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.
प्रश्न 1 किसी एक विषय पर निबन्ध लिखिए जो लगभग 400 शब्दों से कम न हो :
(i) निस्वार्थ भाव से की गई सहायता से असीम आनंद तथा संतोष प्राप्त होता है। किसी ऐसी ही एक घटना का वर्णन कीजिए जब आपने अपनी परेशानियों की परवाह किए बिना किसी जरूरतमंद व्यक्ति की मदद की थी। यह भी स्पष्ट कीजिए कि इस अनुभव से आपके जीवन पर क्या प्रभाव पड़ा ?
(ii) “जल ही जीवन है। जल के बिना सुनहरे कल की कल्पना करना व्यर्थ है।” वर्तमान युग में जल संकट की समस्या किस प्रकार विकराल रूप लेती जा रही है ? जल संरक्षण की आवश्यकता तथा इसके विभिन्न उपायों पर प्रकाश डालते हुए अपने विचार प्रस्तुत कीजिए।
(iii) आपके विद्यालयी जीवन का यह अन्तिम वर्ष है। आज आपका विदाई समारोह आयोजित किया गया है। इतने वर्षों का मित्रों एवं अध्यापकों का साथ छूटने वाला है। इन बीते वर्षों के न भूलने वाले खट्टे-मीठे अनुभव लिखिए।
(iv) “मनुष्य के नैतिक उत्थान का जिम्मेदार परिवार एवं समाज है” — विषय के पक्ष या विपक्ष में अपने विचार व्यक्त कीजिए।
(v) विश्व के मानचित्र पर भारत की एक नई पहचान उभर रही है, इसका कारण है “आज का जागरूक भारत”व्याख्या कीजिए।
(vi) निम्नलिखित में से किसी एक पर मौलिक कहानी लिखिए :
(a) “बीती ताहि बिसार दे आगे की सुध लेय।’
(b) एक मौलिक कहानी लिखिए जिसका अन्तिम वाक्य हो :
…………….. और अपने घर सकुशल पहुँचने पर हमने चैन की साँस ली।
प्रश्न 2 निम्नलिखित अवतरण को पढ़कर, अन्त में दिए गए प्रश्न के उत्तर अपने शब्दों में लिखिए :
पुराने समय की बात है, एक गाँव में दो किसान रहते थे। दोनों ही बहुत गरीब थे, दोनों के पास थोड़ी-थोड़ी ज़मीन थी, दोनों उसमें ही मेहनत करके अपना और अपने परिवार का गुजारा करते थे।
अकस्मात् कुछ समय पश्चात दोनों की एक ही दिन, एक ही समय पर मृत्यु हो गयी। यमराज दोनों को एक साथ भगवान के पास ले गए। भगवान ने उन्हें देख के उनसे पूछा, “तुम्हारे इस जीवन में क्या कमी थी ?” भगवान की बात सुनकर उनमें से एक किसान बड़े गुस्से से बोला, “हे भगवन् ! आपने इस जन्म में मुझे बहुत घटिया जिन्दगी दी थी। आपने कुछ भी नहीं दिया था मुझे। पूरी ज़िन्दगी मैंने बैल की तरह खेतों में काम किया, जो कुछ भी कमाया वह सब पेट भरने में लगा दिया, न ही मैं कभी अच्छे कपड़े पहन पाया और न ही कभी अपने परिवार को अच्छा खाना खिला पाया। जो भी पैसे कमाता था, कोई आकर मुझसे लेकर चला जाता था और मेरे हाथ में कुछ भी नहीं आया। देखो, कैसी जानवरों जैसी ज़िन्दगी जी है मैंने ।”
उसकी बात सुनकर भगवान कुछ समय मौन रहे और पुन: उस किसान से पूछा, “तो अब तुम क्या चाहते हो, इस जन्म में मैं तुम्हें क्या बनाऊँ ?”
भगवान का प्रश्न सुनकर वह किसान पुन: बोला, “भगवन् ! आप कुछ ऐसा कर दीजिए, कि मुझे कभी किसी को कुछ भी देना ना पड़े। मुझे तो केवल चारों तरफ से पैसा ही पैसा मिले।”
अपनी बात कहकर वह किसान चुप हो गया। भगवान ने उसकी बात सुनी और कहा, “तथास्तु ! तुम अब जा सकते हो, मैं तुम्हें ऐसा ही जीवन दूँगा जैसा तुमने मुझसे माँगा है।”
उसके जाने के बाद भगवान ने दूसरे किसान से पूछा, “तुम बताओ, तुम्हारे जीवन में क्या कमी थी ?” उस किसान ने भगवान के सामने हाथ जोड़ते हुए कहा, “हे भगवन् । आपने मुझे सबकुछ दिया, मैं आपसे क्या माँगूं। आपने मुझे एक अच्छा परिवार दिया, मुझे कुछ ज़मीन दी जिस पर मेहनत से काम करके मैंने अपने परिवार को एक अच्छा जीवन दिया। खाने के लिए आपने मुझे और मेरे परिवार को भरपेट भोजन दिया। मैं और मेरा परिवार कभी भूखे पेट नहीं सोया। बस एक ही कभी थी मेरे जीवन में, जिसका मुझे पूरी ज़िन्दगी अफ़सोस रहा और आज भी है। मेरे दरवाजे पर कभी कुछ भूखे और प्यासे लोग आते थे भोजन माँगने के लिए परन्तु कभी-कभी भोजन न होने के कारण मैं उन्हें खाना नहीं दे पाता था और वे मेरे द्वार से भूखे ही लौट जाते थे। ऐसा कहकर वह चुप हो गया।’
भगवान ने उसकी बात सुनकर उससे पूछा, “तो अब क्या चाहते हो तुम, इस जन्म में मैं ङ्केतुम्हें क्या बनाऊँ ? किसान ने हाथ जोड़ते हुए भगवान से विनती की, हे प्रभु ! आप कुछ ऐसा कर दें कि मेरे द्वार से कोई भूखा-प्यासा ना जाए।” भगवान ने कहा, “तथास्तु ! तुम जाओ तुम्हारे द्वार से कभी कोई भूखा-प्यासा नहीं जाएगा।”
अब दोनों का पुन: उसी गाँव में एक साथ जन्म हुआ। दोनों एक साथ बड़े हुए। पहला व्यक्ति जिसने भगवान से कहा था कि उसे चारों तरफ से केवल धन मिले और उसे कभी किसी को कुछ देना ना पड़े, वह व्यक्ति उस गाँव का सबसे बड़ा भिखारी बना। अब उसे किसी को कुछ देना नहीं पड़ता था और जो कोई भी आता उसकी झोली में पैसे डालकर ही जाता था।
कि उसके द्वार से कभी कोई भूखा-प्यासा न जाए, वह उस गाँव का सबसे अमीर आदमी बना।
ईश्वर ने जो दिया है उसी में संतुष्ट रहना बहुत ज़रूरी है। अक्सर देखा जाता है कि सभी लोगों को हमेशा दूसरों की चीजें ज्यादा पंसद आती हैं और इसके चक्कर में वे अपना जीवन भी अच्छे से नहीं जी पाते। हर बात के दो पहलू होते हैं—सकारात्मक और नकारात्मक, अब ये हमारी सोच पर निर्भर है कि हम चीज़ों को नकारात्मक रूप से देखते हैं या सकारात्मक रूप से। अच्छा जीवन जीना है, तो अपनी सोच को अच्छा बनाना होगा। चीज़ों में कमियाँ निकालने की बजाय भगवान ने जो दिया है उसका आनंद लेना और हमेशा दूसरों के प्रति सेवा भाव रखना होगा ! जिस दिन हमारी सोच बदलेगी, जीवन के प्रति हमारा दृष्टिकोण भी बदल जाएगा।
प्रश्न :
(i) दोनों किसान कहाँ रहते थे ? उन दोनों में क्या समानताएँ एवं क्या विषमताएँ थीं ?
(ii) पहले किसान को अपने जीवन से क्या शिकायत थी ? वह दूसरे जन्म में क्या बनना चाहता था ?
(iii) दूसरे किसान ने भगवान से अपने लिए क्या माँगा और क्यों ?
(iv) दोनों किसानों का पुनर्जन्म किस रूप में हुआ ? अब उनका जीवन कैसा था ?
(v) इस गद्यांश से हमें क्या शिक्षा मिलती है ?
प्रश्न 3 निम्नलिखित वाक्यों को शुद्ध करके लिखिए :
(i) श्याम तेजी से दौड़ता है।
(ii) वह मेरे शब्दों पर ध्यान नहीं देता।
(iii) उसने गीत की दो-चार लड़ियाँ गाईं।
(iv) हत्यारे को मृत्युदण्ड की सजा मिली।
(v) हम हमारे देश के लिए जान दे देंगे।
(b) निम्नलिखित मुहावरों को वाक्यों में प्रयुक्त कीजिए :
(i) हाथ तंग होना।
(ii) चुल्लू भर पानी में डूब मरना।
(iii) आसमान सिर पर उठाना।
(iv) कान भरना।
(v) इधर-उधर की हाँकना।
SECTION B
गद्य संकलन (Gadya Sanklan)
प्रश्न 4 “जैसे भी हो, इस बार बेटू को अपने साथ लेकर ही जाना होगा। यही हाल रहा तो इसकी जिंदगी चौपट हो जाएगी। यह भी कोई ढंग है भला।”
(i) उक्त कथन कौन, किससे और किस संदर्भ में कह रहा है ?
(ii) श्रोता उक्त कथन सुनकर धर्म-संकट में क्यों था ?
(iii) बेटू के आ जाने से अम्मा का जीवन किस तरह बीतता था ?
(iv) ‘मजबूरी’ कहानी के माध्यम से कहानीकारा पाठकों का ध्यान किस ओर आकृष्ट कर रही है?
प्रश्न 5 “म्लेच्छों ने मुझे मुलतान की लूट में पकड़ लिया। मैं उनकी कठोरता में जीवित रहकर बराबर उनका विरोध ही करती रही।” कथन के आधार पर इरावती की व्यथा का वर्णन करते हुए उसका चरित्र-चित्रण कीजिए।
प्रश्न 6 “गौरी एक चरित्र प्रधान कहानी है”। कहानी के आधार पर गौरी की देशभक्ति एवं त्याग का वर्णन करते हुए बताइए कि गौरी का योगदान सीताराम जी की तुलना में कहीं कम नहीं था।
काव्य मंजरी (Kavya Manjari)
प्रश्न 7 क्या हवाएँ थीं कि उजड़ा प्यार का वह आशियाना,
कुछ ना आया काम तेरा, शोर करना गुल मचाना,
माना कि उन शक्तियों के साथ चलता जोर किसका
किन्तु ऐ निर्माण के प्रतिनिधि, तुझे होगा बताना
जो बसे हैं, वो उजड़ते हैं, प्रकृति के जड़ नियम से,
पर किसी उजड़े हुए को, फिर बसाना कब मना है ?
है अँधेरी रात पर दीवा जलाना कब मना है ?
(ii) ‘प्यार का आशियाना’ कैसे उजड़ गया ? मनुष्य का शोरगुल मचाना काम क्यों नहीं आया ?
(iii) ‘निर्माण के प्रतिनिधि’ किसे कहा गया है और क्यों ? ‘प्रकृति का जड़ नियम’ क्या है ? समझाइए।
(iv) प्रस्तुत कविता से कवि क्या सन्देश देना चाहते हैं ? समझाकर लिखिए।
प्रश्न 8 ‘एक फूल की चाह’ कविता के माध्यम से कवि सियारामशरण गुप्त जी ने छुआछूत जैसी सामाजिक कुरीति पर कुठाराघात किया है। – सिद्ध कीजिए।
प्रश्न 9 ‘आ: धरती कितना देती है’ का मूल प्रतिपाद्य लिखिए। प्रस्तुत कविता द्वारा कवि ने क्या सन्देश दिया है?
‘सारा आकाश’ (Saara Akash)
प्रश्न 10 “तूने मुझे बचा लिया, वरना सच कहता हूँ कि पागल हो जाता। तू नहीं जानता, हमारे घर की हालत क्या है।” मेरी समझ में नहीं आ रहा था कि कैसे अपनी कृतज्ञता को व्यक्त करूँ। मेरी आँखें भर आईं।
(i) उपन्यास तथा उपन्यासकार का नाम लिखिए। यह किस प्रकार का उपन्यास है ?
(ii) उपर्युक्त कथन का वक्ता कौन है ? वक्ता किसके प्रति आभारी है और क्यों ?
(iii) वक्ता ने श्रोता से कितने रुपये उधार लिए और उन रुपयों से किसके लिए क्या खरीदा ? उसके बाद वक्ता जब घर पहुंचा तो घरवालों की क्या प्रतिक्रिया हुई ?
(iv) श्रोता का चरित्र चित्रण कीजिए।
प्रश्न 11 ‘सारा आकाश’ उपन्यास के आधार पर समर के बाबूजी का चरित्र-चित्रण कीजिए।
प्रश्न 12 ‘‘सारा आकाश’ राजेन्द्र यादव द्वारा लिखित एक उद्देश्यपूर्ण रचना है।’ – उपन्यास के आधार पर इस कथन की व्याख्या कीजिए।
‘आषाढ़ का एक दिन’ (Aashad Ka Ek Din)
प्रश्न 13 विलोम क्या है ? एक असफल कालिदास। और कालिदास ? एक असफल विलोम। हम कहीं एक-दूसरे के बहुत निकट पड़ते हैं।
(i) वक्ता और श्रोता का परिचय दीजिए।
(ii) प्रस्तुत संवाद का प्रसंग स्पष्ट कीजिए।
(iii) उपर्युक्त पंक्तियों के आधार पर वक्ता का दृष्टिकोण स्पष्ट कीजिए।
(iv) उपर्युक्त संवाद के आधार पर बताइए कि विलोम और कालिदास के बीच कैसे संबंध थे ?
प्रश्न 14 “अम्बिका भावनाओं में नहीं यथार्थ में जीती है।” ‘आषाढ़ का एक दिन’ नाटक के आधार पर अम्बिका की चारित्रिक विशेषताओं का वर्णन कीजिए।
प्रश्न 15 प्रियंगुमंजरी मल्लिका को अपने साथ चलने के लिए क्यों कहती है ? मल्लिका की इस पर क्या प्रतिक्रिया थी ?
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