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1 Simple Learn and Remember Equations 1. A combination of constants and variables connected by the sign of fundamental operations of addition, subtraction, multiplication and division is called the algebraic expression.. An equation is a condition on variable such that two expressions in the variable should have equal value.. Solution of equation is the value of the variable f which satisfied the equation.. An equation remains same when the expression on the left and on the right are interchanged.. F balancing solving the equation, we can: (i) Add same number to both sides. (ii) Subtract same number from both sides. (iiii) Multiply and divide both sides by same non-zero numbers. 6. We can transpose any term from one side of an equation to other side of the equation by changing the sign. TEXTBOOK QUESTIONS SOLVED Exercise.1 (Page No. 81-8) Q1. Complete the last column of the table. i I Sol. S.No. Equation Value Say, whether the Equation is satisfied. (YeslNo) (i) x+=o x= No (ii) x+=o x=o No (iii) x+=o x=- Yes (iv) x-7=1 x=7 No (v) x-7=1 x=8 Yes (vi) x = x=o No (vii) x = x= Yes '

2 66 MATHEMATiCS-VII r (viii) x = x = - No (i:y) -I m - = x=-6 No (x) m = m=o No (xi) m = m=6 Yes Q. Check whether the value given in the brackets is a solution to the given equation not: (a) n + = 19 (n = 1) (b) 7n + = 19 (n = - ) (c) 7n + = 19 (n = ) (d) p - = 1 (p = 1) (e) p - = 1 (p = - ) (f) p - = 1 (p = 0) Sol. (a) n + = 19 (n = 1) Putting n = 1 in L.H.S. of the given equation, L.H.S. = 1 + 6*19 = RH.S. L.H.S. * RH.S. Thus, n = 1 is not the solution of the given equation. (b) 7n + = 19 (n = - ) Putting n = - in L.H.S. of the given equation, L.H.S. = 7(- ) *19 = RH.S. L.H.S. * RH.S. Thus, n = (- ) is not the solution of the given equation. (c) 7n + = 19 (n = ) Putting n = in L.H.S. of the given equation, L.H.S. = 7() = 19 = RH.S. L.H.S. = RH.S. Thus, n = is the solution of the given equation. (d) p - = 1 (p = 1) Putting p = 1 in L.H.S. of the given equation, L.H.S. = (1) - - SIMPLE EQUATIONS = RH.S. L.H.S. * RH.S. Thus, p = 1 is not the solution (e) p - = 1 (p = - ) Puttingp L.H.S. = (- ) of the given equation. = (- ) in L.H.S. of the given equation, = RH.S. L.H.S. * RH.S. Thus, p = (- ) is not the solution (f) p - = 1 (p = 0) of the given equation. Putting p = 0 in L.H.S. of the given equation, L.H.S. = (0) * 1 = RH.8. L.H.S. * RH.S. Thus, p = 0 is not the solution of the given equation. Q. Solve the following equations by trial and err method: (i) p+=17 (ii) m-1= Sol. (i) We have, p + = 17 (a) Trial method: p + = 17 Putting p =-, (-) + = = 17 Putting p=-, (-)+=17 Putting p =-1, (-1)+= 17 Putting p = 0, (0) + = 17 Putting p = 1, (1) + = 17 = -1 *17-10+=17-8*17 -+=17 -*17 0+=17 *17 + = 17 7*17

68 MATHEMATiCS-VII (b) Putting p =, ()+ = 17 10 + = 17 1*17 Putting p =, () + = 17 1 + = 17 17 = 17 Thus, p = is the required solution of the given equation.

3 68 MATHEMATiCS-VII (b) Putting p =, ()+ = = 17 1*17 Putting p =, () + = = = 17 Thus, p = is the required solution of the given equation. Err method: p + = 17 p = 17- p = 1 1 Therefe, p = - = Thus, p = is the required solution of the given equation. (ii) We have, m - 1 = (a) Trial method: m -1 = Putting m = -, (- ) - 1 = = -0* Putting m = - 1, (- 1)- 1 = = -17* Putting m = 0, (0)-1 = = 1-1* 1 Putting m = 1, (1)- 1 = - 1 = 1-11* Putting m =, ()- 1 = 6-1 = -8* Putting m =, ()- 1 = 9-1 = -*-- Putting m =, ()- 1 = 1-1 = -* Putting m =, ()- 1 = 1-1 = 1* Putting m = 6, (6)- 1 = 18-1 = = Thus, m = 6 is the required solution of the given equation. SIMPLE EQUATIONS (b) Err method: m -1 = m = + 1 m = Therefe, m = - = 6 Thus, m = 6 is the required solution of the given equation. Q. Write equations f the following statements: (i) The sum of numbers x and is 9. (ii) subtracted from y is 8. (iii) Ten times a is 70. (iv) The number b divided by gives 6. (v) Three-fourth of t is 1. (vi) Seven times m plus 7 gets you 77. (vii) One-fourth of a number x minus gives. (viii) H you take away 6 from 6 times y, you get 60. (ix) H you add to one-third of z, you get 0. Sol. (i) The sum of number x and = x + Accding to condition, x + = 9. (ii) is subtracted from y = y - (iii) Accding to condition, y - = 8. Ten times a = loa Accding to condition, loa = 70. (iv) The number b divided by = b + = Accding to condition, = 6. (v) Three-fourth of t = " t Accding to condition, (vi) Seven times ofm = 7m Accding to condition, + t = 1. 7m plus 7 = 7m + 7 7m + 7 = 77. (vii) One-fourth of a number x = minus = - x Accding to condition, " - =. 69

4 70 (viii) 6 times y = 6y Take away 6 from 6y = 6y - 6 Accding to condition, 6y- 6 = 60. (ix) One-third of z = z z Add to " = " + z Accding to condition, " + = 0. MATHEMATics-VII Q. Write the following equations in statement fms: (i) p + = 1 (ii) m - 7 = (iii) m = 7 (iv) m = (v) m = 6 (vi) p + = (vii) p - = 18 (viii) + = 8 Sol. (i) The sum of numbers p and is 1. (ii) 7 subtracted from m is. (iii) Two times m is 7. (iv) The number m is divided by gives. (v) Three-fifth of the number m is 6. (vi) Three times p plus gets. (vii) If you take away from timesp, you get 18. (viii) If you added to halfofp, you get 8. Q6. Set up an equation in the following cases: (i) Irfan says that he has 7 tnarbles me than five times the marbles Parmit has. Irfan has 7 marbles. (Take m to be the number of Parmit's marbles.) (ii) Laxmi's father is 9 years old. He is years older than three times Laxmi's age. (Take Laxmi's age to be y years.) - (iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest sce is 87. (Take the lowest sce to be I.) (iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees.) SIMPLEQUATIONS 71 Sol. (i) Let m be the number of Parmit's marble. (ii) (iii) (iv) Accding to the given conditions, five times ofparmit's marble and add 7 marbles. m+7=7. Let the age of Laxmi be y years. Accding to the given conditions, three times of Laxmi's y + = 9. Let the lowest marks be l. Accding to given conditions, age and add years. the twice the lowest marks plus 7 get 87. l + 7 = 87. Let the base angle of the isosceles triangle bnb. Accding to given conditions, the a vertex angle is twice either base angle = b. c b We know that, a + b + c = 180 b+b+b=180 Exercise. (Page No. 86) (By angle sum property b = 180. (:. Zb e Zc) of triangle) Q1. Give first the step you will use to separate the variable and then solve the equation: (a) x-i = 0 (b) x + 1 = 0 (d) x + 6 = (e) y - = - 7 )y+= ) y+=- Sol. (a) x - 1 = 0 (c) x-i = (f) y - = Now, adding 1 to both sides, we get x-1+1=0+1 x = 1, which is the required (b) x + 1 = 0 Now, subtracting 1 from both sides, we get = x+1-1=0-1 x = - 1, which is the required

7 MATHEMATiCS-VII (c) x-i = Now, adding 1to both sides, we get x-l+l=+1 x = 6, which is the required (d) x + 6 = Now, subtracting 6 from both sides, we get x+6-6=-6 x = -, which is the required (e) y

5 7 MATHEMATiCS-VII (c) x-i = Now, adding 1to both sides, we get x-l+l=+1 x = 6, which is the required (d) x + 6 = Now, subtracting 6 from both sides, we get x+6-6=-6 x = -, which is the required (e) y - =-7 (f) Now, adding to both sides, we get y-+=-7+ y = -, which is the required y-= Now, adding to both sides, we get y-+=+ y = 8, which is the required (g)y+= Now, subtracting from both sides, we get y+-=- y = 0, which is the required (h) y + = - Now, subtracting from both sides, we get y+-=-- y = - 8, which is the required Q. Give first the step you will use to separate the variable and then solve the equation: (a) l = (d)x= (e) 8y = 6 a 7 (g) - =- 1 b P (b) - = 6 (c) - = 7 (h) 0t = z (f) - = - SIMPLE EQUATIONS Sol. (a) l = Now, dividing by on both sides, we get b (b) - = 6 Now, multiplying l = l = 1, which is the required by on both sides, we get b - x=6x b = 1, which is the required p (c) - = 7 Now, multiplying P -x7=x7 7 both sides by 7, we get p = 8, which is the required (d) x = Now, dividing both sides by, we get (f) x x = ""' which is the required (e) 8y = 6 Now, dividing both sides by 8, we get By 6 8=8 y = %, which is the required z - Now, multiplying both sides by, we 6 t. z -x=-x 1 hi h i h. d I. z = ""' w lc IS t e require so ution. 7

6 7 MATHEMATiCS-VII (g) a 7 " = 1 Now, multiplying both sides by, we get a 7 -x=-x 1 (h) 0t = -10 a = '!..., which is the required. Now, dividing both sides by 0, we get 0t = 0-1 t = ' which is the required Q. Give the steps you will use to separate the variable and then solve the equation: (a) n - = 6 (c) 0p = 0 Sol. (a) n - = 6 Step I. Adding to both sides, we get n - + = 6 + n=8 Step 11. Dividing both sides by, we get (b) m + 7 = 17 n 8 = Step I. Subtracting (b) m+7=17 (d) p = n = 16, which is the required m = from both sides, we get m= 10 Step 11. Dividing both sides by, we get m 10 = m =, which is the required SIMPLE EQUATIONS 7 0p (c) - =0 Step I. Multiplying both sides by, we get 0p - x=0x 0p = 10 Step 11. Dividing both sides by 0, we get p Cd) 10 = 6 0p _ p = 6, which is the required Step I. Multiplying both sides by 10, we get p x 10 = 6 x p=60 Step 11. Dividing both sides by, we get p _ 60 - p = 0, which is the required Q. Solve the following equations: (a) lop = 100 (b) lop + 10 = 100 (c) J!.. = -p p -(d) - = (e) - =6 (g) s + 1 = 0 (h) s = 0 (j) q-6=0 (k) q+6=0 Sol. (c) lop = 100 Now, dividing both sides by 10, we get lop _ (f) s = - 9 (i) q = 6 (l) q + 6 = 1. p = 10, which is the required Cb) lop + 10 = 100 Now, subtracting 10 from both sides, we get lop = lop = 90

76 MATHEMATiCS-VII Then, dividing both sides by 10, we get p (c) - = Now, multiplying lop 90 10 = 10 p = 9, which is the required both sides by, we get.

7 76 MATHEMATiCS-VII Then, dividing both sides by 10, we get p (c) - = Now, multiplying lop = 10 p = 9, which is the required both sides by, we get.!!... x=x p = 0, which is the required (d) - p = Now, multiplying (e) : = 6 Now, multiplying get both sides by (- ), we get -: x (- ) = x (- ) p = - 1, which is the required by and divided by on both sides, we p 6x -x-=-- p=x p = 8, which is the required (n s =-9 Now, dividing both sides by, we get (g) s + 1 = 0 Now, subtracting s -9 = s = -, which is the required 1 from both sides, we get s = = -1, Then, dividing both sides by, we get s -1 =-- SIMPLE EQUATIONS 77 8 = -, which is the required (h) s = 0 Now, dividing both sides by, we get s 0 = s = 0, which is the required (i) q = 6 Now, dividing both sides by, we get (j) q-6=0 q 6 =" q =, which is the required Now, adding 6 to both sides, we get q-6+6=0+6 q=6 Then, dividing both sides by, we get (k) q + 6 = 0 Now, subtracting q 6 =" q =, which is the required 6 from both sides, we get q+6-6=0-6 q=-6 Then, dividing both sides by, we get q -6 -=- q = -, which is the required Cl) q + 6 = 1 Now, subtracting 6 from both sides, we get q = 1-6 q = 6 Then, dividing both sides by, we get q 6 =" q =, which is the required

78 MATHEMATiCS-VII Exercise. (Page No. 89) Q1. Solve the following equations: 7 (a) y + - = - q (d) - + 7 = 19 (g) 7m + - = 1 (j) b _ =. 7 Sol. (a) y + " = 7 y= - -- (Transposing % to RH.S.) 7- y=-- y=- y = 16 16 (Dividing both sides by.

8 78 MATHEMATiCS-VII Exercise. (Page No. 89) Q1. Solve the following equations: 7 (a) y + - = - q (d) = 19 (g) 7m + - = 1 (j) b _ =. 7 Sol. (a) y + " = 7 y= - -- (Transposing % to RH.S.) 7- y=-- y=- y = (Dividing both sides by.) v = - y = 8, which is the required (b) t + 8 = 10 t=10-8 (b) t+ 8 = 10 (e) -x=10 (n (h) 6z + 10 = - (Transposing a (c) - + = -X= -. 1 (l.) - = - 8 to RH.S.) t = - 18 t = - 8 (Dividing both sides by.) -18 t = --' which is the required a (c) - + = a - =- (Transposing to RH.S.) (d) 1+ 7 = a - =-1 a=-lx (Multiplying both sides by.) a = -, which is the required 1=-7 (Transposing 7 to RH.S.) SIMPLE EQUATIONS 79 1=_ q=-x (Multiplying both sides by.) q = - 8, which is the required (e) -x = 10 x = 10 x (n (Multiplying both sides by.) x= 0 x = 0 (Dividing both sides by.) x =, which is the required -x= - x x=-- x=- 19 (g) 7m + - = 1 (Multiplying both sides by.) 1 x=-x- (Dividing both sides by.) x = %, which is the required 19 7m = m= m=-x- 7 1 m = "' which is the required (h) 6z + 10 = - 6z = z =-1 7 7m = - -1 z = -6- (Transposing 19 to RH.S.) (Dividing both sides by 7.) (Transposing 10 to RH.S.) (Dividing both sides by 6.) z = -, which is the required (i) 1 =

80 MATHEMATiCS-VII l= - x (Multiplying both sides by.) l =- (j) b -= b - =+ b - =8 1 l=-x- (Dividing both sides by.) 1 = ±, which is the required 9 b = 8 x (Transposing to RH.S.) (Multiplying both sides by.

9 80 MATHEMATiCS-VII l= - x (Multiplying both sides by.) l =- (j) b -= b - =+ b - =8 1 l=-x- (Dividing both sides by.) 1 = ±, which is the required 9 b = 8 x (Transposing to RH.S.) (Multiplying both sides by.) b = b = - = 1 (Dividing both sides by.) b = 1, which is the required Q. Solve the following equations: (a) (x + ) = 1 (b) (n - ) = 1 (c) (n - ) = - 1 (d) -(-y)=7 (e) -(-x)=9 (r) (-x)=9 (g) + (p -1) = (h) - (p -1) =. Sol. (a) (x + ) = 1 1 (x + ) = (Dividing both sides by.) x+=6 x = 6 - (Transposing to.r.h.s.) (b) (n-)=1 x =, which is the required (n - ) = (Dividing both sides by.) n-=7 n = 7 + (Transposing to RH.8.) n = 1, which is the required (c) (n - ) = - 1 SIMPLE EQUATIONS 81 n-=-- n-=-7-1 n = (Transposing to RH.S.) n = -, which is the required (d) - ( - y) = 7 - ( - y) = (-y) = -+y= (Dividing both sides by.) (Transposing to RH.S.) -(-y)=- (Dividing both sides by.) y = + (Transposing to RH.S.) y =, which is the required (e) -(-x)=9 (f) -(-x)=" x-=" x=-- (Multiplying both sides by.) 9 x= - + (Transposing to RH.S.) x = 1: ' which is the required (-x)=9 9 (-x)=- 9 -x = x=-- (Multiplying both sides by.) (Transposing 1 -x=" x = - 1, which is the required (g) + (p - 1) = to RH.S.) (p - 1) = - (Transposing to RH.S.) (P-1)= 0

8 MATHEMATIcs-VII 0 (P-l)= - =6 p-l=6 p = 6 + 1 (Transposing 1 to RH.S.) p = 7, which is the required (h) - (p- 1) = - (p- 1) = - - (p- 1) = - 0 (P-l)= 0 0 (p-1)=- (Dividing both sides by.

10 8 MATHEMATIcs-VII 0 (P-l)= - =6 p-l=6 p = (Transposing 1 to RH.S.) p = 7, which is the required (h) - (p- 1) = - (p- 1) = - - (p- 1) = - 0 (P-l)= 0 0 (p-1)=- (Dividing both sides by.) p-l=6 (Dividing both sides by.) (Transposing to RH.S.) (Multiplying both sides by(-i).) p=6+1 (Transposing p = 7, which is the required Q. Solve the following equations: (a) =(p-) (b) -=(p-) (c) - 16 = - ( - p) (d) 10 = + (t + ) (e) 8 = + (t + ) (f) 0 = 16 + (m - 6). Sol. (a) = (p- ) (b) 1 to RH.S.) - =p- (Dividing both sides by.) - +=p -=(p-) p= P=-- (Transposing to L.H.S.) (Interchanging 1 p = 'which is the required - - =(p-) - - +=p sides.) (Dividing both sides by.) (Transposing to L.H.S.) SIMPLE EQUATIONS (c) -16=-(-p) - p= p=--- p =, which is the required - ( - p) = - 16 (Interchanging sides.) -16 -p = - (Multiplying both sides by (-).) 16 -p= -- -P= 6 =p: Cd) 10 = + (t + ) (t + ) = 10 (t + ) = 10 - (t + ) = 6 t+= (Interchanging a sides.) (Transposing to RH.S.) p = 6, which is the required t = - (Transposing to RH.S.) t = 0, which is the required (e) 8 = + (t + ) +(t+)=8 (t + ) = 8 - (t + ) = t+= - t+=8 (Interchanging sides.) (Transposing to RH.8.) 6 t+=" (Dividing both sides by.) (Interchanging sides.) (Transposing to RH. S.) (Dividing both sides by.)

8 MATHEMATiCS-VII t = 8 - (Transposing to RH.S.) t =, which is required (n 0 = 16 + (m - 6) 16 + (m - 6) = 0 (m - 6) = - 16-16 m-6= -- m-6=- m = - + 6 (Transposing 6 to RH.S.) m =, which is the required *Q.

11 8 MATHEMATiCS-VII t = 8 - (Transposing to RH.S.) t =, which is required (n 0 = 16 + (m - 6) 16 + (m - 6) = 0 (m - 6) = m-6= -- m-6=- m = (Transposing 6 to RH.S.) m =, which is the required *Q. (a) Construct equations starting with x =. (b) Construct equations starting with x = -. Sol. (a) Equations are (i) x = Multiplying both sides by 10, Adding to both sides, (ii) x = (Interchanging sides.) (Transposing 16 to RH.S.) (Dividing both sides by.) 10x = 0 10x + = x + =...(i) x Dividing both sides by, " = "...(ii) (iii) x = Multiplying both sides by, x = x = 10 Subtracting from both sides, x - = 10-7 x- = 7...(iii) Thus, equations (i), (ii) and (iii) may required equations. (b) Given x = - (i) x = - Multiplying both sides by, (ii) x = - Multiplying both sides by, Adding 7 to both sides, (iii) x = - x = - x- x =-6 x=-x x =-6 x+7=-6+7 x + 7 = 1... (i)...(ii) SIMPLE EQUATIONS Multiplying both sides by, x = - x x =-6 Adding 10 to both sides, x + 10 = x + 10 =...(iii) Thus, equations (i), (ii) and (iii) may required equations. Exercise. (Page No. 91) Ql. Set up equations and solve them to find the unknown numbers in the following cases: (a) Add to eight times a number; you get 60. (b) One-fifth of a number minus gives. (c) If I take three-fourth of a number and add to it, I get 1. (d) When I subtracted 11from twice a number, the result was 1. (e) Munna subtracts thrice the number of notebooks he has from 0, he finds the result to be 8. (f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by, she will get 8. (g) Anwar thinks of a number. If he takes away 7 from. 11 " of the number, the result IS ". Sol. (a) Let the number be x. Accding to the given conditions, 8x + = 60 8x = 60 - (Transposing to RH.S.) 8x =6 x = 7 Thus, the required number is 7. (b) Let the number be y. Accding to the given conditions, 1:-= 6 x=- 8 8 (Multiplying both sides by 8.)

12 86 MATHEMATiCS-VII SIMPLE EQUATIONS 87 Z =+ (Transposing to RH.S.) y - =7 y=7x (Multiplying both sides by.) y = Thus, the required number is. (c) Let the number be z. Accding to the given conditions, -z + = 1 -z = 1- (Transposing to RH.S.) -z= 18 z = 18 x (Multiplying both sides by.) z = 7 7 z= - (Dividing both sides by.) z= Thus, the required number is. (d) Let the number be x. Accding to the given conditions, x-11 = 1 x = x=6 (Transposing 11 to RH.S.) 6 (Dividing both sides by.) X= - x = 1 Thus, the required number is 1. (e) Let the number be rn. Accding to the given conditions, 0-rn = 8 -rn = 8-0 (Transposing 0 to RH.S.) -rn =- Q. rn=- [Dividing both sides by and multiplying both sides by (-1).] m 1 Thus, the required number is 1. (f) Let the number be n. Accding to the given conditions, n+19 = 8 n + 19 = 8 x (Multiplying both sides by.) n + 19=0 n = 0-19 (Transposing 19 to R.H.S.) n = 1 Thus, the required number is 1. (g) Let the number be x. Accding to the given conditions, 11 -x-7= X= - +7 (Transposing 7 to RH.S.) x x= x= X= Thus, the required number is. Solve the following: x -=- (Multiplying both sides by.) (Dividing both sides by.) (a) The teachen tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest sce is 87. What is the lowest sce? (b)' In an isosceles triangle, the base angles are equal. The vertex angle is 0. What are the base angles of the triangle? (Remember, the sum of theee angles of a triangle is 180 ).

88 MATHEMATiCS-VII Sol. (c) Sachin sced twice as many runs as Rahul. Together, their runs fell two sht of a double century. How many runs did each one sce? (a) Let the lowest marks be y.

13 88 MATHEMATiCS-VII Sol. (c) Sachin sced twice as many runs as Rahul. Together, their runs fell two sht of a double century. How many runs did each one sce? (a) Let the lowest marks be y. Given, twice the lowest marks = y and add 7 = y + 7 Accding to the given conditions, y + 7 = 87 y = 87-7.(Transposing 7 to RH.S.) y = (Dividing both sides by.) r= - y = 0 A Thus, the required lowest marks is 0. (b) Let the base angle of the triangle be b. Given, a = 0, b = c in isosceles triangle. We know that, a + b + c = 180 (By angle sum property 0 + b + b = 180 of a triangle.) 0 + b = 180 b = (Transposing 0 to RH.S.) b = b=- (Dividing both sides by.) b =70 Thus, the base angles of the isosceles triangle is 70 each. (c) Let the sce of Rahul be x runs and Sachin's sces x. Given, two sht of a double century = 198. Accding to the given conditions, x+x= 198 x= X=- (Dividing both sides by.) x = 66 Thus, Rahul's sce = 66 runs and Sachin's sce = x 66 = 1 runs. SIMPLE EQUATIONS Q. Solve the following: (i) Irfan says that he has 7 marbles me than five times the marbles Parmit has. Irfan has 7 marbles. How many marbles does Parmit have? (ii) Laxmi's father is 9 years old. He is years older than three times Laxmi's age. What is Laxmi's age? (iii) People of Sundargram planted a total of 10 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two me than three times the number of fruit trees. What was the number of fruit trees planted? Sol. (i) Let the number of marbles Parmit has be m. Accding to the given conditions, m + 7 = 7 ' m = 7-7 (Transposing 7 to RH.S.) m = 0 0 m= (Dividing both sides by.) m = 6 Thus, Parmit has 6 marbles. (ii) Let the age of Laxmi be y years. Given, her father years older than three times Laxmi's age = y + Accding to the given conditions, y + = 9 y = 9 - (Transposing to RH.S.) y = y = (Dividing both sides by.) y = 1 Thus, the age of the Laxmi is 1 years. (iii) Let the number of fruit trees be t. The number of non-fruits tree be t + Accding to the given conditions, t + t + = 10 t + = 10 89

90 MATHEMATICS-VII t = 10 - (Transposing to R.H.S.) t = 100 t = 10 (Dividing both sides by.) t = Thus, the number of fruit trees are. Q. Solve the following riddle: I am a number, Tell my identity!

14 90 MATHEMATICS-VII t = 10 - (Transposing to R.H.S.) t = 100 t = 10 (Dividing both sides by.) t = Thus, the number of fruit trees are. Q. Solve the following riddle: I am a number, Tell my identity! Take me seven times over, And add a fifty! To reach a triple century, You still need fty! Sol. Let the number be n. Given, seven time a number and add a fifty = 7n + 0. Then need fty to reach triple century i.e., 00. Accding to the given conditions, 7n + 0) + 0 = 00 7n + 90 = 00 7n = (Transposing 90 to R.H.S.) 7n = 10 n = 0 (Dividing both sides by 7.) Of n = 0 Thus, the required number is 0. OD

Simple Equations. Chapter A MIND-READING GAME! 4.2 SETTING UP OF AN EQUATION

Simple Equations. Chapter A MIND-READING GAME! 4.2 SETTING UP OF AN EQUATION SIMPLE EQUATIONS 77 Simple Equations Chapter 4 4.1 A MIND-READING GAME! The teacher has said that she would be starting a new chapter in mathematics and it is going to be simple equations. Appu, Sarita