Sec 2 Maths · Linear equations, inequalities and simultaneous equations

Simultaneous equations: word problems questions with answers

Practice questions on simultaneous equations: word problems, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

At a cinema, 290 tickets were sold. Standard tickets cost $15 each and senior citizen tickets cost $12 each. The tickets sold for $3,915 altogether. Find the number of standard tickets and the number of senior citizen tickets sold.

Step 1 of 6

Let x be the number of standard tickets and y the number of senior citizen tickets. Count the tickets

x + y = 290

... (1)

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Questions

Question 1Tickets: a count and a total
At a funfair, 275 tickets were sold. Adult tickets cost $15 each and child tickets cost $4 each. The tickets sold for $2,585 altogether. Find the number of adult tickets and the number of child tickets sold.
Show the worked solution
1. Step 1: Let x be the number of adult tickets and y the number of child tickets. Count the tickets $x + y = 275$ ... (1)
2. Step 2: Add up the money from the tickets (in dollars) $15x + 4y = 2585$ ... (2)
3. Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 4$ : $4x + 4y = 1100$ ... (3)
4. Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
5. Step 5: Now the y-terms cancel. Solve for x $(2) - (3)$ : $11x = 1485$ $x = \dfrac{1485}{11}$ $= 135$
6. Step 6: Substitute your value of x into (1) to find y $135 + y = 275$ $y = 140$ Check in (2): $15(135) + 4(140)$ $= 2585$ , the right-hand side of (2).135 adult tickets and 140 child tickets were sold.
Question 2Two purchases
1 curry puff and 1 packet of soya bean milk cost $4.40. 4 curry puffs and 5 packets of soya bean milk cost $19.60. Find the cost of one curry puff and the cost of one packet of soya bean milk.
Show the worked solution
1. Step 1: Let one curry puff cost x dollars and one packet of soya bean milk cost y dollars. The first purchase $x + y = 4.4$ ... (1)
2. Step 2: The second purchase $4x + 5y = 19.6$ ... (2)
3. Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 5$ : $5x + 5y = 22$ ... (3)
4. Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
5. Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $x = 2.4$
6. Step 6: Substitute your value of x into (1) to find y $2.4 + y = 4.4$ $y = 2$ Check in (2): $4(2.4) + 5(2)$ $= 19.6$ , the right-hand side of (2).One curry puff costs $2.40 and one packet of soya bean milk costs $2.
Question 3Coins (work in cents)
A box contains 46 coins. All the coins are 20-cent coins or 50-cent coins. The total value of the coins is $13.70. Find the number of 20-cent coins and the number of 50-cent coins.
Show the worked solution
1. Step 1: Let x be the number of 20-cent coins and y the number of 50-cent coins. Count the coins $x + y = 46$ ... (1)
2. Step 2: Add up their value. Work in cents $20x + 50y = 1370$ ... (2)
3. Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 50$ : $50x + 50y = 2300$ ... (3)
4. Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
5. Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $30x = 930$ $x = \dfrac{930}{30}$ $= 31$
6. Step 6: Substitute your value of x into (1) to find y $31 + y = 46$ $y = 15$ Check in (2): $20(31) + 50(15)$ $= 1370$ , the right-hand side of (2).There are 31 20-cent coins and 15 50-cent coins.
Question 4Ages
Mr Kumar is 3 times as old as his son. In 8 years' time, Mr Kumar will be 2 times as old as his son. Find their present ages.
Show the worked solution
1. Step 1: Let Mr Kumar be x years old and his son y years old now. Write their present ages as an equation $x = 3y$ $x - 3y = 0$ ... (1)
2. Step 2: Write both ages at that time, then the "times as old" equation $x + 8 = 2(y + 8)$ $x + 8 = 2y + 16$ $x - 2y = 8$ ... (2)
3. Step 3: Eliminate x: add or subtract the equations?The $x$ -terms have the same sign, so subtract.
4. Step 4: Now the x-terms cancel. Solve for y $(2) - (1)$ : $y = 8$
5. Step 5: Substitute your value of y into (1) to find x $x - 3(8) = 0$ $x = 24$ Check in (2): $(24) - 2(8)$ $= 8$ , the right-hand side of (2).Mr Kumar is 24 years old and his son is 8 years old.
Question 5A two-digit number
The sum of the digits of a two-digit number is 5. When the digits are reversed, the new number is 9 less than the original number. Find the original number.
Show the worked solution
1. Step 1: Let the tens digit be x and the units digit be y. The digits add up to the total $x + y = 5$ ... (1)
2. Step 2: Write the original number and the reversed number in terms of x and yOriginal number: $10x + y$ . Reversed number: $10y + x$ .
3. Step 3: Use the difference between the two numbers $(10y + x) - (10x + y) = -9$ $-9x + 9y = -9$ ... (2)
4. Step 4: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 9$ : $9x + 9y = 45$ ... (3)
5. Step 5: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
6. Step 6: Now the y-terms cancel. Solve for x $(3) - (2)$ : $18x = 54$ $x = \dfrac{54}{18}$ $= 3$
7. Step 7: Substitute your value of x into (1) to find y $3 + y = 5$ $y = 2$ Check in (2): $-9(3) + 9(2)$ $= -9$ , the right-hand side of (2).
8. Step 8: Write the original number $10(3) + 2 = 32$ The original number is $32$ .
Question 6Sides and angles of a figure
$ABCD$ is a parallelogram with $\angle DAB = (4x + y - 10)^\circ$ , $\angle BCD = (2x + 35)^\circ$ and $\angle ABC = (4x + 4y + 5)^\circ$ . Form two equations in $x$ and $y$ and solve them. Hence find $\angle ABC$ .
Show the worked solution
1. Step 1: Opposite angles of a parallelogram are equal: ∠DAB = ∠BCD. Simplify $4x + y - 10 = 2x + 35$ $2x + y = 45$ ... (1)
2. Step 2: Angles between parallel sides add up to 180° (co-interior): ∠DAB + ∠ABC = 180°. Simplify $(4x + y - 10) + (4x + 4y + 5) = 180$ $8x + 5y = 185$ ... (2)
3. Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 5$ : $10x + 5y = 225$ ... (3)
4. Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
5. Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $2x = 40$ $x = \dfrac{40}{2}$ $= 20$
6. Step 6: Substitute your value of x into (1) to find y $2(20) + y = 45$ $y = 5$ Check in (2): $8(20) + 5(5)$ $= 185$ , the right-hand side of (2).
7. Step 7: Put x and y into the expression for ∠ABC $\angle ABC = 105^\circ$
Question 7Find the fraction
When $4$ is added to the numerator of a fraction, the fraction becomes $\dfrac{8}{9}$ . When $2$ is subtracted from the denominator of the original fraction, it becomes $\dfrac{4}{7}$ . Let the fraction be $\dfrac{x}{y}$ . Find the fraction.
Show the worked solution
1. Step 1: Write each fact as an equation, cross-multiply and simplify $\dfrac{x + 4}{y} = \dfrac{8}{9}$ $9(x + 4) = 8y$ $9x - 8y = -36$ ... (1) $\dfrac{x}{y - 2} = \dfrac{4}{7}$ $7x = 4(y - 2)$ $7x - 4y = -8$ ... (2)
2. Step 2: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(2) \times 2$ : $14x - 8y = -16$ ... (3)
3. Step 3: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
4. Step 4: Now the y-terms cancel. Solve for x $(3) - (1)$ : $5x = 20$ $x = \dfrac{20}{5}$ $= 4$
5. Step 5: Substitute your value of x into (1) to find y $9(4) - 8y = -36$ $-8y = -72$ $y = \dfrac{-72}{-8}$ $= 9$ Check in (2): $7(4) - 4(9)$ $= -8$ , the right-hand side of (2).
6. Step 6: Write the fractionThe fraction is $\dfrac{4}{9}$ .

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Worked example

Step 1 of 6

Let x be the number of standard tickets and y the number of senior citizen tickets. Count the tickets

x + y = 290

... (1)

Questions

Question 1Tickets: a count and a total

At a funfair, 275 tickets were sold. Adult tickets cost $15 each and child tickets cost $4 each. The tickets sold for $2,585 altogether. Find the number of adult tickets and the number of child tickets sold.

Show the worked solution

Step 1: Let x be the number of adult tickets and y the number of child tickets. Count the tickets $x + y = 275$ ... (1)
Step 2: Add up the money from the tickets (in dollars) $15x + 4y = 2585$ ... (2)
Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 4$ : $4x + 4y = 1100$ ... (3)
Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 5: Now the y-terms cancel. Solve for x $(2) - (3)$ : $11x = 1485$ $x = \dfrac{1485}{11}$ $= 135$
Step 6: Substitute your value of x into (1) to find y $135 + y = 275$ $y = 140$ Check in (2): $15(135) + 4(140)$ $= 2585$ , the right-hand side of (2).135 adult tickets and 140 child tickets were sold.

Question 2Two purchases

1 curry puff and 1 packet of soya bean milk cost $4.40. 4 curry puffs and 5 packets of soya bean milk cost $19.60. Find the cost of one curry puff and the cost of one packet of soya bean milk.

Show the worked solution

Step 1: Let one curry puff cost x dollars and one packet of soya bean milk cost y dollars. The first purchase $x + y = 4.4$ ... (1)
Step 2: The second purchase $4x + 5y = 19.6$ ... (2)
Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 5$ : $5x + 5y = 22$ ... (3)
Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $x = 2.4$
Step 6: Substitute your value of x into (1) to find y $2.4 + y = 4.4$ $y = 2$ Check in (2): $4(2.4) + 5(2)$ $= 19.6$ , the right-hand side of (2).One curry puff costs $2.40 and one packet of soya bean milk costs $2.

Question 3Coins (work in cents)

A box contains 46 coins. All the coins are 20-cent coins or 50-cent coins. The total value of the coins is $13.70. Find the number of 20-cent coins and the number of 50-cent coins.

Show the worked solution

Step 1: Let x be the number of 20-cent coins and y the number of 50-cent coins. Count the coins $x + y = 46$ ... (1)
Step 2: Add up their value. Work in cents $20x + 50y = 1370$ ... (2)
Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 50$ : $50x + 50y = 2300$ ... (3)
Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $30x = 930$ $x = \dfrac{930}{30}$ $= 31$
Step 6: Substitute your value of x into (1) to find y $31 + y = 46$ $y = 15$ Check in (2): $20(31) + 50(15)$ $= 1370$ , the right-hand side of (2).There are 31 20-cent coins and 15 50-cent coins.

Question 4Ages

Mr Kumar is 3 times as old as his son. In 8 years' time, Mr Kumar will be 2 times as old as his son. Find their present ages.

Show the worked solution

Step 1: Let Mr Kumar be x years old and his son y years old now. Write their present ages as an equation $x = 3y$ $x - 3y = 0$ ... (1)
Step 2: Write both ages at that time, then the "times as old" equation $x + 8 = 2(y + 8)$ $x + 8 = 2y + 16$ $x - 2y = 8$ ... (2)
Step 3: Eliminate x: add or subtract the equations?The $x$ -terms have the same sign, so subtract.
Step 4: Now the x-terms cancel. Solve for y $(2) - (1)$ : $y = 8$
Step 5: Substitute your value of y into (1) to find x $x - 3(8) = 0$ $x = 24$ Check in (2): $(24) - 2(8)$ $= 8$ , the right-hand side of (2).Mr Kumar is 24 years old and his son is 8 years old.

Question 5A two-digit number

The sum of the digits of a two-digit number is 5. When the digits are reversed, the new number is 9 less than the original number. Find the original number.

Show the worked solution

Step 1: Let the tens digit be x and the units digit be y. The digits add up to the total $x + y = 5$ ... (1)
Step 2: Write the original number and the reversed number in terms of x and yOriginal number: $10x + y$ . Reversed number: $10y + x$ .
Step 3: Use the difference between the two numbers $(10y + x) - (10x + y) = -9$ $-9x + 9y = -9$ ... (2)
Step 4: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 9$ : $9x + 9y = 45$ ... (3)
Step 5: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 6: Now the y-terms cancel. Solve for x $(3) - (2)$ : $18x = 54$ $x = \dfrac{54}{18}$ $= 3$
Step 7: Substitute your value of x into (1) to find y $3 + y = 5$ $y = 2$ Check in (2): $-9(3) + 9(2)$ $= -9$ , the right-hand side of (2).
Step 8: Write the original number $10(3) + 2 = 32$ The original number is $32$ .

Question 6Sides and angles of a figure

$ABCD$ is a parallelogram with $\angle DAB = (4x + y - 10)^\circ$ , $\angle BCD = (2x + 35)^\circ$ and $\angle ABC = (4x + 4y + 5)^\circ$ . Form two equations in $x$ and $y$ and solve them. Hence find $\angle ABC$ .

Show the worked solution

Step 1: Opposite angles of a parallelogram are equal: ∠DAB = ∠BCD. Simplify $4x + y - 10 = 2x + 35$ $2x + y = 45$ ... (1)
Step 2: Angles between parallel sides add up to 180° (co-interior): ∠DAB + ∠ABC = 180°. Simplify $(4x + y - 10) + (4x + 4y + 5) = 180$ $8x + 5y = 185$ ... (2)
Step 3: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(1) \times 5$ : $10x + 5y = 225$ ... (3)
Step 4: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 5: Now the y-terms cancel. Solve for x $(3) - (2)$ : $2x = 40$ $x = \dfrac{40}{2}$ $= 20$
Step 6: Substitute your value of x into (1) to find y $2(20) + y = 45$ $y = 5$ Check in (2): $8(20) + 5(5)$ $= 185$ , the right-hand side of (2).
Step 7: Put x and y into the expression for ∠ABC $\angle ABC = 105^\circ$

Question 7Find the fraction

When $4$ is added to the numerator of a fraction, the fraction becomes $\dfrac{8}{9}$ . When $2$ is subtracted from the denominator of the original fraction, it becomes $\dfrac{4}{7}$ . Let the fraction be $\dfrac{x}{y}$ . Find the fraction.

Show the worked solution

Step 1: Write each fact as an equation, cross-multiply and simplify $\dfrac{x + 4}{y} = \dfrac{8}{9}$ $9(x + 4) = 8y$ $9x - 8y = -36$ ... (1) $\dfrac{x}{y - 2} = \dfrac{4}{7}$ $7x = 4(y - 2)$ $7x - 4y = -8$ ... (2)
Step 2: Neither add nor subtract gets rid of y yet: multiply so the y-terms are the same size $(2) \times 2$ : $14x - 8y = -16$ ... (3)
Step 3: Eliminate y: add or subtract the equations?The $y$ -terms have the same sign, so subtract.
Step 4: Now the y-terms cancel. Solve for x $(3) - (1)$ : $5x = 20$ $x = \dfrac{20}{5}$ $= 4$
Step 5: Substitute your value of x into (1) to find y $9(4) - 8y = -36$ $-8y = -72$ $y = \dfrac{-72}{-8}$ $= 9$ Check in (2): $7(4) - 4(9)$ $= -8$ , the right-hand side of (2).
Step 6: Write the fractionThe fraction is $\dfrac{4}{9}$ .