SECTION D - QUANTITATIVE APTITUDE - CHAPTER 2

(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:8)(cid:10) (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) LLLLLEEEEEAAAAARRRRRNNNNNIIIIINNNNNGGGGG OOOOOBBBBBJJJJJEEEEECCCCCTTTTTIIIIIVVVVVEEEEESSSSS After studying this chapter, you will be able to: (cid:1) Understand the concept of equations and its various degrees – linear, simultaneous, quadratic and cubic equations; (cid:1) Know how to solve the different equations using different methods of solution; and (cid:1) Know how to apply equations in co-ordinate geometry. 22222.....11111 IIIIINNNNNTTTTTRRRRROOOOODDDDDUUUUUCCCCCTTTTTIIIIIOOOOONNNNN Equation is defined to be a mathematical statement of equality. If the equality is true for certain value of the variable involved, the equation is often called a conditional equation and equality sign ‘=’ is used; while if the equality is true for all values of the variable involved, the equation is called an identity. x+ 2 x+ 3 FFFFFooooorrrrr EEEEExxxxxaaaaammmmmpppppllllleeeee::::: + =3 holds true only for x=1. 3 2 x +2 x + 3 5x+13 So it is a conditional. On the other hand, + = 3 2 6 is an identity since it holds for all values of the variable x. Determination of value of the variable which satisfies an equation is called solution of the equation or root of the equation. An equation in which highest power of the variable is 1 is called a Linear (or a simple) equation. This is also called the equation of degree 1. Two or more linear equations involving two or more variables are called Simultaneous Linear Equations. An equation of degree 2 (highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation. FFFFFooooorrrrr EEEEExxxxxaaaaammmmmpppppllllleeeee::::: 8x+17(x–3) = 4 (4x–9) + 12 is a Linear equation 3x2 + 5x +6 = 0 is a quadratic equation. 4x3 + 3x2 + x–7 = 1 is a Cubic equation. x+2y = 1, 2x+3y = 2 are jointly called simultaneous equations. 22222.....22222 SSSSSIIIIIMMMMMPPPPPLLLLLEEEEE EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN A simple equation in one unknown x is in the form ax + b = 0. Where a, b are known constants and a ¹ 0 NNNNNooooottttteeeee::::: A simple equation has only one root. 4x 14 19 EEEEExxxxxaaaaammmmmpppppllllleeeee::::: -----1 = x + . 3 15 5 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: By transposing the variables in one side and the constants in other side we have (cid:1)(cid:2)(cid:1) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India (cid:2)(cid:3) (cid:4)(cid:2)(cid:3) (cid:4)(cid:6) (cid:9)(cid:10)(cid:11) (cid:8)(cid:4)(cid:2)(cid:7)(cid:3) (cid:4)(cid:6)(cid:1)(cid:5) (cid:12)(cid:3) (cid:10)(cid:2) (cid:2) (cid:2) – = +1 or or . (cid:1) (cid:4)(cid:5) (cid:5) (cid:4)(cid:5) (cid:5) (cid:4)(cid:5) (cid:5) 24x15 x = = 12 5x6 EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((AAAAA))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) ooooorrrrr (((((ddddd))))) 1. The equation –7x + 1 = 5–3x will be satisfied for x equal to: a) 2 b) –1 c) 1 d) none of these x + 4 x - 5 2. The Root of the equation + = 11 is 4 3 a) 20 b) 10 c) 2 d) none of these x 2 3. Pick up the correct value of x for = 30 45 (cid:4) a) x= 5 b) x=7 c) x=1 d) none of these (cid:1) x + 24 x 4. The solution of the equation = 4 + 5 4 a) 6 b) 10 c) 16 d) none of these 5. 8 is the solution of the equation (cid:3)(cid:1)(cid:2) (cid:3)(cid:1)(cid:4)(cid:11) x+ 4 x-5 (cid:1) (cid:2)(cid:13) a) + =11 b) (cid:10) (cid:6) 4 3 (cid:3)(cid:1)(cid:10)(cid:2) (cid:3) (cid:3)(cid:8)(cid:4)(cid:5) (cid:3)(cid:1)(cid:5) (cid:2) (cid:2)(cid:1) (cid:1) (cid:2) (cid:2) c) d) (cid:5) (cid:2) (cid:4)(cid:11) (cid:5) (cid:15)(cid:1)(cid:4)(cid:4) (cid:15)(cid:1)(cid:4) (cid:15)(cid:1)(cid:14) (cid:8) (cid:2) 6. The value of y that satisfies the equation is (cid:12) (cid:6) (cid:2) (cid:4) a) –1 b) 7 c) 1 d) – (cid:14) 7. The solution of the equation (p+2) (p–3) + (p+3) (p–4) = p(2p–5) is a) 6 b) 7 c) 5 d) none of these 12x+1 15x- 1 2x- 5 8. The equation = + is true for 4 5 3x- 1 a) x=1 b) x=2 c) x=5 d) x=7 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) x 1 x 1 9. Pick up the correct value x for which − + − =0 0.5 0.05 0.005 0.0005 a) x=0 b) x=1 c) x=10 d) none of these IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss::::: 1. The denominator of a fraction exceeds the numerator by 5 and if 3 be added to both the (cid:1) fraction becomes . Find the fraction (cid:2) (cid:3) (cid:3)(cid:16)(cid:1) (cid:1) (cid:17) Let x be the numerator and the fraction be . By the question or (cid:3)(cid:16)(cid:5) (cid:3)(cid:16)(cid:5)(cid:16)(cid:1) (cid:2) 4x+12 = 3x+24 or x = 12 (cid:4)(cid:10) (cid:18) The required fraction is (cid:4)(cid:14) 2. If thrice of A’s age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A’s present age. Let x years be A’s present age. By the question 2x–3(x–6) = x or 2x–3x+18 = x or –x+18 = x or 2x = 18 or x=9 ∴ A’s present age is 9 years. 3. A number consists of two digits the digit in the ten’s place is twice the digit in the unit’s place. If 18 be subtracted from the number the digits are reversed. Find the number. Let x be the digit in the unit’s place. So the digit in the ten’s place is 2x. Thus the number becomes 10(2x)+x. By the question 20x+x–18 = 10x + 2x or 21x–18 = 12x or 9x = 18 or x = 2 So the required number is 10 (2 × 2) + 2 = 42. 4. For a certain commodity the demand equation giving demand ‘d’ in kg, for a price ‘p’ in rupees per kg. is d = 100 (10 – p). The supply equation giving the supply s in kg. for a price (cid:1)(cid:2)(cid:19) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India p in rupees per kg. is s = 75( p – 3). The market price is such at which demand equals supply. Find the market price and quantity that will be bought and sold. Given d = 100(10 – p) and s = 75(p – 3). Since the market price is such that demand (d) = supply (s) we have 100 (10 – p) = 75 (p – 3) or 1000 – 100p = 75p – 225 - 1225 or – 175p = ∴p = =7. - 175 So market price of the commodity is Rs. 7 per kg. ∴ the required quantity bought = 100 (10 – 7) = 300 kg. and the quantity sold = 75 (7 – 3) = 300 kg. EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((BBBBB))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. The sum of two numbers is 52 and their difference is 2. The numbers are a) 17 and 15 b) 12 and 10 c) 27 and 25 d) none of these 2. The diagonal of a rectangle is 5 cm and one of at sides is 4 cm. Its area is a) 20 sq.cm. b) 12 sq.cm. c) 10 sq.cm. d) none of these 3. Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. The parts are. a) (20,36) b) (25,31) c) (24,32) d) none of these 4. The sum of the digits of a two digit number is 10. If 18 be subtracted from it the digits in the resulting number will be equal. The number is a) 37 b) 73 c) 75 d) none of these numbers. 5. The fourth part of a number exceeds the sixth part by 4. The number is a) 84 b) 44 c) 48 d) none of these 6. Ten years ago the age of a father was four times of his son. Ten years hence the age of the father will be twice that of his son. The present ages of the father and the son are. a) (50,20) b) (60,20) c) (55,25) d) none of these 7. The product of two numbers is 3200 and the quotient when the larger number is divided by the smaller is 2.The numbers are a) (16,200) b) (160,20) c) (60,30) d) (80,40) 8. The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator the fraction increases by unity. The fraction is. (cid:5) (cid:4) (cid:14) (cid:1) a) b) c) d) (cid:14) (cid:1) (cid:6) (cid:5) (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:20) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 9. Three persons Mr. Roy, Mr. Paul and Mr. Singh together have Rs. 51. Mr. Paul has Rs. 4 less than Mr. Roy and Mr. Singh has got Rs. 5 less than Mr. Roy. They have the money as. a) (Rs. 20, Rs. 16, Rs. 15) b) (Rs. 15, Rs. 20, Rs. 16) c) (Rs. 25, Rs. 11, Rs. 15) d) none of these 10. A number consists of two digits. The digits in the ten’s place is 3 times the digit in the unit’s place. If 54 is subtracted from the number the digits are reversed. The number is a) 39 b) 92 c) 93 d) 94 11. One student is asked to divide a half of a number by 6 and other half by 4 and then to add the two quantities. Instead of doing so the student divides the given number by 5. If the answer is 4 short of the correct answer then the number was (a) 320 (b) 400 (c) 480 (d) none of these. (cid:4) 12. If a number of which the half is greater than th of the number by 15 then the number is (cid:5) (a) 50 (b) 40 (c) 80 (d) none of these. 22222.....33333 SSSSSIIIIIMMMMMUUUUULLLLLTTTTTAAAAANNNNNEEEEEOOOOOUUUUUSSSSS LLLLLIIIIINNNNNEEEEEAAAAARRRRR EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNNSSSSS IIIIINNNNN TTTTTWWWWWOOOOO UUUUUNNNNNKKKKKNNNNNOOOOOWWWWWNNNNNSSSSS The general form of a linear equations in two unknowns x and y is ax + by + c = 0 where a b are non-zero coefficients and c is a constant. Two such equations a x + b y + c = 0 and 1 1 1 a x + b x + c = 0 form a pair of simultaneous equations in x and y. A value for each 2 2 2 unknown which satisfies simultaneously both the equations will give the roots of the equations. 22222.....44444 MMMMMEEEEETTTTTHHHHHOOOOODDDDD OOOOOFFFFF SSSSSOOOOOLLLLLUUUUUTTTTTIIIIIOOOOONNNNN 11111. EEEEEllllliiiiimmmmmiiiiinnnnnaaaaatttttiiiiiooooonnnnn MMMMMeeeeettttthhhhhoooooddddd::::: In this method two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknowns and then solving for the other unknown. EEEEExxxxxaaaaammmmmpppppllllleeeee 11111: SSSSSooooolllllvvvvveeeee::::: 2x + 5y = 9 and 3x – y = 5. SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::2x + 5y = 9 …….. (i) 3x – y = 5 ………(ii) By making (i) x 1, 2x + 5y = 9 and by making (ii) x 5, 15x – 5y = 25 __________________________________ Adding 17x = 34 or x = 2. Substituting this values of x in (i) i.e. 5y = 9 – 2x we find; 5y = 9 – 4 = 5 ∴y = 1 ∴x = 2, y = 1. (cid:1)(cid:2)(cid:21) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 22222..... CCCCCrrrrrooooossssssssss MMMMMuuuuullllltttttiiiiipppppllllliiiiicccccaaaaatttttiiiiiooooonnnnn MMMMMeeeeettttthhhhhoooooddddd::::: Let two equations be: a x + b y + c = 0 1 1 1 a x + b y c = 0 2 2 + 2 We write the coefficients of x, y and constant terms and two more columns by repeating the coefficients of x and y as follows: 1 2 3 4 b c a b 1 1 1 1 b c a b 2 2 2 2 x y 1 and the result is given by: = = (b c −b c ) (c a −c a ) (a b −a b ) 1 2 2 1 1 2 2 1 1 2 2 1 (cid:19) (cid:21) -(cid:19) (cid:21) (cid:21) (cid:20) -(cid:21) (cid:20) (cid:4) (cid:10) (cid:10) (cid:4) (cid:4) (cid:10) (cid:10) (cid:4) so the solution is : x = (cid:20) (cid:19) -(cid:20) (cid:19) y =(cid:20) (cid:19) -(cid:20) (cid:19) . (cid:4) (cid:10) (cid:10) (cid:4) (cid:4) (cid:10) (cid:10) (cid:4) EEEEExxxxxaaaaammmmmpppppllllleeeee 22222::::: 3x + 2y + 17 = 0 5x – 6y – 9 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: 3x + 2y + 17 = 0 ....... (i) 5x – 6y – 9 = 0 ........(ii) MMMMMeeeeettttthhhhhoooooddddd ooooofffff eeeeellllliiiiimmmmmiiiiinnnnnaaaaatttttiiiiiooooonnnnn: By (i) x3 we get 9x + 6y + 51 = 0 ...... (iii) Adding (ii) & (iii) we get 14x + 42 = 0 (cid:2)(cid:10) or x = – = – 3 (cid:4)(cid:2) Putting x = –3 in (i) we get 3(–3) + 2y + 17 = 0 (cid:13) or, 2y + 8 = 0 or, y = – = – 4 (cid:10) So x = –3 and y = –4 MMMMMeeeeettttthhhhhoooooddddd ooooofffff cccccrrrrrooooossssssssss-----mmmmmuuuuullllltttttiiiiipppppllllliiiiicccccaaaaatttttiiiiiooooonnnnn:::::3x + 2y + 17 = 0 5x – 6y – 9 = 0 (cid:1) y 1 (cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:5)(cid:4)(cid:8)(cid:9)(cid:3)(cid:4)(cid:5)(cid:10)(cid:7) = 17×5 – 3(– 9) = 3(– 6)– 5×2 (cid:3) (cid:15) (cid:8) or, (cid:13)(cid:2) = (cid:4)(cid:4)(cid:10) = (cid:4)(cid:2)(cid:11) (cid:3) (cid:15) (cid:8) or = = (cid:1) (cid:2) (cid:4)(cid:8) or x = –3 y = –4 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 22222.....55555 MMMMMEEEEETTTTTHHHHHOOOOODDDDD OOOOOFFFFF SSSSSOOOOOLLLLLVVVVVIIIIINNNNNGGGGG SSSSSIIIIIMMMMMUUUUULLLLLTTTTTAAAAANNNNNEEEEEOOOOOUUUUUSSSSS LLLLLIIIIINNNNNEEEEEAAAAARRRRR EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN WWWWWIIIIITTTTTHHHHH TTTTTHHHHHRRRRREEEEEEEEEE VVVVVAAAAARRRRRIIIIIAAAAABBBBBLLLLLEEEEESSSSS EEEEExxxxxaaaaammmmmpppppllllleeeee 11111::::: SSSSSooooolllllvvvvveeeee for x, y and z: 2x–y + z = 3 x + 3y – 2z = 11 3x – 2y + 4z = 1 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: (((((aaaaa))))) MMMMMeeeeettttthhhhhoooooddddd ooooofffff eeeeellllliiiiimmmmmiiiiinnnnnaaaaatttttiiiiiooooonnnnn 2x – y + z = 3 .......(i) x + 3y – 2z = 11 .... (ii) 3x – 2y + 4z = 1 .... (iii) By (i) × 2 we get 4x – 2y + 2z = 6 …. (iv) By (ii) + (iv), 5x + y = 17 ….(v) [the variable z is thus eliminated] By (ii) × 2, 2x + 6y – 4z = 22 ….(vi) By (iii) + (vi), 5x + 4y = 23 ....(vii) By (v) – (vii), –3y = – 6 or y = 2 Putting y = 2 in (v) 5x + 2 = 17, or 5x = 15 or, x = 3 Putting x = 3 and y = 2 in (i) 2 × 3 – 2 + z = 3 or 6–2+z = 3 or 4+z = 3 or z = –1 So x = 3, y = 2, z = –1 is the required solution. (Any two of 3 equations can be chosen for elimination of one of the variables) (((((bbbbb))))) MMMMMeeeeettttthhhhhoooooddddd ooooofffff cccccrrrrrooooossssssssss mmmmmuuuuullllltttttiiiiipppppllllliiiiicccccaaaaatttttiiiiiooooonnnnn We write the equations as follows: 2x – y + (z – 3) = 0 x + 3y + (–2z –11) = 0 By cross multiplication (cid:1) (cid:12) (cid:8) = = (cid:4)(cid:8)(cid:3)(cid:4)(cid:2)(cid:13)(cid:5)(cid:4)(cid:5)(cid:8)(cid:8)(cid:7)(cid:5)(cid:4)(cid:14)(cid:3)(cid:13)(cid:5)(cid:4)(cid:5)(cid:14)(cid:7) (cid:3)(cid:13)(cid:5)(cid:4)(cid:5)(cid:14)(cid:7)(cid:4)(cid:2)(cid:3)(cid:4)(cid:2)(cid:13)(cid:5)(cid:4)(cid:5)(cid:8)(cid:8)(cid:7) (cid:2)×(cid:14)(cid:5)(cid:4)(cid:8)(cid:3)(cid:4)(cid:8)(cid:7) x y 1 = = 20-z 5z+19 7 (cid:1)(cid:2)(cid:23) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India (cid:10)(cid:11)(cid:8)(cid:22) (cid:5)(cid:22)(cid:1)(cid:4)(cid:6) x = y = (cid:14) (cid:14) Substituting above values for x and y in equation (iii) i.e. 3x - 2y + yz = 1, we have 20−z 5z+19 3  – 2  + 4z = 1  7   7  or 60–3z–10z–38 + 28z = 7 or 15z = 7–22 or 15z = –15 or z = –1 20-(-1) 21 5(-1)+19 14 Now x= = =3 , y= = =2 7 7 7 7 Thus x = 3, y = 2, z = –1 EEEEExxxxxaaaaammmmmpppppllllleeeee 22222::::: Solve for x, y and z : 1 1 1 2 3 4 3 2 1 + + =5, - - =-11, + - =-6 x y z x y z x y z 1 1 1 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We put u = v = w = and get x y z u+v+w = 5 ........ (i) 2u–3v–4w = –11........ (ii) 3u+2v–w = –6 ........ (iii) By (i) + (iii) 4u+3v = –1 ........ (iv) By (iii) x 4 12u+8v–4w = –24 ......... (v) By (ii) – (v) –10u–11v = 13 or 10u + 11v = –13 .......... (vi) By (iv) × 11 44x+33v = –11 …..…(vii) By (vi) × 3 30u + 33v = –39 ……..(viii) By (vii) – (viii) 14u = 28 or u = 2 Putting u = 2 in (iv) 4 × 2 + 3v = –1 or 8 + 3v = –1 or 3v = –9 or v = –3 Putting u = 2, v = –3 in (i) or 2–3 + w = 5 or –1 + w = 5 or w = 5+1 or w = 6 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 1 1 1 1 1 1 Thus x = = y = – = z = = is the solution. u 2 v - 3 w 6 EEEEExxxxxaaaaammmmmpppppllllleeeee 33333::::: SSSSSooooolllllvvvvveeeee fffffooooorrrrr xxxxx yyyyy aaaaannnnnddddd zzzzz::::: xy xz yz = 70, = 84, = 140 x+y x+z y+z SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We can write as x+y 1 1 1 1 + = = or ....... (i) xy 70 x y 70 x+z 1 1 1 1 = or + = ........ (ii) xz 84 z x 84 y+z 1 1 1 1 = + = or ....... (iii) yz 140 y z 140 1 1 1 1 1 1 14  + +  By (i) + (ii) + (iii), we get 2   = + + =  x y z  70 84 140 420 1 1 1 7 1 + + or = = ……(iv) x y z 420 60 1 1 1 4 By (iv)–(iii) = - = or x = 105 x 60 140 420 1 1 1 2 = - = By (iv)–(ii) or y = 210 y 60 84 420 1 1 1 By (iv)–(i) = - or z = 420 z 60 70 Required solution is x = 105, y = 210, z = 420 EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((CCCCC))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. The solution of the set of equations 3x + 4y = 7, 4x – y = 3 is a) (1, –1) b) (1, 1) c) (2, 1) d) (1, –2) x y 2. The values of x and y satisfying the equations + =2, x + 2y = 8 are given by the pair. 2 3 a) (3, 2) b) (–2, –3) c) (2, 3) d) none of these (cid:1)(cid:2)(cid:25)(cid:26) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India x y + =2 3. , x + y = p + q are satisfied by the values given by the pair. p q a) (x=p, y=q) b) (x=q, y=p) c) (x=1, y=1) d) none of these 4. The solution for the pair of equations 1 1 9 1 1 4 + - = , is given by 16x 15y 20 20x 27y 45 1 1 1 1 (a) ,  (b)  ,  (c) (3 4) (d) (4 3) 4 3 3 4 4 5 x+y 3 - = + 5. Solve for x and y: and 3xy = 10 (y–x). The values of x and y are given by x y xy 10 the pair. a) (5, 2) b) (–2, –5) c) (2, –5) d) ( 2, 5) x+y 5 = 6. The pair satisfying the equations x + 5y = 36, is given by x-y 3 a) (16, 4) b) (4, 16) c) ( 4, 8) d) none of these. 7. Solve for x and y : x–3y =0, x+2y = 20. The values of x and y are given as a) x=4, y=12 b) x=12, y=4 c) x=5, y=4 d) none of these 8. The simultaneous equations 7x–3y = 31, 9x–5y = 41 have solutions given by a) (–4, –1) b) (–1, 4) c) (4, –1) d) ( 3, 7) 9. 1.5x + 2.4 y = 1.8, 2.5(x+1) = 7y have solutions as 1 2 a) (0.5, 0.4) b) (0.4, 0.5) c) ( , ) d) ( 2, 5) 2 5 10. The values of x and Y satisfying the equations 3 2 2 3 2 + =3 , + =3 are given by x+y x-y x+y x-y 3 1 a) (1, 2) b) (–1, –2) c) (1, ) d) ( 2, 1) 2 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:25)(cid:25) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((DDDDD))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) aaaaasssss ttttthhhhheeeee sssssooooollllluuuuutttttiiiiiooooonnnnn tttttooooo ttttthhhhheeeee gggggiiiiivvvvveeeeennnnn ssssseeeeettttt ooooofffff eeeeeqqqqquuuuuaaaaatttttiiiiiooooonnnnnsssss ::::: 1. 1.5x + 3.6y = 2.1, 2.5 (x+1) = 6y a) (0.2, 0.5) b) (0.5, 0.2) c) (2, 5) d) (–2, –5) x y x y 2. + + 1= + = 28 5 6 6 5 a) (6, 9) b) (9, 6) c) (60, 90) d) (90, 60) x y z 3. = = 7x + 8y + 5z= 62 4 3 2 a) (4, 3, 2) b) (2, 3, 4) c) (3, 4, 2) d) (4, 2, 3) x y y z z x = 20 = 40 4. , , =24 x+ y y +z z+ x a) (120, 60, 30) b) (60, 30, 120) c) ( 30, 120, 60) d) ( 30, 60, 120) 5. 2x + 3y + 4z = 0, x + 2y – 5z = 0, 10x + 16y – 6z = 0 a) (0,0,0) b) ( 1, –1, 1) c) ( 3, 2, –1) d) (1, 0, 2) 1 1 1 6. (x+y) + 2z = 21, 3x – (y+z) = 65, x + (x+y–z) = 38 3 2 2 a) (4,9,5) b) (2,9,5) c) (24, 9, 5) d) (5, 24, 9) 4 5 x+y 3 - = + 7. 3 xy = 10 (y–x) x y xy 10 a) (2, 5) b) (5, 2) c) (2, 7) d) (3, 4) x y+0.03 y x+0.03 8. + = + = 2 0.01 0.05 0.02 0.04 a) (1, 2) b) (0.1, 0.2) c) (0.01, 0.02) d) (0.02, 0.01) x y yz z x 60 =110, =132, = 9. y -x z-y z+x 11 a) ( 12, 11, 10) b) (10, 11, 12) c) (11, 10, 12) d) (12, 10, 11) 10. 3x–4y+70z = 0, 2x+3y–10z = 0, x+2y+3z = 13 a) (1, 3, 7) b) (1, 7, 3) c) (2, 4, 3) d) (–10, 10, 1) (cid:1)(cid:2)(cid:25)(cid:1) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 22222.....66666 PPPPPRRRRROOOOOBBBBBLLLLLEEEEEMMMMMSSSSS LLLLLEEEEEAAAAADDDDDIIIIINNNNNGGGGG TTTTTOOOOO SSSSSIIIIIMMMMMUUUUULLLLLTTTTTAAAAANNNNNEEEEEOOOOOUUUUUSSSSS EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNNSSSSS IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss ::::: 11111. If the numerator of a fraction is increased by 2 and the denominator by 1 it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2 it becomes 1/2 . Find the fraction SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x/y be the required fraction. x+ 2 x- 4 1 =1, = By the question y +1 y-2 2 Thus x + 2 = y + 1 or x – y = –1 ......... (i) and 2x – 8 = y–2 or 2x – y = 6 ......... (ii) By (i) – (ii) –x = –7 or x = 7 from (i) 7–y = –1 or y = 8 So the required fraction is 7/8. 22222..... The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x years be the present age of the man and sum of the present ages of the two sons be y years. By the condition x = 3y .......... (i) and x + 5 = 2 ( y+5+5) ..........(ii) From (i) & (ii) 3y + 5 = 2 (y+10) or 3y + 5 = 2y + 20 or 3y – 2y = 20 – 5 or y = 15 x = 3 × y = 3 × 15 = 45 Hence the present age of the main is 45 years 33333..... A number consist of three digit of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297 find the number. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the number be 100x + y. we have x + y = 9……(i) Also 100y + x = 100x + y + 297 …………………………….. (ii) From (ii) 99(x – y) = –297 or x – y = –3 ….……………………………………………… (iii) (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:25)(cid:18) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) Adding (i) and (ii) 2x = 6 x = 3 ∴ from (i) y = 6 ∴ Hence the number is 306. EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((EEEEE))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. Monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenses are in the ratio 7 : 9. If each saves Rs. 50 per month find their monthly incomes. a) (500, 400) b) (400, 500) c) (300, 600) d) (350, 550) 2. Find the fraction which is equal to 1/2 when both its numerator and denominator are increased by 2. It is equal to 3/4 when both are incresed by 12. a) 3/8 b) 5/8 c) 3/8 d) 2/3 3. The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages. Find his present age. a) 60 yeas b) 52 years c) 51 years d) 50 years. 4. A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number. a) 54 b) 53 c) 45 d) 55 5. The wages of 8 men and 6 boys amount to Rs. 33. If 4 men earn Rs. 4.50 more than 5 boys determine the wages of each man and boy. a) (Rs. 1.50, Rs. 3) b) (Rs. 3, Rs. 1.50) c) (Rs. 2.50, Rs. 2) d) (Rs. 2, Rs. 2.50) 6. A number consisting of two digits is four times the sum of its digits and if 27 be added to it the digits are reversed. The number is : a) 63 b) 35 c) 36 d) 60 7. Of two numbers, 1/5th of the greater is equal to 1/3rd of the smaller and their sum is 16. The numbers are: a) (6, 10) b) (9, 7) c) (12, 4) d) (11, 5) 8. Y is older than x by 7 years 15 years back X’s age was 3/4 of Y’s age. Their present ages are: a) (X=36, Y=43) b) (X=50, Y=43) c) (X=43, Y=50) d) (X=40, Y=47) 9. The sum of the digits in a three digit number is 12. If the digits are reversed the number is increased by 495 but reversing only of the ten’s and unit digits in creases the number by 36. The number is a) 327 b) 372 c) 237 d) 273 (cid:1)(cid:2)(cid:25)(cid:19) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 10. Two numbers are such that twice the greater number exceeds twice the smaller one by 18 and 1/3 of the smaller and 1/5 of the greater number are together 21. The numbers are : a) (36, 45) b) (45, 36) c) (50, 41) d) (55, 46) 11. The demand and supply equations for a certain commodity are 4q + 7p = 17 and q 7 p = + . respectively where p is the market price and q is the quantity then the 3 4 equilibrium price and quantity are: 3 1 3 (a) 2, (b) 3, (c) 5, (d) None of these. 4 2 5 22222.....77777 QQQQQUUUUUAAAAADDDDDRRRRRAAAAATTTTTIIIIICCCCC EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN An equation of the form ax2 + bx + c = 0 where x is a variable and a, b, c are constants with a ≠ 0 is called a quadratic equation or equation of the second degree. When b=0 the equation is called a pure quadratic equation; when b ≠ 0 the equation is called an affected quadratic. EEEEExxxxxaaaaammmmmpppppllllleeeeesssss::::: i) 2x2 + 3x + 5 = 0 ii) x2 – x = 0 iii) 5x2 – 6x –3 = 0 The value of the variable say x is called the root of the equation. A quadratic equation has got two roots. How to find out the roots of a quadratic equation: ax2 + bx +c = 0 (a ≠ 0) b c or x2 + x + =0 a a b b2 b2 c or x2 + 2 x + = – 2a 4a2 4a2 a  b  2 b2 c or x+  = –  2a 4a2 a b ± b2– 4ac or x + = 2a 2a -b± b2-4ac or x = 2a (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:25)(cid:20) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) LLLLLeeeeettttt ooooonnnnneeeee rrrrroooooooooottttt bbbbbeeeee ααααα aaaaannnnnddddd ttttthhhhheeeee ooooottttthhhhheeeeerrrrr rrrrroooooooooottttt bbbbbeeeee βββββ 2 2 2 2 -b+ b -4ac -b- b -4ac -b+ b -4ac-b- b -4ac Now a + b = + = 2a 2a 2a -2b -b = = 2a a b coefficient of x Thus sum of roots = – = – 2 a coeffientof x −b+ b2 −4ac  −b− b2 −4ac      c Next αβ=  2a   2a  =     a c constant term So the product of the roots = = 2 a coefficientofx 22222.....88888 HHHHHOOOOOWWWWW TTTTTOOOOO CCCCCOOOOONNNNNSSSSSTTTTTRRRRRUUUUUCCCCCTTTTT AAAAA QQQQQUUUUUAAAAADDDDDRRRRRAAAAATTTTTIIIIICCCCC EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN For the equation ax2 + bx + c = 0 we have b c or x2 + x+ = 0 a a  b c or x2 – −  x+ = 0  a a or x2 – (Sum of the roots) x + Product of the roots = 0 22222.....99999 NNNNNAAAAATTTTTUUUUURRRRREEEEE OOOOOFFFFF TTTTTHHHHHEEEEE RRRRROOOOOOOOOOTTTTTSSSSS -b± b2-4ac x = 2a i) If b2–4ac = 0 the roots are real and equal; ii) If b2–4ac >0 then the roots are real and unequal (or distinct); iii) If b2–4ac <0 then the roots are imaginary; iv) If b2–4ac is a perfect square (≠ 0) the roots are real, rational and unequal (distinct); v) If b2–4ac but not a perfect square the rots are real, irrational and unequal. Since b2 – 4ac discriminates the roots b2 – 4ac is called the discriminant in the equation ax2 + bx + c = 0 as it actually discriminates between the roots. (cid:1)(cid:2)(cid:25)(cid:21) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India NNNNNooooottttteeeee::::: (a) Irrational roots occur in pairs that is if (m + n ) is a root then (m – n ) is the other root of the same equation. c (b) If one root is reciprocal to the other root then their product is 1 and so =1 a i.e. c = a (c) If one root is equal to other root but opposite in sign then. b their sum = 0 and so = 0. i.e. b = 0. a EEEEExxxxxaaaaammmmmpppppllllleeeee 11111 ::::: Solve x2 – 5x + 6 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: 1st method : x2 – 5x + 6 = 0 or x2 –2x –3x +6 = 0 or x(x–2) – 3(x–2) = 0 or (x–2) (x–3) = 0 or x = 2 or 3 2nd method (By formula) x2 – 5x + 6 = 0 Here a = 1 b = –5 c = 6 (comparing the equation with ax2 + bx+c = 0) 2 -b± b -4ac -(-5)± 25-24 x = = 2a 2 5±1 6 4 = = and , ∴ x = 3 and 2 2 2 2 EEEEExxxxxaaaaammmmmpppppllllleeeee 22222::::: Examine the nature of the roots of the following equations. i) x2 – 8x + 16 = 0 ii) 3x2 – 8x + 4 = 0 iii) 5x2 – 4x + 2 = 0 iv) 2x2 – 6x – 3 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: (i) a = 1 b = –8 c = 16 b2 – 4ac = (–8)2 – 4.1.16 = 64 – 64 = 0 The roots are real and equal. (ii) 3x2 – 8x + 4 = 0 a = 3 b = –8 c = 4 b2 – 4ac = (–8)2 – 4.3.4 = 64 – 48 = 16 > 0 and a perfect square The roots are real, rational and unequal (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:25)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) (iii) 5x2 – 4x + 2 = 0 b2 – 4ac = (–4)2 – 4.5.2 = 16–40 = –24 < 0 The roots are imaginary and unequal (iv) 2x2 – 6x – 3 = 0 b2 – 4ac = (–6)2 – 4.2 (–3) = 36 + 24 = 60 > 0 The root are real and unequal. Since b2 – 4ac is not a perfect square the roots are real irrational and unequal. IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss::::: 11111..... If œ and ß be the roots of x2 + 7x + 12 = 0 find the equation whose roots are ( œ + ß )2 and (œ - ß)2. SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Now sum of the roots of the required equation = (α+β)2 +(α−β) 2 =(−7) 2 +(α+β) 2 −4αβ = 49 + (–7)2 – 4x12 = 49 + 49 – 48 = 50 Product of the roots of the required equation = (α−β)2.(α−β)2 = 49 (49–48) = 49 Hence the required equation is x2 – (sum of the roots) x + product of the roots = 0 or x2 – 50x + 49 = 0 α2 β2 22222..... If α,β be the roots of 2x2 – 4x – 1 = 0 find the value of + β α −(−4) −1 SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::. α+β= =2, αβ= 2 2 α2 β2 α3 +β3 (α+β)3 −3αβ(α+β) ∴ + = = β α αβ αβ  1 23 −3− .2  2  1 = – 22 −   2 (cid:1)(cid:2)(cid:25)(cid:23) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 33333..... Solve x : 4x – 3.2 x+2 + 2 5 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::4x – 3.2x+2 + 2 5 = 0 or (2x)2 – 3.2 x. 2 2 + 32 = 0 or (2 x)2 – 12. 2 x + 3 2 = 0 or y2 – 12y + 32 = 0 (taking y = 2 x) or y2 – 8y – 4y + 32 = 0 or y(y – 8) – 4(y – 8) = 0 ∴ (y – 8) (y – 4) = 0 either y – 8 = 0 or y – 4 = 0 ∴ y = 8 or y = 4. ⇒ 2 x = 8 = 2 3 or 2 x = 4 = 2 2 ⇒ x = 3 or x = 2. 2  1  1 1 44444. Solve x−  +2x+  =7 .  x  x 4 2  1  1 1 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: x−  +2x+  =7 .  x  x 4 2  1  1 29 x−  +2x+  = .  x  x 4 2 2  1  1 29 or x+  −4+2x+  =  x  x 4 2 2 [as (a – b) = (a + b) – 4ab)] 45 1 or p2 + 2p - = 0 Taking p = x+ 4 x or 4p2 + 8p – 45 = 0 or 4p2 + 18p – 10p – 45 = 0 or 2p(2p + 9) – 5(2p + 9) = 0 or (2p – 5) (2p + 9) = 0. 9 5 ∴Either 2p + 9 = 0 or 2p – 5 = 0 ⇒ p = - or p = 2 2 1 9 1 5 ∴Either x+ = - or x+ = x 2 x 2 i.e. Either 2x2 + 9x +2 = 0 or 2x2 – 5x + 2 = 0 -9± 81-16 5± 25-16 i.e. Either x = or,x- 4 4 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:25)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) -9± 65 1 i.e. Either x = or x = 2 . 4 2 55555..... Solve 2x–2 + 23–x = 3 SSSSSooooollllluuuuutttttiiiiiooooonnnnn: 2x–2 + 23–x = 3 or 2x. 2–2 + 23. 2–x = 3 2x 23 or + =3 22 2x t 8 or + =3 when t = 2x 4 t or t2 + 32 = 12t or t2 – 12t + 32 = 0 or t2 – 8t – 4t + 32 = 0 or t(t–8) – 4(t–8) = 0 or (t–4) (t–8) = 0 ∴ t = 4, 8 For t = 4 2x = 4 = 22 i.e. x = 2 For t = 8 2x = 8 = 23 i.e. x = 3 66666..... If one root of the equation is 2- 3 form the equation. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: other roots is 2 + 3 ∴ sum of two roots = 2 – 3+ 2 + 3= 4 Product of roots = (2 – 3)(2 + 3) = 4 – 3 = 1 ∴ Required equation is : x2 – (sum of roots)x + (product of roots) = 0 or x2 – 4x + 1 = 0. 77777..... If α β are the two roots of the equation x2 – px + q = 0 form the equation α β and whose roots are . β α SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As α, β are the roots of the equation x2 – px + q = 0 α + β = – (– p) = p and α β = q. α β α2 +β2 (α+β)2 −2αβ p2 −2q α β + .= = = . Now ; and = 1 β α αβ αβ q β α (cid:1)(cid:2)(cid:1)(cid:26) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India p2−2q   ∴ Required equation is x2 –   x + 1 = 0 q   or q x2 – (p2 – 2q) x + q = 0 88888. If the roots of the equation p(q – r)x2 + q(r – p)x + r(p – q) = 0 2 1 1 = + are equal show that . q p r SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since the roots of the given equation are equal the discriminant must be zero ie. q2(r – p)2 – 4. p(q – r) r(p – q) = 0 or q2 r2 + q2 p2 – 2q2 rp – 4pr (pq – pr – q2 + qr) = 0 or p2q2 + q2r2 + 4p2r2 + 2q2pr – 4p2qr – 4pqr2 = 0 or (pq + qr – 2rp)2 = 0 ∴ pq + qr = 2pr p q+qr q (p+r) 1 1 2 = 1 or, . =1or, + = or 2pr 2 pr r p q EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((FFFFF))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. If the roots of the equation 2x2 + 8x – m3 = 0 are equal then value of m is (a) – 3 (b) – 1 (c) 1 (d) – 2 2. If 22x + 3 – 32. 2 x + 1 = 0 then values of x are (a) 0, 1 (b) 1, 2 (c) 0, 3 (d) 0, – 3 1 3. The values of 4+ 1 4+ 1 4+ 4+.....2 (a) 1± 2 (bb)2± 5 (c)2± 3 (d) none of these 4. If αβ be the roots of the equation 2x2 – 4x – 3 = 0 the value of ∝2 + β2 is a) 5 b) 7 c) 3 d) – 4 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:1)(cid:25) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 5 If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the 2 b bc squares of their reciprocals then + is equal to ac a 2 a) 2 b) –2 c) 1 d) –1 6. The equation x2 –(p+4)x + 2p + 5 = 0 has equal roots the values of p will be. a) ± 1 b) 2 c) ± 2 d) –2 7. The roots of the equation x2 + (2p–1)x + p2 = 0 are real if. a) p > 1 b) p < (cid:0)4 c) p > 1/4 d) p < 1/4 8. If x = m is one of the solutions of the equation 2x2 + 5x – m = 0 the possible values of m are a) (0, 2) b) (0, –2) c) (0, 1) d) (1, –1) 9. If p and q are the roots of x2 + 2x + 1 = 0 then the values of p2 + q2 becomes a) 2 b) –2 c) 4 d) – 4 10. If L + M + N = 0 and L M N are rationals the roots of the equation (M+N–L) x2 + (N+L–M)x + (L+M–N) = 0 are a) real and irrational b) real and rational c) imaginary and equal d) real and equal α2 β2 11. If µ and β are the roots of x2 = x+1 then value of − is β α a) 2 5 b) 5 c) 3 5 d) –2 5 p q 12. If p ≠ q and p2 = 5p – 3 and q2 = 5q – 3 the equation having roots as and is q p a) x2 – 19x + 3 = 0 b) 3x2 – 19x – 3 = 0 c) 3x2 – 19x + 3 = 0 d) 3x2 + 19x + 3 = 0 13. If one rot of 5x2 + 13x + p = 0 be reciprocal of the other then the value of p is a) –5 b) 5 c) 1/5 d) –1/5 EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((GGGGG))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. A solution of the quadratic equation (a+b–2c)x2 + (2a–b–c)x + (c+a–2b) = 0 is a) x = 1 b) x = –1 c) x = 2 d) x = – 2 2. If the root of the equation x2–8x+m = 0 exceeds the other by 4 then the value of m is a) m = 10 b) m = 11 c) m = 9 d) m = 12 (cid:1)(cid:2)(cid:1)(cid:1) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 3. The values of x in the equation 7(x+2p)2 + 5p2 = 35xp + 117p2 are a) (4p, –3p) b) (4p, 3p) c) (–4p, 3p) d) (–4p, –3p) 6x 6(x+1) 4. The solutions of the equation + =13 are x+ 1 x a) (2, 3) b) (3, –2) c) (–2, –3) d) (2, –3) 5. The satisfying values of x for the equation 1 1 1 1 = + + are x+ p+ q x p q a) (p, q) b) ( –p, –q) c) (p, –p) d) ( –p, q) 6. The values of x for the equation x2+9x+18 = 6–4x are a) (1, 12) b) (–1, –12) c) (1, –12) d) (–1, 12) 7. The values of x satisfying the equation (2x2+5x–2 ) – (2x2+5x–9) = 1 are a) (2, –9/2) b) (4, –9) c) (2, 9/2) d) (–2, 9/2) 8. The solution of the equation 3x2–17x + 24 = 0 are 2 2 2 a) (2, 3) b) (2, 3 ) c) (3, 2 ) d) (3, ) 3 3 3 2 2 3(3x +15) 2x +96 9. The equation + 2x2 + 9 = + 6 6 7 has got the solution as a) (1, 1) b) (1/2, –1) c) (1, –1) d) (2, –1) (cid:1) (cid:1)  −m  +m 10. The equation  x2 −  x + m = 0 has got two values of x to satisfy the equation  2   2  given as  2m   m   2l   l  a) 1,  b) 1,  c) 1,  d) 1,   l −m   l −m   l −m   l −m  22222.....1111100000 PPPPPRRRRROOOOOBBBBBLLLLLEEEEEMMMMMSSSSS OOOOONNNNN QQQQQUUUUUAAAAADDDDDRRRRRAAAAATTTTTIIIIICCCCC EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN 11111..... Difference between a number and its positive square root is 12; find the numbers? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the number be x. Then x – x= 12 …………… (i) (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:1)(cid:18) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) ( x) 2 – x –12 = 0. Taking y = x, y2 – y – 12 = 0 or (y – 4) (y + 3) = 0 ∴ Either y = 4 or y = – 3 i.e. Either x= 4 or x= – 3 If x= – 3 x = 9 if does not satisfy equation (i) so x =4 or x=16. 22222..... A piece of iron rod costs Rs. 60. If the rod was 2 metre shorter and each metre costs Re. 1.00 more, the cost would remain unchanged. What is the length of the rod? 60 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the length of the rod be x metres. The rate per meter is Rs. . x 60 New Length = (x – 2); as the cost remain the same the new rate per meter is x-2 60 60 As given = +1 x-2 x 60 60 or – =1 x-2 x 120 or = 1 x(x-2) or x2 – 2x = 120 or x2 – 2x – 120 = 0 or (x – 12) (x + 10) = 0. Either x = 12 or x = –10 (not possible) ∴ Hence the required length = 12m. 33333..... Divide 25 into two parts so that sum of their reciprocals is 1/6. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: let the parts be x and 25 – x 1 1 1 By the question + = x 25-x 6 25-x+x 1 = or x(25-x) 6 or 150 = 25x – x2 or x2–25x+150 = 0 or x2–15x–10x+150 = 0 or x(x–15) – 10(x–15) = 0 or (x–15) (x–10) = 0 (cid:1)(cid:2)(cid:1)(cid:19) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India or x = 10, 15 So the parts of 25 are 10 and 15. EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((HHHHH))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. Te sum of two numbers is 8 and the sum of their squares is 34. Taking one number as x form an equation in x and hence find the numbers. The numbers are a) (7, 10) b) (4, 4) c) (3, 5) d) (2, 6) 2. The difference of two positive integers is 3 and the sum of their squares is 89. Taking the smaller integer as x form a quadratic equation and solve it to find the integers. The integers are. a) (7, 4) b) (5, 8) c) (3, 6) d) (2, 5) 3. Five times of a positive whole number is 3 less than twice the square of the number. The number is a) 3 b) 4 c) –3 d) 2 4. The area of a rectangular field is 2000 sq.m and its perimeter is 180m. Form a quadratic equation by taking the length of the field as x and solve it to find the length and breadth of the field. The length and breadth are a) (205m, 80m) b) (50m, 40m) c) (40m, 50m) d) none 5. Two squares have sides p cm and (p + 5) cms. The sum of their squares is 625 sq. cm. The sides of the squares are (a) (10 cm, 30 cm) (b) (12 cm, 25 cm) (c) 15 cm, 20 cm) (d) none of these 6. Divide 50 into two parts such that the sum of their reciprocals is 1/12. The numbers are a) (24, 26) b) (28, 22) (c) (27, 23) (d) (20, 30) 7. There are two consecutive numbers such that the difference of their reciprocals is 1/240. The numbers are (a) (15, 16) (b) (17, 18) (c) (13, 14) (d) (12, 13) 8. The hypotenuse of a right–angled triangle is 20cm. The difference between its other two sides be 4cm. The sides are (a) (11cm, 15cm) (b) (12cm, 16cm) (c) (20cm, 24cm) (d) none of these 9. The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are a) (15, 30) b) (32, 13) c) (36, 9) d) (25, 20) 10. The sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right angle triangle is formed. The side of the equilateral triangle is (a) 17 units (b) 16 units (c) 15 units (d) 18 units (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:1)(cid:20) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 11. A distributor of apple Juice has 5000 bottle in the store that it wishes to distribute in a month. From experience it is known that demand D (in number of bottles) is given by D = –2000p2 + 2000p + 17000. The price per bottle that will result zero inventory is (a) Rs. 3 (b) Rs. 5 (c) Rs. 2 (d) none of these. 12. The sum of two irrational numbers multiplied by the larger one is 70 and their difference is multiplied by the smaller one is 12; the two numbers are (a) 3 2, 2 3 (b)5 2, 3 5 (c)2 2, 5 2 (d) none of these. 22222.....1111111111 SSSSSOOOOOLLLLLUUUUUTTTTTIIIIIOOOOONNNNN OOOOOFFFFF CCCCCUUUUUBBBBBIIIIICCCCC EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN On trial basis putting some value of x to check whether LHS is zero then to get a factor. This is a trial and error method. With this factor to factorise the LHS and then to get values of x . IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss : 11111..... Solve x3 – 7x + 6 = 0 Putting x = 1 L.H.S is Zero. So (x–1) is a factor of x3 – 7x + 6 We write x3–7x +6 = 0 in such a way that (x–1) becomes its factor. This can be achieved by writing the equation in the following form. or x3–x2+x2–x–6x+6 = 0 or x2(x–1) + x(x–1) – 6(x–1) = 0 or (x–1)(x2+x–6) = 0 or (x–1)(x2+3x–2x–6) = 0 or (x–1){ x(x+3) – 2(x+3) } = 0 or (x–1)(x–2)(x+3) = 0 ∴ or x = 1 2 –3 22222..... Solve for real xxxxx::::: x3 + x + 2 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: By trial we find that x = –1 makes the LHS zero. So (x + 1) is a factor of x3 + x + 2 We write x3 + x + 2 = 0 as x3 + x2 – x2 – x + 2x + 2 = 0 or x2(x + 1) – x(x + 1) + 2(x + 1) = 0 or (x + 1) (x2 – x + 2) = 0. Either x + 1 = 0 or x2 – x + 2 = 0 i.e. x = –1 1± 1-8 1± -7 i.e. x = = 2 2 (cid:1)(cid:2)(cid:1)(cid:21) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 1± -7 As x = is not real, x = –1 is the required solution. 2 EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222 (((((IIIII))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. The solution of the cubic equation x3–6x2+11x–6 = 0 is given by the triplet : a) (–1, 1 –2) b) (1, 2, 3) c) (–2, 2, 3) d) (0, 4, –5) 2. The cubic equation x3 + 2x2 – x – 2 = 0 has 3 roots namely. (a) (1, –1, 2) b) (–1, 1, –2) c) (–1, 2, –2) d) (1, 2, 2) 3. x 4 – 4 x + 5 are the factors of the left–hand side of the equation. (a) x3 + 2x2 – x – 2 = 0 (b) x3 + x2 – 20x = 0 (c) x3 – 3x2 – 4x + 12 = 0 (d) x3 – 6x2 + 11x – 6 = 0 4. The equation 3x3 + 5x2 = 3x + 5 has got 3 roots and hence the factors of the left–hand side of the equation 3x3 + 5x2 – 3x – 5 = 0 are (a) x – 1, x – 2, x – 5/3 (b) x – 1, x +1, 3x + 5 (c) x + 1, x – 1, 3x - 5 (d) x – 1, x + 1, x – 2 5. Factorise the left hand side of the equation x3 + 7x2 – 21x – 27 = 0 and the roots are as a) (– 3, – 9, – 1) b) (3, – 9, – 1) c) (3, 9, 1) d) (– 3, 9, 1) 6. The roots of x3 + x2 – x – 1 are a) (– 1, – 1, 1) b) (1, 1, – 1) c) (– 1, – 1, – 1) d) (1, 1, 1) 7. The satisfying value of x3 + x2 – 20x = 0 are (a) (1, 4, – 5) (b) (2, 4, – 5) (c) (0, – 4, 5) (d) (0, 4, – 5) 8. The roots of the cubic equation x3 + 7x2 – 21x – 27 = 0 are (a) (–3, –9, –1) (b) (3, –9, –1) (c) (3, 9, 1) (d) (–3, 9, 1) 9. If 4x 3 +8x 2–x–2=0 then value of (2x+3) is given by a) 4, –1, 2 (b) –4, 2, 1 (c) 2, –4, –1 (d) none of these. 10. The rational root of the equation 2x3 – x2 – 4x + 2 = 0 is 1 1 (a) (b) – (c) 2 (d) – 2. 2 2 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:1)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 22222.....1111122222 AAAAAPPPPPPPPPPLLLLLIIIIICCCCCAAAAATTTTTIIIIIOOOOONNNNN OOOOOFFFFF EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNNSSSSS IIIIINNNNN CCCCCOOOOO-----OOOOORRRRRDDDDDIIIIINNNNNAAAAATTTTTEEEEE GGGGGEEEEEOOOOOMMMMMEEEEETTTTTRRRRRYYYYY IIIIInnnnntttttrrrrroooooddddduuuuuccccctttttiiiiiooooonnnnn::::: Co-ordinate geometry is that branch of mathematics which explains the problems of geometry with the help of algebra Distance of a point from the origin. y (x, y) p x o x P (x, y) is a point. By Pythagora’s Theorum OP2 =OL2 + PL2 or OP2 = x2 + y2 So Distance OP of a point from the origin O is x2+y2 DDDDDiiiiissssstttttaaaaannnnnccccceeeee bbbbbeeeeetttttwwwwweeeeeeeeeennnnn tttttwwwwwooooo pppppoooooiiiiinnnnntttttsssss P(x, y) 1 1 (cid:1)(cid:2)(cid:1)(cid:23) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India By Pythagora’s Theorem PQ2=PT2 +QT2 or PQ2 = (x –x )2 + (y –y )2 = (x –x )2 + (y –y )2 2 1 2 1 1 2 1 2 or PQ = (x -x )2 + (y -y )2 1 2 1 2 So distance between two points (x y ) and (x y ) is given by (x -x )2 + (y -y )2 . 1 1 2 2 1 2 1 2 22222.....1111133333 EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNN OOOOOFFFFF AAAAA SSSSSTTTTTRRRRRAAAAAIIIIIGGGGGHHHHHTTTTT LLLLLIIIIINNNNNEEEEE ) y , 2 x (2 P 2 (I) The equation to a straight line in simple form is generally written as y=mx+c …… (i) where m is called the slope and c is a constant. y -y 2 1 If P (x y ) and P (x y ) be any two points on the line the ratio is known as the 1 1, 1 2 2, 2 x -x 2 1 slope of the line. We observe that B is a point on the line y = mx+c and OB is the length of the y-axis that is intercepted by the line and that for the point B x=0. Substituting x=0 in y=mx+c we find y=c the intercept on the y axis. This form of the straight line is known as slope–intercept form. Note : (i) If the line passes through the origin (0, 0) the equation of the line becomes y = mx (or x=my) (ii) If the line is parallel to x–axis, m=0 and the equation of the line becomes y = c (or x = b b is the intercept on x–axis) (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:1)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) (iii) If the line coincides with x–axis, m=0, c=0 then the equation of the line becomes y=0 which is the equation of x–axis. Similarly x=0 is the equation of y–axis. (II) Let y = mx + c ………….. (i) be the equation of the line p p 1 2. Let the line pass through (x , y ). So we get 1 1 y1 = mx +c ...(ii) 1 By (i) – (ii) y–y = m(x–x ) … (iii) 1 1 which is another from of the equation of a line to be used when the slope(m) and any point (x y ) on the line be given. This form is called pppppoooooiiiiinnnnnttttt–––––ssssslllllooooopppppeeeee fffffooooorrrrrmmmmm..... 1 1 (III) If the line above line (iii) passes through another point (x , y ). we write 2 2 y –y = m(x –x ) 2 1 2 1 y-y x-x 1 = 1 by (iii) – (iv) y -y x -x 2 1 2 1 y −y   2 1 (y– y ) =  (x– x ) 1  x −x  1 2 1 Which is the equation of the line passing through two points (x , y ) and (x , y ) 1 1 2 2 (IV) We now consider a straight line that makes x-intercept = a and y-intercept = b Slope of the line y -y b-0 b 2 1 = =- = x -x 0-a a 2 1 (cid:1)(cid:2)(cid:18)(cid:26) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India If (x, y) is any point on this line we may also write the slope as y-0 y = x-a x-a y b Thus =- x-a a y x-a x or =- =- +1 a a a x y Transposing + = 1 a b x y The form + = 1 is called intercept form of the equation of the line and the same is to be a b used when x–intercept and y–intercept be given. Note:(i) The equation of a line can also be written as ax+by+c = 0 (ii) If we write ax+by+c = 0 in the form y = mx+c −a −c −a we get y=   x +   giving slope m=  .  b   a   b  (iii) Two lines having slopes m and m are parallel to each other if and only if m = m 1 2 1 2 and perpendicular to each other if and only if m m = –1 1 2 (iv) Let ax + by + c = 0 be a line. The equation of a line parallel to ax + by + c = 0 is ax + by + k = 0 and the equation of the line perpendicular to ax + by + c = 0 is bx– ay +k = 0 Let lines ax + by + c = 0 and a1x+b1y+c1 = 0 intersect each other at the point (x y ). 1, 1 So ax + by + c = 0 1 1 a1x1 + b1y1 + c1 = 0 x y 1 By cross multiplication = = bc'-b'c ca'-ac' ab'-a'b bc'-b'c ca'-c'a x = . y = 1 ab'-a'b 1 ab'-a'b EEEEExxxxxaaaaammmmmpppppllllleeeee ::::: Let the lines 2x+3y+5 = 0 and 4x–5y+2 = 0 intersect at (x y ). To find the point 1 1 of intersection we do cross multiplication as 2x + 3y + 5 = 0 1 1 4x + 5y + 2 = 0 1 1 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18)(cid:25) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) x y 1 1 = 1 = 3×2-5×5 5×4-2×2 2×5-3×4 Solving x =19/2 y =–8 1 1 (V) The equation of a line passing through the point of intersection of the lines ax + by + c = 0 and a x + b y + c = 0 can be written as ax+by+c+K (a x+b y+c) = 0 when 1 1 1 1 K is a constant. (VI) The equation of a line joining the points (x y ) and (x y ) is given as 1 1 2 2 y-y x-x 1 = 1 y -y x -x 2 1 2 1 If any other point (x y ) lies on this line we get 3 3 y -y x -x 3 1 = 3 1 y -y x -x 2 1 2 1 or x y – x y – x y + x y = x y – x y – x y + x y =0 2 3 2 1 1 3 1 1 3 2 3 1 1 2 1 1 or x y –x y +x y – x y + x y – x y = 0 1 2 1 3 2 3 2 1 3 1 3 2 or x (y –y ) + x (y –y ) + x (y –y ) = 0 1 2 3 2 3 1 3 1 2 which is the required condition of collinearity of three points. IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss::::: 11111..... Show that the points A(2, 3) B(4, 1) and C(–2, 7) are collinear. SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Using the rule derived in VI above we may conclude that the given points are collinear if 2(1–7)+4(7–3)–2(3–1)=0 i.e. if –12+16–4=0 which is true. So the three given points are collinear 22222..... Find the equation of a line passing through the point (5, –4) and parallel to the line 4x+7y+5 = 0 SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Equation of the line parallel to 4x+7y+5 = 0 is 4x+7y+K = 0 Since it passes through the point (5, –4) we write 4(5) + 7(–4) + k = 0 or 20 – 28 + k = 0 or –8 + k = 0 or k = 8 The equation of the required line is therefore 4x+7y+8 = 0. (cid:1)(cid:2)(cid:18)(cid:1) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 33333..... Find the equation of the straight line which passes through the point of intersection of the straight lines 2x+3y = 5 and 3x+5y = 7 and makes equal positive intercepts on the coordinate axes. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: 2x+3y–5 = 0 3x+5y–7 = 0 By cross multiplication x y 1 = = -21+25 -15+14 10-9 x y or = =1 4 -1 So the point of intersection of the given lines is (4, –1) Let the required equation of line be x y + = 1(*for equal positive intercepts a=b) a b ∴ x + y = a Since it passes through (4, –1) we get 4 – 1 = a or a = 3 The equation of the required line is therefore x + y = 3. 44444..... Prove that (3, 1) (5, –5) and (–1, 13) are collinear and find the equation of the line through these three points. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: If A (3, 1) B (5, – 5) and C (–1, 13) are collinear we may write 3(–5–13) +5(13–1) –1(1+5) = 0 or 3(–18) +5(12) – 6 = 0 which is true. Hence the given three points are collinear. As the points A, B, C are collinear, the required line will be the line through any of these two points. Let us find the equation of the line through B(5, – 5) and A (3, 1) Using the rule derived in III earlier we find y +5 x-5 y+5 x-5 = or, = 1+5 3-5 6 -2 or y + 5 +3(x – 5) = 0 or 3x + y = 10 is the required line. 55555..... Find the equation of the line parallel to the line joining points (7, 5) and (2, 9) and passing through the point (3, –4). (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18)(cid:18) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Equation of the line through the points (7, 5) and (2, 9) is given by y – 5 x – 7 = 9 – 5 2 – 7 or –5y + 25 = 4x–28 or 4x+5y–53 = 0 Equation of the line parallel to 4x+5y–53=0 is 4x+5y+k = 0 If it passes through (3, –4) we have 12–20+k = 0 i.e. k=8 Thus the required line is 4x+5y+8 = 0 66666..... Prove that the lines 3x – 4y + 5 = 0, 7x – 8y + 5 = 0 and 4x + 5y = 45 are concurrent. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let (x y ) be the point of intersection of the lines 1 1 3x – 4y +5 = 0 …………… (i) 7x – 8y + 5 = 0 …………... (ii) Then we have 3x – 4y + 5 = 0 1 1 7x – 8y + 5 = 0 1 1 x y 1 20 20 Then 1 = 1 = ∴ x = =5. y = =5. 1 1 -20+40 35-15 -24+28 4 4 Hence (5, 5) is the point of intersection. Now for the line 4x + 5y = 45 we find 4.(5) + 5.5 = 45; hence (5, 5) satisfies the equation 4x+5y=45. Thus the given three lines are concurrent. 77777..... A manufacturer produces 80 T.V. sets at a cost Rs. 220000 and 125 T.V. sets at a cost of Rs. 287500. Assuming the cost curve to be linear find the equation of the line and then use it to estimate the cost of 95 sets. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since the cost curve is linear we consider cost curve as y = Ax + B where y is total cost. Now for x = 80 y = 220000. ∴ 220000 = 80A +B …..(i) and for x = 125 y=287500 ∴ 287500 = 125A +B ……(ii) Subtracting (i) from (ii) 45A = 67500 or A = 1500 From (i) 220000 – 1500 ´ 80 = B or B = 220000 – 120000 = 100000 Thus equation of cost line is y = 1500x + 100000. For x = 95 y = 142500 + 100000 = Rs. 242500. ∴ Cost of 95 T.V. set will be Rs. 242500. (cid:1)(cid:2)(cid:18)(cid:19) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((JJJJJ))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) (((((ddddd))))) 1. The equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is a) 2x – y = 1 b) 3x + 2y = 19 c) 12x – y – 31 = 0 d) none of these. 2. The equation of the straight line passing through the points (–5, 2) and (6, –4) is a) 11x+6y+8 = 0 b) x+y+4 = 0 c) 6x+11y+8 = 0 d) none of these 3 The equation of the line through (–1, 3) and parallel to the line joining (6, 3) and (2, –3) is a) 3x–2y+9 = 0 b) 3x+2y–7 = 0 c) x+y–7 = 0 d) none of these 4. The equation of a straight line passing through the point (–2, 3) and making intercepts of equal length on the ones is (a) 2x+y+1 = 0 b) x–y+5 c) x–y+5 = 0 d) x+y-1=0 5. If the lines 3x – 4y – 13 = 0 8x – 11y – 33 = 0 and 2x – 3y + = 0 are concurrent then value of λ is (a) 11 (b) 5 (c) –7 (d) none of these 6. The total cost curve of the number of copies of a particular photograph is linear. The total cost of 5 and 8 copies of a photograph are Rs.80 and Rs.116 respectively. The total cost for 10 copies of the photograph will be (a) Rs. 100 (b) Rs. 120 (c) Rs. 120 (d) Rs. 140 7. A firm produces 50 units of a product for Rs.320 and 80 units for Rs.380.Considering the cost curve to be a straight–line the cost of producing 110 units to be estimated as (a) 400 (b) 420 (c))))) 440 (d) none of these. 8. The total cost curve of the number of copies photograph is linear The total cost of 5 and 18 copies of a photographs are Rs.80 and 116 respectively. Then the total cost for 10 copies of the photographs is (a) Rs. 140 (b) 93.85 (c) 150 (d) Rs. 130 22222.....1111144444 GGGGGRRRRRAAAAAPPPPPHHHHHIIIIICCCCCAAAAALLLLL SSSSSOOOOOLLLLLUUUUUTTTTTIIIIIOOOOONNNNN TTTTTOOOOO LLLLLIIIIINNNNNEEEEEAAAAARRRRR EEEEEQQQQQUUUUUAAAAATTTTTIIIIIOOOOONNNNNSSSSS 1. Drawing graphs of straight lines From the given equation we tabulate values of (x, y) at least 2 pairs of values and then plot them in the graph taking two perpendicular axis (x, y axis). Then joining the points we get the straight line representing the given equation. EEEEExxxxxaaaaammmmmpppppllllleeeee11111 ::::: Find the graph of the straight line having equation 3y = 9 – 2x (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18)(cid:20) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 9 – 2x SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We have 2x + 3y = 9. We tabulate y = x 0 3 3 y 3 1 A(0,3) B(3,1) Here AB is the required straight line shown in the graph. EEEEExxxxxaaaaammmmmpppppllllleeeee 22222 ::::: Draw graph of the straight lines 3x +4y = 10 and 2x – y = 0 and find the point of intersection of these lines. 10 – 3x SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: For 3x + 4y = 10 we have y = ; we tabulate x 2 10 –10 4 y 1 –5 10 For 2x – y = 0 we tabulate x 0 1 3 y 0 2 6 y 0 = y x - 2 2) 1, (3, 6) ( (2, 1) x x o (10, -5) 3x + 4 y = 10 y (cid:1)(cid:2)(cid:18)(cid:21) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India From the graph, the point of intersection is (1, 2) EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee (((((22222KKKKK))))) CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb)))))((((( (((((ccccc))))) (((((ddddd))))) 1. A right angled triangle is formed by the straight line 4x+3y=12 with the axes. Then length of perpendicular from the origin to the hypotenuse is (a) 3.5 units (b))))) 2.4 units (c) 4.2 units (d) none of these. 2. The distance from the origin to the point of intersection of two straight lines having equations 3x–2y=6 and 3x+2y=18 is (a)3 units (b) 5 units (c) 4 units (d) 2 units. 3. The point of intersection between the straight lines 3x + 2y = 6 and 3x – y = 12 lie in (a) 1st quadrant (b) 2nd quadrant (c) 3rd quadrant (d) 4th quadrant. (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((AAAAA))))) 1. b 2. a 3. c 4. c 5. b 6. d 7. a 8. d 9. c EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((BBBBB))))) 1. c 2. b 3. a 4. b 5. c 6. a 7. d 8. d 9. a 10. c 11. c 12. a EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((CCCCC))))) 1. b 2. c 3. a 4. a 5. d 6. a 7. b 8. c 9. b 10. d EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((DDDDD))))) 1. a 2. c 3. a 4. d 5. a 6. c 7. a 8. c 9. b 10. d EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((EEEEE))))) 1. b 2. a 3. d 4. c 5. b 6. c 7. a 8. a 9. c 10. b 11. a EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((FFFFF))))) 1. d 2. d 3. b 4. b 5. a 6. c 7. d 8. b 9. a 10. b 11. d 12. c 13. b EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((GGGGG))))) 1. b 2. d 3. a 4. d 5. b 6. b 7. a 8. c 9. c 10. a EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((HHHHH))))) 1. c 2. b 3. a 4. b 5. c 6. d 7. a 8. b 9. c 10. a 11. a 12. c EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((IIIII))))) 1. c 2. b 3. b 4. b 5. b 6. a 7. d 8. b 9. a 10. c EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((JJJJJ))))) 1. c 2. c 3. a 4. d 5. c 6. d 7. c 8. b EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 22222(((((KKKKK))))) 1. b 2. b 3. d (cid:1)(cid:2)(cid:18)(cid:23) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India AAAAADDDDDDDDDDIIIIITTTTTIIIIIOOOOONNNNNAAAAALLLLL QQQQQUUUUUEEEEESSSSSTTTTTIIIIIOOOOONNNNN BBBBBAAAAANNNNNKKKKK 1. Solving equation x2- (a+b) x + ab = 0 are, value(s) of x (A) a, b (B) a (C) b (D) None 2. Solving equation x2- 24x + 135 = 0are, value(s) of x (A) 9, 6 (B) 9, 15 (C) 15, 6 (D) None x b a b 3. If + = + the roots of the equation are b x b a (A) a, b2/a (B) a2, b/a2 (C) a2, b2/a (D) a, b2 6x+2 2x2-1 10x-1 4. Solving equation + = we get roots as 4 2x2+2 4x (A) ±1 (B) +1 (C) -1 (D) 0 5. Solving equation 3x2 - 14x + 16 = 0 we get roots as 8 (A) ±1 (B) 2 and (C) 0 (D) None 3 6. Solving equation 3x2 - 14x + 8 = 0 we get roots as (A) ±4 (B) ±2 (C) 4 2/3 (D) None 7. Solving equation (b-c)x2 + (c-a)x + (a-b) = 0 following roots are obtained a-b b-c (A) , 1 (B) (a-b)(a-c), 1 (C) , 1 (D) None b-c a-b x 1-x 8. Solving equation 7 + 8 = 15 following roots are obtained 1-x x 49 64 1 1 49 1 1 64 (A) , (B) , (C) , (D) , 50 65 50 65 50 65 50 65  x 1−x  9. Solving equation 6  + = 15 following roots are obtained  1−x x  4 9 -4 -9 4 5 6 7 (A) , (B) , (C) , (D) , 13 13 13 13 13 13 13 13 10. Solving equation z2 -6z + 9 = 4 z2 -6z + 6 following roots are obtained (A) 3+2 3, 3-2 3 (B) 5, 1 (C) all the above (D) None (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:18)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) x+ 12p-x p+1 = 11. Solving equation following roots are obtained x- 12p-x p-1 (A) 3p (B) both 3pand -4p (C) only -4p (D) -3p 4p 12. Solving equation (1+x)2/3 + (1-x)2/3 = 4(1–x2)1/3 are, values of x 5 5 5 15 (A) (B) – (C) ± (D) ± 3 3 3 3 3 13. Solving equation (2x+1)(2x+3)(x-1)(x-2)=150 the roots available are 1± 129 7 7 (A) (B) , –3 (C) – , 3 (D) None 4 2 2 14. Solving equation (2x+3)(2x+5)(x-1)(x-2)=30 the roots available are 1 -1± 105 1 11 9 (A) 0,1 , –11 , 9 (B) 0,- , (C) 0,- ,- ,- (D) None 2 4 4 2 4 2 4 4 15. Solving equation z+ z= 6 the value of z works out to 25 1 2 1 2 (A) (B) (C) (D) 5 5 25 25 16. Solving equation 10 5 the following values of z are obtained z -33z +32=0 (A) 1, 2 (B) 2, 3 (C) 2, 4 (D) 1, 2, 3 17. When 2z+1+ 3z+4=7the value of z is given by (A) 1 (B) 2 (C) 3 (D) 4 18. Solving equation 2 2 2 following roots are obtained x -9-x+18+ x 2x-15= x -4x+3 2± 94 2± 94 8 8 (A) 3, (B) (C) 4, - (D) 3, 4- 3 3 3 3 19. Solving equation y 2 +4y-21+ y 2 -y-6= 6y 2 -5y-39 following roots are obtained (A) 2, 3, 5/3 (B) 2, 3, -5/3 (C) -2, -3, 5/3 (D) -2, -3, -5/3 4 3 2 20. Solving equation 6x +11x -9x -11x+6=0 following roots are obtained 1 -1± 37 1 -1± 37 1 5 -7 (A) ,-2, (B) - , 2 , (C) , -2 , , (D) None 2 6 2 6 2 6 6 (cid:1)(cid:2)(cid:19)(cid:26) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India x-bc x-ca x-ab 21. If + + =a+ b+c the value of x is d + c c+a a+b (A) a2+b2+c2 (B) a(a+b+c) (C) (a+b)(b+c) (D) ab+bc+ca x+2 x-2 x-1 x+3 22. If - = - then the values of x are x-2 x+2 x+3 x-3 (A) 0,± 6 (B) 0,± 3 (C) 0,±2 3 (D) None x – a x – b b a 23. If + = + then the values of x are b a x – a x – b 2 2 2 2 2 2 a + b a +b a + b (A) 0,(a+b),(a-b) (B) 0,(a+b), (C) 0,(a-b), (D) a+b a+b a+ b 2 2 2 x-a -b c 24. If + =2 the value of is 2 2 2 c x-a -b 1 1 (A) a2+b2+c2 (B) -a2-b2-c2 (C) (D) - a2+b2+c2 a2+b2+c2 2  1  1 25. Solving equation x−  −6x+ +12=0 we get roots as follows  x  x (A) 0 (B) 1 (C) -1 (D) None 2  1  1 26. Solving equation x−  −10x− +24=0 we get roots as follows  x  x (A) 0 (B) 1 (C) -1 (D) (2± 5),(3± 10) 2  1  1  27. Solving equation 2 x−  −5x+ +2+18=0 we get roots as under  x  x  (A) 0 (B) 1 (C) -1 (D) -2± 3 28. If α β are the roots of equation x2-5x+6=0the equation with roots (α + β) and (α - β) is (A) x2-6x+5=0 (B) 2x2-6x+5=0 (C) 2x2-5x+6=0 (D) x2-5x+6=0 29. If α β are the roots of equation x2-5x+6=0 the equation with roots (α2 + β) and (α + β2) is (A) x2-9x+99=0 (B) x2-18x+90=0 (C) x2-18x+77=0 (D) None 30. If α β are the roots of equation x2-5x+6=0 the equation with roots (αβ+α+β) and (αβ-α-β) is (A) x2-12x+11=0 (B) 2x2-6x+12=0 (C) x2-12x+12=0 (D) None (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:19)(cid:25) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 31. The condition that one of ax2+bx+c=0 the roots of is twice the other is (A) b2=4ca (B) 2b2=9(c+a) (C) 2b2=9ca (D) 2b2=9(c-a) 32. The condition that one of ax2+bx+c=0 the roots of is thrice the other is (A) 3b2=16ca (B) b2=9ca (C) 3b2=-16ca (D) b2=-9ca p 2 b 33. If the roots of ax2+bx+c=0 are in the ratio q then the value of (ca) is 2 2 (p+q) (p+q) (p -q) (p -q) (A) (p q) (B) (pq) (C) (p q) (D) (pq) 34. Solving 6x+5y-16=0 and 3x-y-1=0 we get values of x and y as (A) 1, 1 (B) 1, 2 (C) -1, 2 (D) 0, 2 35. Solving x2+y2-25=0 and x-y-1=0 we get the roots as under (A) ±3 ±4 (B) ±2 ±3 (C) 0, 3, 4 (D) 0, -3, -4 x y 5 36. Solving + - =0 and x+y-5=0 we get the roots as under y x 2 (A) 1, 4 (B) 1, 2 (C) 1, 3 (D) 1, 5 1 1 1 1 37. Solving + -13=0 and + -5=0 we get the roots as under x2 y2 x y 1 1 1 1 1 1 1 1 (A) , (B) , (C) , (D) , 8 5 2 3 13 5 4 5 38. Solving x2+xy-21=0 and xy-2y2+20=0 we get the roots as under (A) ±1, ±2 (B) ±2, ±3 (C) ±3, ±4 (D) None 39. Solving x2+xy+y2=37 and 3xy+2y2=68 we get the following roots (A) ±3 ±4 (B) ±4 ±5 (C) ±2 ±3 (D) None 40. Solving 4x.2y=128 and 33x+2y=9xy we get the following roots 7 7 (A) , (B) 2, 3 (C) 1, 2 (D) 1, 3 4 2 41. Solving 9x=3y and 5x+y+1=25xy we get the following roots (A) 1, 2 (B) 0, 1 (C) 0, 3 (D) 1, 3 (cid:1)(cid:2)(cid:19)(cid:1) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 42. Solving 9x+3y-4z=3 x+y-z=0 and 2x-5y-4z=-20 following roots are obtained (A) 2, 3, 4 (B) 1, 3, 4 (C) 1, 2, 3 (D) None 43. Solving x+2y+2z=0 3x-4y+z=0 and x2+3y2+z2=11 following roots are obtained (A) 2, 1, -2 and -2, -1, 2 (B) 2, 1, 2 and -2, -1, -2 (C) only 2, 1, -2 (D) only -2, -1, 2 44. Solving x3-6x2+11x-6=0 we get the following roots (A) -1, -2, 3 (B) 1, 2, -3 (C) 1, 2, 3 (D) -1, -2, -3 45. Solving x3+9x2-x-9=0 we get the following roots (A) ±1, -9 (B) ±1, ±9 (C) ±1, 9 (D) None 46. It is being given that one of the roots is half the sum of the other two solving x3-12x2+47x-60=0 we get the following roots: (A) 1, 2, 3 (B) 3, 4, 5 (C) 2, 3, 4 (D) -3, -4, -5 47. Solve x3+3x2-x-3=0 given that the roots are in arithmetical progression (A) -1, 1, 3 (B) 1, 2, 3 (C) -3, -1, 1 (D) -3, -2, -1 48. Solve x3-7x2+14x-8=0 given that the roots are in geometrical progression (A) ½, 1, 2 (B) 1, 2, 4 (C) ½ , -1, 2 (D) -1, 2, -4 49. Solve x3-6x2+5x+12=0 given that the product of the two roots is 12 (A) 1, 3, 4 (B) -1, 3, 4 (C) 1, 6, 2 (D) 1, -6, -2 50. Solve x3-5x2-2x+24=0 given that two of its roots being in the ratio of 3:4 (A) -2, 4, 3 (B) -1, 4, 3 (C) 2, 4, 3 (D) -2, -4, -3 51. The points (-3 4), (2, 4) and (1, 2) are the vertices of a triangle which is (A) right angled (B) isosceles (C) equilateral (D) other 52. The points (2, 3), (-5, 2) and (-6, -9) are the vertices of a triangle which is (A) right angled (B) isosceles (C) equilateral (D) other 53. The points (2, 3), (-5, 2) and (-4, 9) are the vertices of a triangle which is (A) right angled (B) isosceles (C) equilateral (D) other 54. The points (2, 7), (5, 3) and (-2, 4) are the vertices of a triangle which is (A) right angled (B) isosceles (C) equilateral (D) other 55. The points (1, -1) (- 3,- 3) and (-1, 1) are the vertices of a triangle which is (A) right angled (B) isosceles (C) equilateral (D) other (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:19)(cid:18) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 56. The points (2, -1) (-2, 3) (3, 4) and (-3, -2) are the vertices of a (A) square (B) rhombus (C) parallelogram (D) rectangle 57. The points ( 1 ,- 3 ) (- 3 , 1 ) (- 1 ,- 3 ) and ( 3 ,- 1 ) are the vertices of a triangle 2 2 2 2 2 2 2 2 which is (A) square (B) rhombus (C) parallelogram (D) rectangle 58. The points (2, -2) (-1, 1) (8, 4) and (5, 7) are the vertices of a (A) square (B) rhombus (C) parallelogram (D) rectangle 59. The points (2, 1) (3, 3) (5, 2) and (6, 4) are the vertices of a (A) square (B) rhombus (C) parallelogram (D) rectangle 60. The co-ordinates of the circumcentre of a tringle with vertices (3 -2) (-6 5) and (4 3) are (3 3 ) (3 -3 ) (A) - , (B) , (C) (-3, 3) (D) (3, -3) 2 2 2 2 61. The centroid of a triangle with vertices (1, -2) (-5, 3) and (7, 2) is given by (A) (0, 0) (B) (1, -1) (C) (-1, 1) (D) (1, 1) 62. The ratio in which the point (11, -3) divides the joint of points (3, 4) and (7, 11) is (A) 1:1 (B) 2:1 (C) 3:1 (D) None 63. The area of a triangle with vertices (1, 3) (5, 6) and (-3, 4) in terms of square units is (A) 5 (B) 3 (C) 8 (D) 13 64. The area of a triangle with vertices (0, 0) (1, 2) and (-1, 2) is (A) 2 (B) 3 (C) 1 (D) None 65. The area of the triangle bounded by the lines4x+3y+8=0 x-y+2=0 and 9x-2y-17=0 is (A) 18 (B) 17.5 (C) 17 (D) None 66. The area of the triangle with vertices (4, 5) (1, -1) and (2, 1) is (A) 0 (B) 1 (C) -1 (D) None 67. The area of the triangle with vertices (-3, 16) (3, -2) and (1, 4) is (A) 0 (B) 1 (C) -1 (D) None 68. The area of the triangle with vertices (-1, 1) (3, -2) and (-5, 4) is (A) 0 (B) 1 (C) -1 (D) None 69. The area of the triangle with vertices (p, q+r) (q, r+ p) and (r, p+q) is (A) 0 (B) 1 (C) -1 (D) None (cid:1)(cid:2)(cid:19)(cid:19) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 70. The area of the quadrilateral with vertices (1, 7) (3, -5) (6, -2) and (-4, 2) is (A) 50 (B) 55 (C) 56 (D) 57 71. The centroid of the triangle with vertices (p-q, p-r) (q-r, q- p) and (r-p, r-q) is located at (A) (1, 1) (B) (-1, 1) (C) (1, -1) (D) the origin 72. A lotus over a pond is 1" above the water level. With cool breeze it immersed 7" apart. The depth of the pond in terms of inches is (A) 25 (B) 24 (C) 26 (D) None 73. Points (p, 0) (0, q) and (1, 1) are collinear if (A) 1 p + 1 q =1 (B) 1 p - 1 q =1 (C) 1 p + 1 q =0 (D) 1 p - 1 q =0 74. The gradient or slope of the line where the line subtends an angle q with the X-axis is (A) Sin θ (B) Cos θ (C) Tan θ (D) Cosec θ 75. The equation of the line passing through (5, -3) and parallel to the line is (A) 2x-3y+19=0 (B) 2x-3y-14=0 (C) 3x+2y-19=0 (D) 3x+2y+14=0 76. The equation of the line passing through (5, -3) and perpendicular to the line 2x-3y+14=0 is (A) 3x+2y-9=0 (B) 3x+2y+14=0 (C) 2x-3y-9=0 (D) 2x-3y-14=0 77. The orthocenter of the triangle bound by lines 3x-y=9 x-y=5 and 2x-y=8 is (A) (0, 0) (B) (-6, 1) (C) (6, -1) (D) (-6, -1) 78. The equation of the line passing through points (1, -1) and (-2, 3) is given by (A) 4x+3y-1=0 (B) 4x+3y+1=0 (C) 4x-3y-1=0 (D) 4x-3y+1=0 79. The equation of the line passing through (2, -2) and the point of intersection of 2x+3y-5=0 and 7x-5y-2=0 is (A) 3x-y-4=0 (B) 3x+y-4=0 (C) 3x+y+4=0 (D) None 80. The equation of the line passing through the point of intersection of 2x+3y-5=0 and 7x-5y-2=0 and parallel to the lines 2x-3y+14=0 is (A) 2x-3y+1=0 (B) 2x-3y-1=0 (C) 3x+2y+1=0 (D) 3x+2y-1=0 81. The equation of the line passing through the point of intersection of 2x+3y-5=0 and 7x-5y-2=0 and perpendicular to the lines 2x-3y+14=0 is (A) 3x+2y+5=0 (B) 3x+2y-5=0 (C) 2x-3y+5=0 (D) 2x-3y-5=0 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:19)(cid:20) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) 82. The lines x-y-6=0, 6x+5y+8=0 and 4x-3y-20=0 are (A) Concurrent (B) Non Concurrent (C) Perpendicular to each other (D) Parallel to each other 83. The lines 2x-y-3=0 3x-2y-1=0 and x-3y+2=0 are (A) Concurrent (B) Non Concurrent (C) Perpendicular to each othe (D) Parallel to each other 84. The triangle bound by the lines y = 0, 3x+y-2 = 0 and 3x-y+1 = 0 is (A) right angled (B) isosceles (C) equilateral (D) other 85. The equation of the line passing through (-1 1) and subtending an angle of 45° with the line 6x+5y-1=0 is (A) x+11y-10=0 (B) 11x-y+12=0 (C) both the above (D) None 86. The equation of the line passing through (-1, 1) and subtending an angle of 60° with the line 3x+y-1=0 is (A) y-1=0 (B) 3x-y+( 3+1) (C) both the above (D) None 87. The line joining (-8, 3) and (2, 1) and the line joining (6, 0) and (11, -1) are (A) perpendicular (B) parallel (C) concurrent (D) intersecting to each other at the angle of 45° 88. The lining joining (-1, 1) and (2, -2) and the line joining (1, 2) and (2, k) are parallel to each other for the following value of k (A) 1 (B) 0 (C) -1 (D) None 89. The equation of the second line in question No. (88) is (A) x+y+3=0 (B) x+y+1=0 (C) x+y-3=0 (D) x+y-1=0 90. The lining joining (-1, 1) and (2, -2) and the line joining (1, 2) and (2, k) are perpendicular to each other for the following value of k (A) 1 (B) 0 (C) -1 (D) 3 91. The equation of the second line in question No. (90) is (A) x-y-1=0 (B) x-y+1=0 (C) x-y-3=0 (D) x-y+3=0 92. A factory products 300 units and 900 units at a total cost of Rs.6800/- and Rs.10400/- respectively. The liner equation of the total cost line is (A) y=6x+1,000 (B) y=5x+5,000 (C) y=6x+5,000 (D) None (cid:1)(cid:2)(cid:19)(cid:21) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India 93. If in question No. (92) the selling price is Rs.8/- per unit the break-even point will arise at the level of ______units. (A) 1500 (B) 2000 (C) 2500 (D) 3000 94. If instead in terms of question No. (93) if a profit of Rs.2000/- is to be earned sale and production levels have to be elevated to ________units. (A) 3000 (B) 3500 (C) 4000 (D) 3700 95. If instead in terms of question No. (93) if a loss of Rs.3000/- is budgeted the factory may maintain production level at ________units. (A) 1000 (B) 1500 (C) 1800 (D) 2000 96. A factory produces 200 bulbs for a total cost of Rs.800/- and 400 bulbs for Rs.1200/-. The equation of the total cost line is (A) 2x-y+100=0 (B) 2x+y+400=0 (C) 2x-y+400=0 (D) None 97. If in terms of question No.(96) the factory intends to produce 1000 bulbs the total cost would be Rs._____. (A) 1400 (B) 1200 (C) 1300 (D) 1100 98. If an investment of Rs.1000 and Rs.100 yield an income of Rs.90 Rs.20 respectively for earning Rs.50 investment of Rs._______will be required. (A) less than Rs.500(B) over Rs.500 (C) Rs.485 (D) Rs.486 99. The equation in terms of question No.(98) is (A) 7x – 9y + 1100 = 0 (B) 7x – 90y + 1000 = 0 (C) 7x – 90y + 1100 = 0 (D) 7x – 90y – 1100 = 0 100.If an investment of Rs.60000 and Rs.70000 respectively yields an income of Rs.5750 Rs.6500 an investment of Rs.90000 would yield income of Rs.__________. (A) 7500 (B) 8000 (C) 7750 (D) 7800 101.In terms of question No.(100) an investment of Rs.50000 would yield income of Rs.______. (A) exactly 5000 (B) little over 5000 (C) little less than 5000 (D) at least 6000 102.The equation in terms of question No.(100) is (A) 3x + 40y + 25,000 = 0 (B) 3x – 40y + 50,000 = 0 (C) 3x – 40y + 25,000 = 0 (D) 3x – 40y – 50,000 = 0 (cid:5)(cid:16)(cid:14)(cid:17)(cid:15) (cid:1)(cid:2)(cid:19)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9) AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS 1) A 18) A 35) A 52) D 69) A 86) C 2) B 19) B 36) A 53) B 70) C 87) B 3) A 20) A 37) B 54) A, B 71) D 88) A 4) A 21) D 38) C 55) C 72) B 89) C 5) B 22) D 39) A 56) B 73) A 90) D 6) C 23) D 40) A 57) A 74) C 91) B 7) A 24) A 41) A 58) D 75) A 92) C 8) A 25) B 42) C 59) C 76) A 93) C 9) A 26) D 43) A 60) A 77) B 94) B 10) C 27) D 44) C 61) D 78) A 95) A 11) A 28) A 45) A 62) B 79) B 96) C 12) C 29) C 46) B 63) C 80) A 97) A 13) A 30) A 47) C 64) A 81) B 98) D 14) B 31) C 48) B 65) B 82) A 99) C 15) C 32) A 49) B 66) A 83) B 100) B 16) A 33) A 50) A 67) A 84) C 101) A 17) D 34) B 51) D 68) A 85) C 102) B (cid:1)(cid:2)(cid:19)(cid:23) (cid:3)(cid:4)(cid:5)(cid:5)(cid:4)(cid:6)(cid:7) (cid:8)(cid:9)(cid:4)(cid:10)(cid:11)(cid:3)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7) (cid:14)(cid:12)(cid:15)(cid:14) Copyright -The Institute of Chartered Accountants of India