Areas Related To A Circle Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 13.2
  1. An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at…
  2. The radius of a circular ground is 63 m. There is 7 m wide road inside the…
  3. A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10…
  4. The length of a minute hand of a circular dial is 10 cm. Find the area of the…
  5. The radius of a field in the form of a sector is 21 m. The cost of constructing…
  6. The length of a side of a square field is 20 m. A cow is tied at the corner by…
  7. A chord of a circle of radius 42 cm subtends an angle of measure 60 at the…
  8. A chord of a circle, of length 10 cm, subtends a right angle at the centre.…
Exercise 13.3
  1. A rectangle whose length and breadth are 12 cm and 5 cm respectively is…
  2. ABCD, square park, has each side of length 80 m. There is a flower bed at each…
  3. What will be the cost of covering the white portion in figure 13.21 with a…
  4. ABCD is a square plate of 1 m length. As shown in figure circles are drawn with…
  5. bar {oa} and bar {ob} are two mutually perpendicular radii of a circle of…
Exercise 13
  1. The area of a circular park is 616 m2. There is a 3.5 m wide track around the…
  2. A man is cycling in such a way that the wheels of the cycle are making 140…
  3. If a chord bar {ab} of (O, 20) subtends right angle at O, find the area of…
  4. There are two arcs {apb} of 6 (O, OA) and {aqb} of 6 (M, MA) as shown…
  5. The length of a side of a square garden ABCD is 70 m. A minor segment of 6…
  6. ABCD is a square of side 20 cm. Find the area of blue coloured region formed by…
  7. On a circular table top of radius 30 cm a design is formed leaving an…
  8. Figure 13.27 shows a kite formed by a square PQRS and an isosceles right…
  9. In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area…
  10. A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find…
  11. If an arc of a circle subtends an angle of measure 0 at the centre, then the…
  12. The area of a sector is given by the formula ….(r is the radius and 1 is the…
  13. bar {oa} and bar {ob} are the two mutually perpendicular radii of a circle…
  14. A sector subtends an angle of measure 120 at the centre of a circle having…
  15. If the area and the circumference of a circle are numerically equal, then r =…
  16. The length of an arc subtending an angle of measure 60 at the centre of a…
  17. The area of a minor sector of (O, 15) is 150. The length of the corresponding…
  18. If the radius of a circle is increased by 10 %, then corresponding increase…
  19. If the ratio of the area of two circles is 1: 4, then the ratio of their…
  20. The area of the largest triangle inscribed in a semi-circle of radius 8 is ………
  21. If the circumference of a circle is 44 then the length of a side of a square…
  22. The length of minute hand of a clock is 14 cm. If the minute hand moves from…
  23. The length of minute hand of a clock is 15 cm. If the minute hand moves for…

Exercise 13.2
Question 1.

An arc of a circle whose radius is 21 cm subtends an angle of measure 120 at the centre. Find the length of the arc and area of the sector.


Answer:

Given radius of the circle (r)= 21 cm


Measure of angle (θ)= 120o


Length of the arc = 



= 44 cm


Area of the sector formed by the arc = 



= 462 cm2




Question 2.

The radius of a circular ground is 63 m. There is 7 m wide road inside the ground as shown in figure 13.10. The blue coloured portion of the road, shown in figure 13.10 is to be repaired. If the rate of repair work of the road costs Rs. 25 per m2, find the total cost of repair.




Answer:


Given the rate of repair work per m2 = Rs. 25


The radius of the outer circle (R1) = 63 cm


radius of the inner circle (R2) = 63 – 7


= 56 cm


(∵ the distance between the inner circle and outer circle is 7 cm)


Let the measured angle is 60o


area of the repair road = () – ()


= () – (


= 436.15 m2


cost required to repair the road = 436.15 × 25


= Rs. 10,903.9445 (approximately)



Question 3.

A regular hexagon of side 10 cm is cut from a plane circular sheet of radius 10 cm as shown in the figure 13.11. Find the area of the remaining part of the sheet. ( = 1.73) (= 3.14)




Answer:

Given radius of the circle (r) = 10 cm


Side of the hexagon (a) = 10 cm


we can divide the hexagon into 6 equilateral triangles as follows.



the side of an equilateral triangle will be 10 cm


we have to find the area of the shaded region as shown in fig. 13.11


area of the shaded region = area of circle - area of hexagon


= area of circle - area of 6 right equilateral triangles


= π r2 – 6 ()


= 3.14 × 102 – 6 ( × 102)


= 54.5 cm2



Question 4.

The length of a minute hand of a circular dial is 10 cm. Find the area of the sector formed by the present position and the position after five minute of the minute hand. (π = 3.14)


Answer:

length of the minute hand (r) = 10 cm


(In clocks the length of minute hand will be radius of that clock dial)


The sector formed by the minute hand after 5 minutes will be as follows.



θ =  = 30o


Area of the sector formed by minute hand = 



= 26.166 cm2



Question 5.

The radius of a field in the form of a sector is 21 m. The cost of constructing a wall around the field is Rs. 1875 at the rate of Rs. 25 per meter. If it costs Rs. 10 per m2 to till the field, what will be the cost of tilling the whole field?


Answer:

Radius of the sector (r) = 21 m


cost for constructing a meter wall = Rs. 25


cost of constructing a wall around the field = Rs. 1875


∴ Total length of wall =  = 75 m


perimeter of the sector = 75 m


l + r + r = 75 m


l + 2(21) = 75 m


l = 33 m


∴ Arc length of CDB (L) = 33m


Area of the field = 



= 346.5 m2


Cost of tiling 1 m2 = Rs. 10


∴ Cost of tiling 346.45 m2 = 10 × 346.5


= Rs. 3465



Question 6.

The length of a side of a square field is 20 m. A cow is tied at the corner by means of a 6 m long rope. Find the area of the field which the cow can graze. Also find the increase in the grazing area, if length of the rope is increased by 2 m. (π = 3.14)


Answer:

length of side of a square field (a) = 20 m


length of the rope (r1) = 6 m



since it is square the center angle (θ) = 90o


area of the gazing field = 



= 28.26 m2


If the length of rope is increased by 2m


The radius of the gazing field (r2) = 6m + 2m = 8m


area of the gazing field = 



= 50.24 m2


increase in gazing field = 50.24 – 28.26


= 27.98 m2



Question 7.

A chord of a circle of radius 42 cm subtends an angle of measure 60 at the centre. Find the area of the minor segment of the circle. ( = 1.73)


Answer:

radius of the circle (r) = 42 cm


angle of measure (θ) = 60o


area of the segment = 



= 923.16 cm2


since the center angle of measure is 60o that is a equilateral triangle.


side of the triangle (a) = 42 cm


area of the equilateral triangle = 



= 762.93 cm2


area of minor segment (BDC) = area of segment – area of triangle


= 923.16 – 762.93


= 160.23 cm2




Question 8.

A chord of a circle, of length 10 cm, subtends a right angle at the centre. Find the areas of the minor segment and the major segment formed by the chord. (π = 3.14)


Answer:

Length of the chord = 10 cm


Angle of measure = 90o



since triangle ABC is a right-angled triangle


AB, AC is same as the radius of the circle


BC2 = AB2 + AC2


102 = r2 + r2


100 = 2 r2


50 = r2


r = √50


Area of segment = 



= 39.25 cm2


Area of the triangle = AB× AC


× (√50) × (√50)


= 25 cm2


Area of minor segment = Area of segment - Area of the triangle


= 39.25 – 25


= 14.25 cm2


Area of major segment = Area of circle - Area of minor segment


= π r2 – 14.25


= 3.14 × (√50)2 – 14.25


= 142.82 cm2




Exercise 13.3
Question 1.

A rectangle whose length and breadth are 12 cm and 5 cm respectively is inscribed in a circle. Find the area of the blue coloured region, as shown in the figure 13.19.



Figure 13.19


Answer:

given length (l) = 12 cm


breadth (b) = 5 cm


we can notice from the fig. that the Hypotenuse of ADB is same as diameter of the circle.


from the triangle ABD


BD2 = AD2 + AB2


BD = √(52 + 122)


(∵ AD is breadth, AB is length of the rectangle)


BD = √(25 + 144)


= √169


= 13 cm


∴ diameter of the circle = 13 cm


Radius of the circle (r) = 6.5 cm


Area of the shade region = Area of circle – area of rectangle


= π r2 – b × h


= π × 6.52 – 5 × 12


= 72.732 cm2



Question 2.

ABCD, square park, has each side of length 80 m. There is a flower bed at each corner in the form of a sector of radius 7 m, as shown in figure 13.20. Find the area of the remaining part of the park.



Figure 13.20


Answer:

side of a square park (a) = 80 cm


radius of the sectors (r) = 7 cm


we can notice that all the sectors can form a circle of radius 7 cm


so, the area of remaining part of park =


area of square – area of circle


= a2 - π × r2


= 802 - π × 72


= 6,240.06 cm2



Question 3.

What will be the cost of covering the white portion in figure 13.21 with a silver foil if the rate is Rs. 100 per m2? (π = 3.14)



Figure 13.21


Answer:

Given side of a square (a) = 12 cm


Radius of each sector (r) = 12 cm


since it is square the angle (θ) = 90o


let divide the square into 2 parts. (BAPC, DAQC)


Area of part – 1


The area of the sector BAPC = 


 × 122 × 90


= 113.04 cm2


Area of square = a2


= 122


= 144 cm2


Area of part 1 (APC) = Area of square - The area of the sector BAPC


= 144 – 113.04


= 30.96 cm2


∵ The shapes are identical


The area of part 2 (AQC) = 30.96 cm2


Area of the colored region = area of PART 1+ area of part 2


= 30.96 + 30.96


= 61.92 cm2


Cost for covering 1 m2 silver foil = 100


Cost for covering 61.92 m2 silver foil = 61.92 × 100


= 6192



Question 4.

ABCD is a square plate of 1 m length. As shown in figure circles are drawn with their center at A, B, C, D respectively, each with radius equal to 42 cm. The blue coloured part at each corner, as shown in the C figure 13.22 is cut. What is the area of the remaining portion of the plate?



Figure 13.22


Answer:

Side of square plate (a) = 1 m = 100cm


radius of the segment (r) = 42 cm


we can notice that all the sectors can form a circle of radius 42 cm


the area of the remaining portion of the plate =


Area of square plate - area of the circle


= a2 - π r2


= 1002 - π × 422


= 4,458.23 cm2


=0.4458 m2



Question 5.

 and  are two mutually perpendicular radii of a circle of radius 10.5 cm. D and OD = 6 cm. Find the area of blue coloured region shown in figure 13.23.



Figure 13.23


Answer:

given OA = 10.5 cm = radius (r)


OD = 6 CM


we can see that ODA is a right-angled triangle


area of triangle = OA× OD


10.5 × 6


= 31.5 cm2


we can see that segment OACB is  of the circle.


so, the area of OACB =  × r2


 × 10.52


= 86.54625 cm2


area of blue colored region = area of OACB - area of triangle


= 86.54625 – 31.5


= 55.04625 cm2




Exercise 13
Question 1.

The area of a circular park is 616 m2. There is a 3.5 m wide track around the park running parallel to the boundary. Calculate the cost of fencing on the outer circle at the rate of Rs. 5 per meter.


Answer:

Area of the circular park = 616 m2


width of the track = 3.5 m



Area of the circular park = 616 m2


π × r2 = 616


r2 = 196


r = √196


r = 14 m


∴ the radius of the circular park (r1) = 14 m


the radius of the park with track (r2) = 14 + 3.5


= 17.5 m


Circumference of the track = 2π (r2)


= 2×  (17.5)


= 110 m


Cost of fencing per meter = Rs. 5


Cost of fencing 110 m = 110 × 5


= 550



Question 2.

A man is cycling in such a way that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, then how much distance will he cover in 2 hours?


Answer:

Speed of the cycle = 140RPM


Diameter of the wheel = 60 cm


No. of revolution in 2 hours = 120 × 140


= 16,800


Distance covered in 2 hours = Revolutions × Diameter


= 16,800 × 60


= 1008000 cm


= 10080 m


= 10.8 km



Question 3.

If a chord  of (O, 20) subtends right angle at O, find the area of the minor segment.


Answer:

Given radius = 20cm


if we connect the chord to the center point that forms a segment.


the centre measure angle (θ) =90o


we can observe a right-angled triangle OAB



Area of segment = 



= 314 cm2


Area of the triangle = OA× OB


× (20) × (20)


= 200 cm2


Area of minor segment = Area of segment - Area of the triangle


= 314 – 200


= 114 cm2



Question 4.

There are two arcs  of  (O, OA) and of  (M, MA) as shown in figure 13.24. Find the area enclosed by two arcs.



Figure 13.24


Answer:

from the figure we can notice all these


1. OAB is an equilateral triangle


(∵ All the sides are equal)


2. OAPB is a minor segment with Radius 14 cm, the angle (θ) between them is 60o


3. ABQ is a semi-circle with centre as M radius 7 cm


Area of the segment = 



= 102.62 CM2


Area of the triangle =  × OA2


 × 142


= 84.870 CM2


Area of the semi-circle =


Area of the segment - Area of the triangle


= 102.62 - 84.870


= 17.75 cm2


Area of the semi-circle =  × r2


 × 72


= 76.96 cm2


area enclosed by two arcs =


Area of the semi-circle - Area of the semi-circle


= 76.96 – 17.75


= 59.20 cm2 (approximately)



Question 5.

The length of a side of a square garden ABCD is 70 m. A minor segment of  (O, OA) is drawn on each of two opposite sides for developing lawn, as shown in figure 13.25. Find the area of the lawn.



figure 13.25


Answer:

side of a square garden ABCD (a) = 70 m


From the fig. in the below we can see that


1. OAXD is a segment


2. OAD is an right-angled triangle.


3. AXD is a minor segment


OA, OD are the radii of the segment


from the triangle OAD


AD2 = OA2 + OD2


702 = r2 + r2


2r2 = 4900


r2 = 2450


r = √2450


Area of minor segment = Area of segment - Area of triangle



 -  × OA2


 -  ×


= 1924.22 – 1225


=699.22 cm2


= 700 cm2


since AXD, BYC are same


∴ Area of lawn = 2 × area of minor segment


= 2 × 700


= 1400 cm2



Question 6.

ABCD is a square of side 20 cm. Find the area of blue coloured region formed by the semi-circles drawn on each side as shown in figure 13.26. ( = 3.14)



figure 13.26.


Answer:

side of a square (a) = 20 cm


we can see the semi-circle in the square having 10 cm as radius (r)


area of the colored region =


area of 4 semi-circles - area of the square


= 4 × ( × r2) - a2


= 2 (π × 102) - 202


= 228 cm2



Question 7.

On a circular table top of radius 30 cm a design is formed leaving an equilateral triangle inscribed in a circle. Find the area of the design. (π = 3.14)


Answer:

radius of the circle = 30 cm


Area of the circle = π r2



= 3.14 × 302


= 2826 cm2


PQR is an equilateral triangle


∠ QPR = 60O


∠ QOR = 2 × ∠ QPR = 120o


In OQR triangle,


let OS is perpendicular to QR


In triangle OQR as OQ = OR


∠ QOS = ∠ QPR = 60O


In triangle OSQ ∠S = 90o


sin 60o = 


 = 


QS = 15 √3 cm


Now,


QR = 2 QS = 2 (15 √3) = 30√3 cm


Area of equilateral triangle =  × QR2


 × (30√3)2


= 1167.75 CM2


Area of design = Area of circle - area of triangle PQR


= 2826 – 1167.75


= 1658.25 CM2



Question 8.

Figure 13.27 shows a kite formed by a square PQRS and an isosceles right triangle ARB whose congruent sides are 5 cm long.  is an arc of a  (R, 42). Find the area of the blue coloured region.



Figure 13.27


Answer:

Side of square (r) = 42 cm


In triangle RAB,


RA = RB = 5 cm


In minor sector RQCS angle is 90o


Area of the shaded region = Area of minor sector RQCS + Area of RAB


 +  × RA × RB


 +  × 5 × 5


= 1384.74 + 12.5


= 1372.24 cm2



Question 9.

In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area of the blue coloured region. (π = 3.14)



Figure 13.28


Answer:

side of square (a) = 8 cm


from the fig. we can see that  of circle is placed at four parts of the square. if we join them a circle is formed with radius (r1) 2 cm


Another circle is placed at centre of the square with radius (r2) 2 cm


Area of colored region = Area of square - Area of circles at 4 corners + Area of circle at centre


= a2 – (4 × ( × r12) + (π r22))


= a2 – (π r12 + π r22)


= 82 – (3.14 (22 + 22))


= 38.88 cm2



Question 10.

A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find the area of the blue coloured region shown in figure 13.29. (= 3.14)



Figure 13.29


Answer:

from the given figure


PQR is a right-angle triangle


PQ = 8 cm


QR = 6 cm


PR2 = PQ2 + QR2


PR2 = 82 + 62


PR = √100 = 10 cm


Radius of the incircle (r) =  =  =  = 2 CM


area of coloured region = Area of triangle - area of circle


 × PQ × QR - π r2


 × 8 × 6 - π 22


= 24 – 12.566


= 11.433 cm2



Question 11.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If an arc of a circle subtends an angle of measure 0 at the centre, then the area of the minor sector is…..

A. 

B. 

C. 

D. 


Answer:

we know that area of minor sector = 


So, option(d) is the correct answer.


Question 12.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a sector is given by the formula ….

(r is the radius and 1 is the length of an arc.)

A.  rl

B. r2/ l

C. rl

D. rl


Answer:

we know that when arc length (l) and radius (r) of sector is given


Area of the sector is =  rl


So, option(a) is the correct answer.


Question 13.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

and are the two mutually perpendicular radii of a circle having radius 9 cm. The area of the minor sector corresponding to AOB is cm2( = 3.14)

A. 63.575

B. 63.585

C. 63.595

D. 63.60


Answer:

Given radius (r)= 9 cm


∵ They are perpendicular to each other θ = 90o


area of the minor sector = 




= 63.585 cm2


So, option(b) is the correct answer.


Question 14.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

A sector subtends an angle of measure 120 at the centre of a circle having radius of 21 cm. The area of the sector is ……….. cm2.

A. 462

B. 460

C. 465

D. 470


Answer:

Given radius (r)= 21 cm


angle(θ) = 120o


area of the minor sector = 




= 461.58 cm2


= 462 cm2


So, option(a) is the correct answer.


Question 15.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the area and the circumference of a circle are numerically equal, then r = …………

A. π

B. 

C. 1

D. 2


Answer:

Area of the circle = πr2


circumference of the circle = 2πr


given, Area = Circumference


2πr = πr2


2 = r


So, option(d) is the correct answer.


Question 16.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of an arc subtending an angle of measure 60 at the centre of a circle whose area is 616 is ……….

A. 

B. 66

C. 

D. 33


Answer:

Area of circle = 616cm2


πr2 = 616


r2 = 196


r = 14 cm


Given θ = 60o


area of the minor sector = 




 cm2


So, option(c) is the correct answer.


Question 17.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of a minor sector of (O, 15) is 150. The length of the corresponding arc is (π = 3.14)

A. 30

B. 20

C. 90

D. 15


Answer:

Radius of the minor sector= 15 cm


Area of minor sector = 150


Area of a minor sector =  r l


150 = × 15 × l


l = 20 cm


So, option(b) is the correct answer.


Question 18.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the radius of a circle is increased by 10 %, then corresponding increase in the area of the circle is ………. (π = 3.14)

A. 19 %

B. 10 %

C. 21 %

D. 20 %


Answer:

r be the radius of the circle


Area of the circle = πr2


10% increase in radius = 1.1r


Area of circle with new radius = π(1.1r)2


= 1.21πr2


increase in area = 1.21πr2 - πr2


= 0.21 πr2


∴ 21% increase in the area of circle


So, option(c) is the correct answer.


Question 19.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the ratio of the area of two circles is 1: 4, then the ratio of their circumference …..

A. 1: 4

B. 1: 2

C. 4: 1

D. 2: 1


Answer:

let r1, r2 are the radii of two circles


given,


πr12: πr22 = 1 : 4


r12: r22 = 1 : 4


r1 : r2 =1 : 2


r2 = 2 r1


circumference of 1st circle = 2πr1


circumference of 2nd circle = 2πr2 = 2π(2r1) = 4πr1


ratio of circumference = 2πr1 : 4πr1 = 1 : 2


So, option(b) is the correct answer.


Question 20.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The area of the largest triangle inscribed in a semi-circle of radius 8 is ……

A. 8

B. 16

C. 64

D. 256


Answer:

Given radius of semi-circle = 8 cm


For a largest triangle in semi-circle


base will be its diameter of semi-circle


height will be radius of semi-circle


so, its base = 2 × radius = 2 × 8 = 16 cm


height = 8 cm


Area of the triangle =  × base × height


 × 16 × 8


= 64 cm2


So, option(c) is the correct answer.


Question 21.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

If the circumference of a circle is 44 then the length of a side of a square inscribed in the circle is ….

A. 

B. 

C. 

D. 


Answer:

Given circumference = 44 cm


2πr = 44


πr = 22


r = 7 cm


Given that square inscribed in a circle.


so, diagonal of square will be equal to diameter of circle.


diagonal = 2 × radius = 2 × 7 = 14 cm


diagonal = √2 × side of square


14 = √2 × side


side = 7√2 cm


Question 22.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 14 cm. If the minute hand moves from 2 to 11 on the circular dial, then area covered by it is cm2.

A. 154

B. 308

C. 462

D. 616


Answer:

The angle moved by minute hand = 90o


length of minute hand (r) = 14 cm


Area of covered by minute hand = 




= 153.93 cm2


= 154 cm2


So, option(a) is the correct answer.


Question 23.

Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:

The length of minute hand of a clock is 15 cm. If the minute hand moves for 20 minutes on a circular dial of a clock, the area covered by it is …….. cm2. (= 3.14)

A. 235.5

B. 471

C. 141.3

D. 706.5


Answer:

The angle moved by minute hand = 120o


length of minute hand (r) = 15 cm


Area of covered by minute hand = 




= 235.619 cm2


= 235.5 cm2


So, option(a) is the correct answer.


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HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

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