KVPY (SA) Mock Test - 4

Total = Average x No. of terms
Here, total sum of x + y should be equal to 8.
Possible pairs of x and y are (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0)

Hence, total 9 pairs satisfy the condition.

Suppose number of students in a class=100

So, the number of students who like cricket = n(c) = 71

Number of students who like football = n(f) = 76, and

Number of students who like tennis = n(t) = 82

From the above three, we can find the number of students who at least like two games

n(C ∪ F) = 56, n(F ∪ T) = 58 and n(T ∪ C) = 53

So, n(C ∩ F ∩ T) = n(C ∪ F ∪ T) - n(C) - n(F) - n(T) + n(C ∩ F) + n(F ∩ T) + n(T ∩ C)

⇒ n(C ∩ F ∩ T) = 100 - 71 - 76 - 82 + 56 + 58 + 53 = 38

Hence, all the three sports are liked by at least 38% students.

Relative speed = 54 – 46 = 8 kmph
Relative speed = 8 x 5/18 = 20/9 m/s
Distance to be covered = 210 + 270 = 480 m
Time = 480 x 9/20 = 216 seconds

T₂₀ = T₁₉ + (T₂₀ with largest side 20)
Largest side is 20 and another side is a and b.

∴ a + b > 20

Suppose a = 2, then the number of possible values of b is 1, b has only b = 19
and for a = 3, the number of possible values of b is 2, and b = 18, 19
for a = 4, the number of possible values of b is 3, and b = 17, 18, 19
for a = 5, the number of possible values of b is 4, and b = 16, 17, 18, 19
for a = 6, the number of possible values of b is 5, and b = 15, 16, 17, 18, 19
for a = 7, the number of possible values of b is 6, and b = 14, 15, 16, 17, 18, 19
for a = 8, the number of possible values of b is 7, and b = 13, 14, 15, 16, 17, 18, 19
for a = 9, the number of possible values of b is 8, and b = 12, 13, 14, 15, 16, 17, 18, 19
for a = 10, the number of possible values of b is 9, and b = 11, 12, 13, 14, 15, 16, 17, 18, 19
for a = 11, the number of possible values of b is 8, and b = 12, 13, 14, 15, 16, 17, 18, 19
for a = 12, the number of possible values of b is 7, and b = 13, 14, 15, 16, 17, 18, 19
for a = 13, the number of possible values of b is 6, and b = 14, 15, 16, 17, 18, 19
for a = 14, the number of possible values of b is 5, and b = 15, 16, 17, 18, 19
for a = 15, the number of possible values of b is 4, and b = 16, 17, 18, 19
for a = 16, the number of possible values of b is 3, and b = 17, 18, 19
for a = 17, the number of possible values of b is 2, and b = 18, 19
for a = 18, the number of possible values of b is 1, and b = 19
Hence, the number of possible triangles with largest side = 2 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) + 9 = 81
Hence, T₂₀ - T₁₉ = 81

A and B can do 1/25 ^th of the work in 1 day.
B and C can do 1/30 ^th of the work in 1 day.
From this, we cannot say anything about A and C.

(x² + 4x)² - (x + 2)² - 26 = 0
(x² + 4x + 4 - 4)² - (x + 2)² - 26 = 0
((x + 2)² - 4)² - (x + 2)² - 26 = 0
Putting (x + 2)² = y, we get
(y - 4)² - y - 26 = 0
y² - 8y + 16 - y - 26 = 0
y² - 9y - 10 = 0
y = 10, - 1
y can't be negative as y = (x + 2)².
So, (x + 2)² = 10
From this, we will get the two values of x. Hence, the given equation has 2 real roots.

Let the smaller angle be x.
Larger angle = x + 44^o
We know that, the sum of two supplementary angles is 180^o.
Therefore, x + x + 44^o = 180^o
2x = 180^o - 44^o = 136^o
x = 68^o
Smaller angle = 68^o
Larger angle = 68^o + 44^o = 112^o

Let f(x) = 3x¹⁰ + 7x⁶ + 5x⁴ + 2x² + 1
Here, f(x) and f(-x) have no change of sign.
f(x) = 0 has no real root.

As the light entering the eye increases, the pupil decreases in size. Hence, under bright light, the diameter of the pupil cannot be more than what it was under dim light. Hence, it cannot be more than 7 mm.

Refractive index of glass with respect to air = μ_g/μ_a = 3/2
So, refractive index of air with respect to glass = μ_a/μ_g = 2/3

Sound waves cannot be polarised. The reason is that sound waves are longitudinal waves, which means molecules vibration takes place in forward and backward direction, but in light waves which are transverse waves, molecules vibration takes place up and down and side by side. This is the reason why light waves can be polarised and sound waves cannot be polarised.

m = 10 kg , h = 14m, g = 10
w = F x s = mgh
= 10 x 14 x 10 = 1400J

The IUPAC name of methyl alcohol (CH3OH) is methanol.

(i) C - Cl bond length is shorter than C - I bond length because chlorine atom is smaller in size as compared to iodine atom.
(ii) C - Cl bond is more ionic than C - I bond because of greater difference in electronegativities of C and Cl as compared to that of carbon and iodine.
(iii) More ionic character of bond, more will be its strength.
∴ C - C1 bond is stronger than C - 1 bond because C - Cl bond is more ionic than C - L

Since this element contains five valence electrons, it belongs to group 15 of the periodic table.

Mass number = 40
Since the cation (M+) of this element has 18 electrons, Z = 19
Number of neutrons = 40 - 19 = 21
The element is potassium (K).

H₂ is a non-polar molecule as both atoms are of hydrogen and have same value of electronegativity.

A real gas behaves as an ideal gas at a low pressure and high temperature.

The change in the number of moles = 2 - 4 = - 2
So ΔH = ΔE + (- 2) RT = ΔE - 2RT

As the branching increases, the molecules tend to acquire a spherical shape and become less compact. Due to less compact arrangement of molecules in the crystal lattice, they have lower melting points.

Metal 'M' is positioned between sodium and aluminium in the reactivity series. This means that metal M is less reactive than sodium, but more reactive than aluminium. In general, a more reactive metal can displace a less reactive metal from its salt solution. Lead and zinc are less reactive than Al. They are less reactive than M and they will not be able to displace M from its salt solution. Potassium is more reactive than sodium. Therefore, it is more reactive than M and it will be able to displace M from its salt solution.

The average urine sample tests at about pH 6.0.

Nose is the correct option. Olfaction is a chemoreception that forms the sense of smell. Olfaction has many purposes, such as the detection of hazards, pheromones, and food. It integrates with other senses to form the sense of flavour. Olfaction occurs when odorants bind to specific sites on olfactory receptors.

In case of AIDS, the virus affects the immune system, damages its functions and leads to death.

Vitamin C is known as ascorbic acid.

There are five petals, the largest (standard) overlaps the two lateral petals (wings), which in turn overlap the two smallest anterior petals (keel); this type of aestivation is known as vexillary.

Thalamus is the relay centre of brain.

In eukaryotes, cell undergoes cyclic process of growth and division to produce daughter cells.

Both ethylene and calcium carbides are fruit ripeners.

In dogs, panting is a way to release excess metabolic heat. The more overheated a dog becomes, the heavier he will pant.

Plants grow fast in the dark, and do so because they operate on circadian cycles. Tall spindly plant is a byproduct of "Shade Avoidance Syndrome," a scientific term for a plant's tendency to increase its production of the growth hormone auxin, allowing the plant to grow and stretch more rapidly toward sunlight to improve its conditions.

Cuboidal epithelium consists of cube-like cells which are square in section, but the free surface appears hexagonal. These are found in kidney tubules and glands. These tissues provide mechanical strength to the part where they occur and produce secretions.

Vasopressin and oxytocin are the only two hormones released from posterior pituitary .

Ethylene is a gas that is formed by the breakdown of methionine which is present in all the cells.

2,4-D is a synthetic auxin, which is a class of plant hormones, which results in uncontrolled growth and eventually death in susceptible plants. So, it is used as a systemic herbicide which selectively kills most broad leaf weeds by causing uncontrolled growth in them but leaves of most grasses such as cereals, lawn turf and grassland are relatively unaffected.

Third stage of prophase I of meiosis, during which homologous chromosomes are short and thick.

|x² - 4| = 4 + x²
Case I: If x² - 4 ≥ 0
i.e. x ≤ -2 and x ≥ 2
Then, |x² - 4| = x² - 4
⇒ The given equation becomes x² - 4 = 4 + x².
⇒ -4 = 4; which is not possible.
Case II: If x² - 4 < 0
⇒ -2 < x < 2
Then, |x² - 4| = - (x² - 4)
Thus, the given equation becomes - (x² - 4) = 4 + x².
Or 2x² = 0
⇒ x = 0
Hence, the given equation has only one solution.

Since it is a right-angled triangle, and 3, 4 and 5 are a Pythagorean triplet;

s = (a + b + c)/2 = 12/2 = 6

Substitute these values and check which condition satisfies this.

Let us take choice (3).

L.H.S. (s - a)(s - b) = (6 - 5)(6 - 4) = 2

R.H.S. s(s - c) = 6(6 - 3) = 18

L.H.S. < R.H.S.

It satisfies the given condition.

The substances with a lower value of reduction potential are stronger reducing agents.
Hence, the increasing order of the strengths as reducing agents is:

Cl₂, Fe³⁺, H₂, Zn

C₆H₅NH₂ is least basic as phenyl group withdraws the electron density from the nitrogen atom.

Empirical formula: CH₂
Empirical formula weight = 14
Mass of 1 mole = 42 g
∴Molecular weight = 42
N = 42/14 = 3
Molecular formula = (CH₂)₃ = C₃H₆

The type of epithelium which lines the inner surfaces of fallopian tubes, bronchioles and small bronchi is known as ciliated epithelium.

All the female eggs will have X chromosome with the white-eye mutation while the sperm will posses either a normal X chromosome or a Y chromosome. By using Punnett square, we can predict the outcome of this cross. Female progenies will receive X chromosome from both the sperm and egg cells. All females receive the dominant, red-eyed allele from their fathers and the recessive, white-eyed allele from their mothers.

Chromosomes are mainly concerned with hereditary transmission.

Stipule refers to outgrowths borne on either side (sometimes just one side) of the base of a leafstalk. A stipule is "interpetiolar" if it is located in between the petioles, and generally one stipule from each leaf is fused together, so it appears that there's just one stipule between each leaf. This type of stipule is present in Gardenia, which is a genus of flowering plants in the coffee family, Rubiaceae.

Erythroblastosis foetalis are the conditions where newborn infants suffer from anaemia, jaundice, enlarged spleen or liver and in severe cases, stillbirth or death soon after birth.