Straight Line Drawn from the Vertex of a Triangle to the Base

Straight Line Drawn from the Vertex of a Triangle to the Base Theorem & Proof with Examples

This article provides you with the concept of a Straight line drawn from the vertex of a triangle to the base. Children are aware of the topics triangles and straight lines. Here, we will learn and prove how a triangle bisects the opposite side by a straight line with the help of the midpoint theorem. A triangle is a polygon shape formed by three line segments with three midpoints respectively.

By the end of this page, kids will perceive how a line bisects the other side. Without any delay let us discuss the concept and prove the statement. On this page, we have also given a glimpse on the topics called median and the centroid of a triangle for a better understanding of the Straight Line Drawn from the Vertex of a Triangle to the Base.

Also, Refer:

Median

A median is defined as the line segment joining a vertex of the opposite side of a triangle and bisects the side. In other words, a straight line drawn from one vertex to the opposite of the vertex of a triangle is called a median and every triangle has three medians. Learn the concept of the median of raw data and gain proper knowledge on how to solve it easily.

The Centroid of a Triangle

A triangle is formed by joining three dots by a line segment connected to each other and forms a triangle. A triangle has three medians with three angles. A centroid is defined as “the point of intersection of three medians of a triangle”.

For example, if we take a triangle ABC we have three midpoints D, E, and F respectively. Here, O is the point of intersection of three medians and O is a centroid of a triangle. The below figure gives how a centroid of a triangle looks like.

Centroid of a triangle

Properties of a Centroid

The properties of a centroid are as follows

  • It is a point of congruency.
  • Centroid is the intersection of three medians and bisects on the opposite side.
  • A centroid is always inside of a triangle.
  • It divides each median in the ratio of 2:1.

Statement

A straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the midpoints of the other two sides of the triangle.

Triangle with midpoints

To Prove:

Consider a triangle PQR.

Given M, N, and O are the midpoints of PQ, PR, and QR respectively.

We have to prove that MN bisects PO i.e., PX = XO.

Construction:

From the above △PQR, join the midpoints M, N, and O of a triangle and name the intersection point of a triangle as X.

Traingle

The above figure is the constructed triangle PQR, to prove the theorem statement.

Proof of the Theorem:

Given PQR is a triangle.

Let M be the midpoint of PQ, N be the midpoint of PR, and O be the midpoint of QR.

O is the midpoint of QR then PO is the straight line that bisects MN at point O.

Since M and N are the midpoints of the sides PQ and PR of a △PQR, then

MN ∥ QR by the midpoint theorem.

since MN ∥ QR, then MX ∥ QO.

In △PQO, we know that M is the midpoint of PQ and X be the midpoint of PO.

By converse of midpoint theorem

MX ∥ QO

⇒ PX=XO.

Thus, MN bisects PO.

Hence, the statement Straight Line Drawn from the Vertex of a Triangle to the Base is proved.

FAQ’s on Prove that a Line Drawn from the Vertex to its Base is a Straight Line

1. What do you call the straight line joining the vertex with the center of the base?
In mathematics, a triangle is a line segment that joins the vertex to the midpoint of the opposite side, it bisects the side. Every triangle has three medians from each vertex and all three midpoints intersect each other and it is known as the centroid of a triangle.

2. What is called the line joining vertex and the midpoint of the base of the triangle?
The line segment joining a vertex of a triangle to the midpoint of a triangle is known as the median. Every triangle has three medians and three altitudes.

3. What is the base of a triangle?
The base of a triangle is the bottom line of a triangle, and it is one of the three sides of a triangle. But in a triangle, one side is a base side and the remaining two sides are the height and the hypotenuse side of a triangle.

4. What is the vertex of a triangle?
The vertex or vertices are the corners of a triangle. There will be three vertices for every triangle.

Simple Interest Worksheet

Free Printable Simple Interest Worksheet | Finding Simple Interest Worksheet with Answers PDF

Calculating Simple Interest is an essential skill for anyone who maintains a bank account or wants to apply for a loan. Free Printable Simple Interest Worksheet PDF will improve your homeschool math skills as well as helps you to become better at calculations. Solutions provided for all the Questions in the Simple Interest Word Problems Worksheet will make it easy for you to learn the concept. Learn the Formulae, Tips & Tricks associated with the Simple Interest Concept easily by solving the problems on Simple Interest.

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Calculating Simple Interest Worksheet

I. Find the amount and S.I of the following
a) If p=7000, R=5%, T=3 years
b) If p= 9000, R=3%, T=5 years

Solution:

Given that,
p=7000, R=5%, T=3 years
We know that S.I=ptr/100
S.I=7000 × 3 × 5/100
S.I=105000/100
S.I=1050
Amount=P+S.I
Amount=7000 + 1050=8050
Therefore, S.I is Rs 1050 and the Amount is Rs 8050.
b)If p= 9000, R=3%, T=5 years
Given that,
p= 9000, R=3%, T=5 years
We know that S.I=ptr/100
S.I=9000 × 5 × 3/100
S.I= 1350
Amount=Principle+S.I
Amount=9000 + 1350
=10350
Therefore, S.I is Rs 1350, Amount is Rs 10350.


II. a)If Principle=6000, S.I=250. Find the amount?
b) If S.I=300, Amount=4800. Find the Principle?
c) If Principle=1700,Amount=2000. Find the S.I?

Solution:

a) Given that,
Principle=Rs 6000, S.I=Rs 250
We know that Amount=Principle + S.I
Amount= 6000 +  250
Amount=Rs 6250
Therefore, Amount=Rs 6250.
b) Given that,
S.I=Rs 300, Amount=Rs 4800
We know that Amount=Principle + S.I
Principle=Amount-S.I
principle= 4800 –  300
=Rs 4500
Therefore, Principle is Rs 4500.
c) Given that,
Principle=Rs1700,
Amount=Rs 2000
We know that Amount=principle + Interest
Interest=Amount- Principle
Interest=2000 – 1700
Interest=Rs 300
Therefore, Simple Interest is Rs 300.


III. Find Time when, Principal is Rs 5000, Interest is Rs 200, Rate is 2% p.a?

Solution:

Given p=5000,
I=Rs 160
R=3%
We know that S.I=ptr/100
T = SI×100 / P × R
T=200 × 100/5000 × 3
T=20000/5000 × 3
T=12
Therefore, Time=12 years.


Iv. Find Rate when Principal is Rs 7000, Interest is Rs 700 and Time is 4 years?

Solution:

Given that,
Principal=7000
Interest=700
Time=4 years
We know that S.I=ptr/100
r=SI×100 / P × t
r=700 ×100 /7000 ×4
r=70000/28000
r=2.5
Therefore, rate=2.5%.


V. Find the principal which gives an Amount of Rs 3500 at the rate of 8% for 5 years?

Solution:

Given that,
Amount=3500
rate=8%
T=5 years
We know that S.I=ptr/100
p=SI×100 / t × r
p= 3500 ×100 / 5 × 8
p=350000/40
p=Rs 8750
Therefore, the Principal is Rs 8750.


VI. Sarath deposited Rs 10000 for 2 years and 6 months at the rate of 5% p.a. Find the amount at maturity?

Solution:

Given that,
Sarath deposited p=Rs10000
T=2 years 6 months
r=5%
First, the compound interest for 2 years will be calculated and then the amount after 2 years will be considered as principal and on this principle, the simple interest for 6 months will be calculated.
Compound interest for 3 years
A = P (1 + R/100)ⁿ
A=10000(1 + 5/100)2
=10000(1+1/20)2
=10000(21/20)2
=Rs 1102.5
Now for the next 6 months, the principle will be Rs 1102.5.
Simple Interest = (P*R*T)/100
S.I=1102.5 × 5 × 6/100
S.I=Rs 330.75
Amount=Principle + interest
Amount=1102.5 + 330.75
Amount=Rs 1433.25
Hence, the maturity amount will be Rs 1433.25.


VII. Arun deposited Rs 15000 at a rate of 12% (p.a.) for 175 days. Find the amount he got back after 175 days.

Solution:

Given that,
P=Rs 15000,
r=12%,
T=175 days=5 months 15 days=5 1/2=11/2
Simple Interest = (P×R×T)/100
=15000 × 12 × 11/100 × 2 × 12
=825
Amount=Principle + Interest
=15000 + 825
=15825
Therefore, Amount is Rs 15825.


VIII. Kishore deposited Rs. 500 per month for 12 months in a banks’ recurring deposit account. If the bank pays interest at the rate of 9% per annum, find the amount (in Rs.) he gets on maturity?

Solution:

Given that,
p=500,
T=12 months
R=9%
Equivalent principal for 12 months = 100× n(n+1)/2
=500 × 12(13)/2
=39000
Interest=PRT/100
=39000 × 9 ×1/100 × 12
=292.5
Maturity amount=P × T+I
=500 × 12 + 292.5
=6000+292.5
=6292.5
Therefore, the maturity amount is Rs 6292.5.


VIII. Mahesh borrowed Rs 2000 at the rate of 5% (p.a.) for 5 years and 5 months. Find the amount he paid back?

Solution:

principle = Rs2000
time = 5.5 years
rate of interest = 5%
Simple Interest = P×T×R ÷ 100
= 2000×5.5×5 ÷ 100
= 550
Amount = Principle + Simple Interest
= 2000+550
= 2550
Therefore, Amount is Rs 2550.


IX. Find Principal when, Time is 2 years, Interest is Rs 400 and Rate is 3% p.a?

Solution:

Given that,
Time=2 years,
I=Rs 400
Rate=3%
Pricipal=(100 × Interest)/(Rate × Time)
=(100 × 400)/(3 × 2)
=40000/6
=6666.66
Therefore, Principle is Rs 6666.66.


X. In what time will Rs 500 amount to Rs 570 at the rate of 5% p.a. simple interest?

Solution:

Given that,
Principle = 500
Amount = 570
A = SI + P so,
SI = A – P
= 570-500
= 70
Now, T = SI * 100/P*R
= 70*100/500*5
= 7000/2500
= 2.8 years
Therefore, Time taken will be 2.8 years.


 

 

 

 

 

 

 

Area and Perimeter of a Semi Circle

Area and Perimeter of a Semi Circle – Definition, Formula, Examples | How to find the Area and Perimeter of a Semicircle?

Area and Perimeter of Semi-circle Problems help the students to explore their knowledge of semi-circle Problems. Solve all the Problems to learn the formula of the Area of the semi-circle and the Perimeter of a semi-circle. To know the definition, Problems with Solutions, Formulas of Semi-circle you can visit our website. We have given the complete Semi-circle concept along with examples. Check out the Area and Perimeter of Semi-circle Problems and know the various strategies to solve problems in an easy and understandable way.

Semi-Circle – Definition

A semi-circle is defined as a half-circle that is formed by cutting a whole circle into two halves along with a diameter line. The Semi-circle has only one line of symmetry which is the reflection symmetry. A line segment is known as the diameter of a circle cuts the circle into two equal semicircles. It is also referred to as half-disk. A semi-circle will be half of the circle that is 360 degrees, the arc of the semicircle always measures 180 degrees. The below figure shows the Semi-circle.

Area and Perimeter of a Semi-Circle – Definitions and Formulas

Area of a Semi-Circle: The area of a semicircle is half of the area of a circle. We know that the area of a circle is πr2. So, the area of a semicircle is,
Area of a Semi-circle = 1/2πr2
where r will be the radius.
The value of π is constant. So, the value is 3.14 or 22/7.

The perimeter of a Semi-Circle: The perimeter of a semicircle is defined as the total length of its boundary. It is also known as the Circumference of a Semicircle. It is calculated as adding the length of the diameter and half the circumference of a circle. The perimeter of a circle unit is expressed in linear units like inches, feet, meters or centimeters and etc.
The Perimeter of a semicircle is, πr+2r.
Where r is the radius and π is a constant value that is 3.14.

The perimeter of the semicircle is P = Half of the Circumference of the original circle + Length of the Diameter.
The Circumference of a circle is 2πr.
The half of the circumference of the circle is 1/2 x the circumference of a circle.
=1/2 x 2πr = πr =π(d/2).
The length of the diameter is d = 2r.
In terms of radius, the circumference of the semicircle is P = C = πr + d = πr + 2r = (π + 2)r
In terms of Diameter, the circumference of a semicircle is P = C = π(d/2) + d.

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Solved Problems on Area and Perimeter of a Semicircle

Problem 1: Find the area of a semicircle. If the perimeter of a semicircle is 122 units. Consider the π value is  22/7.

Solution: 
Given the values in the question,
The perimeter of a semicircle is 122 units.
We know that the area of a semicircle is 1/2(πr2).
Using the perimeter of a semicircle value, find the radius (r) value.
The perimeter of a semicircle is πr + 2r = 122units.
πr + 2r = 122 units.
(22/7+2)r = 122 units
(36/7)r = 122 units
r = (7/36) x 122 = 23.7 units.
Now, we will find the area of a semicircle.
A = 1/2(πr2).
A = (22/7 × (23.72))/2 = (3.14)x(561.6)/2 = 881.7 square units.
Therefore, the area of the semicircle is 881.7 sq. units.

Problem 2: If the radius of a semicircle is 4units, then using the semicircle formula find its perimeter?

Solution: 
As given in the question, the radius of the semicircle is 4units.
Now, we have to find the perimeter of a semicircle using the formula.
The perimeter or circumference of a semicircle is, πr + d = πr + 2r
Substitute the radius (r) value in the above formula, we get
P = πr + 2r =(π+ 2)r = (3.14 + 2)(4) = 20.56 units
Thus, the perimeter of a semicircle is 20.56 units.

Problem 3: Using the below figure, find the Perimeter and Area of a semicircle?

Solution: 
Given in the question, the figure consists of radius value, r = 2cm.
Using the radius value, find the area and perimeter of a semicircle.
We know that the Area of a semicircle is (πr2)/2.
So, the value is, A = {(22/7) (2)2} /2 cm2.
A = (3.14) (4) /2 cm2= 6.28cm2
Now, the perimeter of a semicircle is (π + 2)r.
Substitute the values in the above formula. We get,
P = (π + 2)r = (3.14 + 2)(2) = (5.14)(2) = 10.28 cm.
Thus, the perimeter and area of a semicircle are 6.28cm2 and 10.28 cm.

Problem 4: Find the area of semicircle using the below figure. The below figure consists of a semicircle and equilateral triangle.

Solution: 
As given in the question, the figure consists of a semicircle and triangle.
The radius of a semicircle is 4cm.
We know that the area of a semicircle is 1/2(πr2).
Substitute the given values in the above formula, we get
Now, we will the area of a semicircle.
A = 1/2(πr2).
A = (22/7 × (42))/2 = (3.14)x(16)/2 = 25.12 cm2.
Therefore, the area of the semicircle is cm2.

FAQ’S on Area and Perimeter of a Semicircle

1. What are the steps for finding the perimeter of a semicircle?

The steps to determine the perimeter of a semicircle are given below:

  • Step1: First, find the product of π and the radius (r) of the semicircle.
  • Step 2: Next, find the diameter of the semicircle.
  • Step 3: Then Add the values obtained in the above two steps.
  • Step 4: Thus the value obtained is the perimeter of the semicircle.

2. What is the difference between circumference and Perimeter of Semicircle?

The perimeter of a semicircle and the circumference of a semicircle mean the same. Both are referred to the total length of the boundary of a semicircle. Therefore, the circumference of a semicircle is another name for the perimeter of a semicircle.

3. Is a semicircle is half the circle?

Yes, a semicircle is half the circle. It means a circle can be divided into two semicircles.

4. What is the shape of a semicircle? 

The shape of a semicircle is obtained by cutting a circle into two equal parts along its diameter and the full arc of a semicircle is always measured 180 degrees. An example of a semicircle shape is a protractor.

5. What is the angle of the semicircle? 

The angle of the semicircle is 90 degrees, the angle is made by the triangle in a semicircle is a right angle.

Worksheet on Word Problems on Linear Equation

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

Worksheet on Word Problems on Linear Equation will be of much help in practicing problems on the linear equations. If you would like to know more about the linear equation concept then the Linear Equations Word Problems Worksheet with Answers can be of utmost help. Linear Equation Word Problems Worksheet PDF can be a great resource to practice a variety of word problems. Step by Step Solutions provided for all the problems makes it easy for you to understand the concept in a fun and engaging manner.

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Linear Equations Word Problems Worksheet with Answers PDF

I. Two third of a number is 52. Find 50% of the number?

Solution:

Given that,
Two third of the number is 52.
Let the number be x.
2/3(x)=52
x=52 × 3/2
=78
50% of the number=1/2(78)
=39
Therefore, 50% of the number is 39.


II. The difference between the two numbers is 12. The ratio of the two numbers is 5 : 3. What are the two numbers?

Solution:

Given that,
The difference between the two numbers is =12
The ratio of the two numbers is =5 : 3
Let the two numbers be x,y.
x-y=12 –>(1)
x/y=5/3 —>(2)
x=5/3y
substitute x in (1)
5/3y-y=12
2y/3=12
2y=36
y=18
Substitute y in Eqn(1)
x-y=12
x-18=12
x=18+12=30
Therefore, Two numbers are 30, 18.


III. Sum of the two numbers is 70. If one exceeds the other by 10, Find the numbers?

Solution:

Given that,
Sum of the two numbers is= 70
Let x be one of the two numbers.
Then, the other number is (x + 10).
So, we have
x + (x + 10) = 70
x + x + 10 = 70
2x + 10 = 70
Subtract 10 from each side.
2x + 10 – 10 = 70 – 10
2x = 60
Divider each side by 2.
2x/2 = 60/2
x = 30
x + 10 = 30 + 10
x + 10 = 40
Hence, the two numbers are 30 and 40.


IV. The perimeter of a rectangular swimming pool is 120 m. Its length is 4 m more than thrice its width. What are the length and the width of the pool?

Solution:

Given that,
The perimeter of a rectangular swimming pool is =120 m
Let l be the length and w be the width of the swimming pool.
Also given  Length is 4 m more than thrice its width.
So, the length is l = 4w + 4
The perimeter of the swimming pool is 120 m.
2l + 2w = 120
Plug l = 4w + 4
2(3w + 4) + 2w = 120
6w+8+2w=120
8W+8=120
Subtract 8 from each side
8w+8-8=120-8
8w=112
w=112/8=14 m
Therefore, length=4w+4
=4(14)+4
=56+4
=60 m
Therefore, the length and width of the pool are 14 m and 60 m.


V. Vijaya is 5 years older than her younger sister. After 10 years, the sum of their ages will be 45 years. Find their present ages?

Solution:

let the age of younger sister be X and Vijaya=x+5
after10 years younger sister age=x+10and Vijaya=x+10+5
x+10+x+15=45
2x+25=45
x=45-25/2
=20/2
=10
The age of Vijaya=x+5 years=10+5=15 years.
The age of younger sister=10years.
Therefore, the age of Vijaya is 15 years, and her younger sister is 10 years.


VI. In a class of 64 students, the number of boys is 3/5 of the number of girls. Find the number of boys and girls in the class?

Solution:

Let the number of girls be x.
So, 3/5x + x = 64
3x+5x/5=64
8x = 320
x=320/8=40
The number of girls is 40
So the number of boys is 3/5(40)=24
Therefore, the number of boys and girls in the class are 24 and 40.


VII. Among two supplementary angles, the measure of the larger angle is 42 more than the measure of the smaller. Find their measures?

Solution:

Given that,
The measure of the larger angle is 42 more than the measure of the smaller.
Let us assume supplementary be x0 and (180−x)0.
Let the larger angle be x0.
Then,
(180−x)+42=x
222-x=x
222=2x
x=222/2=111
Larger angle=1110.
Smaller angle=(108-42)=660.
Therefore, the Larger angle is 1110 and the smaller angle is 660.


VIII. In an isosceles triangle, the base angles are equal and the vertex angle is 60°. Find the measure of the base angles?

Solution:

Let the base angle be x.
Another base angle =x.
Angle sum property of triangle
2x+60=1800
2x=180-60
2x=120
x=120/2=600
First base angle=600=second base angle.
Therefore, the measure of base angles is 600.


Ix. The sum of two consecutive even numbers is 30. Find the numbers?

Solution:

Given that,
The sum of two consecutive even numbers is =30
Let’s assume the two consecutive even numbers are x and x+2.
According to the question,
The Sum of x and x+2 =30.
We have,
x+(x+2)=30
2x+2=30
2x=30-2
2x=28
x=28/2=14
x+2=14+2=16
Therefore, the two consecutive numbers are 14 and 16.


X. The denominator of a fraction is greater than the numerator by 6. If the numerator is increased by 3 and the denominator is decreased by 1, the number obtained is 4/5, find the fraction?

Solution:

Let the numerator of the rational number be x.
So as per the given condition, the denominator will be x + 6.
The rational number will be x/x+6
According to the given condition,
x+3/x+6-1=4/5
x+3/x+5=4/5
5(X+3)=4(x+5)
5x+15=4x+20
5x+15-4x-20=0
x-5=0
x=5
The rational number will be x/x+6
=5/5+6
=5/11
Therefore, the rational number is 5/11.


XI. Laxman’s father is 66 years old. he is 6 years older than six times adman’s age. what is adman’s age?

Solution:

Let Adam’s father’s present age be f.
Let Adam’s present age be a.
f=66
f=6+6a
66=6+6a
60=6a
a=60/6=10
Hence, Adam’s age is 10years.


XII. Convert the following statements into equations.
(a) 4 added to a number is 10.
(b) 2 subtracted from a number is equal to 14.
(c) 5 times a number decreased by 2 is 13.
(d) 4 times the sum of the numbers x and 5 is 40.

Solution:

(a) Let the number be x.
4+x=10
(b) Let the number be x.
x-2=14
(c) Let the number be x.
5x-2=13
(d) 4(x+5)=40


 

 

 

 

 

 

Mixed Problems using Unitary Method

Mixed Problems Using Unitary Method | Unitary Method Questions and Answers

Mixed Problems using Unitary Method includes questions on direct variation and inverse variation. Solve the Mixed Problems using Unitary Method available here on a regular basis and learn how to approach. Make the most out of the Mixed Questions using Unitary Method and get a grip on the concept. Enhance your problem-solving skills and become proficient in the concept of Unitary Method by availing the Mixed Problems using Unitary Method available. If you have any difficulty you can check our solutions and assess where you went wrong.

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Mixed Questions using Unitary Method with Solutions PDF

I. The cost of 10 pens is 200 Rs. Find:
i.The number of pens that can be purchased with 320 Rs.
ii. The cost of 5 pens?

Solution:

Given that,
The cost of 10 pens is= Rs 200
The cost of 1 pen=200/10=20
The no. of pens that can be purchased with Rs 20 is=1
The no. of pens that can be purchased with Rs 1 is=1/20
The no. of pens that can be purchased with Rs 320=1/20 × 320=16
The cost of one pen=Rs 20
The cost of 5 pens=5 × Rs 20=Rs 100.
Hence, no. of pens that can be purchased with Rs 320 is 16 and the cost of 5 pens is Rs 100.


II. 10 potters can make 130 pots in 7 days. How many potters will be required to make 160 pots in 5 days?

Solution:

10 potters can make 130 pots in 7 days.
1 potter can make 130 pots in 7 × 10 days.
1 potter can make 1 pot in (7 × 10)/130 days.
Let the number of potters required be x, then;
x potters can make 1 pot in (7 × 10)/( 130 × x) days
x potters can make 160 pots in (7 × 10 × 160)/(130 × x ) days
But the number of days given = 5
According to the problem;
(7 × 10 × 160)/(130 × x ) = 5
11200/130x = 5
650x = 11200
x = 11200/650
x = 17
Therefore, 17 potters are required to make 160 pots in 5 days.


III. The cost of 12 bananas is Rs 48. Find the cost of 20 bananas?

Solution:

Given that,
The cost of 12 bananas is= Rs 48
The cost of 1 banana is=Rs 48/12=Rs 4
The Cost of 20 bananas=20 ×Rs 4=Rs 80
Therefore, the cost of 20 bananas is Rs 80.


IV. If 30 students can consume a stock of food in 60 days, then for how many days the same stock of food will last for 20 students?

Solution:

Given that,
30 students can consume a stock of food in =60 days
Let the food be consumed by 20 students in x days.
By inverse Proportion,
30 × 60= 20 × x
1800=20x
x=1800/20
x=90
Hence, The food is consumed by 20 students in 90 days.


V. A car runs 120 km on 10 litres of fuel, how many kilometres will it run on 15 litres of fuel?

Solution:

No. of km a car travel on 10 litres = 120 km
No. of km a car travel on 1 litre = 120/10 = 12 km
No. of km a car travel on 15 litres = 15 x 10 = 150 km
Therefore, the car will run 150 kilometres on 15 litres of fuel.


VI. If 4 buses can carry 260 people, find out the total number of people which 10 buses can carry?

Solution:

Given that,
No. of people carried by 4 buses= 260
No. of people carried by 1 bus=260/4=65
No. of people carried by 10 buses=10 x 65=650
Hence, 10 buses can carry 650 people.


VII. A truck covers a particular distance in 2 hours with a speed of 70 miles per hour. If the speed is increased by 10 miles per hour, find the time taken by the truck to cover the same distance?

Solution:

This is a situation of inverse variation.
Because more speed —–> less time
Given that,  Time = 2 hours and Speed = 70 mph
We know that Distance = Time ⋅ Speed
Distance = 2 ⋅ 70 = 140 miles
If the given speed of 70 mph is increased by 10 mph, then the new speed will be 80 mph.
We also know that Time = Distance / Speed
Time = 140 / 80
Time = 1.75 hours
If the speed is increased by 10 mph, the time taken by the truck is 1.75 hours.


VIII. If 5 men can complete the work in 12 hours, how many men will be able to do the work in 5 hours?

Solution:

This is a situation of inverse variation.
Because fewer hours —–> more men
In 12 hours, Work can be completed by 5 men
No. of hours taken by 1 man to complete the work is
= No. of hours ⋅ No. of men
= 12 ⋅ 5
= 60 hours
No. of men required to complete the work in 5 hours is= 60 /5
= 12 men
Therefore, 12 men will be able to complete the work in 10 hours.


IX. A man has enough money to buy 12 kg of apples at Rs 50 per kg. How much can he buy, if the price is increased by Rs 2 per kg?

Solution:

This is a situation of inverse variation.
Because
more price —–> fewer kgs of apples
The cost of 12 kg of apples at Rs 50 per kg is
= 12 ⋅ 50
= Rs 600
So, the person has Rs 600
If the price is increased by Rs 2 per kg, then the new price per kg is
= Rs 52
No. of kgs of apples he can buy with Rs 600 is
= 600 / 52
= 11.53 kg
If the price is increased by Rs 2 per kg, the person can buy 11.53 kg of apples.


X. A can do a piece of work in 10 days while B can do it in 15 days. With the help of C, they finish the work in 5 days. At what time would C alone do it?

Solution:

Let “x” be the no. of days taken by C to complete the work.
A’s 1 day work = 1/10
B’s 1 day work = 1/15
C’s 1 day work = 1/x
(A + B + C)’s 1 day work = 1/10 + 1/15 + 1/x
L.C.M of (10, 15, x) = 30x.
Then, we have (A + B + C)’s 1 day work =(3x/30x)+(2x/30x)+(30/30x)
(A + B + C)’s 1 day work =(5x+30)/30x —>(1)
Also given A, B and C together can do the work in 5 days.
So, we have (A + B + C)’s 1-day work = 1/5 ——(2)
From (1) and (2), we get
(5x+30)/30x=1/5
5(5x+30)=30x
25x+150=30x
5x=150
x=150/5=30
Therefore, C alone can complete the work in 30 days.


XI. A can do a piece of work in 10 days and B alone can do it in 20 days. They worked together on it for 3 days and then A left. How long would B take to finish the remaining work?

Solution:

A’s 1 day work = 1/10
B’s 1 day work = 1/20
(A + B)’s 1 day work = 1/10 + 1/20
L.C.M of (10, 20) = 20.
Then, we have
(A + B)’s 1 day work = 2/20 + 1/20
(A + B)’s 1 day work = 3/20
Then, the amount of work completed by A and B together in 3 days is= 3 ⋅ 3/20
= 9/20
The amount of work left for B to complete is= 11/20
Number of days that B will take to finish the work is
= Amount of work/part of the work done in 1 day
= (11/20) / (1/20)
= (11/20) ⋅ (20/1)
= 11
So, no. of days taken by B to finish the remaining work is 11 days.


XI. A book contains 100 pages and each page has 20 lines. How many pages will the book contain if every page has 10 lines per page?

Solution:

This is a situation of inverse variation.
Because,
less lines ——–> more pages
20 lines ——–> 100 pages
1 line ——–> 20 ⋅ 100 = 2000 pages
10 lines ——–> 2000 / 10 = 200 pages
Therefore, If the book has 10 lines per page, then it will contain 200 pages.


 

 

 

 

 

 

Four Triangles which are Congruent to One Another

Four Triangles which are Congruent to One Another | How do you Prove 4 Triangles are Congruent?

Four Triangles which are Congruent to One Another: Initially, you should aware of the congruence of triangles as it aids you to perceive the concept of congruence from a different point of view. The meaning of congruence in mathematics is when two figures are similar to each other based on their shape and size. The term congruence is used to specify an object and its mirror image. Two objects are said to be congruent if they lie over each other.

Let us discuss the congruency of triangles and also three line segments that join the middle points of the sides of a triangle, and divide it into four triangles that are congruent to one another. From here you can view the statement and how to prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are congruent to each other.

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Congruent – Definition & Meaning

In mathematics, the word congruent means ‘exactly equal’ in terms of size and shape, and it is also defined as those figures and shapes that can be flipped or rotated and slide they remained to be consistent with the other shapes. All shapes can be reflected to concur with similar shapes.

Two sizes or shapes are congruent if they have only the same size and shape. In other words, two triangles are said to be congruent if pairs of their corresponding sides and their corresponding angles are equal.

Congruency of Triangles

A triangle is a polygon made of three line segments forming with three angles. Congruence in two or more triangles depends on the measurements and length of their sides and angles, the three sides of a triangle define its size and the three angles of a triangle define its shape and then the triangles are said to be congruent.

Therefore, two triangles are superimposed side to side and angle to angle. The symbol is used to denote the congruency in triangles is ” ≅”.

Congruent Triangles

In the above figure, the triangles PQR and ABC are congruent to each other. Symbolically, we write as  ΔPQR ≅ ΔABC.

Statement

The line segments joining the middle points of the sides of a triangle are divided into four triangles and are congruent to each other.

Prove:

Here we will show the three line segments D, E, and F joining the middle points and forming four triangles of a ΔABC that are congruent to one another respectively.

ΔAFE ≅ ΔFBD ≅ ΔEDC ≅ ΔDEF

Construction:

Now, we construct the triangle ABC with the midpoints D, E, and F respectively.

Midpoints of a Triangle

Proof:

The line segments joining the sides of a triangle are half to the third side by midpoint theorem. D, E, and F are midpoints of BC, AC, and AB respectively.

DE = ½AB ⇒ DE = AF = BF → (1)

EF = ½BC ⇒ EF = BD = CD → (2)

DF = ½AC ⇒ DF = AE = EC → (3)

Now in ΔDEF and ΔAFE

DE = AF

DF = AE

EF= FE

So, by SSS theorem of congruence

ΔDEF ≅ ΔAFE.

Similarly,

ΔDEF ≅ ΔFBD and ΔDEF ≅ ΔEDC

Therefore,

ΔAFE ≅ ΔFBD ≅ ΔEDC ≅ ΔDEF.

Hence, four triangles formed by joining the midpoints of a triangle are congruent to each other.

Examples of Four Triangles which are Congruent to One Another

Example 1:
If two triangles ABC and DEF have the measurements such that, AB=4.5 cm, BC=2.1 cm, AC=3 cm, DE=2.1 cm, EF=4.5 cm, and DF=3 cm. Find the triangles ABC and DEF are congruent to each other or not?
Solution: 
Given ΔABC and ΔDEF
AB = EF = 4.5 cm,
BC = DE = 2.1 cm, and
AC = DF = 3 cm
By SSS theorem of congruence,
ΔABC ≅ ΔDEF
Hence, both the triangles are congruent to one another.

Example 2: 
Prove that the given triangles are congruent triangles or not based on the rule of congruence.
Example of Congruence Triangles
Solution:
In ΔPQR and ΔSRQ,
Let us try to notice any three parts of one triangle equal to the corresponding parts of the other triangle.
Now, we find that PQ = SR, PR = SQ, and QR are the common line segment for both the triangles.
Thus, by SSS rule of congruence, ΔPQR and ΔSRQ are congruent triangles.
Therefore, ΔPQR ≅ ΔSRQ.

Example 3: 
Identify the type of congruence in two triangles given
∆ABC, AB=5 cm, BC=4 cm, ∠B=40° and ∆DEF, DE=4 cm, EF=5 cm, ∠E=40°.
Solution: 
Given ∆ABC and ∆DEF
AB = EF = 5 cm,
BC = DE = 5 cm, and
∠B = ∠E =40°.
By SAS rule of congruency,
∆ABC ≅ ∆DEF.

FAQs on Congruent Triangles in Maths

1. What are congruent triangles?
Two triangles are said to be congruent if all the three corresponding sides and all the three corresponding angles are equal in size and measure.

2. How to prove that the four triangles are congruent or not?
By the rule of Angle-Side-Angle, you can easily prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are congruent to each other. If two angles and a non-included side of one triangle are equal to two angles and a non-included side of another triangle, then the triangles are congruent.

3. What are the rules applied for congruency?
The rules applied for congruent triangles are as follows:
(i) SSS Criterion: Side-Side-Side
(ii) SAS Criterion: Side-Angle-Side
(iii) ASA Criterion: Angle-Side-Angle
(iv) AAS Criterion: Angle-Angle-Side
(v) RHS Criterion: Right angle- Hypotenuse-Side

Practice Test on Simple Interest

Practice Test on Simple Interest | Simple Interest Questions and Answers

Simple Interest Practice Test available here is designed to test your knowledge of the concept. Practice the questions from the Simple Interest Practice Test and learn how to solve problems on finding the simple interest, rate, principal, amount, etc. Start answering the different questions framed on the topic and get an idea of how to approach when you come across similar kinds of problems. Step by Step Explanations provided for all the Simple Interest Questions makes it easy for you to understand the concept as well as to enhance math skills.

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Simple Interest Questions with Solutions

I. Find the Principal when S.I. = Rs 192 Rate = 6% per annum Time = 4 years?

Solution:

Given that,
S.I. = Rs 150,
Rate = 5% per annum, Time = 4 years
We know that S.I=ptr/100
150=p × 4 × 5/100
150=p × 20/100
150 × 100/20=p
p=Rs 750
Therefore, the Principal is Rs 750.


II. Find the Rate when Principal = Rs 350 Time = 2¹/₂ years S.I. = Rs 140?

Solution:

Given that,
Principal = Rs 350 Time = 2¹/₂ years S.I. = Rs 140
We know that S.I=ptr/100
140=350 × 2.5 × r/100
14000/350 × 2.5=r
r=100
Therefore, r is 100%.


III. Find the simple interest on Rs 8000 from 20 March 2020 to 7 September 2020 at 10% per annum?

Solution:

Given that,
r=10% per annum
We need to divide 10 by 365 since there are 365 days in a year.
Hence, that is 0.0274% per day
In decimal, we have to divide this percentage by 100, so, that is:
0.0274/100 = 0.000274
Daily interest on the principal is thus the multiplication of this value with the principal (8000), that is:
0.000274 * 8000 = 2.192
Now, we have to calculate the no. of days?
From 20 March till 7 September would be:
20 March – 31 March = 11 days
1 April-30 April=30 days
1May- 30 May=31 days
1 June – 30 June = 30 days
1 July – 31 July = 31 days
1 Aug – 31 Aug = 31 days
1 Sep – 7 Sep = 7 days
Total Days is 11 + 30 + 31 + 30 + 31 + 31+ 7  = 171 days
Thus, total simple interest for 171 days would be =2.192 * 171 = Rs 374.832
Hence, The simple interest is 374.832.


IV. What sum of money will earn an interest of Rs 250 in 3 years at the rate of 12% per annum?

Solution:

Given that,
S.I=Rs 250
T=3 years
R=12%
We know that S.I=ptr/100
principal=S.I * 100/t*r
=250 * 100/3*12
=25000/36=Rs 694.4
Therefore, the principal is Rs 694.4.


V. A man loses Rs 300 yearly when the yearly rate of interest tumbles from 15% to 10%. Find the capital?

Solution:

Given that,
r=15%,t=1
I=prt/100
=p× 15 × 1/100
=15p/100
r1=10%,t1=1
I1=prt/100
=p× 10 × 1/100
=10p/100
15p/100 – 10p/100 =300
p/100(15 – 10)=300
p/100 (5)=300
p=300 × 100/5
=Rs 6000
Therefore, Principal=Rs 6000.


VI. In what time will the simple interest be 3/7 of the principal at 9 percent per annum?

Solution:

Given that,
Simple interest is 3/7of the principal at 9 percent per annum.
i.e. Simple interest=3p/7
Rate = 9%
By Putting all the values in the formula SI=ptr/100,we get
3p/7=p × 9 ×time/100
time=3p × 100/7 × 9p
Time=300p/63p
Time=4.7
To convert 4.7 into months multiply it by 12.
4.7 × 12=56.35 i.e. 56 months 1 week 3 days.
Therefore in 56 months 1 week 3 days, the simple interest is 3/7 of the principal at 9 percent per annum.


VII. The simple interest on sum money is 3/5 of the principal and the number of years is equal to the rate percent per annum. Find the rate per annum?

Solution:

Let the principal be P.
The rate of interest be R%
Also given, Time=R
Simple interest =3p/5
We know that SI=ptr/100
3P/5=P × R × R/100
3P/5=PR2/100
3/5=R2/100
R2=3/5 × 100
R2=60
R=\(\sqrt{ 60 }\)
R=7.74
Therefore, the rate per annum is 7.74%.


VIII. A sum of money doubles itself in 6 years. What is the rate of interest?

Solution:

Let the sum of money that’s principal be x.
Time years
Amount after 6 years becomes double=Therefore,
Simple interest=Amount -principal=


IX. A certain sum amounts to Rs 1600 in 2 years to Rs 2200 in 4 years at simple interest. Find the sum and rate percent per annum?

Solution:

Given:
In 2 years Amount= Rs 1600…………………….(1)
In 4 years Amount =Rs 2200…………………….(2)
Let the sum =P
Rate percent = R%
From equations 1 and 2 we can say that simple interest in 2 years= 600
hence P+600=1600
p=1600-600
∴ P=Rs 1000
We know that SI=ptr/100
600=1000*R*2/100
20R=600
R=30
Therefore, the Rate of Interest is 30%.


X. Rs 10,000 were lent each to Rajesh and Rakesh at 15% per annum for 2¹/₂ years and 5 years respectively. Find the difference in the interest paid by them?

Solution:

Given that,
Rs 10,000 were lent each to Ron and Rob at 15% per annum for 3¹/₂ years and 5 years respectively.
Rajesh,
I = PRT / 100
= (10,000 × 15 × 5/2) / 100
= Rs 3750
Rakesh,
I = PRT / 100
= (10000 × 15 × 5) / 100
= Rs 7500
Difference in Interest=Rs 7500 – Rs 3750=Rs 3750
Therefore, Rakesh paid Rs 3750 more than Rajesh.


XI. In how much time will Simple interest on a certain sum of money at 12¹/₂% per annum be 3/4 for itself?

Solution:

let Principal= x
then SI = (3/4) x
We know that SI = (PRT)/100
( 3/4)x =( x × T × 25/2 ) /100
3/4x = x × T × 25/200
T=3/4 × 200/25
T = 6 years
Therefore, The time taken is 6 years.


XII. Find the time when
(a) Principal = Rs 700       Rate = 7.5% p.a       S.I. = Rs 105
(b) Principal = Rs 800       Rate = 15% p.a.       S.I. = Rs 960

Solution:

(a) Given that,
Principal = Rs 700, Rate = 7.5% p.a, S.I. = Rs 105
we know that S.I=prt/100
105=700 × 7.5 × T/100
105 × 100=700 × 7.5 × T
10500=5250 × T
T=10500/5250
= 2 years
Therefore, Time is 2 years.

(b) Given that,
Principal = Rs 800
Rate = 15% p.a.
S.I. = Rs 960
we know that S.I=prt/100
960=800 × 15 × T/100
960 × 100=800 × 15 × T
96000/12000=T
T=8 years
Therefore, Time is 8 years.


XIII. Find the simple Interest and Amount
(a) Principal = Rs 500 Rate = 10¹/₂% p.a. Time = 5 months
(b) Principal = Rs1000 Rate = 15% p.a. Time = 3 years

Solution:

Given that,
Principal = Rs 500
Rate = 10¹/₂% p.a.
Time = 5 months
We know that S.I=ptr/100
S.I= 500 × 5 × 21/2/100
S.I=500 × 5 × 21/200
S.I=Rs 262.5
Amount=Pricipal+S.I
=Rs 500 + Rs 262.5
=Rs 762.5
Therefore, S.I is Rs 262.5, Amount is Rs 762.5.
b) Given that,
Principal = Rs1000,
Rate = 15% p.a.,
Time = 3 years
We know that S.I=ptr/100
S.I=1000 × 3 × 15/100
=Rs 450
We know that Amount=Principal +Simple interest
=1000 + 450
=Rs 1450
Therefore, Simple Interest is Rs 450, Amount is Rs 1450.


 

Problems on Unitary Method Using Inverse Variation

Problems on Unitary Method Using Inverse Variation | Unitary Method Inverse Variation Questions

If we were to find the ratio of one quantity to another quantity then we use the unitary method. Unitary Method can be solved using two different methods like Direct Variation, Inverse Variation. In this article, we have covered everything on Problems on Unitary Method using Inverse Variation.

Solve the various models of questions on the unitary method using inverse variation and enhance your accuracy and problem-solving skills in the exam. Master the topic of Unitary Method Inverse Variation by attempting the word problems on the unitary method using inverse variation here on a frequent basis.

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Unitary Method using Inverse Variation Questions with Solutions

I. An auto-rickshaw takes 4 hours to cover a distance of 32 km. If its speed is increased by 4 km/h, the time taken by it to cover the same distance is

Solution:

Given that,
Distance =32 km
Time =4hr
Speed =Distance/Time
=32/4
=8 km/hr
New speed =8+4=12 km/hr
The new time is taken by auto-rickshaw
36/12h=36×60/12
=180 min
=2h 30 min
Hence, The time taken by auto-rickshaw to cover the same distance is 2h 30 minutes.


II. A company undertook a software project to complete a part of a project in 8 months with a team of 400 persons. Later on, it was required to complete the project in 4 months. How many extra persons should he employ to complete the work?

Solution:

In 8 months, a part of the project can be completed by 400 persons.
In 1 month, the work can be completed by 8 × 400 = 3200 persons.
In 4 months, the work can be completed by 3200/4 = 800 persons.
Extra Persons required=800 – 400=400
Hence, the Extra persons required is 400 persons.


III. 16 workers can complete work in 42 days. How many workers will complete the same work in 24 days?

Solution:

This is a situation of inverse variation, now we solve using the unitary method.
To complete the work in 42 days, workers required = 16
To complete the work in 1 day, the worker required = (42 × 16)
To complete the work in 24 days workers required = (42 × 16)/24
=16
Therefore, to complete the work in 24 days, 16 workers are required.


IV. If 30 men can do a piece of work in 25 days, then 15 men will complete the same work in how many days?

Solution:

This is a situation of inverse variation, now we solve using the unitary method.
Given that,
30 men can do the work in =25 days.
1 man can do the work in =(30 × 25) days.
15 men can do the work in days= (30 × 25)/15
=50days
Therefore, 15 men can do the work in 50 days.


V. Ravi starts for his school at 8:05 a.m. on his bicycle. If he travels at a speed of 15 km/h, then he reaches his school late by 5 minutes but on traveling at 20 km/.h he reaches the school 10 minutes early. At what time does school start?

Solution:

Let the total distance =x km
Let the Time be taken =t min
With the speed of bicycle 15 km/hr, then he reaches by 5 min
x/15=t+5/60
x/15=t+1/12 –>(1)
With speed 20 km/hr then reaches his school 20 min early-
x/20=t-10/60
x/20=t-1/6 —>(2)
Subtracting 2 from 1, we get
x/15-x/20=1/12+1/6
4x-3x/60=1+2/12
x/60=1/4
x=15
Put x=15 in eq 1
1=t+1/12
t=1-1/12
t=11/12h
11/12*60=55 min
Timing of the school=8:05+55= 9:00 a.m.
Therefore, The school starts at 9:00 a.m.


VI. Sindhu starts at 8:00 AM by bicycle to reach school. She cycles at the speed of 16 km/hour and reaches the school at 8:15 AM. By how much should she increase the speed so that she can reach the school at 8:10 AM?

Solution:

This is a situation of inverse variation, now we solve using the unitary method.
In 15 minutes the same distance is covered at the speed of 16 km/hr.
In 1 minute the same distance is covered at the speed of (16 × 15) km/hr.
In 10 minutes the same distance is covered at the speed of (16 × 15)/10 km/hr.
Therefore, in 10 minutes the same distance is covered at the speed of 24 km/hr.


VII. A car covers a particular distance in 4 hours with a speed of 60 miles per hour. If the speed is increased by 20 miles per hour, find the time taken by the car to cover the same distance.

Solution:

This is a situation of inverse variation.
Because
more speed —–> less time
Given that,
Time is taken by the car to cover the particular distance = 4 hours and
Speed of the car = 60 mph
The formula to find the distance is
Distance = Time ⋅ Speed
Distance = 4 ⋅ 60 = 240 miles
If the given speed of 60 mph is increased by 20 mph, then the new speed will be 80 mph.
The formula to find the time is,
Time = Distance / Speed
Time = 240 / 80
Time = 3 hours
If the speed is increased by 20 mph, the time taken by the truck is 3 hours.


VIII. A man has enough money to buy 12 kg of apples at Rs150 per kg. How much can he buy, if the price is increased by Rs 2 per kg?

Solution :

This is a situation of inverse variation.
Because
more price —–> lesser pounds of apples
Cost of 12 kg of apples at Rs 150 per kg is= 12 ⋅ 150
= Rs 1800
So, the person has Rs1800.
If the price is increased by Rs 2 per kg, then the new price per kg is
= Rs 152
No. of pounds of apples he can buy with Rs 1800 is
= 1800 / 152
= 11.84
If the price is increased by Rs 2 per kg, the person can buy 11.84 kg of apples.


IX. If 5 men can paint a house in 12 hours, how many men will be able to paint it in 10 hours?

Solution:

This is a situation of inverse variation.
Because
fewer hours —–> more men
In 12 hours, the house can be painted by 5 men
No. of hours taken by 1 man to paint the house is
= No. of hours ⋅ No. of men
= 12 ⋅ 5= 60 hours
No. of men required to paint the house in 10 hours is
= 60 / 10
= 6 men
Therefore, 6 men will be able to paint the house in 10 hours.


X. Dev can complete a work in 7 days working 6 hours per day. If he works 3 hours per day, how many days will he take to complete the work?

Solution:

This is a situation of inverse variation.
Because
fewer hours per day —–> more days to complete the work
6 hours per day ——–> 7 days to complete the work
1 hour per day ———> 7 ⋅ 6 = 48 days
3 hours per day ———> 48 / 3 = 16 days
Dev can complete the work in 16 days working 3 hours per day.


XI. A man can type 8 pages of a book every day and complete it in 40 days. How many days will he take to complete it, if he types 16 pages every day?

Solution:

This is a situation of inverse variation.
Because
more pages per day—–> less days to complete the book
8 pages per day ——> 40 days
1 page per day ——–> 8 ⋅ 40 = 320 days
16 pages per day ——> 320 / 16 = 20 days
The man will complete the book in 20 days if he types 16 pages per day.


XII. If a builder can construct a house by using 30 people in 150 days, how many people are required to complete the work in 50 days?

Solution:

Given that,
30 people can construct a house in 150 days.
150 days —–> 30 people
1 day    ——-> (30. 150) people
50 days ——-> (30.150)/50= 90
Therefore, 90 people are required to complete the work in 50 days.


 

 

Problems on Unitary Method using Direct Variation

Problems on Unitary Method using Direct Variation | Unitary Method using Direct Variation Questions with Solutions PDF

Unitary Method is all about finding the value of a single unit. Here in this article, we have covered Problems on Unitary Method using Direct Variation. Refer to them and try to solve on your own and then cross-check with the solutions provided to assess your preparation standards. Unitary Method using Direct Variation Questions with Solutions PDF can be quite handy to finish your homework or clear your queries after the class. Solve different types of problems on direct variation using unitary method and become proficient in the concept.

Do Refer:

Unitary Method Direct Variation Sums

I. If 4 men and 2 women can earn Rs 320 in a day, find how much 6 men and 12 women will earn in a day?

Solution:

This is a situation of direct variation.
More men can earn more in a day.
In a day 4 men can earn money= Rs 320
1 man can earn =Rs 320/4 and
6 men can earn Rs 320/4 × 6 = Rs 480
Also, 2 women can earn Rs 320 and
1 woman can earn Rs 320/2 = Rs 160
12 women can earn = Rs 160 × 12 = Rs 1920
Therefore, 6 men and 12 women can earn Rs(480 + 1920) = Rs 2400.


II. A labour gets Rs 900 for 12 days of work. How many days should he work to get Rs 2400?

Solution:

This is also a situation of direct variation as money is received for working more days.

Rs 900 is earned by labour in 12 days.

Rs 1 is earned by labour in 12/900 days.

Rs2400 is earned by labour in 12/900 × 2400 days=30

Therefore, Rs 2100 is earned by labour in 30 days.


III. If Ajay saves 960 Rs in 12 days, how much amount will he save in 25 days?

Solution:

Given that,
Ajay saves money in 12 days=Rs 960
Saving of Ajay in 1 day=960/12
​=80 Rs
Saving of Ajay in 25 days=80 × 25 =2000 Rs
Therefore, Ajay saves Rs 2000 in 25 days.


IV. 350 apples can be placed in 25 cartoons. How many apples can be placed in 16 cartoons?

Solution:

Given that,
No.of apples placed in 25 cartoons is 350
No.of apples can be placed in 1 cartoon =350/25
​=14 apples
Hence, No. of apples can be placed in 16 cartoons =16 × 14= 224 apples.


V. In a business, if A can earn Rs 8000 in 2 years, At the same rate, find his earning for 5 years.

Solution :

This is a situation of direct variation.
Because more time —–> more earning
Given that,
Earning for 2 years is Rs 8000
Earning for 1 year is
= 8000 / 2
= Rs4000
Then, earning for 5 years is
= 5 ⋅ 4000
= Rs 20,000
So, the earning for 5 years is Rs 20,000.


VI. Rakesh would like to hire a call taxi for 700 miles trip. If the cost of the taxi is Rs 5.50 per mile, what is the total cost for his trip?

Solution:

This is a situation of direct variation.
Because
more miles —–> more cost
Given that,
Cost for one mile = Rs 5.50
The cost of 300 miles is
= 5.50 ⋅ 300
= Rs 1650
Hence, the total cost for the trip is Rs 1650.


VII. Janci ordered 500 gm of ghee for Rs 495. Then she reduced her order to 280 units. How much money does Janci have to pay for 280 units?

Solution:

This is a situation of direct variation.
Because fewer units —–> less cost
Cost of 500 gm = Rs 495
The cost of 1 unit is
= 495 / 500
= 0.99
The cost of 280 units is
= 0.99 ⋅ 280
= Rs 277
So, Janci has to pay Rs 277 for 280 units.


VIII. If Dev sells 2 litres of juice for Rs 40, how much money will he earn by selling 70 litres of juice?

Solution :

This is a situation of direct variation. Because
more gallons of juice —–> amount received will be more
Given that, 2 litres of juice cost Rs 40
The cost of one gallon of juice is= 40 / 2
= Rs 20
The Cost of 70 litres of juice is
= 70 ⋅ 20
= Rs 1400
Hence, Dev will earn Rs1400 by selling 70 litres of juice.


IX. In 3 days, Sarath raised Rs 2,379 for cancer research. How much money will he raise in 20 days?

Solution:

This is a situation of direct variation. Because
more number of weeks —-> amount raised will be more
Given that, Sarath raised Rs 2,379 in 20 days
Amount raised in one day is
= 2379 / 3
= Rs 793
Amount raised in 20 days is= 793 ⋅ 20
= Rs 15860
Hence, the money raised in 20 weeks is Rs15860.


X. Ramya can type 690 words in 30 minutes. How many words will she be able to type in 10 minutes with the same efficiency?

Solution:

The number of words typed in 30min=690.
The number of words typed in 1min=690/30=23
Therefore, the number of words typed in 10min=10×23=230.
Hence, Ramya will be able to type 230 words in 10 minutes.


XI. 75 plastic boxes cost Rs 3,750. Find the cost of 24 such plastic boxes?

Solution:

This is a situation of direct variation.
Because
less number of plastic boxes —–> cost will be less
Given that, 75 plastic boxes cost Rs 3,750
The cost of one plastic box is
= 3750 / 75
= Rs 50
Then, the cost of 24 plastic boxes is
= 24 ⋅ 50
= Rs 1200
Therefore, the cost of 24 plastic boxes is Rs 1200.


XII. 20 m of cloth costs Rs 900. Find the cost of 65 m of the same cloth?

Solution:

Given that,
The cost for 20 m of cloth is= Rs 900
The cost for 1 m of cloth is =900/20
=45 Rs
Thus, the cost for 65 m of cloth =65 × 45= Rs 2925


 

 

Problems using Unitary Method

Problems Using Unitary Method | Unitary Method Questions and Answers

Are you facing difficulties in solving Problems using Unitary Method? If that is the case you don’t have to fret anymore as we are here to help you learn how to solve problems using unitary method. We have covered different models of questions framed on the topic so that you can master the topic.

The Unitary Method Examples provided include questions on finding unit value from a gross value, determining the required quantity using unit quantity, etc. Have a quick recap of the concept of the Unitary Method by answering the Word Problems on Unitary Method available here.

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Unitary Method Questions with Solutions

I. A bicycle factory produces 1428 bicycles in a week. How many bicycles will it produce in 20 days?

Solution:

Given that,
A bicycle factory produces bicycles in a week=1428
No. of bicycles produced in a day=1428/7=204 cycles
No. of bicycles produced in 20 days=204 × 20=4080
Therefore, No. of bicycles produced in 20 days is 4080.


II. 5 workers can make 80 toys in 8 days. How many workers will be required to make 100 toys in 4 days?

Solution:

5 workers can make 80 toys in 8 days.
1 worker can make 80 toys in 8 × 5 days.
1 worker can make 1 toy in (8 × 5)/100 days.
Let the number of workers required be x, then;
x workers can make 1 toy in (8 × 5)/( 80 × x) days
x workers can make 100 toys in (8 × 5 × 100)/(80 × x ) days
But the number of days given = 4
According to the problem,
(8 × 5 × 100)/(80× x ) = 4
4000/80x = 4
320x = 4000
x = 4000/320
x = 12
Therefore, 12 workers are required to make 100 toys in 4 days.


III. The weight of 75 articles is 5 kg. What is the weight of 180 such articles? How many such articles weigh 8 kg?

Solution:

The weight of 75 articles is= 5 kg
1 article will weigh= 5/75 kgs
So 180 such articles will weigh= 5/75 x 180
=12 kgs
Now, we have to find out how many such articles will weigh up to 8 kg..
If 5 kg weight is of 75 articles
Then 1 kg will be the weight of =75/5
=15
And 8 kg will be the weight of 15 x 8 articles
= 120
Hence, 120 articles will weigh up to 8 kg.


IV. The cost of 5 kgs of rice is Rs. 450. Find the cost of 40 kgs?

Solution:

Given that,
The cost of 5 kgs of rice is=Rs 450
The cost of 1 kg of rice is=450/5=90
The cost of 40 kgs of rice is=90 x 40=3600
Hence, the cost of 40 kgs of rice is Rs 3600.


V. 5 men can build the wall in 10 days. How many men will build the wall in 2 days?

Solution:

Given that,
5 men can build the wall in 10 days
We know that more men will take less time to build the wall. This is a case of inverse proportion.
In inverse proportion,
x1y1=x2y2
5 x  10=x. 2
50=2x
x=50/2=25
Therefore, 25 men will build the wall in 2 days.


VI. A, B, and C can do a piece of work in 20 days, 30 days, and 60 days respectively. At what time will they all together finish it ?

Solution:

A’s 1 day work = 1/20
B’s 1 day work = 1/30
C’s 1 day work = 1/60
(A + B + C)’s 1 day work = 1/20 + 1/30 + 1/60
L.C.M of (20, 30, 60) = 60.
Then, we have
(A + B + C)’s 1 day work = 3/60 + 2/60 + 1/60
(A + B + C)’s 1 day work = 6/60
(A + B + C)’s 1 day work = 1/10
So, A, B, and C can together finish the work in 10 days.


V. A worker is paid Rs.850 for 5 days’ work. If he works for 18 days, how much will he get?

Solution:

Given that,
A worker is paid money for 5 days’ work=Rs 850
A worker paid money for 1 day of work=850/5=170
For 18 days work=170 x 18= 3060
Hence, A worker will get money of Rs 3060.


VI. If 24 men can do a piece of work in 30 days, then in how many days will 40 men complete the same work?

Solution:

Given that,
24 men can do a piece of work in 30 days.
1 man can do the same work in 30/24 days.
Therefore, 40 men can do it in 30×40/24 days
=50 days
Hence,  40 men can do it in 50 days.


VII. A car travels 200 km in 3 hours.
(a) How long will it take to travel 900 km?
(b) How far will it travel in 7 hours?

Solution:

Given that,
A car travels 200 km in 3 hours.
(a) To travel 1 km, Time taken will be=3/200 hours
To travel 900 km, Time taken will be=3 × 900/200=13.5 hours
Hence, For traveling 900 km the time taken will be 13.5 hours.
(b) In 3 hours, the car travels = 200 km
In 1 hour, the car will travel = 200/3 km
In 7 hours, the car will travel = (200*7)/3
= 466.6 km
Therefore, In 7 hours the car will travel 466.6 km.


VIII. In a camp, there are provisions for 100 persons for 25 days. If 40 more persons join the camp, find the number of days the provision will last?

Solution:

Given that,
No. of persons in the camp=100
No. of days provisions are provided=25
No. of persons will join the camp=40
If no. of persons increase provisions decrease. So it is inversely proportional.
100 × 25=40 × x
2500=140x
x=2500/140
x=17
Hence, Provisions will last for 17 days.


IX. The annual rent of a shop is Rs 48000. Find the rent for 7 months?

Solution:

Given that,
The annual rent of a shop is= Rs 48000
Rent for 1 month=Rs 48000/12=Rs 4000
Rent for 7 months=7 ×Rs 4000=Rs 28000
Hence, Rent for 7 months is Rs 28000.


X. 15 workers harvest the crops in the field in 7 hours. How many workers will be required to do the same work in 5 hours?

Solution:
Given that,
15 workers harvest the crops in the field = 7 hours
No. of workers harvest the crops in the field in one hour=15 × 7=105
No. of workers harvest the crops in the field in 5 hours= 105/5=21
Therefore, No. of workers who harvest the crops in the field in 5 hours is 21.

XI. 3 buses carry 249 passengers. How many passengers can be carried by 7 such buses?

Solution:

Given that,
No. of passengers can be carried by 3 buses= 249
No. of passengers can be carried by1 bus=249/3=83
No. of passengers can be carried by 7 such buses=83 ×7=581
Hence, 581 passengers can be carried in 7 such buses.


XII. Rajesh can walk 120 km in 15  hours. What distance can he walk in for 4 hours?

Solution:

Given that,
No. of km Rajesh can walk in 15  hours=120 km
No. of km Rajesh can walk in one hour=120/15=8 km
In 4 hours, Rajesh can walk=8 × 4=32 km.
Hence, Rajesh can walk 32 km in 4 hours.


XIII. A train covers 680 km in 8 hours. What distance will it cover in 9 hours?

Solution:

Given that,
In 8 hours the train covers distance=680
The train covers the distance in 1hour = 680/8=85
In 9 hours the train covers distance=85 x 9=765 km
Hence, The distance covered in 9 hours is 765 km.