• Word Problems For First Grade

SIMULTANEOUS EQUATIONS WORDED PROBLEMS WORKSHEET #1. Solve each problem by forming a pair of simultaneous equations: 1. 1.9 Solving Linear Equations - Age Problems Objective: Solve age problems by creating and solving a linear equa-tion. An application of linear equations is what are.

Practice test on word problems involving simultaneous linear equations using either the method of elimination or the method of substitution for finding the value of the two linear variables. A fraction is such that if numerator is multiply by 3 and the denominator is reduce by 2 we get 3/5 but if the numerator is increased by 4 and the denominator is doubled we get 5/14. Find the fraction. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 1, the fraction becomes 7/6.

Find the fraction. A number consists of two digits whose sum if 5. When the digits are reversed, the number becomes greater by 9.

Find the number. The sum of a two digit number and the number obtained by reversing the digit is 110. The difference between the digits is 4. Find the number. Seven times a 2 – digit number is equal to 4 times the number obtained by reversing the digits. The difference between the digits is 1.

Find the number. Point A and B are 50 km part on a highway. A car starts from A and another car start from B at the same time. If they travelled in the same direction, they meet in 5 hours but if they move towards each other they meet in 1 hour. Find their speeds.

A boat goes 6 km upstream and 57 km downstream in 5 hours. In 9 hours it can go 21 km upstream and 38 km downstream. Determine the speed of stream and that of boat in still water. The distance between two stations is 340 km. Two trains starts simultaneously from these stations on parallel tracks to cross each other. If the speed on one of them is greater than the other by 5 km/hr and the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increase by 2 units. If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units. Find the length and breadth of the rectangle.

Raga travel 150 km to his home partly by bus and partly by auto-rickshaw. He takes 2 hours if travels 30 km by bus and rest by auto-rickshaw. He takes 5 minutes longer if he travel 50 km by bus and the rest by auto-rickshaw. Find the speed of bus and the auto-rickshaw. 30kmph, 20 kmph 7.

Word Problems For First Grade

Speed of boat 11 kmph, speed of steam 8 kmph 8. 80 kmph, 75 kmph 9. L = 15, b = 8 10. Speed of bus = 60 kmph, speed of auto-rickshaw 80 kmph ● Simultaneous Linear Equations ● Simultaneous Linear Equations - Worksheets.

Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations. Problems of different problems with the help of linear simultaneous equations: We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations. In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols. Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem. Worked-out examples for the word problems on simultaneous linear equations: 1. The sum of two number is 14 and their difference is 2.

Find the numbers. Solution: Let the two numbers be x and y.

(i) x - y = 2. (ii) Adding equation (i) and (ii), we get 2x = 16 or, 2x/2 = 16/2 or, x = 16/2 or, x = 8 Substituting the value x in equation (i), we get 8 + y = 14 or, 8 – 8 + y = 14 - 8 or, y = 14 - 8 or, y = 6 Therefore, x = 8 and y = 6 Hence, the two numbers are 6 and 8. In a two digit number.

The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number. Solution: Let the digit in the units place is x And the digit in the tens place be y. Then x = 3y and the number = 10y + x The number obtained by reversing the digits is 10x + y. If 36 is added to the number, digits interchange their places, Therefore, we have 10y + x + 36 = 10x + y or, 10y – y + x + 36 = 10x + y - y or, 9y + x – 10x + 36 = 10x - 10x or, 9y - 9x + 36 = 0 or, 9x - 9y = 36 or, 9(x - y) = 36 or, 9(x - y)/9 = 36/9 or, x - y = 4.

(i) Substituting the value of x = 3y in equation (i), we get 3y - y = 4 or, 2y = 4 or, y = 4/2 or, y = 2 Substituting the value of y = 2 in equation (i),we get x - 2 = 4 or, x = 4 + 2 or, x = 6 Therefore, the number becomes 26. 3. If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions. Solution: Let the fraction be x/y. If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have (x + 2)/(y + 2) = 9/10 or, 10(x + 2) = 9(y + 2) or, 10x + 20 = 9y + 18 or, 10x – 9y + 20 = 9y – 9y + 18 or, 10x – 9x + 20 – 20 = 18 – 20 or, 10x – 9y = -2. (i) If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have (x – 3)/(y – 3) = 4/5 or, 5(x – 3) = 4(y – 3) or, 5x – 15 = 4y – 12 or, 5x – 4y – 15 = 4y – 4y – 12 or, 5x – 4y – 15 + 15 = – 12 + 15 or, 5x – 4y = 3. (ii) So, we have 10x – 9y = – 2.

(iii) and 5x – 4y = 3. (iv) Multiplying both sided of equation (iv) by 2, we get 10x – 8y = 6.

(v) Now, solving equation (iii) and (v), we get 10x – 9y = -2 10x – 8y = 6 - y = - 8 y = 8 Substituting the value of y in equation (iv) 5x – 4 × (8) = 3 5x – 32 = 3 5x – 32 + 32 = 3 + 32 5x = 35 x = 35/5 x = 7 Therefore, fraction becomes 7/8. 4. If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son. Solution: Let father’s age be x years Son’s ages = y years Then 2y + x = 56 (i) And 2x + y = 82 (ii) Multiplying equation (i) by 2, (2y + x = 56 × 2)we get.