Answers & Solutions

DATE : 13/05/2018

BOOKLET CODE

13-15

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472

Time : 120 Mins.

Answers & Solutions

Max. Marks : 100

for NTSE (Stage-II) - 2018 SCHOLASTIC APTITUDE TEST (For Students of Class X)

INSTRUCTIONS TO CANDIDATES : Read the following instructions carefully before you open the question booklet. 1.

There are 100 questions in this test. All are compulsory. The question numbers 1 to 40 belong to Sciences, 41 to 60 pertain to Mathematics and 61 to 100 are on Social Science subjects.

2.

Please follow the instructions given on the answer sheet for marking the answers.

3.

Write your eight-digit Roll Number as allotted to you in the admission card very clearly on the test-booklet and darken the appropriate circles on the answer sheet as per instructions given.

4.

Write down and darken Booklet Number in the appropriate circles on the answer sheet as per instructions given.

5.

Since the time allotted for this question paper is very limited and all questions carry equal marks, you should make the best use of it by not spending too much time on any one question.

6.

Rough work can be done anywhere in the booklet but not on the answer sheet/loose paper.

7.

Every correct answer will be awarded one mark.

8.

There will be NO NEGATIVE marking.

1

NTSE (Stage-II) - 2018 (Scholastic Aptitude Test)

1.

Under which condition stated below, the sixcarbon glucose molecule is broken down into three-carbon molecules pyruvate and lactic acid?

Answer ( 4 )

(1) aerobic condition in muscle cells

5.

S o l . Edward Jenner found that the cow pox infection protects the person from subsequent infection from small pox.

(2) anaerobic condition in yeast cells

Four important events given below may have led to the origin of life on the earth.

(3) aerobic condition in mitochondria

I.

(4) anaerobic condition in muscle cells

II. Availability of water

Formation of amino acids and nucleotides

Answer (4)

III. Organization of cells

S o l . During vigorous muscular activity there is lack of oxygen supply and thus anaerobic respiration takes place in muscle cells. During this process lactic acid is formed which gets accumulated in the muscle cells leading to cramps.

IV. Formation of complex molecules

2.

Select the correct sequence of events. (1) I, II, III and IV (2) II, I, IV and III (3) I, IV, II and III

Which among the following is the correct sequence regarding the flow of impulse in a neuron?

(4) II, III, I and IV Answer ( 2 )

(1) Dendrite  Axon  Cell body

S o l . Origin of life on the earth evolved in the following sequences :- First, water is available, secondly formation of amino acids and nucleotides occurs, thirdly formation of complex molecules occurs, lastly organization of cells occurs.

(2) Axon Cell body  Dendrite (3) Axon Dendrite  Cell body (4) Cell body Axon  Nerve terminal Answer ( 4 )

6.

S o l . Transmission of impulse through neurons is in the form of electrical signals and occurs in one direction only, i.e., Dendrite  Cell body  Axon  Nerve terminal 3.

I.

(2) Cerebrum

(3) Cerebellum

(4) Hypothalamus

Energy transfer in the biotic world always proceeds from the autotrophs.

II. Energy flow is unidirectional.

In a hypertensive patient, the systolic pressure increased to 150 mm of Hg. Which part of the brain would be involved in the regulation of blood pressure? (1) Medulla

Read the following statements carefully.

III. Energy availability is maximum at the tertiary level. IV. There is loss of energy from one trophic level to the other. Select the relevant statements for the forest ecosystem

Answer ( 1 )

(1) I, II and IV

(2) I, II and III

Sol. All the involuntary actions including blood pressure, salivation and vomiting are controlled by the medulla in the hind brain.

(3) I, III and IV

(4) II, III and IV

4.

Answer ( 1 ) S o l . Flow of energy in an ecosystem is always unidirectional i.e., the energy which passes from the autotrophs to the herbivores does not revert to the autotrophs. There is a loss of energy from one trophic level to the other.

Edward Jenner's contribution for the eradication of small pox is (1) his proposition that small pox had possibly spread throughout the world from India and China. (2) his discovery procedure.

of

7.

transformation

(3) his finding that rubbing of the skin crust of small pox victims on the arm of a healthy person, would develop resistance against small pox.

In a highly pesticide polluted pond, which of the following aquatic organisms will have the maximum amount of pesticide per gram of body mass? (1) Lotus (2) Fishes

(4) his finding that the cow pox infection protects the person from subsequent infection from small pox.

(3) Spirogyra (4) Zooplanktons 2


Answer ( 2 )

10.

S o l . Pesticides gets accumulated at each trophic level. As, fishes occupy the last trophic level in pond. So, fishes will have the maximum amount of pesticide per gram of body mass. 8.

Raw banana has bitter taste, while ripe banana has sweet taste. It happens because of the conversion of (1) starch to sugar (2) sucrose to fructose

A farmer made an observation in a backwater paddy field of coastal Kerala that the paddy plants wilt during noon onwards everyday but appear normal next morning. What would be the possible reason for wilting?

(3) amino acids to sugar (4) amino acids to protein Answer ( 1 )

(1) The rate of water absorption is less than the rate of transpiration in the afternoon.

S o l . During ripening, there is breakdown of starch to simple sugars (such as glucose, fructose, sucrose) which increases sweetness of banana.

(2) The rate of water absorption is more than the rate of transpiration in the afternoon.

11.

(3) The changes in the rate of water absorption and transpiration are not associated with wilting. (4) The rate of water absorption is not related to the rate of transpiration.

In the flowering plants sexual reproduction involves several events beginning with the bud and ending in a fruit. These events are arranged in four different combinations. Select the combination that has the correct sequence of events. (1) Embryo, zygote, gametes, fertilization

Answer ( 1 )

(2) Gametes, fertilization, zygote, embryo

S o l . When rate of water absorption is less than the rate of transpiration, plant cells loose water and thus plants wilt in the afternoon.

(3) Fertilization, zygote, gametes, embryo

9.

(4) Gametes, zygote, embryo, fertilization Answer ( 2 )

Observe the experimental sets [A] & [B]. [A]

S o l . The male and female gametes fuse to form zygote. This process is called fertilization. Zygote then divides to form embryo.

[B]

Test tube

12.

Water Cotton Plug Plant

Day 1

Day 5

Observe the test tube A & B. From the list given below, choose the combination of responses of shoot and root that are observed in B.

In pea plants, Round (R) and Yellow (Y) features of seeds are dominant over wrinkled (r) and green (y) features. In a cross between two plants having the same genotypes (RrYy), the following genotypic combinations of offspring are noticed. A. RrYY

B. Rryy

C. rrYy

D. rryy

The phenotypic features of A, B, C and D are given below in an order in four combinations. Select the correct combination of characters that corresponds to the genotypes,

(1) Positive phototropism and positive geotropism

(1) Round & yellow; round & green; wrinkled & yellow; wrinkled & green.

(2) Negative phototropism and positive geotropism

(2) Round & green; wrinkled & yellow; wrinkled & green; round & yellow.

(3) Positive phototropism and negative geotropism

(3) Wrinkled & green; round & yellow; wrinkled & yellow; round & green.

(4) Only negative phototropism

(4) Wrinkled & yellow; round & green; wrinkled & yellow; round & yellow.

Answer ( 1 ) S o l . Shoot bends towards light which shows positive phototropism, while root bends towards earth in response to gravity which shows positive geotropism.

Answer ( 1 ) S o l . In heterozygous condition, dominant trait (i.e., round and yellow) will be expressed. So, phenotypes will be 3


S o l . According to ideal gas equation,

A – RrYY : Round and Yellow

13.

B – Rryy : Round and Green

PV = nRT

C – rrYy : Wrinkled and Yellow

Volume and temperature are constant (given)

D – rryy : Wrinkled and Green

 Pn

Eukaryotic organisms have different levels of organization. Select the combination where the levels are arranged in the descending order.

Number of moles of O2 (n1 ) 

Number of moles of He (n2 ) 

(1) DNA, chromosome, cell, nucleus, tissue (2) Tissue, cell, nucleus, chromosome, DNA

So, the pressure of He gas would be greater than that of O2 gas

(4) Tissue, cell, chromosome, nucleus, DNA 16.

Answer ( 2 ) S o l . Chromosomes, present in nucleus, are made up of DNA and proteins. Nucleus is the control room of the cell. A group of cells that are similar in structure and/or work together to achieve a particular function forms a tissue.

At 298 K and 1 atm pressure, a gas mixture contains equal masses of He, H2, O2 and NH3. Which of the following is correct for their average molecular velocities? (1) He > H2 > NH3 > O2 (2) He < H2 < O2 < NH3 (3) H2 < He < NH3 < O2

The gaseous byproduct of a process in plants is essential for another vital process that releases energy. Given below are four combinations of processes and products.

(4) O2 < NH3 < He < H2 Answer ( 4 )

Choose the correct combination

Sol.

Gases

(1) Photosynthesis and oxygen

Molecular masses (u)

(2) Respiration and carbon dioxide

Hydrogen (H2)

2

(3) Transpiration and water vapour

Helium (He)

4

(4) Germination and carbon dioxide

Ammonia (NH3)

17

Oxygen (O2)

32

Answer ( 1 ) S o l . Oxygen is a byproduct of photosynthesis which occurs in plants. Oxygen is essential for another vital process in organisms which releases energy, i.e., respiration. 15.

100  25 moles 4

As n2 > n1

(3) Nucleus, cell, DNA, chromosome, tissue

14.

100  3.125 moles 32

Average molecular velocity =

8 RT M

According to the above relation, greater the molecular mass lesser will be the average molecular velocity.

100 grams of oxygen (O2) gas and 100 grams of helium (He) gas are in separate containers of equal volume at 100°C. Which one of the following statement is correct?

 The correct order is : O2 < NH3 < He < H2 17.

(1) Both gases would have the same pressure (2) The average kinetic energy of O 2 molecules is greater than that of He molecules (3) The pressure of He gas would be greater than that of the O2 gas (4) The average kinetic energy of He and O2 molecules is same

A test tube along with calcium carbonate in it initially weighed 30.08 g. A heating experiment was performed on this test tube till calcium carbonate completely decomposed with evolution of a gas. Loss of weight during this experiment was 4.40 g. What is the weight of the empty test tube in this experiment? (1) 20.08 g

(2) 21.00 g

(3) 24.50 g

(4) 2.008 g

Answer ( 1 )

Answer ( 3 ) 4


S o l . On thermal decomposition of calcium carbonate.

Answer ( 4 ) S o l . a lead nitrate + b aluminium chloride 

 CaCO3   CaO  CO2 100 g 56 g 44 g

c aluminium nitrate + d lead chloride a Pb (NO3)2 + b AlCl3  c Al(NO3)3 + d PbCl2 On balancing the above equation

44 g CO2 is formed from 100 g CaCO3 4.40 g CO2 is formed from

3Pb(NO3)2 + 2AlCl3  2Al(NO3)3 + 3PbCl2

100  4.4 = 10 g CaCO3 44

So, a = 3, b = 2, c = 2, d = 3 20.

Initial weight of (CaCO3 + test tube) = 30.08 g Calculated weight of CaCO3 = 10 g Weight of empty test tube = 30.08 – 10 = 20.08 g 18.

Match List-I (Mixture to be separated) with the List-II (Method used) and select the correct option using the codes given below.

A. Petroleum products

(1) Au < Ag < Cu < Al (2) Al < Cu < Ag < Au (3) Au < Cu < Al < Ag (4) Ag < Cu < Al < Au Answer ( 1 ) S o l . Number of -particles deflected  Z2

List-II (Method used)

List-I (Mixture to be separated)

Where Z = Atomic number

I. Chromatography

B. Camphor and rock salt II. Centrifugation C. Cream from milk

Ag

Cu

Al

Z

79

47

29

13

Au < Ag < Cu < Al 21.

(2) A-II, B-IV, C-III, D-I

The correct order of increasing pH values of the aqueous solutions of baking soda, rock salt, washing soda and slaked lime is (1) Baking soda < Rock Salt < Washing Soda < Slaked lime

(3) A-IV, B-III, C-I, D-II (4) A-IV, B-III, C-II, D-I

(2) Rock Salt < Baking Soda