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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
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
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
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Cyclic test 6 eye lesson sep 27
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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
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
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.
Actually ac is answer
28.(b) reduction.
29.a
30.a
31.a
32.a
33.a
34.c
35.a
28.(b) reduction.
29.a
30.a
31.a
32.a
33.a
34.c
35.a
36.d
37.b
38.d
39.a
40.b
Cyclic 1 acids and bases
Lesson
1.c means 3
2.3
3.1
4.3
5.3
6.2
7.3
8.1
9.2
10.1
11.2
12.3
13.3
14.2
15.4 shade
but. Actual ans is is 2
16.3
17.2
18.a
19.b
20.d
21.b
22.d
23.c
24.a
25.a
26.a
27.a
question not there
28.c
29.2
30.3
31 3
32.2
33.1
34.4
35.1
36.2.
37.3
38.3
39.1
40.1
Test no 23 key
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Csa and tsa and volumes of cylinders
Sunday 1 November 2020
1st..1m
2nd.44039.28cm2
3rd.68.75 rs
4.ans.7.48m2
5.15:22
6.17.6 m3
7.3.432 kgs
8.3cm and 140.3 cm3
9.0.4708 m2
10.1051776 cc
3rd.68.75 rs
4.ans.7.48m2
5.15:22
6.17.6 m3
7.3.432 kgs
8.3cm and 140.3 cm3
9.0.4708 m2
10.1051776 cc
Test no 30 9th class key
Sunday 1 November
Csa tsa and volumes of cone
1.3768cm2
2..204.1 cm2
3.8cm
4.1079cm
5.301.71cm2
6.3:1
7.3m,5m
8.,10cm
9.9:5
10.3:1
Key for 2nd cyclic science
Light and eye lesson key
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