Page 1 :
MATHEMATICS IN REAL LIFE–8, Chapter 8 : Linear Equations in One Variable, EXERCISE 8.1, 1. x – 3 = 8, , , , 7., x =8+ 3, , (By transposing – 3), , , , x=8, , , , 20, 5x, =, 5, 5, , , , x=4, , , , 2x, 4, =, 2, 2, , , , x=2, , (By transposing 6), , (On dividing both sides by 2), , 5. 7x – 5 = 16, , 7x = 16 + 5, , 7x = 21, , , 7x, 21, =, 7, 7, , , , x=3, , (By transposing – 5), , (On dividing both sides by 7), , 3x, = 18, 2, 3x, × 2 = 18 × 2, 2, 3x = 36, 3x, 36, =, (On dividing both sides by 3), 3, 3, , x, ×5=7×5, 5, (On multiplying both sides by 5), x = 35, , 4, 20, +x=, 7, 7, , , , x=, , 4, 20, 4, –, (By transposing ), 7, 7, 7, , , , x=, , 20  4, 7, , , , x=, , 16, 7, , 9. 7 = 2x – 5, or, 2x – 5 = 7, , 2x = 7 + 5, , 2x = 12, , , 2x, 12, =, 2, 2, , , , x=6, , (By transposing – 5), , (On dividing both sides by 2), , 10. 1.6x + 2 = 5.2, , 1.6x = 5.2 – 2, , 1.6x = 3.2, , , , Multiplying both sides by 2, we get, , , , , 8., , 4. 2x + 6 = 10, , 2x = 10 – 6, , 2x = 4, , , , , , (By transposing 7), , 3. 5x = 20, Dividing both sides by 5, we get, , 6., , x, =7, 5, , x = 11, , 2. x + 7 = 15, , x = 15 – 7, , ANSWER KEYS, , 1.6 x, 3.2, =, 1.6, 1.6, , (By transposing 2), , (On dividing both sides by 1.6), , x=2, , 11. 13 + 2a = 39, , 2a = 39 – 13, , 2a = 26, , , 2a, 26, =, 2, 2, , , , a = 13, , (By transposing 13), , (On dividing both sides by 2), , x = 12, Mathematics In Real Life-8, , 1

Page 2 :
12. 16a – 4 = 28, , 16a = 28 + 4, , 16a = 32, , , 16a, 32, =, 16, 16, , , , a=2, , m, –m= –8–6, 2, , (By transposing – 4), , m, = – 14, 2, , , (On dividing both sides by 16), ,  m, (– 2) , = (– 14) × (– 2),  2 , , , , (On multiplying both sides by – 2), , EXERCISE 8.2, 1. 2x – 5 = 3x + 5, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, 2x – 3x = 5 + 5, , – x = 10, , , , , 10x, 5, =, (On dividing both sides by 10), 10, 10, , x=, , 1, 2, , 3, 4, =m+, 5, 5, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, , 2m – m =, m=, , 4, 3, +, 5, 5, 7, 5, , , , , 5., , (On dividing both sides by 5), , m =  4 (On multiplying both sides by – 1), , m, +6=m–8, 2, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, , 2, , x, =4–x, 3, , –, , , , , , , , 7. 3t +, or, , x, +x=4–7, 3, ,  x  3x, =–3, 3, 2x, =–3, 3, , 2x, × 3 = – 3 × 3 (On multiplying both sides by 3), 3, 2x = – 9, , 9, 2x, =, 2, 2, x=, , (On dividing both sides by 2), , 9, 2, , 7, 2, =, + 5t, 5, 5, , 2, 7, + 5t = 3t +, 5, 5, , Transposing variables on L.H.S. and constant terms, on R.H.S., we get, , 4. 3m + 4 = 8m + 24, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, 3m – 8m = 24 – 4, , – 5m = 20, 5 m, 20, =, 5, 5, –m =4, , m = 28, , Transposing variables on L.H.S. and constant terms, on R.H.S., we get, , , , 3. 2m –, , , , 6. 7 –, , x =  10 (On multiplying both sides by – 1), , 2. 9x + 15 = 20 – x, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, 9x + x = 20 – 15, , 10x = 5, , , , , 5t – 3t =, , , , , , 7, 2, –, 5, 5, , 5, 5, 2t = 1, , 2t =, , 2t, 1, =, (On dividing both sides by 2), 2, 2, t=, , 1, 2, , 8. 3.5x – 9 = 2.4x + 13, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, , Answer Keys

Page 3 :
3.5x – 2.4x = 13 + 9, , 1.1x = 22, , 2., , 22, 1.1x, =, (On dividing both sides by 1.1), 1.1, 1.1, , , , ,  3(3x + 5),  9x + 15,  9x + 10x, , 19x, , x = 20, , 9. 11x –, , 4, 5, = 2x +, 3, 3, , Transposing variables on L.H.S. and constant terms, on R.H.S., we get, 11x – 2x =, , 5, 4, +, 3, 3, , , , 3, 9x, =, 9, 9, , , , x=, , (On dividing both sides by 9), , 3., , x=, , , , x=0, , 13, 2, =, – 5x, 14, 7, Transposing variables on L.H.S. and constant terms, on R.H.S., we get, 12x + 5x =, 17x =, , 13, 2, +, 14, 7, , 17, 14, , x=, , 2x  3, 9, =, 3x  2, 7, , , , 13 x, 3, =, 13, 13, , , , x=, , , , (On dividing both sides by 13), , 3, 13, , =, , 4x – 1 = 5x + 2, – 1 – 2 = 5x – 4x, –3= x, , (By cross multiplication), (By transposing), , 4., , =, Mathematics In Real Life-8, ,  13, = 1 = R.H.S.,  13, , 45, 9, =, = R.H.S., 35, 7, , (3x  4)  2 x, 9, =–, (2  5x)  7 x, 58, , , 3x  4  2 x, 9, =–, 2  5x  7 x, 58, , , , x4, 9, =–, 2  12 x, 58, , x=3, , 4(  3)  1,  12  1, Verification: L.H.S. =, =, 5(  3)  2,  15  2, , (By transposing), , 6  39, 45, 13, =, = 13, 35, 9  26, 13, 13, , 1, 14, , 4x  1, =1, 5x  2, , , , , (By cross multiplication), , 6, 3, Verification: L.H.S. =, = 13, 9, 3, 2, 3   2, 13,  13 , , EXERCISE 8.3, 1., , 5, = R.H.S., 3, ,  3, 2   3,  13 , , 17, 17 x, =, 14  17, 17, (On dividing both sides by 17), , , , (By transposing), , 3(0)  5, 05, =, 3  2(0), 30, ,  7(2x + 3) = 9(3x + 2),  14x + 21 = 27x + 18,  21 – 18 = 27x – 14x, , 13x = 3, , 3 = 13x, , 1, 3, , (By cross multiplication), , 0, =0, 19, , , , =, , 10. 12x –, , , , 5(3 – 2x), 15 – 10x, 15 – 15, 0, , 9x =, , , , , , =, =, =, =, , Verification: L.H.S. =, , 9, 3, 9x = 3, , , , x  5, 5, =, 3  2x, 3, , , , , 58(x + 4) = –9(2 – 12x), (By cross multiplication), 58x + 232 = – 18 + 108x, 3

Page 4 :
, , , , 58x – 108x = – 18 – 232, (By transposing), – 50x = – 250,  50 x,  250, =,  50,  50, (On dividing both sides by –50), , , , (3  5  4)  2 × 5, (2  5 × 5)  7 × 5, , =, , (15  4)  10, 19  10, =, (2  25)  35,  23  35, , =, , 9, 9, =–, = R.H.S.,  58, 58, , , , , 3m – 3 + 30 = 4m + 2, – 3 + 30 – 2 = 4m – 3m, (By transposing), 25 = m, , , , m = 25, , 1, 1, (25 – 1) + 5 =, × 24 + 5, 2, 2, = 12 + 5 = 17, , Verification: L.H.S. =, , R.H.S. =, , 1, × 51 = 17, 3, L.H.S. = R.H.S., , 5x, =–5, 3x  4, 5x, 5x,  5x + 15x, , 20x, , , , = – 5(3x + 4), = – 15x – 20, = – 20, = – 20, , (By cross multiplication), (By transposing), ,  20, 20x, =, (On dividing both sides by 20), 20, 20, x=1, , Verification: L.H.S. =, , 5 × (  1), 5, =, 3 × (  1)  4, 34, , 5, =, = – 5 = R.H.S., 1, 4, , y=, , 3, 11, , (By cross multiplication), (By transposing), , (On dividing both sides by 11),  3, 2,  11 , , 6, 11, Verification: L.H.S. =, =, 9,  3, 1, 3, 1, 11,  11 , 6, 6, 11, =, = 11,  9  11, 2, 11, 11, , =, 8., , 6, = – 3 = R.H.S., 2, , 1.2 x  3, 9, =, 3.2 x  6, 8, , , , , , , 8(1.2x + 3) = 9(3.2x + 6), (By cross multiplication), 9.6x + 24 = 28.8x + 54, 24 – 54 = 28.8x – 9.6x, (By transposing), – 30 = 19.2x, x=, , 1, 1, (2 × 25 + 1) = (50 + 1), 3, 3, , =, , , , 2y = – 3(3y + 1), 2y = – 9y – 3, 2y + 9y = – 3, 11y = – 3, , 1, 1, (m – 1) + 6 × 5 = 6 × (2m + 1), 2, 3, 3(m – 1) + 30 = 2(2m + 1), , , , 6., , , , 1, 1, (m – 1) + 5 =, (2m + 1), 3, 2, Multiplying both sides by 6, the L.C.M. of 2 and 3,, we get, , 6×, , 2y, =–3, 3y  1, , , , , , x=5, , Verification: L.H.S. =, , 5., , 7., , 30, 19.2, , (On dividing both sides by 19.2), , , x=, ,  300, 192, , , , x=, ,  25, 16, ,   25 , 1.2 × , 3,  16 , ,  30, 3, Verification: L.H.S. =, = 16,  80,   25 , 6, 3.2 × , 6, 16,  16 ,  30  48, 18, 16, =, = 16, 16,  80  96, 16, 16, , =, , 18, 9, =, = R.H.S., 16, 8, Answer Keys

Page 5 :
9., , 4y  8, 5, =, 5y  8, 6, , 12., ,  6(4y + 8) = 5(5y + 8),  24y + 48 = 25y + 40,  48 – 40 = 25y – 24y, , 8 =y, , , 10., , (By cross multiplication), , =, , 40, 5, =, = R.H.S., 48, 6, , xb, xb, =, ab, ab,  (x + b) (a + b) = (a – b) (x – b), (By cross multiplication), 2, , ax + bx + ab + b = ax – ab – bx + b2, , , , , , 34  17 x 5x  60, , =8, 1  7x, 7x  1, , , , 34  17 x 5x  60, , =8, 1  7x, 1  7x, , , , 34  17 x  (5 x  60), =8, (1  7 x), , , , 34  17 x  5x  60, =8, (1  7 x), , , , 22 x  26, =8, 1  7x, , y=8, 4×88, 32  8, =, 5×8  8, 40  8, , , , , , (By transposing), , Verification: L.H.S. =, , , , 17(2  x) 5( x  12), , =8, 1  7x, 7x  1, , , , ax + bx – ax + bx = – ab + b2 – ab – b2, (By transposing), 2bx = – 2ab, 2ab, x=, 2b, (On dividing both sides by 2b), , – 22x – 26 = 8(1 – 7x), (By cross multiplication), – 22x – 26 = 8 – 56x, – 22x + 56x = 8 + 26, (By transposing), 34x = 34, , , , , , , x=a, , x=1, , (On dividing both sides by 34), ab,  ( a  b), Verification: L.H.S. =, =, =–1, ab, ab, , R.H.S. =, 11., , Verification: L.H.S. =, , ab,  ( a  b), =, =–1, ab, ab, , 17(2  1) 5(1  12), , 1 7 ×1, 7 ×1  1, , 7x  4, 4, =–, x2, 3, , =, , 17 × 1 5  13, 17 65, , =, , 17, 7 1, 6 6, ,  3(7x + 4) = –,  21x + 12 = –,  21x + 4x = –, , 25x = –, , =, ,  17  65, 48, =, = 8 = R.H.S., 6, 6, , , , , 4(x + 2) (By cross multiplication), 4x – 8, 8 – 12, (By transposing), 20, 20, x =–, (On dividing both sides by 25), 25, x=, , 4, 5, ,  4,  28, 7, 4, 4,  5 , Verification: L.H.S. =, = 5, 4, 4, 2, 2, 5, 5,  28  20, 8, 5, =, = 5,  4  10, 6, 5, 5, 8, 4, =, =, = R.H.S., 6, 3, Mathematics In Real Life-8, , 13., , 4x  3, 0.05, =, 0.25, 10 x  9, , , , 4x  3, 1, 5, =, =, 5, 10 x  9, 25, ,  5(4x – 3),  20x – 15,  20x – 10x, , 10x, , =, =, =, =, , , , x=, , , , x=, , 1(10x – 9), 10x – 9, – 9 + 15, 6, , (By cross multiplication), (By transposing), , 6 3, =, 10 5, (On dividing both sides by 10), , 3, 5, , 5

Page 6 :
, , 3, 4   3, 5, , 12, 3, Verification: L.H.S. =, = 5, 30, 3, 9, 10    9, 5, 5, , 20x – 30x = – 10, (By transposing), , , , – 10x = – 10, , , , x=1, , (On dividing both sides by – 10), , 12  15, 12, 3, 3, 5, = 5, =, = 5, 69, 3, 3, , Verification: L.H.S. =, , 3, 1, =, 5(  3), 5, , =, , 0.05, 5, 1, =, =, 0.25, 25, 5, L.H.S. = R.H.S., , R.H.S. =, , 14., , 16., , 3, 1, =, 2 x  (3  4 x), 3, , , , 3, 1, =, 6x  3, 3, , , , , , , , , 3×3, 9, 9+3, 12, 6x, , , , = 6x – 3 (By cross multiplication), = 6x – 3, = 6x, (By transposing), = 6x, = 12, , , , , , 6, , 20, = 10 = R.H.S., 2, , 2, x 2 – 11x + 30 = x – 7x + 12, , , , (Cancelling x2 from both sides), – 11x + 30 = 7x + 12, , , , , – 11x + 7x = 12 – 30, – 4x = – 18, , , , x=, , , , 3, 3, =, 2 × 2  (3  4 × 2), 4  (3  8), , (By transposing), ,  18, 9, =, 4, 2, , =, , 3, 3, =, 4  (  5), 45, , =, , 3, 1, =, = R.H.S., 9, 3, , 5x  30  30  15x, = 10, 3x  1, 20 x, = 10, 3x  1, , 20x = 10(3x – 1), (By cross multiplication), 20x = 30x – 10, , x=, , 9, 2, , 9  10, 1, 9, , 5, 2 =– 1, 2, 2, Verification: L.H.S. =, =, =, 3, 9, 96, 3, 3, 2, 2, 2, 98, 9, 1, 4, 1, 2, R.H.S. = 2, =, = 2 = , 9, 9  12, 3, 3, 6, , 2, 2, 2, , 5( x  6)  15(2  x), = 10, 3x  1, , , =, , (On dividing both sides by – 4), (On dividing both sides by 6), , 15., , 5 × 7  15 × 1, 35  15, =, 31, 2, , (x – 5)(x – 6) = (x – 3)(x – 4), (By cross multiplication), , x=2, , Verification: L.H.S. =, , =, , x5, x4, =, x3, x6, , , 1, 3, =, 3, 2x  3  4x, , 5(1  6)  15(2  1), 3×11, , L.H.S. = R.H.S., 17., , 2, 7, =, x9, 4x  3, , , , , , , 2(4x – 3) = – 7(x + 9), (By cross multiplication), 8x – 6 = – 7x – 63, 8x + 7x = – 63 + 6, (By transposing), 15x = – 57, , Answer Keys

Page 7 :
, , x=, , , , x=, ,  57,  19, =, 15, 5, (On dividing both sides by 15), ,  19, 5, , 2, 2, 2, =, =, 26,  19  45,  19, 9, 5, 5, 5, , Verification: L.H.S. =, , 10, 5, =, 26, 13, , =, , 7, , R.H.S. =, ,   19 , 4, 3,  5 , , =, , , , – 6x = 3, , , , x=, , , , x=, , 3, 2, Verification: L.H.S. =, = 2, 5,   1, 1, 5, 1, , 2,  2 , , 7,  76, 3, 5, , 34, 7, , 2 =1, 2, =, =, 52, 7, , 2, 2, ,   1, 3, 1,  2 , , 3, 1, R.H.S. =, = 2, 5,   1, 2, 5, 2, , 2,  2 , , 7×5, 5, =, 13, 91, L.H.S. = R.H.S., , =, , 18., , 2  3y, 4, =, 1  5y, 3,  3(2 + 3y), , 6 + 9y,  9y – 20y, , – 11y, , , =, =, =, =, , y=, , 4(1 + 5y), 4 + 20y, 4–6, –2, 2, 11, , L.H.S. = R.H.S., , (By transposing), , EXERCISE 8.4, , (On dividing both sides by – 11), , 22  6, 6, 11, 11, Verification: L.H.S. =, =, =, 10, 11  10, 2, 1, 1  5 , 11,  11 , 11, , 1. Let the number to be added be x. Therefore,, 3, 7, 3× ,  +x=,  4 , 7, , 2, , , ,  21, 3, +x=, 4, 7, , , , x=, , 3, 21, 21, +, (By transposing, ), 7, 4, 4, , , , x=, , 12  147, 28, , 3x  2, 3x  1, =, 5x  1, 5x  2, , , , x=, , 159, 28, , , , , Hence, the required number is, , 28, 28, 4, = 11 =, =, = R.H.S., 21, 21, 3, 11, , 19., , 32, 1, , 2 =1, 2, =, =, 54, 1, , 2, 2, , (By cross multiplication), , 2, 2  3 ,  11 , , 1, 2,   1, 3, 2,  2 , , 7, 7, =,  76  15,  91, 5, 5, , =, , 3, 1, =, 6, 2, (On dividing both sides by – 6), , , , , , (3x – 2)(5x + 2) = (5x – 1) (3x + 1), 15x2 + 6x – 10x – 4 = 15x2 + 5x – 3x – 1, 15x2 – 4x – 4 = 15x 2 + 2x – 1, , (Cancelling 15x2 from both sides), – 4x – 4 = 2x – 1, – 4x – 2x = – 1 + 4 (By transposing), , Mathematics In Real Life-8, , 159, ., 28, , 2. The lengths of sides of a triangle are 2x, 2x + 3 and, 2x + 5., Perimeter of triangle = Sum of the lengths of sides, , 56 = 2x + 2x + 3 + 2x + 5, , 6x + 8 = 56, 7

Page 8 :
, , , 6x = 56 – 8, 6x = 48, , (By transposing), , 48, =8, 6, x =8, Hence, lengths of sides of triangle are 2 × 8 = 16,, 2 × 8 + 3 = 19 and 2 × 8 + 5 = 21., , , , x=, , 3. Let the required number be x. Then according to the, given condition,, 8x – 3x = 35, , 5x = 35, , , 35, =7, 5, x =7, , x=, , , Hence, the required number is 7., 4. Let the two numbers be 4x and 7x., , 7x – 4x = 9, , 3x = 9, , x =3, Hence, the required numbers are 4 × 3 = 12, 7 × 3 = 21., 5. Let the required number be x., Therefore,, according to condition,, x x x,  , = 169, 3 4 2, , , , 4 x  3x  6 x, = 169, 12, ( L.C.M. of 3, 4 and 2 is 12), , 13x, = 169, 12, , 13x = 169 × 12, 169  12, , x=, 13, , x = 13 × 12, , x = 156, The required number is 156., 6. Let the number of boys in the class be x., , 8, , , , , 5x  3x, = 96, 5, , , , 8x, = 96, 5, , 96  5, 8, x = 12 × 5 = 60, , x=, , , Hence, number of boys = 60, , 3 × 60, = 36., 5, 7. Let the three consecutive integers be x, (x + 1) and, (x + 2). Therefore, according to condition,, , Number of girls =, , x + (x + 1) + (x + 2) = 48, , 3x + 3 = 48, , 3x = 48 – 3, , 3x = 45, , , , (By transposing), , 45, = 15, 3, x + 1 = 15 + 1 = 16, x + 2 = 15 + 2 = 17, , x=, , Hence, the required three consecutive integers are, 15, 16 and 17., 8. The ratio of ages of A and B = 5 : 7, Let the ages of A and B are 5x and 7x respectively., After 4 years, age of A = (5x + 4) years, and age of B = (7x + 4) years, According to the condition., (5x + 4) + (7x + 4) = 56, , 12x + 8 = 56, , 12x = 56 – 8, (By transposing), , 12x = 48, , , x=, , 9. Let the required number be x., Then, according to the given condition,, , , , 3x, 5, , Total number of students in the class = x +, 3x, x+, = 96, 5, , , , 8x = 96 × 5, , 48, =4, 12, Hence, present age of A = 5 × 4 = 20 years, present age of B = 7 × 4 = 28 years, , , , Therefore, number of girls in the class =, , , , (By transposing), , 21, =3, 7, Hence, the required number is 3., , , 3x, 5, , 7x + 7 = 28, 7x = 28 – 7, 7x = 21, x=, , 10. Let the present age of Ashwin be x years., After 20 years, Ashwin’s age = (x + 20) years., Now, as per condition,, (x + 20) = 3x, , 3x – x = 20, (By transposing), , 2x = 20, , Answer Keys

Page 9 :
20, = 10, 2, Hence, present age of Ashwin = 10 years., 11. Let the present age of grandson be x years., Then grandfather’s age = 8x years., According to the condition,, 8x = x + 63, , 8x – x = 63, (By transposing), , 7x = 63, , , x=, , 63, =9, 7, Hence, age of grandson = 9 years, Age of grandfather 8 × 9 = 72 years, , , , x=, , 12. Let number of `1 coins in piggy bank be x. Therefore, Number of 50 p coins = 3x,  Total amount = `1 × x + 50 paise × 3x, = x rupees + 150x paise, = 100x paise + 150x paise, = 250x paise, According to the condition,, 250x paise = `45, , 250x paise = 4500 paise, , 250x = 4500, , , x=, , 4500, = 18, 250, , Hence, number of `1 coins = 18 and number of 50, paise coins 3 × 18 = 54., 13. The ratio of present ages of Aman and Naman = 4 : 5., Therefore,, Let Aman’s present age = 4x years and, Naman’s present age = 5x years., After 8 years, Aman’s age = (4x + 8) years, Naman’s age = (5x + 8) years, According to the question,, , 4x  8, 5, =, 6, 5x  8, , 6(4x + 8) = 5(5x + 8), (By cross multiplication), , 24x + 48 = 25x + 40, , 24x – 25x = 40 – 48, (By transposing), , –x =– 8, , x =8, Hence, Aman’s present age = 4 × 8 = 32 years, and Naman’s present age = 5 × 8 = 40 years, 14. Let the ages of Mohan and Sohan be 5x and 7x., If Mohan is 9 years older, then his age = 5x + 9, If Sohan is 9 years younger, then his age = 7x – 9, Mathematics In Real Life-8, , According to condition,, , 5x  9, 2, =, 7x  9, 1, , , 5x + 9 = 2(7x – 9), (By cross multiplication), , , , 5x + 9 = 14x – 18, , , , 5x – 14x = – 18 – 9, , , , – 9x = – 27, , , , x =3, , (By transposing), , Hence, Mohan’s present age = 5 × 3 = 15 years, Sohan’s present age = 7 × 3 = 21 years., 15. Let son’s age be x years., Then, father’s age = 3x years., After 15 years,, Son’s age = (x + 15) years, Father’s age = (3x + 15) years, According to condition,, (3x + 15) = 2(x + 15), , , 3x + 15 = 2x + 30, , , , 3x – 2x = 30 – 15, , , , x = 15, , (By transposing), , Hence, Son’s age = 15 years, Father’s age = 3 × 15 = 45 years., 16. Let the digit at ones place be x., Then, digit at tens place = 3x, The two digit number formed by these digits, = (3x × 10) + x = 31x, And, number formed by reversing the digits, = 10 × x + 3x = 13x, According to the question,, 31x + 13x = 88, , , 44x = 88, , , , x=, , 88, =2, 44, , Hence, the required number = 31 × 2 = 62., 17. Let the digit at units place be x., Then, digit at tens place = 2x, Therefore, the number = 10 × 2x + x, = 20x + x, = 21x, The number formed by reversing the digits, = 10 × x + 2x, = 12x, 9

Page 10 :
According to the question,, 21x – 18 = 12x, , 21x – 12x = 18, , 9x = 18, , , x=, , (By transposing), , 18, =2, 9, , x4,  The fraction =, x, , According to the condition,, , ( x  4)  3, 1, =, x5, 4, , , , x7, 1, =, x5, 4, , 3x – 2x = 100, , , , x = 100, , Then, length = x + 4,  Area of rectangle = length × breadth, = (x + 4) × x, = (x2 + 4x) sq. cm., If length and breadth are increased by 3 cm each,, then,, new length of rectangle = x + 4 + 3 = (x + 7) cm, new breadth of rectangle = (x + 3) cm., New area of rectangle = (x + 7) × (x + 3) sq. cm, = (x2 + 10x + 21) sq. cm., According to given condition,, , , , x 2 + 10x + 21 = x 2 + 4x + 81, , (Cancelling x2 from both sides), , , 10x + 21 = 4x + 81, , , , 10x – 4x = 81 – 21, , , , 6x = 60, , , , x = 10, , length of rectangle = 10 + 4 = 14 cm., (By cross multiplication), , Hence, the required fraction =, , 22. Let the speed of steamer in still water be x km/hr., Speed of steam = 2 km/hr.,  Speed of steamer during downstream = (x + 2) km/hr., Speed of steamer during upstream = (x – 2) km/hr., , (By transposing), , Now, the distance covered by the steamer during, downstream in 4 hours = 4 × (x + 2) km., The distance covered by the steamer during, upstream in 5 hours = 5 × (x – 2) km., distance, ), time,  Distance covered in upstream = distance covered, in downstream, , 5 (x – 2) = 4(x + 2), , ( speed =, , 7, ., 11, , 20. Let one number be x., Then, other number = x + 50., Therefore, as per given condition,, , 10, , , , Hence, breadth of rectangle = 10 cm, , , 4(x – 7) = x + 5, , 4x – 28 = x + 5, , 4x – x = 5 + 28, , 3x = 33, , x = 11, Nemerator = 11 – 4 = 7, Denominator = 11, , 2, x, =, 3, x  50, , 3x = 2x + 100, , (x2 + 10x + 21) = (x2 + 4x) + 81, , Hence, the required number is 31 × 2 = 62., 19. Let the denominator of the fraction be x., Therefore, numerator of the fraction = x – 4, , x43, 1, =, x5, 4, , , , 21. Let the breadth of a rectangle be x., ,  36, x=, =2,  18, , , , 3x = 2(x + 50) (By cross multiplication), , Hence, the required numbers are 100 and 150., , Hence, the required number = 21 × 2 = 42., 18. Let the digit at units place be x., Then, digit at tens place = 3x, Therefore, the number = 10 × 3x + x, = 30x + x = 31x, The number formed by reversing the digits, = 10 × x + 3x, = 13x, According to the question,, 13x = 31x – 36, , 13x – 31x = – 36, (By transposing terms), , – 18x = – 36, , , , , , , , , 5x – 10 = 4x + 8, 5x – 4x = 8 + 10 (By transposing terms), x = 18, , Hence speed of steamer in still water = 18 km/hr., Answer Keys

Page 11 :
, , MULTIPLE CHOICE QUESTION, 1., , , , 3x  5, 1, =, 2x  1, 3, , , , , , 3(3x + 5) = 1(2x + 1), (By cross multiplication), 9x + 15 = 2x + 1, 9x – 2x = 1 – 15, 7x = – 14, , , , 2., , x =– 2, , (By cross multiplication), , x =5+3, , (By transposing – 3), , , x =8, Hence, option (c) is correct., 7., , x3, 5, =, x3, 6, , , 6(x – 3) = 5(x + 3), (By cross multiplication), , , , 6x – 18 = 5x + 15, , Hence, option (a) is correct., , , , 6x – 5x = 15 + 18 (By transposing terms), , 2, 7, x+1=, 5, 5, , , , , , , 2, 2, x=, 5, 5, , , , x =1, , , , (By transposing – 1), , x2 + 15 = (x + 1)2 – 6, , , x =3, ,  The required number is 10., Hence, option (b) is correct., 5. Let the number be x. Therefore,, 5x – 7 = 2x + 8, , 5x – 2x = 8 + 7 (By transposing terms), , 3x = 15, , x =5, , 15 = 2x – 5, 2x = 15 + 5 (By transposing terms), , , , , 2x = 20, x = 10, , Hence, option (b) is correct., 9. Let the number be x. Therefore,, 35 – x = x – 27, x + x = 35 + 27, (By transposing terms), , , 2x = 62, , , , x = 31, , The number is 31., Hence, option (a) is correct., 10. Let present age be x years. Then age after 14 years, = (x + 14) years, Age four years before = (x – 4) years., Therefore,, 3(x – 4) = x + 14, , ,  The required number is 5., Hence, option (b) is correct., , Mathematics In Real Life-8, , , , x 2 + 15 = x 2 + 2x + 1 – 6, (Cancelling x2 from both sides), , The number is 10., , x = 10, , x3, –2=–1, 5, x3, , =– 1+ 2, 5, x3, , =1, 5, , 8. Let the number be x. Its successor = (x + 1)., Therefore,, ,  The two numbers are 6 and 15., Hence, option (c) is correct., 4. Let the number be x. Therefore,, 4x + 8 = 5x – 2,  4x – 5x = – 2 – 8, (By transposing terms), , – x = – 10, , , x = 33, , Hence, option (a) is correct., 2, 7, x=, –1, 5, 5, , Hence, option (c) is correct., 3. Let the two numbers be 2x and 5x. Therefore,, 2x + 5x = 21, , 7x = 21, , 6., , x–3=5, , , , 3x – 12 = x + 14, 3x – x = 14 + 12, (By transposing terms), , , (Transposing – 2), , 2x = 26, , 26, = 13 years., 2, Hence, option (c) is correct., , , x=, , 11

Page 12 :
11., , x x x,   = 14, 2 3 4, , , , , , 6x  4x  3x, = 14, 12, , ( L.C.M. of 2, 3 and 4 is 12), 7x, = 14, 12, , , , 7x = 168, , , , x=, , 168, = 24, 7, , Hence option (d) is correct., 12., , x8, x3, =, 3, 2, , , , 2(x – 8) = 3(x – 3), (By cross multiplication), , 2x – 16 = 3x – 9, , 2x – 3x = – 9 + 16, (By transposing terms), , –x =7, , x =– 7, Hence, option (b) is correct., , MENTAL MATHS CORNER, Fill in the blanks:, 1. The equation 5x + 8 = 0 is an example of linear, equation., 2. The value of the variable for which an equation is, true is called the solution or root of the equation., 3. A number which when added to three times of itself, gives 20 is 5.,  Let the number be x. Therefore,, x + 3x = 20, , 4x = 20, , , 20, x=, =5, 4, x =5, , The number is 5., 4. In a linear equation, the degree of variable is 1., 5. A number when multiplied by 4 exceeds itself by 24., The number is 8.,  Let the number be x. Therefore,, 4x – x = 24, , 3x = 24, , 12, , x=, , 24, =8, 3, , The number is 8., 6. 0.4x + 0.5 = 0.3x + 0.6, then x is 1.,  0.4x + 0.5 = 0.3x + 0.6, , 0.4x – 0.3x = 0.6 – 0.5, (By transposing terms), , 0.1x = 0.1, , x =1, 7. The sum of two numbers which are in the ratio 5 : 7, is 120. Then the numbers are 50 and 70., Let the numbers be 5x and 7x., , 5x + 7x = 120, , 12x = 120, , , x=, , 120, 12, , x = 10,  The numbers are 50 and 70., 8. If the value of x is 5, then 3x + 5 = 20., , 3 × 5 + y = 20, , y = 20 – 15 = 5, 9. Two consecutive natural numbers whose sum is 55, are 27 and 28., Let two consecutive numbers be x and x + 1.,  x + (x + 1) = 55, , 2x + 1 = 55, , 2x = 55 – 1, , 2x = 54, , x = 27, , x + 1 = 27 + 1 = 28, 10. The general form of linear equation is ax + b = c,, where, a  0., , REVIEW EXERCISE, 1. Let the first number be x., Then the other number = 43 – x., Therefore,, x – (43 – x) = 13, , x – 43 + x = 13, , 2x – 43 = 13, , 2x = 13 + 43, , 2x = 56, 56, = 28, 2, Hence the required numbers are 28 and 43 – 28 = 15., , , , x=, , Answer Keys

Page 13 :
2. Let the present ages of husband and wife be 4x, and 3x., Thirty years later,, Husband’s age = (4x + 30) years, Wife’s age = (3x + 30) years, As per condition,, (4x + 30) = (3x + 30) + 10, , 4x + 30 = 3x + 40, , 4x – 3x = 40 – 30 (By transposing terms), , x = 10,  Husband’s present age = 4 × 10 = 40 years, Wife’s present age = 3 × 10 = 30 years, 3. Let the three consecutive multiples of 3 be 3x,, (3x + 3), (3x + 6). Then,, 3x + (3x + 3) + (3x + 6) = 333, , 9x + 9 = 333, , 9x = 333 – 9, , 9x = 324, , , 324, x=, = 36, 9, , , , x = 36, , 5x  4 5x, , 2, 4, 54  4 54, , =, 2, 4, 20  4 20, , =, 2, 4, 16, 5, =, 2, , R.H.S. =, , = 8 + 5 = 13, Hence L.H.S. = R.H.S., 5., , 5x  4 5x, 3, , (x + 2) + 4 =, 2, 4, 2, Multiply both sides by 4, the L.C.M. of 2 and 4, we get, 5x  4, 3, 5x, 4×, (x + 2) + 4 × 4 = 4 ×, +4×, 4, 2, 2, , , , , , , To verify:, , 6(x + 2) + 16 = 2(5x – 4) + 5x, 6x + 12 + 16 = 10x – 8 + 5x, 6x – 10x – 5x = – 8 – 16 – 12, (By transposing), – 9x = – 36, x =4, L.H.S. =, , 3, (x + 2) + 4, 2, , =, , 3, (4 + 2) + 4, 2, , =, , 3, ×6+4, 2, , =3×3+4, = 9 + 4 = 13, , Mathematics In Real Life-8, , , , 7, 2 x  7  5x, =, 6, 9x  3  4x, , , , 7, 7x  7, =, 6, 5x  3, , , , The three multiples are 3 × 36 = 108, 3 × 36 + 3 = 111, and 3 × 36 + 6 = 114, 4., , 2 x  (7  5x), 7, =, 9 x  (3  4 x), 6, , 6., , 6(7x – 7) = 7(5x – 3), (By cross multiplication), , , , , 42x – 42 = 35x – 21, 42x – 35x = – 21 + 42, , , , 7x = 21, , (By transposing), , 21, 7, , , , x=, , , , x=3, , 3, 2, 1, , =, 5x 3x, 10, , , , , 9  10, 1, =, 15x, 10, 1, 1, =, 10, 15x, 15x = – 10, , , , x=, , 10, 15, , , , x=, , 2, 3, , (By cross multiplication), , 7. (a + 3) (a – 3) – a(a + 5) = 6,  (a2 – 32) – (a2 + 5a) = 6, [ (a + b) (a – b) = (a2 – b2)], , , , , a 2 – 9 – a 2 – 5a = 6, – 5a – 9 = 6, – 5a = 6 + 9, , (Transposing – 9), , 13

Page 14 :
, , , , , – 5a = 15, 15, =–3, 5, , , , a=–, , , , a=3, , 39x – 54 = 24, 39x = 24 + 54 (Transposing – 54), , , , 39x = 78, , , , x=, , , , x=2, , 78, 39, , 8. Let the number be x. Therefore,, , , 4, 2, x = x + 10, 5, 3, , , , 4x, 2x, –, = 10, 5, 3, , , , 12x  10x, = 10, 15, , , , a=, , 12, 18, , , , 2x, = 10, 15, , , , a=, , 2, 3, , , , (By transposing, , 2x, ), 3, , 2x = 150, 150, = 75, 2, , , , x=, , , , x = 75, , Hence, the required number is 75., 9. Let three consecutive integers be x, (x + 1) and, (x + 2). Then,, x + (x + 1) + (x + 2) = 54, , 3x + 3 = 54, , 3x = 54 – 3 (Transposing + 3), , 3x = 51, , x = 17, , x + 1 = 17 + 1 = 18, x + 2 = 17 + 2 = 19, Hence, required three consecutive integers are, 17, 18, 19., 10., , 3x  2 2x  3, 2, , =, –x, 3, 4, 3, 3x  2 2x  3, 2, , +x=, (By transposing –x), 3, 4, 3, 3(3x  2)  4(2 x  3)  12 x, 2, , =, 12, 3, ( L.C.M. of 4 and 3 is 12), , , , 2, 9 x  6  8x  12  12 x, =, 3, 12, 13x  18, 2, , =, 3, 12, , 3(13x – 18) = 2 × 12, (By cross multiplication), , , , 14, , 11. 15(a – 4) – 2(a – 9) + 5(a + 6) = 0,  15a – 60 – 2a + 18 + 5a + 30 = 0,  18a – 12 = 0, , 18a = 12, (By transposing – 12), , 12. Let the two numbers be 3x and 5x., Now, as per condition,, , 3x  10, 5, =, 5x  10, 7,  7(3x + 10),  21x + 70,  21x – 25x, , – 4x, , , =, =, =, =, , x=, , 5(5x + 10) (By cross multiplication), 25x + 50, 50 – 70, (By transposing terms), – 20, ,  20, =5, 4, , Hence, the required two numbers are 3 × 5 = 15 and, 5 × 5 = 25., , HOTS QUESTIONS, 1. Let three consecutive multiples of 9 be 9x, 9(x + 1), and 9(x + 2)., Therefore, 9x + 9(x + 1) + 9(x + 2) = 999,  9x + 9x + 9 + 9x + 18 = 999, , 27x + 27 = 999, , 27x = 999 – 27, (Transposing + 27), , 27x = 972, , , x=, , 972, = 36, 27, , , 9x = 9 × 36 = 324,  9(x + 1) = 9 × (36 + 1) = 333,  9(x + 2) = 9 × (36 + 2) = 342, Hence, three required consecutive multiples of 9 are, 324, 333 and 342., , Answer Keys