Students can get various questions related to simultaneous linear equations word problems on this page. Practice all the questions to get a clear idea of the topic. You can also find a detailed solution for each problem on the Practice Test on Word Problems Involving Simultaneous Linear Equations page. Solve the questions and verify your answers to get good marks in the exam.<\/p>\n
Follow these steps and instructions while solving the system of linear equations and get the answers easily.<\/p>\n
Example 1.<\/strong><\/p>\n In a rectangle, if the length is increased and the width is reduced each by 2 cm then the area is reduced by 26 sq.cm . If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 30 sq.cm . Find the area of the rectangle?<\/p>\n Solution:<\/strong><\/p>\n Let the length and width of the rectangle is x and y.<\/p>\n Then, the rectangle area = xy<\/p>\n According to the question,<\/p>\n If the length and width of the rectangle is reduced by 2 cm, then the area is reduced by 26 cm.<\/p>\n (x + 2)(y – 2) = x . y – 26<\/p>\n xy – 2x + 2y – 4 = xy – 26<\/p>\n xy – 2x + 2y – xy = -26 + 4<\/p>\n 2y – 2x = -22<\/p>\n 2(y – x) = -22<\/p>\n y – x = -22\/2<\/p>\n y – x = -11 —– (i)<\/p>\n If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 30 sq.cm<\/p>\n (x – 1) . (y + 2) = x . y + 30<\/p>\n xy + 2x – y – 2 = xy + 30<\/p>\n xy + 2x – y – xy = 30 + 2<\/p>\n 2x – y = 32 —– (ii)<\/p>\n Add equation (i) and equation (ii)<\/p>\n y – x + (2x – y) = -11 + 32<\/p>\n y – x + 2x – y = 21<\/p>\n x = 21<\/p>\n Substitute x = 21 in equation (i)<\/p>\n y – 21 = -11<\/p>\n y = -11 + 21<\/p>\n y = 10<\/p>\n So, Area of the rectnagle = x . y<\/p>\n = 21 x 10<\/p>\n = 210 sq. cm.<\/p>\n Hence, the rectangle area is 210 sq. cm.<\/p>\n Example 2.<\/strong><\/p>\n One number is five times the other number. The difference between the two numbers is 32. Find the two numbers?<\/p>\n Solution:<\/strong><\/p>\n Let the two numbers be x and y.<\/p>\n Given that, one number is five times the other number.<\/p>\n x = 5y ——- (i)<\/p>\n The difference between the two numbers is 32.<\/p>\n x – y = 32 ——- (ii)<\/p>\n Substitute x = 5y in equation (ii)<\/p>\n 5y – y = 32<\/p>\n 4y = 32<\/p>\n y = 32\/4<\/p>\n y = 8<\/p>\n Putting y = 8 in equation (i)<\/p>\n x = 5 x 8<\/p>\n x = 40<\/p>\n Hence, the two numbers are 40, 8.<\/p>\n Example 3.<\/strong><\/p>\n I am four times old as my grand-daughter. After five years, I will be 2\u00bd times as old as my grand-daughter. Find my present age and the present age of my grand-daughter?<\/p>\n Solution:<\/strong><\/p>\n Let my present age of is x, the present age of my grand-daughter is y.<\/p>\n According to the question,<\/p>\n I am four times old as my grand-daughter<\/p>\n x = 4y —– (i)<\/p>\n After five years, I will be 2\u00bd times as old as my grand-daughter.<\/p>\n (x + 5) = 2\u00bd(y + 5)<\/p>\n x + 5 = 5\/2(y + 5)<\/p>\n 2(x + 5) = 5(y + 5)<\/p>\n 2x + 10 = 5y + 25<\/p>\n 2x – 5y = 25 – 10<\/p>\n 2x – 5y = 15 —– (ii)<\/p>\n Put x = 4y in equation (ii)<\/p>\n 2(4y) – 5y = 15<\/p>\n 8y – 5y = 15<\/p>\n 3y = 15<\/p>\n y = 5<\/p>\n Put y = 5 in equation (i)<\/p>\n x = 4(5)<\/p>\n x = 20<\/p>\n Therefore, my present age is 20 years, my grand daughter’s present age is 5 years.<\/p>\n Example 4.<\/strong><\/p>\n The difference between the two numbers is 15. Two times the smaller number added to the larger number gives 30. Find the two numbers?<\/p>\n Solution:<\/strong><\/p>\n Let the smaller number is x, largest number is y.<\/p>\n Given that, the difference between numbers is 15.<\/p>\n y – x = 15 —– (i)<\/p>\n Two times the smaller number added to the larger number gives 30<\/p>\n 2x + y = 30 —— (ii)<\/p>\n Subtract equation (ii) from equation (i)<\/p>\n 2x + y – (y – x) = 30 – 15<\/p>\n 2x + x = 15<\/p>\n 3x = 15<\/p>\n x = 15\/3<\/p>\n x = 5<\/p>\n Put x = 5 in equation (i)<\/p>\n y – 5 = 15<\/p>\n y = 15 + 5<\/p>\n y = 20<\/p>\n Therefore, the two numbers are 5, 20.<\/p>\n Example 5.<\/strong><\/p>\n If I double a number and add three times a second number, the answer is 1. If I multiply the first number by 3 and take away twice the second number, the answer is 8. Find the numbers?<\/p>\n Solution:<\/strong><\/p>\n Let the two numbers be x, y.<\/p>\n As per the first condition in the question,<\/p>\n Double a number and add three times the second number, the answer is 1.<\/p>\n 2x + 3y = 1 —- (i)<\/p>\n Second condition is multiply the first number by 3 and take away twice the second number, the answer is 8.<\/p>\n 3x – 2y = 8 —– (ii)<\/p>\n x = (8 + 2y)\/3<\/p>\n Substituting x = (8 + 2y)\/3 in equation (i)<\/p>\n 2(8 + 2y)\/3 + 3y = 1<\/p>\n (16 + 4y)\/3 + 3y = 1<\/p>\n (16 + 4y + 9y)\/3 = 1<\/p>\n 16 + 13y = 3<\/p>\n 13y = 3 – 16<\/p>\n 13y = -13<\/p>\n y = -13\/13<\/p>\n y = -1<\/p>\n Substitute y = -1 in equation (ii)<\/p>\n 3x – 2(-1) = 8<\/p>\n 3x + 2 = 8<\/p>\n 3x = 8 – 2<\/p>\n 3x = 6<\/p>\n x = 6\/3<\/p>\n x = 2<\/p>\n So the numbers are 2, -1.<\/p>\n Example 6.<\/strong><\/p>\n The numerator of a fraction is 4 smaller than its denominator. If both the numerator and denominator are increased by 1, the fraction is 5\/8. Find the original fraction?<\/p>\n Solution:<\/strong><\/p>\n Let the fraction be x\/y.<\/p>\n Given that, the numerator of a fraction is 4 smaller than its denominator.<\/p>\n x – 4 = y<\/p>\n x – y – 4 = 0 —- (i)<\/p>\n If both the numerator and denominator are increased by 1, the fraction is 5\/8<\/p>\n (x + 1) \/ (y + 1) = 5\/8<\/p>\n 8(x + 1) = 5 (y + 1)<\/p>\n 8x + 8 = 5y + 5<\/p>\n 8x – 5y + 8 – 5 = 0<\/p>\n 8x – 5y + 3 = 0 —- (ii)<\/p>\n Multiply the equation (i) by 5.<\/p>\n 5(x – y – 4) = 0<\/p>\n 5x – 5y – 20 = 0 —– (iii)<\/p>\n Subtract equation (iii) from the equation (ii)<\/p>\n 8x – 5y + 3 – (5x – 5y – 20) = 0<\/p>\n 8x – 5y + 3 – 5x + 5y + 20 = 0<\/p>\n 3x + 23 = 0<\/p>\n x = -23\/3<\/p>\n Put x = -23\/3 in equation (i)<\/p>\n -23\/3 – y – 4 = 0<\/p>\n y = -23\/3 – 4<\/p>\n y = (-23 – 12)\/3<\/p>\n y = -35\/3<\/p>\n So, the fraction is (-23\/3) \/ (-35\/3) = 23\/35.<\/p>\n Example 7.<\/strong><\/p>\n The sum of ages of a mother and her son two years ago was 40. In two years’ time from now, the age of the mother will be three times that of her son by then. Find their ages after 7 years’ time?<\/p>\n Solution:<\/strong><\/p>\n The present ages of a mother and her son is x and y respectively.<\/p>\n The sum of ages of a mother and her son two years ago was 40<\/p>\n (x – 2) + (y – 2) = 40<\/p>\n x + y – 4 = 40<\/p>\n x + y = 44 ——- (i)<\/p>\n In two years time from now, the age of the mother will be three times that of her son by then<\/p>\n (x + 2) = 3(y + 2)<\/p>\n x + 2 = 3y + 6<\/p>\n x – 3y + 2 – 6 = 0<\/p>\n x – 3y – 4 = 0 —— (ii)<\/p>\n Subtract equation (ii) from equation (i)<\/p>\n x – 3y – 4 – (x + y) = 0 – 44<\/p>\n x – 3y – 4 – x – y = -44<\/p>\n -4y = -44 + 4<\/p>\n -4y = -40<\/p>\n y = 40\/4<\/p>\n y = 10<\/p>\n Put y = 10 in equation (i)<\/p>\n x + 10 = 44<\/p>\n x = 44 – 10<\/p>\n x = 34<\/p>\n So, the present ages of mother and son are 34 years, 10 years.<\/p>\n The mother age after 7 years is 34 + 7 = 41 years, son age after 7 years is 10 + 7 = 17 years.<\/p>\n Example 8.<\/strong><\/p>\n The Sum of the cost price of the two products is $50. The Sum of the selling price of the same two products is $52. If one is sold at 20% profit and another one is sold at 20% loss, find the cost price of each product.<\/p>\n Solution:<\/strong><\/p>\n Let x, y are the cost prices of two products.<\/p>\n Given that, the sum of the cost prices of the two products is $50.<\/p>\n x + y = 50 —— (i)<\/p>\n One product is sold at 20% profit and another one is sold at 20% loss.<\/p>\n Assume that x is sold at 20% profit. Then its selling price is = 120% of x = 1.2x<\/p>\n Then another product y is sold at 20% loss. Then its selling price = 80% of y = 0.8y<\/p>\n The sum of the selling price of the two products is $52.<\/p>\n 1.2x + 0.8y = 52<\/p>\n Multiply both sides by 10.<\/p>\n 12x + 8y = 520<\/p>\n 3x + 2y = 130 —- (ii)<\/p>\n Multiply the equation (i) by 2 and subtract it from equation (iii).<\/p>\n 2x + 2y = 100<\/p>\n 3x + 2y – (2x + 2y) = 130 – 100<\/p>\n 3x + 2y – 2x – 2y = 30<\/p>\n x = 30<\/p>\n Put x = 30 in equation (i)<\/p>\n 30 + y = 50<\/p>\n y = 50 – 30<\/p>\n y = 20<\/p>\n So, the cost prices of the two products are $30 and $20.<\/p>\n Example 9.<\/strong><\/p>\n Points A and B are 50 km part on a highway. A car starts from A and another car starts from B at the same time. If they traveled in the same direction, they meet in 5 hours but if they move towards each other they meet in 1 hour. Find their speeds?<\/p>\n Solution:<\/strong><\/p>\n Given that,<\/p>\n The distance between two points = 50 km<\/p>\n Speed = distance \/ time<\/p>\n When they travel in the same direction,<\/p>\n Suppose take B travels for x km, then A will traveled for 50 + x<\/p>\n Speed of car A = (50 + x) \/ 5<\/p>\n Speed of car B = x \/ 5<\/p>\n When they travel in the opposite direction,<\/p>\n Suppose they meet when A travels for y km, then B will have traveled 50\u2212y km in the same time.<\/p>\n Speed of car A = y\/1<\/p>\n Speed of car B = (50 – y) \/ 1<\/p>\n Equating the speeds of the cars in both cases,<\/p>\n Speed of car A,<\/p>\n (50 + x) \/ 5 = y<\/p>\n 50 + x = 5y<\/p>\n x – 5y + 50 = 0 —— (i)<\/p>\n Speed of car B,<\/p>\n x\/5 = (50 – y)<\/p>\n x = 5(50 – y)<\/p>\n x = 250 – 5y<\/p>\n x + 5y – 250 = 0 —- (ii)<\/p>\n Subtracting equation (ii) from equation (i)<\/p>\n x + 5y – 250 – (x – 5y + 50) = 0<\/p>\n x + 5y – 250 – x + 5y – 50 = 0<\/p>\n 10y – 300 = 0<\/p>\n 10y = 300<\/p>\n y = 300\/10<\/p>\n y = 30<\/p>\n Substitute y = 30 in equation (i)<\/p>\n x – 5(30) + 50 = 0<\/p>\n x – 150 + 50 = 0<\/p>\n x – 100 = 0<\/p>\n x = 100<\/p>\n Therefore, the speed of car A is 100 km\/hr and car B is 30 km\/hr.<\/p>\n Example 10.<\/strong><\/p>\n In the triangle, the sum of two angles is 90\u00b0 which is the measure of the third angle. Also, the difference between these 2 angles is 10\u00b0, find the measure of these two unknown angles?<\/p>\n Solution:<\/strong><\/p>\n Let the measures of two angles of a triangle are x, y.<\/p>\n The sum of two angles is 90\u00b0.<\/p>\n x + y = 90\u00b0\u00a0 —– (i)<\/p>\n The difference between these 2 angles is 10\u00b0<\/p>\n x – y = 10\u00b0 —– (ii)<\/p>\n Add equation (i) and equation (ii)<\/p>\n x + y + x – y = 90 – 10<\/p>\n 2x = 80<\/p>\n x = 40\u00b0<\/p>\n Put x = 40\u00b0 in equation (ii)<\/p>\n 40\u00b0 – y = 10\u00b0<\/p>\n 40\u00b0 – 10\u00b0 = y<\/p>\n 30\u00b0 = y<\/p>\n Therefore, the unknown two angles of the triangle are 40\u00b0, 30\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":" Students can get various questions related to simultaneous linear equations word problems on this page. Practice all the questions to get a clear idea of the topic. 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