# CBSE Class 10 Mathematics Previous Year Question Paper 2018 All India Set - 3

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Q 1. What is the value of (cos267° - sin23°)?

Q 2. In an AP. if the common difference (d) = -4, and the seventh term (a7) is 4, then find the first term.

Q 3. Given that ∆ABC ∼ ∆PQR .

Q 4. What is the HCF of smallest prime number and the smallest composite number?

Q 5. Find the distance of point P(x, y) from the origin.

Q 6. If x = 3 is one root of the quadratic equation x- 2kx - 6 = 0, then find the value of k.

Q 7. Two different dice are tossed together. Find the probability:

1. Of getting a doublet
2. Of getting a sum 10, of the numbers on the two dice.

Q 8.  Find the radio in which P(4, m ) divides the line segment joining the points A(2,3) And B(6,-3). Hence find m.

Q 9. An integer is chosen at random between 1 and 100. Find the probability that it is:

1. Divisible by 8
2. Not divisible by 8

Q 10. In the given figure □ABCD is a rectangle. Find the values of x and y.

Q 11. Find the sum of first 8 multiples of 3.

Q 12. Given that  is irrational, prove that  is an irrational number.

Q 13. If A (-2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR

If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of quadrilateral ABCD.

Q 14. Find all zeros of the polynomial (2x- 9x+ 5x+ 3x - 1) if two of its zeros are  and .

Q 15. Find HCF and LCM of 404 and 96 and verify HCF × LCM = Product of the two given numbers.

Q 16. Prove that the lengths of tangents drawn from an external point to circle are equal.

Q 17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

OR

If the area of two similar triangle are equal, prove that they are congruent.

Q 18. A plane left 30 minute late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Q 19.  The table below shows the salaries of 280 persons:

 Salary(In thousand) No.of persons 5-10 49 10-15 133 15-20 63 20-25 15 25-30 6 30-35 7 35-40 4 40-45 2 45-50 1

Calculate the median salary of the data.

Q 20.  A wooden article was made by scooping out of hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm. find the total surface area of the article.

OR

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Q 21. Find the area of the shaded region in Fig.3, where areas drawn with centres  A,B,C and D intersect in pairs at mid-points P,Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12cm (Use 𝜋 =3.14)

Q 22. If 4 tan 𝛳=3, evaluate

OR

If tan 2A = cot (A-18°), where 2A is an acute angle, find the value of A.

Q 23.  As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (Use =1.732)

Q 24. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

1. The area of the metal sheet used to make the bucket.
2. Why we should avoid the bucket made by ordinary plastic? (Use = 3.14)

Q 25.  Prove that:

Q 26. The mean of the following distribution is 18. Find the frequency f of the class 19-21.

 Class 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency 3 6 9 13 f 5 4

OR

The following distribution gives the daily income of 50 workers of a factory:

 Daily income(In) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10

Convert the distribution above to a less than type of cumulative frequency distribution and draw its ogive.

Q 27. A motor boat whose speed is 18km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

OR

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km /hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

Q 28. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers.

Q 29. Draw a triangle ABC with BC = 6 cm AB = 5 cm and ∠ ABC = 60° Then construct a triangle whose sides are  of the corresponding sides of the ∆ABC.

Q 30.  In an equilateral ∆ABC, D is a point on side BC such that BD =  . Prove that 9[AD]= 7[AB]2

OR

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the sequence on the other two side

Q 1. What is the value of (cos267° - sin23°)?

Solution:

Cos267° - Sin223°

= Cos(90° - 23°) - sin23°    (∵ cos= sin2(90° - 𝛳))

= sin223° - sin223°

= 0

Q 2. In an AP. if the common difference (d) = -4, and the seventh term (a7) is 4, then find the first term.

Solution:

Given that d = -4, a= 4 , n = 7

a= 4

Now, an= a + (n-1) d
∴ a+ (7-1) (-4) =4

∴ a - 6 × 4 = 4

∴ a - 24 = 4

∴ a = 28

∴ First term of the given A.P. is 28.

Q 3. Given that ∆ABC ∼ ∆PQR .

Solution:

Given that ∆ABC ∼ ∆PQR

Using theorem "If two triangles are similar then the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Q 4. What is the HCF of smallest prime number and the smallest composite number?

Solution:

Smallest prime number is 2 and smallest composite number is 4.

2 = 2 × 1

4 = 2 × 2

∴HCF of 2 and 4 is 2.

Q 5. Find the distance of point P(x, y) from the origin.

Solution:

According to the question,

Origin O (0,0) and point P (x, y)

Distance of a point P(x, y) from O (0, 0) =

Q 6. If x = 3 is one root of the quadratic equation x- 2kx - 6 = 0, then find the value of k.

Solution:

Let p(x) = x2 – 2kx – 6

x = 3 is one root of given quadratic equation.

∴ p (3) = 0

∴ 3- 2k  × 3 - 6 = 0

∴ 9 - 6k - 6 = 0

∴ 3 = 6k

Q 7. Two different dice are tossed together. Find the probability:

1. Of getting a doublet
2. Of getting a sum 10, of the numbers on the two dice.

Solution:

When two dice are tossed together, the sample space S is given by

1.  Let A be the event of getting a doublet.
∴ A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
∴ n(A) = 6

2. Let B be the event of getting a sum of 10 of the numbers on two dice.
∴ B = {(5,5), (6,4), (4,6)}
∴ n(B) = 3

Q 8.  Find the ratio in which P(4, m ) divides the line segment joining the points A(2,3) And B(6,-3). Hence find m.

Solution:

Let P (4, m) divides the line segment joining the points A (2, 3) and B (6, –3) in the ratio t : 1.

CO-ordinates P are

Given co-ordinates of point P are (4, m).

∴ The ratio in which point P divides the line segment joining the points A and B is 1:1.

Q 9. An integer is chosen at random between 1 and 100. Find the probability that it is:

1. Divisible by 8
2. Not divisible by 8

Solution:

According to the question sample space S is as follows:
s = {2, 3, 4, 5… 99}

∴ n(s) = 98

1.  Let A be the event that chosen integer is divisible by 8.
∴ A = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
∴ n(A) = 12

Thus, probability that a chosen integer is divisible by 8 is  .
2. Probability that a chosen integer is not divisible by 8 = P(A’) = 1 - P(A) =  .

Q 10. In the given figure □ABCD is a rectangle. Find the values of x and y.

Solution:

Since □ABCD is a rectangle,

⇒ AB = CD and AD = BC

∴ x + y = 30 … (i)

and x – y = 14  …(ii)

∴ 2x = 44
∴ x = 22

Put x = 22 in (i)

∴ 22 + y = 30

∴ y = 30 - 22

∴ y = 8

∴ The value of x and y are 22 and 8 respectively.

Q 11. Find the sum of first 8 multiples of 3.

Solution:

First 8 multiples of 3 are {3, 6, 9, 12, 15, 18, 21, 24}.

This form an A. P. with

first term, a = 3,

Common difference, d = 6 - 3 = 3,

Last term, l = 24

And n = 8

∴ Sum of first 8 multiples of 3 is 108.

Q 12. Given that  is irrational, prove that  is an irrational number.

Solution:

Let us assume that  is a rational number.

Hence, we can find co-prime p and q (q ≠ 0) such that

But  is irrational.

Thus, our assumption is wrong.

Hence,  is an irrational.

Q 13. If A (-2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR

If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of quadrilateral ABCD.

Solution:

Diagonals of parallelogram bisect each other

So midpoint of AC= midpoint of BD

After comparing the coordinates, we get

∴ The coordinates are A (-2, 1), B (1, 0), C (4, 1) and D (1, 2) Now using distance formula,

As we know, opposite sides of a Parallelogram are equal in length.

OR

We will join B and D so that we will be getting two triangle ABD and BCD.

Then we will find their individual areas and add them together

To find the Area of a Quadrilateral ABCD.

Q 14. Find all zeros of the polynomial (2x- 9x+ 5x+ 3x - 1) if two of its zeros are  and

Solution:

Now   and  are the two zeroes of the given polynomial

So the product  will be a factor of the given polynomial

∴ f(x) = (x- 4x + 1) (2x- x + 1)

∴ 2x- 9x+ 5x+ 3x - 1 = (x- 4x + 1) (2x- x - 1)

Hence, the other zeroes of f(x) are the zeroes of the polynomial 2x- x - 1.

∴ 2x-x - 1 = 2x- 2x + x - 1 = (2x + 1) (x - 1)

So, 2x- 9x+ 5x+ 3x - 1 = (x- 4x + 1) (2x- x - 1)

=

Hence the roots of the Polynomial f(x) are  and 1.

Q 15. Find HCF and LCM of 404 and 96 and verify HCF x LCM = Product of the two given numbers.

Solution:

We will find the Prime factors of 404 and 96.

∴ 404 = 2 × 2 × 101

∴ 96 = 2 × 2 × 2 × 2 × 2 × 3

So,

LCM (404, 96) = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

HCF = 2 × 2 = 4

Here,

LCM × HCF = 9696 × 4 = 38784...............i)

Product of numbers = 404 × 96 = 38784.......ii)

∴ LCM × HCF = Product of numbers (from i and ii)

Hence verified.

Q 16. Prove that the lengths of tangents drawn from an external point to circle are equal.

Solution:

Here

PA and PB are tangents to the circle with centre O,

And AO and OB are the radii of the Circle.

In ∆OPA and ∆OPB

∠OAP = ∠OBP…...............each 90° (radius and tangent are⊥ at their point of contact)

OA = OB           ...............(radii of the same circle)

OC = OC           ...............(Common)

∆OPA ≅ ∆OPB.................. (by RHS Theorem)

∴ PA = PB........................ (CPCT)

Hence Proved

Q 17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

OR

If the area of two similar triangle are equal, prove that they are congruent.

Solution:

Let sides of a square be x units

∴ Diagonal length =  units

Since, Area of an Equilateral triangle =

∴ Area of an Equilateral triangle described on side =   ........i)

∴ Area of equilateral triangle described on diagonal =   .........ii)

OR

Suppose two similar triangles are ∆ABC and ∆PQR.

Given,

Area of ∆ABC = Area of ∆PQR
Now using properties of similar triangles, we get

Also, we know that ratio of areas of two Similar Triangles is equal to the ratio of the square of their corresponding sides.

∴ AB = PQ

So from i)

∴ AB = PQ

∴ AC = PR

∴ BC = QR

∆ABC ≅ ∆PQR… (SSS test)

Q 18. A plane left 30 minute late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Solution:

Here,
Distance =1500km

As we know,

Let the usual speed of plane be x km/hr.

After increasing the speed by 100 km/hr it's speed becomes (x+100) km/hr.

Given that difference in speed is 30 mins which is  hours

∴ T- T=

∴ x+ 100x - 300000 = 0

∴ x+ 600x - 500x - 300000 = 0

∴ (x + 600) (x - 500) = 0

∴ x = -600 or x = 500

Speed can’t be negative

∴ x = 500 km/hr

Therefore, the usual speed of plane is 500km/hr.

Q 19.  The table below shows the salaries of 280 persons:

 Salary(In thousand) No.of persons 5-10 49 10-15 133 15-20 63 20-25 15 25-30 6 30-35 7 35-40 4 40-45 2 45-50 1

Calculate the median salary of the data.

Solution:

 Salary Frequency CF 5-10 49 49 10-15 133 182 15-20 63 245 20-25 15 260 25-30 6 266 30-35 7 273 35-40 4 277 40-45 2 279 45-50 1 280

The cumulative frequency which is greater and nearest to 140 is 182.

So,

Medium Class is 10-15

Thus,

l = 10

h = 5

N = 280

cf = 49

f = 133

Median for the grouped date is given by,

Q 20.  A wooden article was made by scooping out of hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm. find the total surface area of the article.

OR

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution:

Radius of cylindrical part = Radius of Hemispherical Part = 3.5 cm

Height of cylindrical part = h =10 cm

Here,

Surface Area of the article = Curved surface Area of Cylinder + 2 x Curved surface Area of Hemisphere

= 2𝜋rh + 2 x 2𝜋r2

= 2𝜋 x 3.5 x 10 + 2 x 2𝜋 x 3.5 x 3.5

= 70𝜋 + 49𝜋

=119 x 𝜋

= 374cm2

OR

Volume of rice (heap) = volume of cone =

r = 12m, h = 3.5m

Area of canvas cloth = curved surface area of the cone = 𝜋rl

Area of canvas cloth = 𝜋rl =

Therefore Area of canvas required to cover the heap of the rice is 471.42m2

Q 21. Find the area of the shaded region in fig.3, where areas drawn with centres  A,B,C and D intersect in pairs at mid-points P,Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12cm (Use 𝜋 =3.14)

Solution:

According to the question,

Here side of the square = 12 cm and and  𝛳=90° (angle of the square)

∴ Area of shaded region = Area of the square ABCD - 4(Area of quadrant)

= 144 - 113.04

= 30.96 cm2

∴ Area of shaded region is 30.96 cm2.

Q 22. If 4 tan 𝛳=3, evaluate

OR

If tan 2A = cot (A-18°), where 2A is an acute angle, find the value of A.

Solution:

Let ∆XYZ is a right angle triangle at Y and ∠Z = 𝛳

Given, 4tan 𝛳 = 3

Let XY=3k and YZ = 4k

By Pythagoras Theorem, we get

∴ XZ2  = XY+ YZ2

∴  XZ= (3k) 2 + (4k) 2

∴ XZ= 9k+ 16k2

∴ XZ= 25k2

∴ XZ = 5k

Substituting the values in ,  , we get

OR

According to the question,

tan2A = cot (A-18°)

∴ tan2A = tan [90° - (A - 18°)]   [∵cot A = tan (90°-A)]

∴ tan2A = tan (90° - A + 18°)

∴ 2A = (90° - A + 18°)

∴ 2A + A = 108°

∴3A = 108°

∴ A = 36°

Q 23.  As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (Use =1.732)

Solution:
Let A and B be the two ships.

Let d be the distance between the two ships.
i.e. AB = d metres.
Suppose that the observer is at the point P.
It is given that PC = 100 m.
Let h be the distance (in metres) from B to C.

In right ∆PCA,

In right ∆PCB,

Substituting the value of h in (1), we get

d + 100 =

⟹ d = 100 - 100

⟹ d =  = 100 x 0.732 = 73.2 m

Thus, the distance between two ships is 73.2 m.

Q 24. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

1.  The area of the metal sheet used to make the bucket.
2. Why we should avoid the bucket made by ordinary plastic? (Use ∏= 3.14)
Solution:

Height of the cone, h = 24 cm

Upper radius of the cone, R = 15 cm

Lower radius of the cone, r = 5cm

Slant height of the cone,

Now, area of metal sheet used to make the bucket

= Total surface area of the bucket

𝜋(R+r)l + 𝜋r2

= 𝜋 (15 + 5) x 26 + 𝜋 x (5)2

= 𝜋 [520 + 25]

= 𝜋 x 545

= 3.14 x 545
= 1711.3 cm3

Plastics are not biodegradable. That is, plastic material mostly end as harmful waste that pollutes the environment and causes health problems, we should avoid using plastic.

Q 25.  Prove that:

Solution:

Q 26. The mean of the following distribution is 18. Find the frequency f of the class 19-21.

 Class 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency 3 6 9 13 f 5 4

OR

The following distribution gives the daily income of 50 workers of a factory:

 Daily income(In) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10

Convert the distribution above to a less than type of cumulative frequency distribution and draw its ogive.

Solution:

We have,

 Class interval Frequency fi Mid-value Xi fi x xi 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 F 20 20f 21-23 5 22 110 23-25 4 24 96 ∑fi = 40 + f ∑fixi = 704 + 20f

Now, Mean =

⟹ 18 (40 + f) = 704 + 20f

⟹ 720 + 18f = 704 + 20f

⟹ 2f = 16

⟹ f = 8

Thus, the missing frequency is 8.

OR

Less Than Series:

 Class interval Cumulative frequency Less than 120 12 Less than 140 26 Less than 160 34 Less than 180 40 Less than 200 50

Now, we mark the upper class limits along X-axis and the cumulative frequencies along Y-axis.

Thus, we plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) to get 'less than type’ ogive.

Q 27. A motor boat whose speed is 18km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

OR

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km /hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?

Solution:

Let x be the speed of the stream.

∴ Speed of the boat while going upstream= (18 - x) km/ hr

And, speed of the boat while going downstream= (18 + x) km/ hr

Time taken by the boat to go upstream =

Time taken by the boat to go downstream =

It is given that time taken for going upstream is 1 hour more than the time taken for going downstream.

Thus, we have

⟹ 48x = 324 - x2

⟹ x+ 48x - 324 = 0

⟹ x+ 54x - 6x - 324 = 0

⟹ x(x + 54) - 6(x + 54) =0

⟹ (x - 6) (x + 54) = 0

⟹ x - 6 = o or x + 54 = 0

⟹ x = 6 or x = -54

As speed cannot be negative, the speed of the stream is 6km/hr.

OR

Let the original average speed of the train be x km/hr.

Time taken by train to cover a distance of 63 km =  hours

Time taken by train to cover a distance of 72 km =   hours

According to given condition,

⟹ 63x + 378 + 72x = 3x+ 18x

⟹ 135x + 378 = 3x+ 18x

⟹ 3x- 117x - 378 = 0

⟹ x- 39x - 126 = 0

⟹ x- 42x + 3x - 126 = 0

⟹ x(x-42) + 3(x-42) = 0

⟹ (x - 42) (x + 3) = 0

⟹ x - 42 = 0 or x + 3 = 0

⟹ x = 42 or x = -3

Since, speed cannot be negative, we reject x = - 3.

Hence, x = 42

Thus, the original average speed of the train is 42 km/hr.

Q 28. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers.

Solution:

Let the four consecutive numbers in an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d).

Sum of the numbers = 32

⟹ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⟹ 4a = 32

⟹ a = 8

It is given that,

⟹ 960 - 135d= 448 - 7d2

⟹ 128d= 512

⟹ d= 4

⟹ d = ∓2

⟹ a - 3d = 8 - 3(2) = 8 - 6 = 2

⟹ a - d = 8 - 2 = 6

⟹ a + d = 8 + 2 = 10

⟹ a + 3d = 8 + 3(2) = 8 + 6 = 14

Thus, the four numbers are 2, 6, 10, 14.

Q 29. Draw a triangle ABC with BC = 6 cm AB = 5 cm and ∠ ABC = 60° Then construct a triangle whose sides are  of the corresponding sides of the ∆ABC.

Solution:

Steps of construction:

1. Draw a line segment BC = 6cm
2. At B, construct ∠XBC = 60°
3. With B as centre and radius 5 cm, draw an arc intersecting XB at A.
4. Join AC to obtain ∆ABC.
5. Below BC, make an acute ∠CBY.
6. Along BY, mark off 4 points
B1, B2, B3, B4 such that BB= B1B= B2B= B3B4
7. Join B4C
8. From point B3, draw a line parallel to B4C intersecting BC at C’
9. From point C’, draw a line parallel to CA intersecting AB at A’
10. Thus, ∆A’BC’ is the required triangle.

Q 30.  In an equilateral ∆ABC, D is a point on side BC such that BD =  . Prove that 9[AD]= 7[AB]2

OR

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

Solution:

Let ABC be an equilateral triangle and let D be a point on BC such that BD = .

Construction: Draw AE ⊥BC. Join AD.

∆ABC is an equilateral triangle.

⟹ E is the mid-point of BC

∴ BE = CE =

In right-angle ∆AEB, by Pythagoras theorem,

AB= AE+ BE2   ............(i)

In right-angle ∆AEB, by Pythagoras theorem,

Subtracting (ii) from (i), we have

AB- Ad= (AE+ BE2) - (AE+ DE2)

OR

Consider the following figure:

Given: In ∆ABC, ∠ABC = 90°

To prove: AC= AB+ BC2

Construction: Draw seg BD ⊥ hypotenuse AC and A-D-C

Proof:

In ABC, seg BD ⊥ hypotenuse AC.........(Construction)

∴ ∆ABC ∼ ∆ADB................................(Similarity in right angled triangles)

..................................(Corresponding sides of similar triangles)

∴ AB= AC x DC...............................(i)

Similarly, ∆ABC ∼ ∆BDC....................(Similarity in right angled triangles)

..................................(Corresponding sides of similar triangles)

∴ BC= AC x DC   ............................(ii)

AB+ BC= AC x AD + AC x DC.........[Adding equations (i) and (ii)]

= AC x AC

=AC2

∴ AC= AB+ BC2

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