# CBSE Class 10 Mathematics Previous Year Question Paper 2019 All India Set-3

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Q 1. In Figure 1, PS = 3 cm. QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, ∠PQ ⏊ RQ and RQ = 9 cm. Evaluate tan θ.

OR

If tan ⍺ = , find the value of sec ⍺.

Q 2. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

Q 3. Find the value(s) of x, if the distance between the point A(0,0) and B(x, -4) is 5 units.

Q 4. Find after how many places of decimal the decimal form of the  number will terminate.

OR

Express 429 as a product of its prime factors.

Q 5. Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3).

Q 6. Find the sum of the first 10 multiples of 3.

Q 7. If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.

OR

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. When is the minimum distance each should walk so that each can cover the same distance in complete steps?

Q 8. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

Q 9. Using completing the square method, show that the equation x2 – 8x + 18 = 0 has no solution.

Q 10. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7?

Q 11. Solve the following pair of linear equations:

3x + 4y = 10

2x – 2y = 2

Q 12. Points A(3, 1), B(5, 1) C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

OR

Point P and Q trisect the line segment joining the point A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of point P and Q.

Q 13. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days: 0 - 6 6-12 12-18 18-24 24-30 30-36 36-42 Number of students: 10 11 7 4 4 3 1

Q 14. In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

OR

Prove that opposite side of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

Q 15. A, B and C are interior angles of a triangle ABC. Show that

(i)

(ii)

OR

Q 16. Prove that   is an irrational number.

OR

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

Q 17. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations.

Q 18. Water in a cannel, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

Q 19. The perpendicular from A on side BC of a ∆ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.

OR

AD and PM are medians of triangles ABC and PQR respectively where ∆ ABC ∼ ∆ PQR. Prove that .

Q 20. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle.

Q 21. Find the value of k so that area of triangle ABCD with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Q 22.  If  and -3 are the zeroes of the polynomial ax2 + 7x + b, then find the values of a and b.

Q 23. Change the following distribution to a ‘more than type’ distribution.

Hence draw the ‘more than type’ ogive for this distribution.

 Class interval: 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Frequency: 10 8 12 24 6 25 15

Q 24. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given = 1.732)

Q 25. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

OR

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Q 26. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the A.P. is zero.

OR

The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.

Q 27. In Figure3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find

(a) the total surface area of the block.

(b) the volume of the block formed.

OR

A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use 𝜋 = 3.14)

Q 28. Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides  are  times the corresponding sides of the first triangle.

Q 29. Prove that :

Q 30. A motorboat whose speed in still water is 9 km/h, goes 15 km downstream and comes to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.

Q 1. In Figure 1, PS = 3 cm. QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, ∠PQ ⏊ RQ and RQ = 9 cm. Evaluate tan θ.

OR

If tan ⍺ = , find the value of sec ⍺.

Solution:

In the figure we have a right angled triangle ΔPSQ

So on applying Pythagoras theorem we get

∴ PQ2 = PS2 + QS2

In ΔPQR

OR

Q 2. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:

It is given that a > b,

AB is the tangent to the smaller circle,

So ΔABO is right angled triangle

∴ AO2 = AB2 + BO2

Now we know that the perpendicular drawn from the center to a chord bisects the chord so

The length of chord will be 2AB =

Q 3. Find the value(s) of x, if the distance between the point A(0,0) and B(x, -4) is 5 units.

Solution:

A(0, 0) and B(x, -4)

The distance AB is given by distance formula

Q 4. Find after how many places of decimal the decimal form of the  number will terminate.

OR

Express 429 as a product of its prime factors.

Solution:

Now power of 5 is 4 and that of 2 is 3

∴  4 > 3, so the decimal will terminate after 4 places

OR

∴ 429 = 3 × 11 × 13

Q 5. Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3).

Solution:

Q 6. Find the sum of the first 10 multiples of 3.

Solution:

The multiples of 3 are

3, 6, 9, 12…

The above series is in arithmetic progression

∴ a = 3 and d = 3

We need sum of 10 multiples

Q 7. If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.

OR

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. When is the minimum distance each should walk so that each can cover the same distance in complete steps?

Solution:

65 = 5 × 13

117 = 3 × 3 × 13

HCF = 13

Given that

HCF = 65n - 117

∴ 13 = 65n - 117

∴  130 = 65n

∴ n = 2

OR

We have to find the LCM of 30 cm, 36 cm and 40 cm to get the required minimum distance. Because we are asked the minimum distance

Now,

30 = 3 × 2 × 5,

36 = 2 × 3 × 2 × 3

40 = 2 × 2 × 2 × 5

∴ LCM (30, 36, 40) = 23 × 32 × 5 = 360

Minimum distance each should walk 360 cm. so that, each can cover the same distance in complete steps.

Q 8. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

Solution:

A die is thrown,

So sample space S is given by

∴ S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

Let A be the event such that composite number shows up

∴ A = {4, 6}

∴ n(A) = 2

(i) Probability of getting composite number

Let B be the event such that prime number shows up

∴ B = {2, 3, 5}

∴ n(B) = 3

(ii) Probability of getting prime number

Q 9. Using completing the square method, show that the equation x2 – 8x + 18 = 0 has no solution.

Solution:

Which is not possible, so the given quadratic equation has no roots.

Q 10. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7?

Solution:

Poonam selects a card in random from 7 to 40

So the sample space S is

∴ S = {7, 8, 9, 10, ……40}

∴ n(S) = 34

Let A be the event of getting multiple of 7

∴ A = {7, 14, 21, 28, 35}

∴ n(A) = 5

Probability of getting multiple of 7

Q 11. Solve the following pair of linear equations:

3x + 4y = 10

2x – 2y = 2

Solution:

3x + 4y = 10    …. (i)

2x – 2y = 2       …. (ii)

Multiply equation (ii) by 2, we get

4x – 4y = 4       …. (iii)

Adding equation (i) and (iii), we get

∴ 7x = 14

→ x = 2

Put x = 2 in the equation (ii), we get

∴ 2(2) – 2y = 2

→ 2y = 4 – 2

→ y =

Hence the solutions is x = 2 and y = 1.

Q 12. Points A(3, 1), B(5, 1) C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

OR

Point P and Q trisect the line segment joining the point A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of point P and Q.

Solution:

Points A (3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of parallelogram

We know that diagonals of parallelogram bisect each other

The M be the midpoint of both AC and BD

So by midpoint formula

OR

P and Q trisect AB, given that A (-2, 0) and B (0, 8)

Now P will divide AB in ratio 1: 2

So by using section formula we get

Now P is the midpoint of AQ

Let coordinates of Q be (a, b)

So by using midpoint theorem

Q 13. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days: 0 - 6 6-12 12-18 18-24 24-30 30-36 36-42 Number of students: 10 11 7 4 4 3 1

Solution:

We will use direct method to find the mean:

 No. of days No. of students Class marks(xi) fixi 0-6 10 30 6-12 11 99 12-18 7 105 18-24 4 84 24-30 4 108 30-36 3 99 36-42 1 39 Σf1 = 40 Σf1x1 = 564

Hence, the number of days a student was absent is 14.1

Q 14. In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

OR

Prove that opposite side of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

Solution:

Let TP = x.

In ΔPRT and ΔQRT

PT = TQ …(tangents drawn from external point are equal in length)

RT ….common side

PTR = QTR…(the line joining the center of circle and the external point bisects the angle between the tangents, which are drawn from that external point)

ΔPRT ΔQRT…( SAS test )

PRT = QRT…( C.P.C.T.)

Also,

PRT = QRT = 90o…(linear pair)
We know that the perpendicular drawn from the centre of the circle to a chord bisects it.
Therefore, PR = RQ = 4 cm and OR  PQ
Using Pythagoras theorem in right angled triangle PRO, we have

Now, again using Pythagoras theorem in triangle PRT we have

(PT)2 = (PR)2 + (RT)2

→ x2 = 42 + RT2 ....(i)

Since we know that the radius drawn to tangent is perpendicular to the tangent so, we have
In right triangle OPT,

Again using (i), we have

OR

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S.
Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒ ∠7 = ∠8
Similarly we get,
∠4 = ∠3

∠5 = ∠6

∠1 = ∠2

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º
⇒ ∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360º
⇒ 2 ∠1 + 2 ∠8 + 2 ∠4 + 2 ∠5 = 360º
⇒ 2(∠1 + ∠8) + 2(∠4 + ∠5) = 360º
⇒ (∠1 + ∠8) + (∠4 + ∠5) = 180º
⇒ ∠AOD + ∠COB = 180º
Similarly, we can prove that ∠AOB + ∠COD = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Q 15. A, B and C are interior angles of a triangle ABC. Show that

(i)

(ii)

OR

Solution:

(i) In triangle ABC, sum of all interior angles is 180°

(ii) Using equation (a), we have

OR

Q 16. Prove that   is an irrational number.

OR

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

Solution:

Prove that  is an irrational number.
Let us assume that  is a rational number.
That is, we can find integers a and b (b ≠ 0) such that
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
(Squaring on both the sides) …. (i)
Therefore, a2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,

This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that  is rational.
So, we conclude that  is irrational.

OR

Let the largest number which divides the given numbers be x.

Since, it leaves the remainders 1, 2 and 3 while dividing 1251, 9377 and 15628
So, the numbers which are divisible by x are:
1251 – 1 =1250,  9377 – 2 =9375  and  15628 – 3 = 15625
Here, x is the largest number dividing these 3 numbers so x becomes the GCD or HCF
Now, 1250 = 54 × 2
9375 = 55 × 3 and 15625 = 56
Therefore x = HCF (1250, 9375, 15625) = 5= 625.
Hence, the required number is 625.

Q 17. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations.

Solution:

The given equations are:
x - y + 1 = 0…… (i)
3x + 2y - 12 = 0…… (ii)
To draw the graph of these equations, let us find some points lying on them:
For equation (i)

 x 0 1 -1 2 y 1 2 0 3

Plot these points i.e. (0, 1), (1, 2), (-1, 0) and (2, 3) on the graph
Now, draw a straight line connecting them and extend the line.

For equation (ii)

 x 0 1 4 2 y 6 0 3

Plot these points i.e. (0, 6),  , (4, 0) and (2, 3) on the graph
Now, draw a straight line connecting them and extend the line.

Clearly, the two lines intersect at a point (2, 3).
Thus, x = 2 and y = 3 satisfy both the equations.

Q 18. Water in a cannel, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

Solution:

Speed of flowing of water from canal is 10 km/h

This means the length of water flows in 1 hour = 10 km
→ length of water flows in 30 minutes = 5 km

Now,
volume of water flowing from canal = length of water flows in 30 minutes × breadth of canal × depth of canal

= 5000 m × 6 m × 1.5 m = 45000 m³

Let the area of field = x m²

Then, volume of water irrigates into the field = area of field × height of water during irrigation

= x m² × 8× 10⁻² m = 0.08x m³

Now, volume of water flowing from canal = volume of water in field

45000 m³ = 0.08x m³

x = 562500 m²

We know, 1 hectare = 10000 m²

So, area of field in hectare = 562500/10000 = 56.25 hectares.

Q 19. The perpendicular from A on side BC of a ∆ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.

OR

AD and PM are medians of triangles ABC and PQR respectively where ∆ ABC ∼ ∆ PQR. Prove that .

Solution:

Given that in ΔABC, we have

AD ⊥ BC and BD = 3CD
Now, using Pythagoras theorem in right angle triangles ADB and ADC, we have

Subtracting equation (ii) from equation (i), we get

OR

Given: AD and PM are medians of triangle ABC and PQR respectively.

Q 20. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle.

Solution:

Radius of the circle = OA = OB = 14 cm

Let θ be the angle at the centre = 60°
Now in  ΔAOB,

By the angle sum property of triangle,

Therefore, all the angles of a triangle AOB is  60°
So, ΔAOB is equilateral

→ OA = OB = AB

Area of minor segment = Area of sector – Area of triangle AOB

Q 21. Find the value of k so that area of triangle ABCD with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Solution:

Given: A(k + 1, 1), B(4, -3) and C(7, - k)
Area of triangle ABD is 6 square units
We know that the area of triangle is given by

Q 22.  If  and -3 are the zeroes of the polynomial ax2 + 7x + b, then find the values of a and b.

Solution:

Solving (i) and (ii) simultaneously, we get

a = 3
Substituting the value of a in (ii), we get

b = -6

Q 23. Change the following distribution to a ‘more than type’ distribution.

Hence draw the ‘more than type’ ogive for this distribution.

 Class interval: 20-30 30-40 40-50 50-60 60-70 70-80 80-90 Frequency: 10 8 12 24 6 25 15

Solution:

 Intervals f cf(more than type) 20-30 10 100 (30, 100) 30-40 8 90 (40, 90) 40-50 12 82 (50, 82) 50-60 24 70 (60, 70) 60-70 6 46 (70, 46) 70-80 25 40 (80, 40) 80-90 15 15 (90, 15) 100

Q 24. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given = 1.732)

Solution:

From the figure, DE is the tower.
When sun’s altitude is 60˚ then ∠DPE = 60˚ and length of shadow is PE.
Shadow is 40 m more when angle changes from 60˚ to 30˚
FP = 40 m.
As tower is a vertical ground, ∠DEF is 90˚.
In right angled triangle DEF,

Q 25. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

OR

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Solution:

Given : ΔPQR in which XY || QR, XY intersects PQ and PR at X and Y respectively.

To prove :

Construction : Join RX and QY and draw YN perpendicular to PQ and XM perpendicular to PR.

Proof:

Since the area of triangles with same base and between same parallel lines are equal, so
∴ ar(ΔQXY) = ar(ΔRXY)  …..(vii)
As ΔQXY and ΔRXY are on same base XY and between same parallel lines XY and QR.
Therefore, from (v), (vi) and (vii) we get

OR

Given : A right angle triangle ABC at B.
To prove : AC2 = AB2 + BC2
Construction : Draw BD perpendicular to AC.
Proof :
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

∴ AD × AC + CD × AC = AB2 + BC2
∴ AC(AD + CD) = AB2 + BC2
∴ AC × AC = AB2 + BC2
∴ AB2 + BC2 = AC2

Q 26. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the A.P. is zero.

OR

The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.

Solution:

Let the mth term of an Arithmetic Progression be tm and nth term be tn.
Also, a and d be the first term and common difference respectively.
According to the question,
mtm = ntn
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ am + m2d – md = an + n2d – nd = 0
∴ a(m – n) + (m2 – n2)d – (m – n)d = 0
∴ a(m – n) + (m + n)(m – n)d – (m – n)d = 0
∴ (m – n)[a + (m + n)d – d] = 0
It is given that m ≠ n.
∴ a + (m + n – 1)d = 0
∴ tm + n = 0

OR

Let the three numbers of an arithmetic progression be a – d, a, a + d.
According to the question,
a – d + a + a + d = 18
∴ 3a = 18
∴ a = 6
Hence, three numbers are 6 – d, 6, 6 + d
According to the question,
∴  (6 – d)(6 + d) = 5d
∴ 36 – d2 = 5d
∴ d2 + 5d – 36 = 0
∴  (d + 9)(d – 4) = 0
∴ d = -9 and d = 4
If d = -9 then the three numbers are 6 + 9, 6, 6 – 9 i. e. 15, 6, -3.
If d = 4 then the three numbers are 6 – 4, 6, 6 + 4 i. e. 2, 6, 10.

Q 27. In Figure3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find

(a) the total surface area of the block.

(b) the volume of the block formed.

OR

A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use 𝜋 = 3.14)

Solution:

Let the side of the cube be a.
∴ a = 6 cm
Let radius of the hemisphere be r.
∴ r = 2.1 cm
(a)
The total surface area of the block
= Total surface area of a cube – base area of hemisphere + Curved surface area of hemisphere
= 6a2 – πr2 + 2πr2
= 6 × 62 +πr

= 216 + 13.86

= 229.86 cm2

(b)
Volume of the clock
= Volume of a cube + volume of hemisphere

= 216+19.404
= 235.404 cm3

OR

Volume of the bucket = 12308.8 cm3, r1 = 20 cm and r2 = 12 cm, h = ?

Hence, height of the bucket = 15 cm, area of the metal sheet used = 2160.32 cm2.

Q 28. Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides  are  times the corresponding sides of the first triangle.

Solution:

Steps of construction :

1. Draw a line segment AB of 5 cm. Taking A and B as centres, draw two arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm and 7 cm respectively.
2. Draw a ray AX making acute angle with the line AB on opposite side of vertex C.
3. Locate 7 points A1, A2, A3, A4, A5, A6, A7 (as 7 is greater between 5 and 7) on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
4. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.
5. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

Q 29. Prove that :

Solution:

Q 30. A motorboat whose speed in still water is 9 km/h, goes 15 km downstream and comes to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.

Solution:

Let T be the time for downstream and t be the time for upstream, travel.
T + t = 3 hrs 45 min = 15/4 hrs   ….(i)
For downstream, distance/velocity = time

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