DATE : 14/05/2017
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Answers & Solutions
Max. Marks : 100
f o r fo
NTSE (Stage-II) - 2017 SCHOLASTIC APTITUDE TEST (For Students of Class X)
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2.
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1
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
SCHOLASTIC APTITUDE TEST 1.
Small cut pie Small pieces ces of soft soft ste stems ms are are plac placed ed in in growt growth h medium with following plant hormones. Which combination of plant hormones will show slowest growth?
Answer (3) Sol. Many of the involuntary actions are controlled by mid-brain and hind-brain. [As mentioned in NCERT text book].
(1) Au Auxin xin + Cyto Cytokin kinin in
A and B are the parts of fore-brain.
(2) Gib Gibber berell ellins ins + Auxi Auxin n (3)) Gibb (3 Gibberel erellins lins + Cyto Cytokinin kinin
C is the part of hind-brain.
(4) Ab Absci scisic sic acid acid + Auxi Auxin n
D is the part of mid-brain.
Answer (4 (4))
4.
Sol. Abscisic acid is growth inhibiting hormone whereas auxin, gibberellin and cytokinin are growth promoting hormones.
2.
An anim animal al kept kept in a jar jar has has the the foll follow owing ing fe featu atures res.. I.
It is is bila bilate tera rall lly y symm symmet etri rica cal. l.
II.
It has has coe coelo lomi mic c cavi cavity ty..
Option (1), (2) and (3) include all growth promoting hormones whereas option (4) includes both growth promoting and growth inhibiting hormones. Therefore, this combination of hormones will show slowest growth.
III. Th The e body body is segme segment nted. ed.
Which one of the fol Which follow lowing ing dem demons onstra trates tes the characteristics of cardiac muscle cells?
(1) Ar Arth thro ropo poda da
IV.. It has IV has jointed jointed appen appendage dages. s. To which phylum does the animal belong to?
(2) An Anne neli lida da
(1)) Invol (1 Involunta untary ry and multi multinucle nucleated ated
(3) Pla Platyh tyhelm elmint inthes hes
(2)) Unbr (2 Unbranche anched d and and uninu uninucleat cleated ed (4) Mo Moll llus usca ca
(3)) Cyl (3 Cylindr indrical ical and and uninuclea uninucleated ted
Answer (1)
(4)) Unb (4 Unbranc ranched hed and and invol involunta untary ry
Sol. Annelida - The animals do not have jointed appendages.
Answer (3 (3)) Sol. Cardiac muscle cells are involuntary, uninucleated and cylindrical.
Platyhelminthes - They are acoelomates.
Cardiac muscles are composed of branched muscle fibres. 3.
Mollusca - They are not segmented. 5.
From the From the give given n figur figure, e, iden identif tify y the the part part of huma human n brain brain controlling most of the involuntary actions.
Read the Read the foll followi owing ng stat stateme ements nts and sel select ect the cor correc rectt option. Statement I : Nostoc and and Bacteria are prokaryotes.
A
Statement II : Penicillium and Spirogyra are fungi. (1)) Only stat (1 statemen ementt I is true true D D
(2)) Only stat (2 statemen ementt II is true true
B
(3)) Bot (3 Both h statemen statements ts I and II II are true true
C
(4)) Bot (4 Both h stateme statements nts I and and II are false false Answer (1) (1) A & B
(2) B & C
(3) C & D
(4) D & A
Sol. Penicillium is a fungus but Spirogyra is a green filamentous alga. 2
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
6.
You find a herbaceous flowering plant growing in your school garden having leaves with parallel venation. Choose the correct additional features the given plant would be possessing. I.
It has no secondary vascular tissues.
II.
Its flower possesses three sepals.
9.
(3) II and IV
(4) III and IV
(2) Anabaena
(3) Rhizobium
(4) Trichoderma
Sol. Trichoderma is used as a biopesticide. Azoll a, Anabaena, Rhizobium are used as biofertilizers.
IV. Its embryo has 2 cotyledons. (2) I and III
(1) Azolla
Answer (4)
III. It possesses tap root.
(1) I and II
Which of the following organisms is used as a biopesticide?
10. A tall plant (TT) is crossed with a dwarf plant (tt). All F1 plants showed tall phenotype.
Answer (1)
Which of the following correctly defines a test cross?
Sol. Presence of tap root and two cotyledons in seeds are the characteristic features of dicot plant.
(1) TT (F1) × Tt (P)
(2) Tt (F 1) × Tt (P)
(3) tt (F1) × Tt (P)
(4) Tt (F 1) × tt (P)
7.
Varieties of vegetables such as cabbage, broccoli and cauliflower have been produced from a wild cabbage species. Such process of producing new varieties of living organisms is called (1) Natural selection
(2) Artificial selection
(3) Speciation
(4) Genetic drift
Answer (4) Sol. Tt(F1) × tt(P) Test cross involves the breeding of an individual with a phenotypically recessive individual, in order to determine the zygosity of the former by analyzing proportions of offspring phenotypes.
Answer (2)
Here,
Sol. Artificial selection : It is the process by which man selects the traits useful to him for improving the qualities of domesticated plants and animals
P generation
Natural Selection : It is the process whereby organisms better adapted to their environment tend to survive and produce more offspring. F generation 1
–
Wings of bat
II.
Wings of bat
–
Wings of butterfly
III. Forelimbs of horse
–
Wings of butterfly
IV. Wings of bird
–
Wings of bat
(1) I and II
(2) II and IV
(3) III and IV
(4) II and III
(All tall plants)
Tt(F1 generation) × tt(P generation) 11. Which one of the following pairs of causative agent and type of disease are correct?
Which of the following are pairs of analogous organs? Forelimbs of horse
Dwarf plant (tt)
Test cross will be,
Genetic drift : Change in gene frequency by chance in a small population due to random sampling.
I.
×
Tt
Speciation : It is the splitting of one species into two or more species or transformation of one species into a new species over a period of time due to geographical barrier or reproductive isolation.
8.
Tall plant (TT)
I.
Leishmania - Sleeping sickness
II.
Nematode - Elephantiasis
III. Trypanosoma - Kala azar IV. Staphylococcus - Acne (1) I & II
(2) II & III
(3) II & IV
(4) III & IV
Answer (3) Sol. I.
Answer (2)
II.
Sol. Analogous organs : Organs which are similar in function but differ in origin.
Leishmania causes kala-azar. Elephantiasis is caused by filarial worm, a nematode.
In I - Given organs are similar in origin
III. Trypanosoma causes sleeping sickness.
In III - Both have different origin and function
IV. Staphylococcus causes acne. 3
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
12. Pancreatic juice contains more than one enzyme. Which among the following combination is correct? (1) Pepsin and lipase
(2) Amylase and pepsin
(3) Pepsin and trypsin
(4) Trypsin and lipase
Answer (2) Sol. M2O3 Valency of metal M = 3 Valency of nitrogen = 3
Answer (4)
Formula of metal nitride =
Sol. Trypsin, amylase, pancreatic lipase are constituents of pancreatic juice but pepsin is a constituent of gastric juice secreted by peptic/chief/zymogenic cells.
(2) Angiosperm
(3) Gymnosperm
(4) Pteridophyte
3
3
Formula = MN 16. A solution is a homogeneous mixture of two or more substances. Which of the following is a solution? (1) Milk
(2) Smoke
(3) Brass
(4) Face Cream
Answer (3)
Answer (4)
Sol. "Brass". It is a homogeneous mixture of copper and zinc.
Sol. Motile sperms are produced in bryophytes and pteridophytes.
17. 1.80 g of glucose is dissolved in 36.00 g of water in a beaker. The total number of oxygen atoms in the solution is
In bryophytes, haploid (gametophytic) phase is dominant. In pteridophytes, diploid (sporophytic) phase is dominant. 14. At every 20 minutes, one bacterium divides into two. How many bacteria will be produced after two hours, if one starts with 10 bacteria? 5
(1) 2 × 10
5
5
(2) 2 × 10
6
6
6
(3) 2 × 10
Nitrogen (N)
Valency
13. You discover a new species of a plant. You also discover that it produces motile sperms and dominant generation has diploid cells. It belongs to (1) Bryophyte
Metal (M)
(1) 12.405 × 1023
(2) 12.405 × 1022
(3) 6.022 × 1023
(4) 6.022 × 1022
Answer (1) Sol. Number of atoms of oxygen from glucose (C6H12O6) = Number of moles of C6H12O6 × Number of atoms of oxygen in one molecule × 6.022 × 1023
(4) 2 × 10
Answer (3)
=
Sol. 2 hours = 2 × 60 min = 120 min
1.8 180
6 6.022 1023
= 0.01 × 6 × 6.022 × 10 23
Time for 1 generation is 20 min
= 0.3613 × 1023 120 So in 120 min, bacterial cell will divide 6 times i.e., 20
Number of atoms of oxygen from water = Number of moles of H2O × Number of atoms of oxygen in a molecule × 6.022 × 1023
No. of generations n = 6 So, 2n = 2 6 = 64
36
Initial bacterial cells = 10
=
6
So, 2 × 10 bacteria will be produced after 2 hours from 10 bacteria.
18
1 6.022 1023
= 2 × 1 × 6.022 × 10 23 = 12.044 × 1023
15. The metal (M) forms an oxide, M2O3. The formula of its nitride will be
Total number of atoms of oxygen in solution
(1) M2N3
(2) MN
= 0.3613 × 1023 + 12.044 × 10 23
(3) M2N
(4) M3N2
= 12.405 × 1023 4
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
18.
35
Cl and 37Cl are the two isotopes of chlorine, in the ratio 3 : 1 respectively. If the isotope ratio is reversed, the average atomic mass of chlorine will be
Sol. Metal 'M' is Cu and its sulphide is Cu2S('X') i.e., Cuprous sulphide 2Cu2S (X)
(1) 35.0 u
3O2 2Cu2O 2SO2 (Y)
(2) 35.5 u
2Cu2 O Cu2 S (Y) ( X) (cuprous oxide) (cuprous sulphide)
(3) 36.0 u (4) 36.5 u
21. Which one of the following statement is incorrect about graphite and diamond?
Answer (4) Sol. Isotopes of chlorine are 35Cl and
37
Cl.
(1) Graphite is smooth and slippery
According to question, 35Cl and 37Cl ratio will be (1:3)
Average atomic mass of chlorine =
=
6Cu SO2
(2) Diamond is good conductor of heat
35 1 37 3
(3) Graphite is a good conductor of electricity
4
(4) Physical and chemical properties of graphite and diamond are different
146 4
Answer (4) = 36.5 u Sol. Physical properties will differ while the chemical properties of both graphite and diamond will remain same. (Since both are allotropes and allotropes differ in their physical properties)
19. The turmeric solution will turn red by an aqueous solution of (1) Potassium acetate
22. The functional groups present in the following compound are
(2) Copper sulphate (3) Sodium sulphate
O
(4) Ferric chloride Answer (1) Sol.
CH3COOK
CH COO + K –
+
3
From weak acid (CH3COOH)
C
OH
O
C
CH3
O
From strong base (KOH)
(1) Alcohol, ketone and ester
Solution will be basic in nature. In basic
(2) Ester and carboxylic acid
medium, turmeric solution turns red.
(3) Carboxylic acid and ketone
20. A metal 'M' of moderate reactivity is present as its sulphide 'X'. On heating in air, 'X' converts into its oxide 'Y' and a gas evolves. On heating 'Y' and 'X' together, the metal 'M' is produced. 'X' and 'Y' respectively are
(4) Ester and alcohol Answer (2)
(1) 'X' = Cuprous sulphide, 'Y' = Cuprous oxide
O
(2) 'X' = Cupric sulphide, 'Y' = Cupric oxide
C
OH
O
C
Carboxylic acid group Sol.
(3) 'X' = Sodium sulphide, 'Y' = Sodium oxide (4) 'X' = Calcium sulphide, 'Y' = Calcium oxide
O
Answer (1) 5
CH
3
Ester group
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
23. A part of the modern periodic table is presented below in which the alphabets represent the symbols of elements.
25. The schematic diagram is given below.
Group 1
Heat
A (Solid)
Table 2
14
15
16
B + HCl (vapour) (vapour)
Cool
17 Heat NaOH (aq)
Period 2
M
3
A
4
E
5
G
C (Gas)
Q
J
R L
conc. HCl
H2O
E (aq) (acidic solution)
D Shake well
T X Which of the following is an incorrect statement?
Consult the above part of the periodic table to predict which of the following is a covalent compound? (1) RQ2
(2) AT
(3) JQ
(4) JX2
(1) A and E are chemically same (2) A and D are chemically same (3) D and E are chemically same
Answer (1)
(4) C and E are chemically same
Sol. Elements of the same group will form covalent compounds. Q and R belong to the same group. Rest will form ionic compounds.
Answer (4) A
24. A compound 'X' reacts with a compound 'Y', to produce a colourless and odourless gas. The gas turns lime water milky. When 'X' reacts with methanol in the presences of concentrated H2SO4, a sweet smelling substance is produced. The molecular formula of the compound 'X' is
Sol.
heat
NH4Cl Solid
B NH3 + HCl vapour vapour
heat NaOH (aq) C NH3 Gas
(1) C2H4O
conc. HCl
D NH4Cl
H2O
E NH4 Cl aqueous acidic solution
(2) C2H4O2 C and E are not chemically same; since C is ammonia and E is ammonium chloride solution.
(3) C2H6O (4) C2H6O2
26. Which of the following is a feasible reaction?
Answer (2)
(1) Ba(s) + K2SO4(aq) BaSO4(aq) + 2K(s)
Sol. X + Y Colourless gas (It turns lime water milky i.e., it is CO 2)
(2) Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s)
X + CH3OH Sweet smelling compound
(3) Mg(s) + Na2SO4(aq) MgSO4(aq) + 2Na(s)
Inference : Compound is ester (4) Cu(s) + MgSO4(aq) CuSO4(aq) + Mg(s)
X should be a carboxylic acid
Answer (2)
This is an esterification reaction CH3COOH or C2H4O2 + CH3OH
conc. H SO –H O 2
Sol. The order of reactivity :
4
2
K > Ba > Na > Mg > Zn > Cu > Ag
O CH C 3
O
The reaction that is feasible is
CH
3
Methyl ethanoate (ester)
Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s) 6
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
28. The velocity-time graph of an object moving along a straight line is shown below.
27. Some ice pieces kept at a temperature –5°C are heated gradually to 100°C in a beaker. The temperatures of the contents are plotted against time. The correct plot is
Velocity
(1)
) C ° ( e r u t a r e p m e T
100°C
0
3 1 2 Time Which one of the following graphs represents the acceleration (a) - time (t ) graph for the above motion?
4°C 0°C –5°C Time
a
(2)
) C ° ( e r u t a r e p m e T
(1)
100°C
0°C –5°C
0
1
2
3 t
a
Time
(2) 0 1
) C ° (
(3)
(4)
e r u t a r e p m e T
) C ° ( e r u t a r e p m e T
a
(3)
0°C –5°C
0
Time
t
100°C
a
(4) 0
Time
Latent heat
Sol.
of vaporisation (Boiling)
Latent heat of fusion (melting) Time
1
2
3 t
Answer (1) Sol. For t = 0 to t = 1 velocity is linearly increasing so acceleration is constant. For t = 1 to t = 2 velocity is constant so acceleration is zero. For t = 2 to t = 3 velocity is linearly decreasing so acceleration is constant and opposite to velocity (i.e., negative).
0°C –5°C
Answer (3)
) C ° ( e r u t a r e p m e T
3 t
2
100°C
100°C
n o i t a r e l e c c a
0°C
0
–5°C
7
2 1
3
t
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
29. To read a poster on a wall, a person with defective vision needs to stand at a distance of 0.4 m from the poster. A person with normal vision can read the poster from a distance of 2.0 m. Which one of the following lens may be used to correct the defective vision?
31. A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is ‘v ’, the diameter of the jet is ‘d’ and the density of water ‘’, then the force exerted on the wall by the jet of water is (1)
(1) A concave lens of 0.5 D (2) A concave lens of 1.0 D
(2)
(3) A concave lens of 2.0 D (4) A convex lens of 2.0 D
(3)
Answer (3) Sol. u = – 2 m
(4)
v = –0.4 m 1
f
1
–
v
1
f
1
d
2
d
2
v 2
d
2
v 2
d
2
v 2
4
4
8
2
v
Answer (2)
u
–
1 0.4
Sol. Force exerted = Rate of change in linear momentum
1 2
P = – 2 D 30. A ball released from rest at time t = 0 hits the ground. It rebounds inelastically with a velocity 5 ms–1 and reaches the top at t = 1.5 s. What is the net displacement of the ball from its initial position after 1.5 s?
P
t
m v t
A l v t
d 2 ∵ A 4
F = Av 2 F =
2
2
d v
4
32. Two blocks A and B of masses 8 kg and 2 kg respectively, lie on a horizontal frictionless surface as shown in the figure. They are pushed by a horizontally applied force of 15 N. The force exerted by B on A is
(g = 10 ms–2) t = 0
t = 1.5 s
A 8 kg
15 N
B 2 kg
(1) 1.25 m
(2) 3.75 m
(1) 1.5 N
(2) 3.0 N
(3) 5.00 m
(4) 6.25 m
(3) 4.5 N
(4) 6.0 N
Answer (2)
Answer (2)
Sol. Ball rebounds with velocity = 5 m/s
Sol.
Time taken =
5 10
0.5 s
a
15 N
A
8 kg
Time taken to fall on ground = 1.5 – 0.5 (8 + 2)a = 15 N
=1s Initial height = Final height =
1 2
15
10 12 5 m
55 2 10
B
2 kg
Acceleration of the system (a) = a
1.25 m
N
Displacement = 5 – 1.25 = 3.75 m
2 kg
N = 2 × a = 2 × 1.5 = 3 N 8
10
= 1.5 ms–2
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
35. A source produces sound waves under water. Waves travel through water and then into air. Which of the following statements about the frequency (f ) and the wavelength () is correct as sound passes from water to air?
33. A beaker half-filled with water is put on a platform balance which is then set to zero. A 800 g mass is immersed partially in water using a spring balance as shown in the figure. If the spring balance reads 300 g, what will be the reading on the platform balance?
(1) f remains unchanged but decreases (2) f remains unchanged but increases (3) remains unchanged but f decreases (4) remains unchanged but f increases Answer (1) Sol. Velocity of sound in water > velocity of sound in air. 36. The diameter of a wire is reduced to one-fifth of its original value by stretching it. If its initial resistance is R , what would be its resistance after reduction of the diameter? (1)
R
(2)
625
(3) 25 R (2) 300 g
Answer (4)
(3) 500 g
(4) 800 g
r 1 Sol. New resistance = r 2
Sol. Reading of spring balance = 300 g
r 1 = R and
Actual weight = 800 g
Buoyancy = 500 g Reading of the platform = Buoyancy
34. An object falls a distance H in 50 s when dropped on the surface of the earth. How long would it take for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the earth? (Neglect air resistance.) (4) 100.0 s
(2) The kinetic energy after collision is
g
2
R 4
(4) There will be a loss of kinetic energy equal to Mm
2
m
u
2 M
g 2
Answer (4) Sol. Let final velocity be v . By conservation of momentum
2
mu = (M + m)v
Let the time taken be t to fall same distance 2
2
m u
m
1
1
2
M m u 2
2 M
H g 50 50
2
1
1
GM 2
1
(3) There will be a loss of kinetic energy equal to
Answer (3) Sol.
5
(1) The kinetic energies before and after collision are same.
= 500 g
(3) 70.7 s
R
37. An object of mass ‘m’ moving along a straight line with a velocity ‘u ’ collides with a heavier mass ‘M ’ and gets embedded into it. If the compound system of mass ( m + M ) keeps moving in the same direction then which of the given options is true?
800 g – 300 g = Buoyancy
(2) 50.0 s
R
R = (5) R = 625R
Actual weight – Reading = Buoyancy
(1) 35.4 s
r 2
4
4
Actual weight – Buoyancy = Reading
25
(4) 625 R
(1) 200 g Answer (3)
R
g
1
2
2
v
t 2 g 50 50
t = 70.7 s
Initial KE = 9
mu
(M m) 1 2
mu 2
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
When charge is doubled
2
Mu Final KE = mv = (M + m) 2 2 M m 1
=
1
2
2
1
a
m Let time taken be t 1
2
m u
2 (M m) 1
Loss in KE =
2
mu
2
1 2
1
2
2
m u
=
=
2
mu
2
1
2qE
m
2 (M m)
1
2qE
m 1 M m
t 1
1
qE
2
m
t12 t 2
40. AB is a long wire carrying a current l 1 and PQRS is a rectangular loop carrying current I 2 (as shown in the figure).
2
mu M
2 (M m)
B
38. A vessel is filled with oil as shown in the diagram. A ray of light from point O at the bottom of vessel is incident on the oil-air interface at point P and grazes the surface along PQ. The refractive index of the oil is close to
Q
P
I 2
Q
P
oil
S
A
17 cm
Which among the following statements are correct? (a) Arm PQ will get attracted to wire AB, and the arm RS will get repelled from wire AB.
O
12 cm
(1) 1.41
(2) 1.50
(3) 1.63
(4) 1.73
(b) Arm PQ will get repelled from wire AB, and arm RS attracted to wire AB. (c) Forces on the arms PQ and RS will be unequal and opposite.
Answer (4)
(d) Forces on the arms QR and SP will be zero. 2
Sol. Value of hypotenuse =
12
=
17
433
2
cm
1 sin
433 12
(1) Only (a)
(2) (b) and (c)
(3) (a) and (c)
(4) (b) and (d)
Answer (3) Sol. B
For critical angle, 1.73
Q
39. A charged particle placed in an electric field falls from rest through a distance d in time t . If the charge on the particle is doubled, the time of fall through the same distance will be (1) 2t
(2) t
t
t
(3)
R
I 1
air
t 2
(4)
2
I 2 I 1 P
2
F m
1
1
2
2
d a t2
As current in arm PQ and wire AB are in same direction so PQ will be attracted to wire AB and current in RS in opposite direction to current in wire AB so wire AB will be repelled.
qE m
qE m
S
A
Answer (3) Sol. Acceleration a
R
Magnetic force on wire PQ will be greater than the force on the wire RS as PQ is near to the wire AB.
t2 10
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
41. The sum of all the possible remainders, which can be obtained when the cube of a natural number is divided by 9, is
p
12
3
( p 3)
p
p
(1) 5
(2) 6
p2 = 36
–3 p = (– p + 3) p
(3) 8
(4) 9
p = 6
–3 p = – p2 + 3 p
–6 p = – p2
Answer (4) Sol. Cube of any positive integer can be written in the form of 9m, 9m + 1 and 9m + 8,
p2 – 6 p = 0 p = 0, 6
All possible remainders are 0, 1 and 8 p = 6
Required sum = 9
44. Two quadratic equations x 2 – bx + 6 = 0 and x 2 – 6 x + c = 0 have a common root. If the remaining roots of the first and second equations are positive integers and are in the ratio 3 : 4 respectively, then the common root is
42. When a polynomial p( x ) is divided by x – 1, the remainder is 3. When p( x ) is divided by x – 3, the remainder is 5. If r ( x ) is the remainder when p( x ) is divided by ( x – 1)( x – 3), then the value of r (–2) is (1) –2
(2) –1
(3) 0
(4) 4
(1) 1
(2) 2
(3) 3
(4) 4
Answer (3) Answer (2)
Sol. p( x ) = ( x – 1) q( x ) + 3
...(i)
p( x ) = ( x – 3) g ( x ) + 5
...(ii)
Sol. Let the common root be According to the question, other roots will be 3k and 4k ,
Multiply (i) by ( x – 3) and (ii) by ( x – 1) and subtract (i) from (ii), we get p( x )
x 2 – bx + 6 = 0 has roots and 3k
Product of roots = × 3 k = 6
1 ( x 1)( x 3)[q( x ) g( x)] 2x 4 2
When p( x ) is divided by ( x – 1)( x – 3) then the remainder will be r ( x ) = x + 2, r (–2) = 0 43. For what value of p, the following pair of linear equations in two variables will have infinitely many solutions?
3k
k
6
2
x 2 – 6 x + c = 0
Sum of roots = + 4 k = 6
px + 3y – ( p – 3) = 0
2 6
4
12 x + py – p = 0 (1) 6
(2) –6
(3) 0
(4) 2
8
6
2 – 6 + 8 = 0
Answer (1)
( – 2) ( – 4) = 0
Sol.
= 4, 2 px + 3y – ( p – 3) = 0
For = 2, other roots are 3 and 4
12 x + py – p = 0
For = 4, other roots are not integral
For infinitely many solutions
Common root will be 2. 11
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
45. First term of an arithmetic progression is 2. If the sum of its first five terms is equal to one-fourth of the sum of the next five terms, then the sum of its first 30 terms is (1) 2670
(2) 2610
(3) –2520
(4) –2550
Let the side of square be a cm. Area of square = a2 PM is the altitude of equilateral PQR ∵ O is the centre of the circle
Sol. a = 2 (Given)
a
PM
According to the question
4
4
x
2d
1 (5a 35d ) 4
1 (a 7d ) 4
a
1 2 2d (a 7d ) 4
d
S30
n
2
2
(a 5d a 6d a 7d a 8d a 9d )
5a 10d
x
x
3a 4
6a
4 3 3
a
2 3
Area of PQR = [∵ a = 2]
3
6
4
3 a
2
4
2 x
2
Area of square = k (Area of equilateral )
[2a (n 1)d ]
a
30 [2 2 (30 1)( 6)] 2
3 3a 2 k 16
2
k
16 3 3
Option (1) is correct. 47. Let AP be a diameter of a circle of radius r and PT be the tangent to the circle at the point P such that the line AT intersects the circle at B. If PT = 8 units and BT = 4 units then r is equal to
46. A circle C is drawn inside a square S so that the four sides of S are tangents to C . An equilateral triangle T is drawn inside C with its vertices on C . If the area of S is k times the area of T , Then the value of k is 16
(1)
units
4 3
(2) 4 units
16
(2)
3 3
32
4
3
(3)
32
(4)
3 3
3
units
(4)
2 3
units
Answer(1)
3 A
Answer (1)
x
P
Sol.
r
S O
Sol. C
Option (4) is correct.
(3)
= –2550
(1)
[PO : OM = 2 : 1]
4
Let side of equilateral PQR be x cm 3
1
4
a
2
a
3a
1 (a6 a7 a8 a9 a10 ) a1 a2 a3 a4 a5 4 a + a + d + a + 2d + a + 3d + a + 4 d
2
OM
Answer (4)
a
PO
Q
M
r
B
4 u n i t s
R
T P
T
12
8 units
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
PT be the tangent and TA be the secant r
2
PT = TB TA
2
A
52 4
r
64 = 4 (4 + x ) x = 12 units
r
Now, APT is a right triangle, AP 2 + PT 2 = AT 2 2
2
(2r ) + (8) = (16)
10
cm
O a B 2
2
Area of ABCD =
2
1 2
BD2 A
(2r )2 = 256 – 64 4r 2 = 192 r = 48
=
4 3
B O
2
r =
a
1 2
(2r )2
5 cm2
48. If the quadratic equation x 2 + bx + 72 = 0 has two distinct integer roots, then the number of all possible values for b is (1) 12
(2) 9
(3) 15
(4) 18
C
D
units
50. If the discriminants of two quadratic equations are equal and the equations have a common root 1, then the other roots (1) Are either equal or their sum is 2
Answer (1)
(2) Have to be always equal
Sol. Given quadratic equation is x 2 + bx + 72 = 0
(3) Are either equal or their sum is 1
Let its roots be and
(4) Have their sum equal to 1
+ = – b and = 72
Answer (1)
Now, = 1 × 72 or (–1) × (–72)
Sol. Let other roots of two quadratic equation be and
= 2 × 36 or (–2) × (–36)
The quadratic equations would be
= 3 × 24 or (–3) × (–24) = 4 × 18 or (–4) × (–18) = 6× 12 or (–6) × (–12)
( + + 2)( – ) = 4( – )
49. If the area of a square inscribed in a semicircle is 2 cm2, then the area of the square inscribed in a full circle of the same radius is
( – )[ + – 2] = 0 Either, = or + = 2
2
(1) 5 cm
(2) 10 cm
cm2
Hence, option (1) is correct
(4) 25 cm2
51. Three circular wires are attached in series such that, if one wire is rotated, other two also get rotated. If
Answer (1)
4
Sol. Let the side of the square be a cm and radius of semi-circle be r cm Area = 2 cm2 = a2
the diameter of a wire is
...(i)
(OA)2 = (OB)2 + ( AB)2
r
5a
2
4
5
times that of immediate
left wire and the left most wire rotates at the speed of 32 revolutions per minute, then the number of revolutions made by right most wire per minute will be
In, OAB,
2
…(ii)
( + 1)2 – ( + 1)2 = 4( – )
So, option (1) is correct.
5 2
x 2 – ( + 1) x + = 0
( + 1)2 – 4 = ( + 1)2 – 4
Hence, total possible values of b will be 12
(3)
…(i)
Now, according to the question
= 8 × 9 or (–8) × (–9)
2
x 2 – ( + 1) x + = 1
[From equation ...(i)] 13
(1) 40
(2) 49
(3) 50
(4) 60
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
Answer (3)
53. Shyam wants to make a solid brick shape structure from 400 wooden cubes of unit volume each. If the sides of the solid brick have the ratio 1 : 2 : 3, then the maximum number of cubes, which can be used will be
Sol. 5r 4
4r 5
r
O1
r
O2
O3
According to the question, if the radius of right most circle is
(1) 400
(2) 288
(3) 300
(4) 384
Answer (4)
4r
, then the radius of the circle with centre
5
O2 and O1, will be r and
5r 4
Sol. Let the sides of solid brick be x , 2 x and 3 x respectively
Volume of solid = n × volume of unit cube
Let, the right most circle rotates x times per minute
6 x 3 = n × 1
Hence, 5r
32 2
4
x 2
4r
3 x
n
6
5
The only possible value of n = 384 satisfying the given condition
x = 50 revolutions per minute 52. Let ABC be an equilateral triangle. If the co-ordinates of A are (1, 2) and co-ordinates of B are (2, –1), then
The correct option is (4)
(1) C cannot lie in the first quadrant
54. Positive integers from 1 to 21 are arranged in 3 groups of 7 integers each, in some particular order. Then the highest possible mean of the medians of these 3 groups is
(2) C cannot lie in the second quadrant (3) C is the origin (4) C cannot lie in the third quadrant Answer (2)
(1) 16
(2) 12.5
(3) 11
(4) 14
A (1, 2) Answer (4) Sol. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 and 21
Sol. B (2, –1) AB
10
For getting maximum medians, the desired three groups must be
C (x, y ) units
1, 2, 3, 18, 19, 20, 21
and ( x – 2)2 + (y + 1) 2 = 10 2
2
( x – 1) + (y – 2) = 10
...(i)
4, 5, 6, 14, 15, 16, 17 and
...(ii)
7, 8, 9, 10, 11, 12, 13
From (i) and (ii) we get x = 3 y
...(iii)
Mean of the medians of the above groups
and using (iii) and (i), we get 2y 2 – 2y – 1 = 0 y
y
1
=
3
2 1 2
18 14 10 3
= 14
So, option (4) is correct. 3
or
1–
55. On dividing 2272 as well as 875 by a 3-digit number N , we get the same remainder in each case. The sum of the digits of N is
3 2
( x , y ) lies either in Ist or in IIIrd quadrant But cannot lie in IInd quadrant Option (2) is correct 14
(1) 10
(2) 11
(3) 12
(4) 13
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
4 x + 3y = 0
Answer (1) Sol.
∵
Point (, –)
A 3-digit number gives the same remainder on dividing 2272 as well as 875
4 – 3 = 0
3-digit number must be a factor of
From (i) & (ii) we get
(2272 – 875) = 1397 = 11 × 127
Required 3-digit number is 127 Sum of all the digits of 127
= 1 + 2 + 7 = 10 Option (1) is correct.
5
. The co-ordinates of the point, which lies
in the fourth quadrant at a unit distance from the origin and on perpendicular to l, are
3 5
4
(1) , –
4 5
(2) , –
5
(3) (3, –4)
sin
3
4 3 ( , – ) , – 5 5
(1) 1
(2) 5
(3) 1, 9
(4) 9
Since p( x ) is exactly divisible by ( x + 2)
5
p(–2) = 4 – 10k + k 2 + 5
(4) (4, –3)
k 2 – 10k + 9 = 0 k = 1 or k = 9 When k = 1. p( x ) = x 2 + 5 x + 6
5
y
5
Sol. Let p( x ) be x 2 + 5 kx + k 2 + 5
3
3
Answer (4)
Answer (1) Sol.
5
,
57. The value(s) of k for which x 2 + 5kx + k 2 + 5 is exactly divisible by x + 2 but not by x + 3 is(are)
3
4
Hence option (1) is correct.
56. A line l passing through the origin makes an angle with positive direction of x - axis such that sin
...(ii)
tan
3
= ( x + 2) ( x + 3) l
For k =1, polynomial p( x ) is also divisible by ( x + 3)
4
k 1 y = mx
x
y
O
3 4
The value of k is 9
x ( , – )
x
Hence, option (4) is correct. 58. If cos4 + sin2 = m, then
4y = 3 x
y
(1) 1 m 2
3 x – 4y = 0 Point lies in 4th quadrant
(3)
Point will be (, –)
3 4
(4)
Answer (3)
3 + 4 = 0
Sol. Given cos4 + sin2 = m
4 5
2
m 1
3
m 1
Two lines are perpendicular
3
1
(2)
4
m
13 16
cos4 + sin2 = m 1
3 + 4 = 5
m = cos4 + 1 – cos2 ...(i)
m=
cos
4
– cos
2
Equation of 2nd line will be
1 4
–
1 4
1
[using perfect square method] y
–
4 3
x
2
1 2 m = cos – 2
3y = – 4 x 15
3 4
NTSE (Stage-II) - 2017 Scholastic Aptitude Test ∵
0 cos 2 1 1
–
2
cos2 –
0
cos2
3
cos2
4
3 4
Sol.
–
–
1 2
1
1 2
2
2 1
45 m
A
1 4
2
2
E
3 4
1 4
3
F
D
4
2 0 m
(90° – )
B
m1
C
Hence, option (3) is correct. Let two vertical poles be AE and BF are of 45 m and 20 m heights respectively erected at A and B.
59. Cost of 2 apples, 3 bananas and one coconut is 26. Also the cost of 3 apples, 2 bananas and two coconuts is ` 35. Then the cost of 12 apples, 13
Now, let AD = x and BD = y .
bananas and 7 coconuts is
BC = AC = x + y
`
(1)
` 172
(2)
` 148
(3)
` 143
(4)
` 126
[∵ ABC is an equilateral triangle]
In AED and BFD,
Answer (2)
45
Sol. Let the cost of 1 apple, 1 banana and 1 coconut be A, B and C respectively.
x
According to the question, ...(i)
3 A + 2B + 2C = 35
...(ii)
tan
y
2 A + 3B + C = 26
20
4 x
y
...(i)
9
In AEC and BFC
Multiplying equation (i) by 3 and (ii) by 2,
20
and add them, we get
x
y
tan
12 A + 13B + 7C = 148 Option (2) is correct
and
60. ABC is a field in the form of an equilateral triangle. Two vertical poles of heights 45 m and 20 m are erected at A and B respectively. The angles of elevation of the tops of the two poles from C are complementary to each other. There is a point D on AB such that from it, the angles of elevation of the tops of the two poles are equal. Then AD is equal to (1)
(3)
17
5
m
12
20
5 13
m
(2)
(4)
20
10
17
y
x
20 x
y
tan(90 – )
45
x
y
tan tan(90 – )
=1 x + y = 30 From (i) and (ii), we get
m
13
10
45
x
m
20
10
m
13
12
So, option (2) is correct
Answer (2) 16
...(ii)
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
61. Arrange the developments related to European history in a chronological sequence. I.
Napoleon invaded Italy.
II.
Unification of Italy.
Direction (Questions 65-72) Read the statements and select the correct answer from the options given below.
III. Unification of Germany. IV. Vienna Settlement.
1.
Statements I is true, Statement II is false.
2.
Statement I is false, Statement II is true.
3. Both Statements are true, and Statement II provides explanation to Statement I.
(1) I, III, II and IV (2) I, II, IV and III
4.
(3) I, IV, II and III
Both Statements are true, but Statement II does not provide explanation to Statement I.
(4) I, II, III and IV 65. Statement I : During the years of the Great Depression the economic crisis was worse in Germany.
Answer (3) 62. Which of the following statements about Liberals in 19th century Europe are correct? I.
They favoured the Catholic Church.
II.
They opposed dynastic rule with unlimited power.
Statement II : The President of the Weimar Republic had the power to impose emergency. Answer (4) 66. Statement I : The Forest Act of 1878 categorized some forests as reserved forests'.
III. They were democrats. IV. They did not want any voting rights for women. (1) I, II and III
(2) I, II and IV
(3) II and IV
(4) III and IV
Statement II : They were considered the best forests for people's use.
Answer (3)
Answer (1)
63. Which of the following statements are correct? I.
In the beginning Bombay was under the Portuguese control.
67. Statement I : Shifting cultivation was widely prevalent in different parts of India in the 19th century.
II.
Control of Bombay passed onto the French in the 17th century.
Statement II : More and more people took to shifting cultivation when forest laws were enacted.
III. The Marathas replaced the French in Bombay.
Answer (1)
IV. Bombay became the capital of the Presidency in early 19th century.
68. Statement I : Cricket emerged as a colonial game.
(1) I, II and IV
(2) I and IV
(3) I, II and III
(4) II, III and IV
Statement II : Cricket was started in England. Answer (3)
Answer (2)
69. Statement I : Mahatma Gandhi wished everyone had clothes to wear.
64. Which of the following statements are correct? I.
The Chinese introduced printing
II.
The Buddist missionaries introduced printing in Japan.
Statement II : He wanted everyone to wear the single Ioin cloth as he did. Answer (2)
III. The Chinese developed printing to facilitate their expanding trade. IV. Printing reached Europe through Italy. (1) I, II and III
(2) I, II and IV
(3) II, III and IV
(4) I and IV
70. Statement I : The Spanish conquest of America was not a conventional military conquest. Statement II : One of the most powerful weapon was the spread of smallpox. Answer (3)
Answer (2) 17
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
Identify the state with population growth rate marked by 'X' in the given graph.
71. Statement I : The silk routes led to trade and cultural links between distant parts of the world. Statement II : Early Christian missionaries travelled to Asia through this route.
(1) Goa
(2) Kerala
(3) Sikkim
(4) Nagaland
Answer (4)
Answer (2)
72. Statement I : The French used forced labour in IndoChina for building canals.
76. River Alaknanda forms confluences (Prayags) in Uttarakhand. Match the codes given in Figure with Table (Prayags) and select the correct answer using the code given below.
Statement II : Vietnam became a major exporter of rice in the world.
Figure
Answer (4) 73. Match List I (Layers of Atmosphere) and List II (Characteristics) and select the correct answer using the code given below. List I (Layers of Atmosphere) A. Ionosphere Stratosphere B. Exosphere C. Troposphere D.
A
A
IV.
L
A K
N
A
C
I.
Karn Prayag
II.
Rudra Prayag
III.
Nand Prayag
IV.
Vishnu Prayag
B
A
Contains Ozone Reflects radio Waves Fall in Temperature Extremely low air density
II. III.
D
D N
List II (Characteristics) I.
Table (Prayags)
(1) A-II, B-I, C-III, D-IV
(2) A-II, B-III, C-I, D-IV
(3) A-III, B-II, C-I, D-IV
(4) A-III, B-I, C-II, D-IV
(1) A-II, B-III, C-IV, D-I
(2) A-II, B-I, C-IV, D-III
Answer (1)
(3) A-II, B-III, C-I, D-IV
(4) A-III, B-I, C-IV, D-II
77. Match List-I (Original Rock) with List-II (Metamorphic Rock) and select the correct answer using the code given below:
Answer (2) 74. Which of the following statements are correct? I.
Rann of Kachchh is formed by the recession of the sea
II.
Kuchaman, Sambhar and Didwana are salt water lakes
III. The land to the east of Aravallis is known as Bagar IV. The fertile flood plains formed by small streams in Rajasthan are known as Rohi
List-I (Original Rock)
List-II (Metamorphic Rock)
A.
Granite
I.
Diamond
B.
Coal
II.
Marble
C.
Limestone
III.
Slate
D.
Shale
IV.
Gneiss
(1) A-III, B-IV, C-II, D-I
(2) A-III, B-II, C-IV, D-I
(3) A-IV, B-II, C-I, D-III
(4) A-IV, B-I, C-II, D-III
Answer (4)
(1) I, II and IV
(2) I, III and IV
(3) II, III and IV
(4) I, II, III and IV
78. Observe the given map.
Answer (1)
INDIA NORTH EASTERN REGION
75. Observe the graph given below: Growth of Population 1951-2011
40 % n i e t a R h t w o r G l a d a c e D
India X
35 30
24.76
25
24.66 23.87 21.54
24.8
20 15
26.29
17.64 19.24
21.64
14.32
10
9.43
5
4.86
0 1 6 9 1 1 5 9 1
1 7 9 1 1 6 9 1
1 8 9 1 1 7 9 1
1 9 9 1 1 8 9 1
1 0 0 2 1 9 9 1
1 1 0 2 1 0 0 2
Which one of the following statement is NOT true about the shaded state indicated on the map? 18
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
(1) Society predominantly follows right of female ultimogeniture
81. Match List-I (Industries) with List-II (Important Centers) and select the correct answer using the codes given below.
(2) The state is an example of areas with karst topography
List-I (Industries)
(3) The state is a major producer of potatoes in India
A.
Cotton textile
I.
Ludhiana
B.
Hosiery
II.
Rishra
C.
Jute
III.
Coimbatore
D.
Silk textile
IV.
Mysuru
(4) Some parts of the state receive extremely high rainfall Answer (3) 79. Match List-I (Mineral Oil Refineries) with List-II (States) and select the correct answer using the codes given below.
List-I (Mineral Oil Refineries) Numaligarh
I.
Punjab
B.
Bathinda
II.
Andhra Pradesh
C. Tatipaka
III. Madhya Pradesh
D. Bina
IV. Assam
(1) A-I, B-III, C-IV, D-II
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-II, C-I, D-IV
(4) A-III, B-I, C-II, D-IV
Answer (4)
List-II (States)
A.
List-II (Important Centers)
82. Which one of the following island is closest to the equator? (1) Minicoy
(2) Car Nicobar
(3) Little Nicobar
(4) Great Nicobar
Answer (4)
(1) A-IV, B-II, C-III, D-I
83. Which of the following characteristics are true about plantation agriculture?
(2) A-IV, B-I, C-II, D-III (3) A-II, B-I, C-IV, D-III (4) A-IV, B-III, C-II, D-I
I.
Generally plantation agriculture is considered as an example of subsistence farming.
II.
Generally single crop is grown on a large area in plantation agriculture.
Answer (2) 80. ‘Slash and Burn Agriculture’ is known by specific name in different states of India. Match the shades states marked in the given map with codes given in the Table (Different names of Slash and Burn Agriculture) and select the correct answer using the code given below.
III. It has an interface of agriculture and industry. IV. It uses capital intensive inputs. (1) I and IV
(2) III and IV
(3) I, II and III
(4) II, III and IV
Answer (4)
Table (Different Names of Slash and Burn Agriculture)
84. Match List-I (Vegetation zones) with List-II (Mean Annual Temperature Range) and select the correct answer using the code given below.
I. Bringa
List-I (Vegetation Z ones)
II. Waltre III. Dahiya
A. Alpine
IV. Kuruwa
List-II (Mean Annual Temperature Range) I.
Above 24°C
B.
Temperate
II.
17°C to 24°C
(1) A-III, B-IV, C-II, D-I
C.
Tropical
III.
Below 7°C
(2) A-III, B-II, C-IV, D-I
D.
Sub-tropical
IV.
7°C to 17°C
(3) A-I, B-IV, C-II, D-III
(1) A-III, B-I, C-II, D-IV
(2) A-III, B-I, C-IV, D-II
(4) A-I, B-II, C-IV, D-III
(3) A-III, B-IV, C-I, D-II
(4) A-I, B-II, C-III, D-IV
Answer (4)
Answer (3) 19
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
85. 'In a democracy, the will of the people is supreme.' Which of the following statement concerning democracy in India best reflects this?
(1) A-IV, B-I, C-II, D-V (2) A-IV, B-I, C-II, D-III (3) A-V, B-IV, C-II, D-III
(1) The President appoints the Prime Minister who is the leader of the political party possessing a majority in the Lok Sabha.
(4) A-V, B-II, C-III, D-IV Answer (1)
(2) An assembly of elected representatives exercises political authority on behalf of the people.
88. Which of the following statements about the federal system in India are true?
(3) In case of a difference between the two Houses of Parliament, the final decision is taken in a joint session of the two Houses.
I.
The Constitution of India provides for a three-fold distribution of legislative powers between the Union and the State Governments.
(4) The permanent executive has more powers than the political executive.
II.
Both the Union and the State Governments can legislate on residuary subjects.
Answer (2)
III. The Parliament cannot on its own change the power-sharing arrangement between the Union and the State Governments.
86. Which of the following statements about the Panchayati Raj Institutions after the Constitutional Amendment in 1992 are false? I.
IV. The High Courts have no role in resolving disputes about the division of powers between the Union and the State Governments.
Seats are reserved for the Scheduled Castes, Scheduled Tribes, and Other Backward Classes in the elected bodies of the Panchayati Raj Institutions.
II.
(1) I and III (2) II, III and IV
Elections to the Panchayati Raj Institutions are supervised by the Election Commission of India.
(3) I, III and IV
III. Elections to the Panchayati Raj Institutions are held regularly after every five years.
(4) I, II and IV Answer (3)
IV. Half of the seats in all the States are reserved for women. (1) I and III
(2) I and II
(3) III and IV
(4) II and IV
89. Which of the following group of States/Union Territories have only one Lok Sabha constituency? (1) Arunachal Pradesh, Sikkim, Lakshadweep
Answer (4)
(2) Goa, Meghalaya, Andaman and Nicobar Islands
87. Match List I (Political Systems) with List II (Nations) and select the answer using the codes given below. List I (Political Systems)
(3) Chandigarh, Sikkim, Mizoram (4) Manipur, Dadra and Nagar Haveli, Puducherry
List II (Nations)
A. Federal, Presidential, Republic
I.
B. Federal, Parliamentary, Republic
II. United Kingdom
C. Unitary, Parliamentary, Monarchy
III. Germany
D. Presidential cum Parliamentary, Republic
IV. United States of America
Answer (3)
India
90. Which of the following statements best reflects the 'socialist' feature of the Preamble to the Constitution of India? (1) There are no unreasonable restrictions on how the citizens express their thoughts. (2) The traditional social inequalities have to be abolished. (3) Government should regulate the ownership of land and industry to reduce socio-economic inequalities. (4) No one should treat a fellow citizen as inferior.
V. France
Answer (3) 20
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
91. Which of the following statements about the Indian judiciary is true?
96. Match List-I (Type of Unemployment) with List-II (Characteristics) and select the correct answer using the codes given below
(1) India has an integrated judiciary.
List-I (Type of Unemployment)
(2) The Judiciary in India is subordinate to the Executive.
List-II (Characteristics)
(3) The Supreme Court is more powerful than Parliament.
A. Seasonal
I.
(4) The Chief Justice of India is appointed by the Prime Minister.
B. Frictional
II. An absence of demand for a certain type of workers
Answer (1)
C. Disguised
92. Which of the following Fundamental Rights includes the Right to Education?
III. Occurs when moving from one job to another
D. Structural
IV. Actual contribution by the additional labour is nil
E. Cyclical
V. Job opportunities during certain months in the year
(1) Right to Equality (2) Right to Freedom (3) Cultural and Educational Rights
Oc cu rs dur in g bo om or recession in the economy
(1) A(V), B(III), C(IV), D(II), E(I)
(4) Right to Constitutional Remedies
(2) A(IV), B(V), C(III), D(I), E(II)
Answer (2) 93. Which of the following is NOT an indicator of economic development?
(3) A(I), B(II), C(III), D(IV), E(V) (4) A(V), B(IV), C(III), D(II), E(I)
(1) Increased per capita income
Answer (1)
(2) Decreased infant mortality
97. Suppose Indian Farmers sell wheat at
(3) Increased life expectancy at birth (4) Decreased women participation in job market Answer (4) 94. The poverty line in Dinanagar is set at ` 100 per capita per day. Five Hundred people live in Dinanagar of whom 50 earn ` 30 per capita per day and another 25 earn ` 80 per capita per day each. Everybody else earn more than ` 100 per day per capita. What is the minimum amount that the government of Dinanagar will have to spend to completely eradicate poverty? (1)
` 3000
(2)
` 3500
(3)
` 4000
(4)
` 4500
95. The local telephone company sells me a landline connection only if I purchase a handset from them as well. Which of the following rights does this practice violate under the Consumer Protection Act 1986?
50 per kg
and the international price of wheat is ` 40 per kg. What is the minimum rate of import duty Government of India must impose on imported wheat so that it does not adversely affect Indian farmers in the domestic market? (1) 10%
(2) 20%
(3) 25%
(4) 30%
Answer (3) 98. The wage rate of a worker in a country is ` 300 per day. Which of these person(s) would you consider unemployed? A. Ramu is willing to work at cannot find work.
Answer (3)
`
`
300 a day, but
B. Suresh is willing to work only at more, and cannot find work.
`
400 a day or
C. Shanti stays at home because she has young children to look after.
(1) Right to represent
(1) Ramu
(2) Right to information
(2) Suresh
(3) Right to choose
(3) Ramu and Suresh
(4) Right to seek redressal
(4) Ramu and Shanti
Answer (3)
Answer (1) 21
NTSE (Stage-II) - 2017 Scholastic Aptitude Test
99. Which of the following can be used as collateral in Indian banks to borrow money?
Number of Families
Total amount of land owned and operated by each group (in hectares)
A
100
300
B
180
300
Answer (3)
C
30
300
100. The total agricultural land in a village is 1200 hectares. This is distributed among 320 families who form four groups in the following pattern. It is assumed that the land is distributed equally within each group. Identify the group of small farmers.
D
10
300
(1) Bank Passbook
(2) Credit Card
(3) Own House
(4) Passport
Group
(1) A
(2) B
(3) C
(4) D
Answer (2)
22
