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| Problems on Simultaneous Equations |
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| If one number is thrice the other and their sum is 60, find the numbers. |
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| Let the numbers be x and y. |
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x is 3 times y  x = 3y …(1) |
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| Sum of x and y is 60 |
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 x + y = 60 …(2) |
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| Putting the value of x from (1) in (2), we get, |
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| 3y + y = 60 |
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4y = 60  y = 15 |
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| Substituting y = 15 in (1), we get, |
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| x = 3 x 15 |
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| = 45 |
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 The required numbers are 15 and 45. |
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Find the fraction which becomes  when the denominator is increased by 5 and is equal to  when the numerator is diminished by 4. |
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Let the fraction by  |
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When y is increased by 5, fraction becomes  |
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  or 2x = y + 5 |
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 2x - y = 5 …(1) |
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When x is diminished by 4, fraction becomes  |
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  |
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| or 3x - 12 = y |
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 3x - y = 12 …(2) |
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| 2x - y = 5 …(1) |
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| Subtracting (1) from (2), we get x = 7 |
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| Substituting x = 7 in (1), we get, |
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2  7 - y = 5 |
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 - y = 5 - 14 |
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 - y = - 9 |
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 y = 9 |
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 The fraction is  . |
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| Six years hence a man's age will be three times his son's age, and three years ago he was nine times as old as his son. Find their present ages. |
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| Let the present age of the man be x years, and the present age of his son be y years. |
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| 6 years hence their ages will be (x + 6) years and (y + 6) years. |
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| x + 6 = 3(y + 6) |
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| x + 6 = 3y + 18 |
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| x - 3y = 18 - 6 |
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| x - 3y = 12 …(1) |
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| 3 years ago, their ages were (x - 3) years and (y - 3) years. |
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| x - 3 = 9(y - 3) |
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| x - 3 = 9y - 27 |
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| Subtracting (1) from (2) -6y = -36 |
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| y = 6 |
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| Substituting y = 6 in (1), we get |
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x - 3  6 = 12 |
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| or x - 18 = 12 |
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| x = 12 + 18 = 30 |
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 The present age of the man is 30 years and the present age of his son is 6 years. |
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| The sum of the digits of a two digit number is 7. If the digits are reversed, the new number increased by 3 equals four times the original number. Find the original number. |
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| Let ten's digit be x and unit's digit be y. Then, |
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| Sum of the digits = x + y |
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| The number is = 10x + y |
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| These steps will be repeated in digit sums |
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 Sum of the digits is 7, |
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 x + y = 7 …(1) |
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| When the digits are reversed, the new number becomes |
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| = 10y + x |
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| New number increased by 3 = 4 times the original number |
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 (10y + x) + 3 = 4 (10x + y) |
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| or 10y + x + 3 = 40x + 4y |
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| 10y + x - 40x - 4y = -3 |
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| -39x + 6y = -3 |
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| - 13x + 2y = -1 (Dividing by 3) …(2) |
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| Multiplying (1) by 2, we get, |
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| 2x + 2y = 14 …(3) |
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| Subtracting (3) from (2), |
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-15x = - 15  x = 1 |
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| Substituting x = 1 in (1) |
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1 + y = 7  y = 6 |
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| The number is 10x + y = 10 x 1 + 6 |
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| = 16 |
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| A boat goes 40 km down stream in 2 hours and returns in 4 hours. Find the speed of the boat in still water and the speed of the current. |
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| Let the speed of the boat in still water = x km/h, and |
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| The speed of the current be = y km/h. |
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| Time taken to go down stream = 2 hours |
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| Time x Speed = Distance |
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| 2(x + y) = 40 |
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| x + y = 20 (Divided by 2) …(1) |
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| Again, the speed of the boat upstream |
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| = (x - y) km/h |
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| Time take to go up stream = 4 hours |
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| Distance covered = 40 km |
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 4(x - y) = 40 |
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| x - y = 10 (Dividing by 4) …(2) |
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| Adding (1) and (2), we get |
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| x + y = 20 …(1) |
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| x - y = 10 …(2) |
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2x = 30  x = 15 |
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| Substituting x = 15 in (1), |
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15 + y = 20  y = 5 |
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 The speed of the boat in still water is 15 km/h and the speed of the current is 5 km/h. |
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| A lady has 50 cents and $ 2 coins in her purse. She has 90 coins in all and their total value is
$105. How many $ 2 coins does she have? |
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| Let the number of 50 cents coins be x and |
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| the number of $ 2 coins be y. |
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 x + y = 90 …(1) |
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| The total amount is $ 105 |
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  …(2) |
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| From (2) |
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| x + 4y = 210 …(3) |
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| x + y = 90 …(1) |
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| 3y = 120 Subtracting (1) from (2) |
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 y = 40 |
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 The lady has 40 coins of
$ 2. |
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