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

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]² = 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

 

 

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) =  . 

 

 

Since □ABCD is a rectangle,

⇒ AB = CD and AD = BC

∴ x + y = 30 … (i)

and x – y = 14  …(ii)

Adding (i) and (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 (2x² - 9x³ + 5x² + 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) = (x² - 4x + 1) (2x² - x + 1)

∴ 2x⁴ - 9x³ + 5x² + 3x - 1 = (x² - 4x + 1) (2x² - x - 1)

Hence, the other zeroes of f(x) are the zeroes of the polynomial 2x² - x - 1.

∴ 2x² -x - 1 = 2x² - 2x + x - 1 = (2x + 1) (x - 1)

So, 2x⁴ - 9x³ + 5x² + 3x - 1 = (x² - 4x + 1) (2x² - 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

∴ T₁ - T₂ =  

 

∴ x² + 100x - 300000 = 0

∴ x² + 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  = XY² + YZ²

∴  XZ² = (3k) ² + (4k) ²

∴ XZ² = 9k² + 16k²

∴ XZ² = 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) 

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

⟹ x² + 48x - 324 = 0

⟹ x² + 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 = 3x² + 18x

⟹ 135x + 378 = 3x² + 18x

⟹ 3x² - 117x - 378 = 0

⟹ x² - 39x - 126 = 0

⟹ x² - 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.

 

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 - 135d² = 448 - 7d²

⟹ 128d² = 512

⟹ d² = 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 BB₁ = B₁B₂ = B₂B₃ = 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.

 

OR

Consider the following figure:

 

Given: In ∆ABC, ∠ABC = 90°

To prove: AC² = AB² + 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)

∴ AB² = AC x DC...............................(i)

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

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

∴ BC² = AC x DC   ............................(ii)

AB² + BC² = AC x AD + AC x DC.........[Adding equations (i) and (ii)]

              = AC [AD + DC]

               = AC x AC

               =AC²

∴ AC² = AB² + BC²