Worksheet on Word Problems on Linear Equation | Linear Equations Word Problems Worksheet

Find the solution for any typical problem of linear equations by practicing our Worksheet on Word Problems on Linear Equation. You can find out different types of equations from simple to complex ones. We are offering solutions for every equation along with an explanation. Practice using the below Linear Equations Word Problems Worksheet and learn how to find a solution for a given linear equation problem quickly.

Linear Equations Problems with Solutions

1. Convert the following statements into equations?
(a) 5 added to a number is 9.
(b) 3 subtracted from a number is equal to 12.
(c) 5 times a number decreased by 2 is 4.
(d) 2 times the sum of the numbers K and 7 is 13.

Solution:

(a) The given statement is 5 added to a number is 9.
Let’s consider a number as ‘K’.
Then, adding 5 with K should be 9. That is 5 + K = 9.
By converting the given statement into an equation we will get the equation like 5 + K = 9.
(b) The given statement is 3 subtracted from a number is equal to 12.
Let’s consider a number as ‘K’.
Then, subtracting 3 from K should be 12. That is K – 3 = 12.
By converting the given statement into an equation we will get the equation as K – 3 = 12.
(c) The given statement is5 times a number decreased by 2 is 4.
Let’s consider a number as ‘K’.
Five times a number is equal to 5K.
5Kis decreased by 2 and that is 5K – 2.
Then, by decreasing 2 from 5K should be 4. That is5K – 2 = 4.
By converting the given statement into an equation we will get the equation as 5K – 2 = 4.
(d)The given statement is 2 times the sum of the number K and 7 is 13.
Let the sum of the number K and 7 is K + 7.
Now, 2 times the sum of K and 7 is 2 ( K + 7 ) which is equal to 13.
By converting the given statement into an equation we will get the equation as
2 ( K + 7 ) = 13.


2. A number is 12 more than the other. Find the numbers if their sum is 48?

Solution:

The given statement is a number is 12 more than the other and if their sum is 48.
Let’s consider the first number as ‘K’.
If another number is 12 more than K and that is K + 12.
So, the sum of two numbers is
K + K + 12 = 48.
2K + 12 = 48.
2K = 48 – 12 = 36.
K = 36 ÷ 2 = 18
So first number K is 18 then other number is 12 + K = 12 + 18 = 30.
The sum of the two numbers 18 + 30 = 48.

The two numbers are 18 and 30.


3. Twice the number decreased by 22 is 48. Find the number?

Solution:

The given statement is twice the number decreased by 22 is 48.
Let’s consider the number as ‘K’.
Twice the number is 2K.
2K is decreased by 22, that is 2K – 22.
So 2K – 22 = 48.
2k = 48 + 22 = 70.
K = 70 ÷ 2 = 35.

That is the number is K = 35.


4. Seven times the number is 36 less than 10 times the number. Find the number?

Solution:

The given statement is Seven times the number is 36 less than 10 times the number.
Let’s consider a number as ‘K’.
Seven times the number is 7K.
Ten times the number is 10K.
36 less than ten times a number is 10K – 36.
7K should be equal to 10K – 36.
That is 7K = 10K – 36.
36 = 10K – 7K.
36 = 3K.
K = 36 ÷ 3 = 12.

The number is 12.


5. 4/5 of a number is more than 3/4 of the number by 5. Find the number?

Solution:

The given statement is 4/5 of a number is more than 3/4 of the number by 5.
Let’s consider a number as ‘K’.
4 / 5 of a number is 4K / 5.
3 / 4 of a number is 3K / 4.
4K / 5 is more than 3K / 4 by 5.
That is 3K / 4 = 4K / 5 + 5.
By using LCM on RHS.
3K / 4 = ( 4K + 25 ) / 5.
By using cross multiplication on L.H.S and R.H.S for the above equation.
( 3K ) X 5 = ( 4K + 25 ) X 4
15K = 16K + 100.
K = 100.
The number is 100.


6. The sum of two consecutive even numbers is 38. Find the numbers?

Solution:

The given statement is the sum of two consecutive even numbers is 38.
Let’s consider the consecutive even numbers are K + 2 and K + 4.
Sum of two consecutive even numbers is ( K + 2 ) + ( K + 4 ) = 38.
2K + 6 = 38.
2K = 38 – 6 = 32.
K = 32 ÷ 2 = 16.
K + 2 = 16 + 2 = 18.
K + 4 = 16 + 4 = 20
So the two consecutive even numbers are 18 and 20.


7. The sum of three consecutive odd numbers is 51. Find the numbers?

Solution:

The given statement is the sum of three consecutive odd numbers is 51.
Let’s consider the consecutive odd numbers as K + 1, K + 3 and K + 5.
The Sum of three consecutive odd numbers is
( K + 1 ) + ( K + 3 ) + ( K + 5 ) = 51.
3K + 9 = 51.
3K = 51 – 9 = 42.
K = 42 ÷ 3 = 14.
K + 1 = 15.
K + 3 = 17.
K + 5 = 19.
The consecutive odd numbers are 15, 17, and 19.


8. Rene is 6 years older than her younger sister. After 10 years, the sum of their ages will be 50 years. Find their present ages?

Solution:

The given statement Rene is 6 years older than her younger sister. After 1 0 years, the sum of their ages will be 50 years.
Let’s consider younger sister age as ‘K’.
Rene is 6 years older than her younger sister, that is 6 + K.
After 10 years the sum of Rene and Younger sister age will be equal to 50 can be written as
( K + 10 ) + ( K + 6 + 10 ) = 50.
2K + 26 = 50.
2K = 50 – 26 = 24.
K = 24 ÷ 2 = 12.
Younger sister age is 12.
Rene is 6 years older than her younger sister that is 12 + 6 = 18.


9. The length of a rectangle is 10 m more than its breadth. If the perimeter of a rectangle is 80 m, find the dimensions of the rectangle?

Solution:

The given statement is the length of a rectangle is 10m more than its breadth and the perimeter of a rectangle is 80m.
Let’s consider the breadth of a rectangle as ‘K’.
b = K.
The length of a rectangle is 10m more than the breadth that is K + 10.
l = K + 10.
The perimeter of rectangle is 2 ( l + b ) = 80m.
2 ( K + 10 + K ) = 80m.
2K + 10 = 80 ÷ 2 = 40m.
2K = 40 – 10 = 30m
K = 30 ÷ 2 = 15m.
The breadth of a rectangle K = 15m.
Length of a rectangle K + 10 = 15 + 10 = 25m.


10. A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find the length and breadth of the plot?

Solution:

The given statement is a 300 m long wire is used to fence a rectangular plot whose length is twice its width.
Let’s consider the width of the rectangle as ‘K’.
b = K.
The length of the rectangle is twice its width that is 2K.
l = 2K.
Perimeter of a rectangle is 2 ( l + b ) = 300.
2 ( K + 2K ) = 300m.
3K = 300 ÷ 2 = 150m.
K = 150 ÷ 3 = 50m.
The width of the rectangle is K = 50m.
The length of rectangle is 2K = 2 X 50 = 100m.


11. The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2, find the fraction?

Solution:

The given statement is the denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.
Let’s consider the numerator as ‘K’.
The denominator of the fraction is greater than the numerator by 8 that is K + 8.
So fraction is K / K + 8.
The numerator is increased by 17 and the denominator is decreased by 1 which is
( K + 17 ) / ( K + 8 – 1 ) = 3 / 2.
By applying cross multiplication for above equation we get
2 ( K + 17 ) = 3 ( K + 7 ).
2K + 34 = 3K + 21.
34 – 21 = 3K – 2K.
K = 13.
The numerator of the fraction is K = 13.
Denominator of fraction is K + 8 = 13 + 8 = 21.
So fraction is 13 / 21.


12. A sum of $2700 is to be given in the form of 63 prizes. If the prize is either $100 or $25, find the number of prizes of each type?

Solution:

The given statement is a sum of $2700 is to be given in the form of 63 prizes. If the prize is either $100 or $25.
Let’s consider ‘K’ as the number of the $100 prizes and ‘L’ as the number of the $25 prize.
100K + 25L = $2700 ——– (1).
Total number of prizes is 63 that is K + L = 63 ——- (2).
Multiply the equation (2) with 100 on both sides. That is
100K + 100L = $6300 ——– (3).
Subtract the equation (1) and (3).
100K + 25L = $2700.
100K + 100L = $6300.
(-)—–(-)——-(-)——————-
– 75L = – $3600.
L = 3600 ÷ 75 = 48.
Substitute the L value in equation number (2).
That is K + 48 = 63.
K = 63 – 48 = 15.
Finally, the number of prizes are 15 and 48.


13. In a class of 42 students, the number of boys is 2/5 of the girls. Find the number of boys and girls in the class?

Solution:

The given statement isa class of 42 students, the number of boys is 2/5 of the girls.
Let’s consider the number of girls in the class is ‘K’.
So, the number of boys is 2 / 5 of the girls. That is 2K/ 5.
The total number of students in the class is 42. That is
2K / 5 + K = 42.
2K + 5K = 42 X 5 = 210.
7K = 210.
K = 210 ÷ 7 = 30.
The total number of girls in the class is K = 30
The total number of boys in the class is 2K / 5 = 2 X 30 / 5 = 2 X 6 = 12.


14. Among the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller. Find their measures?

Solution:

The given statement is the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller.
Let’s consider the small-angle as ‘K’ and the larger angle as ‘L’.
‘K’ and ‘L’ are supplementary angles. That is K + L = 180° ———(1).
The larger angle is 36° more than the smaller angle. That is L = K + 36°.
Substitute the larger angle value in equation (1).
Then, K + K + 36° = 180°.
2K = 180° – 36° = 144°.
K = 144° ÷ 2 = 72°.
L = K + 36° = 72° + 36° = 108°.
The measures of the angles are 72° and 108°.


15. My mother is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age. Find my age and my mother’s age?

Solution:

The given statement is mother’s age is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age.
Let’s consider my age as ‘K’.
My mother’s age is 12 years more than twice my age and that is 2K + 12.
After 8 years.
My age is K + 8.
My mother’s age is 2K + 12 + 8.
My mother’s age will be 20 years less than three times my age.
3(K + 8) – 20 = 2K + 20.
3K + 24 – 20 = 2K + 20.
3K + 4 = 2K + 20.
3K – 2K = 20 – 4 = 16.
My age is K = 16 years.
My mother’s age is 2K + 12 = 2(16) + 12 = 44 years.


16. In an isosceles triangle, the base angles are equal and the vertex angle is 80°. Find the measure of the base angles?

Solution:

The given statement is in an isosceles triangle, the base angles are equal and the vertex angle is 80°.
As per the given information, the vertex angle is 80°.
The base angles are equal.
Triangle is K + K + 80° = 180°.
2K = 180° – 80° = 100°.
K = 100 ÷ 2 = 50°.
So, the measures of the base angles are 50°.


17. Adman’s father is 49 years old. He is 5 years older than four times Adman’s age. What is Adman’s age?

Solution:

The given statement is an Adman’s father is 49 years old. He is 5 years older than four times Adman’s age.
Let’s consider an Adman’s age as ‘K’
Adman’s father is 49 years old.
Adman’s father is 5 years older than four times Adman’s age. That is
4K + 5 = 49.
4K = 49 – 5 = 44.
K = 44 ÷ 4 = 11 years
So, an Adman’s age is 11 years.


18. The cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is $12.80, find the cost of each?

Solution:

The given statement is the cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is $12.80.
Let’s consider the cost of an eraser as ‘K’.
The cost of a pencil is 25 cents more than the cost of an eraser. That is K + 25.
The cost of 8 pencils and 10 erasers is $12.80.
1 dollar = 100 cents. So, $12.80 is 1280 cents.
8(K + 25) + 10K = 1280 cents.
8K + 200 + 10K = 1280 cents.
18K = 1280 – 200 = 1080 cents.
K = 1080 ÷ 18 = 60 cents.
The cost of an eraser is 60 cents.
Cost of pencils is K + 25 = 60 + 25 = 85 cents.


19. Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other?

Solution:

The given statement is Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other.
By dividing the 36 into two parts
K / 5 = (36 – K) / 7.
By cross multiplying the above terms, we will get
7K = (36 – K)5.
7K = 36 X 5 – 5K.
7K + 5K = 180.
12K = 180.
K = 180 ÷ 12 = 15.
The first part is 15 and the second part is 36 – 15 = 21.


20. The length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle. Find the length and breadth of the given rectangle?

Solution:

The given statement is the length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle.
Let’s consider the breadth of the rectangle as ‘K’.
b = K.
Then, the length of the rectangle exceeds its breadth by 3 cm. That is K + 3
l = K + 3.
If the length and breadth are each increased by 2cm is K + 2 and K + 3 + 2.
The area of a rectangle is 70sq. cm
Area of the rectangle is (l X b) = 70 sq. cm.
(K + 2) X (K + 5) = K(K + 3) + 70.
K^2 + 5K + 2K + 10 = K^2 + 3K + 70.
7K + 10 = 3K + 70.
4K = 60.
K = 15.
L = K + 3 = 15 + 3 = 18.
The length of the rectangle is 18 and the breadth of the rectangle is 15.


Worksheet on Significant Figures | Significant Figures Practice Worksheets

Take the guidance on the concept of Significant Figures by taking the help of Worksheet on Significant Figures. Practice all the problems present in the Significant Figures Worksheets and get a good grip on the topic. Test your subject knowledge by taking the practice tests available on our website. Also, you can easily enhance your problem-solving skills by solving Worksheet on Significant Figures on a daily basis. Assess your preparation standards and concentrate on the areas you are facing difficulty.

Significant Figures Examples with Answers

Check different problems impose on Significant Figures and get a grip on every concept available in the Significant Figures.

1. Find the Number of a Significant Figure in Each of the Following
(a) 8.4
(b) 173.6 m
(c) 407 g
(d) 4.68 m
(e) 8.1165 kg
(f) 0.054 km
(g) 0.00343 l
(h) 93.040 mg
(i) 30.030300 g
(j) 40.00 km
(k) 3.600 ml
(l) 70.002

Solution:

(a) There are 2 significant figures available in 8.4. They are 8, 4.
(b) There are 4 significant figures available in 173.6 m. They are 1, 7, 3, and 4.
(c) There are 3 significant figures available in 407 g. They are 4, 0, and 7.
(d) There are 3 significant figures available in 4.68 m. They are 4, 6, and 8.
(e) There are 5 significant figures available in 8.1165 kg. They are 8, 1, 1, 6, and 5.
(f) There are 2 significant figures available in 0.054 km. They are 5, and 4.
(g) There are 3 significant figures available in 0.00343 l. They are 3, 4, and 3.
(h) There are 5 significant figures available in 93.040 mg. They are 9, 3, 0, 4, and 0.
(i) There are 8 significant figures available in 30.030300 g. They are 3, 0, 0, 3, 0, 3, 0, and 0.
(j) There are 4 significant figures available in 40.00 km. They are 4, 0, 0, and 0.
(k) There are 4 significant figures available in 3.600 ml. They are 3, 6, 0, and 0.
(l) There are 5 significant figures available in 70.002. They are 7, 0, 0, 0, and 2.


2. Round off each of the following correct up to 3 significant figures
(a) 57.3628 g
(b) 6.31874 kg
(c) 44.422 km
(d) 60.001 cm
(e) 0.0023596 m
(f) 0.0024030 l

Solution:

(a) Given that 57.3628 g. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 6 is greater than 5. So, the digit 3 becomes 4, and the digits 6, 2, and 8 disappear.
Therefore, 57.3628 g = 57.4 g rounded off to 3 significant figures.
(b) Given that 6.31874 kg. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decimal.
The digit 8 is greater than 5. So, the digit 1 becomes 2, and the digits 8, 7, and 4 disappear.
Therefore, 6.31874 kg = 6.32 kg rounded off to 3 significant figures.
(c) Given that 44.422 km. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 2 is less than 5. So, the digit 4 remains 4, and the digits 2, and 2 disappear.
Therefore, 44.422 km = 44.4 km rounded off to 3 significant figures.
(d) Given that 60.001 cm. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 0 is less than 5. So, the digit 0 remains 0, and the digits 0, and 1 disappear.
Therefore, 60.001 cm = 60.0 cm rounded off to 3 significant figures.
(e) Given that 0.0023596 m. It has 5 significant figures. To round off the given number to 3 significant digits, we need to round it off to 5 places after the decimal.
The digit 9 is greater than 5. So, the digit 5 becomes 6, and the digits 9, and 6 disappear.
Therefore, 0.0023596 m = 0.00236 m rounded off to 3 significant figures.
(f) Given that 0.0024030 l. It has 5 significant figures. To round off the given number to 3 significant digits, we need to round it off to 5 places after the decimal.
The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3, and 0 disappear.
Therefore, 0.0024030 l = 0.00240 l rounded off to 3 significant figures.


3. Round off
(a) 16.367 g correct to the three significant figures.
(b) 0.00949 mg correct to the two significant figures.
(c) 0.005618 g correct to the one significant figures.
(d) 28.303 mm correct to the four significant figures.
(e) 33.422 km correct to the four significant figures.
(f) 4.0832 kg correct to the two significant figures.
(g) 0.004628 mm correct to the one significant figures.

Solution:

(a) Given that 16.367 g. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 6 is greater than 5. So, digit 3 becomes 4, and the digits 6, and 7 disappear.
Therefore, 16.367 g = 16.4 g rounded off to 3 significant figures.
(b) Given that 0.00949 mg. It has 3 significant figures. To round off the given number into 2 significant digits, we need to round it off to 4 places after the decimal.
The digit 9 is greater than 5. So, the digit 4 becomes 5, and the digit 9 disappear.
Therefore, 0.00949 mg = 0.0095 mg rounded off to 2 significant figures.
(c) Given that 0.005618 g. It has 4 significant figures. To round off the given number into 1 significant digit, we need to round it off to 3 places after the decimal.
The digit 6 is greater than 5. So, the digit 5 becomes 6, and the digits 6, 1, and 8 disappear.
Therefore, 0.005618 g = 0.006 g rounded off to 3 significant figures.
(d) Given that 28.303 mm. It has 5 significant figures. To round off the given number into 4 significant digits, we need to round it off to 2 places after the decimal.
The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3 disappear.
Therefore, 28.303 mm = 28.30 mm rounded off to 4 significant figures.
(e) Given that 33.422 km. It has 5 significant figures. To round off the given number to 4 significant digits, we need to round it off to 2 places after the decimal.
The digit 2 is less than 5. So, digit 2 remains 2, and the digits 2 disappear.
Therefore, 33.422 km = 33.42 km rounded off to 4 significant figures.
(f) Given that 4.0832 kg. It has 5 significant figures. To round off the given number to 2 significant digits, we need to round it off to 1 place after the decimal.
The digit 8 is greater than 5. So, the digit 0 becomes 1, and the digits 8, 3, and 2 disappear.
Therefore, 4.0832 kg = 4.1 kg rounded off to 2 significant figures.
(g) Given that 0.004628 mm. It has 4 significant figures. To round off the given number to 1 significant digit, we need to round it off to 3 places after the decimal.
The digit 6 is greater than 5. So, the digit 4 becomes 5, and the digits 6, 2, and 8 disappear.
Therefore, 0.004628 mm = 0.005 mm rounded off to 1 significant figure.


4. Round off
(a) $ 3067.665 to the nearest cents.
(b) 0.00588 m to the nearest cm.
(c) 18.0333 kg to the nearest g.
(d) $ 49.63 to the nearest dollar.

Solution:

(a) Given that $ 3067.665. It has 7 significant figures. To round off the $ 3067.665 to the nearest cents, we need to round it off to 3 places after the decimal.
The digit 5 is equal to 5. So, digit 6 becomes 7, and the digit 5 disappear.
Therefore, $ 3067.665 = $ 3067.67
(b) Given that 0.00588 m. It has 3 significant figures. To round off the 0.00588 m to the nearest cm, we need to round it off to 3 places after the decimal.
The digit 5 is equal to 5. So, the digit 5 becomes 6, and the digit 8, and 8 disappear.
Therefore, 0.00588 m = 0.01 m.
(c) Given that 18.0333 kg. It has 6 significant figures. To round off the 18.0333 kg to the nearest g, we need to round it off to 3 places after the decimal.
The digit 3 is less than 5. So, the digit 3 remains 3, and the digits 3 disappear.
Therefore, 18.0333 kg = 18.033 kg.
(d) Given that $ 49.63. It has 4 significant figures. To round off the $ 49.63 to the nearest dollar, we need to round it off to the nearest number.
The digit $ 49.63 is closer to $ 50.
Therefore, $ 49.63 = $ 50.


Worksheet on Three Dimensional Figures | Three Dimensional Shapes Practice Worksheets

We are here to help all the people who wish to learn the Three Dimensional Figures concepts. We have given detailed explanations and also different problems on 3-Dimensional objects along with explanations. Also, students can easily improve their preparation level by practicing with Worksheet on Three Dimensional Figures. Use the Three Dimensional Figures Worksheets and try to answer the questions on your own. Solving the 3D Figures Problems regularly helps you to improve your accuracy and speed in the final exams.

Objective Questions on Three Dimensional Figures

1. Mention the number of faces for each of the given figures

(i) Cuboid
(ii) Tetrahedron
(iii) Triangular prism
(iv) Cube
(v) Square pyramid

Solution:

1. (i) 6
(ii) 4
(iii) 5
(iv) 6
(v) 5
Explanation:
(i) A Cuboid is made up of six rectangles, each of the rectangles is called the face.
(ii) A tetrahedron has 4 faces all are triangles.
(iii) A triangular prism consists of 2 triangular faces and 3 rectangular faces. Therefore, total the triangular prism has 5 faces.
(iv) A cube consists of 6 faces.
(v) A square pyramid consists of faces one of which is a square face and the rest four are triangular faces.


2. Write down the number of edges of each of the following figures
(i) Tetrahedron
(ii) Triangular prism
(iii) Cube
(iv) Rectangular pyramid

Solution:

1. (i) 6
(ii) 9
(iii) 12
(iv) 8
Explanation:
(i) A tetrahedron has 6 edges.
(ii) A triangular prism consists of 9 edges. If PQRSTV is a triangular prism, then the 9 edges of the triangular prism are PQ, QR, RP, ST, TV, VS, PS, QT, RV.
(iii) A cube consists of 12 edges. If PQRS is a cube, then the 12 edges of the cube are PQ, QR, RS, SP, EF, FG, GH, HE, PE, SH, QF, RG.
(iv) A rectangular pyramid consists of 8 edges. If the rectangular pyramid OPQRS has 8 edges, then PQ, QR, RS, SP, OP, OQ, OR, and OS are edges.


3. Write down the number of vertices of each of the following figures
(i) Tetrahedron
(ii) Square pyramid
(iii) Cuboid
(iv) Triangular prism

Solution:

1. (i) 4
(ii) 5
(iii) 8
(iv) 6
Explanation:
(i) A tetrahedron has 4 vertices.
(ii) A square pyramid consists of 5 vertices. If OPQRS is a square pyramid having O, P, Q, R, S as its vertices.
(iii) A cuboid has 8 vertices. If PQRS is a cuboid, then the 8 vertices of the cuboid are P, Q, R, S, E, F, G, H.
(iv) A triangular prism consists of 6 vertices. If PQRS is a triangular prism, then the 6 vertices of the triangular prism are P, Q, R, S, T, V.


4. Fill in the blanks
(i) A triangular pyramid is called a …………….. .
(ii) The point at which three faces of a figure meet is known as its ……………..
(iii) A cube has …………….. vertices, …………….. edges and …………….. faces.
(iv) A cuboid is also known as a rectangular …………….

Solution:

1. (i) tetrahedron
(ii) vertex
(iii) 8, 12, 6
(iv) prism


5. Which of the following has a single vertex?
(i) Sphere and Cylinder
(ii) Sphere
(iii) Cylinder
(iv) Cone

Solution:

The answer is Cone. A cone has only a single vertex. The cone decreases smoothly from the circular flat base to the top point.


6. Which of the below statements are not correct?
(i) Each face of a polyhedron is a polygon.
(ii) The faces of a polyhedron are all flat.
(iii) A polyhedron is a three-dimensional figure.
(iv) A cube and a cone are both polyhedrons.

Solution:

(iv) A cube and a cone are both polyhedrons.


7. A ………………… is a space figure with six faces, 8 vertices, and its opposite sides are parallel.
(i) Triangular Prism
(ii) Pyramid
(iii) Rectangular Prism
(iv) Cylinder

Solution:

(iii) Rectangular Prism

A Rectangular Prism is a space figure with six faces, 8 vertices, and its opposite sides are parallel.


8. How many faces does a triangular prism have?
(i) Five
(ii) Four
(iii) Three
(iv) Two

Solution:

(i) Five
A triangular prism has Five faces.


Worksheet on Simplification of Algebraic Expressions | Simplifying Expressions and Equations Worksheet

Detail worksheet on simplification is here. Check simplification example problems and solutions here. Refer step by step procedure to find the accurate solution on fundamental expressions and simplifications. Simplify all the algebraic expressions and rules to solve problems.

Solve many mock questions based on the simplification of algebraic expressions and integers from Simplification of Algebraic Expressions Worksheet to get perfection and crack the exam easily. Before solving the below problem, check our previous articles to know the formulae, rules, and methods. Follow the below sections to know various model questions and solutions.

Problem 1:

After striking a floor, a certain ball rebounds 4/5th of the height from which it has fallen. What is the total distance that it travels before coming to rest if it is gently dropped from a height of 120 meters?

Solution: 

As given in the question,

The height at which the ball rebounds = 4/5

The height at which the ball drops = 120 meters

Let the distance be n

The total distance it travels to come to rest position = 120 + 120 + n/1 – n

= 120 + 120 + (4/5 )/1-(4/5)

= 120 + 96/ 1-4/5

= 1080 kms

Therefore, the total distance is 1080 kms.


Problem 2:

8 years hence the sum of A’s and B’s age is 70 years. 4 years ago, the ratio between the sum of ages of A and B together and C’s age was 2:1. What is the present age of C?

Solution: 

As given in the question,

The sum of A’s and B’s age is 70 years.

A + B = 70

As given, the ratio between the sum of ages of A and B together and C ‘s age was 2:1

The equation will be A + B – 8/C-4 = 2/1

23 = C-4

C = 17

Therefore, the present age of 17 years


Problem 3:

The income of a company doubles after one every year. If the initial income was Rs. 4 lakhs. What would be the income after 5 years?

Solution: 

As given in the question,

Incomes double every year

The initial income = Rs 4 lakhs

After the first year, the income will become = 4 * 2 = 8 lakhs

After the 2 nd year, the income will become = 8 * 2 = 16 lakhs

After the 3 rd year, the income will become = 16 * 2 = 32 lakhs

After the 4th year, the income will become = 32 * 2 = 64 lakhs

After the 5th year, the income will become = 64 * 2 = 128 lakhs

Hence, the income after 5 years = 128 lakhs


Problem 4:

Man earns Rs 20 on the first day and spends Rs 15 on the next day. He again earns Rs 20 on the third day and spends Rs 15 on the next day. If he continues to save like this, how soon will he have Rs 60 in his hand?

Solution: 

As given in the question,

The amount of money man earns = Rs 20

The amount of money man spent = Rs 15

Therefore, for every 2 days, he saves Rs 5.

For the first 2 days, the amount he saves = Rs 5

For the first 4 days, the amount he saves = Rs 10

For 8 days, he saves = Rs 20

For 16 days, he saves = Rs 40

Therefore, he saves Rs 60 in 18 days

Thus, he takes 18 days to save Rs 60.

Hence, the final solution is 18 days.


Problem 5:

If Raju have 24 hamburgers and he eats half of them in one hour, then his dog eats half of what was left in one hour, how many hamburgers do he have left?

Solution: 

As given in the solution,

No of hamburgers Raju have = 24

Amount of burger he eats in one hour = 1/2

Amount of burger his dog eats = 1/2

Therefore, the no of burgers left = 24 * 1/2 * 1/2

= 24 * 1/4 = 6

Hence, Raju has 6 hamburgers left with him.

Thus, the final solution is 6 hamburgers.


Practice Questions & Worksheet on Simplification

Problem 6:

Number of working days of a school is 5 days per week and I spend 5 hours in class each day.. Boring!!! How much time do I spend in class each week?

Solution: 

As given in the question,

No of hours I spent in each day = 5

No of working days = 5

Therefore, to find the total time spend in class each week, we apply the simplification rule of multiplication.

Hence, the total time spend = 5 * 5 = 25

I attend 25 hours of class in each week.

Thus, the final solution is 25 hours.


Problem 7:

Find the three consecutive even integers such that 6 times the sum of first and the third is 24 and greater than 11 times the second.

Solution: 

As given in the question,

Let the three consecutive even integers = x, x+2,x+4

Also given, 6 times the sum of first and the third is 24 and greater than 11 times the second

Therefore, 6(x+x+4)=11(x+2)+24

6(2x+4)=11x+22+24

2x+24 = 11x + 46

On further simplification, x= 22

Then, x+2 = 24

and x+4 = 26

Therefore, the three consecutive even integers are 22,24 and 26

Hence, The final solution is 22,24 and 26


Problem 8:

Leonard found four consecutive integers such that 3 times the sum of the first and fourth was 114 less than the product of -5 and the sum of the first two. What were the integers?

Solution: 

Let the four consecutive integers be x, x+1,x+2,x+3

As given in the question,

3 times the sum of the first and fourth was 114 less than the product of -5 and the sum of the first two

=3(x+x+3) = -5(x+x+1)-144

=3(2x+3) = -5(2x+1)-144

=6x+9 = -10x-5-144

=6x+9 = -10x-119

=16x+9 = -119

=16x/16 = -128/16

= x=-8

As considered, the consecutive integers are x, x+1,x+2,x+3

Therefore, the integers are -8,-7,-6,-5


Problem 9:

The sum of 2 numbers is 36 and their product is 248. What will be the sum of their reciprocals?

Solution: 

As per the given question,

A+B=36

A * B = 248

To find the sum of their recriprocals, we have to simplify the equation 1/A+1/B

=B+A/BA = 36/248 = 9/62

Therefore the sum of their reciprocals = 9/62


Problem 10:

In a poultry farm having hens and pigs, Rohan can see 84 heads and 282 legs. How many hens are there?

Solution: 

As per the given question,

Every hen has 1 head and 2 legs

Every pig has 1 head and 4 legs

Therefore, hens + pigs = 84 is the first equation

2(Hens) + 4 (Pigs) = 282 is the second equation

On simplifying both the equations, we get

4Hens + 4Pigs – 2Hens – 4Pigs =336-282

2Hens = 54

Hens = 27

Therefore, 27 hens are present in the poultry farm.

The final solution is 27 hens.


Problem 11:

The summation of 5 consecutive numbers is found out to be 335. If we add the largest and smallest number what will we get?

Solution: 

Let the numbers be x,x+1,x+2,x+3,x+4,x+5 = 335

5x + 10 = 335

5x = 325

x = 65

Therefore, the numbers are 65,66,67,68,69

To find the sum of largest and smallest number, we have to add 65 and 69

Therefore, the solution is 134

Hence, the final solution of adding the largest and smallest number is 134.


Problem 12:

Since Raj was not paying attention in class, instead of multiplying M by 3/4, he divided by 3/4. This led to the difference of 14 between the two answers. What is the value of M?

Solution: 

As given in the question,

The number that has to be multiplied = 3/4

The number Raj divided = 3/4

The difference of two answers = 14

To find the value of M, we simplify the equation as

M*3/4 – M/3/4 = 14

3M/4 – 4M/4 = 14

9M-16M/12 = 14

-7M/12 = 14

M=-24

Hence, the final solution is -24


Problem 13:

Raman has 2 urns. Both these urns have the same pebbles. If 20 pebbles from urn B are shifted to urn A, then the number of pebbles in both urns get interchanged. But if 10 pebbles from urn A are put into urn B, then the number of pebbles in urn B is twice the number of pebbles in urn A. How many pebbles do A and B respectively have?

Solution: 

As given in the question,

No of urns Raman has = 2

Also given,

20 pebbles from urn B are shifted to urn A, then the number of pebbles in both urns get interchanged and 10 pebbles from urn A are put into urn B, which are twice the number of pebbles in urn A

On simplifying the equation, we get

B-20 = A

2(A-10) = (B+10)

2A-20 = B+10

2A-B = 30

B-A+2A-B = 20+30

A = 50

Therefore, B = A + 20 = 50 + 20 = 70

No of pebbles A and B have are 50 and 70

Therefore, the final solution is 50 and 70 respectively.


Problem 14:

On exchanging the digits in unit’s and ten’s place, the difference between original and new numbers become 27. The digits in the unit place are 2 times the digit in the hundred’s place. The digit in the ten’s place is 3 times the digit in the hundred’s place. What is 75% of the original number?

Solution: 

As given in the question,

U = 2H

H = 3H

(3*100+4*10+7*1)

(H*100+T*10+U*1)

100H +30H + 2H = 132H

100H + 10T + U

100H + 20H + 3H = 123H

132H -123H = 27

9H = 23

H = 3

Also given,

U = 6

T = 9

HTU = 396

Therefore, 75% of the original number = 396

Thus, the final solution is 396


Problem 15:

In an area 2% of families have 5 children each. But 8% have no children at all. Amongst the rest of the families 18% have 4 children and 27% have only one child. How many families live in the area, if 297 families have either 2 or 3 children each?

Solution: 

As given in the question,

Percentage of families having 5 children = 2%

Percentage of families having no children = 8%

Total percentage of families = 2% + 8% = 10%

Consider there are 100% families

Therefore, no of families with above data = 100=10 = 90 families

Percentage of families having 4 children = 18%

Percentage of families having 1 child = 27%

Total no of families in those 90 families = 45% of 90

To find the families with 2 or 3 children

100%-45% = 55% of 90

55/100* 90 = 99/2 = 44.5 families

For 100% of families, 44.5 families have children

To find the families who have either 2 or 3 children

= 297 * 100/49.5 = 600

Therefore, 600 families have either 2 or 3 children.


Problem 16:

A group wanted to renovate their club. Each member contributed an amount equal to twice the number of members in the club. But the government contributed same amount as the number of members. If each member had contributed the same amount as the number of numbers and the government had given the twice amount of the members, then they would have Rs.210 less. How many members are there?

Solution: 

M*2M+M

(2M2 + M) – (M2 + 2M)=210

M2 – M = 210

(M – 15) (M + 14) = 0

M * M = 2M

M2 + 2M

M-15 = 0

M = 15

Therefore, there were 15 members in the group who wanted to renovate their club.

The final solution is 15 members.


Worksheet on Fundamental Operations | Four Fundamental Operations Worksheet

Worksheet on Fundamental Operations and solved examples are here. Check the practice test papers and problems involving various fundamental operations. Know the methods and applications of four fundamental properties by solving Questions on 4 Fundamental Operations. Refer to Problems on addition, subtraction, multiplication, division operations. You can find Step by Step Solutions to all the Problems in the Fundamental Operations Worksheet. Go through the below sections to know various problems involving fundamental operation methods.

Problem 1: 

Two lakh sixty-three thousand nine hundred fifty-three visitors visited the trade fair on Sunday, four lakh thirty-three thousand visited on Monday, and three lakh twenty thousand six hundred fifty-six visited on Tuesday. How many visitors in all visited the trade fair in three days?

Solution:

As per the question,

Visitors visited the trade fair on Sunday = 2,63,953

The visitors visited the trade fair on Monday = 4,33,000

Visitors visited the trade fair on Tuesday = 3,20,656

To find the no of visitors visited the trade fair in three days, we apply rule of addition

Therefore, no of visitors = 2,63,953 + 4,33,000 + 3,20,656 = 10,17,609

Hence, 10,17,609 visitors visited the trade fair in three days.

Thus, the final solution is 10,17,609


Problem 2:

The weight of 298 bags of wheat is 29204 kg. Find the weight of such 125 bags of wheat?

Solution:

As per the question,

The weight of wheat bags = 29204 kg

No of bags = 298

To find the weight of each bag, we apply the law of division rule.

Therefore, the weight of each bag = 29204/298 = 98 kgs

To find the weight of 125 bags, we apply the law of multiplication

Therefore, the weight of 125 bags = 125 * 98 = 12250

Thus, 125 bags weighs 12250 kgs.

Hence, the final solution is 12250 kgs.


Problem 3:

In a garment manufacturing unit, vests are packed in packets of 6 pieces. Ten such packets are then bundled in a carton. How many cartons are required to pack 540 vests?

Solution:

As given in the question,

Total vests = 540

Vests in a packet = 6

Packets in cartons = 10

Number of packets = 540/6 = 90

Number of cartons = 90/10 = 9

Therefore, 9 cartons are required to pack 540 vests.

Hence, the final solution is 9 cartons.


Problem 4:

A bookseller sold 56.248 copies in the first year. The sales doubled in the second year. How many copies of the books were sold in the first two years?

Solution:

As given in the question,

Sales of copies in the first year = 56, 248

Sales of copies in the second year = 56,248 * 2 = 1,12,496

Now, to find the number of copies sold in the first two years , we apply addition rule

Therefore, the number of copies in first 2 years = 56,248 + 1,12,496 = 1,68,744

Thus, the final solution is 1,68,744 copies


Problem 5:

There were 25,000 participants in a mention. One-fourth of them were above 50 years of age. How many of the participants were in the age group of 50 years or less?

Solution:

As given in the question,

Total number of participants = 25,000

Participants above the age of 50 years = 1/4 of 25,000 = 6250

To find the  participants age of 50 years or less, we apply law of subtraction

Therefore, participants age of 50 years = 25000 – 6250 = 18,750

Hence, 18,750 participants were in the age group of 50 years or less.

Thus, the final solution is 18,750 participants.


Problem 6:

The total production of natural rubber in India during three years was 49400 kgs. If the production during two years was respectively. 152870 kgs and 165850 kgs. Find the production of natural rubber during third year?

Solution:

As given in the question,

Total production in three years = 494000 kgs

Total production in two years = 152870 kg + 165850 kgs = 318720 kgs

Production during third year = Total production – Production in two years

= 494000 – 318720 = 175280 kgs

Production of natural rubber during third year = 175280 kgs

Hence, 175280 kgs is produced during third year.

Thus, the final solution is 175280 kgs.


Problem 7:

Ashok packs 580368 apples in 428 boxes. How many apples will he pack in 515 boxes?

Solution:

As per the question,

Ashok packs no of apples = 580368

No of boxes = 428

To find apples in one box = 580368 / 428 = 1356 apples

For 515 boxes = 1356 * 515 = 698340 apples

Therefore, he will pack 6,98,340 apples in 515 boxes.

Thus, the final solution is 6,98,340 apples


Problem 8:

A factory produced 1188440 bulbs in one year. How many bulbs did it produce in the month of August?

Solution:

As given in the question,

Production in one year = 1188440 bulbs

Production in one day = 1188440 / 365 = 3256 bulbs

Number of days in the month of August = 31

Production in the month of August = 3256 * 31 = 1,00,936

Therefore, 1,00,936 bulbs were produced in the month of August.


Problem 9:

Fernando opened a pizza box. Inside there was 3/4 of a pizza. Fernando ate 1/2 of what was remaining. How much of a pizza did Fernando pizza?

Solution:

As given in the question,

The amount of pizza = 3/4

Amount of pizza Fernando ate = 1/2

Therefore, the amount of pizza = 3/4 * 1/2 = 3/8

Each pizza has 6 slices, hence amount of pizza Fernando pizza

= 1/2 * 6 = 3

Thus, 3 parts of pizza Fernando ate

Hence, the final solution is 3 parts.


Problem 10:

Two brothers had a total of 24 oranges. The first brother ate 1/3 of the oranges. The second brother ate 1/4 of the oranges. How many oranges did they eat together?

Solution:

As per the question,

No of oranges both the brothers ate = 24

The amount of oranges first brother ate = 1/3

The amount of oranges second brother ate = 1/4

Amount of oranges first brother ate alone = 24 * 1/3 = 24/3 = 8

Amount of oranges second brother ate alone = 24 * 1/4 = 24/4 = 6

Therefore, two brothers ate = 8+6 = 14 oranges

Thus, the final solution is 14 oranges.


Problem 11:

The price of the cycle is Rs.5699. Find the price of 17 cycles?

Solution:

As given in the question,

The price of each cycle = Rs.5699

To find the price for 17 cycles, we have to apply the fundamental operation of multiplication.

Therefore, the price of 17 cycles = 5699 * 17 = 96883

Hence, the total price for 17 cycles is Rs. 96,883

Thus, the final solution is Rs. 96,883


Problem 12:

An employee earns Rs. 65596 in the month of March. How much is he earning in a single day?

Solution:

As given in the question,

An employee earns in March month = 65596

No of days present in March = 31

To find the earnings of each day separately, we have to apply the fundamental rule of division

Therefore, his earnings in March for each day = 65596/31 = 2116

Hence, he earns Rs. 2116 per day.

Thus, the final solution is Rs. 2116


Problem 13:

At a furniture store, I bought 2 tables and 4 chairs. If each table cost $79 and each chair cost $29. How much did I spend in all?

Solution:

As given in the question,

No of chairs I bought = 4

No of tables I bought = 2

The price of each table = $79

The price of each chair = $29

The total price of tables = 2 * 79 = 158

The total price of chairs = 4 * 29 = 116

The final amount of money spent = 158 + 116 = $274

Therefore, $274 was spent on buying chairs and tables.

Thus, the final solution is $274


Problem 14:

In an aquarium, there were 3 large fish tanks and 5 small fish tanks. Each large tank had 82 fish inside, and each small tank had 20 fish inside. How many fish does the aquarium have?

Solution:

As given in the question,

No of large fish tanks = 3

Number of small fish tanks = 5

No of fish inside the large tank = 82

Number of fish inside the small tank = 20

Total no of fishes in the large tanks = 3 * 82 = 246

Total no of fishes in the small tanks = 5 * 20 = 100

Therefore, the total number of fish in the aquarium = 346

Thus, there are 346 fish in the aquarium.

Hence, the final solution is 346 fish.


Problem 15:

At the grocery store, Meena bought 4 bags of potatoes and 3 block of cheese. Each bag of potatoes cost $17. Each block of cheese cost $14. How much did I spend in all?

Solution:

As given in the question,

No of bags of potatoes = 4

No of blocks of cheese = 3

The price of potatoes = 17

The price of cheese = 14

The total no of potatoes now available = 4 * 17 = 68

The total amount of cheese now available = 3 * 14 = 42

Therefore the total amount spent = 68 + 42 = 110

Hence, the final solution is $110


Problem 16:

Your basket team made 10 three-pointers, 8 two-point field goals, and 12 one-point foul shots. Your opponents made 8 three-pointers, 12 two-point field goals, and 11 one-point foul shots. Who won? By how many points did they win?

Solution:

As given in the question,

The basketball team makes pointers = 10

No of pointers = 3

Total points for the team = 10 * 3 = 30

No of field goals = 8

No of points = 2

Total no of field goals = 8 * 2 = 16

No of foul shots = 1

No of points = 1

Total points for the basket team = 30 + 16 + 12 = 58

The points for the opponent team is as follows.

No of three-pointers = 8

No of points = 3

Total three-pointers = 8 * 3 = 24

No of two-point field goals = 12

No of points = 2

Total two-point field goals = 12 * 2 = 24

No of foul shots = 11

No of points = 1

Total one-point foul shots = 1 * 11 = 11

Total points for opponent team = 24 + 24 + 11 = 59

Therefore, considering the total points of the team

Basketball team scored 58 points and the opponent team scored 59 points in total.

Therefore, the Opponent team has greater points and hence they win.

No of points by which the opponent team wins = 59-58 = 1

Therefore, the opponent wins by 1 point.

Hence the final solution is opponent team wins and it wins by 1 point.


Problem 17:

Latha saved Rs.16785 last year. Her father gave her Rs. 4325 more. How much money does she have now?

Solution:

The amount of money Latha saved = Rs. 16785

The amount of money her father gave her = Rs. 4325

To find the total money she saved, we apply the fundamental operation of addition

Therefore, the amount of money she saved = 16785 + 4325 = 21110

Thus, she saved Rs. 21110

Hence, the final solution is Rs. 21110


Problem 18:

Vinay has 15,180. If he spends Rs. 1690 for shopping. How much will he be left with?

Solution:

As per the solution,

The amount of money Vinay spent = Rs. 15,180

The amount of money he spends for shopping = RS. 1690

To find the amount he left with, we apply the fundamental operation of subtraction.

Therefore, the amount of money he spent with = 15,180 – 1,690 = 13,490

Thus, the final solution is Rs. 13, 490


Problem 19:

If 4095 sample copies of a book were distributed among 365 dollars. How many books did each dealer get?

Solution:

No of copies of a book = 4095

Amount of dollars = 365

To find the no of books each dealer get, we apply the fundamental law of division

Therefore, no of books dealer get = 4095/365 = 11 books

Thus, the final solution is 11 books.


Problem 20:

A company produces 19216 bikes every month. How many bikes will be produced in a year?

Solution:

As given in the question,

No of bikes company produces = 19.216

No of months in a year = 12

To find the production of bikes for a year, we apply the fundamental operation of multiplication.

Therefore, no of bikes in a year = 19216 * 12 = 230592

Thus, 230592 bikes are produced in a year.

Hence, the final solution is 230592 bikes.


Worksheet on Division of Integers | Division of Integers Worksheet

In the given Worksheet on Division of Integers, you can find Problems on Integers Division. Find Step by Step Solutions for all the problems. Refer to all types of questions involved in the division of integers. Check solved examples and know the procedure followed to solve the problems. Go through the below sections, to know the practice tests and example problems. Try to solve as many times as possible so that your accuracy and speed will be increased. Thus, you can attempt the exam with confidence and answer the questions on Division of Integers easily.

Question 1:

If the temperature is dropping at a rate of 6 degrees per hour. How many hours will it take for the temperature to drop 30 degrees?

Solution: 

As given in the question,

The rate at which the temperature is dropping = 6 degrees per hour.

To find the hours it takes to drop the temperature to 30 degrees, we have to go for the division law of integers.

Therefore, change in temperature = -30/-6

= 5 hours

Thus, it takes 5 hours for the temperature to drop by 30 degrees per hour.

The final solution is 5 hours


Question 2:

If a stock is losing value at a rate of $8 per day. How many days before the stock has lost a value of $48?

Solution: 

The rate at which the stock is losing value = $8

To find the days that stock lost the value = $48, we have to apply the division of integers.

Therefore, no of days = -48/-8 = 6

Thus, it takes 6 days before the stock has lost the value of $48.

The final solution is $48


Question 3:

If a group of hikers is descending at a rate of 15 feet per hour down a mountain. How many hours will it take for the group of hikers to descend 105 feet?

Solution: 

The rate at which the hikers are descending = 15 feet per hour

To find time for a group of hikers to descend 105 feet, we apply the law of division.

Therefore, the time = -105/-15 = 7

Hence, the group of hikers descend 15 feet in 7 hours.

The final solution is 7 hours


Question 4:

If a bucket full of water is evaporating at a rate of 3 cm per hour. How long before the bucket has evaporated a total of 18cm?

Solution: 

The rate at which the bucket full of water is evaporated = 3 cm per hour

To find the time of the bucket evaporated a total of 18cm, we apply the division of integers.

Therefore, the total time = -18/-3 = 6

Thus, it takes 6 hours for the bucket to evaporate 18cm.

Hence, the final solution is 6 hours.


Question 5:

The product of 2 integers is 270. If one of the integers is (-18). Find the other one?

Solution: 

Let the other integer be x

As given in the question, the product of 2 integers is 270

x * (-18) = 270

x = 270/18

x = -15

Therefore, the final solution is -15


Question 6:

Find an integer which when multiplied by 4 and then divided by 9 becomes (-28)?

Solution: 

Let the integer be x

x * 4 / 9 = -28

4x / 9 = -28

4x = -28 * 9

x = -28 * 9 / 4

x = -63

Therefore, the integer which when multiplied by 4 and then divided by 9 becomes (-28) is -63


Question 7:

If the quotient obtained on dividing on integer by -9 is 8, Find the integer?

Solution: 

Let the integer be x

x / -9 = -8

x = -8 * -9

x = 72


Question 8:

A shopkeeper earns a profit of Rs 1 on selling one pen and suffers a loss of 30 paise on selling one pencil in a particular month, he incurs a lot of Rs. 5. In that month, he sold 40 pens. How many pencils did he sell in that period?

Solution: 

Let the no of pencils he sell in that month be x

No of pens he sell in that month = 40

40 * (1) + x * (-30/100) = -5

40 – 3x/10 = -5

40 + 5 = 3x/10

45 = 3x/10

x = 10 * 45/3

x = 150


Question 9:

The population of a small town is changing at rate of -255 people per year. How long will it take for the change in population to be -2040 people?

Solution: 

The rate of the population changing per year = -255

To find the change in population to be -2040 people, we have to apply division rule.

Therefore, the years for the change of population = -2040/-255 = 8

Hence, the total years = 8 years

The final solution is 8 years.


Question 10:

During a six-hour period, the temperature dropped 18 degrees. How much did the temperature change per hour?

Solution: 

Time of the period = 6 hours

Drop in the temperature = 18

To find the temperature change, apply the division rule.

Therefore, the total change = -18/6 = -3 degree

Hence, the temperature changed by 3 degrees per hour.


Question 11:

A stock decreased in value by $80 during the five days. How much did the stock decrease each day?

Solution: 

As given in the question,

The value at which the stock decreased = $80

No of days = 5

To find the stock decreased each day we apply division of integers.

Therefore, the stock decreased = -80/5 = -16


Question 12:

The outside temperature is -20degree F and rising at a rate of 5 degrees per hour. How long will it be before the temperature reaches 0 degrees F?

Solution: 

The temperature outside = -20 degree F

The rate of rising the temperature = 5 degrees per hour

To find how long will it be before the temperature reaches 0 degrees F, we apply the division of integers.

Therefore, the time taken = 20/5 = 4

Hence, it takes 4 hours to reach 0 degrees F


Question 13:

Judges in some figure skating competitions must give a mandatory 5 point deduction for each jump missed during the technical part of the competition. Marisa has participated in 5 competitions this year and has been given a total of -20 points for jumps missed. How many jumps did she miss?

Solution: 

Total no of points = -20

No of participants in the competition = 5

Therefore, total number of jumps missed = -20/(-5) = 4 jumps

The final solution is 4 jumps


Question 14:

Miranda is an excellent spinner who averages +3 points on her spins during competitions. Last year her total spin points equaled +21.

About how many spins did she successfully complete?

Solution: 

No of average spins = 3

Total spin points = 21

Therefore, no of spins she completed = 21/3 = 7 spins

Hence, the final solution is 7 spins.


Question 15:

The temperature dropped 32 degrees F in 4 hours. Suppose the temperature dropped by an equal amount each hour. What integer describes the change?

Solution: 

As given in the question,
Temperature drop = 32 degrees F
Time taken = 4 hours

Therefore, to find the integer that describes the change, we apply the division rule.

Hence, the time taken in 4 hours = -32/4 = -8 degrees F

The final solution is -8 degree F


Question 16:

A stock market fell 60 points over a period of 4 days. What was the average change in the stock market per each day?

Solution: 

As given in the question,

No of points fell in stock market = 60

Time of period = 4 days

To find the average change in the stock market per day, we apply the division method.

Therefore, the average change = 60/4 = -15

Hence, the final solution is every 15 points change in the stock market per day.


Question17:

In a lab, a substance was cooled by 36 degrees over a period of 6 hours at a constant rate. What was the change in temperature each hour?

Solution: 

As given in the question,

The temperature at which it was cooled = 36 degrees

Period of time = 6 hours

To find the change in temperature each hour, we apply the law of division rule.

Therefore, the change in temperature = 36/6 = 6

Hence, for every hour there will be a change of 6 degrees.

Thus, the final solution is 6 degrees.


Question 18:

The outside temperature is -20 degrees F and raising at the rate of 5 degrees per hour. How long will it be before the temperature reaches 0 degrees F?

Solution: 

As per the given solution,

The outside temperature = -20 degree F

The rate at which the temperature is rising = 5 degrees per hour

To find the temperature to reach 0 degrees F, we apply the division law of integers

Therefore, the change in temperature = 20/5 = 4 hours

Hence, it takes 4 hours before the temperature reaches 0 degree F

Thus, the final solution is 4 hours


Question 19:

Judges in some figure skating competitions must give a mandatory 5 point deduction for each jump missed during the technical part of the competition. Marisa has participated in 5 competitions this year and has been given a total of -20 points for jumps missed. How many jumps did she miss?

Solution: 

As given in the question,No of deduction points = 5 pointsNo of competitions Marisa participated in = 5Total points for jumps missed = -20

To find the jumps she missed, we apply the division rule.

Therefore, the jumps missed = -20/-5 = 4

Hence, Marisa missed 4 jumps in 5 competitions

Thus, the final solution is 4 jumps.


Question 20:

Mirinda is an excellent spinner who average +3 points on her spins during competitions. Last year her total spin points equaled +21. About how many spins did successfully complete?

Solution: 

As given in the question,

The average points on spin = +3

Total points equalled = +21

To find the spins that successfully completed, we apply the division law of integers.

Therefore, the spins completed = 21/3 = 7

Thus, the final solution is 7 spins.


Worksheet on Different Types of Quadrilaterals | Types of Quadrilaterals Worksheets

Solve all problems available on Worksheet on Different Types of Quadrilaterals to get good marks in the exam. You can learn different methods to solve a single problem on the Quadrilateral Practice Worksheets. Check out Free Quadrilateral Worksheets on our website and get a grip on complete Quadrilateral concepts in minutes. Practice using the Types of Quadrilaterals Worksheets available and be familiar with various questions. For your convenience, we even provided step by step solutions to all the problems making it easy for you to understand the problems.

Solved Problems on Different Types of Quadrilaterals

1. Construct a parallelogram PQRS in which PQ = 5.2 cm, QR = 4.7 cm and PR = 7.6 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 5.2 cm, QR = 4.7 cm and PR = 7.6 cm.
1. Draw a line segment of length 5.2 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 7.6 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4.7 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point P as a center, draw an arc with a radius of 4.7 cm.
5. By taking the point R as a center, draw an arc with a radius of 5.2 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

PQRS is a required parallelogram.
worksheet on Different Types of Quadrilaterals 1


2. Construct a parallelogram PQRS in which PQ = 4.3 cm, PS = 4 cm and QS = 6.8 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 4.3 cm, PS = 4 cm and QS = 6.8 cm.
1. Draw a line segment of length 4.3 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 4 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 6.8 cm. Mark the point as S where the two arcs cross each other. Join the points Q and S as well as P and S.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point Q as a center, draw an arc with a radius of 4 cm.
5. By taking the point S as a center, draw an arc with a radius of 4.3 cm.
6. Mark the point as R where the two arcs cross each other. Join the points R and S as well as R and Q.

PQRS is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 2


3. Construct a parallelogram ABCD in which BC = 6 cm, AB = 4 cm and ∠ABC = 60°.

Solution:

Steps of Construction:
Given that a parallelogram ABCD in which BC = 6 cm, AB = 4 cm and ∠ABC = 60°.
1. Draw a line segment of length 4 cm and mark the ends as A and B.
2. Take point B as a center and make a point by taking 60º using a protector.
3. Next, take point B as a center and draw an arc by taking the radius 6 cm. Mark the point as C where the point and arc cross each other. Join the points C and B.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point C as a center, draw an arc with a radius of 4 cm.
5. By taking point A as a center, draw an arc with a radius of 6 cm.
6. Mark the point as D where the two arcs cross each other. Join the points D and C as well as D and A.

ABCD is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 4


4. Construct a parallelogram PQRS in which QR = 5 cm, ∠PQR = 120° and RS = 4.8 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which QR = 5 cm, ∠PQR = 120° and RS = 4.8 cm.
1. Draw a line segment of length 5 cm and mark the ends as Q and R.
2. Take point R as a center and make a point by taking 120º using a protector.
3. Next, take point R as a center and draw an arc by taking the radius 4.8 cm. Mark the point as S where the point and arc cross each other. Join the points S and R.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point S as a center, draw an arc with a radius of 5 cm.
5. By taking point Q as a center, draw an arc with a radius of 4.8 cm.
6. Mark the point as P where the two arcs cross each other. Join the points P and S as well as P and Q.

PQRS is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 5


5. Construct a parallelogram PQRS, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side?

Solution:

Steps of Construction:
Given that a parallelogram PQRS, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm.
1. Draw a line segment of length 4.4 cm and mark the ends as P and Q.
2. Make the diagonals half to get the exact vertices of a parallelogram. Take point P as a center and draw an arc by taking the radius 2.8 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 3.5 cm. Mark the point as O where the point and arc cross each other. Join the points P and O, Q and O.
4. Extend the line PO with a radius of 3.5 cm and mark it as R.
5. Extend the line QO with a radius of 2.8 cm and mark it as S.
6. Join the points P and S as well as Q and R, R and S.

PQRS is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 6


6. Construct a parallelogram PQRS in which PQ = 6.5 cm, PR = 3.4 cm, and the altitude PL from P is 2.5 cm. Draw the altitude from R and measure it?

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 6.5 cm, PR = 3.4 cm, and the altitude PL from P is 2.5 cm.
1. Draw a line segment of length 6.5 cm and mark the ends as P and Q.
2. Take point P as a center and draw a perpendicular line PX by taking the radius 2.5 cm and mention that point as L. Draw a parallel line to PQ at a point L.
3. Next, take point P as a center and draw an arc by taking the radius of 3.4 cm. Mark the point as R where the point and arc cross each other. Join the points P and R, Q and R.
4. Take point R as a center and draw an arc by taking the radius of 6.5 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, P and S.

PQRS is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 7


7. Construct a parallelogram PQRS, in which diagonal PR = 3.8 cm, diagonal QS = 4.6 cm, and the angle between PR and QS is 60°.

Solution:

Steps of Construction:
Given that a parallelogram PQRS, in which diagonal PR = 3.8 cm, diagonal QS = 4.6 cm, and the angle between PR and QS is 60°.
1. Draw a line segment of length 3.8 cm and mark the ends as P and R.
2. Bisect the line PR and point it as O.
3. Next, take point O and make a point by taking the angle 60° by taking the O as a center.
4. Take point O as a center and draw an arc by taking the radius of 2.3 cm on both sides of O and name them as Q and S. Join the points PQ, QR, RS, SP.

PQRS is a required parallelogram.
Worksheet on Different Types of Quadrilaterals 7


8. Construct a rectangle PQRS whose adjacent sides are 11 cm and. 8.5 cm.

Solution:

Steps of Construction:
Given that a rectangle PQRS whose adjacent sides are 11 cm and. 8.5 cm.
1. Draw a line segment of length 11 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 90º using a protector and make the point as E.
Rectangle: All angles of a rectangle are 90º.
3. Next, take point P as a center and draw an arc by taking the radius 8.5 cm. Mark the point as S where the point and arc cross each other. Join the points S and P.
4. Take point Q as a center and draw an arc by taking the radius of 8.5 cm.
5. Also, take point S as a center and draw an arc by taking the radius of 11 cm.
Mark the point as R where the two arcs cross each other. Join the points S and R, Q and R.

PQRS is a required rectangle.
Worksheet on Different Types of Quadrilaterals 8


9. Construct a square, each of whose sides measures 5.4 cm.

Solution:

Steps of Construction:
Given that a square, each of whose sides measures 5.4 cm.
1. Draw a line segment of length 5.4 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 90º using a protector and make the point as E.
Square: All angles of a square are 90º.
3. Next, take point P as a center and draw an arc by taking the radius 5.4 cm. Mark the point as S where the point and arc cross each other. Join the points S and P.
4. Take point Q as a center and make a point by taking 90º using a protector. Take point Q as a center and draw an arc by taking the radius of 5.4 cm.
5. Also, take point S as a center and draw an arc by taking the radius of 5.4 cm. Mark the point as R where the two arcs cross each other. Join the points S and R, Q and R.

PQRS is a required square.
Worksheet on Different Types of Quadrilaterals 9


10. Construct a square, each of whose diagonals measures 5.6 cm.

Solution:

Steps of Construction:
Given that a square, each of whose diagonals measures 5.6 cm.
1. Draw a line segment of length 5.6 cm and mark the ends as P and R.
2. Bisect the line PR and take half of its radius 2.8 cm. Take point P as a center and draw an arc by taking the radius 2.8 cm.
3. Next, take point R as a center and draw an arc by taking the radius 2.8 cm. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.
4. Similarly, draw two arcs with 2.8 cms and make the point as Q.
5. Join the points P and Q, S and Q.

PQRS is a required square.
Worksheet on Different Types of Quadrilaterals 10


11. Construct a rectangle ABCD in which BD = 3.6 cm and diagonal AD = 6 cm. Measure the other side of the rectangle.

Solution:

Steps of Construction:
Given that a rectangle ABCD in which BD = 3.6 cm and diagonal AD = 6 cm.
1. Draw a line segment of length 3.6 cm and mark the ends as B and D.
2. Take point B as a center and make a point by taking 90º using a protector and make the point as E.
Rectangle: All angles of a rectangle are 90º.
3. Next, take point C as a center and draw an arc by taking the radius 6 cm. Mark the point as A where the point and arc cross each other. Join the points A and B, A and D.
4. Similarly, draw two arcs and make the point as C.
5. Join the points A and C, D and C.

ABCD is a required rectangle.
Worksheet on Different Types of Quadrilaterals 11


12. Construct a rhombus PQRS when the length measures of the diagonals are 8 cm and 6 cm.

Solution:

Steps of Construction:
Given that a rhombus PQRS when the length measures of the diagonals are 8 cm and 6 cm.
1. Draw a line segment of length 8 cm and mark the ends as P and R.
2. Draw perpendicular bisector XY of PR meeting PR at O.
3. Next, From O cut off OS = 1/2 × 6 cm = 3 cm along OX and OQ = 1/2 × 6 cm =3 cm along OY.
4. Join PQ, QR, RS, and SP.

PQRS is a required rhombus.
Worksheet on Different Types of Quadrilaterals 12


13. Construct a rhombus PQRS in which PQ = 4 cm and diagonal PR is 6.5 cm.

Solution:

Steps of Construction:
Given that a rhombus PQRS in which PQ = 4 cm and diagonal PR is 6.5 cm.
1. Draw a line segment of length 4 cm and mark the ends as P and Q.
2. Take the point Q as a center and draw an arc by taking the radius 4 cm.
3. Take the point P as a center and draw an arc by taking the radius 6.5 cm and name it as R.
4. Join PR and QR.
5. Take the point R as a center and draw an arc by taking the radius 4 cm.
6. Take the point P as a center and draw an arc by taking the radius 4 cm. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.

PQRS is a required rhombus.
Worksheet on Different Types of Quadrilaterals 13


14. Draw a rhombus ABCD whose side is 7.2 cm and one angle is 60°.

Solution:

Steps of Construction:
Given that a rhombus ABCD whose side is 7.2 cm and one angle is 60°.
1. Draw a line segment of length 7.2 cm and mark the ends as A and B.
2. Take point A as a center and make a point by taking 60º using a protector.
3. Next, take point A as a center and draw an arc by taking the radius 7.2 cm. Mark the point as D where the point and arc cross each other. Join the points D and A.
4. Take point D as a center and draw an arc by taking the radius 7.2 cm. Take point B as a center and draw an arc by taking the radius of 7.2 cm.
5. Mark the point as C where the two arcs cross each other. Join the points C and B, C and D.

ABCD is a required rhombus.
Worksheet on Different Types of Quadrilaterals 14


15. Construct a trapezium PQRS in which PQ = 6 cm, QR = 4 cm, RS = 3.2 cm, ∠Q = 75° and SR ∥ PQ.

Solution:

Steps of Construction:
Given that a trapezium PQRS in which PQ = 6 cm, QR = 4 cm, RS = 3.2 cm, ∠Q = 75° and SR ∥ PQ.
1. Draw a line segment of length 6 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 75º using a protector.
3. Next, take point Q as a center and draw an arc by taking the radius 4 cm. Mark the point as R where the point and arc cross each other. Join the points Q and R.
4. RS || PQ, so angle Q + angle R = 180º, angle C = 105º as they are interior angles.
5. Take point R as a center and make a point by taking 105º using a protector.
6. Next, take point R as a center and draw an arc by taking the radius 3.2 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, S and P.

PQRS is a required rhombus.
Worksheet on Different Types of Quadrilaterals 15


16. Draw a trapezium PQRS in which PQ ∥ SR, PQ = 7 cm, QR = 5 cm, PS = 6.5 cm and ∠Q = 60°.

Solution:

Steps of Construction:
Given that a trapezium PQRS in which PQ ∥ SR, PQ = 7 cm, QR = 5 cm, PS = 6.5 cm and ∠Q = 60°.
1. Draw a line segment of length 7 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 60º using a protector.
3. Next, take point Q as a center and draw an arc by taking the radius 5 cm. Mark the point as R where the point and arc cross each other. Join the points Q and R.
4. RS || PQ, so angle Q + angle R = 180º, angle C = 120º as they are interior angles.
5. Take point R as a center and make a point by taking 120º using a protector.
6. Next, take point R as a center and draw an arc by taking the radius 6.5 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, S and P.

PQRS is a required rhombus.
Worksheet on Different Types of Quadrilaterals 16


Quadrilateral Worksheet with Solutions | Quadrilateral Questions and Answers

If you need help solving Quadrilateral questions you can take the help of worksheets present here. Have a glance at the Quadrilateral Worksheet during your practice and take your preparation to the next level. Allot time to practice all the questions available on Printable Worksheet on Quadrilaterals and improve the areas you are lagging. For complete guidance make use of the Quadrilateral Practice Worksheets and clear all your doubts related to them.

We have provided various problems according to the new updated syllabus in the following sections. Solve all of them and check your answers to clear your doubts. For a better understanding of the concept, we have given the step by step explanation for each question.

List of Quadrilateral Worksheets

Check out the below list to know the depth concepts available in Quadrilateral concepts.

Solved Problems on Quadrilateral Worksheet

1. Fill in the blanks:
(i) A quadrilateral has …………… diagonals.
(ii) A quadrilateral has …………… angles.
(iii) How many sides present in a quadrilateral?
(iv) A quadrilateral has …………… vertices, no three of which are……………….
(v) A diagonal of a quadrilateral is a line segment that joins two ……………… vertices of the quadrilateral.
(vi) The sum of the angles of a quadrilateral is ……………….

Solution:

(i) Two
(ii) Four
(iii) Four
(iv) four, collinear
(v) Opposite
(vi) 360°


2. In the adjoining figure, PQRS is a quadrilateral.

Quadrilateral Worksheets

(i) How many pairs of opposite sides are there? Name them.
(ii) How many pairs of adjacent sides are there? Name them.
(iii) Also, find how many pairs of adjacent angles are there? Name them.
(iv) How many diagonals are there? Name them.
(v) How many pairs of opposite angles are there? Name them.

Solution:

(i) two; (PQ, SR), (PS, QR)
(ii) four; (PQ, QR), (QR, RS), (RS, SP), (SP, PQ)
(iii) four; (∠P, ∠Q), (∠Q, ∠R), (∠R, ∠S), (∠S, ∠P)
(iv) two; (PR, QS)
(v) two; (∠P, ∠R), (∠Q, ∠S)


3. Prove that the sum of the angles of a quadrilateral is 360°.

Solution:

In the quadrilateral PQRS,

  • ∠PQR, ∠QRS, ∠RSP, and ∠SPQ are the internal angles.
  • PR is a diagonal
  • PR divides the quadrilateral into two triangles, ∆PQR and ∆PSR

We know that the sum of internal angles of a quadrilateral is 360°, that is, ∠PQR + ∠QRS + ∠RSP + ∠SPQ = 360°.

let’s prove that the sum of all the four angles of a quadrilateral is 360 degrees.
The sum of angles in a triangle is 180°. Now consider triangle PSR,
∠S + ∠SPR + ∠SRP = 180° (Sum of angles in a triangle)
Now consider triangle PQR,
∠Q + ∠QPR + ∠QRP = 180° (Sum of angles in a triangle)
On adding both the equations obtained above we have,
(∠S + ∠SPR + ∠SRP) + (∠Q + ∠QPR + ∠QRP) = 180° + 180°
∠S + (∠SPR + ∠QPR) + (∠QRP + ∠SRP) + ∠Q = 360°
We see that (∠SPR + ∠QPR) = ∠SPQ and (∠QRP + ∠SRP) = ∠QRS.
Replacing them we have, ∠S + ∠SPQ + ∠QRS + ∠Q = 360°
That is, ∠S + ∠P + ∠R + ∠Q = 360°.

Or, the sum of angles of a quadrilateral is 360°. This is the angle sum property of quadrilaterals.


4. The three angles of a quadrilateral are 74°, 56°, and 106°. Find the measure fourth angle?

Solution:

Given that the angles of a quadrilateral are 74°, 56°, and 106°.
The sum of the angles of a quadrilateral is 360°.
The quadrilateral consists of four angles.
Let the fourth angle is x.
x + 74° + 56° + 106° = 360°
x = 360° – 74° – 56° – 106°
x = 124°

The fourth angle is 124°.


5. The angles of a quadrilateral are in the ratio 2 : 4 : 6 : 8. Find the measure of each of these angles?

Solution:

Given that the angles of a quadrilateral are in the ratio 2 : 4 : 6 : 8.
Let the common ratio is x.
2x + 4x + 6x + 8x = 360°
20x = 360°
Divide the above equation by 20 on both sides.
20x/20 = 360°/20
x = 18°
The angle of quadrilaterals are
2x = 2 . 18° = 36°
4x = 4 . 18° = 72°
6x = 6 . 18° = 108°
8x = 8 . 18° = 144°
The measure of each of these angles is 36°, 72°, 108°, 144°.


6. A quadrilateral has three acute angles, each measuring 70°. Find the measure of the fourth angle?

Solution:

Given that a quadrilateral has three acute angles, each measuring 70°.
Sum of four angles in any Quadrilateral =360°
Three angles each =70°
70° + 70° + 70° + fourth angle =360°
Fourth angle =360° – 210°
= 150°

Therefore, the fourth angle is 150°.


7. Three angles of a quadrilateral are equal and the measure of the fourth angle is 90°. Find the measure of each of the equal angles?

Solution:

Given that three angles of a quadrilateral are equal and the measure of the fourth angle is 90°.
Let each unknown angle is x. The sum of the angles of a quadrilateral are equal to 360°
x + x + x + 90° = 360°
3x + 90° = 360°
3x = 360° – 90°
3x = 270°
Divide the above equation by 3 on both sides.
3x/3 = 270°/3
x = 90°

The three angles are 90°, 90°, and 90°.


8. Two angles of a quadrilateral measure 65° and 55° respectively. The other two angles equal. Find the measure of each of these equal angles?

Solution:

Given that two angles of a quadrilateral measure 65° and 55° respectively. The other two angles equal.
The sum of the angles of a quadrilateral is equal to 360 degrees.
Let the other two angles be x.
So, x + x + 65° + 55° = 360°
2x + 120° = 360°
2x = 360 – 120° = 240°
x = 240°/2
x = 120 degrees

Hence, the Other two angles are 120 degrees each.


9. Two angles of a quadrilateral measure 45° and 75° respectively. The other two angles equal. Find the measure of each of these equal angles?

Solution:

Given that two angles of a quadrilateral measure 45° and 75° respectively. The other two angles equal.
The sum of the angles of a quadrilateral is equal to 360 degrees.
Let the other two angles be x.
So, x + x + 45° + 75° = 360°
2x + 120° = 360°
2x = 360 – 120° = 240°
x = 240°/2
x = 120 degrees

Hence, the Other two angles are 120 degrees each.


Worksheet on Integers Multiplication | Multiplication of Integers Worksheet with Answers

Worksheet on Multiplication of Integers is here to guide the candidates to know about the concept completely. To determine the integer multiplication method, we will follow various rules, properties, and methods. Before going to check the Multiplying Integers Worksheets, check all the rules, methods, and formulae used in multiplying the integers.

When the two integers with the same sign multiply, then the result value will always be positive. When the two integers with different signs multiply, then the value of the results will be negative. In the below sections, find the Multiplication of Integers Worksheet, practice questions, step by step procedure to solve the problems.

Question 1:

Henry made 3 withdrawals of $2 each from his savings account. What was the change in his balance?

Solution: 

As given in the question,

Henry made withdrawals = 3

The amount he withdrew = $2

As he withdrew the amount, it will be negative

Change in his balance = 3(-2) = -6

Therefore, the total change in the amount is -$6

Thus, the final solution is his savings account was declined by $6


Question 2:

Lisa plays a video game and she loses points in it. She loses 5 points 4 times?

Solution: 

As given in the question,

Lisa plays a video game and loses points = 5

Number of times she loses points = 4


Question 3:

Daniel has caddied 14 times over the past two months and earned a total of $196. How much does he earn totally in 8 months?

Solution: 

As given in the question,

No of times Daniel caddied= 14

The total amount of money he earned for 2 months = $196

For every month, he earned $196

Therefore, for every 1 month, he earned = $98

Thus, for every 8 months the amount he earned = $98*8

=$784

Hence, the final solution is $784


Question 4:

There are 63 groups at the confirmation meeting. If each group had three people, how many people were at the meeting?

Solution: 

As given in the question,

No of groups at confirmation meeting = 63

Number of people each group have = 3

No of people at the meeting = 63*3

=189

Therefore, the no of people at the meeting = 189

Thus, the final solution is 189 members were at the meeting.


Question 5:

Galapagos tortoises can nap sleep 16 hours a day. How many total hours of sleep could a group of 19 tortoises get in one day?

Solution: 

Galapagos tortoises can sleep for hours = 16

No of tortoises = 19

Total no of hours = 16*19 = 304 hours

Therefore, the total no of hours of sleep a group of 19 tortoises can get = 304 hours.

The final solution is 304 hours.


Question 6:

The stationary shop has 163 packets of pens. Each packet has 15 pens. How many pens are there?

Solution: 

No of packets of pens = 163

No of pens in each packet = 15

Total no of pens = 163*15 = 2445 pens

Therefore, the total no of pens in 163 packets = 2445 pens

Hence, the final solution is 2445 pens


Question 7:

The zookeeper wants to give each monkey 12 bananas. There are 53 monkeys. How many bananas would he need?

Solution: 

No of bananas for each monkey = 12

No of monkeys = 53

Total.no of bananas he need = 12*53

=636 bananas


Question 8:

One packet can hold 56 chocolates. How many chocolates will 17 packets hold?

Solution: 

No of chocolates a packet can hold = 56

To find the chocolates in 17 packets, we multiply 17 with 56

Therefore, no of choclates = 17*56 = 952

Thus, no of chocolates a packet can hold = 952 chocolates


Question 9:

Mother bought 8 T-Shirts at rupees 475 each. How much money did she pay in all?

Solution: 

No of t-shirts = 8

Amount of money for each shirt = Rs. 475

Total amount of money she paid = 475*8 = Rs. 3800

Therefore, the total amount of money = Rs. 3800

Hence, the final solution is 3800 rupees


Question 10:

Jay has 2140 Philippines stamps. He has 4 times as many foreign stamps as Philippine stamps. How many foreign stamps does he have?

Solution: 

No of Philippines Stamps Jay has = 2140

No of foreign stamps he has = 4 times

Total no of foreign stamps he has = 2140*4 = 8560

Hence, the final solution is 8560 stamps.


Question 11:

Manny bought 18 boxes of marbles. Each box had 555 marbles. How many marbles were there in all?

Solution: 

No of marble boxes Manny bought = 18

No of marbles each box has = 555

Total number of marbles present = 555*18

= 9990

Therefore, the final solution is 9990 marbles.


Question 12:

A baseball team has 9 players. In a tournament, there are 28 teams. How many players are there in all?

Solution: 

No of players = 9

Number of teams = 28

No of players in a team = 28*9 = 252

Therefore, the total number of players in the tournament = 252

Hence, the final solution is 252 players.


Question13:

A tray of eggs holds 12 eggs. If you have 63 full trays, how many eggs would you have?

Solution: 

No of eggs a tray holds = 12

No of full trays = 63

Total number of eggs present = 12*63 = 756

Therefore, the total number of eggs = 756 eggs

Thus, the final solution is 756 eggs.


Question 14:

A store owner was buying uniforms for his employees. If each of his three stores needed eight uniforms. How many uniforms would he need?

Solution: 

No of stores = 3

No of uniforms = 8

Total no of uniforms he need = 3*8 = 24

Therefore, the total number of uniforms = 24

Hence, the final solution is 24 uniforms.


Question 15:

John bought 2 boxes of books at a yard sale. If each box had five books. How many books did he buy?

Solution: 

No of boxes John bought = 2

No of books in each box = 5

Total books = 2*5 = 10

Therefore, the total number of books he bought = 10 books

Hence, the final solution is 10 books.


Question 16:

An employee earns eight dollars an hour at a construction site. If he works 8 hours in 1 week. How much money he would earn?

Solution: 

Amount of money employee earns at a construction site = $8

The time he works in a week = 8 hours

Amount of money he earned = 8*8 = 64

Therefore, the total amount of money he earned at the construction site = $64

Thus, the final solution is $64


Question17:

A pet store sold 5 gerbils in one week. If each of the gerbils costs 8 dollars, how much money will they have made?

Solution: 

No of gerbils a pet store sold = 5

Cost of each gerbils = $8

The total amount of money = 5*8 = 45

Therefore, the total amount of money the pet store have made = $45

Thus, the final solution is $45


Question 18:

The pupils of Mrs. Luna went on an educational tour at Manila Ocean Park. The Entrance Fee is $280 per child. There were 37 pupils in the class. How much did Mrs. Luna pay for the entrance fee if the pupils?

Solution: 

The entrance fee of Manila Ocean Park = $280

No of pupils = 37

Total amount = 280*37 = 10,360

The total amount of money Luna paid for the entrance for 37 pupils = $10,360

Thus, the final solution is $10, 360


Question 19:

In class, there are 40 children. Each child has 4 pencils. How many pencils are there in all?

Solution: 

No of children = 40

No of pencil each child has = 4

Total pencils = 40*4 = 160

Therefore, the total no of pencils = 160 pencils

Thus, the final solution is 160 pencils.


Question 20:

There are 20 oranges in each tree in an orchard. The orchard has 6 orange trees in all. How many oranges are there in an orchard?

Solution: 

No of oranges = 20

Number of trees = 6

No of oranges = 20*6 = 120

Total number of oranges for all trees = 120 oranges

Thus, the total number of oranges = 120

Hence, the final solution is 120 oranges.


Worksheet on Ratio and Proportion | Ratio and Proportion Worksheet with Answers

Practicing from Worksheet on Ratio and Proportion helps students to think more of the concept. Solve Ratio and Proportion Question and Answers available to score better grades in the exam. The Questions in this Worksheet are based on Expressing Ratios in their Simplest Form, Simplification of Ratios, Comparison of Ratios, Arranging Ratios in Ascending and Descending Order, Mean Proportional Between Numbers, etc.

Solve as many times as possible in order to be familiar with the types of Ratio and Proportion Questions. Answering the Problems over here helps you get a good grip on the entire concept. In addition, you will learn the tips and tricks on how to solve ratio and proportion problems using different methods.  For better understanding, we even listed solutions for each and every problem making it easier for you to cross-check whether your answers are correct or not.

1. Express each of the following ratios in the simplest form

(a) 5.6 m to 28 cm

(b) 6 hours to a day

(c) 20 liters to 15 liters

(d) 170 : 240

Solution:

(a) 5.6 m to 28 cm

1 m = 100cm

5.6 m = 5.6*100 = 560 cm

Ratio of 5.6m to 20 cm = 560 cm: 20 cm

= 140:5

= 28:1

(b) 6 hours to a day

In one day there are 24 hrs

6 hrs to a day = 6 hrs: 24 hrs

= 1 :4

(c) 20 liters to 15 liters

= 20 liters :15 liters

=  4:3

(d) 170 : 240

= 170/240

= 17/24

Therefore, ratio of 170:240 in its simplified form is 17/24


2. Simplify the following ratios

(a) 1/4 : 1/3 : 1/6
(b) 3.6 : 5.4
(c) 3²/₃ : 4¹/₂

Solution:

(a) 1/4 : 1/3 : 1/6

LCM of 4, 3, 6 is 12

Expressing them in terms of a least common factor we have

1/4 = 3*3/4*3 = 9/12

1/3 = 1*4/3*4 = 4/12

1/6 = 1*2/6*2 = 2/12

Therefore 1/4:1/3:1/6 in simplified form is 9:4:2

(b) 3.6 : 5.4

Simplifying it we get the ratio as under

Dividing with GCD(3.6, 5.4) i.e. 0.9 we get the simplified form

= (3.6/0.9):(5.4/0.9)

= 4:6

(c) 3²/₃ : 4¹/₂

= 11/3:9/2

LCM of (3, 2) is 6

Expressing the given ratio in terms of LCM we get the equation as follows

11/3 = 11*2/3*2 = 22/6

9/2 = 9*3/2*3 = 27/6

Therefore, ratio 3²/₃ : 4¹/₂ in simplified form is 22:27


3. Compare the following ratios

(a) 5 : 2 and 4 : 3
(b) 1/3 : 1/5 and 1/5 ∶ 1/6

Solution:

(a) 5 : 2 and 4 : 3

Express the given ratio as fraction we get

5:2 = 5/2

4:3 = 4/3

Find the LCM(2, 3) i.e. 6

Making the denominator equal to 6 we get

5/2 = 5*3/2*3 = 15/6

4/3 = 4*2/3*2 = 8/6

5:2 > 4:3

(b) 1/3: 1/5 and 1/5 ∶ 1/6

1/3:1/5

Finding LCM of 3, 5 we get the LCM as 15

Expressing the ratios given in terms of the LCM as a common denominator

1/3 = 1*5/3*5

= 5/15

1/5 = 1*3/5*3 = 3/15

thus it becomes 5:3

1/5 ∶ 1/6

Finding LCM of 5, 6 we get the LCM as 30

Expressing the ratios given in terms of the LCM as Common Denominator

1/5 = 1*6/5*6 = 6/30

1/6 = 1*5/6*5 = 5/30

Therefore, it becomes 6:5

Therefore, 1/3: 1/5 < 1/5 ∶ 1/6


4. In the ratio 3 : 5, the consequent is 20. Find the antecedent?

Solution:

Let the Antecedent and Consequent be 3x and 5x

We know Consequent = 20

5x =20

x= 20/5

= 4

Antecedent = 3x

= 3*4

= 12

Therefore, Antecedent is 12.


5. Divide 2000 among A, B, C in the ratio 2 : 3 : 5?

Solution:

Let the numbers be 2x, 3x, 5x

Sum = 2000

2x+3x+5x = 2000

10x = 2000

x = 2000/10

x = 200

Since the sum is to be split among A, B, C in the ratio of 2:3:5

we get A’s Share = 2x

B’s Share = 3x

C’s Share = 5x

A’s Share = 2*200

= 400

B’s Share = 3*200

= 600

C’s Share = 5*200

= 1000

Therefore, A, B, C’s Share in the amount of 2000 are 400, 600, 1000 respectively.


6. Determine whether the ratios form a Proportion or not

(a) 50 cm : 1 m = $80 : $160

(b) 200 ml : 2.5 l = $4 : $20

Solution:

(a) 50 cm : 1 m = $80 : $160

50 cm: 1 m

We know 1m = 100 cm

= 50 cm: 100 cm

= 1:2

$80 : $160

= 1:2

Since both the ratios are equal they are said to be in Proportion

(b)200 ml : 2.5 l = $4 : $20

1 liter = 1000 ml

2.5 l = 2.5*1000

= 2500

200 ml: 2500 ml

= 2:25

$4:$20

= 1:5

Since both the ratios aren’t equal given values doesn’t form a Proportion.


7. Find the value of x in each of the following

(a) 4, 5, x, 48

(b) 7, 21, 30, x

(c) x, 28, 24, 4

Solution:

(a) 4, 5, x, 48

We know Product of Means = Product of Extremes

4*48 = 5*x

5x = 192

x = 192/5

(b) 7, 21, 30, x

We know Product of Means = Product of Extremes

7*x = 21*30

x = (21*30)/7

= 90

(c) x, 28, 24, 4

We know Product of Means = Product of Extremes

x*4 = 28*24

x = (28*24)/4

= 168


8. Find the fourth proportional to 54, 27, 18, x

Solution:

Product of Extremes = Product of Means

54*x = 27*18

x = (27*18)/54

= 9


9. Find the third proportional to

(a) 9, 6, x

(b) 6, 12, x

Solution:

To find third proportional we write the expression as

9:6 = 6:x

9*x = 6*6

x = 36/9

= 4

(b) 6, 12, x

6:12 = 12:x

6*x = 12*12

x = 144/6

x= 24


10. Find the mean proportional between

(a)  5 and 20

(b) 1.6 and 0.4

Solution:

Mean Proportional between two numbers is defined as the square root of the product of two numbers

(a)  5 and 20

= √(5*20)

=√100

= 10

Mean Proportional of 5 and 20 is 10

(b) 1.6 and 0.4

Mean Proportional between two numbers is defined as the square root of the product of two numbers

= √(1.6*0.4)

= v6.4

= 0.8

Mean Proportional of 1.6 and 0.4 is 0.8