AIIMIIT education by naveen rachuri: key for maths second Cyclic feb 2022 key paper

Key for 2nd cyclic test noble

Test no 7

8th cbse key compound interest

Test no 6 10th Science key and maths key

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Then 8th key

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Key and solutions are there

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Test no 5

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Science half yearly 2 key dated ictober 29

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1 B
2 B
3 A
4 A
5 B
6 B
7 A
8 A
9 A
10 B
11 C
12 B
13 B
14 D
15 C
16 A
17 A
18 B
19 C
20 D
21 A
22 A
23 B
24 B

25 C
26 B
27 C
28 C
29 B
30 D
31 B
32 C
33 C
34 B
35 B
36 B
37 A
38 C
39 D
40 C
41 A
42 D
43 C
44 B
45 D
46 A
47 B
48 A
49 B
50 C

Science half yearly key paper

Hakf yearly maths 1 key

Cyclic 7

Science

9th is C not d

Mistake is there in cyclic 7 9th

32nd is a

Co2 is not released during day

As it is used for photosynthesis

Maths

Cyclic test 6 eye lesson sep 27

Cyclic 6 probability key

Cyclic 5 refraction key

33 is d concave lens

Check once

Cyclic 5 trigonometry key

Cyclic 5 trigonomtry

Cyclic maths 4 key

Cyclic 4 science key

Reflection key

10th Science cyclic3 key

metals and non metals

10th cyclic 3 maths key

Key for cyclic 3

1.Answer: (c) Equilateral

2.Answer: (c) √3/4 a2

Area of an equilateral triangle with side length a = √3/4 a2

3.Answer: (b) 3

Explanation: By midpoint theorem,

DE=½ BC

DE = ½ of 6

DE=3 cm

4.Answer: (c) 10 cm

Explanation: Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.

By Pythagoras theorem,

(16/2)2+(12/2)2=side2

82+62=side2

64+36=side2

side=10 cm

5.Answer: (d) 108 sq.cm

Solution: Let A1 and A2 are areas of the small and large triangle.

Then,

A2/A1=(side of large triangle/side of small triangle)

A2/48=(3/2)2

A2=108 sq.cm.

6.Answer: (a) 30 cm

Solution: Perimeter of triangle = sum of all its sides

P = 30+40+x

100=70+x

x=30 cm

7.Answer: (d) 18 cm

Explanation: ABC ~ DEF

AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6 cm

AC = (12.4)/6 = 8 cm

Perimeter = AB+BC+AC

= 4+6+8

=18 cm

8.Answer: (a) 4.33 cm

Explanation: The height of the equilateral triangle ABC divides the base into two equal parts at point D.

Therefore,

BD=DC= 2.5 cm

In triangle ABD, using Pythagoras theorem,

AB2=AD2+BD2

52=AD2+2.52

AD2 = 25-6.25

AD2=18.75

AD=4.33 cm

9.Answer: (b) ∠B=∠D

If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if ∠B=∠D.

10.Answer: (d) 16: 81

Explanation: Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81

11.Answer: (d) Isosceles triangles

Explanation:

All circles, squares, and equilateral triangles are similar figures.

12.Answer: (c) BD . CD = AD2

Explanation:

In ΔADB and ΔADC,

∠D = ∠D = 90°

∠DBA = ∠DAC

By AAA similarity criterion,

ΔADB ~ ΔADC

BD/AD = AD/CD

BD.CD = AD2

13.Answer: (a) ΔPQR ~ ΔCAB

Explanation:

Given that, in triangles ABC and PQR, AB/QR = BC/PR = CA/PQ

If sides of one triangle are proportional to the side of the other triangle, and their corresponding angles are also equal, then both the triangles are similar by SSS similarity. Therefore, ΔPQR ~ ΔCAB

14 .Answer: (b) similar but not congruent

Explanation:

In ΔABC and ΔDEF,

∠B = ∠E, ∠F = ∠C and AB = 3 DE

By AA similarity criterion,

ΔABC ~ ΔDEF

AB = 3DE

⇒ AB/DE = 3

⇒ AB/DE = BC/EF = AC/DF = 3

For triangles to be congruent, the ratio of sides must be 1

Therefore, triangles are similar but not congruent.

15.Answer: (a) 16

Explanation:

Given,

ΔABC ~ ΔPQR

and BC/QR = 1/4

Ratio of area of similar triangles is equal to the square of its corresponding sides.

So, ar(ΔPRQ)/ar(ABC) = (QR/BC)2 = (4/1)2 = 16

16.Answer: (b) DE = 12 cm, ∠F = 100°

Explanation:

Given,

ΔABC ~ ΔDFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm

In triangle ABC,

∠A + ∠B + ∠C = 180°

∠B = 180° – 30° – 50° = 100°

Since ΔABC ~ ΔDFE, the corresponding angles are equal.

Thus, ∠D = ∠A = 30°

∠F = ∠B = 100°

∠E = ∠C = 50°

And

AB/DF = AC/DE

5/7.5 = 8/DE

DE = (8 × 7.5)/5 = 12 cm

17.Answer: (c) AB/DE = BC/EF = CA/FD

Explanation:

If two triangles are similar, i.e. when ΔABC ~ ΔDEF, then

(i) their corresponding angles are equal and

∠A = ∠D, ∠B = ∠E, ∠C = ∠F and

(ii) their corresponding sides are in the same ratio (or proportion).

AB/DE = BC/EF = CA/FD

18.Answer: (d) ASA

19.Answer: (b) (4,-2)

Explanation: x2–2x –8 = x2–4x + 2x –8

= x(x–4)+2(x–4)

= (x-4)(x+2)

Therefore, x = 4, -2.

20.Answer: (a) 3x2-3√2x+1

Explanation: Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;

x2–(α+β)x +αβ

= x2 –(√2)x + (1/3)

= 3x2-3√2x+1

21.Answer: (b) 4

Explanation: Degree is the highest power of the variable in any polynomial.

22.Answer: (b) b-a+1

Explanation: Since one zero is -1, hence;

P(x) = x3+ax2+bx+c

P(-1) = (-1)3+a(-1)2+b(-1)+c

0 = -1+a-b+c

c=1-a+b

Product of zeroes, αβγ = -constant term/coefficient of x3

(-1)βγ = -c/1

c=βγ

βγ = b-a+1

23.Answer: (a) Zero of p(x)

Explanation: Let p(x) = mx+n

Put x = a

p(a)=ma+n=0

So, a is zero of p(x).

24.Answer: (a) Intersects x-axis

25.Answer: (d) At most n zeroes

Explanation: Maximum number of zeroes of a polynomial = Degree of the polynomial

26.Answer: (d) More than 3

Explanation: The polynomials x2-3x-10, 2x2-6x-20, (1/2)x2-(3/2)x-5, 3x2-9x-30, have zeroes as -2 and 5.

27.Answer: (b) ±3√3

Explanation: x2-27 = 0

x2=27

x=√27

x=±3√3

28.Answer: (a) -b/a

Explanation:

Let α be the third zero.

Given that two zeroes of the cubic polynomial are 0.

Sum of the zeroes = α + 0 + 0 = -b/a

α = -b/a

29.Answer: (b) -10

Explanation:

Given that 2 is the zero of the quadratic polynomial x2 + 3x + k.

⇒ (2)2 + 3(2) + k = 0

⇒ 4 + 6 + k = 0

⇒ k = -10

30.Answer: (c) (x²/2) – (x/2) – 6

Explanation:

Let the given zeroes be α = -3 and β = 4.

Sum of zeroes, α + β= -3 + 4 = 1

Product of Zeroes, αβ = -3 × 4 = -12

Therefore, the quadratic polynomial = x² – (sum of zeroes)x + (product of zeroes)

= x² – (α + β)x + (αβ)

= x² – (1)x + (-12)

= x² – x – 12

Dividing by 2,

= (x²/2) – (x/2) – 6

31.Answer: (b) both negative

Explanation:

Given quadratic polynomial is x2 + 99x + 127.

By comparing with the standard form, we get;

a = 1, b = 99 and c = 127

a > 0, b > 0 and c > 0

We know that in any quadratic polynomial, if all the coefficients have the same sign, then the zeroes of that polynomial will be negative.

Therefore, the zeroes of the given quadratic polynomial are negative.

32.Answer: (c) -2, -5

Explanation:

x2 + 7x + 10 = x2 + 2x + 5x + 10

= x(x + 2) + 5(x + 2)

= (x + 2)(x + 5)

Therefore, -2 and -5 are the zeroes of the given polynomial.

33.Answer: (b) two real and unequal roots

If the discriminant of a quadratic polynomial, D > 0, then the polynomial has two real and unequal roots.

34.Answer: (a) degree of p(x) < degree of g(x)

Explanation:

We know that, p(x)= g(x) × q(x) + r(x)

Given that, q(x) = 0

When q(x) = 0, then r(x) = 0

So, now when we divide p(x) by g(x),

Then p(x) should be equal to zero.

If r(x) = 0, then the degree of p(x) < degree of g(x).

35.Answer: (a) g(x) × q(x) + r(x)

By division algorithm of polynomials, p(x) = g(x) × q(x) + r(x).

36.Answer: (c) -d/a

The product of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is -d/a.

37.Answer: (a) Three

If the graph of a polynomial intersects the x-axis at three points, then

38.Answers: c) ‘a’ is a non zero real number and b and c are any Polynomials.

39.Answers: d) D = 0

40.Answers: b) 1/4

10th cbse Cyclic test 2 science key

Answer

Chemical reactions lesson

1. b) Except (b) all other reactions involve compounds.
2. (d) CuSO4 +H2S -> CuS + H2SO4
It is a Double decomposition
3.(c) Iron is more reactive than copper, hence Cu will not displace iron from iron sulphate, hence no reaction will take place.
4.(b) Zn + 2HCl -> ZnCl2 + H2
5.(c) Na2CO3+ 2HCI -> 2NaCI + H2O + CO2
6.(c)
7.(c) Cu +4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O
8.(b)
9.(d)
10.(b) Here H2S is oxidizing in to H2O, hence behave as a reducing agent.
11.(a)
12.(b)
13.(d)
14. (b) Na2CO3 + 2HCI ->2NaCI + CO2 + H2O
15.(b) The reaction represents a neutralization reaction in which base (NaOH) reacts with an acid (HNO3) to form salt (NaNO3) and water (H2O).
16.(d)
17.(a) Write silver chloride in sunlight turns to grey.
18.(b)
19.(a) 2Cu + O2 -> 2CuO
Black
20.(a) 4P + 3O2 ->2P2O3
(Oxidation)
Or 4P + 5O2 ___ 2P2O5
(Oxidation)
21.(c) During rusting iron get oxidized to form rust (Fe2O3 x H2O)
22.(c) Gold is very least reactive hence does not corrode at all.a
23.(a)
24.(c)
N the given reaction H2 is undergoing oxidation, hence behave as reducing agent.
25.(c)
It shows oxidation and reduction (redox) properties.
26.(a) Oxidation and reduction both occur so the reaction is redox.
27.(a,) Cu is above of Ag in electrochemical series and thus
Cu + 2Ag+-> Cu2+ + 2Ag reaction occurs.