity are known as coelomates. (iii) Pseudocoelomates: The pseudocoelom is a fully functional body cavity, which is not lined by mesoderm. It is scattered between the ectoderm and the endoderm. The pseudocoelom is also known as the false coelom since it is not well organised as the true coelom. (b) Earthworm is used in the process of vermicomposting. They belong to phylum Annelida. Some of the salient features of annelids are as follows: • They are bilaterally symmetrical and their body is segmented. • They have three germ layers. Thus, they are triploblastic animals. • They possess true organs inside their body structures. • True body cavity or coelom is present. 22. (i) We know that
Follow Us on
www.meritnation.com
22. An object will float in water if its density is less than that of water. Otherwise, it will sink. The given object will sink in water as its density (2 g/cm3 ) is more than that of water (1 g/cm3 ). (ii) When a body floats in water, the upthrust it experiences is equal to the weight of the body. Hence, upthrust (U ) = Weight of the body U = mg
(Where, m = mass of the body and g = acceleration due to gravity) Mass (m) = Density (d ) × Volume (V ) Therefore, U = d × V × g ...(1) Upthrust is also equal to the weight of water displaced. Volume of water displaced is ,
V being the volume of the object.
23.(i)According to Bohr’s model, an electron revolving in a particular orbit does not lose energy. Therefore, when the electron is present in its energy level, it will keep on revolving in the same orbit without losing energy. Hence, the atom becomes stable. (ii) Figure B shows the correct arrangement. In the process of formation of Mg2+ , Mg loses the two electrons that are present in its valence shell. (iii) The electronic structure of calcium atom is as follows:
It will form a cation by losing the two electrons present in its N-shell.
Follow Us on
www.meritnation.com
24. (i) Scientifically work is said to be done when the following two conditions are satisfied: (a) An object must get displaced by the action of a force. (b) Displacement of the object should not be normal to the direction of applied force. (ii)Applied force, F = 4 N Displacement, s = 12 m ∴Work done, w = Fs = 4 × 12 = 48 J (iii)Mass, m = 120 kg Height, h = 194 cm = 1.94 m Acceleration due to gravity, g = 9.8 m/s2 Time, t = 4 s
Hence, the power applied by the weightlifter is 570.36 W.
Section B 25. Rhizoids are root-like structures found in Funaria (Moss). Hence, the correct option is (c). 26. Pileus, gills and annulus are found in Agaricus (mushroom). Pileus is an umbrella-like structure found on fleshy stalk, gills are plate-like structure found on the lower side of pileus and annulus is a ring-like structure found on stalk. Hence, the correct option is (b). 27. Compound eyes and jointed appendages are the characteristic features of arthropods, such as Periplaneta americana (cockroach). Hence, the correct option is (d). 28. Formalin is an aqueous solution of formaldehyde, which is used for preserving specimens. Hence, the correct option is (b). 29. Parallel venation is generally found in monocots like onion. Hence, the correct option is (a). 30. Mosquito larva is also known as wriggler. Hence, the correct option is (d).
Follow Us on
www.meritnation.com
31. According to the law of reflection of sound, Angle of incidence = Angle of reflection So, graph between the angle of incidence and the angle of reflection of sound wave must be a straight line passing through the origin. Hence, the correct option is (c). 32. The time period of the wave is T. Its region of minimum density will again be minimum after half of the time period T. Hence, the correct option is (b). 33. Density of a body is given as Density = If the volume is reduced to half, without affecting its volume, then the density of the solid will become twice the initial density. Hence, the correct is option (c). 34. The correct sequence of doing this experiment is given below. (I) Note the weight of the brass block using a spring balance. (I I) Note the level of water in the measuring cylinder. (I II) Immerse the block in water. (I V) Note the water level with the brass block completely immersed in the measuring cylinder. Hence, the correct option is (b). 35. In a liquid, the weight loss by a body is equal to the weight of the liquid displaced by the body.
36. To make the body sink in water, the child will have to overcome the force of buoyancy. This is because the buoyant force experienced by the body is equal to the weight of the liquid displaced by the balloon. Volume of the water displace by the balloon = 200 cm3
Follow Us on
www.meritnation.com
37. Since pressure is inversely proportional to the area of contact of the body with the surface, pressure will be more for the face having less surface area. Hence, the correct option is (d). 38. When we push a stone over wood using our thumb, it does not goes into the wood because the area of the wood is large and thus pressure is small. Whereas, for a pin area of its tip is small therefore pressure exerted by the pin on the wood is larger and pin easily go inside the wood. Hence, the correct option is (b). 39. In the experiment conducted to find out the speed of a pulse propagated through a stretched string, the string should be long, soft and flexible. Hence, the correct option is (c). 40. Length of the string = L Distance covered by the pulse in two journeys = 2L Time taken by the pulse to complete two journeys = 50 s
Velocity of the pulse in the string =
m/s
Hence, the correct option is (c).
41.According to the law of conservation of mass, mass of reactants is equal to the mass of products. ∴ Mass of Mg + Mass of O2 = Mass of MgO Mass of MgO = (7.2 + 16) g = 23.2 g Hence, the correct option is (d). 42.Mass of zinc discs increased due to the deposition of copper on them. As a result, the mass of the solution decreased. The given experiment shows the validity of the law of conservation of mass. Hence, the correct option is (d).
Follow Us on
www.meritnation.com
MATHEMATICS (Set-1) QUESTIONS
Follow Us on
www.meritnation.com
MATHEMATICS (Set-1) Time allowed: 3 hours
Maximum marks: 90
General instructions: (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. (iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 10 questions of 4 marks each. (iv) Use of calculators is not permitted.
Follow Us on
www.meritnation.com
Section A Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices have been provided, of which only one is correct. Select the correct choice.
1. Which of the following following equations is parallel to y-axis? y-axis? A. x = y − − 2 B. y = = 16 C. y = x + + 5 D. x = = 7 2.
Which of the following equations satisfies the values of x and y in the given table? A. y = x − − 2 B. x = y − − 3 C. x = y + + 3 D. y = x + + 2
3. In the the given given figure, A B C D is a rhombu rhombuss and and PQ C B is a rectan rectangle. gle.
If AC = 12 cm and B D = 16 cm, then what is the measur measuree of PB? A. 8 cm B. 9.6 cm C. 10 cm D. 12.8 cm 5. The ratio of the radii of the bases bases and the heights of two cylinders are 2 : 3 and 6 : 5 respectively. What is the ratio of their curved surface areas? A. 8 : 15 B. 2 : 3 C. 4 : 5 D. 16 : 25
Follow Us on
www.meritnation.com
6. What is the volume volume of a cylinder whose height and curved surface area are 5 cm and 60π cm2 respectively?
A. 125π cm3 B. 150π cm3 C. 180π cm3 D. 216π cm3 7. If the mean of the data data set {y − 1, y − 2, y, y + 5, 2y − 4} is 14, then what is the value of y? A. 6 B. 9 C. 12 D. 15 8. A coin is tossed 100 times and head appears 58 times. The probability of getting getting a head is A.
B.
C.
D.
Section B Question numbers 9 to 14 carry 2 marks each.
9. Find ∠QRS.
Follow Us on
www.meritnation.com
10. Prove that the diagonals of a rectangle divide it into two congruent triangles. 11. A cylinder and a cone have the same radius and the height of the cone is thrice the height of the th e cylinder. What is the ratio of their volumes? 12. A coin is tossed a certain number of times. In this experiment, the ratio of number of heads obtained to the number of tails obtained is 3 : 7. 7. What is the probability of getting a head in the next throw? 13. The following data is in ascending order. 2, 7, 6 x , 7 x , 40, 46 The median of this data is 19.5. Find the value of x .
14. The following table represents the heights of 90 students in a class. Using this table, answer the following questions:
(a) Find the probability that the height of a randomly chosen student is more more than or equal to 130 cm, but less than 160 cm. (b) Find the probability that the height of a randomly chosen student is more than 180 cm.
Follow Us on
www.meritnation.com
Section C Question numbers 15 to 24 carry 3 marks each.
15. If the point (−1, 3) is the point of intersection of the lines, 2 x + 3 y − k = 0 and kx + py + 1 = 0, then find the value of p. 16. Two lines are drawn parallel to x -axis; one is above the x -axis at a distance of p units, and the other is below the x -axis at a distance of q units. In the same way, two more lines are drawn parallel to y - axis such that both are to the left of y -axis at a distance of p and q units respectively (p > q). What is the area of the figure enclosed by these lines? 17. In the given figure, AB CD and AB EF are parallelograms.
Prove that (i) area (AB CD) = area (ABE F)
18. The given figure shows a circle with centre O. A, B, C, and D are points on the circle such that AB = 16 cm. If CD = 30 cm, then find the distance of the chord CD from the centre.
Follow Us on
www.meritnation.com
19.
In the given figure, ABCD is a rhombus and CM is perpendicular to BD. The lengths of BD and CM are 22 cm and 10 cm respectively. What is the length of diagonal AC? 20. Draw a line segment PQ of length 6.9 cm and construct an angle of 15° at point P. 21. A rectangular sheet of dimensions 10 cm × 12 cm is folded to form a cylinder of height 10 cm. Find the radius of the cylinder.
22. A road roller of diameter 120 cm and length 175 cm covers an area of 1.155 hectare on the ground. Find the number of complete rotations taken by the roller to cover the area. (1 hectare = 10,000 m2 ) 23. The marks obtained by 8 students in a class are as follows: 74, 46, 57, 33, 10, 29, 91, 92 If marks of a student are chosen at random, then find the probability that they are more than the mean marks. 24. The mean of the data {155, 160, (6x − 5), 149, 150, 147, 2 (3x + 1), 148, 144} is 150. Find the median.
Section D Question numbers 25 to 34 carry 4 marks each.
25. What is the area of the figure enclosed by the line 3 x + 4 y = 24, x -axis and y -axis? 26. If −6 ≤ x ≤ 4 and −2 ≤ y ≤ 6, then what are the integral solutions of the equation. x − 2 y + 10 = 0?
Follow Us on
www.meritnation.com
27. In the figure, WXYZ is a parallelogram. Find the values of x, y , and z.
28. In the given figure, ∠ACD = ∠CDP, BC = 12 cm, and PC = 28 cm.
If the area of the quadrilateral ABCD is 420 cm2 , then find the measure of AQ. 29.
The given figure shows a circle where AB and CD are its two equal chords. KL and MN are the perpendicular bisectors of AB and CD respectively. Show that KL = MN and BN = DL.
Follow Us on
www.meritnation.com
30. The points M, N, and P divide the side AB of Δ ABC into four equal parts. The lines MX, NY, and PZ are drawn parallel to side BC .
31. Construct an isosceles right-angled triangle ABC whose perimeter is 10 cm. 32. A cuboid of dimension 6 cm × 4 cm × 4 cm is cut into cubes of side 2 cm. Find the percentage increase in the surface area of the cuboid. 33. The internal measure of the floor of a cuboidal room is 15 m ×12 m. If the cost of whitewashing the walls and roof at the rate of Rs 6.50 per m2 is Rs 2574, then find the height of the room. 34. There were 50 questions in an exam. For each correct answer, 4 marks were awarded. For each wrong answer, 2 marks were reduced and for each non-attempted answer, 1 mark was deducted. If a student scores 120 marks by giving 35 correct answers, then find the probability that (a) he answers a question incorrectly (b) he leaves a question unanswered
Follow Us on
www.meritnation.com
MATHEMATICS (Set-1) SOLUTIONS
Follow Us on
www.meritnation.com
MATHEMATICS (Set-1) SOLUTIONS Answers: (1-mark)
Section A
1. An equation of the form x = a, where a is a constant, is parallel to y -axis. ∴ x = 7 is parallel to y -axis. The correct answer is D. 2. If x = 3, y = 3 + 2 = 5 If x = 5, y = 5 + 2 = 7 If x = 7, y = 7 + 2 = 9 If x = 9, y = 9 + 2 = 11 If x = 11, y = 11 + 2 = 13 If x = 13, y = 13 + 2 = 15 Thus, the rule followed in the given table is y = x + 2 The correct answer is D. 3. It is given that D and F are the mid-points of sides AB and C A respectively. We know that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side. ∴DF || BC Similarly, we can show that DE||AC and EF||A B. Therefore, ADEF, BDFE, and DFC E are parallelograms. Now, DF is a diagonal of parallelogram ADEF. ∴Δ D EF ≅ Δ FAD [Diagonal of a parallelogram divides the parallelogram into two congruent triangles] Similarly, Δ D EF ≅ Δ E D B and Δ D EF ≅ Δ C FE Hence, all four triangles i.e., Δ D EF, Δ FAD, Δ ED B, and Δ CFE are congruent.
The correct answer is D. 4. In a rhombus, diagonals bisect each other at right angle. ∴Δ OBC is right-angled at point O, where
On applying Pythagoras theorem to Δ OBC, we obtain B C2 = O B2 + Oc2 2 2 2 ⇒ B C = (8 cm) + (6 cm) 2 2 2 ⇒ BC = 64 cm + 36 cm 2 2 ⇒ BC = 100 cm ⇒ B C = 10 cm
Follow Us on
www.meritnation.com
ABCD and PQ CB are parallelograms lying on the same base BC and between the same parallels PD and BC. ∴ Area (PQC B) = Area (ABC D)
5. Let r1and r2 be the radii and h1 and h2 be the heights of the two cylinders.
Thus, the ratio of the surface areas of the two cylinders is 4 : 5. The correct answer is C.
Follow Us on
www.meritnation.com
6. Let the radius of the base of the cylinder be r . Height of the cylinder, h = 5 cm It is given that the curved surface area of the cylinder is 60π cm2. 2 ∴ 2πrh = 60 π cm 2 ⇒ 2 × r × 5 cm = 60 cm ⇒ r = 6 cm The volume of a cylinder is given by πr2h. 3 3 ∴ Volume of cylinder = (π × 62 × 5) cm = 180π cm The correct answer is C. 7.
8. Total number of trails = 100 Number of times head appeared = 58
The correct answer is B.
Follow Us on
www.meritnation.com
Section B (2 mark)
9. We know that angles in the same segment are equal. ∴ ∠ SQR = ∠ SPR ⇒ ∠ SQ R = 70° In Δ Q RS, ∠QSR + ∠SQR + ∠QR S = 180° ⇒ 30° + 70° + ∠QR S = 180° ∴ ∠QR S = 80° Thus, the measure of ∠QRS is 80°. 10. Let PQRS be a rectangle and PR be a diagonal. Now, the parallel lines, PQ and SR, are cut by a transversal PR.
∴ ∠ QPR
= ∠PR S and ∠RPS = ∠QRP (Alternate interior angles) Now, in Δ P Q R and Δ P SR, ∠QPR = ∠PRS ∠QRP = ∠RPS PR = PR (Common) ∴ Δ P QR ≅ Δ RSP (By SAS congruency criterion) Similarly, we can prove that the diagonal QS divides the rectangle PQRS into two congruent triangles. Thus, a diagonal of a rectangle divides it into two congruent triangles.
11. Let the radius of the cylinder be r . ⇒ Radius of cone = r Let the height of the cylinder be h. ∴ Height of cone = 3h
Follow Us on
www.meritnation.com
12. Since the ratio of the number of heads obtained to the number of tails obtained is 3 : 7, the number of heads and the number of tails are 3 x and 7 x , respectively. Therefore, total number of times the coin was tossed = 3 x + 7 x = 10x Thus, probability of getting a head in the next throw
13. The given data is 2, 7, 6 x , 7 x , 40, 46. Here, n = 6 (even) We know that, when n is even,
14. (a) Number of students with height more than or equal to 130 cm, but less than 160 cm = 10 + 20 + 36 = 66 ∴ Probability that height of a randomly chosen student is more than or equal to 130 cm, but less than 160 cm
Follow Us on
www.meritnation.com
14. (b) There is no student with height more than 180 cm.
(3 marks)
Section C
15. It is given that the point (−1, 3) is the point of intersection of the lines, 2 x + 3 y − k = 0 and kx + py + 1 = 0 Therefore, the point (−1, 3) lies on the lines, 2 x + 3 y − k = 0 and kx + py + 1 = 0 Since the point (−1, 3) lies on the line, 2 x + 3 y − k = 0, x = −1 and y = 3 satisfies the equation, 2 x + 3 y − k = 0. ∴ 2 (−1) + 3 × 3 − k = 0 ⇒ 7 − k = 0 ⇒ k = 7 As the point (−1, 3) also lies on the line, kx + py + 1 = 0, x = −1 and y = 3 satisfies the line, kx + py + 1 = 0 ∴ k (−1) + p(3) + 1 = 0 ⇒ 7 (−1) + p × 3 + 1 = 0 [∵k = 7] ⇒ 3p − 6 = 0 ⇒ 3p = 6 ⇒ p = 2 Thus, the value of p is 2. 16. It is given that the line parallel to x -axis is p units above the axis. Therefore, the equation of the line is, y = p. Similarly, the equation of the line at a distance of q units below the x -axis is, y = −q. The equations of the lines to the left of y-axis are x = −p and x = −q. It is given that p > q; therefore, the line x = −p will lie farther from the y-axis than the line x = −q. Thus, the four given lines can be plotted as:
It can be seen that the figure enclosed by these lines is a rectangle ABCD. AB = CD = p − q BC = AD = p + q ∴ Area of rectangle AB CD = AB × B C = (p − q) (p + q) sq. units = (p2 − q2) sq. units
Follow Us on
www.meritnation.com
17. (i) We know that parallelograms on the same base and between the same parallels are equal in area. Parallelograms ABCD and ABEF are on the same base AB and between the same parallels. ∴ area (AB CD) = area (AB EF) … (1) (ii) We know that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of triangle is half the area of parallelogram. Δ A FG and parallelogram AB EF are on the same base AF and between the same parallels AF and BE.
18.
Let M be the foot of the perpendicular drawn from centre O to chord CD. Join OB and OD It is known that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Follow Us on
www.meritnation.com
19. Area of Δ BC D = (½) × B D × CM = (½) × 22 cm × 10 cm = 110 cm2 It is known that the diagonals of a parallelogram divide it into two congruent triangles. Also, it is known that congruent figures have equal area. ∴ ar (Δ B C D) = ar (Δ A BD) 2 ⇒ ar (Δ A BD) = 110 cm ar (AB CD) = ar (Δ A B D) + ar (Δ B C D) = 110 cm2 + 110 cm2 = 220 cm2 It is known that the area of a rhombus is half the product of its diagonals. ∴ ar (ABC D) = (½) × AC × B D 2 ⇒ 220 cm = (½) × AC × BD 2 ⇒ 220 cm = (½) × AC × 22 cm ⇒ AC = 20 cm Thus, the length of diagonal AC is 20 cm. 20. The below given steps are followed to construct an angle of 15°. Step I: Draw a line segment PQ of length 6.9 cm. Step II: Taking P as centre and with some radius, draw an arc of a circle, which intersects PQ at R. Step II I: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S. Step IV: Taking R and S as centres and with radius more than (½)R S, draw arcs to intersect each other at T. Join PT. Step V: Let PT intersect the arc RS at U. Taking U and R as centres and with radius more than (½)RU, draw arcs to intersect each other at V. Join PV, ∠VPQ = 15° is the required angle.
21.
Height of cylinder = 10 cm Circumference of the base of cylinder = 12 cm Let r be the radius of the circular base of cylinder. ∴ 2πr = 12 cm
Follow Us on
www.meritnation.com
22. Diameter of the road roller = 120 cm ∴ Radius of
the road roller,
Length of the road roller, h = 175 cm = 1.75 m Area of the ground covered by the roller in 1 complete rotation = Lateral surface area of the cylinder = 2πrh
= 6.6 m2 Total area covered by the road roller on the ground = 1.155 hectare = 1.155 ×10000 m2 = 11550 m2 Hence, number of complete rotations taken by the roller on the ground
Follow Us on
www.meritnation.com
23.
24. We know that,
Thus, the data is as follows. 155, 160, 145, 149, 150, 147, 152, 148, 144 Arranging the data in ascending order, 144, 145, 147, 148, 149, 150, 152, 155, 160 Here, n = 9
Follow Us on
www.meritnation.com
Section D (4marks)
25. The given equation is 3x + 4y = 24.
It can be seen that the figure enclosed by the line 3 x + 4 y = 24, x -axis and y -axis is a triangle
Thus, the required area is 24 sq. units.
Follow Us on
www.meritnation.com
26. The given equation is x − 2y + 10 = 0.
Now, it is given that −2 ≤ y ≤ 6. Therefore (4, 7) is not a solution. Thus, the integral solutions of the given equation for the given range of x and y are: (−6, 2), (−4, 3), (−2, 4), (0, 5) and (2, 6)
Follow Us on
www.meritnation.com
27. We know that in a parallelogram, the opposite sides are of equal lengths. ∴ WX = YZ ⇒ 5 z − 8 = 3 z − 2 ⇒ 5 z − 3 z = −2 + 8 ⇒ 2 z = 6 ⇒ z = 3 Now, ∠XYZ + ∠ XY P = 180°(Linear pair angles) ⇒ ∠ XYZ + 84° = 180° ⇒ ∠ XYZ = 96° Now, applying angle sum property in Δ XYZ, we obtain ∠ XYZ + ∠YZX + ∠ ZXY = 180° ⇒ 96° + 32° + ∠ ZXY = 180° ⇒ ∠ ZXY = 180° − 128° = 52° Now, the parallel lines WX and YZ are cut by transversal XZ. ∴ ∠ ZXY = ∠XZW (Alternate interior angles) ∴ x = 52° Now, ∠ZWX = ∠XY Z ( opposite angles of a parallelogram are equal) ∴ y = 96° ( ∠XYZ = 96°) 28. To solve the given problem, AP is joined first. Let AP intersect DC at O.
It is given that ∠ACD = ∠CDP ∴AC||DP [∠ ACP and ∠ CDP are pair of alternate interior angles] It is observed that Δ ACD and Δ ACP are lying on same base AC and between the same parallels AC and DP. ∴ar (Δ ACD) = ar (Δ A CP) ⇒ ar (Δ A CD) + ar (Δ A B C) = ar (Δ A CP) + ar (Δ A BC) ⇒ ar (AB CD) = ar (Δ A BP) 2 2 ⇒ ar (Δ A B P) = 420 cm [ar (ABCD) = 420 cm (Given)]
Thus, the measure of AQ is 21 cm.
Follow Us on
www.meritnation.com
29. It is known that, the perpendicular bisector of a chord always passes through the centre. ∴ Both K L and MN pass through the centre of the circle. Thus, the point of intersection of KL and MN is the centre (say, O) of the circle.
It is also known that, equal chords are equidistant from the centre. ...(1) ∴ OK = OM Comparing Δ O K N and Δ O ML, [Each measure a right angle] ∠ OKN = ∠ OML OK = OM [From (1)] [Vertically opposite angles] ∠ KON = ∠ MOL ∴ Δ O KN ≅ Δ OM L [ASA congruency rule] The corresponding parts of congruent triangles are equal. ...(2) ∴ ON = OL KN = ML ...(3) Adding (1) and (2), we have OK + OL = OM + ON ⇒ KL = MN ...(4) It is given that, KL and MN are perpendicular bisectors of AB and CD respectively.
BK = MD ...(5) [AB = CD] Subtracting (3) from (5), we have BK − KN = MD − ML ⇒ BN = DL ... (6) Hence, from (4) and (6), KL = MN and BN = DL. ⇒
Follow Us on
www.meritnation.com
30. In Δ A NY, M is the mid-point of side AN and MX||N Y ∴ By converse of mid-point theorem, X is the mid-point of side AY. ∴ AX = XY … (1) In Δ A BC, N is the mid-point of AB and NY||B C ∴ By converse of mid-point theorem, Y is the mid-point of side AC. ∴ AY = YC ⇒ AX + XY = YZ + ZC ⇒ 2AX = YZ + ZC [From (1)]
Hence, proved. 31. The angles of an isosceles right-angled triangle are 90°, 45° and 45°.
In Δ AB C, let ∠A = 90°, ∠B = ∠C = 45°. It is given that the perimeter of Δ A BC is 10 cm. The steps of construction are as follows: (i) Draw a line segment XY of length of 10 cm. (ii) At points X and Y, draw angles ∠LXY and ∠MYX of measures 45° each. (iii) Draw the bisectors of ∠LXY and ∠MYX such that they intersect each other at A. (iv) Draw the perpendicular bisectors of AX and AY to intersect XY at points B and C respectively. (v) Join AB and AC. Δ A BC is the required triangle.
Follow Us on
www.meritnation.com
32.
It can be seen from the figure that 12 cubes of size 2 cm can be cut from the given cuboid. Total surface area of original cuboid = 2 (l × b + b × h + l × h) = 2 (6 × 4 + 4 × 4 + 6 × 4) cm2 = 2 (24 + 16 + 24) cm2 = 128 cm2 Surface area of cube = 6 × (2 cm)2 = 24 cm2 Surface area of 12 cubes = 12 × 24 cm2 = 288 cm2 ∴ Percentage increase in total surface area =
Thus, the percentage increase in the surface area of cuboid is 125%. 33. Internal length of the room, l = 15 m Internal breadth of the room, b = 12 m Let the internal height of the room be h m. Now, surface area of 4 walls and ceiling = 2h(l + b) + lb = [2h(15 + 12) + 15 × 12] m2 = (54 h + 180) m2 It is given that the cost of whitewashing the walls and roof at the rate of Rs 6.50 per m2 is Rs 2574. ∴ Surface area of 4 walls and ceiling
= 396 m2 ⇒ 54 h + 180 = 396 ⇒ 54 h = 396 − 180 = 216
Hence, the height of the room is 4 m.
Follow Us on
www.meritnation.com
34. (a) There were 50 questions. Since the student answers 35 questions correctly, ∴ Number of questions that he answered incorrectly or did not answer at all = 50 − 35 = 15 Let the number of questions that he answered incorrectly be x . ⇒ He did not answer (15 − x ) numbers of questions. Thus, he scored [35 × 4 − 2 × x − 1 × (15 − x )] marks = (125 − x ) marks According to the given information, 125 − x = 120 ⇒ x = 125 − 120 = 5 ∴ He answered 5 questions incorrectly.
Follow Us on
www.meritnation.com
MATHEMATICS (Set-2) QUESTIONS
Follow Us on
www.meritnation.com
MATHEMATICS (Set-2) Time allowed: 3 hours
Maximum marks: 90
General instructions: (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. (iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 10 questions of 4 marks each. (iv) Use of calculators is not permitted.
Follow Us on
www.meritnation.com
Section A Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices have been provided, of which only one is correct. Select the correct choice.
1. A. a unique solution B. exactly two solutions C. exactly four solutions D. infinitely many solutions 2.
3. In parallelogram PQRS, the vertices are taken in order. If ∠P:∠Q = 7:3, then what is the measure of ∠R? A. 126° B. 116° C. 64° D. 54° 4.
What are the respective values of x and y? A. 30° and 50° B. 25° and 70° C. 55° and 60° D. 25° and 55°
Follow Us on
www.meritnation.com
5. How many cubical boxes of edge 50 cm can be kept in a cuboidal box of length 6 m, breadth 4 m, and height 4.5 m? A. 320 B. 568 C. 692 D. 864 6. If a solid sphere is cut into two hemispheres, then by what percent is the surface area increased? A. 25% B. 50% C. 75% D. 0% 7. From a group of 2 boys and 3 girls, two children are selected. What are the sample space of the given random experiment?
8. Which of the following statements is correct? A. The mean, median, and mode of a data set are always equal. B. A data set can have more than one mode. C. A data set can have more than one median. D. A data set can have more than one mean.
Section B Question numbers 9 to 14 carry 2 marks each.
9.
In the given figure, X and Y are the mid-points of the sides AB and AC respectively. XY is extended to a point D such that XY = Y D and C D is joined. Show that BX||CD.
Follow Us on
www.meritnation.com
10.
What is the measure of ∠ACB? 11. A cylindrical can of length 12 cm and base radius 7 cm is to be covered with a plastic sheet. What is the area of the plastic sheet required? 12. The mean of n observations is 35. If 5 is subtracted from the thrice of each observation, then find the mean of new observations. 13. In the past 50 cricket matches between two teams A and B,team A had won 25 matches and lost 20 matches and the remaining matches were drawn. What is the experimental probability that the next cricket match between the two teams will result in a draw? 14. A frequency distribution table is given as:
Is the class intervals in the given frequency distribution table continuous? If not, then write them in continuous form.
Follow Us on
www.meritnation.com
Section C Question numbers 15 to 24 carry 3 marks each.
15.
16. Draw the graph of the equation 3( x + y − 9) + 2( x − 1) + ( y − 2) = 5( x + y − 5) on a Cartesian plane? 17. In the given figure, ABCD is a rectangle of dimension 20 cm and 26 cm and ACEF is a parallelogram. Find the area of the parallelogram ACEF.
18.
19.
In the given figure, ABCD is a rectangle. What are the values of x and y?
Follow Us on
www.meritnation.com
20. Construct an angle of 112.5° using a compass. 21. If the volume of a sphere is 4851 cm3 , then find its surface area.
22. A cuboidal tank of dimensions 15 m × 10 m × 8 m is filled with water up to 3 m. A pipe is connected to the tank in order to fill it. Water flows into the tank through the pipe at the rate of 50 L/s. How much time will the pipe take to fill the entire tank? 23. Find the missing observation in the following data (table), if its mean is 16.8.
24. Riya had some notes of denominations Rs 100, Rs 50 and Rs 10. The ratio of number of notes of these denominations is 2 : 3 : 5. She went to the market, where she paid Rs 180 in such a way that she had given the minimum number of notes. After paying Rs 180 she selected a note at random. What is the probability that it was a Rs 100 note?
Section D Question numbers 25 to 34 carry 4 marks each.
25.
26. When a chair is sold at a loss of 10% and a table is sold at a profit of 5%, total profit of Rs 5 is occurred in the whole transaction. If x and y are respectively the cost of a chair and a table, then how can this situation be expressed mathematically. 27.
The given figure shows a rhombus ABCD and a parallelogram BEFC. If C DFB, then what is the value of z?
Follow Us on
www.meritnation.com
28.
ABCD is a parallelogram. L and M are the mid-points of the sides CD and AB respectively. X and Y are any points on the sides AD and BC respectively. Prove that: Area of quadrilateral LXMY =
Area of parallelogram ABCD
29. Prove that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at the remaining part of the circle. Using this result, find the value of x in the given figure?
30.
The given figure shows a parallelogram AB CD. Side DC is extended to a point E such that N is the mid-point of AE. DL and CM are perpendiculars on AE. What is the ratio of the areas of Δ A D L and Δ C NM? 31. Construct a right triangle whose base is 7.5 cm and the difference between its hypotenuse and the remaining side is 4.5 cm. 32. A metallic cylinder of radius 8 cm and height 20 cm is melted and recasted into 15 equal small spheres. What is the diameter of each sphere?
Follow Us on
www.meritnation.com
33. The volume of a cylinder is 1232 cm2 and the circumference of its base is 44 cm. If it is open at one end, then what is its surface area? 34. The results of a random survey of the number of school dropouts of various age groups of a certain locality are displayed in the given frequency table.
Follow Us on
www.meritnation.com
MATHEMATICS (Set-2) SOLUTIONS
Follow Us on
www.meritnation.com
MATHEMATICS (Set-2) SOLUTIONS Answers: (1-mark)
Section A
1.
2.
3. In a parallelogram, opposite angles are equal. It is given that PQRS is a parallelogram. Therefore, ∠P = ∠R and ∠Q = ∠S. Since ∠P:∠Q = 7:3, ∠P = ∠R = 7 x , ∠Q = ∠S = 3 x In a quadrilateral, the sum of all the interior angles is 360°. ∴∠P + ∠Q + ∠R + ∠S = 360° ⇒ 7 x + 3 x + 7 x + 3 x = 360° ⇒ 20 x = 360°
Thus, the measure of ∠R is 126°. The correct answer is A. Follow Us on
www.meritnation.com
4. In Δ COA, O C = OA (Radii) ⇒ ∠CAO = ∠ACO ∴ ∠ CAO + ∠ ACO + ∠ COA = 180° ⇒ y + y + 70° = 180° ⇒ 2y = 180° − 70° = 110° ⇒ y = 55° ∴ ∠ ACO = 55° We know that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.
Now, ∠ACO = ∠ACB + ∠BCD ⇒ 55° = x + 30° ⇒ x = 55° − 30° = 25° Thus, the respective values of x and y are 25° and 55°. The correct answer is D.
5. Length (l) of the cuboidal box = 6 m = 600 cm Breadth (b) of the cuboidal box 4m = 400 cm Height (h) of the cuboidal box = 4.5 m = 450 cm 3 ∴ Volume of the cuboidal box = l × b × h = (600 × 400 × 450) cm Length of the edge of the cubical box, a = 50 cm 3 3 ∴ Volume of the cubical box = a = (50 × 50 × 50) cm ∴ Number of cubical boxes that can be kept in the cuboidal box
Thus, 864 cubical boxes can be kept inside the cuboidal box. The correct answer is D. 6. Let r be the radius of the sphere. Surface area of sphere, S1 = 4πr 2 … (1) When the sphere is cut into two hemispheres, radius of each hemisphere = r Surface area of each hemisphere = 3πr 2 Surface area of both hemispheres, S2 = 2 ×3πr 2 = 6πr 2 … (2) Increase in surface area = S2 − S1 = 6πr 2 − 4πr 2 = 2πr 2 Percentage increase = 50% Thus, if a sphere is cut into two hemispheres, the surface area is increased by 50%. The correct answer is B.
Follow Us on
www.meritnation.com
7. Let boy 1 is B1, boy 2 is B2 , girl 1 G1 , girl 2 is G2 and girl 3 is G3 . From a group of 2 boys and 3 girls, two children are selected., we can get ten possible outcomes. These can be tabulated as : B1B2, B1G1, B1G2, B1G 3, B 2G 1, B 2G 2, B 2G 3, G 1G 2, G 1G 3, G 2G 3 ∴ Sample space = {B1B2, B1G1, B1G2, B1G3, B 2G 1, B 2G 2, B 2G 3, G 1G 2, G 1G 3, G 2G 3} The correct answer is A. 8. The mode of a set of observations is the observation that occurs most often. Therefore, a data set can have more than one mode. For example, for the data set {9, 9, 10, 11, 11}, the modes are 9 and 11. The correct answer is B.
Section B (2 mark)
9. Given, X and Y are the mid-points of AB and AC respectively. It is known that, the line joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
XD = XY + YD = XY + XY (XY = YD) = 2XY = BC [Using (1)] It is also known that, a quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel. ∴ BCDX is a parallelogram. ⇒ BX||CD Thus, the result is proved. 10. In Δ A OB, AO = BO ⇒ ∠OAB = ∠ OBA (Angles opposite to equal sides are equal) ⇒ ∠ OAB = 60° ∴ ∠AOB = 60° We know that the angle subtended by the arc at the centre is double the angle made by it at any point on the remaining part of the circle.
Thus, the measure of ∠ACB is 30°.
Follow Us on
www.meritnation.com
11. Base radius of the can, r = 7 cm Height of the can, h = 12 cm Total area to be covered = 2πrh + 2πr 2
Thus, the required area is 836 cm2. 12. Let the n observations be x1, x2 , x3, … and xn.
Follow Us on
www.meritnation.com
13. Total number of cricket matches the two teams had played = 50 Number of matches won by team A = 25 Number of matches lost by team A = 20 Therefore, number of matches which were drawn = 50 − (25 + 20) = 5
14. It is observed that the class intervals of the given data are not continuous because the upper class limit of any class interval does not coincide with the lower class limit of the next class interval.
The continuous form of the given data is:
Follow Us on
www.meritnation.com
Section C (3 mark)
15. The final velocity of bicycle can be represented by the following equation:
v = u + at where, v = Final velocity
Follow Us on
www.meritnation.com
16. The given equation is 3( x + y − 9) + 2( x − 1) + ( y − 2) = 5( x + y − 5).
Thus, the given equation can be written as 0 × x + y + 6 = 0. For any value of x , the value of y will be −6. The graph of the given equation can be plotted as:
17. The dimensions of the rectangle ABCD are 20 cm and 26 cm. 2 ∴Area of rectangle AB CD = Length × Breadth = 20 cm × 26 cm = 520 cm We know that the diagonal of a rectangle divides it into two congruent triangles. Therefore, Δ A B C and Δ ADC are congruent. ∴Area (Δ AB C)
= Area (Δ A DC)
In the given figure, it is seen that Δ A BC and parallelogram ACEF lie on the same base AC and between the same parallels AC and EF.
Thus, the area of parallelogram ACEF is 520 cm2 .
Follow Us on
www.meritnation.com
18. Given, L is the mid point of the chord PQ. It is known that, the line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord. ∴ O L ⊥PQ ⇒ OL ⊥XY It is also known that, the perpendicular drawn from the centre of a circle to a chord bisects the chord. ∴ XL = LY ...(1) PL = LQ [Given] ... (2) Subtracting (1) from (2), we have PL −XL = LQ − LY ⇒ PX = QY ...(3) From (1) and (3), we have
Hence, the result is proved. 19. It is known that each angle of a rectangle is 90°. ∴ ∠ BA D = 90° ⇒ ∠BAC + ∠CAD = 90° ⇒ 38° + ∠CAD = 90° ⇒ ∠CAD = 52° It is also known that the diagonals of a parallelogram bisect each other and a rectangle is a parallelogram. Also, diagonals of rectangle are equal. ∴ OA = OD ⇒ ∠OAD = ∠ODA ⇒ 52° = x ⇒ x = 52° Using exterior angle property in Δ O AD: ∠COD = ∠ OAD + ∠ODA ⇒ y = 52° + 52° ⇒ y = 104° Thus, the values of x and y are 52° and 104° respectively. 20. Therefore we first draw angles of 90° and 135°, and then bisect the angle between them. The steps of the construction are as follows: Step 1: Draw a line AB and mark a point C on it. Step 2: Draw an angle of 90° at C using a compass. The angle obtained is ∠QCB = 90º and ∠ACQ = 180° − 90º = 90º. Step 3: With M and Q as centres, and radius more than half of MQ, draw arcs intersecting each other at R. The angle obtained is ∠RCB = 135º.
Follow Us on
www.meritnation.com
Step 4: With K and Q as centres and radius more than half of KQ, draw arcs intersecting each other at P. Join PC.
Thus, ∠PC B = 112.5° is the required angle.
21. Let the radius of the sphere be r . It is given that the volume of the sphere is 4851 cm3 .
Thus, the surface area of the sphere is 1386 cm2 .
Follow Us on
www.meritnation.com
22. The dimensions of the cuboidal tank are 15 m × 10 m × 8 m. It is given that the tank is already filled with water up to 3 m. Hence, amount of water to be filled in the tank = 15 m × 10 m × (8 − 3) m = (15 × 10 × 5) m3 = 750 m3 = 750 × 1000 L (1 m3 = 1000 L) = 750000 L The rate at which water is flowing into the tank from the pipe is 50 L/s.
Therefore, time required to fill the entire tank
23. Let the missing observation be x .
Follow Us on
www.meritnation.com
24. Let the numbers of Rs 100, Rs 50 and Rs 10 notes be 2 x , 3 x and 5 x respectively.
Section D (4mark)
25. Let the numerator of the fraction be x and the denominator be y.
According to the given information,
Follow Us on
www.meritnation.com
Four different solutions of equation (1) have been shown in the table as:
On plotting and joining the points (−2, −4), (1, −2), (4, 0) and (7 (7,, 2), a straight line representing equation (1) is obtained. o btained. This can be done as:
Now, the value of x for for y = 4 can be found by first drawing a perpendicular from y = = 4 to the line, and then, from that point, by drawing a perpendicular to the x -axis. -axis. The point where the perpendicular intersects the x -axis -axis is the value of x corresponding to y = = 4. It can be seen that, x = = 10 for y = = 4.
Follow Us on
www.meritnation.com
26. The costs of a chair and a table are Rs x and and Rs y respectively. respectively. x + y ). Total To tal cost price of a chair and a table = Rs ( x ). When the chair is sold at 10% loss, then the selling price of the chair becomes becomes Rs ( x − − 10% of x )
27. It can be observed from the figure that DC||BF and BC acts as a transversal (Alternate interior interior angles) angles) ∴ ∠ DCB = ∠ C B F = 45° (Alternate It is known that in a parallelogram, adjacent angles are supplementary. 180° ∴ ∠ CBE + ∠FE B = 180° ⇒ ∠CB E = 180º − ∠ F E B ⇒ ∠CB E = 180° − 110° 110° ⇒ ∠CB E = 70° Now, ∠CBF + ∠FBE = ∠C B E 45° + ∠FB E = 70° 70° ⇒ ∠FB E = 25° 25° (Alternate nate interior interior angles) ∠ CFB = ∠ F B E (Alter ⇒ ∠CF B = 25° 25° ⇒ z = 25° Thus, the value of z is 25°.
Follow Us on
www.meritnation.com
28. Join LM.
Since Sin ce AB CD is a par paralle allelog logram ram,, AB || CD and and AB = CD.
AM || CD and AM = LD It is known that if in a quadrilateral, a pair of opposite sides is parallel and equal, then the quadrilateral is a parallelogram. ∴ AMLD is a parallelogram. oppositee sides of a paralle opposit parallelogram logram are paralle parallel) l) ∴ A D || L M ( ⇒ AD || LM || BC It is known that if a triangle triangle and a parallelogram lie on the same base and between the same same parallels, then the area of triangle is half the area of parallelogram. Δ L M Y and parallelogram parallelogram LM B C lie on the same base base L M and between between the same same parallels parallels LM and BC. ⇒
Follow Us on
www.meritnation.com
29. Let PQ be an arc of the circle subtending ∠PO Q at the centre and ∠PAQ at the remaining part of the circle.
We have to prove that ∠ POQ = 2∠ PAQ Let us join OA by extending it to a point B. By exterior angle property in Δ AOQ, we obtain ∠ BOQ = ∠OAQ + ∠ OQA … (I) ∴ OA = OQ (Radii) ∴ ∠ OAQ = ∠ OQA … (ii) Using equation (ii) in (I), we obtain ∠ BOQ = ∠OAQ + ∠ OAQ ∠ BOQ = 2∠ OAQ … (iii) Similarly, ∠BO P = 2 ∠OAP … (iv) Adding (iii) and (iv), we obtain ∠ BOQ + ∠BO P = 2 [∠ OAQ + ∠ OAP] ∠ PO Q = 2∠PAQ Thus, the angle subtended by an arc at the centre of the circle is double the angle subtended by that arc at the remaining part. Using the above result, we obtain ∠AOC = 2∠ABC
ABCD is a cyclic quadrilateral. ∴ ∠ABC + ∠AD C = 180° 65° + x = 180° x = 180° − 65° x = 115°
Follow Us on
www.meritnation.com
30. Since ABCD is a parallelogram, AD || BC. N is given as the mid-point of AE. ∴ AN = NE ... (1) According to the converse of mid-point theorem, the line drawn through the mid-point of a side of a triangle and parallel to another side bisects the third side. ∴ C is the mid-point of DE. In Δ DEL, C is the mid-point of DE and CM || DL. Therefore, by applying the converse of mid-point theorem, it is obtained that M is the mid-point of LE. ∴ L M = ME ... (2) From equations (1) and (2): AN + ME = LM + NE ⇒ AL + LN + ME = L N + NM + NM + ME ⇒ AL = 2 N M ... (3) According to the mid-point theorem, the line segment joining the mid-points of two sides of a triangle is half the third side. ∴ M C = (½)DL ... (4)
31. Let Δ A BC be right-angled at B in such that B C = 7.5 cm and AC − A B = 4.5 cm. The steps of construction are as follows: (i) Draw a line segment BC of length 7.5 cm. (ii) At point B, draw ∠XB C = 90°. (iii) Extend XB to Y. (iv) Draw an arc taking B as the centre and radius equal to 4.5 cm such that it intersects BY at point D. (v) Join CD.
Follow Us on
www.meritnation.com
(vi) Draw the perpendicular bisector of CD intersecting XB at point A. (vii) Join AC. Thus, Δ A BC is the required triangle.
32. Radius of metallic cylinder , R = 8 cm Height of metallic cylinder, H = 20 cm 2 3 ∴ Volume of the metallic cylinder = π R H = π × 82 × 20 = (π × 64 × 20) cm Let r be the radius of each small sphere.
Follow Us on
www.meritnation.com
33. Let the radius and height of the cylinder be r and h respectively. Circumference of the base = 2πr = 44 cm
Follow Us on
www.meritnation.com
34. It is observed that intervals are not equally spaced in the given data. The areas of rectangles in a histogram are proportional to their corresponding frequencies. Since the intervals are not equally spaced in this case, the width of the rectangles will vary, and the bar graph will not be correct. In order to draw the histogram correctly, the new lengths of the rectangles proportional to the minimum class size, i.e., 2 (= 10 − 8) need to the calculated. This can be done as:
Now, the required histogram can be constructed by taking the new lengths of the rectangles along the vertical axis, and the age along the horizontal axis. This can be drawn as:
Follow Us on
www.meritnation.com
