Q.1. If the quadratic equation px2– 2√5px + 15 = 0 has two equal roots, then find the value of p.
Answer. The given quadratic equation can be written as px2– 2√5px + 15 = 0
a = p, b = -2√5p, c = 15
For equal roots, D = 0
D = b2 – 4ac
0 = (– 2√5p)2 – 4 ×p × 15
0 = 4 ×5p2 – 60p
0 = 20p2 – 60p
p = 60p / 20p = 3
∴ p = 3
Q.2. In Figure 1, a tower AB is 20 m high and BC, its shadow on the ground, is 20-√3 m long. Find the Sun’s altitude.
Answer.
Q.3. Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.
Answer. Total outcomes = 6n = 62 = 36
Possible outcomes having the product of the two numbers on the top of the dice as 6 are (3 × 2, 2 × 3, 6 × 1, 1 × 6), i.e., 4
P(Product of two numbers is 6) = 4/36 = 1/9
Q.4. In Figure 2, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.
Answer.
Section – B
Q.5. In Figure 3, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.
Answer.
Q.6. In Figure 4, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm2, then find the lengths of sides AB and AC.
Answer.
Q.7. Solve the following quadratic equation for x: 4x2 + 4bx -(a2 – b2) = 0
Answer.
Q.8. In an AP, if S5+ S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms.
Answer.
Q.9. The points A(4,7), B(p,3) and C(7,3) are the vertices of a right triangle, right-angled at B. Find the value of p.
Answer.
Q.10. Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.
Answer.
Section – C
Q.11. The 14th term of an AP is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.
Answer.
Q.12. Solve for x:
Answer.
Q.13. The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 1500 √3 m, find the speed of the plane in km/hr.
Answer.
Q.14. If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/5 AB, where P lies on the line segment AB.
Answer.
Q.15. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3 . If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer.
Q.16. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment. [Use π = 22/7 ]
Answer.
Q.17. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs 100 per sq. m, find the amount, the associations will have to pay. [Use π = 22/7 ] What values are shown by these associations?
Answer.
Q.18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.
Answer.
Q.19. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per 100 sq. cm. [Use π = 3.14]
Answer. Let the side of cuboidal block (a) = 10cm
Let the radius of hemisphere be r
Side of cube = Diameter of hemisphere
Largest possible diameter of hemisphere = 10cm
Q.20. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π = 22/7 ]
Answer.
Section – D
Q.21. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.
Answer.
Q.22. Find the 60th term of the AP 8,10,12,…, if it has a total of 60 terms and hence find the sum of its last 10 terms.
Answer.
Q.23. A train travels at a certain average speed for a distance of 54 km and then travels a . distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
Answer.
Q.24. Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively
To Prove : Lengths of tangents are equal i.e. PQ = PR
Construction: Join OQ, OR and OP
Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]
So, ∠OQP = 90°
Hence ΔOQP is right triangle.
Q.25. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Answer.
Q.26. Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.
Answer.
Q.27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.
Answer.
Q.28. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is
(i) a card of spade or an ace. (ii) a black king.
(iii) neither a jack nor a king. (iv) either a king or a queen.
Answer.
Q.29. Find the values of k so. that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.
Answer.
Q.30. In Figure 5, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).
Answer.
Q.31. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. [Use π = 22/7 ]
Answer.
SET II
Q.10. If A(4, 3), B(-l, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
Answer.
Q.18. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is 1256 cm2. [Use π= 3.14]
Answer.
Q.19. Solve for x:
Answer.
Q.20. The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.
Answer.
Q.28. A bus travels at a certain average speed for a distance of 75 km and then travels a distance of 90 km at an average speed of 10 km/h more than the first speed. If it takes 3 hours to complete the total journey, find its first speed.
Answer.
Q.29. Prove that the tangent at any point of a circle is perpendicular to the radius through the I point of contact.
Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively
To Prove : Lengths of tangents are equal i.e. PQ = PR
Construction: Join OQ, OR and OP
Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]
So, ∠OQP = 90°
Hence ΔOQP is right triangle
Q.30. Construct a right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD, the perpendicular from B on AC. Draw the circle through B, C and D and construct the tangents from A to this circle.
Answer.
Q.31. Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.
Answer.A(k + 1, 1), B(4, -3) and C(7, -k)
Area of ΔABC = ½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
6 = ½ [(k+1)(-3 + k) + 4(-k-1) + 7(1+3)]
12 = [-3k + k2 -3 + k-4k-4 + 28]
12 = [ k2 -6k + 21]
⇒ k2 -6k + 21-12 ⇒ k2 -6k + 9
⇒ k2 -3k -3k + 9 ⇒ k(k-3)-3(k-3) = 0
⇒ k-3 = 0 ⇒ k-3 = 0
⇒ k = 3 ⇒ k = 3
Solving get k = 3
SET III
Q.10. Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0
Answer.
Q.18. The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.
Answer.
Q.19. Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP=2/5 AB.
Answer.
Q.20. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is 2/5, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Answer.
Q.28. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
Answer.
Q .29. Arithmetic Progressions, 12,19,… has 50 terms. Find its last term. Hence find the sum of its last 15 terms.
Answer.
Q.30. Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are 5/7 times the corresponding sides of ∆ABC.
Answer.
Q.31. Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Q.1. In Fig. 1, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.
Answer.
∠ACB = 90° …………..[Angle in the semi-circle
In ΔABC, ∠CAB + ∠ACB + ∠CBA = 180°
30° + 90° + ∠CBA = 180°
∠CBA = 180° – 30° – 90° = 60° [Angle-sum-property of a Δ]
∠PCA = ∠CBA ………….[Angle in the alternate Segment]
∴ ∠PCA = 60°
Q. 2. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?
Answer. As we know, a2 – a1 = a3 -a2
2k -1-(k+9) = 2k +7 – (2k -1)
2k -1- k – 9 = 2k +7 – 2k + 1
k – 10 = 8
∴ k = 8 + 10 = 18
Q 3. A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.
Answer.
Let AC be the ladder
Cos60° = AB/AC
½ = 2.5/AC
∴ Length of ladder, AC = 5cm
Q. 4. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Q. 5. If -5 is a root of the quadratic equation 2×2+ px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.
Answer. 2x2 + px – 15 = 0
Since (-5) is a root of the given equation
∴ 2(-5)2 + p(-5) – 15 = 0
= 2(25) – 5p – 15 = 0
= 50 – 15 = 5p
= 35 = 5p
= p = 7 ——(i)
p(x2+x) + k ⇨ px2 + px + k = 0
Here, a = p, b = p, c = k
D = 0 (Roots are equal)
b2 – 4ac = 0 , (p)2 – 4(p)k = 0
(7)2 – 4(7)k = 0
49 – 28k = 0
∴ k = 49/28 = 7/4
Q. 6. Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. Find the coordinates of P and Q.
Answer.
Q. 7. In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that: AB + CD = BC + DA.
Answer.
AP = AS
BP = BQ
CR = CQ
DR = DS
[∴ Tangents drawn from an external point are equal in length]
By adding (i) to (iv)
(AP + BP) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (BQ + CQ) + (AS + DS)
∴ AB + CD = BC + AD (Hence Proved)
Q. 8. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
Answer.
Q. 9. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Q. 10. In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°.
Answer.
Section – C
Q 11. In Fig. 4, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)
Answer.
Q 12. In Fig. 5, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs 500/sq. metre. (Use π = 22/7 )
Answer.
Q 13. If the point P(x, y) is equidistant from the points A (a + b,b – a) and B(a -b,a + b), prove that bx = ay.
Answer.
Q 14. In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (Use π= 22/7 )
Answer.
Q 15. If the ratio of the sum of first n terms of two A.P’s is (7n + 1) : (4n + 27), find the ratio of their mth terms.
Answer.
Q 16. Solve for x:
Answer.
Q 17. A conical vessel, with bash radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use π= 22/7)
Answer.
Q 18. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 (5/9) cm. Find the diameter of the cylindrical vessel.
Answer.
Q 19. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill.
Answer.
Q 20. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (ii) at least two tails.
Answer.
Section – D
Q 21. Due to heavy floods in a State, thousands were rendered homeless. 50 schools collectively offered to the State Government to provide place and the canvas for 1,500 tents to be fixed by the Government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? (Use π= 22/7)
Answer.
Q 22. Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively
To Prove : Lengths of tangents are equal i.e. PQ = PR
Construction: Join OQ, OR and OP
Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]
So, ∠OQP = 90°
Hence ΔOQP is right triangle
Q. 23. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.
Answer.
Q 24. In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of DO’/CO.
Answer. Given: two equal circles, with centres O and O’, touch each other at point X. OO’ is produced to meet the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC.
Q 25. Solve for x:
Answer.
Q 26. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3= 1.73)
Answer.
Q 27. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.
Answer.
Q 28. In Fig. 8, the vertices of ∆ABC are A(4, 6), B(l, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB= AE/AC= 1/3 .Calculate the area of ∆ADE and Calculate the area compare it with area of ∆ABC.
Answer.
Q 29. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16.
Answer.
Q 30. In Fig. 9, is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is r :
Answer.
Q 31. A motor boat whose speed is 24 km/h in still water takes 1 hr more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer.
SET II
Q 10.Solve for x:
Answer.
Q 18. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Answer.
Q 19. If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.
Answer.
Q 20. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is: (i) a black King (ii) a card of red colour (iii) a card of black colour
Answer.
Q 28. Draw an isosceles ∆ABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are 3/4 of the corresponding sides of ∆ABC.
Answer.
Q 29. Prove that the tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.
Answer. Given: XY is a tangent at point P to the circle with centre O.
To prove: OP⏊XY
Construction: Take a point Q on XY other than P and join OQ.
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.
∴ OQ > OP
This happen with every point on the line XY except the point P.
OP is the shortest of all the distances of the point O to the points of XY
∴ OP⏊XY
Q 30. As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30° to 60°. Find the distance travelled by the ship during the period of observation. (Use √3 = 1.73)
Answer.
Q 31. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park.
Answer.
SET III
Q 10. Solve for x:
Answer.
Q 18. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card (i) is divisible by 9 and is a perfect square (ii) is a prime number greater than 80.
Answer.
Q 19. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.
Answer. Let three consecutive natural numbers are x, x+1,x+2
According to the question, (x+1)2 -[(x+2)2 –x2 ] = 60
⇒ x2+1+2x -[x2+4+4x –x2 ] = 60
⇒ x2 + 1 + 2x – x2– 4 – 4x + x2 = 60
⇒ x2 – 2x – 63 = 0
⇒ x2 – 9x +7x – 63 = 0
⇒ x(x –9) + 7(x – 9) = 0
⇒ (x –9)(x + 7) = 0
⇒ x –9 = 0 , x = 9
⇒ x + 7 = 0 , x = -7
Natural No’s can not be -ve, ∴ x = 9
∴ Numbers are 9, 10, 11
Q 20. The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that S1+ S3 = 2S2.
Answer.
Q 28. Two pipes running together can fill a tank in 11 (1/9) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.
Answer.
Q 29. From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation.
Answer.
Q 30. Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are 4/5 of the corresponding sides of first triangle.
Answer.
Q 31. A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.
Answer. x can be any one of 1,4,9, or 16, i.e. 4 ways y can be any one of 1,2,3 or 4 ways
Q.1. What is the common difference of an A.P. in which a21 – a7 = 84 ?
Solution: Given, a21 – a7 = 84
⇒ (a + 20d) – (a + 6d) = 84
⇒ a + 20d – a – 6d = 84
⇒ 20d – 6d = 84
⇒ 14d = 84
Hence common difference = 6
Q. 2.If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP.
Solution: Given, ∠APB = 60°
∠APO = 30°
In right angle ΔOAP,
OP/OA = cosec 30°
⇒ OP/a = 2
⇒ OP = 2a.
Q. 3.If a tower 30 m high, casts a shadow 10√3 m long on the ground, then what is the angle of elevation of the sun?]
Solution: In ΔABC,
tan θ = AB/ BC
⇒ tan θ = 30/10√3 = √3
⇒ tan θ = tan 60°
⇒ θ = 60°
Hence angle of elevation is 60°.
Q. 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0-18. What is the number of rotten apples in the heap?
Solution: Total apples = 900
P(E) = 0.18
No. of rotten apples / Total No. of apples = 0.18
No. of rotten apples / 900 = 0.18
No. of rotten apples = 900 × 0.18 = 162
Section – B
Q. 5. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.
Solution: Given equation is px2 – 14x + 8 = 0
Let one root = α
then other root = 6α
Sum of roots = -b/a
α+6α=-(-14)/p
7α=14/p or α= 2/p ……….(1)
Product of roots = c/a
(α)(6α)=8/p
6α2=8/p ……….(2)
Putting value of α from eq. (i),
⇒6×(2/p)2 = 8/p
⇒6×4/p2 = 8/p
⇒24p = 8p2
⇒8p2-24p = 0
⇒8p(p-3) = 0
⇒ Either 8p = 0
p = 0
or (p-3) = 0
p = 3
For p=0, given condition is not satisfied
ஃ p=3
Q 6.Which term of the progression 20, 19¼ , 18½ , 17 ¾, … is the first negative term ?
Solution: Given, A.P. is 20, 19¼ , 18½ , 17 ¾, …
Q. 7. Prove that the tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.
Solution: Given, a circle of radius OA and centred at O with chord AB and tangents PQ & RS are drawn from point A and B respectively.
Draw OM ⊥ AB, and join OA and OB.
In ∆OAM and ∆OMB,
OA = OB (Radii)
OM = OM (Common)
∠OMA = ∠OMB (Each 90°)
∆OAM = ∆OMB (By R.H.S. Congurency)
∠OAM = ∠OBM (C.PC.T.)
Also, ∠OAP = ∠OBR = 90° (Line joining point of contact of tangent to centre is perpendicular on it)
On addition,
∠OAM + ∠OAP = ∠OBM + ∠OBR
⇒ ∠PAB = ∠RBA
⇒ ∠PAQ – ∠PAB = ∠RBS – ∠RBA
⇒ ∠QAB = ∠SBA
Hence Proved
Q. 8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA
Solution: Given, a quad. ABCD and a circle touch its all four sides at P, Q, R, and S respectively.
To prove: AB + CD = BC + DA
Now, L.H.S. = AB + CD
= AP + PB + CR + RD
= AS + BQ + CQ + DS (Tangents from same external point are always equal)
= (AS + SD) + (BQ + QC)
= AD + BC
= R.H.S.
Hence Proved.
Q 9.A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.
Solution: Let co-ordinate of P (0, y)
Co-ordinate of Q (x, 0)
Q. 10.If the distances of P(x, y), from A(5, 1) and B(-1, 5) are equal, then prove that 3x = 2y.
Q 12.The first term of an A.E is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.
Solution: Given, a = 5, an = 45, Sn = 400
We have, Sn = ⇒ 400 = n/2 [5 + 45]
⇒ 400 = n/2 [50]
⇒ 25n = 400
⇒ n = 16
Now, an = a + (n – 1) d
⇒ 45 = 5 + (16 – 1)d
⇒ 45 – 5 = 15d
⇒ 15d = 40
⇒ d = 8/3
So n = 16 and d = 8/3
Q 13.On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.
Solution:
Let height AB of tower = h m.
Q14. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Solution: Given, no. of white balls = 15
Let no. of black balls = x
Total balls = (15 + x)
According to the question,
P(Blackball) = 3 × P(White ball)
⇒ x/15+x = 3 × 15/15+x
⇒ x = 45
No. of black balls in bag = 45
Q 15.In what ratio does the point (2411, y) the line segment joining the points P(2, -2) and Q(3, 7) ? Also, find the value of y.
Solution: Let point R divides PQ in the ratio k : 1
Q 16. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semi-circle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.
Solution: Given, radius of large semi-circle = 4.5 cm
Q17. In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. [Use π = 227]
Solution: Angle for shaded region = 360° – 60° = 300°
Area of shaded region
Q 18.Water in a canal, 5-4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation ?
Solution: Width of canal = 5.4 m
Depth of canal = 1.8 m
Length of water in canal for 1 hr = 25 km = 25000 m
Volume of water flown out from canal in 1 hr = l × b × h = 5.4 × 1.8 × 25000 = 243000 m3
Volume of water for 40 min = 243000 × 40 60 = 162000 m3
Area to be irrigated with 10 cm standing water in field = Volume/ Height
= (162000×100)/10 m2
= 1620000 m2
= 162 hectare
Q 19.The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution: Slant height of frustum ‘l’ = 4 cm
Perimeter of upper top = 18 cm
Q 20. The dimensions of a solid iron cuboid are 4.4 m × 2.6 m × 1.0 m. It is melted and recast into hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.
Solution: Inner radius of pipe ‘r’ = 30 cm
The thickness of pipe = 5 cm
Outer radius ‘R’ = 30 + 5 = 35 cm
Now, Volume of hollow pipe = Volume of Cuboid
Section – D
Q. 21.Solve for x:
Solution:
Q 22.Two taps running together can fill a tank in 3 1/13 hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank ?
Solution: Let tank fill by one tap = x hrs
other tap = (x + 3) hrs
Together they fill by (3) 1/13 = 40/13 hrs
Either x – 5 = 0 or 13x + 24 = 0
x = 5, x = -24/13 (Rejected)
One tap fill the tank in 5 hrs
So other tap fill the tank in 5 + 3 = 8 hrs
Q 23.If the ratio of the sum of the first n terms of two A.P.S is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
Solution:
Ratio of the sum of first n terms of two A.P.s are
Hence ratio of 9th terms of two A.P.s is 24 : 19
Q 24.Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Solution: Given, a circle with centre O and external point P. |
Two tangents PA and PB are drawn.
To Prove: PA = PB
Construction: Join radius OA and OB also join O to P.
Proof: In ∆OAP and ∆OBP,
OA = OB (Radii)
∠A = ∠B (Each 90°)
OP = OP (Common)
∆AOP = ∆BOP (RHS cong.)
PA = PB [By C.PC.T.]
Hence Proved.
Q 25.In the given figure, XY and XY are two parallel tangents to a circle with centre O and another tangent AB with a point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Solution: Given, XX’ & YY’ are parallel.
Tangent AB is another tangent which touches the circle at C.
To prove: ∠AOB = 90°
Construction: Join OC.
Proof: In ∆OPA and ∆OCA,
OP = OC (Radii)
∠OPA = ∠OCA (Radius ⊥ Tangent)
OA = OA (Common)
∆OPA = ∆OCA (CPCT)
∠1 = ∠2 …(i)
Similarly, ∆OQB = ∆OCB
∠3 = ∠4 …(ii)
Also, POQ is a diameter of circle
∠POQ = 180° (Straight angle)
∠1 + ∠2 + ∠3 + ∠4 = 180°
From eq. (i) and (ii),
∠2 + ∠2 + ∠3 + ∠3 = 180°
⇒ 2(∠2 + ∠3) = 180°
⇒ ∠2 + ∠3 = 90°
Hence, ∠AOB = 90°
Hence Proved.
Q 26.Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are 3 4 times the corresponding sides of the ∆ABC.
Draw an angle 45° at B and 30° at C. They intersect at A.
Draw an acute angle at B.
Divide angle ray in 4 equal parts as B1, B2, B3 and B4.
Join B4 to C.
From By draw a line parallel to B4C intersecting BC at C’.
Draw another line parallel to CA from C’ intersecting AB ray at A. Hence, ∆A’BC’ is required triangle such that ∆A’BC’ ~ ∆ABC with A’B = ¾ AB
Q 27. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use √3 = 1.732]
Solution: Let aeroplane is at A, 300 m high from a river. C and D are opposite banks of river.
Q 28. If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k.
Solution: Since A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear points, so area of triangle = 0.
Q. 29. Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and
(ii) even product.
Solution: When two different dice are thrown together
Q. 30. In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.
Solution: Area of Shaded region = Area of a rectangle – Area of a semi-circle
Q. 31.In a rain-water harvesting system, the rainwater from a roof of 22 m × 20 m drains into a cylindrical tank having a diameter of base 2 m and height 35 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.
Solution: Volume of water collected in system = Volume of a cylindrical tank
Set II
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Section – B
Q. 10.Which term of the A.P. 8, 14, 20, 26,… will be 72 more than its 41st term?
Solution: A.P. is 8, 14, 20, 26,….
a = 8, d = 14 – 8 = 6
Let an = a41 + 72
a + (n – 1)d = a + 40d + 72
⇒ (n – 1) 6 = 40 × 6 + 72 = 240 + 72 = 312
⇒ n – 1 = 52
⇒ n = 52 + 1 = 53rd term
Section – C
Q. 18.From a solid right circular cylinder of height 24 cm and radius 0.7 cm, a right circular cone of the same height and same radius is cut out. Find the total surface area of the remaining solid.
Solution: Given, Height of cylinder ‘h’ = 2.4 cm,
Radius of base ‘r’ = 0.7 cm
Q. 19.If the 10th term of an A.E is 52 and the 17th term is 20 more than the 13th term, find the A.P.
Solution: Given, a10 = 52;
a17 = a13 + 20
⇒ a + 16d = a + 12d + 20
⇒ 16d = 12d + 20
⇒ 4d = 20
⇒ d = 5
Also, a + 9d = 52
⇒ a + 9 × 5 = 52
⇒ a + 45 = 52
⇒ a = 7
Therefore A.E = 7, 12, 17, 22, 27,….
Q. 20. If the roots of the equation (c^2 – ab) x^2 – 2(a^2 – bc) x + b^2 – ac = 0 in x are equal, then show that either a = 0 or a^3 + b^3 + c^3 = 3abc.
Solution:
Section – D
Q. 28. Solve for x:
Solution:
Q 29.A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less on the journey. Find the original speed of the train.
Solution: Let original speed of train = x km/hr
Increased speed of train = (x + 5) km/hr
Distance = 300 km
According to the question,
Q 30.A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.
Solution: Let AB is a tower, the car is at point D at 30° and goes to C at 45° in 12 minutes.
Q. 31.In the given figure, ΔABC is a right-angled triangle in which ∠A is 90°. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.
Solution: In right ΔBAC, by Pythagoras theorem,
Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Section – B
Q. 10.For what value of n, are the terms of two A.Ps 63, 65, 67,…. and 3, 10, 17,…. equal ?
Q. 18.A toy is in the form of a cone of radius 3-5 cm mounted on a hemisphere of the same radius on its circular face. The total height of the toy is 15*5 cm. Find the total surface area of the toy.
Solution: Given, radius of base ‘r’ = 3.5 cm
Total height of toy = 15.5 cm
Height of cone ‘h’ = 15.5 – 3.5 = 12 cm
Q. 19.How many terms of an A.E 9, 17, 25,… must be taken to give a sum of 636?
Solution: A.P. is 9, 17, 25,….,
Sn = 636
a = 9, d = 17 – 9 = 8
Q. 20. If the roots of the equation (a2 + b2) x2 – 2 (ac + bd) x + (c2 + d2) = 0 are equal, prove that a/b = c/d
Solution:
Section – D
Q. 28.Solve for x:
Solution:
Q. 29. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?
Solution: Let B can finish a work in x days
so, A can finish work in (x – 6) days
Together they finish work in 4 days
Now,
⇒ 4 (2x – 6) = x2 – 6x
⇒ 8x – 24 = x2 – 6x
⇒ x2 – 14x + 24 = 0
⇒ x2 – 12x – 2x + 24 = 0
⇒ x(x – 12) – 2(x – 12) = 0
⇒ (x – 12) (x – 2) = 0
Either x – 12 = 0 or x – 2 = 0
x = 12 or x = 2 (Rejected)
B can finish work in 12 days
A can finish work in 6 days.
Q. 30.From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in a same straight line with its base, with angles of depression 30° and 45°. Find the distance between cars.
[Take √3 = 1.732]
Solution: Let AB is a tower.
Cars are at point C and D respectively
Distance between two cars = x + y = 173.2 + 100 = 273.2 m
Q. 31.In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
Solution: Given, C (O, OB) with AC = 24 cm AB = 7 cm and ∠BOD = 90°
∠CAB = 90° (Angle in semi-circle)
Using pythagoras theorem in ∆CAB,
BC2 = AC2 + AB2 = (24)2 + (7)2 = 576 + 49 = 625
⇒ BC = 25 cm
Radius of circle = OB = OD = OC = 25/2cm
Area of shaded region = Area of semi-circle with diamieter BC – Area of ∆CAB + Area of sector BOD
Q.1. Distinguish between ‘Fixed Capital Account’ and ‘Fluctuating Capital Account’ on the basis of credit balance.
Answer: Fixed Capital Accounts always show a credit balance while fluctuating capital accounts may show credit or debit balance.
Q.2. A and B were partners in a firm sharing profits and losses in the ratio of 5 : 3. They admitted C as a new partner. The new profit sharing ratio between A, B and C was 3 : 2 : 3. A surrendered ⅕th of his share in favour of C. Calculate B’s sacrifice.
Q.3. P and Q were partners in a firm sharing profits and losses equally. Their fixed capitals were ₹ 2,00,000 and ₹ 3,00,000 respectively. The partnership deed provided for interest on capital @ 12% per annum. For the year ended 31st March, 2016, the profits of the firm were distributed without providing interest on capital. Pass necessary adjustment entry to rectify the error.
Answer:
Q.4. X Ltd. invited applications for issuing 500, 12% debentures of ₹ 100 each at a discount of 5%. These debentures were redeemable after three years at par. Applications for 600 debentures were received. Pro-rata allotment was made to all the applicants.
Pass necessary journal entries for the issue of debentures assuming that the whole amount was payable with the application.
Answer:
Q.5. Z Ltd. forfeited 1,000 equity shares of ₹ 10 each for the non-payment of the first call of ₹ 2 per share. The final call of ₹ 3 per share was yet to be made. Calculate the maximum amount of discount at which these shares can be reissued.
Answer: The maximum amount of discount at which these shares can be re-issued is ₹5 per share or ₹ 5000.
Q.6. Durga and Naresh were partners in a firm. They wanted to admit five more members to the firm. List any two categories of individuals other than minors who cannot be admitted by them.
Answer: Any two of the following:
• Persons of unsound mind / Lunatics
• Insolvent persons
• Any other individual who has been disqualified by law
Q.7. BPL Ltd. converted 500, 9% debentures of ₹ 100 each issued at a discount of 6% into equity shares of ₹ 100 each issued at a premium of ₹ 25 per share. Discount on issue of 9% debentures has not yet been written off.
Showing your working notes clearly, pass necessary journal entries for conversion of 9% debentures into equity shares.
Answer:
Q.8. Kavi, Ravi, Kumar, and Guru were partners in a firm sharing profits in the ratio of 3: 2: 2: 1. On 1.2.2017, Guru retired and the new profit sharing ratio decided between Kavi, Ravi and Kumar were 3 : 1: 1. On Guru’s retirement, the goodwill of the firm was valued at ₹ 3,60,000.
Showing your working notes clearly, pass necessary journal entry in the books of the firm for the treatment of goodwill on Guru’s retirement.
Answer:
Q.9. Disha Ltd. purchased machinery from Nisha Ltd. and paid to Nisha Ltd. as follows :
(i) By issuing 10,000, equity shares of ₹ 10 each at a premium of 10%.
(ii) By issuing 200, 9% debentures of ₹ 100 each at a discount of 10%.
(iii) Balance by accepting a bill of exchange of ₹ 50,000 payable after one month.
Pass necessary journal entries in the books of Disha Ltd. for the purchase of machinery and making payment to Nisha Ltd.
Answer:
Q.10. Ganesh Ltd. is registered with an authorized capital of ₹ 10,00,00,000 divided into equity shares of ₹ 10 each. Subscribed and fully paid-up capital of the company was ₹ 6,00,00,000. For providing employment to the local youth and for the development of the tribal areas of Arunachal Pradesh the company decided to set up a hydropower plant there. The company also decided to open skill development centers in Itanagar, Pasighat, and Tawang. To meet its new financial requirements, the company decided to issue 1,00,000 equity shares of ₹ 10 each and 1,00,000, 9% debentures of ₹ 100 each. The debentures were redeemable after five years at par. The issue of shares and debentures was fully subscribed. A shareholder holding 2,000 shares failed to pay the final call of ₹ 2 per share. Show the share capital in the Balance Sheet of the company as per the provisions of Schedule III of the Companies Act, 2013. Also, identify any two values that the company wishes to propagate.
Answer:
Values (Any two):
• Providing employment opportunities to the local youth.
• Promotion of development in tribal areas.
• Promotion of skill development in Arunachal Pradesh.
• Paying attention to regions of social unrest.
(Or any other suitable value)
Q.11. Madhu and Neha were partners in the firm sharing profits and losses in the ratio of 3: 5. Their fixed capitals were ₹ 4,00,000 and ₹ 6,00,000 respectively. On 1.1.2016, Tina was admitted as a new partner for 41 to share in the profits. Tina acquired her share of profit from Neha. Tina brought ₹ 4,00,000 as her capital which was to be kept fixed like the capitals of Madhu and Neha. Calculate the goodwill of the firm on Tina’s admission and the new profit sharing ratio of Madhu, Neha, and Tina. Also, pass necessary journal entry for the treatment of goodwill on Tina’s admission considering that Tina did not bring her share of goodwill premium in cash.
Answer:
(a) Calculation of Hidden Goodwill:
Tina’s share = 1⁄4
Tina’s Capital = ₹ 4,00,000
(a) Total capital of the new firm = 4,00,000 × 4 = 16,00,000
(b) Existing total capital of Madhu, Neha and Tina
(b) Calculation of New Profit Sharing ratio : Madhu’s new share
= 3/8 Neha’s new share = 5/8 – 1/4
= 3/8 Tina’s share = 1⁄4
i.e. 2/8 New Ratio = 3:3:2
(c)
Q.12. Ashok, Babu, and Chetan were partners in firm sharing profits in the ratio of 4 : 3 : 3. The firm closes its books on 31st March every year. On 31st December 2016, Ashok died. The partnership deed provided that on the death of a partner his executors will be entitled to the following :
(i) Balance in his capital account. On 1.4.2016, there was a balance of ₹ 90,000 in Ashok’s Capital Account.
(ii) Interest on capital @ 12% per annum.
(iii) His share in the profits of the firm in the year of his death will be calculated on the basis of the rate of net profit on sales of the previous year, which was 25%. The sales of the firm till 31st December 2016 were ₹ 4,00,000.
(iv) His share in the goodwill of the firm. The goodwill of the firm on Ashok’s death was valued at ₹ 4,50,000.
The partnership deed also provided for the following deductions from the amount payable to the executor of the deceased partner :
(i) His drawings in the year of his death. Ashok’s drawings till 31.12.2016 were ₹ 15,000.
(ii) Interest on drawings @ 12% per annum which was calculated as
₹ 1,500. The accountant of the firm prepared Ashok’s Capital Account to be presented to the executor of Ashok but in a hurry he left it incomplete. Ashok’s Capital Account as prepared by the firm’s accountant is given below :
You are required to complete Ashok’s Capital Account.
Answer:
Q.13. A, B, C, and D were partners in a firm sharing profits in the ratio of 3: 2 : 3: 2. On 1.4.2016, their Balance Sheet was as follows :
From the above date, the partners decided to share the future profits in the ratio of 4 : 3: 2: 1. For this purpose, the goodwill of the firm was valued at ₹ 2,70,000. It was also considered that :
(i) The claim against Workmen Compensation Reserve has been estimated at ₹30,000 and fixed assets will be depreciated by ₹ 25,000.
(ii) Adjust the capitals of the partners according to the new profit sharing ratio by opening the Current Accounts of the partners.
Prepare Revaluation Account, Partners’ Capital Account, and the Balance Sheet of the reconstituted firm.
Answer:
Q.14. On 1.4.2015, J.K. Ltd. issued 8,000, 9% debentures of ₹ 1,000 each at a discount of 6%, redeemable at a premium of 5% after three years. The company closes its books on 31st March every year. Interest on 9% debentures is payable on 30th September and 31st March every year. The rate of tax deducted at the source is 10%.
Pass necessary journal entries for the issue of debentures and debenture interest for the year ended 31.3.2016.
Answer:
Q.15. Pass necessary journal entries on the dissolution of a partnership firm in the following cases :
(i) Dissolution expenses were ₹ 800.
(ii) Dissolution expenses ₹ 800 were paid by Prabhu, a partner.
(iii) Geeta, a partner, was appointed to look after the dissolution work, for which she was allowed a remuneration of ₹ 10,000. Geeta agreed to bear the dissolution expenses. Actual dissolution expenses of ₹ 9,500 were paid by Geeta.
(iv) Janki, a partner, agreed to look after the dissolution work for a commission of ₹ 5,000. Janki agreed to bear the dissolution expenses. Actual dissolution expenses of ₹ 5,500 were paid by Mohan, another partner, on behalf of Janki.
(v) A partner, Kavita, agreed to look after the dissolution process for a commission of ₹ 9,000. She also agreed to bear the dissolution expenses. Kavita took over the furniture of ₹ 9,000 for her commission. Furniture had already been transferred to realisation account.
(vi) A debtor, Ravinder, for ₹ 19,000 agreed to pay the dissolution expenses which were ₹ 18,000 in full settlement of his debt.
Answer:
Q.16. C and D are partners in a firm sharing profits in the ratio of 4: 1. On 31.3.2016, their Balance Sheet was as follows :
On the above date, E was admitted for 1⁄4th share in the profits on the following terms :
(i) E will bring ₹ 1,00,000 as his capital and ₹ 20,000 for his share of goodwill premium, half of which will be withdrawn by C and D.
(ii) Debtors ₹ 2,000 will be written off as bad debts and a provision of 4% will be created on debtors for bad and doubtful debts.
(iii) Stock will be reduced by ₹ 2,000, furniture will be depreciated by ₹ 4,000 and 10% depreciation will be charged on plant and machinery.
(iv) Investments of ₹ 7,000 not shown in the Balance Sheet will be taken into account.
(v) There was an outstanding repairs bill of ₹ 2,300 which will be recorded in the books.
Pass necessary journal entries for the above transactions in the books of the firm on E’s admission.
OR
Q.16. Sameer, Yasmin, and Saloni were partners in the firm sharing profits and losses in the ratio of 4 : 3 : 3. On 31.3.2016, their Balance Sheet was as follows :
On the above date, Sameer retired and it was agreed that :
(i) Debtors of ₹ 4,000 will be written off as bad debts and a provision of 5% on debtors for bad and doubtful debts will be maintained.
(ii) An unrecorded creditor of ₹ 20,000 will be recorded.
(iii) Patents will be completely written off and 5% depreciation will be charged on stock, machinery and building.
(iv) Yasmin and Saloni will share future profits in the ratio of 3: 2.
(v) Goodwill of the firm on Sameer’s retirement was valued at ₹ 5,40,000.
Pass necessary journal entries for the above transactions in the books of the firm on Sameer’s retirement.
Answer:
OR
Q.17. VXN Ltd. invited applications for issuing 50,000 equity shares of ₹ 10 each at a premium of ₹ 8 per share. The amount was payable as follows :
On Application: ₹ 4 per share (including ₹ 2 premium)
On Allotment: ₹ 6 per share (including ₹ 3 premium)
On First Call: ₹ 5 per share (including ₹ 1 premium)
On Second and Final Call: Balance Amount
The issue was fully subscribed. Gopal, a shareholder holding 200 shares, did not pay the allotment money and Madhav, a holder of 400 shares, paid his entire share money along with the allotment money. Gopal’s shares were immediately forfeited after allotment. Afterward, the first call was made. Krishna, a holder of 100 shares, failed to pay the first call money and Girdhar, a holder of 300 shares, paid the second call money also along with the first call. Krishna’s shares were forfeited immediately after the first call. The second and final call was made afterward and was duly received. All the forfeited shares were reissued at ₹ 9 per share fully paid up.
Pass necessary journal entries for the above transactions in the books of the company.
OR
Q.17. JJK Ltd. invited applications for issuing 50,000 equity shares of ₹ 10 each at par. The amount was payable as follows :
On Application: ₹ 2 per share
On Allotment: ₹ 4 per share
On First and Final Call: Balance Amount
The issue was oversubscribed three times. Applications for 30% shares were rejected and money refunded. The allotment was made to the remaining applicants as follows :
Category No. of Shares Applied No. of Shares Allotted
I 80,000 40,000
II 25,000 10,000
Excess money paid by the applicants who were allotted shares was adjusted towards the sums due on allotment.
Deepak, a shareholder belonging to Category I, who had applied for 1,000 shares, failed to pay the allotment money. Raju, a shareholder holding 100 shares, also failed to pay the allotment money. Raju belonged to Category II. Shares of both Deepak and Raju were forfeited immediately after allotment. Afterward, the first and final call was made and was duly received. The forfeited shares of Deepak and Raju were reissued at ₹ 11 per share fully paid up.
Pass necessary journal entries for the above transactions in the books of the company.
Answer:
OR
Answer:
PART B
(Analysis of Financial Statements)
Q.18. Normally, what should be the maturity period for a short-term investment from the date of its acquisition to be qualified as cash equivalents?
Answer: Maximum maturity period is 90 days/ 3 months for a short-term investment from the date of acquisition to be qualified as cash equivalents.
Q.19. State the primary objective of preparing a cash flow statement.
Answer: To find out the inflows and outflows of cash and cash equivalents from Operating, Investing, and Financing activities.
Q.20. What is meant by ‘Analysis of Financial Statements’? State any two objectives of such an analysis.
Answer: Analysis of Financial Statements is the process of critical evaluation of the financial information contained in the financial statements in order to understand and make decisions regarding the operations of the firm.
(Or any other suitable meaning)
Objectives of ‘Financial Statements Analysis’: (Any two)
(i) Assessing the earning capacity or profitability of the firm as a whole as well as its different departments so as to judge the financial health of the firm.
(ii) Assessing managerial efficiency by using financial ratios to identify favorable and unfavorable variations in managerial performance.
(iii) Assessing the short-term and the long-term solvency of the enterprise to assess the ability of the company to repay principal amount and interest.
(iv) Assessing the performance of the business in comparison to that of others through inter-firm comparison.
(v) Assessing developments in the future by forecasting and preparing budgets.
(vi) To Ascertain the relative importance of different components of the financial position of the firm.
Q.21. The proprietary ratio of M. Ltd. is 0·80: 1. State with reasons whether the following transactions will increase, decrease, or not change the proprietary ratio :
(i) Obtained a loan from bank ₹ 2,00,000 payable after five years.
(iv) Issued equity shares to the vendors of machinery purchased for ₹ 4,00,000.
Answer:
Transaction
Effect on Quick Ratio
Reasons
(i)
Decrease
No change in Shareholders’ funds but total assets will increase by ₹ 2,00,000
(ii)
No Change
No change in total assets and Shareholders’ funds
(iii)
Decrease
Both Shareholders’ funds and total assets are decreased by same amount
(iv)
Increase
Shareholders’ funds and total assets both are increased
Q.22. Financial statements are prepared following the consistent accounting concepts, principles, procedures, and also the legal environment in which the business organizations operate. These statements are the sources of information on the basis of which conclusions are drawn about the profitability and financial position of a company so that their users can easily understand and use them in their economic decisions in a meaningful way. From the above statement identify any two values that a company should observe while preparing its financial statements. Also, state under which major headings and sub-headings the following items will be presented in the Balance Sheet of a company as per Schedule III of the Companies Act, 2013.
(i) Capital Reserve
(ii) Calls-in-Advance
(iii) Loose Tools
(iv) Bank Overdraft
Answer:
Values (Any two):
• Transparency
• Consistency
• Following rules and regulations / Ethical code of conduct
• Honesty and loyalty towards owners
• Providing authentic information to users
(Or any other suitable value)
Q.23. From the following Balance Sheet of SRS Ltd. and the additional information as of 31.3.2016, prepare a Cash Flow Statement :
Notes to Accounts
Additional Information :
(i) ₹ 50,000, 12% debentures were issued on 31.3.2016.
(ii) During the year a piece of machinery costing ₹ 40,000, on which accumulated depreciation was ₹ 20,000, was sold at a loss of ₹ 5,000.
Answer:
Notes:
Calculation of Net Profit before tax:
Net profit as per statement of Profit & Loss 75,000
Add: Proposed Dividend 1,00,000
Net Profit before tax & extraordinary items 1,75,000
PART B
(Computerized Accounting)
Q.18. What is meant by a ‘Database Report’ ?
Answer: A database report is the formatted result of database queries and contains useful data for decision-making and analysis.
Q.19. What is meant by a ‘Query’ ?
Answer: Queries provide the capability of combined data from multiple tables and placing specific conditions for the retrieval of data. It is another tabular view of the data showing information from multiple tables, resulting in the presentation of the information required, raised in the query.
Q.20. Explain ‘Flexibility’ and ‘Cost of the installation’ as considerations before opting for specific accounting software.
Answer: Flexibility: (It may include the following points)
• Related to data entry, availability, and design of various reports.
• Between users (Accountants)
• Between systems.
Cost of installation and maintenance: (It may include the following points in explanation)
• Ability to afford hardware and software
• Cost-benefit analysis and study of available options
• Training of staff, cost of updating
Q.21. Explain any four sub-groups of the Account Group ‘Profit and Loss’
Q.22. Explain the steps involved in the installation of computerized accounting software.
Answer: Steps in the installation of CPS:
1. Insert CD in the system
2. Select C: E:, or D: drive from my computer
OR
Start > run > type the filename E:\install.exe
3. The default directories of application, data, and configuration will open in a window. Change the setting if you wish by providing desired file name and drive name.
4. Click on install. The installation process will start and a message of successful installation will appear after its completion. The CD can be removed as the application is successfully installed.
Q.23. What is meant by ‘Conditional formatting’? Explain its benefits.
Answer: Conditional formatting means a format change, such as background cell shading or font color i.e. applied to a cell when a specified condition for the data in the cell is true. Conditional formatting is often applied to worksheets to find:
1. Data that is above or below a certain value.
2. Duplicate data values.
3. Cells containing specific text.
4. Data that is above or below average
5. Data that falls in the top ten or bottom ten values
Benefits of using conditional formatting:
1. Helps in answering questions that are important for making decisions.
2. Guides with help of using visuals.
3. Helps in understanding the distribution and variation of critical data.
Deforestation means the process of cutting down and burning the trees in forest and woodland and converting the land to other use. In other words, it is the destruction of forest, removal of vegetation from an area and clearing of trees for various commercial purposes and for fulfilling the personal needs. Deforestation is the quick woodland devastation through the incessant cutting of plants without replanting.
CAUSES OF DEFORESTATION:
1. Globalization
2. Urbanization
3. Over Population
4. Climate
5. Over Grazing
6. Shifting Cultivation
7. Fuel Wood
8. Forest Fires
9. Timber
10. Industry Establishment
11. Encroachment of Forest
12. Forest Diseases
13. Landslides
14. Ravine Formation
Globalization:
Due to Globalization many industries and factories are built which emit carbon di oxide that affects the trees and forests. India and China are the major countries where trees and forests are used to produce products and supplies in various parts of the world.
Urbanization:
As the world progresses, trees are cleared to oblige growing urban regions for the utilization of construction materials, furniture, paper products, material utilized for highways and streets and timberlands. They are cut down to create land for grazing cattle and for growing crops. Trees are also cut down in developing countries to be used as firewood or turned into charcoal, which are used for cooking and heating purposes.
Over Population:
An increase in population increases the products consumption for which the trees are being destroyed. The fundamental needs are asylum and food supplied with the aid of forests where an ideal measure of utilization and development is required.
Over population in countries like China and India are a result where deforestation rate is higher than comparative countries. The considerable demand for housing in the urban sector increases the demand for wood in the construction of the houses. With more demand, greater is the harm done to the forests. As the land area is limited, the only option for the real estate dealers is to buy the forest land for cheap, clear them and make housing sites for the population.
Climate:
Atmosphere influences people as well as trees, streets, and little plants. The major factor is “ACID RAIN”. Waxy outer coating that covers the leaves is weakened by acid rain.When this happens, it allows the acid to seep into the tree that protects the leaves. Instead of water that changes from a liquid to a gas inside the leaves, gas takes the place of the water. This stops the plant from absorbing carbon dioxide for photosynthesis, and the plant then dies.
Another factor of climate is global warming. As the temperature increases than the average temperature, it affects the growth of the plants and soil thus leading to deforestation.
Overgrazing:
Overgrazing not only destroys forests regenerated growth but also makes soil more compact and impervious. Soil becomes less fertile due to destruction of organic matter and the seeds of certain species do not germinate in excessively grazed soils which results in reduction of species. This leads to desertification. Overgrazing also accelerates the soil erosion which results in the removal of minerals and nutrients from the topsoil and adversely affects the soil structure that ultimately lowers the productivity. The uncontrolled and indiscriminate grazing in the forests leads to degradation of forest soil and affect natural regeneration of forests. Due to excess grazing of the cattle in the grazing lands, the topsoil is washed away which makes it useless for any purpose, including grazing. This prompts to clear the forest areas for growing fodder which led to deforestation.
Shifting Cultivation:
In North Eastern India, due to heavy water erosion shifting cultivation is locally called “JHUM”. Numerous ranchers slaughter the woods for farming and business purposes and another timberland territory is devastated when soil fertility is depleted because of continued cropping. This degrades about one million hectare land every year thus leading to deforestation.
Fuel Wood:
Maximum forest habitat destruction is performed for wood-fuel. In India alone, the annual demand for firewood was 235 million cubic meters (according to Forest survey of India, 1987) where 135 million tons of firewood was consumed in rural areas while 23 million tons was consumed in urban areas.
So, fuel wood is a big deforestation cause.
Forest Fires:
Some fires are incidental while the majority of them are deliberate. According to Forest Survey of India (1996), 53.1% forest vegetation was affected by fire which destroyed about 0.5 million hectares of forests annually. Thus, frequent fires are the major cause of deforestation.
Timber:
According to Forest Survey of India 1987, the annual demand of timber was 12 million cubic meters. Thus, the increased demand for timber led to a rapid depletion of forest. Timber and plywood industries are mainly responsible for the destruction of forest trees and deforestation.
Industry Establishment:
For the establishment of factories and industries, precious plants, wild animals, and rare birds are destroyed, and the quality of environment is adversely affected. The forest-based industries such as Resin and Turpentine industry are responsible for the destruction of trees in the hills as raw materials are supplied to these industries thus causing deforestation.
Encroachment of Forest:
It means encroachment by tribal on forest land for agriculture and other purposes. According to Forest Survey of India, about 7 million hectares of forest land has been encroached for agriculture which produced environmental hazards and deforestation.
Forest Diseases:
Many diseases that are caused by parasitic fungi, rusts, viruses and nematodes causes’ death and decay of forest plants. Young seedlings are destroyed due to attack of nematodes. Many diseases such as heart rot, blister rust, oak will, phloem necrosis and Dutch elm disease, etc. damage the forest trees in large numbers.
Landslides:
The landslides occur mainly in the areas where developmental activities are in progress. The construction of roads and railways particularly in hilly terrains, setting up of big irrigation projects have caused enough destruction to forest and accelerated the natural process of denudation. Deforestation due to landslide in the hills is the major concern.
Ravines Formation:
The forested areas and farming grounds at the edges of large stream gorges (Yamuna and Chambal) face a severe soil disintegration risk. Once the ravines are formed, they continue to destroy the vegetational cover.
EFFECTS OF DEFORESTATION:
1. Deforestation results in many effects like loss of animals home, death of animals, environmental changes, seasonal changes, increase in temperature, rise in environmental heat, global warming, increase in greenhouse gases, melting of ice caps and glaciers, increase in sea level, weakening of ozone layer, hole in the ozone layer, death of sea animals, increasing risks of natural disaster like Storms, cyclones, typhoons, floods, droughts and many other adverse shifts that are sufficient to last life on earth.
2. It is affecting the human lives to a great extent by forcing the negative changes to the environment and atmosphere.
3. Deforestation affects human wellbeing and the new ecosystem by means of atmosphere unevenness, rising an Earth-wide temperature boost, soil disintegration, floods, elimination of biodiversity, diminishing levels of fresh oxygen and rising carbon dioxide, expanding air contamination and expanding levels of toxic gases.
All the harmful impacts of deforestation are causing numerous medical issues, and above all else lung and respiratory issues.
4. Deforestation is not only disturbing the human lives by causing several imbalance ecologically and environmentally but also alarming continuously and indicating the need to stop cutting plants for the safety of human lives. A few people do deforestation to accomplish their insatiability of acquiring money from wood. People are cutting plants for their agricultural activities, logging (to make papers, match-sticks, furniture etc.), urbanization (road construction, housing etc.), desertification of land, mining (oil and coal mining), and fires (to get heat) etc.
5. It disrupts the carbon cycle. The forest trees take carbon dioxide as well as the atmosphere. It affects the human lives and causes an imbalance in ecology and environment. The human health is affected by the pollution which is occurred by deforestation. The land pollution, air pollution and global warming are the main reasons for the various effects on human, wildlife, and nature.
6. Population growth and agricultural development has put unprecedented pressure on India’s forests in the past half century. With the simultaneous rise in both the number of cattle and the amount of land under cultivation, livestock owners were forced to move to forest areas to graze their herd. As indicated by the State of the Forest Study 1995, 78 percent of all woods have undergone frequencies of deforestation and 74 percent of forests need recovery.
SOIL EROSION:
Prompt impacts of deforestation incorporate the washing ceaselessly of soil in the rainstorm season. This is because trees are no longer connecting and holding the soil and so mud slides are possible. The earth is leached of minerals by the large amounts of water. The absence of vegetation likewise implies not many creatures will be found in the field. Most of nutrients are stored in the vegetation and the trees, so if these factors receive bad cycle, our eco-system will be destroyed. Once the trees and plants are cut down, essential nutrients are separated easily and are washed out by rainfall. Thus, we would lose the nutrients for our body needs for daily life. In the event that the ground gets dries and splits under the sun’s warmth without the shelter of the trees, we cannot develop any plants in light of the fact that the soil components are lost. According to the statistics, nearly 80% of tropical forest soil is now infertile, and they will cause worse eco-system which will affect the animals who live there and their habitats. It may change their genetic as well.
SOCIAL EFFECTS OF DEFORESTATION:
1. Deforestation has so many social effects on our society. Its impact not only affects humans but also plants, animals, and the surrounding environment. It causes and forces the surrounding to adapt in order to survive such difficult situations.
2. Indigenous people who consider the forests as their primary habitats are rendered homeless when forests are depleted. This can be seen in mostly undeveloped areas where many people use the forest as their primary habitat. The people living in these areas are forced to move while their surroundings are being altered. The cutting down of forest trees forces the people who live around such areas to move and seek shelter elsewhere.
3. People and animals who live in the rainforest areas depends on their natural environment. Individuals who live close to the timberland in these regions typically rely on their indigenous habitat for fundamental things like food, shelter, water and so forth. Cutting down the trees in those areas usually tend to affect all living things and surroundings that forces them to migrate and look for another conducive atmosphere.
4. Social conflicts and struggles over land and other resources results in the loss of lands and people who live there have to migrate to other places in search of land and resources.
FACTS RELATED TO DEFORESTATION IN INDIA:
1. Deforestation could lead to changes in surface conditions, which would increase the intensity and decrease the duration of rainfall, thereby increasing run-off. This causes soil erosion which leads to the riverbeds being silted. This is how floods occur.
2. India is losing 1.5 million hectares (mha) of forests per year, thereby bringing down the total forest area from 74 mha to 40 mha.
3. Deforestation causes loss of top soil to the tune of 12000 million tons.
4. Due to deforestation, India loses Rs 10,000 crores every year in the form of damage by floods.
ISSUES IN DEFORESTATION:
1. On the rural side, the issue is that the people are depended on the forests for wood fuel and that is why their needs are fulfilled as well as the forests are not depleted because of their actions.
2. On the urban side, we must remember that the trend of consumerism has found currency with the urbanites accompanied with total disregard for the environment. So, can the consumerist attitude and the conversationalist attitude co-exist?
3. Corporates are facing increasing pressure from various organizations to take care of the environment, yet there is no social concern or initiative coming from them directly. Can the government be unbiased in its actions?
4. Almost all the forests are owned by the government and it is their responsibility to see that the forests are safeguarded. But due to lack of political will and pressure from other sectors, the government is not able to decide and act on its own. Can the government be unbiased in its actions?
5. The standard of fixing (33% compulsory forest cover) for all the states of the nation is not a viable, as there are various differences among the states, and many of them are solely dependent on the forest resources for their revenue. Can a region-specific target be drawn up, so that the overall forest cover concedes to the total target of 33%?
6. It is noteworthy to mention that the rate of deforestation exceeds the rate of afforestation. As the land area is fixed, and the cleared forests are used for other purposes, can the scheme of afforestation be successful in the near future?
7. There is a close link between agriculture and forestry. With improvements in the agriculture sector, and with the increase in population, there is a continuous need for upgrading the resources. Can the forest resources be experimented with the modern technologies?
OVERVIEW OF WORLDWIDE DEFORESTATION:
Everyday more than thousands of trees are cut down all over the world to use the timber as a major source of fuel, building material and paper products. Urbanization has forced man to acquire huge forest areas. As population grows, the need for agricultural land has also increased over the years. Deforestation has such a large number of detrimental environmental effects. One of the most severe consequences of this is the natural surroundings loss of numerous creature species.
Thus, deforestation can alter the earth’s biodiversity making a lot of rare species even more extinct. Deforestation also plays a major role in global warming and it is also responsible to the contribution of up to 20% of the total greenhouse gases emitted. Trees play a major role in absorbing most of the greenhouse gases like carbon dioxide. As lots of trees are cut down, the concentration of the greenhouse gases in the atmosphere increases thus increasing the temperature of the earth. Another immediate impact of deforestation is expanded soil disintegration. This can also lead to unnatural floods and droughts. Clearing forests can disrupt the normal water flow thus causing abnormal floods and droughts. Plants consume water through roots, discharges into the air that creates mists and downpour.
STATISTIC:
As population grows so does the rise and demand of more forests to be cut down and this leads to deforestation. This is a breakdown of land area per sq. km 2002/2008.
SOLUTION TO DEFORESTATION:
1. Reforestation: Because of public education, new technology and innovation has occurred in most parts of the world that implements reforestation and it very well may be found in nations across Asia.
2. Legislation: Due to new laws and regulations it can be seen that new trees have been planted and old trees are not allowed to be cut down. If this continues there might be a chance to stop deforestation and reverse the whole process completely.
3. Wildlife Sanctuaries: Sanctuaries are very important, not only to save wildlife, but to save trees as well. Sanctuaries go a long way in protecting all wildlife.
4. Cities: All cities should be managed properly. The new projects need to be controlled and planned accordingly and new trees should be planted in the process.
5. Commercial Forest Plantation: There should be a special forest plantations for all the wood that is required by the industries. In this manner the wood can be cut in a controlled and regulated environment.
6. Water Management: Improper water management affects deforestation in a big way. If the wildlife does not have water, then the entire ecosystem will be damaged. The construction of new dams should be planned properly, and area receives abundance of water.
The government must be blamed for the destruction of the forests due to:
1. Their flexibility in allocating the forest land to corporates under political pressure.
2. The distribution of lands to tribal people, on which agriculture cannot be done due to soil variety. This causes the allotted land to be wasted as the cleared land for agriculture can no longer be used for the purpose intended nor can they be used as forest cover as earlier.
OTHER REASONS OF DEFORESTATION:
1. One of the major reasons for the destruction of the forests are the building of dam’s reservoirs. These projects, albeit, intended for the benefit of the people, extends on the reverse side into trouble for the people. Displacement of the masses on one side, the projects cause large areas of virgin forests to be destroyed ruthlessly.
2. The other reason could be attributed to the lack of vigilance of the people who use the forest as camps. Their carelessness may cause forest fires which devours large areas of lush green forests.
UNIVERSAL SOLUTION TO DEFORESTATION:
Forests are an important natural resource for any country and deforestation retards a country’s development. Essential assets can only be accessed by “Afforestation” to fulfill the needs of the growing populace. Afforestation refers to the scheduled of planting trees for food and fodder development. Nurseries play a significant part in growing the cover of forests. As significant for what it is worth for a youngster to attain a nursery through her/his youthful age, so it is for plants to develop under appropriate consideration and security. This prepares them to withstand adverse situations during planting.
There is some hope. Projects with solar powered ovens reduce the need to cut the trees for fuel. Crops best suited for poorer soils is being introduced.
CONCLUSION:
Woodlands are very important for appropriate irrigation, medication, air newness, air contamination reduction, wood obtaining for some reasons and so on. It upsets all the procedures when we cut plants and impacts human lives. Instead of cutting plants to fulfill the need of paper, we should make the habit of recycling the old things as possible and avoid cutting of new plants. As existence without water is beyond the realm of imagination, similarly existence without plants and trees is likewise unrealistic as it is the wellspring of sunlight, natural air, creature environment, shadows, wood and so on.
Babur in his famous book Baburnama tells us about the establishment of Mughal empire in India. He defeated Ibrahim Lodhi in the first battle of Panipat in 1526. He writes,
“Many men like us have taken breath at this spring,
And have passes away in the twinkling of an eye;
We took the world by courage and might,
But we could not take it with us to the grave.”
The word mughals is said to be derived by Mongols. Their ancestral roots are from Turks and Uzbegs. Their lineage is the most successful lineage of Muslim rulers of India. They ruled around 181 years from Babur to Aurangzeb. They have the credit of having an organized and a better administration for the country with supreme authority.
The first half opf the sixteenth century is the evident witness of a new dynasty – the Mughals who moved from central Asia to India. Mughals have left behind a great legacy of administrational structures. They have given the concept of the fusion of Indo-Islamic art, architecture and culture.
Their successor pattern was that of Timurid tradition where the throne was inherited ad shared among brothers. Therefore, all the brothers had a desire to rule the mighty mughal empire. This can be interpreted and inferred as there are many rebellions in the contemporary accounts of historians.
Time line of Mughal empire
The first mughal king was Babur who was followed by Humayun, Akbar, Jahangir, Shah Jahan and Aurangzeb. After Aurangzeb, mughal rulers were not able to keep their kingdom intact. The failure of Bhadur Shah Zafar lead to the establishment of British rule in India.
S. No.
Name of the ruler
Time Period
1.
Babur
1526-1530
2.
Humayun
1530-1540 and 1555-56
3.
Sher Shah
1540-1555
4.
Akbar
1556-1605
5.
Jahangir
1605-1627
6.
Shah Jahan
1628-1658
7.
Aurangzeb
1658-1707
Note- Sher Shah was not a mughal ruler, still he managed to give a stable rule for the country.
Babur (1526-1530 Common Era (CE))
Zahir-ur-din Babur succeeded the throne of Farghana a place in Central Asia in 1494 at a very tender age of 12.
Babur lost his kingdom to many tribes and they moved to India in search of new lands to conquer.
Rana Sanga who ruled Mewar and Daulat Khan Lodhi who was the governor of Bengal invited Babur to throw Ibrahim Lodhi from the throne of Delhi.
The ill disciplinary army of Lodhi was defeated by the small and well trained army of Babur.
However, he received threats from Rana Sanga of Mewar and the Afghan chief of Bengal, Nusrat Shah.
He defeated Rana Sanga in Battle of Khanua in 1527.
The Aghans were defeated in the Battle of Ghagra in 1528.
He also defeated Rajputs in the battle of Chanderi in 1528.
After these wars, Babur was able to lay a string and vast foundation of a powerful mughal empire.
Babur is known for the development of new techniques of warfare.
He is given the share of effectively using canons in a battle.
He invaded India five times.
He died at the age of 48 years in 1536.
Humayun who was his oldest son became his successor.
Reasons for Babur’s success in the first battle of Panipat
Babur introduced many new and strong weapons against Indian army.
He had built a well trained chivalry.
He had a good and effective war planning. His soldiers were arranged in a way where they can move easily from one place to another.
The Mughals were the descendants of two great heredities – Gengis Khan and Timur, the famous ruler of Iran.
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