Class 6 – Page 16 – Samacheer Kalvi

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.2

Miscellaneous Practice Problems

Question 1.
A shopkeeper buys three articles for ₹ 325, ₹ 450, and ₹ 510. He is able to sell them for ₹ 350, ₹ 425, and ₹ 525 respectively. Find the gain or loss to the shopkeeper on the whole.
Solution:
Total cost price of the three articles = Rs 325 + Rs 450 + Rs 510 = Rs 1285
Total selling price of the three articles = Rs 350 + Rs 425 + Rs 525 = Rs 1,300
CP < SP
Profit = SP – CP
= Rs 1,300 – Rs 1,285
= Rs 15

Question 2.
A stationery shop owner bought a scientific calculator for Rs 750. He had put a battery worth Rs 100 in it. He had spent Rs 50 for its outer pouch. He was able to sell it for Rs 850. Find his profit or loss.
Solution:
CP = Rs 750 + Rs 100 + Rs 50
= Rs 900
SP = Rs 850
SP < CP
Loss = CP – SP
= Rs 900 – Rs 850
= Rs 50

Question 3.
Nathan paid ₹ 800 and bought 10 bottles of honey from a village vendor. He sold them in a city for ₹ 100 per bottle. Find his profit or loss.
Solution:
Cost of 10 bottles of honey = Rs 800
CP = Rs 800
Selling price of 10 bottles = Rs 100 × 10
SP = Rs 1000
SP > CP
Profit = SP – CP
= Rs 1,000 – Rs 800
= Rs 200

Question 4.
A man bought 400 metre of cloth for Rs 60,000 and sold it at the rate of Rs 400 per metre. Find his profit or loss.
Solution:
Cost of 400 m of cloth = Rs 60,000
CP = Rs 60,000
Selling price of 400 m of cloth = 400 × Rs 400
SP = Rs 1,60,000
SP > CP
Profit = SP – CP
= Rs 1,60,000 – Rs 60,000
= Rs. 1,00,000

Challenge Problems

Question 5.
A fruit seller bought 2 dozen bananas at ₹ 20 a dozen and sold them at ₹ 3 per banana. Find his gain or loss.
Solution:
Cost price of 2 dozen bananas = 2 × Rs 20
CP = Rs. 40
2 dozen = 24 bananas
Selling price of 2 dozen bananas = Rs 3 × 24
SP = Rs 72
SP > CP
Profit = SP – CP
= Rs 72 – Rs 40
= Rs 32

Question 6.
A store purchased pens at Rs 216 per dozen. He paid Rs 58 for conveyance and sold the pens at the discount of Rs 2 per pen and made an overall profit of Rs 50. Find the M.P. of each pen.
Solution:
Cost of 1 dozen pens = Rs 216 + Rs 58
CP = Rs 274
Discount for each pen = Rs 2
Overall profit = Rs 50
Total discount for 12 pens = Rs 2 × 12
= Rs 24
Selling price of 12 pens = Mp – Discount
SP = Mp – Rs 24
Profit = SP – CP
Rs 50 = Mp – Rs 24 – Rs 274
MP = Rs 50 + Rs 24 + Rs 274
= Rs 348 (1 dozen)
MP of each pen = Rs 348 / 12
= Rs 174/6
= Rs 29

Question 7.
A vegetable vendor buys 10 kg of tomatoes per day at ₹ 10 per kg, for the first three days of a weak. 1 kg of tomatoes got smashed on every day for those 3 days. For the remaining 4 days of the week, he buys 15 kg of tomatoes daily at ₹ 8 per kg. If for the entire week he sells tomatoes at ₹ 20 per kg, then find his profit or loss for the week.
Solution:
First 3 days
Tomatoes purchased for one day = 10 Kg
Tomatoes purchased for 3 days = 3 × 10 Kg = 30 Kg
The cost price of purchasing 30 Kg of Tomatoes = Rs 10 × 30 = Rs 300
Next four days
Tomatoes purchased for one day = 15 Kg
Tomatoes purchased for four days = 4 × 15 Kg = 60 Kg
Cost price of purchasing 60 Kg of Tomatoes = Rs 8 × 60 = Rs 480
Total cost price = Rs 300 + Rs 480 = Rs 780
Tomatoes smashed on the first 3 days = 3 x 1 Kg = 3 Kg
Remaining Tomatoes on the first 3 days = (30 – 3) Kg = 27 Kg
Weight of the remaining tomatoes in that week = (27 + 60) Kg = 87 Kg
The selling price of 1 Kg of Tomato = Rs 20
The selling price of 87 Kg of Tomatoes = Rs 20 × 87
SP = Rs 1,740
SP > CP
Profit = SP – CP = Rs 1,740 – Rs 780 = Rs 960

Question 8.
An electrician buys a used T.V. for Rs 12,000 and a used Fridge for Rs 11,000. After spending Rs 1,000 on repairing the T.V. and Rs 1500 on painting the Fridge, he fixes up the M.P. of T.V. as Rs 15,000 and that of Fridge as Rs 15,500. If he gives Rs 1000 discount on each find his profit or loss.
Solution:
Total cost price of TV and Fridge = Rs 12,000 + Rs 11,000 + Rs 1,000 + Rs 1,500 = Rs 25,500
Selling price of TV = MP – Discount = Rs 15,000 – Rs 1,000 = Rs 14,000
Selling price of Fridge = MP – Discount = Rs 15,500 – Rs 1,000 = Rs 14,500
Total Selling price of TV and Fridge = Rs 14,000 + Rs 14,500 = Rs 28,500
SP > CP
Profit = SP – CP = Rs 28,500 – Rs 25,500 = Rs 3,000

Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.
A piece of wire is 36 cm long. What will be the length of each side if we form
i) a square
ii) an equilateral triangle.
Solution:
Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Question 2.
From one vertex of an equilateral triangle with side 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion?
Solution:

If an equilateral triangle of side 6 cm is removed the perimeter = (40 + 34 + 6 +34) cm = 114 cm
Perimeter of the remaining portion = 114 cm

Question 3.
Rahim and peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with length 40 m and breadth 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Perimeter of the square = (4 × sides) units
side of the square = 50 m
∴ Perimeter = 4 × 50 m = 200 m
Rahim walked twice around the park.
∴ Total distance walked by Rahim = 2 × 200m
Total distance covered by Rahim = 400m
Perimeter of a rectangle = 2 (length + breadth) unit
Length = 40 m
Breadth = 30 m
∴ perimeter = 2 × (40 + 30) m = (2 × 70) m = 140 m
Peter walked around twice
∴ Distance covered by Peter = 2 × 140 m = 280 m
∴ Distance covered by Peter = 280 m
400 m > 280 m
Difference = 480 – 280 = 120 m
∴ Rahim covers 120m more than Peter

Question 4.
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park?
Solution:
Given length of rectangular park is 14m more than its breadth.
Let the breadth be b m .
∴ Length of the park will be l = b + 14 m
Given perimeter = 200 m
2 × (l + b) = 200 m
2 × (b + 14 + b) = 200 m [∵ l = b + 14]
2 × (2b + 14) = 200 m
2b + 14 = \(\frac{200}{2}\) m
2b + 14 = 100 m
2b = 100 – 14 m
2b = 86 m
b = \(\frac{86}{2}\) m
b 43 m
Length Length of the park = 57 m
Area of a rectangle = (length × breadth) unit²
= (57 × 43) m² = 2,451 m²
Area of the park = 2,451 m²

Question 5.
Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per meter.
Solution:Perimeter of a square = (4 × side) units
side = 5 m
∴ Perimeter = (4 × 5) m = 20 m
Each side is fenced with 2 rows of wire
∴ Distance to be fenced = 2 × 20 m = 40 m
Cost of fencing per metre = ₹ 10
∴ Cost of fencing 40 m = 40 × 10 = ₹ 400
Cost of fencing the garden = ₹ 400

Challenge Problems

Question 6.
A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Solution:
Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Question 7.
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Solution:
l = 40 cm, b = 20 cm
Area of the rectangle = l × b sq units
= 40 × 20 cm²
= 800 cm²
a = 10 cm
Area of the square = a × a sq. units
= 10 × 10 cm²
= 100 m²
No of squares formed = \(\frac{800}{100}\) cm²
= 8

Question 8.
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Solution:
Length l = 3b
breadth = b
perimeter = 64 cm
2 (l + b) = 64 cm
2(3b + b) = 64
2 (4b) = 64
8b = 64
b = \(\frac{64}{8}\)
breadth b = 8 cm
length l = 3b
l = 3 × 8
= 24 cm

Question 9.
How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
12 rectangles
(1, 23), (2, 22), (3, 21), (4, 20), (5, 19), (6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:

The perimeter of square B = 4 × 4 m = 16 m
The perimeter of square A = 4 × 2 m = 8 m
The perimeter of square B is twice that of square A.

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Let the side of square is s units then area = (s × s) units²
If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units
Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units² = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units²
Area of the new square is reduced to \(\frac{1}{16}\) times to that of original area.

Question 12.
Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Solution:

Given perimeter of square = perimeter of rectangle
4 × side = 2 (length + breadth)
(4 × 10) m = 2(l + 8)m
\(\frac{4 \times 10}{2}\) = l + 8
20 = l + 8
l = 20 – 8
l = 12 m
∴ length of the rectangle = 12 m
Area of the square plot – side × side = 10 × 10 m² = 100 m²
Area of the rectangular plot = length × breadth = (12 × 8) m² = 96 m²
100 m² > 96 m²
∴ Square plot has greater area by 4m²

Question 13.
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Area of a square = a × a sq.units
= 6 x 6 cm²
= 36 cm²
Area of a rectangle = l × b sq units
= 9 x 6 cm²
= 54 cm²
Area of a triangle = \(\frac{1}{2}\) × b × h sq.units
= \(\frac{1}{2}\) × 4² × 6 cm²
= 12 cm²
Total area of the shaded portion
= (36 + 54 + 12) cm²
= 102 cm²

Question 14.
Find the approximate area of the flower in the given square grid.

Solution:
Approximate area = Number of full squares + Number of more than half squares + \(\frac{1}{2}\) × Number of half squares.
= 10 + 5 + (\(\frac{1}{2}\) × 1) Sq units. = 10 + 5 + \(\frac{1}{2}\) sq. units
= 15 \(\frac{1}{2}\) sq. units = 15.5 sq. units.