Worksheet on Types of Ratios

I. Find the compound ratio of the following ratios.

(i) 2 : 3 and 5 : 12
(ii) m : n and n : p
(iii) (a−b): (a+b),(a−b)2: (a+b)2
Solution:
(i)  Given that,
2 : 3 and 5 : 12
Compound ratio of 2 : 3 and 5 : 12 is
=2*5:3*12
=10:36
=5:18
(ii) Given that,
m : n and n : p
Compound ratio of m : n and n : p
=m*n:n*p
=mn/np
=m/p
(iii) Given that,
(a−b): (a+b),(a−b)2: (a+b)2
Compound ratio will be
=(a-b)(a−b)2: (a+b)(a+b)2
=(a-b)3:(a+b)3
II. Find the compound ratio of each of the following three ratios.
(i) 1 : 2, 4 : 7 and 7 : 8
(ii) a: b, b: c, and c : d
(iii) 2x : 3y, ab : y2 and y : b
Solution:
(i) Given that,
1: 2, 4: 7, and 7: 8

Worksheet on Proportion and Continued Proportion | Free Printable Proportion and Continued Proportion Problems with Solutions Worksheet PDF

The worksheet on proportion and continued proportion help students to practice and acquire more knowledge on the concept of proportion where it also assists them in real-life incidents too. The proportion is a mathematical comparison of two numbers, and also we can compare the four quantities. Proportions are represented by the symbol “::” or “=”. If three quantities are in continued proportion then the ratio between the first and second quantity of ratio is equal to the second and the third quantity of the ratio.

For example, if the three quantities a, b, and c are continued proportion, then a: b = b: c i.e., \(\frac{a}{b}\) = \(\frac{b}{c}\). This free printable proportion and continued proportion worksheet pdf with answers is fun to practice and express the following ratios whether they are proportion and continued proportion or not.

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Worksheet on Proportion and Continued Proportion PDF with Answers

1. Check the following numbers are in proportion or not.
(i) 1.5, 5.5, 2.7, 9.9
(ii) 2.5, 4.5, 1\(\frac{1}{4}\), 2\(\frac{1}{4}\)

Solution: 

(i) Given numbers are 1.5, 5.5, 2.7, 9.9.
For example, to say the numbers are in proportion or not. We have to prove a: b = c: d.
Now, we find 1.5: 5.5 = 2.7: 9.9.
1.5: 5.5 = \(\frac{1.5}{5.5}\) = \(\frac{1.5×10}{5.5×10}\) = \(\frac{15}{55}\) = \(\frac{3}{11}\)
2.7: 9.9 = \(\frac{2.7}{9.9}\) = \(\frac{2.7×10}{9.9×10}\) = \(\frac{27}{99}\) = \(\frac{3}{11}\)
Thus, \(\frac{1.5}{5.5}\) = \(\frac{2.7}{9.9}\).
Therefore, the numbers 1.5, 5.5, 2.7, 9.9 are in proportion.

(ii) Given numbers are 2.5, 4.5, 1\(\frac{1}{4}\), 2\(\frac{1}{4}\).
As you know earlier, the numbers to be in proportion when a: b = c: d.
Now, 2.5: 4.5 = \(\frac{2.5}{4.5}\) = \(\frac{2.5×10}{4.5×10}\) = \(\frac{25}{45}\)= \(\frac{5}{9}\).
1\(\frac{1}{4}\): 2\(\frac{1}{4}\) = \(\frac{5}{4}\): \(\frac{9}{4}\) = \(\frac{5}{4}\)× 4: \(\frac{9}{4}\)× 4 = 5: 9 = \(\frac{5}{9}\).
Hence, \(\frac{5}{9}\) = \(\frac{5}{9}\).
Therefore, the numbers 2.5, 4.5, 1\(\frac{1}{4}\), 2\(\frac{1}{4}\) are in proportion.

2. Find x in the following proportions.
(i) 4: 3 = 8: x
(ii) 2.5: x = 8.1: 1.4
(iii) 1.0 : 2 = x: 4

Solution: 

(i) Given 4: 3 = 8: x
⇒ \(\frac{4}{3}\) = \(\frac{6}{x}\)
⇒ 4x = 8×3
⇒ 4x = 24
⇒ x = \(\frac{24}{4}\)
⇒ x = 6.
Thus, the value of x is 6.

(ii) Given 2.5: x = 8.1: 1.4
⇒ \(\frac{2.5}{x}\) = \(\frac{8.1}{1.4}\)
⇒ \(\frac{25}{10x}\) = \(\frac{81}{14}\)
⇒ 25 × 14 = 81 × 10x
⇒ 350 = 810x
⇒ x = \(\frac{810}{350}\)
⇒ x = 2.3
Hence, the x value is 2.3.

(iii) Given 1.0 : 2 = x: 4
⇒ \(\frac{1.0}{2}\) = \(\frac{x}{4}\)
⇒ \(\frac{10}{20}\) = \(\frac{x}{4}\)
⇒ 10 × 4 = 20x
⇒ 20x = 40
⇒ x = \(\frac{40}{20}\)
⇒ x = 2.
The value of x is 2.

3. Find the following are in continued proportion or not?
(i) 0.2, 2.4, 4.0
(ii) 4, 6, 9

Solution: 

(i) Given numbers are 0.2, 2.4, 4.0.
If we have three numbers to find whether they are continued proportion or not, thus a:b = b:a.
Now, 0.2: 2.4 = 2.4: 4.0
⇒ \(\frac{0.2}{2.4}\) = \(\frac{2.4}{4.0}\)
⇒ \(\frac{2}{24}\) = \(\frac{24}{40}\)
⇒ \(\frac{1}{12}\) ≠ \(\frac{3}{5}\)
Hence, the given numbers are not in continued proportion because 0.2: 2.4 ≠ 2.4: 4.0.

(ii) Given numbers are 4, 6, 9.
As we know, a:b = b:a
Now, 4: 6 = 6:9
⇒ \(\frac{4}{6}\) = \(\frac{6}{9}\)
⇒ \(\frac{2}{3}\) = \(\frac{2}{3}\)
Thus, the given numbers 4, 6, and 9 are in continued proportion because 4: 6 = 6: 9.

4. Find m, if the numbers 8, 4, and m are in continued proportion.

Solution: 

Given the numbers 8, 4, and m are in continued proportion.
8: 4 = 4: m
⇒ \(\frac{8}{4}\) = \(\frac{4}{m}\)
⇒ 8m = 16
⇒ m = \(\frac{16}{8}\)
⇒ m = 2.
Hence, the value of m is 2.

5. Find the fourth proportion of 6, 2, and 9.

Solution: 

Given proportion numbers are 6, 2, and 9.
Let the fourth proportional be p.
Now, according to given problem 6, 2, 9, and p are proportionality.
Thus,
\(\frac{6}{2}\) = \(\frac{9}{p}\)
⇒ 6p = 9 × 2
⇒ 6p = 18
⇒ p = \(\frac{18}{6}\)
⇒ p = 3.
Therefore, the fourth proportional number is 3.

6. Find the third proportion to the following numbers.
(i) 4, 6
(ii) 1\(\frac{2}{3}\), \(\frac{4}{5}\)
(iii) 3.4, 7.2

Solution: 

(i) Given proportion numbers 4, 6.
Now, find the third proportion.
Let the third proportion be k.
Thus,
\(\frac{4}{6}\) = \(\frac{6}{k}\)
⇒ 4k = 6 × 6
⇒ 4k = 36
⇒ k = \(\frac{36}{4}\)
⇒ k = 9.
Therefore, the third proportion number is 9.

(ii) Given proportion numbers are 1\(\frac{2}{3}\), \(\frac{4}{5}\)
Let the third proportion be k.
Thus,
\(\frac{5}{3}\) × \(\frac{5}{4}\) = \(\frac{4}{5}\) × \(\frac{1}{k}\)
⇒ \(\frac{25}{12}\) = \(\frac{4}{5k}\)
⇒ 25 × 5k = 4 × 12
⇒ 125k = 48
⇒ k = \(\frac{48}{125}\)
Hence, the third proportion number is \(\frac{48}{125}\).

(iii) Given proportion numbers are 3.4, 7.2.
Let the third proportion be k.
Now,
\(\frac{3.4}{7.2}\) = \(\frac{7.2}{k}\)
⇒ \(\frac{34}{72}\) = \(\frac{72}{10k}\)
⇒ 34 × 10k = 72 × 72
⇒ 340k = 5184
⇒ k = \(\frac{5184}{340}\)
⇒ k = 15.2
Thus, the third proportion value is 15.2.

7. Find q in the following numbers so that the numbers are proportional.
(i) 10, q, 20, 15
(ii) 8, 16, 24, q
(iii) q, 35, 15, 40

Solution: 

(i) Given numbers are 10, q, 20, 15
To find the proportion numbers, we use a:b = c:d i.e., \(\frac{a}{b}\) = \(\frac{c}{d}\).
Now,
\(\frac{10}{q}\) = \(\frac{20}{15}\)
⇒ 10 × 15= 20q
⇒ 20q = 150
⇒ q = \(\frac{150}{20}\)
⇒ q = 7.5
Thus, the value of q is 7.5.

(ii) Given numbers are 8, 16, 24, q.
Now,
\(\frac{8}{16}\) = \(\frac{24}{q}\)
⇒ \(\frac{1}{2}\) = \(\frac{24}{q}\)
⇒ q = 24 × 2
⇒ q = 48
Hence, the value of q is 48.

(iii) Given numbers are q, 35, 15, 40.
We use the proportion formula, a:b = c:d i.e., \(\frac{a}{b}\) = \(\frac{c}{d}\).
Now,
\(\frac{q}{35}\) = \(\frac{15}{40}\)
⇒ 40q = 15 × 35
⇒ 40q = 525
⇒ q = \(\frac{525}{40}\)
⇒ q = 13.12
Therefore, the value of q for given proportional numbers is 13.12.

Worksheet on Mean Proportional PDF | Free Printable Mean Proportional Problems with Answers Activity Sheet

Getting confused while solving the mean proportional problems then practice more with this Mean Proportional Worksheet pdf and learn the concept of mean proportional efficiently. Mean proportional is also known as Geometric Mean and it is not similar to the arithmetic mean. The Mean Proportion is calculated between two terms of a ratio by taking the square root of the product of those two quantities in terms of ratio.

Let us understand more about mean proportional by taking help from this free printable mean proportional of two numbers worksheet. For example, if a, b, and c are in continued proportion then b is called the mean proportional of a and c. The mean proportion is expressed as b = √ac. This free printable Worksheet on Mean Proportional with Answers PDF helps you understand the problem-solving techniques and feels fun to practice.

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Mean Proportional Worksheet PDF with Solutions

Example 1: 
Find the mean proportional of the following sets of positive integers:
(i) x- y, x³- x²y
(ii) x³y, xy³

Solution: 

(i) Given x- y, x³- x²y
Now, to find the mean proportion.
Let p be the mean proportional between x- y, x³- x²y.
So, x- y: p:: p: x³- x²y
Product of extremes = Product of means.
Now,
p² = (x- y)(x³- x²y)
⇒ p² = x²(x- y)(x- y)
⇒ p² = x²(x- y)²
⇒ p = √(x²(x- y)²)
⇒ p = x(x- y)
Thus, the mean proportion of x- y, x³- x²y is x(x- y).

(ii) Given x³y, xy³
Let m be the mean proportion of x³y, xy³.
So, x³y: m:: m: xy³.
Product of extremes = Product of means.
Product of extremes = x³y × xy³
Product of means = m × m = m²
m² = x³y × xy³
⇒ m² = (x³ × x) × (y³ × y)
⇒ m² = x4 × y4
⇒ m = √(x4 × y4)
⇒ m = √(x²y²)²
⇒ m = x²y²
Hence, the mean proportion of x³y, xy³ is x²y².

Example 2:  
Find the mean proportional of the following:
(i) 8 and 32
(ii) 0.04 and 0.56
(iii) 4 and 25

Solution: 

(i) Given 8 and 32
Let the mean proportion between 8 and 32 is a.
Now, 8: a:: a: 32
We know that, Product of extremes = Product of means.
Here, extremes are 8 and 32 and means are a and a.
So, 8 × 32 = a × a
⇒ a² = 256
⇒ a = √256
⇒ a = 16
Thus, the value of mean proportion ‘a’ is 16.

(ii) Given 0.04 and 0.56
Let the mean proportion be p.
Now, 0.04: p:: p: 0.56
We know that, Product of extremes = Product of means.
The extremes are 0.04 and 0.56 and the means are p and p.
So, 0.04 × 0.56 = p × p
⇒ p² = 0.02
⇒ p = √0.02
⇒ p = 0.14
Therefore, the value of the mean proportion p is 0.14.

(iii) Given 4 and 25
Let the mean proportion between 4 and 25 be x.
Thus, 4: x:: x: 25
We know that, Product of extremes = Product of means.
Here, the extremes are 4 and 25 and the means are x and x.
So, 4 ×25 = x × x
⇒ x² = 100
⇒ x = √100
⇒ x = 10
Hence, the mean proportion of 4 and 25 is 10.

Example 3: 
If b is the mean proportion between a and c, show that a4+ a2b2+ b4/b4+ b2c2+ c4 = \(\frac{a²}{c²}\).

Solution: 

Given b is the mean proportion between a and c, then we have b² = ac.
Now, we have to prove that a4+ a2b2+ b4 / b4+ b2c2+ c4 = \(\frac{a²}{c²}\) i.e., L.H.S = R.H.S.
LHS = a4+ a2b2+ b4/ b4+ b2c2+ c4
Let us substitute b² = ac in LHS.
LHS = a4+ a²(ac)+ (ac)²/(ac)²+ (ac)c²+ c4
⇒ LHS = \(\frac{a²( a²+ ac+ c² )}{c²( a²+ ac+ c² )}\)
⇒ LHS = \(\frac{a²}{c²}\) = RHS
⇒ LHS = RHS
Therefore, a4+ a2b2+ b4/ b4+ b2c2+ c4= \(\frac{a²}{c²}\).

Example 4: 
Find the mean proportion of the following
(i) 4\(\frac{4}{5}\), 2\(\frac{1}{2}\)
(ii) a²b, ab²

Solution: 

(i) Given 4\(\frac{4}{5}\), 2\(\frac{1}{2}\)
Now, change the mixed fraction into proper fraction
4\(\frac{4}{5}\) = \(\frac{24}{5}\)
2\(\frac{1}{2}\) = \(\frac{5}{2}\)
Let m be the mean proportion of \(\frac{24}{5}\), \(\frac{5}{2}\).
Product of extremes = Prouct of means
Here, the extremes are \(\frac{24}{5}\) and \(\frac{5}{2}\), the means are m and m.
\(\frac{24}{5}\) × \(\frac{5}{2}\) = m × m
⇒ m² = \(\frac{24}{2}\)
⇒ m² = 12
⇒ m = √12
⇒ m = 3.46
Thus, the value of mean proportion is 3.46.

(ii) Given a²b, ab²
Let k be the mean proportion of a²b and ab².
So, a²b: k:: k: ab².
In mean proportion, Product of extremes = Product of means.
Now, a²b × ab² = k²
⇒ k² = a³b³
⇒ k² = (ab)¹ × (ab)²
⇒ k = √((ab)¹)²
⇒ k = ab
Therefore, ab is the mean proportion of a²b, ab².

Worksheet on Discount and Markup | Markup and Discount Solved Problems with Examples

Worksheet on Discount and Markup is available on this page. The students can learn how to solve Discount and Markup-related problems here. We have provided the different models of questions asked on Discount and Markup in the Discount and Markup Worksheet. Get free access to practice the questions on Markup and Discount and crack the exams easily. Make use of this page and score well in all competitive exams.

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Discount and Markup Worksheet

1. A product cost in a shop is $100. If the markup set by the shop is 20%. Calculate the markup price set by the shop on that product.

Solution:

Given that the product price by the shop is $100.
The markup set by the store is 20%.
Markup Percentage = 20%
Now, find the markup price set by the shop on the product.
The Markup Price = Markup Percentage of product price
Markup Price = 20% of $100 = \(\frac { 20 }{ 100 } \) × $100 = $20.

Therefore, the Markup Price set by the shop on the product is $20.

2. The store price of a rice cooker is $500. If the markup percent set by the store is 23%. Calculate the markup price set by the store on the rice cooker.

Solution:

Given that the store price of a rice cooker is $500.
Store price = $500
The markup set by the store is 23%.
Markup Percentage = 23%
Now, find the markup price set by the store on the rice cooker.
The Markup Price = Markup Percentage × store’s cost
Now, substitute Store price and Markup Percentage in the above formula.
Markup Price = 23% × $500 = \(\frac { 23 }{ 100 } \) × $500 =  23 × $5 = .$115.

Therefore, the Markup Price set by the store on the rice cooker is $115.

3. The cost price of a ceiling fan is Rs 3,000. The markup percent set by the store owner is 30%. Calculate the markup price and selling price of the ceiling fan.

Solution:

Given that the cost price of a ceiling fan is Rs 3,000.
Cost price = Rs 3,000.
The markup percent set by the store owner is 30%.
Markup Percentage = 30%
Now, find the markup price set by the store owner on the ceiling fan.
The Markup Price = Markup Percentage × store’s cost
Now, substitute Cost price and Markup Percentage in the above formula.
Markup Price = 30% × Rs 3,000 = \(\frac { 30 }{ 100 } \) × Rs 3,000 =  30 × Rs 30 = Rs 900.
Markup Price = Rs 900.
Now, find the selling price set by the store owner on the ceiling fan.
Selling price = store’s cost + markup price
Now, substitute Cost price and Markup price in the above formula.
Selling price = store’s cost + markup price = Rs 3,000 + Rs 900 = Rs 3,900

Therefore, the Markup Price set by the store on the ceiling fan is Rs 900 and the Selling price set by the store on the ceiling fan is Rs 3,900.

4. The cost price of an electronic device is $520. The markup percent set by the store is 20%. Calculate the markup price set by the shopkeeper on the electronic device. Also, find the selling price of the electronic device.

Solution:

Given that the cost price of an electronic device is $520.
Cost price = $520.
The markup percent set by the store owner is 20%.
Markup Percentage = 20%
Now, find the markup price set by the store owner on the electronic device.
The Markup Price = Markup Percentage × store’s cost
Now, substitute Cost price and Markup Percentage in the above formula.
Markup Price = 20% × $520 = \(\frac { 20 }{ 100 } \) × $520 = $104.
Markup Price = $104.
Now, find the selling price set by the store owner on the electronic device.
Selling price = store’s cost + markup price
Now, substitute Cost price and Markup price in the above formula.
Selling price = store’s cost + markup price = $520 + $104 = $624

Therefore, the Markup Price set by the store on the electronic device is $104 and the Selling price set by the store on the electronic device is $520.

5. If the selling price of a carrom board is $600. If the markup set by the store was 20%. Then calculate the cost price of the carrom board. Also, find the markup price for the carrom board.

Solution:

Given that the selling price of a carrom board is $600.
Selling price = $600.
The markup set by the store was 20%.
Markup Percentage = 20%.
Let the Cost Price = c = 100%
c + m = p
100% + 20% = 120%.
c + 20%c = selling price
c[1 + 20%] = selling price
c [1 + \(\frac { 20 }{ 100 } \)] = selling price
c [1 + 0.2] = $600
c (1.2) = $600
c = \(\frac { $600 }{ 1.2 } \)
c = $500.
The cost price of the carrom board is $500.
Now, find the markup price for the carrom board.
Markup price = Selling price – store’s cost price
Substitute the Selling price and store’s cost price in the above formula.
Markup price = $600 – $500 = $100

Therefore, the cost price of the carrom board is $500 and the Markup price of the carrom board is $100.

6. The store price of a chair is $100. If the discount given by the store is 25%, calculate the discount amount offered by the store on the chair.

Solution:

Given that the store price of a chair is $100.
Store price = $100.
The discount given by the store is 25%.
Discount = 25%.
Now, find the discount amount offered by the store on the chair.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 25% x $100 = \(\frac { 25 }{ 100 } \) x $100 = $25

Therefore, the discount amount offered by the store on the chair is $100.

7. The cost price of an oxygen cylinder kept in a sale is Rs 6,000. If the discount offered on the grinder is 40%. Calculate the discount price that is offered on the oxygen cylinder.

Solution:

Given that the cost price of an oxygen cylinder is $6,000.
Cost price = $6,000.
The discount given by the store is 40%.
Discount = 40%.
Now, find the discount amount offered by the store on the oxygen cylinder.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 40% x $6,000 = \(\frac { 40 }{ 100 } \) x $6,000 = $2,400

Therefore, the discount amount offered by the store on the oxygen cylinder is $2,400.

8. The store price of a charger is $70. If the discount offered by the store is 15%. Calculate the discount amount and selling price of the charger.

Solution:

Given that the store price of a charger is $70.
Store price = $70.
The discount given by the store is 15%.
Discount = 15%.
Now, find the discount amount offered by the store on the charger.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 15% x $70 = \(\frac { 15 }{ 100 } \) x $70 = $10.5
Now, find the selling price of the charger.
Selling price = cost/store price – discount amount
Substitute cost/store price and discount amount in the above equation.
Selling price = $70 – $10.5 = $59.5

Therefore, the discount amount offered by the store on the charger is $10.5 and the Selling price on the charger is $59.5.

9. The cost price of a keyboard is $150. If the discount offered by the shopkeeper is 35%. Calculate the discount amount and selling price of the keyboard.

Solution:

Given that the cost price of a keyboard is $150.
Cost price = $150.
The discount offered by the shopkeeper is 35%.
Discount = 35%.
Now, find the discount amount offered by the shopkeeper on the keyboard.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 35% x $150 = \(\frac { 35 }{ 100 } \) x $150 = $52.5
Now, find the selling price of the keyboard.
Selling price = cost/store price – discount amount
Substitute cost/store price and discount amount in the above equation.
Selling price = $150 – $52.5 = $97.5

Therefore, the discount amount offered by the store on the keyboard is $52.5 and the Selling price on the keyboard is $97.5.

10. The selling price of a motor is $375. If the discount offered by the store was 5%. Calculate the cost price of the motor. Also, find the discount amount offered by the store.

Solution:

Given that the cost price of a keyboard is $150.
Cost price = $150.
The discount offered by the shopkeeper is 35%.
Discount = 35%.
Now, find the discount amount offered by the shopkeeper on the keyboard.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 35% x $150 = \(\frac { 35 }{ 100 } \) x $150 = $52.5
Now, find the selling price of the keyboard.
Selling price = cost/store price – discount amount
Substitute cost/store price and discount amount in the above equation.
Selling price = $150 – $52.5 = $97.5

Therefore, the discount amount offered by the store on the keyboard is $52.5 and the Selling price on the keyboard is $97.5.

11. The marked price of a video player is $ 2340. The shopkeeper offers a discount of 30% on it. Find its selling price.

Solution:

Given that the marked price of a video player is $ 2340.
Marked price = $2340.
The discount offered by the shopkeeper is 30%.
Discount = 30%.
Now, find the Discount  offered by the shopkeeper on the video player.
Discount amount = discount percent x cost/store price
Substitute the discount percent and cost/ store price in the above formula.
Discount amount = discount percent x cost/store price
Discount amount = 30% x $2340 = \(\frac { 30 }{ 100 } \) x $2340 = $702
Now, find the selling price of the video player.
Selling price = cost/store price – discount amount
Substitute cost/store price and discount amount in the above equation.
Selling price = $2340 – $702 = $1638

Therefore, the discount amount offered by the store on the video player is $702 and the Selling price on the video player is $1638.

12. Find the rate of discount being given on a chair whose selling price is $273 after deducting a discount of $52 on its marked price.

Solution:

Given that the selling price of a chair is $ 273.
Selling price = $273.
The deducting a discount of $52 on its marked price of a chair.
Discount = $52.
Now, find the marked price on the chair.
Marked price on the chair = discount + selling price
Substitute the discount and selling price in the above formula.
Marked price on the chair = $52 + $273
Marked price on the chair = $325
Now, find the rate of discount of the chair.
Rate of discount = (discount x 100)/ Marked price
Substitute discount and Marked price in the above equation.
Rate of discount = ($52 x 100)/ $325 = 16%

Therefore, the Rate of discount is 16%.

Worksheet on Dividing a Quantity in a Given Ratio | Dividing Quantities in Given Ratio Worksheet Pdf with Answers

Worksheet on Dividing a Quantity in a Given Ratio Concept provides problems on quantities into two parts and three parts of a given ratio. We will discuss the quantities of a given ratio in various aspects to get a clear understanding of the whole concept to the children. As we know, a ratio is a comparison of quantities of the same unit and it can be described as a fraction. Here, the ratio of X and Y is defined as X : Y = \(\frac{X}{Y}\). The quantity X in the ratio is called antecedent and Y is the consequent.

Practice the questions given in the below Dividing quantity in a given ratio worksheet pdf and answer all complex calculations with ease. Dividing quantities in a given ratio worksheet makes you learn the concept in a fun-learning & engaging manner. This Math Dividing a Quantity in a Given Ratio Word Problems Worksheet gives a step-to-step explanation so that you don’t feel bored and difficult to study & solve the calculations.

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Free Printable Worksheet on Dividing a Quantity in a Given Ratio

Example 1:
A number is divided into two parts in the ratio of 4: 9. If the larger part is 270, then find the actual number and the smaller part.

Solution:

The given ratio is 4: 9.
Let the numbers be 4k and 9k
Given larger part of the number is 270.
Now, we find the number by solving 9k = 270.
k = \(\frac{270}{9}\)
⇒ k = 30
Another number is 4k, substitute the k value in 4k.
4k = 4 × 30 = 120.
Thus, the smaller part number is 120 and the actual number is 390.

Example 2:
A bag contains 3 dollars, 50 cents, 4 dollars in the ratio of 5: 4: 2. The total amount is $ 2450. Find the number of each denomination?

Solution:

Let the number of each denomination be 5x, 4x, and 2x respectively.
The amount of 3 dollars = 5x × 300 cents = 1500x cents
The amount of 50 cents = 4x × 50 cents = 200x cents
The amount of 4 dollars = 2x × 400 cents = 800x cents
The total amount given = 2450 × 100 cents =245000 cents.
Now, 1500x + 200x + 800x = 245000
⇒  2500x = 245000
⇒ x = \(\frac{245000}{2500}\)
⇒ x = 98
Now, we substitute the x value in each denomination.
The number of 3 dollars i.e., 5x = 5×98 = 490
The number of 50cents i.e., 4x = 4×98 = 392
The number of 4 dollars i.e., 2x = 2×98 = 196
Therefore, the denominations of each number are 490, 392, and 196.

Example 3:
The sum of the numbers is 250, and the two numbers are in the ratio 4: 6. Find the numbers?

Solution:

Given two numbers in the ratio is 4: 6.
Let the two numbers be 4a and 6a.
Now, 4a+ 6a = 250
⇒ 10a = 250
⇒ a = \(\frac{250}{10}\)
⇒ a = 25
Next, we multiply the value of ‘a’ with the two numbers.
4a = 4×25 = 100
6a = 6×25 = 150
Thus, the numbers are 100 and 150.

Example 4:
Divide 420 into three parts which are in the ratio 3: 6: 5.

Solution:

The given number is 420 and the ratio is 3: 6: 5.
Now, to calculate the sum of the ratios = 3+6+5 = 14
Next, divide the three parts
First part = \(\frac{3}{14}\)×420 = 3×30 = 90
Second part = \(\frac{6}{14}\)×420 = 6×30 = 180
Third part = \(\frac{5}{14}\)×420 = 5×30 = 150
Hence, 420 can be divided into 90, 180, and 150 in the ratio of 3: 6: 5.

Example 5:
Four workers worked for 7 hours, 6 hours, 5 hours, and 6 hours. The total wage of ₹14640 was divided among the four workers according to their number of hours worked. How much did they get for the hours they worked?

Solution:

Let the hours of the workers worked to be in the ratio of 7: 6: 5: 6.
The sum of the ratio = 7+6+5+6 = 24hours
Now, get the amount per hour = Total wage / Hours worked = ₹14640/24 = ₹610 per hour
The wage of the first worker = 7×₹610 = ₹4270
The wage of the second worker = 6×₹610 = ₹3660
The wage of the third worker = 5×₹610 = ₹3050
The wage of the fourth worker = 6×₹610 = ₹3660
Therefore, each worker got the amount of ₹4270, ₹3660, ₹3050, and ₹3660.

Example 6:
Divide 4m 25cm in the ratio 4: 6.

Solution:

Given quantity 4m 25cm
Now, first, we have to convert the given quantity into one unit.
As we know, 1m = 100cm
4m = 4×100 = 400cm
4m 25cm = 400cm + 25cm = 425cm.
Given ratio is 4: 6
The total number of parts = 4x+6x = 10x
Now, 10x = 425
⇒ x = \(\frac{425}{10}\) = 42.5
So,
The quantity of first number 4x = 4×42.5 = 170cm
The quantity of second number 6x = 6×42.5 = 255cm
Hence, the two quantities of the two numbers are 170cm and 255cm.

Example 7:
The length and breadth of a rectangle are in the ratio of 8: 5. If the breadth is 85cm, then find the length of a rectangle?

Solution:

Given the ratio of length and the breadth is 8: 5 and the breadth is 85cm
Let the length be 8k and breadth be 5k.
Now, breadth 5k = 85cm
⇒ k = \(\frac{85}{5}\) = 17cm
Here, length = 8k
⇒ 8k = 8×17 = 136cm
Thus, the length of a rectangle is 136cm.

Example 8:
A certain sum of money is divided into three parts in the ratio 4: 3: 2.  If the first part is ₹448, then find the total amount, the second part, and the third part?

Solution:

Given three parts ratio is 4: 3: 2.
Let the amount of money be 4x, 3x, and 2x.
From the given the first part is ₹448.
4x = ₹448
⇒ x = \(\frac{448}{4}\) = 112
Thus, x = 112
Now,
3x = 3×112 = 336 and 2x = 2×112 = 224
Therefore, the second amount = ₹336
the third amount = ₹224
The total amount of money = First amount + Second amount + Third amount = ₹448 +  ₹336 + ₹224 = ₹1008.
Hence, the total amount of money is ₹1008, the second part is ₹336, and the third part is ₹224.

Perimeter and Area of Irregular Figures

In this article, you will learn how to find the Perimeter and Area of irregular figures. An irregular shape can be of any size and length. Irregular shapes can be seen all around us, for example, a diamond shape, a kite, a leaf, a flower, etc. The Area of irregular shapes means the space occupied by the shape which is measured in square units. The Perimeter of irregular shapes is by adding the length of their sides. Any shape whose sides and angles are not of equal length is termed an irregular shape.

On this page, learn about the definition of the Area and perimeter of irregular figures, how to find the area and perimeter of irregular figures, solve example problems, and so on.

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Definition of Irregular Figures

Irregular Figures is defined as a figure that is not a standard geometric shape. An irregular shape is simply a shape where not every single side is the same length. But some irregular figures are made up of two or more standard geometric shapes. If the shape is irregular then it has angles that are not all the same size. Based on the number of sides or corners we can decide the irregular figure.

How to find Perimeter and Area of Irregular Figures?

The following are the steps for finding the area and perimeter of irregular figures:
How to find Area of Irregular Shapes or Figures?
Step 1: First, divide the compound shape into a basic regular shape.
Step 2: Next, find each basic shape area separately.
Step 3: Then Add all the areas of basic shapes together.
Step 4: Now, write the final answer in square units.

How to find the Perimeter of Irregular Figures?
To find the perimeter of the irregular figure, we can simply add up each of its outer sides length of a shape. To find the perimeter of any shape like rectangle, square, and so on you have to add all the lengths of four sides. Consider ‘A’ is in this case the length of the rectangle and ‘B’ is the width of the rectangle.

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Solved Examples on Perimeter and Area of Irregular Figures

Example 1: The Irregular Figure is given below. Find the area of that figure.

Solution: 
As given in the question, the irregular figure is given.
Now, we can break the given irregular figure. After splitting the figure we have two rectangles.
Next, we will find the area of the two rectangles. The area of the irregular figure is the sum of the areas of two rectangles.
The width of one rectangle is 12 and the length of the rectangle is 4.
Next, the width of the other rectangle is 2, but its length is not given. By using the upper rectangle length we can find the length of the lower rectangle. So the right side of the figure is the length of the upper rectangle plus the length of the lower rectangle.
Since the total length is 10 units, the right side of the upper rectangle is 4 units long. So the length of the lower rectangle will be 6 units.
So the area of the figure is,
The Area of the figure is the Area of the upper rectangle + Area of the lower rectangle
We know that the Area of the rectangle is, length x width (or) breadth.
So, the area of a figure is , lw + lw = 12(4) + 2(6).
Area of the figure is = 48 + 12 = 60sq.units.
Therefore, the total area of the figure is 60 square units.

Example 2: Find the area of the below-given irregular figure.

Solution:
As given in the question, the given figure is an irregular figure.
Now, we can break the given irregular figure. After splitting the figure, we have one triangle and one rectangle.

Next, we will find the area of the irregular figure. The area of the irregular figure is the sum of the areas of two rectangles.
The rectangle has a length of 8 units and a width of 4 units. We need to find the base and height of the triangle.
On both sides of the rectangle 4units, the vertical side of the triangle is 3 units, which is 7- 4 = 3units.
Next, the length of the rectangle is 8units, so the base of the triangle is 3units, which is 8-5= 4units.
Now, we can add the areas then we get the area of the irregular figure.
So, the Area of the figure is the Area of the rectangle + the Area of the triangle.
We know the formulas, the area of the rectangle is, length x width (or) breadth.
The area of the triangle is 1/2bh.
So, the area of a figure is , lw + 1/2bh = 8(4) + 1/2(3)(3).
Area of the figure is = 32 + 4.5 = 36.5sq.units.
Hence, the total area of the given irregular figure is 36.5square units.

Example 3:

 

 

 

 

Worksheet on Ratio of Two or More Quantities | Sharing a Quantity in a Given Ratio Worksheet with Answers PDF

Worksheet on Ratio of Two or More Quantities gives information of ratio in different types and quantities. The ratio indicates their relative sizes and always expresses the ratio numbers in the lowest terms possible. A ratio compares two or more different quantities that have the same units of measure. In this ratio of two or more quantities word problems worksheet pdf, students can get the questions that are relevant to how to divide a quantity into two parts or more than the quantities of a given ratio with detailed solutions.

Practice with these activity sheet questions to have the hold on the concept of a ratio if we have more than two quantities of the same kind. This free printable activity sheet on the ratio of two or more quantities is a fun-learning workbook to solve all the questions. Also, students can able to create the questions themselves in different types by using this ratio of two or more quantities worksheet.

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Free Printable Ratio of Two or More Quantities Worksheet PDF with Answers

Example 1:
Find p: s, if p: q = 4: 6 and r: s = 3: 12.

Solution:

Given ratios are p: q = 4:6 and r: s = 3: 12
Now, to find the ratio of p: s.
As we know, if there is a ratio of quantities a: b and c: d. To get a: d = \(\frac{a}{b}\) × \(\frac{c}{d}\) = \(\frac{a}{d}\).
Similarly, we apply the same formula
p: s = \(\frac{p}{q}\) × \(\frac{r}{s}\)
⇒ p: s = \(\frac{4}{6}\) × \(\frac{3}{12}\)
⇒ p: s = 1: 6
Therefore, the value of the ratio p: s is 1: 6.

Example 2:
If x: y = 5: 4 and y: z = 2: 6, then find (i) x: z and (ii) x: y: z.

Solution:

(i) Given x: y = 5: 4 and y: z = 2: 6
To get x: z, we use the formula, x: z = \(\frac{x}{y}\) × \(\frac{y}{z}\)
So,
x: z = \(\frac{5}{4}\) × \(\frac{2}{6}\)
⇒ x: z = \(\frac{5×2}{4×6}\)
⇒ x: z = \(\frac{10}{24}\)
⇒ x: z = \(\frac{5}{12}\)
Thus, the ratio x: z = 5: 12.

(ii) Given ratio x: y = 5: 4 and y:  = 2: 6
Now, to find the ratio of x: y: z.
Here, the values of y are not the same. To get the ratio of x: y: z, we have to multiply the value of y with a number to get both the values equal.
\(\frac{x}{y}\) = \(\frac{5}{4}\) × \(\frac{1}{1}\) = \(\frac{5}{4}\)
\(\frac{y}{z}\) = \(\frac{2}{6}\) × \(\frac{2}{2}\) = \(\frac{4}{12}\)
Thus, the ratio x: y: z = 5: 4: 12.

Example 3:
Divide ₹ 3876 in the ratio \(\frac{4}{5}\): 1\(\frac{3}{5}\).

Solution:

Given ratio \(\frac{4}{5}\): 1\(\frac{3}{5}\) i.e., \(\frac{4}{5}\): \(\frac{8}{5}\).
Therefore, the ratio becomes 4: 8 = 4+8 = 12.
Let the common multiple be x.
12x = 3876
x = \(\frac{3876}{12}\)
⇒ x = 323
Now, 4x = 4×323 = 1292
8x = 8×323 = 2584
Thus, the ratio is divided into ₹1292 and ₹2584.

Example 4:
If a: b = 3: 7, then find the ratios of the following:
(i) (2a+ 5b): (3a+ 6b)
(ii) (a+ 2b): (7a- b)

Solution:

(i) Given ratio a: b = 3: 7
The given equation is (2a+ 5b): (3a+ 6b).
Now, substitute the values of a and b in the above equation.
(2a+ 5b): (3a+ 6b) = ((2×3)+(5×7)): ((3×3)+(6×7)) = (6+35): (9+42) = 41: 51
Thus, the ratio of the equation (2a+ 5b): (3a+ 6b) is 41: 51.

(ii) Given equation is (a+ 2b): (7a- b)
The values of a and b were given are 3 and 7.
Now, substitute the values in the above equation.
(a+ 2b): (7a- b) = (3+(2×7)): ((7×3)-7) = (3+14): (21-7) = 15: 14
Hence, the ratio is 15: 14.

Example 5:
Find the value of m, if (m-1): (m+3) is the reciprocal ratio of 12: 8.

Solution:

The reciprocal ratio of 12: 8 is 8: 12.
Now, 8: 12 = (m-1): (m+3)
⇒ \(\frac{8}{12}\) = \(\frac{m-1}{m+3}\)
⇒ 8(m+3) = 12(m-1)
⇒ 8m+24 = 12m-12
⇒ 8m- 12m = -12 – 24
⇒ -4m = -36
⇒ m = \(\frac{36}{4}\)
⇒ m = 9.
Therefore, the value of m is 9.

Example 6:
If 4x = 7y = 2z, find the ratio of x: y: z.

Solution:

Given equation is 4x = 7y = 2z.
Now, 4x = 7y
⇒ x = \(\frac{7y}{4}\) —- (i)
7y = 2z
⇒ z = \(\frac{7y}{2}\) —- (ii)
Now,
x: y: z = \(\frac{7y}{4}\): y : \(\frac{7y}{2}\) (from (i) and (ii))
⇒ x: y: z = \(\frac{7}{4}\): 1 : \(\frac{7}{2}\)
⇒ x: y: z = 7: 4: 14
Hence, the ratio of x: y: z is 7: 4: 14.

Example 7:
Divide $ 3570 in the ratio \(\frac{1}{5}\): 1\(\frac{2}{3}\): \(\frac{2}{15}\).

Solution:

Given ratio is \(\frac{1}{5}\): 1\(\frac{2}{3}\): \(\frac{2}{15}\) i.e., \(\frac{1}{5}\): \(\frac{5}{3}\): \(\frac{2}{15}\).
Let the common multiple be k.
Therefore, \(\frac{1k}{5}\): \(\frac{5k}{3}\): \(\frac{2k}{15}\) = 3570
⇒ \(\frac{3k}{15}\): \(\frac{25k}{15}\): \(\frac{2k}{15}\) = 3570
⇒ \(\frac{30k}{15}\) = 3570
⇒ 2k = 3570
⇒ k = \(\frac{3570}{2}\)
⇒ k = 1785
Now, to find the three equal ratios
\(\frac{1k}{5}\) = \(\frac{1×1785}{5}\) = \(\frac{1785}{5}\) =357
\(\frac{5k}{3}\) = \(\frac{5×1785}{3}\) = \(\frac{8925}{3}\) = 2975
\(\frac{2k}{15}\) = \(\frac{2×1785}{15}\) = \(\frac{3570}{15}\) = 238
The ratio is divided into $357, $2975, and $238.

Example 8:
If a: b = 4: 1, b: c = 12: 7, and c: d = 7: 9, then find the triplicate ratio of a: d.

Solution:

Given ratios are a: b = 4: 1, b: c = 12: 7, and c: d = 7: 9
Now, to find the ratio of a: d.
Here,
a: d = \(\frac{a}{b}\) × \(\frac{b}{c}\) × \(\frac{c}{d}\)
⇒ a: d = \(\frac{4}{1}\) × \(\frac{12}{7}\) × \(\frac{7}{9}\)
⇒ a: d = \(\frac{4×12×7}{1×7×9}\)
⇒ a: d = \(\frac{4×4×1}{1×1×}\)
⇒ a: d = 16: 3
The ratio of a: d is 16: 3.
Triplicate ratio = 16³ : 3³ = 4096 : 27.
Therefore, the triplicate ratio of a: d = 4096: 27.

Example 9:
What number should be added to the ratio of 13: 33? So, the ratio becomes 4: 9.

Solution:

Given ratio is 13: 33
Now, we have to add the number so that we can get the ratio of 4: 9.
If we add the number 3 to the ratio of 13: 33
We got, 13+3: 33+3 = 16: 36.
So, the ratio becomes 16: 36 = 4: 9.
Thus, the number 4 should be added to the given ratio.

Worksheet on the Application of Overhead Expenses

Worksheet on the Application of Overhead Expenses will help the students to find various questions on the Application of Overhead Expenses concept. We provide all of our worksheets for free. Take the reference of our worksheets and get the practice experience with them. An in-depth explanation along with answers and images are given on this page. Also, various methods are included to solve different problems on the Application of Overhead Expenses. Therefore, practice every problem and get complete knowledge on the entire concept.

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Application of Overhead Expenses Worksheets

Check out the below problems which are available along with solutions. Practice all the problems on your own to get a grip on the complete concept.

1. A trader from Canada purchases a electrical rice cooker at a price of $8,000 and pays $500 on its maintenance and transportation. He then sells it for $11,000 to a customer. Find the profit/loss faced by him. Also, find the percent for the same.

Solution:

Given that a trader from Canada purchases a electrical rice cooker at a price of $8,000 and pays $500 on its maintenance and transportation.
Therefore, the total cost to purchase an electrical rice cooker is $8,000 + $500 = $8,500
The cost price = $8,500
He then sells it for $11,000 to a customer.
Selling Price = $11,000
Now, subtract the cost price from the Selling Price to know he gained profit or loss.
Selling Price – cost price = $11,000 – $8,500 = $2,500
So, he gained profit of $2,500.
Now, find the Profit percentage.
Profit percentage = (Profit/Cost Price) × 100 = ($2,500/$8,500) × 100 = 29.41%

Therefore, the profit faced by him on selling the electrical rice cooker is $2,500 and the profit percentage is 29.41%.

2. A trader from India buys 2 carrom board sets for $2000 and $2500 respectively. He spends $200 on the overall transportation of the sets. He then sells them in India for $2,500 and $2,300 respectively. Find the overall profit/loss faced by the trader. Also, find the percent for the same.

Solution:

Given that a trader from India buys 2 carrom board sets for $2000 and $2500 respectively. Also, he spends $200 on the overall transportation of the sets.
Therefore, the total cost to purchase of 2 carrom board sets is $2,000 + $2,500 + $200 = $4,700
The cost price = $4,700
He then sells them in India for $2,500 and $2,300 respectively.
Then, the total Selling Price = $2,500 + $2,300 = $4,800
Now, subtract the selling price from the cost Price to know the profit he gained.
Selling Price – cost price = $4,800 – $4,700 = $100
So, he gained profit of $100.
Now, find the Profit percentage.
Profit percentage = (Profit/Cost Price) × 100 = ($100/$4,700) × 100 = 2.12%

Therefore, the profit faced by him on selling the electrical rice cooker is $100 and the profit percentage is 2.12%.

3. A shopkeeper from Canada buys some electronic gadgets from Sydney at a price of $15,000 and takes them back to Canada by paying $1,000. There he sells them for $18,000. Find the overall profit/loss faced by the trader. Also, find the percent for the same.

Solution:

Given that a shopkeeper from Canada buys some electronic gadgets from Sydney at a price of $15,000 and takes them back to Canada by paying $1,000.
Therefore, the total cost to purchase some electronic gadgets is $15,000 + $1,000 = $16,000
The cost price = $16,000
He then sells them for $18,000.
Then, the total Selling Price = $18,000
Now, subtract the selling price from the cost Price to know the profit he gained.
Selling Price – cost price = $18,000 – $16,000 = $2,000
So, he gained profit of $2,000.
Now, find the Profit percentage.
Profit percentage = (Profit/Cost Price) × 100 = ($2,000/$16,000) × 100 = 12.5%

Therefore, the profit faced by him on selling some electronic gadgets is $2,000 and the profit percentage is 12.5%.

4. A trader buys 12 pens for Rs100 each and the overhead expenses were Rs20 per pen. If he makes Rs30 profit on each pen. Determine the selling price of each pen and the overall selling price of the pens.

Solution:

Given that a trader buys 12 pens for Rs100 each and the overhead expenses were Rs20 per pen.
If one pen cost is Rs 100, then 12 pens cost is 12 × Rs 100 = Rs 1,200
The overhead expenses per pen are Rs20. For 12 pens, the overhead expenses are 12 × Rs 20 = Rs 240
Single pen cost price is Rs100 + Rs 20 = Rs 120
The total cost price of pens = Rs1,200 + Rs 240 = Rs 1440
If he makes Rs30 profit on each pen. So, the profit on 12 pens is 12 × Rs 30 = Rs 360
Now, find the selling price of each pen.
Profit = Selling price – Cost price
The selling price of each pen = Profit + Cost price
selling price of each pen = Rs30 + Rs 120 = Rs 150
Now, find the selling price of all pens.
Profit = Selling price – Cost price
The selling price of all pens = Profit + Cost price
selling price of all pens = Rs 360 + Rs 1440 = Rs 1800

Therefore, the selling price of each pen is Rs 150 and the overall selling price of the pens is Rs 1800.

5. A shopkeeper buys 10 bike toys for $200 each. He spends $10 as overhead expenses for each bike toy. He faces a loss of $20 on each of these bike toys. Determine the selling price of each bike toy and the overall selling price of the bike toys.

Solution:

Given that a shopkeeper buys 10 bike toys for $200 each. He spends $10 as overhead expenses for each bike toy.
If one bike toy cost is $200, then 10 bike toys cost is 10 × $200 = $2000
The overhead expenses per bike toy are $20. For 10 bike toys, the overhead expenses are 10 × $ 20 = $ 200
Single bike toy cost price is $200 + $ 20 = $220
The total cost price of bike toys = $2000 + $ 200 = $ 2200
He faces a loss of $20 on each of these bike toys. So, the loss on 10 bike toys is 10 × $20 = $200
Now, find the selling price of each bike toy.
Loss = Cost price – Selling price
The selling price of each bike toy = Cost price – Loss
selling price of each bike toy = $220 – $20 = $200
Now, find the selling price of all bike toys.
Loss = Cost price – Selling price
The selling price of all bike toys = Cost price of all bike toys – Loss on all bike toys
selling price of all bike toys = $ 2200 – $200 = $2000

Therefore, the selling price of each bike toy is $200 and the overall selling price of the bike toys is Rs $2000.

6. A trader bought water bottles for $300 and spent $10 on their packaging. He sells them for $700. Find the profit/ loss percent faced by him.

Solution:

Given that a trader bought water bottles for $300 and spent $10 on their packaging.
Therefore, the total cost to bought water bottles is $300 + $10 = $310
The cost price = $310
He then sells it for $700.
Selling Price = $700
Now, subtract the cost price from the Selling Price to know he gained profit or loss.
Selling Price – cost price = $700 – $310 = $390
So, he gained profit of $390.
Now, find the Profit percentage.
Profit percentage = (Profit/Cost Price) × 100 = ($390/$310) × 100 = 125.80%

Therefore, the profit percent faced by him is 125.80%.

7. A shopkeeper from Canada bought some home appliances for $30,000. He spends $600 on its transportation. He sells them at a profit of 10%. Find the selling price and profit gained by the shopkeeper.

Solution:

Given that a shopkeeper from Canada bought some home appliances for $30,000. He spends $600 on its transportation.
Therefore, the total cost to bought some home appliances is $30,000 + $600 = $30,600
The cost price = $30,600
He sells them at a profit of 10%.
Now, find the profit.
Profit% = Profit/CP × 100
Substitute the Profit% and Cost Price in the above formula.
10 = Profit/$30,600 × 100
Profit = 10/100 × $30,600
Profit = $3,060
Now, find the Selling Price.
Profit = Selling price – Cost price
Substitute the Profit and Cost Price in the above formula.
Profit = Selling price – Cost price
Selling price = Profit + Cost price
Selling price = $3,060 + $30,600 = $33,660

Therefore, the profit gained by the shopkeeper is $3,060 and the Selling price on the home appliances is $33,660.

8. A trader purchases some electronic products from a retailer for Rs 30,000. He spends Rs 500 on the maintenance of the products. He then sells them for Rs 65,000. Determine the profit/loss. Also, find the percent for the same.

Solution:

Given that a trader purchases some electronic products from a retailer for Rs 30,000. He spends Rs500 on the maintenance of the products.
Therefore, the total cost to purchase some electronic products is Rs 30,000 + Rs 500 = Rs 35,000
The cost price = Rs 35,000
He then sells them for Rs 65,000.
Then, the total Selling Price = Rs 65,000.
Now, subtract the selling price from the cost Price to know the profit he gained.
Profit = Selling price – Cost price
Selling Price – cost price = Rs 65,000 – Rs 35,000 = Rs 30,000
So, he gained profit of Rs 30,000.
Now, find the Profit percentage.
Profit percentage = (Profit/Cost Price) × 100 = (Rs 30,000/Rs 35,000) × 100 = 85.71%

Therefore, the profit faced by him on selling some electronic products is Rs 30,000 and the profit percentage is 85.71%.

Graph of Standard Linear Relations Between x, y

A linear equation is also known as the one-degree equation because it has the highest degree as 1. The standard linear relations between x, y is ax + by + c = 0 where a, b, c are the real numbers and x, y are variables of the given equation. The graph of a linear equation with variables x and y forms a straight line. Let us discuss how to draw the graph of standard linear relations between x, y from this article. For better understanding, we have provided examples of problems on the graph of standard linear relations between x, y. By this, the students who are lagging in the concept of coordinate geometry can score good marks in the exams.
graph paper

Quadrants and Convention for Signs of Coordinates

A graph has four quadrants. In quadrant 1 the values of x and y are positive, in Quadrant 2 the x variable value is positive and y variable value is negative, in Quadrant 3 the x variable value and y variable value is negative and in Quadrant 4 the x variable value is negative and y variable value is positive.

  • Quadrant I – (+, +)
  • Quadrant II – (+, -)
  • Quadrant III – (-, -)
  • Quadrant IV – (-, +)

Linear Equation in One Variable

A linear equation with one variable has only x value. The equation of the linear equation in one variable is ax + b = 0 where a and b are the real numbers and x is the variable.
Example: Solve the linear equation in one varible 2x + 5 = 17.
Solution:
Given the linear equation 2x + 5 = 17
2x = 17 – 5
2x = 12
x = 12/2
x = 6

Linear Equation in Two Variable

An equation with two variables has ordered pairs x and y. Substitute the values of x and y to get the ordered pairs to plot the points on the graph.

X variable value X + 6 = y Y value Ordered Pair (x, y)
-2 -2 + 6 = y 4 (-2,4)
-1 -1 + 6  = y 5 (-1,5)
0 0 + 6 = y 6 (0, 6)
1 1 + 6 = y 7 (1, 7)
2 2 + 6 = y 8 (2, 8)

Solved Problems on how to draw Graph of Standard Linear Relations Between x, y

Example 1.
Draw the graph of the linear equation in two variables y = 2x + 3.
Solution:
Given equation is y = 2x + 3
In the given equation
If x = -2 then y = 2(-2) + 3 = -4+3 = 1
If x = -1 then y = 2(-1) + 3 = 1
If x = 0 then y = 2(0) + 3 = 3.
If x = 1 then y = 2(1) + 3 = 5.
If x = 2 then y = 2(2) + 3 = 7
Plot the graph using the points (-2,1), (-1,1), (0,3), (1,5) and (2,7).
Graph of Standard Linear Relation between x, y example_1

Example 2.
Draw the graph of the linear equation in two variables x − y = 2
Solution:
x – y = 2
– y = 2 – x
y = x – 2
If x = 0 then y = 0 – 2 = -2
If x = 1 then y = 1 – 2 = -1
If x = 2 then y = 2 – 2 = 0
If x = 4 then y = 4 – 2 = 2
Then plot the following points in the graph.

X 0 1 2 4
y -2 -1 0 2

Graph of Standard Linear Relation between x, y example_2
Example 3.
Draw the graph of the linear equation in two variables y = 4x + 1.
Solution:
Given equation is y = 4x + 1
In the given equation
If x = -5 then y = 4(-5) + 1 = -20+ 1 = -19
If x = -1 then y = 4(-1) + 1 = -3
If x = 0 then y = 4(0) + 1 = 1.
If x = 1 then y = 4(1) + 1 = 5.
If x = 2 then y = 4(2) + 1 = 9
Plot the graph using the points (-5,-19), (-1,-3), (0,1), (1,5) and (2,9).
Graph of Standard Linear Relation between x, y example_3

Example 4.
Draw the graph of the linear equation in two variables 2x – 3y + 1 = 0
Solution:
Given equation is 2x – 3y + 1 = 0
2x + 1 = 3y
y = 2x + 1/3
In the given equation
If x = -1 then y = 2(-1) + 1/3 = -1/3
If x = 0 then y = 2(0) + 1/3= 1/3
If x = 1 then y = 2(1) + 1/3 = 1
If x = 2 then y = 2(2) + 1/3 = 5/3
Plot the graph using the points (-1,-1/3), (0,1/3), (1,1), (2,5/3).
Graph of Standard Linear Relation between x, y example_4

Example 5.
Draw the graph of the linear equation in two variables x + y = 4
Solution:
x + y = 4
– y = x – 4
y = -x + 4
If x = 0 then y = 0 + 4 = 4
If x = 1 then y = -1 + 4 = -3
If x = 2 then y = -2 + 4 = 2
If x = 4 then y = – 4 + 4 = 0
Then plot the following points in the graph (0,4), (1,-3), (2,2) and (4,0)
Graph of Standard Linear Relation between x, y example_5

FAQs on Graph of Standard Linear Relations Between x, y

1. How do you graph X or Y?

To plot a graph using an equation, first, we construct a table using the two values of x and y by substituting the values of x and y in the equation starting from 0 to so on. Then draw the points on the graph using the x and y values then values of x lie on the x-axis and the values of y lie on the y-axis.

2. What type of graph is XY?

Suppose we have an X-Y Plot. X-Y plots are used to determine the relationships between the two different things. The x-axis is used to measure one variable and the y-axis is used to measure the other variable. If both variables increase at the same time then they have a positive relationship.

3. Which graph has a straight line?

A linear graph has a straight line. It means if any relation gives a single straight line then it is a linear graph. The “linear” stands for a straight line.

Drawing Graph of y = mx + c Using Slope and y-intercept | How do you Graph with Slope and y-Intercept?

Learn how to draw the graph of y = mx + c Using Slope and y-intercept from this page. The formula to draw the graph using the slope and y-intercept is y = mx + c where m is slope and c is the y-intercept. Now in this article, we are going to draw the Coordinate Geometry Graph using slope and y-intercept. Let us see the steps for graphing an equation using slope & y-intercept with example problems.
y = mx+c

How to Draw y=mx+c Graph using Slope, y-intercept?

Follow the steps to draw the graph of the equation y = mx + c using slope and y-intercept.

  1. First, find the y-intercept c of the equation of the line.
  2. Now plot the y-intercept where the y-axis cuts the point at (0,c ).
  3. Next find the slope of the equation, y = mx + c.
  4. Now plot the points on the graph make sure you go up to the right (y-axis) and down to the right if it is negative (y’-axis).
  5. Join the points with a line.

Do Refer:

Plotting a Straight Line y = mx + c Examples

The following graphs show the equation of the line y = mx + c, where the line cuts the y-axis at the point (0, c) which is the distance from the origin to point c.

Example 1.
Find the equation of a line in the form y = mx + c, having a slope of 2 units and an intercept of 4 units?
Solution:
Given that the slope of the line is m = 2
and the y-intercept of the line c = 4.
The slope-intercept form of the equation of a line is y = mx + c is
y = 2x + 4
Therefore the equation of the line is y = 2x + 4.
Drawing Graph of y = mx + c Using Slope and y-intercept_1

Example 2.
Find the equation of a line in the form of y = mx + c and having a slope of 10 units and an intercept of – 6 units?
Solution:
Given that the slope of the line is m = 10
and the y-intercept of the line c = -6.
The slope-intercept form of the equation of a line is y = mx + c is
y = 10x – 6.
Therefore the equation of the line is y = 10x – 6.
Drawing Graph of y = mx + c Using Slope and y-intercept_2

Example 3.
Find the equation of the line using the slope equal to 5 and y-intercept equal to -1/3.
Solution:
Given that the slope of the line is m = 5
and the y-intercept of the line c = -1/3.
The slope-intercept form of the equation of a line is y = mx + c is
y = 5x – 1/3
Therefore the equation of the line is y = 5x – 1/3
Drawing Graph of y = mx + c Using Slope and y-intercept_3

Example 4.
Find the equation of the line using the slope equal to 3/2 and y-intercept equal to 4/5.
Solution:
Given that the slope of the line is m = 3/2
and the y-intercept of the line c = 4/5.
The slope-intercept form of the equation of a line is y = mx + c is
y = 3/2x + 4/5
Therefore the equation of the line is y = 3/2x + 4/5.
Drawing Graph of y = mx + c Using Slope and y-intercept_4

Example 5.
Find the equation of the line using the slope equal to 2/7 and y-intercept equal to 1.
Solution:
Given that the slope of the line is m = 2/7
and the y-intercept of the line c = 1.
The slope-intercept form of the equation of a line is y = mx + c is
y = 2/7x – 1.
Therefore the equation of the line is y = 2/7x – 1.
Drawing Graph of y = mx + c Using Slope and y-intercept_5

FAQs on Graph of y = mx + c Using Slope and y-intercept

1. What is the formula for slope and y-intercept?

The formula for slope and y-intercept is y = mx + c where m is slope and c is the y-intercept.

2. How do you find the y-intercept?

The y-intercept is the point at which the graph crosses the y-axis where the x-coordinate is zero. Make x-coordinate as zero and solve y to find the y-intercept of the equation.

3. How do you find the point-slope form given the slope and y-intercept?

The slope-intercept formula is y = mx + c is used when you know the slope of the line to be examined and the point given is also the y-intercept (0, c).