KVPY Class-XI (SA) - 19-11-2017

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KVPY Class-XI (SA) - 19-11-2017

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Weekly Quiz Competition

Question 1
1 / -0

The number of positive integers n in the set {2, 3,...., 200} such that 1/n has a terminating decimal expansion is

A
16

B
18

C
40

D
100

Solution

The numbers will be 2, 4, 8, 16, 32, 64, 128, 5, 25, 125, 10, 20, 40, 50, 80, 100, 160 and 200.
Question 2
1 / -0

If a, b and c are real numbers, such that a + b + c = 0 and a² + b² + c² = 1, then (3a + 5b - 8c)² + (-8a + 3b + 5c)² + (5a - 8b + 3c)² is equal to

A
49

B
98

C
147

D
294

Solution

Expanding, we get
98(a² + b² + c²) - 98(ab + bc + ca)
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(ab + bc + ca) = -1/2
98(a² + b² + c²) - 98(ab + bc + ca)
= 98(1) - 98 (- 1/2)
= 147
Question 3
1 / -0

Let S be a set of all ordered pairs (x, y) of positive integers, with HCF(x, y) = 16 and LCM(x, y) = 48,000. The number of elements in S is

A
4

B
8

C
16

D
32

Solution

48000 = 16 x 3000
= 16 x [3¹ x 2³ x 5³]
As the HCF is 16, 2³ can be selected in 1 way and 3¹ x 5³ can be selected in (1 + 1)(3 + 1) = 8 ways.
Number of ordered pairs = 8
Question 4
1 / -0

Consider a set A of natural numbers n whose units digit is non-zero, such that if this units digit is erased, then the resulting number divides n. If K is the number of elements in the set A, then

A
K is infinite

B
K is finite, but K > 100

C
25 ≤ K ≤ 100

D
K < 25

Solution

Such numbers are 9 from 11 to 19.
4 i.e. (22, 24, 26, 28)
3 i.e. (33, 36, 39)
2 i.e. (44, 48)
5 i.e. (55, 66, 77, 88, 99)
Total number of elements = 23
K < 25
Question 5
1 / -0

There are exactly twelve Sundays in the period from January 1 to March 31 in a certain year. Then, the day corresponding to February 15 in that year is

A
Tuesday

B
Wednesday

C
Thursday

D
Cannot be determined

Solution

Obviously, 1st January will be Monday as there will be 90 days from January 1 to March 31 (Non-leap year)
If year is a leap year, then days will be 91 = 13 weeks, which is not possible.
∴ 15^th February will be a Thursday.
Question 6
1 / -0

Consider a three-digit number with the following properties.

I. If its digits in units place and tens place are interchanged, then the number increases by 36.
II. If its digits in units place and hundreds place are interchanged, then the number decreases by 198.

Suppose that the digits in tens place and hundreds place are interchanged. Then, the number

A
increases by 180

B
decreases by 270

C
increases by 360

D
decreases by 540

Solution

Let the three digit number be 100a + 10b + c
Given, 100a + 10b + c = 100a + 10c + b - 36
9b - 9c + 36 = 0
c = b + 4
b = c - 4 ...(i)
Also, 100a + 10b + c = 100c + 10b + a + 198
99a - 99c = 198
a = c + 2 ...(ii)
Now, 100a + 10b + c - (100b + 10a + c)
= 90(a - b)
= 90(c + 2 - c + 4) (using (i) and (ii))
= 540
∴ Value decreases by 540.
Question 7
1 / -0

The particles used in the Rutherford's scattering experiment to deduce the structure of atoms

A
had atomic number 2 and were fully ionised

B
had atomic number 2 and were neutral

C
had atomic number 4 and were fully ionised

D
had atomic number 4 and were neutral

Solution

α-particles (ionised helium)
Question 8
1 / -0

The number of completely filled shells for the element ₁₆S³² is

A
1

B
2

C
3

D
4

Solution

The electronic configuration of sulphur is 2, 8, 6.
Question 9
1 / -0

The refractive index of water in a biology laboratory tank varies as 1.33 + 0.002/λ², where λ is the wavelength of light. Small pieces of organic matter of different colours are seen at the bottom of the tank using a travelling microscope. Then, the image of the organic matter appears

A
deeper for the violet pieces than the green ones

B
shallower for the blue pieces than the orange ones

C
at the same depth for both the blue and orange pieces

D
deeper for the green pieces than the red ones

Solution

Since refractive index is inversely proportional to λ, the refractive index for blue is more than that for orange. So, blue light will get more refracted after leaving water; hence it will appear shallower when seen from above.
Question 10
1 / -0

The acidity of compounds I-IV in water

I. Ethanol
II. Acetic acid
III. Phenol
IV. Acetonitrile

follows the order:

A
IV < I < III < II

B
I < II < III < IV

C
IV < I < II < III

D
IV < III < I < II

Solution

Acetic acid is the most acidic due to equivalent resonating structures of the carboxylate ion formed.
Phenol is more acidic as compared to alcohols as the phenoxide ion is more stable (due to resonance) than the alkoxide ion formed.
Question 11
1 / -0

The reddish-brown precipitate formed in the Fehling's test for aldehydes (RCHO) is due to the formation of

A
Cu

B
Cu₂O

C
CuO

D
(RCOO)₂Cu

Solution

The reddish-brown precipitate formed in the Fehling's test for aldehydes (RCHO) is due to the formation of cuprous oxide.
RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O + 3H₂O
Question 12
1 / -0

White phosphorus catches fire in air to produce dense white fumes. This is due to the formation of

A
P₄O₁₀

B
PH₃

C
H₃PO₃

D
H₃PO₂

Solution

4P + 5O₂ → P₄O₁₀
Question 13
1 / -0

The maximum number of electrons that can be filled in the shell with the principal quantum number n = 4 is

A
64

B
26

C
18

D
32

Solution

4s, 4p, 4d and 4f contain total 32 electrons.
Question 14
1 / -0

The oxidation states of P atom in POCl₃, H₂PO₃, and H₄P₂O₆ respectively are

A
+5, +4, +4

B
+5, +5, +4

C
+4, +4, +5

D
+3, +4, +5

Solution

Let oxidation state of P = x

POCl₃
x + (-2) + 3(-1) = 0
x = +5

H₂PO₃
2(+1) + x + 3(-2) = 0
x = +4

H₄P₂O₆
4(+1) + 2x + 6(-2) = 0
x = +4
Question 15
1 / -0

A solution (5 mL) of an acid X is completely neutralised by y mL of 1 M NaOH. The same volume (y mL) of 1 M NaOH is required to neutralise 10 mL of 0.6 M of H₂SO₄ completely. The normality (N) of the acid X is

A
1.2

B
2.4

C
4.8

D
0.6

Solution

(Number of eq.)_NaOH = (Number of eq.) H₂SO₄
⇒(1 × 1) × y = (0.6 × 2) × 10
⇒y = 12 ml

Now, (Number of eq.)_acid = (Number of eq.)_NaOH
⇒ N × 5 = (1 × 1) × 12
⇒ y = 12 ml

Now, (Number of eq.)_acid = (Number of eq.)_NaOH
⇒N × 5 = (1×1) ×12
⇒ N = 12/5 = 2.4
Question 16
1 / -0

According to Watson-Crick model, hydrogen bonding in a double-stranded DNA occurs between

A
adenine and guanine

B
adenine and thymine

C
cytosine and adenine

D
guanine and thymine

Solution

According to Watson-Crick model, hydrogen bonding in a double-stranded DNA occurs between adenine and thymine. Adenine pairs with thymine and guanine pairs with cytosine in DNA.
Question 17
1 / -0

Which of the following statements about mitosis is CORRECT?

A
One nucleus gives rise to 4 nuclei.

B
Homologous chromosomes synapse during anaphase.

C
The centromeres separate at the onset of anaphase.

D
Non-sister chromatids recombine.

Solution

In anaphase, sister chromatids separate from centromeres, so the number of chromosomes becomes double.
Question 18
1 / -0

Gaseous exchange of oxygen and carbon dioxide between alveolar air and capillaries takes place by

A
active transport

B
diffusion

C
carrier-mediated transport

D
imbibition

Solution

Gaseous exchange of oxygen and carbon dioxide between alveolar air and capillaries takes place by diffusion along the concentration gradient.
Question 19
1 / -0

Of the periods listed below, which is the earliest when ostracoderms, the jawless and finless fishes, appeared?

A
Devonian Period

B
Cambrian Period

C
Carboniferous Period

D
Silurian Period

Solution

Ostracoderms are the jawless and finless fishes. They belonged to Silurian period of Paleozoic era.
Question 20
1 / -0

Scurvy is caused by the deficiency of

A
nicotinic acid

B
ascorbic acid

C
pantothenic acid

D
retinoic acid

Solution

Ascorbic acid is required for a variety of biosynthetic pathways. It is required for collagen synthesis during wound healing.
Question 21
1 / -0

Optical activity of DNA is due to its

A
bases

B
sugars

C
phosphate

D
hydrogen bonds

Solution

DNA polymer is made up of nitrogenous base, a sugar, and one or more phosphates. Optical activity results due to the molecular asymmetry. The nucleic acid bases have a plane of symmetry. Hence, they do not induce optical activity. Sugars are asymmetric and cause optical activity of DNA.
Question 22
1 / -0

The monarch butterfly avoids predators such as birds by

A
changing colour frequently

B
flying away from the predator swiftly

C
producing a chemical obnoxious to the predator

D
producing ultrasonic waves

Solution

A prey may have some defence mechanism, like producing a toxic substance, to protect itself from predators.
Question 23
1 / -0

Filariasis is caused by

A
Entamoeba histolytica

B
Plasmodium falciparum

C
Tiypanosoma brucel

D
Wuchereria bancrofti

Solution

Wuchereria bancrofti lives in lymphatic vessels and causes swelling of lower limbs and scrotum, known as filariasis.
Question 24
1 / -0

Which of the following conversions does NOT happen under anaerobic conditions?

A
Glucose to ethanol by Saccharomyces

B
Lactose to lactic acid by Lactobacillus

C
Glucose to CO₂ and H₂O by Saccharomyces

D
Cellulose to glucose by Cellulomonas

Solution

Conversion of glucose to CO₂ and H₂O takes place during aerobic respiration.
Question 25
1 / -0

The number of electrons required to reduce one molecule of oxygen to water during mitochondrial oxidation is

A
4

B
3

C
2

D
1

Solution

O₂ + 4e^- + 4H⁺ → 2H₂O
Question 26
1 / -0

Which of the following molecules is derived from pantothenic acid?

A
Thiamine pyrophosphate

B
Nicotinamide adenine dinucleotide phosphate

C
Flavin adenine dinucleotide phosphate

D
Acetyl-CoA

Solution

Vitamin B₅ is the pantothenic acid that synthesises Co-enzyme A(CoA).
Question 27
1 / -0

Match the disease given in Column I with the principal causal organism in Column II and choose the correct combination:

Column I Column II

(P) AIDS (i) HBV

(Q) Syphilis (ii) Neisseria sp.

(R) Viral hepatitis (iii) Treponema sp.

(S) Gonorrhoea (iv) HIV

A
P - iv, Q - iii, R - i, S - ii

B
P - iv, Q - ii, R - i, S - iii

C
P - i, Q - ii, R - iv, S - iii

D
P - i, Q - iv, R - ii, S - iii
Question 28
1 / -0

Which of the following options lists the primary energy sources for all forms of life on Earth?

A
Light, Inorganic substances

B
Inorganic substances, Organic substances

C
Light, Organic substances

D
N₂, CO₂

Solution

Autotrophs use light for photosynthesis and some bacteria utilise inorganic compounds for chemosynthesis. These organisms are producers in the ecosystem.
Question 29
1 / -0

In an in vitro translation experiment, poly (UC) RNA template produced poly (Ser-Leu), while poly (AG) RNA template produced poly (Arg-Glu) polypeptide. Which of the following options represents correct interpretations of the codon assignments for Ser, Leu, Arg, and Glu?

A
Ser - UCU, Leu - CUC, Arg - AGA, Glu - GAG

B
Ser - CUC, Leu - GAG, Arg - UCU, Glu - AGA

C
Ser - AGA, Leu - UCU, Arg - GAG, Glu - CUC

D
Ser - GAG, Leu - AGA, Arg - CUC, Glu - UCU

Solution

Ser - UCU, Leu - CUC, Arg - AGA, Glu - GAG
Sequence of 3 nitrogenous bases is one codon.
Question 30
1 / -0

A single bacterium is actively growing in a medium that supports its growth to a number of 100 million. Assuming the division time of the bacterium as 3 hours and the life span of non-dividing bacteria as 5 hours, which ONE of the following represents the maximum number of bacteria that would be present at the end of 15 hours?

A
10

B
64

C
24

D
32

Solution

Time = 15/3 = 5 times division occurs.
Number of bacteria = 2⁵ = 32
Question 31
1 / -0

A couple has two sons and two daughters. Only one son is colour blind and the rest of the siblings are normal. Assuming colour blindness is sex-linked, which ONE of the following would be the phenotype of the parents?

A
Mother would be colour blind, father would be normal.

B
Father would be colour blind, mother would be normal.

C
Both the parents would be normal.

D
Both the parents would be colour blind.

Solution

Male child receives X-chromosome from mother only. The other normal son indicates that mother is the carrier. So, phenotypically both the parents will be normal.

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Re-Attempt Weekly Quiz Competition

Column I	Column II
(P) AIDS	(i) HBV
(Q) Syphilis	(ii) Neisseria sp.
(R) Viral hepatitis	(iii) Treponema sp.
(S) Gonorrhoea	(iv) HIV

KVPY Class-XI (SA) - 19-11-2017

Result

KVPY Class-XI (SA) - 19-11-2017

Score

-

Rank

-

SHARING IS CARING

Question 1

16

18

40

100

Solution

Question 2

49

98

147

294

Solution

Question 3

4

8

16

32

Solution

Question 4

K is infinite

K is finite, but K > 100

25 ≤ K ≤ 100

K < 25

Solution

Question 5

Tuesday

Wednesday

Thursday

Cannot be determined

Solution

Question 6

increases by 180

decreases by 270

increases by 360

decreases by 540

Solution

Question 7

had atomic number 2 and were fully ionised

had atomic number 2 and were neutral

had atomic number 4 and were fully ionised

had atomic number 4 and were neutral

Solution

Question 8

1

2

3

4

Solution

Question 9

deeper for the violet pieces than the green ones

shallower for the blue pieces than the orange ones

at the same depth for both the blue and orange pieces

deeper for the green pieces than the red ones

Solution

Question 10

IV < I < III < II

I < II < III < IV

IV < I < II < III

IV < III < I < II

Solution

Question 11

Cu

Cu2O

CuO

(RCOO)2Cu

Solution

Question 12

P4O10

PH3

H3PO3

H3PO2

Solution

Cu₂O

(RCOO)₂Cu

P₄O₁₀

PH₃

H₃PO₃

H₃PO₂

Glucose to CO₂ and H₂O by Saccharomyces