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1.) Convert the given statement into equation: An integer increased by 5 equals 20. |
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| Hint: ViewAssume the integer is x, if it is increased by 5 then, x + 5 = 20 |
2.) Determine the number, if the sum of two odd consecutive numbers is 56. |
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| Hint: ViewAccording to the given condition, we can write, x + x + 2 = 56, 2x + 2 = 56, 2x = 56 – 2, 2x = 54, x = 54/2, x = 27, which is the required odd number. Hence, the other odd number is: x + 2 = 27 + 2 = 29. Therefore, the two odd numbers are 27 and 29. |
3.) The ages of Rahul and Ramya are in the ratio 5: 7. After 4 years, the sum of their ages will be 56 years. Find their present ages. |
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| Hint: ViewAssume that the ages of Rahul and Ramya are 5p and 7p. After 4 years, the ages of Rahul and Ramya will be 5p + 4 and 7p + 4. According to the given condition, we get the following equation: (5p + 4) + (7p + 4) = 56, 12p + 8 = 56, 12p = 56 – 8, 12p = 48, p = 48/12, p = 4 Therefore, Rahul’s present age = 5(4) = 20 Ramya’s present age = 7(4) = 28. Hence, the present age of Rahul and Ramya are 20 and 28, respectively. |
4.) Four-fifths of a number is greater than three-fourths of a number by 8. Find the number. |
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| Hint: Viewlet the number be x. four-fifth of a number = (4/5)x, Three-fourth of a number = (3/4)x. the equation will be (4/5)x = (3/4)x + 8 => (4/5)x - (3/4)x = 8 => [(4 x 4)x - (3 x 5)x]/20 = 8 => (16x - 15x)/20 = 8, x/20 = 8 => x = 8 x 20 = 160. The number is 160 |
5.) Find the number, if 15 is added to three times of the number results in 45. |
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| Hint: ViewAcoording to the equation, 15 + 3x = 45, 3x = 45 - 15, x = 30/3 = 10, so, the number is 10 |
6.) The sum of two consecutive multiples of 4 is 60. Find the number. |
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| Hint: ViewThe two consecutive multiples of 4 is x & (x + 4), sum of the numbers are 60, x + (x + 4) = 60, 2x + 4 = 60, 2x = 60 - 4, 2x = 56, x = 56/2 = 28. so the numbers are 28 and 32. |
7.) (x+2)/(x−2) = 7/2 |
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| Hint: View(x+2)/(x−2) = 7/2 => 2(x + 2) = 7(x - 2) => 2x + 4 = 7x - 14 => 7x - 2x = 14 + 4 => 5x = 18 => x = 18/5 |
8.) (2x+5)/(3x+4) = 5. |
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| Hint: View(2x+5)/(3x+4) = 5 => 2x + 5 = 5(3x + 4) => 2x + 5 = 15x + 20 => 15x + 20 = 2x + 5 => 15x - 2x = 5 - 20 => 13x = -15 => x = -15/13. |
9.) 0.5x – (0.6 – 0.2x) = 0.2 – 0.3x |
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| Hint: View 0.5x – (0.6 – 0.2x) = 0.2 – 0.3x => 0.5x - 0.6 + 0.2x = 0.2 - 0.3x => 0.5x + 0.2x + 0.3x = 0.2 + 0.6 => x = 0.8 |
10.) [(2x−1)/3] - [(2x−2)/5] = 1/3 |
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| Hint: View[(2x−1)/3] - [(2x−2)/5] = 1/3 => [5(2x-1) - 3(2x-2)]/15 = 1/3 => (10x - 5 - 6x + 6)/15 = 1/3 => (4x + 1) = 15/3 => 4x +1 = 5 => 4x = 4 => x = 1. |
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