Cross Multiply Questions are practice problems that are given below to help the student practice problems and gain an advanced level of confidence in solving problems that involve cross multiplication. Before starting to solve the cross multiplication problems, one has to understand what cross multiplication is. And the answer to the question is, Cross Multiply is a technique/method used in the case of fractions/ratios/proportions or rational numbers to solve the unknown and also to simplify the steps further. In this method, at least 1 fraction is involved. Let us consider the two sides as LHS (Left Hand Side) and RHS (Right Hand Side). The numerator of LHS(a1) is multiplied by the denominator of the RHS(b2). And the denominator(b1) of the LHS is multiplied by the numerator of the RHS(a2).
Cross Multiply Rule
The above-mentioned way of multiplying is for simple problems or problems that involve only 1 unknown. In the case of two linear equations with two variables, the steps are
x = (b1c2 – b2c1)/(b2a1 – b1a2)
y = (c1a2 – c2a1)/(b2a1 – b1a2)
where (b2a1 – b1a2) is not equal to zero.
This implies
x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (b2a1 – b1a2)
Refer to the derivation of these results at Cross Multiply – Pair of linear equations with 2 variables
Cross Multiply Questions With Solutions
Question 1:
Simplify x/10 = 2/5
Solution:
Given x/10 = 2/5
By cross multiplying
5x = 20
x = 20/5 = 4
Question 2:
Simplify
x:3 :: 2 : 4
Solution
Given x : 3 :: 2 : 4
When this proportion is expressed in fractions it is
x/3 = 2/4
By cross multiplying
4x = 6
x = 6/4 = 1.5
Question 3:
Simplify for x using cross multiplication
( x+1 ) / 2 = ( x+1) / 3
Solution:
Given ( x+1 ) / 2 = ( x+1 ) / 3
By cross multiplying
3 ( x+1 ) = 2 ( x+1 )
3x + 3 = 2x + 2
3x – 2x = 2 – 3
x = -1
Question 4:
Compare the fractions using cross multiplication and arrange in ascending order
1/2 , 3/4 , 1/4
Solution:
Let us first consider first two fractions 1/2 and 3/4
i) 1/2 and 3/4
By cross multiplying
4 × 1 and 3 × 2
4 and 6
6 > 4
Therefore 3/4 is greater than 1/2
Let us first consider first two fractions 3/4 and 1/4
ii) 3/4 and 1/4
By cross multiplying
4 × 3 = 4 × 1
12 = 4
Therefore 3/4 is greater than 1/4
Let us first consider first two fractions 3/4 and 1/4
iii) 1/4 and 1/2
By cross multiplying
2 × 1 = 4 × 1
2 = 4
Therefore 1/2 is greater than 1/4
We deduced by cross multiplying that 3/4 is greater than 1/2 and 1/4 . Also 1/2 is greater than 1/4
So arranging these fractions into ascending order will be 1/4 , 1/2 , and 3/4.1/4 < 1/2 < 3/4
Question 5:
If 10 candles can burn for 36 hrs, how many hours will it take to burn 25 candles?
Solution:
Given 10 candles can burn for 36 hrs. We need to find out how many hours will 25 candles burn.
Let us express these numbers in a ratios and proportions
Candles: Hours
Therefore, 10 : 36 : : 25 : x
or 10/36 = 25/x
Cross multiplying
10 x = 25 × 36
10 x = 900
x = 900 / 10 = 90
So 25 candles can burn for 90 hours.
Question 6:
A bus travels from Bangalore to Mangalore which is 353 km in 8 hours. How many km does it travel in 3 hours? Solve using the method of cross multiplication
Solution:
Given that a bus travels 353 km in 8 hrs, we need to determine the distance travelled in 3 hours.
Let us assume the speed maintained is constant in this problem.
Expressing this in ratio and proportions
Ratio = Distance : Time
Proportion = 353 : 8 : : x : 3
Expressing in fractions
353 / 8 = x / 3
By cross multiplying.
353 × 3 = 8 × x
1059 = 8x
x = 1059/ 8 = 132.375 km.
So the bus travels 132.375 km in 3 hours
Question 7:
Simplify ( x+1 ) = ( x+3 ) / 7
Solution:
Given ( x+1 ) = ( x+3 ) / 7
By cross multiplying we get
7 × ( x + 1) = ( x + 3 )
7x + 7 = x + 3
7x – x = 3 – 7
6x = -4
x = – 4 / 6
x = -2 / 3
Question 8:
Simplify (2x + 3 ) / 4 = ( 4 x + 9 ) / 3
Solution:
Given (2x + 3 ) / 4 = ( 4 x + 9 ) / 3
By Cross multiplying,
(2x + 3 ) × 3 = ( 4 x + 9 ) × 4
6x + 9 = 16 x + 36
16x + 36 – 6x – 9 = 0
10 x + 27 = 0
x = -27 /10 = -2.7
Question 9:
Solve the given linear equations using a cross multiplication method.
x + y = 2 ; x – y = 3
Solution:
For the give linear equations to be solved by the method of cross multiplication, we need to use the formula
x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (b2a1 – b1a2) where (b2a1 – b1a2) is not equal to zero.
Let us substitute the values of
a1 = 1 , b1 = 1, c1 = -2
a2 = 1 , b2 = -1, c2 = -3
(b1c2 – b2c1) = (1)(-3) – (-1)(-2) = – 3 – 2 = – 5
(c1a2 – c2a1) = (-2)(1) – (-3)(1) = -2 + 3 = 1
(b2a1 – b1a2) = (-1)(1) – (1)(1) = -1 -1 = – 2
Substituting the values in the solution
x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (b2a1 – b1a2)
= x / – 5 = y / 1 = 1 / – 2
For x
x / – 5 = 1 / – 2
Cross multiplying
-2x = -5
Or x = 5 / 2
For y
y / 1 = 1 / – 2
Cross multiplying
y = 1 / -2 = – 1 / 2
Question 10:
Solve the given linear equations using a cross multiplication method.
2x + y = 4 ; x – 2y = 6
Solution:
For the give linear equations to be solved by the method of cross multiplication, we need to use the formula
x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (b2a1 – b1a2) where (b2a1 – b1a2) is not equal to zero.
Let us substitute the values of
a1 = 2 , b1 = 1, c1 = -4
a2 = 1 , b2 = -2, c2 = -6
(b1c2 – b2c1) = (1)(-6) – (-2)(-4) = -6 – 8 = -14
(c1a2 – c2a1) = (-4)(1) – (-6)(2) = -4 + 12 = 8
(b2a1 – b1a2) = (-2)(2) – (1)(1) = -4 -1 = -5
Substituting the values in the solution
x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (b2a1 – b1a2)
= x / -14 = y / 8 = 1 / -5
For x
x / -14 = 1 / -5
Cross multiplying
-5x = -14
Or x = 14 / 5
For y
y / 8 = 1 / -5
Cross multiplying
y = 8 / -5 = -8 / 5
Related Articles
Practice Questions on Cross Multiply
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- Solve for x
2/x = 10/5
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- Simplify using the cross multiplication method
x = 4/(x-4)
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- Solve the system of equations using the Cross Multiply rule.
x + y = 8
x – 2y = 4
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- Solve the system of equations for x and y using the Cross Multiply rule.
2x + y = 10
3x – 2y = 4
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- Solve for x and y using a cross multiplication method.
2x = 10
3x = 4 + 6y
