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Trigonometry Ratios Table 0-360: Trigonometry is a branch of mathematics that deals with the study of the length and angles of a triangle. It is usually associated with a right-angle triangle in which one of the angles is 90 degrees. It has a vast number of applications in the field of mathematics. You can figure out many geometrical calculations much simpler if you are aware of the Trigonometric Functions and Table.

Trigonometric Ratios Table help you find the trigonometric standard angles such as 0°, 30°, 45°, 60°, and 90°. You can find Trigonometric Ratios such as sine, cosine, tangent, cosecant, secant, cotangent, etc. In short, you can write the Trigonometric Ratios as sin, cos, tan, cosec, sec, and cot. You can solve Trigonometry Problems easily if you know the standard values of the Trigonometric Ratios. Thus, remember the standard angle values to make your job easier.

Trigonometric Table has a wide range of applications and it was used ever since before the existence of calculators.  Another Important Application of the Trigonometric Table is in the Fast Fourier Transforms.

Trigonometric Ratios Table for Standard Angles

Trig Values Table: 0 to 360 Degrees

Tricks to Remember Trigonometry Table

It is easy to remember the trigonometry table. If you are aware of the trigonometric formulas remembering the table is quite simple. The Trigonometric Ratios Table is dependant on the Trigonometric Formulas. Try to remember the trigonometric table easily by going through the simple formulas.

• sin x = cos (90° – x)
• cos x = sin (90° – x)
• tan x = cot (90° – x)
• cot x = tan (90° – x)
• sec x = cosec (90° – x)
• cosec x = sec (90° – x)
• 1/sin x = cosec x
• 1/cos x = sec x
• 1/tan x = cot x

How to Create a Trigonometric Ratio Table?

Check out the simple guidelines listed below to create a Trigonometric Table having Values of Standard Angles. They are in the following fashion

Step 1:

Create a table having the top row and list out the angles 0°, 30°, 45°, 60°, 90° and also write trigonometric functions such as sin, cos, tan, cosec, sec, cot.

Step 2: Determine the Value of Sin

In the second step determine the value of sin, divide 0, 1, 2, 3, 4 by 4 under the root.

Step 3: Determine the Value of Cos

Cos is opposite to sin and to find the value of cos divide by 4 in the opposite sequence of sin. For instance, divide 4 with 4 under the root to obtain the value of cos 0°

Step 4: Determine the value of tan

Tan is obtained by dividing sin with cos. To find the value of tan 0° divide the Value of Sin 0° by the Value of Cos 0°.

Step 5: Determine the value of the cot

Value of cot is equal to reciprocal of tan. The value of cot at 0° is obtained by dividing 1 with the value of tan at 0°. In the same way, you can find the value of the cot for all the angles.

Step 6: Determine the value of cosec

Cosec value at 0° is the reciprocal of sin at 0°. You can find all the angles of cosec as such

Step 7: Determine the value of sec

sec values can be obtained by the reciprocal values of cos. Sec value at 0° is the opposite of cos on 0°. In the similar way entire table of values is given.

FAQs on Trigonometric Ratios Table

1. How to find the Trigonometric Functions Values?

All the Trigonometric Functions Values can be found easily using the formulas and they are given as such

• Sin = Opposite/Hypotenuse
• Cos = Adjacent/Hypotenuse
• Tan = Opposite/Adjacent
• Cot = 1/Tan = Adjacent/Opposite
• Cosec = 1/Sin = Hypotenuse/Opposite
• Sec = 1/Cos = Hypotenuse/Adjacent

2. What is Trigonometric Values Table?

Trigonometric Values table is made of trigonometric ratios that are interrelated to each other – sine, cosine, tangent, cosecant, secant, cotangent.

3. What are Trigonometric Ratios?

Trigonometric Ratios is a relationship between measurements of length and angles of a right angle triangle.

Whole Numbers is a part of a number system that includes all the positive integers from 0 to infinity. These numbers are present on the number line and are usually called real numbers. Thus, we can say that Whole Numbers are Real Numbers but not all Real Numbers are Whole Numbers. Complete Set of Natural Numbers including “0” are called Whole Numbers.

Whole Numbers – Definition

Whole Numbers are numbers that don’t have fractions and is a collection of positive integers including zero. It is denoted by the symbol “W” and is given as {0, 1, 2, 3, 4, 5, ………}. Zero on a whole denotes null value or nothing.

• Whole Numbers: W = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10……}
• Natural Numbers: N = {1, 2, 3, 4, 5, 6, 7, 8, 9,…}
• Integers: Z = {….-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,…}
• Counting Numbers: {1, 2, 3, 4, 5, 6, 7,….}

Whole numbers are positive integers along with zero and don’t have fractional or decimal parts. You can perform all the basic operations such as Addition, Subtraction, Multiplication, and Division.

Symbol

The Symbol to denote the Whole Numbers is given by the alphabet W = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,…

• All-natural numbers are whole numbers
• All positive integers including zero are whole numbers
• All whole numbers are real numbers
• All counting numbers are whole numbers

Properties of Whole Numbers

Whole Numbers Properties depend on arithmetic operations such as Addition, Subtraction, Multiplication, Division. When you multiply or add two whole numbers the result will always be a Whole Number. If you Subtract Two Whole Numbers the result may not always be a Whole Number and it can be an Integer too. Division of Whole Numbers can result in a Fraction at times. Let us see few more Properties of Whole Numbers by referring below.

Closure Property: Whole Numbers can be closed under addition or multiplication. If a, b are two whole numbers then a.b and a+b is also a whole number.

Commutative Property of Addition and Multiplication: Sum and Product of Two Whole Numbers will be the same no matter the order in which they are added or multiplied. If a, b are two whole numbers then a+b = b+a, a.b = b.a

Additive Identity: If a Whole Number is added to 0 the result remains unchanged. If a is a whole number then a+0 = 0+a = a

Multiplicative Identity: Whenever you multiply a whole number with 1 the result remains unchanged. Let us consider a whole number “a” then a.1 = 1. = a

Associative Property: If you are grouping the whole numbers and adding or multiplying a set the result remains the same irrespective of the order. If a, b, c are whole numbers then a + (b + c) = (a + b) + c and a. (b.c)=(a.b).c

Distributive Property: If a, b, c are three whole numbers then the distributive property of multiplication over addition is given by a.(b+c) =(a.b)+(a.c), Similarly Distributive Propoerty of Multiplication over Subtraction is given by a.(b-c) = (a.b)-(a.c)

Multiplication by Zero: If you multiply a Whole Number with Zero the result is always zero. i.e. a.0=0.a=0

Division by Zero: If you divide a Whole Number with Zero the result is undefined, i.e. a divided by 0 is not defined.

Difference between Natural Numbers and Whole Numbers

By referring to the below sections you will better understand the difference between Whole Numbers and Natural Numbers.

Solved Examples on Whole Numbers

1. Are 101, 147, 193, 4028 whole numbers?

Yes, 101, 147, 193, 4028 are all whole numbers.

2. Solve 8 × (3 + 12) using the Distributive Property?

We know as per the Distributive Property a.(b+c) =(a.b)+(a.c)

Applying the Input Numbers in the formula we have the equation as such

8 × (3 + 12) = 8*3 +8*12

= 24+108

= 132

FAQs on Whole Numbers

1. Is 0 a Whole Number?

Yes, 0 is a Whole Number.

2. What is the Symbol of Whole Numbers?

The Letter W represents the Whole Numbers.

3. Are all Natural Numbers Whole Numbers?

Yes, all Natural Numbers are Whole Numbers but not all Whole Numbers are Natural Numbers. Natural Numbers begin from 0 and counts till infinity. Whole Numbers begin from 0 and end at infinity.

4. What is the set of whole numbers?

The whole numbers are the natural numbers together with 0. The set of whole numbers is a subset of the integers but does not include the negative integers.

Binary Subtraction is one among the four binary operations in which we perform subtraction of binary numbers i.e. 0 or 1. It is similar to the Basic Arithmetic Operation of Decimals. When we subtract 1 from 0 we need to borrow 1 from the next digit to reduce the digit by 1 and the remainder left here is 1. Go through the entire article to know about Binary Subtraction Rules, Subtraction Table, Tricks and Procedure on How to Subtract Binary Numbers, etc.

What is meant by Binary Subtraction?

Binary Numbers Subtraction is similar to Subtraction of Decimals or Base 10 Numbers. For instance, 1+1+1 is 3 in base 10 whereas in a binary number system 1+1+1 is 11. While Performing Addition and Subtraction in Binary Numbers be careful with borrowing as you might need to do them quite often.

While performing subtraction of several columns of binary digits you need to consider the borrowing. If you subtract 1 from 0 the result will be 1 where 1 is borrowed from the next highest order digit.

Binary Subtraction Table

On Adding Two Binary Numbers 1 and 1 we get the result 10 in which we consider 0 and carry forward 1 to the next higher-order bit. On Subtracting 1 from 1, the result is 0 and nothing will be carry forwarded.

While subtracting 1 from 0 in the case of decimal numbers we borrow 1 from the preceding higher-order number and make it 10 and after subtracting result becomes 9. However, in the case of Binary Subtraction, the result is 0.

Rules for Binary Subtraction

Binary Subtraction is quite simple compared to Decimal Subtraction if you remember the following tips and tricks.

0 – 0 = 0
0 – 1 = 1 ( with a borrow of 1)
1 – 0 = 1
1 – 1 = 0

You can look at the binary subtraction examples provided below for better understanding.

How to Subtract Binary Numbers?

Follow the below-listed steps to perform Binary Subtraction. You will find the Subtraction of Binary Numbers much easier after going through the below steps. They are as follows

• Align the numbers similar to an ordinary subtraction problem. Write the larger number up and the smaller number below it. If the Smaller Digit has few digits align them towards the right same as in decimal subtraction.
• Begin from the right column and perform the subtraction operation of binary numbers. While doing so keep the binary subtraction rules in mind and do accordingly.
• Solve column by column moving from right to left.

Binary Subtraction Examples

1. Find the Value of 1010011 – 001110?

Solution:

Write the given numbers as if you subtracting decimal numbers. Align them to the right and fill them with leading zeros so that both the numbers have the same digits.

1011011

(-)0001010

——————

1010001

Binary Notation

Decimal Notation

The decimal Equivalent of given numbers is

1011011 = 91

001010 = 10

91-10 = 81

2. Find the value of 1100010 – 001000?

Solution:

Write the given numbers as if you subtracting decimal numbers. Align them to the right and fill them with leading zeros so that both the numbers have the same digits.

1100010

(-)0001000

——————

1,011,010

Decimal Notation

The decimal Equivalent of given numbers is

1100010 = 98

1000 = 8

98-8 = 90

Binary Subtraction using 1’s Complement

Go through the below procedure and perform the Binary Subtraction easily. They are as follows

• Firstly, write the 1’s complement of the subtrahend.
• And then add the 1’s complement subtrahend with the minuend.
• If the result has a carryover add the carryover in the least significant bit.
• If it has no carryover take the 1’s complement of resultant and it is negative.

Questions on Binary Subtraction using 1’s Complement

1. (11001)₂  – (1010)₂

Solution:

(11001)₂  = 25

(1010)₂  = 10 – subtrahend

Fill with leading zeros till you have the same number of digits in both the numbers. Firstly, take the 1’s complement of subtrahend i.e. (01010)₂.

1’s complement of the subtrahend is 10101. Add 1 to the 1’s complement of the second number

10101
+   1
↓
10110

Now instead of subtracting add the 1’s complement of the second number to the first one

11001
+ 10110

——————
101111

Remove the leading 1 to obtain the result.

Then, remove the leading zeros as it will not alter the result and you can write the final result of subtraction as such

11001
– 1010

——————

1111

FAQs on Binary Subtraction

1. What are the Rules of Binary Subtraction?

Rules of Binary Subtraction are as follows

0 – 0 = 0
0 – 1 = 1 ( with a borrow of 1)
1 – 0 = 1
1 – 1 = 0

2. How many basic binary subtraction combinations are possible?

There are four possible binary subtraction combinations when subtracting binary digits.

3. What signs does the binary digit 0 and 1 represent?

0 represents the positive sign and 1 represents the negative sign.

Mathematics is the study of numbers, shapes, and patterns. It includes various complex and simple arithmetic topics that help people in their daily life routine. In Maths, Numbers play a major role and they can be of different types like Real Numbers, Whole Numbers, Natural Numbers, Decimal Numbers, Rational Numbers, etc. Today, we are going to discuss one of the main topics of Decimal Numbers. In Decimals, identifying the Decimal Place Values is a fundamental topic and everyone should know the techniques clearly. So, here we will be discussing elaborately the topic of Decimal Place Values Chart.

Let’s get into it.

What is the Place Value of Decimals?

Place value is a positional notation system where the position of a digit in a number, determines its value. The place value for decimal numbers is arranged exactly the identical form of treating whole numbers, but in this case, it is reverse. On the basis of the preceding exponential of 10, the place value in decimals can be decided.

Decimal Place Value Chart

On the place value chart, the numbers on the left of the decimal point are multiplied with increasing positive powers of 10, whereas the digits on the right of the decimal point are multiplied with increasing negative powers of 10 from left to right.

• The first digit after the decimal represents the tenths place.
• The second digit after the decimal represents the hundredths place.
• The third digit after the decimal represents the thousands place.
• The rest of the digits proceed to fill in the place values until there are no digits left.

How to write the place value of decimals for the number 132.76?

• The place of 6 in the decimal 132.76 is 6/100
• The place of 7 in the decimal 132.76 is 7/10
• The place of 2 in the decimal 132.76 is 2
• The place of 3 in the decimal 132.76 is 30
• The place of 1 in the decimal 132.76 is 100.

Examples:

1.  Write the place value of digit 7 in the following decimal number: 5.47?

The number 7 is in the place of hundredths, and its place value is 7 x 10 ^-2 = 7/100 = 0.07.

2. Identify the place value of the 6 in the given number: 689.87?

Given number is 689.87

The place of 6 in the decimal 689.87 is 600 or 6 hundreds. 

3. Write the following numbers in the decimal place value chart.

(i) 4532.079

(ii) 490.7042

Solutions: 

(i) 4532.079

4532.079 in the decimal place value chart.

(ii) 490.7042

490.7042 in the decimal place value chart.

Solving the Profit and Loss Questions can give you an idea of how to solve related problems. Know different methods and formulae involved to calculate the Profit and Loss. Try to solve the Profit and Loss Questions on your own and then verify your solution with ours to know where you went wrong. By solving them regularly you can increase your speed and accuracy thereby attempt the exam well and score better grades in the exam.

Question 1:

If the manufacturer gains 15%, the wholesale dealer gets 20% and the retailer gets 30%, then find the cost of production of a blackboard, the retail price of which is $ 1360?

Solution:

Cost of production of blackboard be ‘X’

115% of 120% of 130% of cost price is

i.e. \(\frac { 115 }{ 120 } \)*\(\frac { 120 }{ 100 } \)*\(\frac { 130}{ 100} \)*1360

Cost of production of a blackboard $ 758.08

Question 2:

A man bought a dog and a kennel for $ 4500. He sold the dog at a gain of 25% and the carriage at a loss of 15%, thereby gaining 3% on the whole. Find the cost of the dog.

Solution:

Cost price (C.P.) of dog ‘X’

Cost price (C.P.) of kennel ‘4500 – X’

25% of x – 15% of (4500 – X) = 3% of 4500

Cost of a dog is $ 1373.68.

Question 3:

Profit earned by selling television for $ 6000 is 25% more than the loss incurred by selling the article for $ 4500. At what price should the article be sold to earn 25% profit?

Solution:

let cost price (C.P.) be ‘X’

By equalizing,

(6000 – X) = \(\frac { 125 }{ 100 } \)*(X-4500)

Desired selling price of the television is $ 6458.3.

Question 4:

A manufacturer undertakes to supply 2200 pieces of a particular component at $ 30 per piece. According to his estimates, even if 6% fail to pass the quality tests, then he will make a profit of 30%. However, as it turned out, 60% of the components were rejected. What is the loss to the manufacturer?

Solution:

Incurred cost = $(\(\frac { 100 }{ 130 } \)*30*(94% of 2200))

Loss = C.P. – S.P.

The loss to the manufacturer is $ 8123.07.

Question 5:

John bought a lorry for a certain sum of money. He spent 15% of the cost on repairs and sold the lorry for a profit of $ 1600. How did he spend on repairs if he made a profit of 30%?

Solution:

Let the Cost price (C.P.) of the lorry be ‘X’

Profit (P) = $16,00

Profit (%) = 30

Profit (%) = \(\frac { 100 }{ CP } \)*P

30 = \(\frac { 100 }{ CP} \)*1600

CP = \(\frac { 100 }{ CP} \)*1600

CP = \(\frac { 100 }{ 30} \)*1600

= $5333.3

The C.P Includes both original price and repairs cost

C.P +0.15C.P = $5333.3

1.15 C.P = $5333.3

C.P = $4637.68

Repairs Cost = Total Cost Price – Cost Price for which john bought the lorry

= $5333.3 – $4637.68

= $ 695.61

Expenditure spend on repairs = $ 695.61.

Question 6:

A boy bought 25 litres of milk at the rate of $ 12 per litre. He got it churned after spending $ 15 and 7kg of cream and 25 lit of toned milk were obtained. If he sold the cream at $ 40 per kg and toned milk at $ 6 per lit, what is his profit % in the transaction?

Solution:

Cost price (C.P.) = $ ((25*12) + 15)

Selling price (S.P.) = $ ((7*40) + (25*6))

Gained profit % on the above transaction is 36.50%.

Question 7:

Arun purchased 150 reams of paper at $ 90 per ream. He spent $ 300 on transportation, paid local tax at the rate of 50 paise per ream, and paid $ 76 to coolie. If he wants to have a gain of 10%, what must be the selling price per ream?

Solution:

Total investment = $ (((150*90) + 300) + ((50/100 * 150) + 76))

Find selling price per ream

Selling price per ream = $ 102.3

Question 8:

A retailer mixes three varieties of dal costing $ 50, $ 25, and $ 30 per kg in the ratio 3: 5: 2 in terms of weight, and sells the mixture at $ 35 per kg. What percentage of profit does he make?

Solution:

Cost price (C.P.) of dal for 10 kgs = $ ((3*50) + (5*25) + (2*30))

Selling price for 10 kgs = $ (10*35)

Profit percentage on transactions is 4.74%.

Question 9:

A man buys chocolates at 3 for $ 2 and an equal number at 5 for $ 3 and sells the whole at 6 for $ 4. His gain or loss percent is?

Solution:

Suppose he buys 7 eggs of each kind

Find C.P. and S.P. for 14 eggs

C.P. = $ ((\(\frac { 2 }{ 3 } \) * 7) + (\(\frac { 3 }{ 5 } \) * 7)) = $ 8.86

Similarly, find S.P.

The gain % obtain is 5.34%.

Question 10:

The manufacturer of a certain item can sell all he can produce at the selling price of $ 65 each. It costs him $ 45 in material and labor to produce each item and he has overhead expenses of $ 3500 per week in order to operate the plant. The number of units he should produce and sell in order to make a profit of at least $ 1200 per week is?

Solution:

Consider he produces ‘x’ items

Cost price (C.P.) = $ (45x + 3500)

Selling price (S.P.) = $ 65x

The number of units produced per week to gain profit is 235.

Looking for help on finding Profit and Loss Concepts? Then, you have come the right way. Here let us observe some fully solved example problems on calculating profit or loss. You can find step by step solutions to all the Profit and Loss Questions available here. Try Practicing from the Profit and Loss Problems and get acquainted with the concepts better. Learn various methods for Calculating Profit and Loss and solve related problems easily. Assess your preparation standards on the concept and concentrate on the areas you are lagging in accordingly.

Question 1:
If a manufacturer allows 40% commission on the retail price of his product, he earns a profit of 9%. What would be his profit percent if the commission is reduced by 25 percent?

Solution:

We need to find out the profit % when the given commission is reduced by 25 percent.
Given data:
According to the question consider
Cost price (C.P.) of the product = $ 100
Then, a commission of the product = $ 40
Therefore selling price (S.P.) = $ (cost price (C.P.) – commission)
= $ (100 – 40)
= $ 60
Given that profit = 9%
Therefore Cost price (C.P.) = \(\frac { 100 }{ 100+gain%} \)* S.P
So,
C.P. = $ \(\frac { 100 }{ 100+9 } \)* 60
= $ \(\frac { 6000 }{109 } \)
Now new commission = $ 15
Therefore new selling price (S.P.) = $ 100 – 15
= $ 75
Gain = S.P. – C.P.
= $ (75 – \(\frac { 6000 }{109 } \))
= $ \(\frac { 2175 }{109 } \)
Gain% = (\(\frac { Profit }{C.P. } \)*100)%

=(\(\frac { 2175 }{109 } \)*\(\frac { 109 }{6000 } \)*100)%

= 36.25 %
Hence, gain % is 36.25.
Question 2:
After getting two successive discounts, a pant with the least price of $ 200 is available at $ 125. If the second discount is 14%, find the first discount.

Solution:
Let the first discount be ‘P%’
Then, 86% of (100 – P) % of 200 = 125

\(\frac { 86 }{ 100 } \)*\(\frac { (100 – P) }{ 100 } \)*200 = 125

100-P = \(\frac {(125*100*100) }{ 200*86 } \)
100 – P = 72.67
P = 100 – 72.67
P = 27.32%
Therefore, first discount price of pant is 27.32%.
Question 3:
A women sells an article at a profit of 20%. If he had bought it at 15% less and sold it for $ 11.50 less, he would have gained 25%. Find the cost price of the article.

Solution:
Given data:
Consider cost price (C.P.) of article be ‘X’
First selling price of article ‘X’ = 120% of ‘X’

= \(\frac { 120 }{ 100 } \)*X
= \(\frac { 6 }{ 5 } \)*X
Cost price of article for ‘X’ at 75% = 75% of ‘X’
=\(\frac { 75 }{ 100 } \)*X

=\(\frac { 3 }{ 4 } \)*X
Second selling price of article ‘X’ = 125% of 3/4 * X
= \(\frac { 125 }{ 100 } \)*\(\frac { 3x }{ 4 } \)

= \(\frac { 15x }{ 16 } \)

As given the article is sold at $ 11.50 less
Therefore, selling prices are equalized to a reduced price

\(\frac { 6x }{5 } \) –\(\frac { 15x }{ 16 } \) = 11.50
\(\frac { 21x }{80 } \) = 11.50
X = $ 43.8
Almost equal to $ 44
Hence, the cost price of an article is given as $ 44.
Question 4:
A dealer sold three – fourth of his articles at a gain of 25% and the remaining at cost price. Find the profit earned by him in the whole transaction.

Solution:
A dealer sold his ¾ th quantity with a gain of 25% and the remaining ¼ that its cost price.
Given data:
Consider cost price (C.P.) of whole articles be ‘X’
Cost price (C.P.) of \(\frac { 3}{ 4} \)th quantity = $ \(\frac { 3x}{ 4} \)
Cost price (C.P.) of \(\frac { 1}{ 4} \)th quantity = $ \(\frac { x}{ 4} \)
Total selling price (S.P.) = $ ((125% of \(\frac { 3x}{ 4} \)) + \(\frac { x}{ 4} \))
= $ (\(\frac { 15x}{ 16} \) + \(\frac { x}{ 4} \))
= $ (\(\frac { 19x}{ 16} \))
Profit / Gain = S.P. – C.P.
= $ (\(\frac { 19x}{ 16} \) – x)
= $ \(\frac { 3x}{ 16} \).
Gain % = (\(\frac { gain}{ C.P. } \)*100)%

= (\(\frac {3x}{ 16 } \)*\(\frac {1}{ x } \)*100)%

= 18.75%.
Hence, the gain % of the article is 18.75%.
Question 5:
A man sold two flats for $ 775,000 each. On one he gains 18% while on the other he losses 18%. How much does he gain or lose in the whole transaction?

Solution:
In this problem he gets an equal amount of profit and loss such cases there is always a loss. Therefore the selling price (S.P.) is immaterial.
Loss % = (\(\frac {common loss and gain %}{ 10 } \))²

= (\(\frac {18 }{ 10 } \))²

= (\(\frac {324 }{ 100 } \))

= 3.24%
The total loss incurred by the person is 3.24%.
Question 6:
Pure petrol costs $ 100 per lit. After adulterating it with kerosene costing $ 50 per lit, a shopkeeper sells the mixture at the rate of $ 96 per lit, thereby making a profit of 20%. In what ratio does he mix the two?

Solution:
Here, we have two different cost prices for different mixtures and one selling price (S.P.).
Given data:
Cost price (C.P.) of petrol = $ 100 per lit
Cost price (C.P.) of kerosene = $ 50 per lit
Selling price (S.P.) of mixture = $ 96 per lit
As we have two cost prices,
Mean cost price = $(\(\frac {100 }{ 120 } \))* 96)

= $ 80 per lit.
Since they asked us to find a ratio it is easy to find out by the allegation rule
Cost price (C.P.) of a unit Cost price (C.P.) of a unit quantity of $ X item quantity of $ Y item
Mean cost
$ M
(M – Y) (X – M)
Similarly using this concept here,
Cost price (C.P.) of a unit Cost price (C.P.) of a unit quantity of $ 100 item quantity of $ 50 item
Mean cost
$ 80
(80 – 50) (100 – 80)
Therefore, required ratio = 30 : 20
= 3 : 2.

Question 7:
Find cost price (C.P.), when
1. Selling price (S.P.) = $ 50, Gain = 18%
2. Selling price (S.P.) = $ 51, Loss = 14%

Solution:
Here, we need to find cost price (C.P.) using below formulae
1. Given data
Selling price (S.P.) = $ 50 & Gain = 18%
C.P. = \(\frac { 100 + Gain%) }{ 100 } \)*S.P.

=$ \(\frac { 100 + 18 }{ 100 } \)*50= $ 59.
2. Given data
Selling price (S.P.) = $ 51, Loss = 14%

C.P. = \(\frac { 100 – Loss% }{ 100 } \)*S.P.

= $ \(\frac { 100 – 14 }{ 100 } \)*51

= $ 43.86.

If you are seeking help on the concept of Ratio and Proportion you can always make use of Worked out Problems on Ratio and Proportion. All the Problems are explained with straightforward description making it easy for you to understand the concept. Solve different problems on Ratio and Proportion available here firstly on your own and cross-check your solutions.

You can find Ratio and Proportion Questions related to Simplification of Ratios, Comparison of Ratios, Arranging Ratios in Ascending Order, Descending Order, Word Problems on Ratio and Proportion, etc. Sample Problems on Ratio and Proportion will help you get a good grip on the concept and its fundamentals too in no time.

Ratio and Proportion Problems and Solutions

1. Two numbers are in the ratio 4 : 5. If the sum of numbers is 72, find the numbers?

Solution:

Let the numbers be 4x and 5x

Since Sum of the Numbers is 72 we have the equation as such

4x+5x = 72

9x= 72

x = 72/9

= 8

Substitute the value of x to obtain the numbers

4x = 4*8  32

5x = 5*8 = 40

Therefore, the Numbers are 32 and 40.

2. If x : y = 3 : 2, find the value of (2x + 4y) : (x + 5y)?

Solution:

We know x:y = 3:2

we can rewrite it as

x/y = 3/2

Given equation (2x + 4y) : (x + 5y)

we can rewrite it as

(2x + 4y)/(x + 5y)

Dividing Numerator and Denominator with y we have the equation as follows

= (2(x/y)+4(y/y))/((x/y)+5(y/y))

Since we know the value of x/y substitute it in the above equation

= (2(1/2)+4(1))/((1/2)+5(1))

= (1+4)/(1/2+5)

= 5/(11/2)

= 10/11

Therefore, value of (2x + 4y) : (x + 5y) is 10/11.

3. The average age of three boys is 36 years and their ages are in the proportion 5 : 6 : 7. Find the age of the youngest boy?

Solution:

From the ratio 5:6:7, the ages of boys are 5x, 6x, 7x

Given Average Age of Boys = 36

5x+6x+7x = 36

18x = 36

x = 2

Age of Youngest Boy = 5x

= 5*2

= 10 years

Therefore, the Age of the Youngest Boy is 10 Years.

4. If 3A = 4B = 5C, find A : B : C?

Solution:

Let us assume a constant k

3A=4B=5C = k

equating them we have

3A= k, 4B = k, 5C = k

A = k/3, B = k/4, C = k/5…….(1)

Finding LCM for the obtained values 3, 4, 5

LCM(3, 4, 5) = 60

Multiplying with 60 the eqn (1) we get the Ratio as Follows

Ratio of A:B:C is 20:15:12

5. What must be added to each term of the ratio 3 : 2, so that it may become equal to 5 : 4?

Solution:

Let the Number to be added be x then (3+x):(2+x) = 5:4

(3+x)/(2+x) = 5/4

(3+x)4 = 5(2+x)

12+4x= 10+5x

12-10 = 5x-4x

x =2

To make the ratio 3:2 to 4:5 you need to add 2.

6. The length of the ribbon was originally 33 cm. It was reduced in the ratio 3:2. What is its length now?

Solution:

Length of Ribbon = 33 cm

Let the Original Length be 3x

Reduced Length be 2x

But 3x = 33 cm

x = 33 cm/3

= 11 cm

Reduced Length = 2x

= 2*11

= 22 cm

Therefore, the Length of the Ribbon is 22 Cm.

7. The ratio of the number of boys and girls is 5 : 3. If there are 15 girls in a class, find the number of boys in the class and the total number of students in the class?

Solution:

Given Ratio of Boys to Girls is 5:3

There are 15 Girls in the Class

Boys/Girls = 5/3

Boys/15 = 5/3

Boys = (5*15)/3

= 25

Number of Students in Class = Boys +Girls

= 25+15

= 40

Therefore, there are 25 Boys and 40 Students in the Class.

8. Find the third proportional of 10 and 20?

Solution:

Let us consider the Third Proportional of 10 and 20 be x

10, 20, and x are in Proportion

10:20 = 20:x

Product of Means = Product of Extremes

20*20 = 10*x

400 =10x

x = 400/10

= 40

Third Proportional of 10, 20 is 40

9. The first, second, and third terms of the proportion are 40, 36, 35. Find the fourth term?

Solution:

Let us consider the fourth term be x

40, 36, 35, x

Product of Means = Product of Extremes

36*35 = 40*x

x = (36*35)/40

= 31.5

Fourth Proportional of 40, 36, 35 is 31.5

10. Arrange the following ratios in Ascending Order

3:2, 4:3, 5 : 6, 1 : 4

Solution:

Given Ratios are 3/2, 4/3, 5/6 and 1/4

Finding the LCM of 2, 3, 6, 4 we get 12

Express the given ratios in terms of common denominator we get

3/2 = (3*6/2*6) = 18/12

4/3 = (4*4/3*4) = 16/12

5/6 = (5*2/6*2) = 10/12

1/4 = (1*3/4*3) = 3/12

Clearly, 3/12<10/12<16/12<18/12

Therefore, 1:4 <5:6<4:3<3:2

Practice Test on Ratio and Proportion helps students to get knowledge on different levels. The Ratio and Proportion Questions and Answers provided range from beginner, medium, hard levels. Practice the Questions here and get to know how to solve different problems asked. All the Ratio and Proportion Word Problems covered are as per the latest syllabus. Master the topic of Ratio and Proportion by practicing the Problems on a consistent basis and score better grades in your exam.

Ratio and Proportion Questions and Answers

1. The ratio of monthly income to the savings in a family is 5 : 3 If the savings be $6000, find the income and the expenses?

Solution:

Let us assume the Income be 5x

whereas savings be 3x

Given Savings = $6000

3x = 6000

x = 6000/3

= 2000

Income = 5x

= 5*2000

= $10,000

Expenses = Income – Savings

= 5x – 3x

= 2x

= 2*2000

= $4000

Therefore, Income and Expenses are $10,000 and $ 4000.

2. Two numbers are in the ratio 7: 4. If 3 is subtracted from each of them, the ratio becomes 5 : 2. Find the numbers?

Solution:

Let us consider the number be x

so 7x:4x

If 3 is subtracted the ratio becomes 5:2 then we have

7x-3:4x-3 = 5:2

equating them ad solving we get the values as

7x-3/4x-3 = 5/2

(7x-3)2 = 5(4x-3)

14x-6 = 20x-15

-6+15 = 20x-14x

9 = 6x

6x = 9

x = 9/6

= 3/2

Therefore the numbers are 7(3/2) and 4(3/2)

= 21/2, 6

3. Two numbers are in the ratio 3 : 5. If their sum is 720, find the numbers?

Solution:

Let us consider the number be x

Therefore two numbers become 3x:5x

Since their Sum = 720

3x+5x = 720

8x = 720

x = 720/8

= 90

Numbers are 3x and 4x

Thus, they become 3(90) and 4(90) i.e. 270 and 360.

4. A sum of money is divided among Rohan and Anand in the ratio 4 : 6. If Anand’s share is $600, find the total money?

Solution:

Let the money be x

Rohan and Anand’s Share = 4x:6x

Anand’s Share = $600

6x = $600

x = $100

Rohan’s Share = 4x

= 4*100

= $400

Total Money = Rohan’s Share + Anand’s Share

= $400+$600

= $1000

Therefore, the Sum of Money is $1000

5. The difference between the two numbers is 33 and the ratio between them is 5 : 2. Find the numbers?

Solution:

Let the number be x

From the given data

we have 5x-2x = 33

3x = 33

x = 11

Numbers are 5x, 2x

thus, they become 5*11 and 2*11

= 55, 22

Therefore, the numbers are 55 and 22.

6.  The ages of A and B are in the ratio 3 : 6. Four years later, the sum of their ages is 53. Find their present ages?

Solution:

Let the Present Ages be 3x and 6x

After four Years Age of A and B Becomes 3x+4 and 6x+4

We know sum of their ages after 4 years = 53

3x+4+6x+4 = 53

9x+8 = 53

9x = 53-8

9x = 45

x =5

Present Ages of A and B is 3x and 6x

thus 3*5 and 6*5 i.e. 15 and 30

Therefore, the Present Ages of A and B are 15 and 30.

7. If 3A = 4B = 5C, find the ratio of A : B : C?

Solution:

Let us assume that 3A = 4B = 5C = k

Equating them we have A = k/3, B = k/4, C = k/5

Therefore, Ratio becomes = k/3:k/4:k/5

LCM of 3, 4, 5 is 60

Thus expressing them in terms of least common multiple we have

A:B:C = 20:15:12

Therefore, Ratio of A:B:C is 20:15:12

8. A certain sum of money is divided among a, b, c in the ratio 3:4:5. of a share is $300, find the share of b and c?

Solution:

Let us consider the sum of money as x

Since it is shared among the ratio of 3:4:5 we have 3x:4x:5x

We know a’s share is 3x = $300

x =$100

Share of b = 4x

= 4*100

= $400

Share of C = 5x

= 5*100

= $500

9. Divide $900 among A, B, C in the ratio 3: 4 ∶ 5?

Solution:

Let us assume the total money as x

Since the sum is to be shared among A, B, C in the ratio of 3:4:5 we have

3x+4x+5x = $900

12x = $900

x = $900/12

=$75

Share of A = 3x

= 3*75

= $225

Share of B = 4x

= 4*75

= $300

Share of C = 5x

= 5*75

= $375

10. Find the first term, if second, third, and fourth terms are 21, 80, 120?

Solution:

Let the Terms be a, a+d, a+2d, a+3d

Given Second Term = 21

a+d = 21

Third Term = 80

a+2d = 80

Fourth Term = 120

a+3d = 120

Using the Eliminating Method

a+d = 21

a+2d = 80

_______

Subtracting them we get the value of d as

-d = -59

d= 59

Substitute the value of d in any of the terms

a+d = 21

a+59 =21

a =21-59

= -38

Ratio and Proportion are mainly explained using fractions. If a fraction is expressed in the form of a:b it is called a ratio and when two ratios are equal it is said to be in proportion. Ratio and Proportion is the fundamental concept to understand various concepts in maths. We will come across this concept in our day to day lives while dealing with money or while cooking any dish. Check out Definitions, Formulas for Ratio and Proportion, and Example Questions belonging to the concept in the further modules.

Quick Links of Ratio and Proportion Topics

If you want to get a good hold of the concept Ratio and Proportion you can practice using the quick links available for various topics in it. You just need to tap on the direct links available and get a good grip on the concept.

• Practice Test on Ratio and Proportion
• Worked out Problems on Ratio and Proportion

What is Ratio and Proportion?

Ratio and Proportion is a crucial topic in mathematics. Find Definitions related to Ratio and Proportion along with examples here.

In Certain Situations comparison of two quantities by the division method is efficient. Comparison or Simplified form of two similar quantities is called ratio. The relation determines how many times one quantity is equal to the other quantity. In other words, the ratio is the number that can be used to express one quantity as a fraction of other ones.

Points to remember regarding Ratios

• Ratio exists between quantities of a similar kind
• During Comparison units of two things must be similar.
• There should be significant order of terms
• Comparison of two ratios is performed if the ratios are equivalent similar to fractions.

Proportion – Definition

Proportion is an equation that defines two given ratios are equivalent to each other. In Simple words, Proportion states the equality of two fractions or ratios. If two sets of given numbers are either increasing or decreasing in the same ratio then they are said to be directly proportional to each other.

Ex: For instance, a train travels at a speed of 100 km/hr and the other train travels at a speed of 500km/5 hrs the both are said to be in proportion since their ratios are equal

100 km/hr = 500 km/5 hrs

Continued Proportion

Consider two ratios a:b and c:d then in order to find the continued proportion of two given ratio terms we need to convert to a single term/number.

For the given ratio, the LCM of b & c will be bc.

Thus, multiplying the first ratio by c and the second ratio by b, we have

The first ratio becomes ca: bc

The second ratio becomes bc: bd

Thus, the continued proportion can be written in the form of ca: bc: bd

Ratio and Proportion Formulas

Ratio Formula

Let us consider, we have two quantities and we have to find the ratio of these two, then the formula for ratio is defined as

a: b ⇒ a/b

a, b be two quantities. In this a is called the first term or antecedent and b is called the second term or consequent.

Example: In the Ratio 5:6 5 is called the first term or antecedent and 6 is called the consequent.

If we multiply and divide each term of the ratio by the same number (non-zero), it doesn’t affect the ratio.

Proportion Formula

Consider two ratios are in proportion a:b&c:d the b, c are called means or mean terms and a, d are known as extremes or extreme terms.

a/b = c/d or a : b :: c : d

Example: 3 : 5 :: 4 : 8 in this 3, 8 are extremes and 5, 4 are means

Properties of Proportion

Check out the important list of properties regarding the Proportion Below. They are as follows

• Addendo – If a : b = c : d, then a + c : b + d
• Subtrahendo – If a : b = c : d, then a – c : b – d
• Componendo – If a : b = c : d, then a + b : b = c+d : d
• Dividendo – If a : b = c : d, then a – b : b = c – d : d
• Invertendo – If a : b = c : d, then b : a = d : c
• Alternendo – If a : b = c : d, then a : c = b: d
• Componendo and dividendo – If a : b = c : d, then a + b : a – b = c + d : c – d

Difference Between Ratio and Proportion

Fourth, Third and Mean Proportional

If a : b = c : d, then:

d is called the fourth proportional to a, b, c.
c is called the third proportion to a and b.
Mean proportional between a and b is √(ab).
Comparison of Ratios
If (a:b)>(c:d) = (a/b>c/d)

The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).

Duplicate Ratios

If a:b is a ratio, then:

• a²:b² is a duplicate ratio
• √a:√b is the sub-duplicate ratio
• a³:b³ is a triplicate ratio

Ratio and Proportion Tricks

Check out the Tricks and Tips to Solve Problems related to Ratio and Proportion. They are as under

• If u/v = x/y, then u/x = v/y
• If u/v = x/y, then uy = vx
• If u/v = x/y, then v/u = y/x
• If u/v = x/y, then (u-v)/v = (x-y)/y
• If u/v = x/y, then (u+v)/v = (x+y)/y
• If u/v = x/y, then (u+v)/ (u-v) = (x+y)/(x-y), it is known as Componendo Dividendo Rule
• If a/(b+c) = b/(c+a) = c/(a+b) and a+b+ c ≠0, then a =b = c

Solved Questions on Ratio and Proportion

1. Are the Ratios 4:5 and 5:10 said to be in Proportion?

Solution:

Expressing the given ratios 4:5 we have 4/5 = 0.8

5:10 = 5/10 = 0.2

Since both the ratios are not equal they are not in proportion.

2. Out of the total students in a class, if the number of boys is 4 and the number of girls being 5, then find the ratio between girls and boys?

Solution:

The ratio between girls and boys is 5:4. The ratio can be written in factor form as 5/4

3. Two numbers are in the ratio 3 : 4. If the sum of numbers is 42, find the numbers?

Solution:

Given 3/4 is the ratio of any two numbers

Let us consider the numbers be 3x and 4x

Given, 3x+4x = 42

7x = 42

x = 42/7

x = 6

finding the numbers we have 3x = 3*6 = 18

4x = 4*6 = 24

Therefore, two numbers are 18, 24

Before, going for examples on simplification, know the rules and methods. Check all the best possible methods to solve the simplification of expressions. Refer to all the formulae, rules, and methods for the simplification of integers. You can go through the entire article to know more on what are the rules to be followed while simplifying expressions. Practice the Simplification Questions available and cross-check your answers here to know where you stand in your preparation level.

Rules of Simplification

While solving the Examples on Simplification keep the below pointers in mind. They are as follows

1. Remove Brackets – If two or more signs occur in the expression, then convert them into one. The brackets will be square, round and curly braces.
2. Group the values into one group i.e., either positive or negative.
3. Whenever there are two integers, the result will give the sign of the greater value.

Examples on Simplification

Question 1.

At 135 feet below sea level, a submarine has started. It dives 239 feet before rising 307 feet. Find the exact depth of the submarine at which it is spent currently?

Solution:

As per the given question,

At 135 feet below sea level, a submarine has started

The level at which it dives = 239 feet

The rise of the sea level =307 feet

To find, the current depth of the submarine, we write the equation as

-135+(-239)+307

=-374+307

=-67

Therefore, the current depth of the submarine is 67 feet below sea level.

The final solution is -67 feet.

Question 2:

Jenny purchases a credit card from a local retailer. She begins with a $200 balance. She then makes the following purchases: lamp $8, rug $63, vacuum $39. After this shopping trip, she loads $147 on her card, then spends $113 on groceries. Express each transaction as an integer, then determine the new balance on the prepaid credit card?

Solution:

As per the given question,

Jenny begins with a balance = $200

She purchases lamp = $8

She purchases rug = $63

She purchases vacuum = $39

She loads on her card = $133

As she has the balance of $200, it is positive, and also she loads $147 to her card, therefore it is also positive. The remaining values are negative.

The new balance on the prepaid credit card = 200 +(-8) + (-63) + (-39) + 147 + (-113)

= 347 + (-223)

=124

Therefore, the new balance on the prepaid credit card = $124

Hence, the final solution = $124

Question 3:

Simplify: 37 – [5 + {28 – (19 – 7)}]

Solution:

Step 1: Innermost grouping symbols removal is the first step to simplify the expression. Consider {28-(19-7)}, to remove the brackets we subtract the equation {28-(19-7)}, here comes as {28-12}
The simplified expression is 37 – [5 + {28 – 12}]
Step 2: Follow the same procedure to remove parentheses in {28-12}, we have to subtract the equation {28-12}, here comes the final equation as
37 – [5 + 16]
Step 3: Now, the equation is further simplified and it contains only square brackets, we should perform all the set of operations within two brackets.
37 – 21
Step 4: In the final step, we subtract the values and the final result will be 16.
Thus the final solution will be 16.

Question 4:

Simplify the equation 15 – (-5) {4 – 7 – 3} ÷ [3{5 + (-3) x (-6)}]

Solution:

Step 1: Innermost grouping symbols removal is the first step to simplify the expression. Consider {5+(-3)x(-6)}, to remove the brackets we multiply the equation {5+18}, here comes the final expression as

15 – (-5) {4 – 7 – 3} ÷[3 {5 + 18}]

Step 2: Follow the same procedure to remove parentheses in {4-7-3}, we have multiplied the equation {4-7-3}, here comes the final result as 4-4

After performing 2 steps, the result equation will be

15 – (-5) x 0 ÷[3 {5 + 18}]

Step 3: Next simplification must be the addition of 5 and 18. The result will be

15 – (-5) x 0 ÷ 3 x 23

Step 4: As the brackets of two sets are removed. Therefore, change the negative signs and rewrite the equation.

15 + (5 x 0) ÷ 3 x 23

Step 5: As all other brackets are removed and the expression contains only and braces. Perform all the operations that are possible within the brackets.

= 15 – (-5) x 0 ÷ 69

Step 6: On further simplification, the result will be

15 – (-5) x 0

Step 7:

Therefore, the final solution is 15

Question 5:

The temperature of the fridge compartment is set at 8 degrees C. The freezer compartment is set at -10 degrees C. What is the difference between the temperature settings?

Solution:

As per the given question,

The temperature of fridge compartment = 8 degrees C

Temperature of freezer compartment = -10 degree C

Difference between temperature = 8-(-10)

=8+10

=18 degrees C

Question 6:

Rekha climbs up 5 stairs every second and then climbs down 2 stairs over the next second. How many seconds will she take to climb 60 steps?

Solution:

As per the given question,

Stairs climbed up by Rekha = 5 in 1 sec

Stairs climbed down by Rekha = 2 in 1 sec

Climbing down is considered as a decrease in value, hence it is negative.

Therefore, stairs climbed = 5+(-2) in 2 secs

Steps climbed = 3 stairs in 2 secs

Thus, the time to climb 1 step in time = 2/3 secs

Thus, the time taken to climb 60 steps = 2 x 60/3

= 40sec

Hence, the final solution is 40 sec.

Question 7:

I start with integer (-8), Add (-12) to it, subtract 10 from the result. Divide the result by (+3) and multiply the answer by (-2). What do you get?

Solution:

Step 1: As given in the question, the integer value is -8

Step 2: Adding -12 to the integer value

i.e., -8-12=-20

Step 3: Subtracting 10 from the value = -20-(10) = -30

Step 4: Dividing by +3

=(-30)/(3) = -10

Step 5: Multiply the answer by -2

(-10) x (-2) = 20

Therefore, the final solution is 20

Question 8:

Arnav has $20. He spent $8 on Monday. He got $5 as pocket money on Tuesday. He gave a $7 loan to his friend on Wednesday. He ate ice cream for $10 on Thursday. He received a reward of $5 on Friday. He repaid the loan of $7 on Saturday. How much money did Arnav totally have on Sunday?

Solution:

As per the given question,

Arnav has $20

He spent on Monday = $8

He got pocket money on Tuesday = $5

He gave a loan to a friend on Wednesday = $7

He ate ice cream on Thursday = $10

Received a reward on Friday = $5

His friend repay the loan on Saturday = $7

Money Arnav has on Sunday = 20+(-8)+5(-7)+(-10)+5+7

=20+5+5+7+(-8)+(-7)+(-10)

=37-8-7-10

=37-25

=12

Therefore, the money Arnav has on Sunday = $12

Question 9:

A Hiker is descending 152m in 8 minutes. What will be his elevation in half an hour?

Solution:

As given in the question,

In 8 mins Hiker descends 152cm.

In 1 min Hiker descends = 152/8 = 19m

Therefore, In 30min Hiker descends = 19 x 30 = 570m

Hence, the final solution is 570m

Question 10:

In an exam, the student gets +4 for each correct answer and -2 for each wrong answer. Rohith’s final score is 68 marks and he attempted 25 questions correctly. How many marks did he lose for wrong answers?

Solution:

Rohith’s final score = 68 marks

Rohith attempted correctly = 25 questions

Let the number of wrong answers be a

Marks for correct answers = 25 x 4 = 100

Marks for wrong answers = a x (-2) = -2a

Total wrong answers = 100 +(-2a) = 68

100-2a=68

-2a=68-100

-2a=-32

a=32/2

a=16

Therefore, Rohith lost 16 marks for wrong answers.