Surender Singh Rothan

Polynomials 2 UNIT II : ALGEBRA 59 Polynomials CBSE Syllabus Z eroes of a polynomial. Relationship between zeroes and coefficients of quadratic polynomials. Statement and simple problems on division algorithm for polynomials with real coefficients. CONTENTS ♦ Concept Flow Map ♦ Introduction ♦ Geometrical Meaning of Zeroes of a Polynomial ♦ Graphs of Polynomials and their Zeroes ♦ Polynomials of Higher Degree ♦ Relationship between Zeroes and Coefficients of a Polynomial ♦ Division Algorithm ♦Very Short Answer Objective Type Questions ♦ NCERT Exemplar Exercises (Solved) ♦ Mind Map ♦Trend Setter (Previous Years’ Board Paper Questions) ♦ Chapter-End Exercises ♦ Chapter Test CONCEPT FLOW MAP Polynomials Geometrical meaning of the zeroes of a polynomial Relationship between zeroes and coefficients of a polynomial Polynomial of degree 1, 2, 3 are called linear, quadratic, cubic polynomials Division algorithm If the value of a polynomial at k is 0, then k is a zero of the polynomial. A polynomial of degree n can have at most n real zeroes 2.1 INTRODUCTION In earlier classes, we have studied about polynomials. In this chapter, we shall discuss some properties of polynomials like zeroes of a polynomial, relationship of zeroes and coefficients of a polynomial with particular reference to quadratic polynomial etc. Let us have a brief look on some terms and facts related to polynomials. 59 Tangents to a Circle 743 7. In the given figure, O is the centre and SAT is a tangent to the circle at A. If ∠BAT = 30°, find ∠ AOB and ∠ AQB. P O B Q A S T 30° 8. In the Fig., if ∠ O′PR = 45°, find out ∠QRO′. O r  12. Two circles C (O, r) and C  O′,  touch 2  internally at the point X. Chord XM of circle C (O, r) intersects the circle r  C  O′,  at Y. Prove that OY bisects XM. 2  13. In the given figure, I is the incentre of ∆ ABC. AI when produced meets the circumcircle of ∆ ABC at D. If ∠ BAC = 66° and ∠ ACB = 80°, calculate : (i) ∠ DBC, (ii) ∠ IBC, (iii) ∠ BID. O P Q R 9. Two circles intersect each other at two points A and B. At A, tangents AP and AQ to the two circles are drawn which intersect other circles at the points P and Q respectively. Prove that AB is the bisector of angle PBQ. 14. In the given figure, PQ is a tangent at a point R to the circle with centre O. If ∠TRQ = 30°, find ∠ PRS. R P A Q 30° T O S P O O Q B 10. Two circles intersect each other at A and B. Their common tangent meets the two circles at C and D. Prove that ∠ CAD + ∠ CBD = 180°. 4 cm A C A D 2 C 15. A point P, outside a circle, is 13 cm away from its centre. A secant from P cuts the circle at Q and R such that PQ = 9 cm and QR = 7 cm. Find the radius of the circle. 16. In the given figure, the diameters of two wheels have measures 2 cm and 4 cm. cm M B O O B 11. Two circles intersect at two points A and B and a straight line T PAQ intersects the circles at A P and Q. If the P Q tangents at P and Q intersect at T, prove that P, B, Q, T are concyclic. B D Determine the lengths of the belts AD and BC that pass around the wheels, if it is given that belts cross each other at right angles. 17. In the given figure, P B 45° AB = AD = DC = PB and ∠ DBC = 45°. Determine : A (i) ∠ ABD, (ii) ∠ APB. Hence or otherwise, prove that AP is parallel to DB. C D Introduction to Trigonometry and Trigonometric Identities EXAMPLE 19 : SOLUTION : 843 tan θ + 4 m , then find the value of . n 4 cot θ + 1 AB m Given, sin θ = = AC n ∴ AB : AC = m : n If sin θ = A nk Let AB = mk and AC = nk. Using Pythagoras theorem, BC = AC2 − AB2 = n2 k2 − m2 k2 C mk B = k n2 − m2 AB mk tan θ= = = BC k n2 − m2 ∴ cot θ = ∴ 1 = tan θ m m 2 n − m2 n2 − m2 m n2 − m2 tan θ + 4 = 4 cot θ + 1 4 n2 − m2 m + 4 n2 − m2 +4 = +1 n2 − m2 2 2 4 n −m +m = m 2 n – m2 m m 12 2 2 EXAMPLE 20 : If cot B = , prove that tan B – sin B = sin4 B sec2 B. 5 base 12 SOLUTION : We have, cot B = = perpendicular 5 So,we draw a right triangle ABC, right-angled at C such that base = BC = 12 units and perpendicular = AC = 5 units. By Pythagoras Theorem, we have A AB2 = BC2 + AC2 AB2 = (12)2 + 52 = 169 ⇒ AB = 169 = 13 AC 5 AC 5 ∴ sin B = = , tan B = = AB 13 BC 12 AB 13 B and sec B = = BC 12 Now, L.H.S. = tan2 B – sin2 B = (tan B)2 – (sin B)2 ⇒  13 5 12 C 2 2 25 25 1   1 =  5  –  5  = = − – 25    144 169   12   13 - × 25 25  169 − 144  = 25  25 × = = 144 ×- × × 169   2 2 4 5 ×5 5 ⇒  L.H.S. = =   ...(i) 2 2 2 (12) × (13) (12) × (13)2 Also, R.H.S. = sin4 B sec2 B = (sin B)4 (sec B)2 4 2 54 × 132 54 ⇒  R.H.S. =  5  ×  13  = =  13   12  134 × 122 (12)2 × (13)2 Thus, L.H.S. = R.H.S. Hence proved. Introduction to Trigonometry and Trigonometric Identities 845 From equations (i) and (ii), we have AC AB BC = = PR PQ QR ⇒ DABC ~ PQR Then by similarity of triangles ∠B = ∠Q Hence proved. From figure, find the values of : (i) sin A (ii) cot A (iii) tan B (iv) sin2 B + cos2 B SOLUTION : In ∆ACD, we have AC2 = AD2 + CD2 EXAMPLE 23 : ⇒  AC2 = 32 + 42 ⇒  AC2 = 25 ⇒ C 4 A AC = 3 6 D B 25 = 5 In ∆BCD, we have ⇒  BC2 = BD2 + CD2 BC2 = 62 + 42 ⇒  BC2 = 36 + 16 ⇒ BC2 = 52 ⇒  BC = 52= 13 × 4= 2 13 CD 4 AD 3 Now : (i) sin A = = (ii) cot A = = AC 5 CD 4 CD 4 2 (iii) tan B = = = DB 6 3 CD 4 2 (iv) sin B == = BC 2 13 13 BD 6 3   cos B == = BC 2 13 13 2 2 ∴ sin2 B + cos2 B =  2  +  3  = 4 + 9 = 13 = 1    13   13  13 13 13 EXAMPLE 24 : SOLUTION : A In figure, AD = DB and ∠B is a right angle. Determine : (i) sin θ (ii) cos θ (iii) tan θ (iv) sin2 θ + cos2 θ b Given, AB = a ⇒  AD + DB = a C [ AD = DB] ⇒  AD + AD = a ⇒ 2AD = a a Thus,  AD = DB = 2 By Pythagoras theorem, we have ⇒ ⇒ AD = a 2 AC2 = AB2 + BC2 b2 = a2 + BC2 ⇒  BC2 = b2 – a2 ⇒ BC = b2 − a2 D a B 19 CHAPTER 2 : Units and Measurement SOLVED EXAMPLES Example 2.6 : Derive the relationship between par sec and SI unit of distance. Solution : By definition, when base line is of length equal to distance between centres of sun and earth, d = 1.5 × 1011 m Parallax angle, q = 1′′ = 4.85 × 10–6 rad d 1.5 × 1011 m Distance, D= = 3.1 × 1016 m = 1 par sec = q 4.85 × 10 −6 rad Example 2.7 : The nearest star to our solar system is 4.29 light years away. Express this distance in par sec. How much parallax would this star show when viewed from two locations of earth six months apart in its orbit around the sun? [NCERT] Solution : 4.29 light year = 4.29 × 9.46 × 1015 m = 4.058 × 1016 m Now, 3.1 × 1016 m = 1 par sec 4.058 ×- × 1016 m = par sec = 1.323 par sec 3.1 × 1016 When the star is viewed from two locations six months apart in earth’s orbit, the baseline is equal to earth’s orbit diameter = 3 × 1011 m. 3 × 1011 m Parallax angle = = 1.51′′ 4.058 × 1016 m 2.4.2 Measurement of Size of Far Off Object by Parallax Method Parallax method can also be employed to measure the size of an astronomical/celestial body. In this case, from the same position on earth, the angle subtended by peripheral ends of far off object at the point of observation is measured with the help of an d A B astronomical telescope. For example, If the diameter D D of a planet (to be found) is equal to ‘d’ and distance between planet and earth is D, then for angle ‘a’ O subtended by the diametrically Earth opposite ends of planet at the FIGURE 2.7 point of observation (Figure 2.7) AB= d= Da The angle (a) subtended by the extreme ends of planet at the point of observation is known as angular diameter of the planet. Hence, if D is a known quantity and ‘a’ is measured, the size of planet can be estimated. SOLVED EXAMPLES Example 2.8 : When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72′′ of arc. Calculate the diameter of Jupiter. [NCERT] Solution : Distance of Jupiter from Earth, D = 824.7 million km = 8.247 × 1011 m Angular diameter, a = 35.72′′ = 35.72 × 4.85 × 10–6 rad = 1.73 × 10–4 rad Diameter of Jupiter, d = Da = 8.247 × 1011 x 1.73 × 10–4 m = 1.427 × 108 m Example 2.9 : Find the radius of sun if its angular diameter is 1920′′ and distance between sun and earth is 1.496 × 1011 m. Solution : Angular diameter, a = 1920 × 4.85 × 10–6 rad = 9.312 × 10–3 rad Distance, D = 1.496 × 1011 m Diameter of sun, d = Da = 1.496 × 1011 × 9.312 × 10–3 = 1.39 × 109 m Example 2.10 : A planet of diameter 2.08 × 108 m is located at a distance of 7.28 × 1011 m from earth at an instant of time. Find the angular diameter of the planet. 2.08 × 108 m Solution : Angular diameter, a = = 2.857 × 10–4 rad 7.28 × 1011 m = 0.587′ 2.4.3 Measurement of Very Small Distance : Size of Molecule As discussed earlier, not only do we measure very large distances, we also come across very small distances (of microscopic length) in our daily life. These small distances are expressed as submultiples of SI unit of length. Table 2.6 gives an account of smaller units of length and their relationship with metre (m). TABLE 2.6 : Smaller units of measuring length S.No. Unit Relation with metre (m) 1. metre (m) 1m 2. centimetre (cm) 10–2 m 3. millimetre (mm) 10–3 m 4. micrometre (mm) 10–6 m 5. nanometre (nm) 10–9 m 6. picometre (pm) 10–12 m 7. femto (fm) 10–15 m 8. angstrom (Å) 10–10 m The measurement of lengths less than 1 micro-metre, i.e., the ones which can’t be measured by a screw gauge, is done with the help of an optical microscope which uses visible light of wavelength of the order of 10–7 m. For particles of size smaller than this, we use an electron microscope, which uses an electron beam instead of visible light. This helps to resolve atomic sizes also. However, the resolution of electron microscope is limited because electrons behave as waves of wavelength of the order 10–10 m. For still smaller sizes, tunnelling microscopy has been developed which has a better limit of resolution than angstrom. Oleic acid molecule Oleic acid has molecules of size of the order 10–9 m. Oleic acid is basically a soapy liquid having large molecular size. For measuring the size of oleic acid molecule, we dissolve 1 cm3 of oleic acid in alcohol and prepare a 24 CONCEPTUAL PHYSICS—XI Example 2.14 : A physical balance has least count equal to 0.1 g. The mean value of mass of the body calculated with this balance is 342.46 g. How will you express this result correctly? Solution : Since least count of the physical balance is 0.1g, we drop down the hundredth place of the obtained value by increasing the tenth’s place by 1 (as hundredth place reads ‘6’ which is greater than 5). Thus, the result is correctly expressed as, mass = (342.5 ± 0.1) g 2.9  C OMBINATION OF ERRORS Once the observations of a quantity from the measuring instrument are taken, we calculate the required quantity by performing certain mathematical operations over it. These involve the basic arithmethic operations too. Some such operations involve addition, subtraction, multiplication, division, finding root or raising to a power etc. For example, to calculate volume of a cylindrical box, we measure the diameter of box and its height. Errors while measuring these quantities also creep into the calculated values. Such errors result from the combination of errors in the measurement of individual quantities. These errors are said to propagate along with the obtained value along with the mathematical operations on observed values. (i)Error on addition of values :  Let observed values be a and b and absolute errors associated with them be ∆a and ∆b respectively. Thus measured values are, (a ± ∆a) and (b ± ∆b) respectively.   Now suppose we derive the quantity c by adding both measured values and wish to calculate the error ∆c associated with this calculated value. Now, c=a+b By addition, c ± ∆c = (a ± ∆a) + (b ± ∆b) ⇒ c ± ∆c = (a + b) ± (∆a + ∆b) ⇒ c ± ∆c = c ± (∆a + ∆b) ⇒ ± ∆c = ± ∆a ± ∆b which is the maximum absolute error in c. (ii)Error on subtraction of values :  Considering the above suppositions, we again calculate c=a–b c ± ∆c = (a ± ∆a) – (b ± ∆b) ⇒ c ± ∆c = (a – b) ± (∆a ± ∆b) ⇒ ± ∆c = ± (∆a + ∆b) ⇒ ± ∆c = ± (∆a + ∆b) We thus conclude that when we add or subtract two or more physical quantities, the absolute error associated with the final result obtained is equal to the sum of absolute errors associated with individual quantities. (iii)Error in the product of two values :  Using the above assumptions, if c = a × b, then (c ± ∆c) = (a ± ∆a)(b ± ∆b) ∆c  c  ∆c  c  ∆c  c  ∆c 1± c  c 1 ± ⇒   c 1 ± ⇒   c 1 ± ⇒  ⇒ ∆a   ∆b   = a 1 ±  b 1 ± b  a     ∆a   ∆b   1± = (ab)  1 ± a   b   ∆a∆b ∆a ∆b   ± ± = (c) 1 ± ab a b    ∆a ∆b  + ≈ 1± b   a [Neglecting ∆a∆b/ab as it is a very small value] ∆c  ∆a ∆b  = ± + c b   a ⇒ Thus, the fractional (relative) error associated with product of two physical quantities is equal to the sum of individual relative error of each quantity. (iv)Error in the quotient of two values :  We here consider a c= b a ± ∆a Again, c ± ∆c = b ± ∆b ∆a  ∆a    a 1 ±  a 1 ± a  ∆c   a  =   c 1 ± ⇒ =  c  ∆b  b  ∆b    b 1 ± 1 ± b  b     a ∆   1 ± a  ∆c   c 1 ±  ⇒ = c c    1 ± ∆b  b   −1 ∆c  ∆a   ∆b    c 1 ± 1± ⇒ = 1 ±    c  a  b    Using Binomial expansion, as ∆b/b << 1, we get by neglecting higher powers of ∆b, ∆c ∆a   ∆b   = 1 ±  1  b  c a    ∆b ∆a ∆a∆b = 1  ± ± b a ab ∆a∆b We again neglect as it is a very small value. ab ∆c ∆b ∆a 1± Thus, = 1 ± c b a or 1± ± ∆c  ∆a ∆b  = ± + c b   a Hence, relative error is equal to sum of individual relative errors of the quantities being divided. 413 CHAPTER 10 : Mechanics of Fluids (Hydrostatics) Hence, pressure in the fluid at rest is same at all points. Pascal’s law is some times also stated as : A fluid exerts pressure equally in all directions. OR A change in external pressure applied to an enclosed fluid is transmitted equally in all directions and acts undiminished at each point of the fluid and the walls of the container. Experimental verification of Pascal’s law Consider a vessel with F2 A2 F1 three openings of cross A1 sectional areas A1, A2 and A3 provided with frictionless and air tight pistons as shown in the Figure 10.9. The vessel is filled with A3 water and the pistons are F3 kept in position. FIGURE 10.9 Let an additional force F1 be applied on the piston A1 to increase the external pressure on water. It is observed that the other two pistons A2 and A3 move outwards. This shows that the change in external pressure applied on water is transmitted in all directions. It is also found that the outward motion of the pistons A2 and A3 can be stopped by applying forces F2 and F3 such that F1 F2 F3 = = A1 A2 A3 It means that the pressure exerted on each piston is the same and transmitted equally in all directions. This verifies Pascal’s law. 10.7 A PPLICATIONS OF PASCAL’S LAW (i) Hydraulic lift : It works on Pascal’s law and used to lift heavy loads. It consists of two cylinders C1 and C2 F1 F2 provided with air tight frictionless pistons of A1 A2 P cross sectional areas A1 l1 l2 C1 and A2. The Cylinders P C2 are connected to each other by a pipe as shown in Figure 10.10. The Oil cylinders and the pipe FIGURE 10.10 are filled with a liquid such as oil. Let the smaller piston (A1) be pushed down by a force F1 \ Pressure exerted by F1 on the liquid, P = F1/A1 According to Pascal’s law, the same pressure is transmitted equally in all directions. \ Force acting on the larger piston F2 = PA2 = F2 A = 2 F1 A1 \ As A2 > A1 \ F1 A2 A1 F2 > F1 Thus, by making A2 > A1, very heavy loads (like car and truck) can be lifted easily by applying small force. Hence, a hydraulic lift acts a force multiplier. (ii) Hydraulic brakes : It also works on Pascal’s law. it consists of a master cylinder ‘M’ filled with brake oil and provided with the air tight piston P which is connected to a brake paddle through a lever system. The master cylinder M is connected to a wheel cylinder W through a tube (T). Cylinder W is provided with pistons P1 and P2 which are connected to brake shoes S1 and S2 as shown in the Figure 10.11. When the brake shoes press against the inner rim of the wheel, the motion of the wheel is retarded. M Tube T P1 Wheel cylinder P2 Lever system P Master cylinder W Brake paddle Brake shoe S1 S2 FIGURE 10.11 When the brake paddle is pressed, the piston of the master cylinder M is pushed inwards. The increase in pressure in the oil at P is transmitted to piston P1 and P2. As a result, P1 and P2 are pushed outward and the brake shoes press against the inner rim of the wheel. Hence, the motion of the wheel get retarded. SOLVED EXAMPLES Example 10.1 : Calculate the pressure at the bottom of an ocean 1000 m deep. Given : Density of ocean water = 1.02 × 103 kg m–3 Atmospheric pressure = 1.01 × 105 Pa, g = 9.8 ms–2 Solution : Pressure at the bottom P = Pa + hrg P = 1.01 × 105 + 1000 × 1.02 × 103 × 9.8 = 1.01 × 107 Pa Example 10.2 : A truck of mass 10,000 kg stands on the bigger piston of area 20 m2 of a hydraulic lift. Find the mass of a block that must be kept on the smaller piston of area 100 cm2 to lift the truck. 666 CONCEPTUAL PHYSICS—XI and a single formula for spherical lenses to describe all different cases. 17.1.2 Reflection by Spherical Mirrors Consider a parallel beam of light incident on (a) concave mirror, and (b) a convex mirror. This is shown in Figs. 17.6(a) and 17.6(b). Here we assume that the parallel rays incident at points which are close to the pole P of the mirror and make small angles with the principal axis. We find that the reflected rays converge to a point F on the principal axis of a concave mirror Fig. 17.6(a). In the case of a convex mirror, the reflected rays appear to diverge from a point F on the principal axis [Fig. 17.6(b)]. The point F is called the principal focus of the mirror. For a parallel beam of incident light making same angle with the principal axis, the reflected rays would converge (or appear to diverge) to a point in a plane through F normal to the principal axis, called the focal plane of the mirror [Fig. 17.6(c)]. C P P F C F (a) (b) F C P (c) Fig. 17.6 Focus of concave and convex mirrors. The distance between the focus F and the pole P of the mirror is called the focal length of the mirror. It is denoted by f. Let us now prove that the focal length f is related to the radius of curvature R of the mirror by the relation f = R/2. Consider a ray parallel to the principal axis striking the mirror M and making an angle q with the normal MC, where C is the centre of curvature of the mirror as is shown in Fig. 17.7(a). Draw MD as the perpendicular from M on the principal axis. M C 2 F (a) M D P 2 P D F C (b) Fig. 17.7 Ray diagram of an incident ray on (a) concave spherical mirror, (b) convex spherical mirror. Then ∠MCP = q and ∠MFP = 2q MD MD Now tan q = and tan 2q = CD FD Since angle q is small for paraxial rays (which are close to the principal axis), MD MD tan q ≈ q and tan 2q ≈ 2q giving =2 FD CD CD CP But FD  FP and  2 2 CP ∴ FP = 2 In other words the focal length, R f = (17.1) 2 17.1.3 The Mirror Equation A Let us consider the M formation of the image by a concave spherical B F P B C mirror of an object AB A f placed on the principal v u N axis of the mirror MN as Fig. 17.8 Ray diagram for image shown in Fig. 17.8. formation by a concave mirror. The image is said to be real if, the rays emanating from a point object actually converge to a point. It is said to be virtual if the rays do not actually meet but appear to diverge from the point when produced backwards. Thus, an image produced through reflection/ refraction has, in fact, point to point correspondence with the object. To locate the position of the image, we can take any two rays emanating from a point on an object, trace their paths and find their point of intersection. To obtain the image of the point due to reflection at a spherical mirror, it is, in practice, convenient to choose any two of the following rays : (i)The ray from the point parallel to the principal axis incident on the mirror, after reflection goes through the focus of the mirror; (ii)The incident ray passing through the centre of curvature of the mirror, on reflection simply traces its path; (iii)The incident ray passing through the focus of the concave mirror, on reflection would be parallel to the principal axis; (iv)The ray incident at any angle at the pole of the mirror would give a reflected ray following the laws of reflection . Figure 17.8 depicts the ray diagram by considering three rays which results in the formation of the image A′B′ (which is real in this case) of an object AB by a concave mirror. When object is between innity and pole, image is formed between P and F behind the mirror. Nature and size : Virtual, erect and diminished. When object is at innity, image is formed at focus behind the mirror. Nature and size : Virtual, erect and highly diminished (point sized) Mirror formula : 1 1 1 f v u Focus : The point at the half way from radius of curvature. All the reected rays pass through focus. Pole (P) : The point at which principal axis meets the mirror. All distances are measured from pole. Centre of curvature (C) : The centre of a hollow sphere of which the mirror is a part. Magnication : v h or u h Focal length : Half of distance of radius of curvature. It is represented by f. Principal axis : An imaginary horizontal line which is supposed to pass through the centre of the mirror. Radius of curvature (R) : Distance between centre of curvature to any point on the mirror. Some important terms Used as rear view mirror Used in torches, search lights, vehicle headlights, dentist mirror, shaving mirror etc. When object is placed between P and F, image is formed behind the mirror. Nature and size : Virtual, erect and enlarged. When object is placed at F, image is formed at innity. Nature and size : Real, inverted and highly enlarged. When object is between C and F, image is formed beyond C. Nature and size : Real, inverted and enlarged. When object is at C, image is formed at C. Nature and size : Real, inverted and same size as that of the object. When object is placed beyond C, image is formed between F and C. Nature and size : Real, inverted and diminished. When object is at innity, image is formed at focus. Nature and size : Real, inverted, highly diminished (point sized). Image formation by concave mirror See on next page Concave mirror Two types of spherical mirror Formation of image by spherical mirror Refraction of Light Convex mirror Image formation from a mirror obeys laws of reection Image formation by convex mirror 2. Incident ray, reected ray and normal all lie in the same plane. 1. Angle of incidence = Angle of reection, i = r Laws of Reection Reection of Light It gives us power to see things. Light travels in straight line. It changes its path when it reects or refracts from a surface. LIGHT Concept Chart 428 Future-track : Science — X Light—Reflection and Refraction 429 18.6 PRACHI (CBSE Accountancy)—XI Note : These errors arises because entries are omitted to be recorded in journal. In other words, these are the errors of omission. Thus, errors can be rectified by making simple journal entry. Illustration 2 : (Two sided errors) Rectify the following errors : 1. Goods purchased from Nandan ` 4,000 were entered in the books as ` 400. 2. Goods worth ` 3,000 returned to Anamika were entered in the books as ` 300. 3. Goods sold to Anand ` 5,000 were entered in the books as ` 500. 4. Goods worth ` 1,500 returned by Green & Co. were entered in the books as ` 150. Solution : Journal Date 1. Particulars Purchases A/c L.F. Dr. (`) Dr. Cr. (`) 3,600 To Nandan 3,600 (For credit purchases entered as ` 400 instead of ` 4,000) 2. Anamika Dr. 2,700 To Purchases Returns A/c 2,700 (For purchases returns entered as ` 300 instead of ` 3,000) 3. Anand Dr. 4,500 To Sales A/c 4,500 (For credit sales entered as ` 500 instead of ` 5,000) 4. Sales Returns A/c Dr. 1,350 To Green & Co. 1,350 (For sales returns entered as ` 150 instead of ` 1,500) Illustration 3 : (Two sided errors) Rectify the following errors : 1.Goods purchased from Nandan ` 4,000 were entered in the books as ` 4,400. 2.Goods worth ` 3,000 returned to Anamika were entered in the books as ` 3,200. 3. Goods sold to Anand ` 5,000 were entered in the books as ` 5,050. 4. Goods worth ` 1,500 returned by Green & Co. were entered in the books as ` 5,100. Solution : Journal Date 1. Particulars Nandan To Purchases A/c L.F. Dr. Dr. (`) Cr. (`) 400 400 (For credit purchases entered as ` 4,400 instead of ` 4,000) 2. Purchases Returns A/c To Anamika Dr. (For purchases returns entered as ` 3,200 instead of ` 3,000) 200 200 18.7 Rectification of Errors 3. Sales A/c Dr. 50 To Anand 50 (For credit sales entered as ` 5,050 instead of ` 5,000) 4. Green and Company Dr. 3,600 To Sales Returns A/c 3,600 (For sales returns entered as ` 5,100 instead of ` 1,500) Illustration 4 : (Two sided errors) Rectify the following errors : 1. Goods purchased from Nandan ` 4,000 were entered in sales book. 2. Goods worth ` 3,000 returned to Anamika were entered in sales return book. 3. Goods sold to Anand ` 5,000 were entered in purchases book. 4.Goods worth ` 1,500 returned by Green & Co. were entered in the purchases return book. 5. Goods returned to Gopal ` 1,000 were entered in sales book. Solution : Journal Entries Date 1. Particulars L.F. Dr. (`) Purchases A/c Dr. 4,000 Sales A/c Dr. 4,000 To Nandan Cr. (`) 8,000 (For credit purchases from Nandan wrongly entered in sales books) 2. Anamika Dr. 6,000 To Purchases Returns A/c 3,000 To Sales Returns A/c 3,000 (For goods returned to Anamika wrongly entered in sales returns book) 3. Anand Dr. 10,000 To Sales A/c 5,000 To Purchases A/c 5,000 (For credit sales to Anand wrongly entered in purchases book) 4. Sales Returns A/c Dr. 1,500 Purchases Returns A/c Dr. 1,500 To Green & Co. 3,000 (For goods returned by Green & Co. wrongly entered in purchases returns book) 5. Sales A/c To Purchases Returns A/c (For goods returned to Gopal wrongly entered in sales book) Dr. 1,000 1,000 18.22 PRACHI (CBSE Accountancy)—XI Solution : Journal Date 1. (i) Suspense A/c To W’s A/c Particulars Dr. L.F. Dr. (`) Cr. (`) 100 100 (For excess amount debited to W’s a/c, now corrected) (ii) Sales A/c To Suspense A/c Dr. 1,000 1,000 (For sales book overcast, now rectified) 2. B To Sales A/c Dr. 150 150 (For cash sales to B wrongly posted to his credit, now credited to sales a/c) 3. Sales Returns A/c Sales A/c To Suspense A/c Dr. Dr. - (For the correction of sales returns wrongly credited to sales a/c, G’s a/c being correctly credited) 4. Sales Returns A/c To C Dr. 1,240 1,240 (For goods returned by C not recorded in the books, now recorded) 5. Suspense A/c To Sales Returns A/c Dr. 100 100 (For sales returns book has been overcast) 6. Bills Receivable A/c Bills Payable A/c To Suspense A/c Dr. Dr. 1,600 1,600 3,200 (For bills receivable received wrongly posted to the credit of bills payable a/c, now corrected) Suspense Account Dr. Particulars To Difference in trial balance ` Particulars 4,260 By Sales A/c (Balancing figure) By Sales Returns A/c Cr. ` 1,000 130 To W’s A/c 100 By Sales A/c To Sales Returns A/c 100 By B/R A/c 1,600 By B/P A/c 1,600 4,460 130 4,460 Note : It will be ascertained after the preparation of Suspense Account that its debit side is short by ` 4,260. Hence, balancing figure will be assumed as the opening balance of Suspense Account. Illustration 18 : (Preparation of suspense account) A ledger keeper could not agree the trial balance. He transferred an amount of ` 185 being excess of the debit side of the suspense account. The following errors were subsequently discovered : 18.23 Rectification of Errors 1.Amount payable to Govind Ram for repairs done to radio ` 150 and radio supplied for ` 950, were entered in the purchases book as ` 1,000. 2.An item of ` 500 relating to prepaid insurance account was omitted to be brought forward from the previous year’s books. 3.A discount of ` 58 allowed to a customer has been credited to his a/c as ` 56. 4.An amount of ` 500 due from Rakesh which has been written off as bad debts in a previous year, now recovered and had been posted to the personal account of Rakesh. 5.Goods purchased for ` 125 have been posted to the debit of the supplier, Vinod & Co. 6.Furniture amounting to ` 650 was bought from Quality Furniture Co. on credit but passed in the purchases book as ` 560. 7. Rest of difference was due to a wrong total in sales a/c in the ledger. Pass rectifying journal entries and prepare suspense account. Solution : Journal Date 1. Repairs A/c Particulars Radio A/c Dr. L.F. Dr. (`) Cr. (`) 150 Dr. 950 To Purchases A/c 1,000 To Govind Ram 100 (For amount payable for repairs and cost of radio wrongly passed through purchases book) 2. Prepaid Insurance A/c Dr. 500 To Suspense A/c 500 (For prepaid insurance omitted to be brought forward) 3. Suspense A/c Dr. 2 To Customer’s A/c 2 (For wrong amount posted to the credit of customer’s a/c) 4. Rakesh Dr. 500 To Bad Debts Recovered A/c 500 (For bad debts recovered wrongly credited to the a/c of Rakesh, now corrected) 5. Suspense A/c Dr. 250 To Vinod & Co. 250 (For credit purchase wrongly debited to Vinod & Co. now corrected) 6. Furniture A/c Dr. 650 To Purchases A/c 560 To Quality Furniture Co. 90 (For furniture bought for ` 650 passed through purchases book as ` 560) 7. Suspense A/c To Sales A/c (For undercasting of sales a/c, now rectified) Dr. 433 433