Mathematics Class VI

Question 1 of 20

1. Question
1 point(s)
A square of length 170 m has an area of?
- 22500
- 29000
- 28900
- 28800
Correct

Incorrect

Hint

Side of the square field = 170 m
∴ Area of the square field = Side x Side
= 170 m x 170 m
= 22500 sq m.
Question 2 of 20

2. Question
1 point(s)
A rectangular park 400 m long and 800 m wide is to be fenced at the rate of ₹4 per meter.
- 8600
- 9400
- 8400
- 9600
Correct

Incorrect

Hint

Length of the park = 400 m
Breadth = 800 m
∴ Perimeter of the park = 2 [length + breadth]
= 2 [400 m + 800 m]
= 2 x 1200 m = 2400 m.
Cost of fencing the rectangular park = 2400 x 4 = ₹9600
Question 3 of 20

3. Question
1 point(s)
Find the perimeter of a square whose side is 30 cm.
- 80 cm
- 120 cm
- 140 cm
- 160 cm
Correct

Incorrect

Hint

Side of the square = 30 cm
∴ Perimeter of the square = 30 cm $\times$ 4 = 120 cm
Question 4 of 20

4. Question
1 point(s)
A square shaped land has each side equal to 220 m. If 3 layers of metal wire has to be used to fence it, what is the length of the wire needed?
- 2800 m
- 2460 m
- 2640 m
- 2840 m
Correct

Incorrect

Hint

It is given that
Each side of a square field = 220 m
We can find the wire required to fence the square field by determining the perimeter = 4 $\times$ each side of a square field
By substituting the values
Perimeter of the square field = 4 $\times$ 220 = 880 m
So, the length of wire which is required to fence three layers is = 3 $\times$ 880 = 2640 m
Hence, the length of wire needed to fence 3 layers is 2640 m.
Question 5 of 20

5. Question
1 point(s)
The dimensions of a photograph are 15 cm $ \times $ 30 cm. What length of cardboard is needed to frame the picture?
- 90 cm
- 100 cm
- 80 cm
- 70 cm
Correct

Incorrect

Hint

It is given that
Dimensions of a photograph = 15 cm × 30 cm
So, the required length of the cardboard can be determined from the perimeter of the photograph = 2 (L + B)
By substituting the values = 2 (15 + 30) = 2 × 45 = 90 cm
Hence, the length of the cardboard required to frame the picture is 90 cm.
Question 6 of 20

6. Question
1 point(s)
Arvind has a garden, 60 m long and 50 m wide. He wants to fix a metal pipe post every 4 meters apart. Each post was 3 m long. What is the total length of the pipes he bought for the posts?
- 135
- 80
- 165
- 110
Correct

Incorrect

Hint

The dimensions of garden are
Length = 60 m
Breadth = 50 m
So, the perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (60 + 50) = 2 × 110 = 220 m
Given:
Arvind fixes a post every 4 meters apart
No. of posts required = $\dfrac{{220}}{4}$ = 55
The length of each post = 3 m
So, the total length of the pipe required = 55 × 3 = 165 m
Hence, the total length of the pipes he bought for the posts is 165 m.
Question 7 of 20

7. Question
1 point(s)
The perimeter of a regular hexagon is 240 cm. How long is each side?
- 30 cm
- 20 cm
- 35 cm
- 25 cm
Correct

Incorrect

Hint

We know that a regular hexagon is a closed polygon having 6 sides of same length.
It is given that
Perimeter of a regular hexagon = 240 cm
It can be written as
Perimeter = 6 × side of the regular hexagon
So, we get
Side of the regular hexagon = $\dfrac{{{\rm{Perimeter}}}}{6}$
By substituting the values
Side of the regular hexagon = $\dfrac{{240}}{6}$ = 40 cm
Hence, the side of the regular hexagon measures 20 cm.
Question 8 of 20

8. Question
1 point(s)
Find the perimeter of a regular pentagon with each side measuring 12 m.
- 48 m
- 58 m
- 60 m
- 68 m
Correct

Incorrect

Hint

We know that a regular pentagon is a closed polygon which has five sides of same length.
It is given that
Side of the regular pentagon = 12 m
So, we get
Perimeter = 5 × side of the regular pentagon
By substituting the values
Perimeter = 5 × 12 = 60 m
Hence, the perimeter of a regular hexagon is 60 m.
Question 9 of 20

9. Question
1 point(s)
A rectangular piece of land measure 1.7 km by 1.3 km. Each side is to be fenced with six rows of wires. What length of the wire is needed?
- 18 km
- 15 km
- 22 km
- 25 km
Correct

Incorrect

Hint

It is given that
Measure of rectangular piece of land = 1.7 km × 1.3 km
We know that
Perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (1.7 + 1.3) = 2 × 3 = 6 km
The above obtained perimeter = one row of wire needed to fence the rectangular piece of land
So, the length of wire needed to fence the land with 6 rows of wire = 6 × 3 = 18 km
Hence, the length of wire needed is 18 km.
Question 10 of 20

10. Question
1 point(s)
The area of a rectangle is 36 cm2 and its breadth is 1.2 cm. Find the length of the rectangle.
- 15 cm
- 17.5 cm
- 30 cm
- 35 cm
Correct

Incorrect

Hint

It is given that area of a rectangle = 36 cm2
Breadth of a rectangle = 1.2 cm
We know that
Area of a rectangle = L × B
It can be written as
L = $\dfrac{{{\rm{Area}}}}{{\rm{B}}} = \dfrac{{36}}{{1.2}}$ = 30 cm
Hence, the length of the rectangle is 30 cm.
Question 11 of 20

11. Question
1 point(s)
The area of a rectangle is 735 cm2 and its one side is 35 cm, find its other side.
- 25 cm
- 20 cm
- 24
- 21 cm
Correct

Incorrect

Hint

It is given that
Area of a rectangle = 735 ${\rm{c}}{{\rm{m}}^2}$
Length of one side = 35 cm
We know that area of a rectangle = Product of length of two sides
So, the other side = $\dfrac{{{\rm{area}}}}{{{\rm{side}}}}$
By substituting the values
Other side = $\dfrac{{735}}{{35}} =$ 21 cm
Hence, the other side of the rectangle is 21 cm.
Question 12 of 20

12. Question
1 point(s)
What will happen to the area of rectangle if its
(i) Length and breadth both are halved.
(ii) Length is trebled and breadth is doubled.
(iii) Length is doubled and breadth is halved.
- Halved, 4 times and no change respectively.
- Quarter, 5 times and no change respectively.
- Quarter, 5 times and 4 times respectively.
- Halved, 5 times and doubled respectively.
Correct

Incorrect

Hint

Solution:
(i) Length and breadth are halved
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length and breadth are halved it becomes half the original value
New length = $\dfrac{{\rm{l}}}{2}$
New breadth = $\dfrac{{\rm{b}}}{2}$
New area of the rectangle = $\dfrac{{\rm{l}}}{2}$ × $\dfrac{{\rm{b}}}{2}$ = $\dfrac{{{\rm{lb}}}}{4}$
Hence, the area of the rectangle becomes quarter of its original area.
(ii) Length is trebled and breadth is doubled
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length is trebled and breadth is doubled, we get
New length = 3l
New breadth = 2b
New area of the rectangle = 3l × 2b = 5lb
Hence, the area of the rectangle becomes 5 times more than the original area.
(iii) Length is doubled and breadth is halved
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length is doubled and breadth is halved, we get
New length = 2l
New breadth = $\dfrac{{\rm{b}}}{2}$
New area of the rectangle = 2l × $\dfrac{{\rm{b}}}{2}$ = lb
Hence, the area of the rectangle does not change.
Question 13 of 20

13. Question
1 point(s)
What will happen to the area of a square if its side is increased four times the original side?
- 4 times
- 8 times
- 16 times
- 12 times
Correct

Incorrect

Hint

Consider s as the original side of the square
We know that original area = s × s = s2
If the side of the square is increased four times, we get
New side = 4s
So, the new area of the square = 4s × 4s = 16s2
Hence, the area becomes 16 times more than that of the original area.
Question 14 of 20

14. Question
1 point(s)
What will happen to the area of rectangle if its-
(i) Length and breadth are doubled
(ii) Length is trebled and breadth is same
(iii) Length is halved and breadth quadrupled.
- 4 times, 3 times and 2 times respectively.
- 5 times, 4 times and 3 times respectively.
- 4 times, 3 times and 3 times respectively.
- 4 times, 4 times and 2 times respectively.
Correct

Incorrect

Hint

Solution:
(i) Length and breadth are doubled
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length and breadth are doubled it becomes two times more than the original value
New length = 2l
New breadth = 2b
New area of the rectangle = 2l × 2b = 4lb
Hence, the area of the rectangle becomes 4 times more than its original area.
(ii) Length is trebled and breadth is same
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length is trebled and breadth is same, we get
New length = 3l
New breadth = b
New area of the rectangle = 3l × b = 3lb
Hence, the area of the rectangle becomes 3 times more than the original area.
(iii) Length is halved and breadth is quadrupled.
Consider l as the initial length and b as the initial breadth
So, the original area = l × b
If the length is halved and breadth is quadrupled, we get
New length = $\dfrac{{\rm{L}}}{2}$
New breadth = 4b
New area of the rectangle = $\dfrac{{\rm{L}}}{2}$ × 4b = 2lb
Hence, the area of the rectangle becomes 2 times more than the original area.
Question 15 of 20

15. Question
1 point(s)
A rectangle has area of 400 cm2 and breadth of 25 cm. What is it’s perimeter.
- 90 cm
- 80 cm
- 72 cm
- 82 cm
Correct

Incorrect

Hint

It is given that
Area of the rectangle = 400 cm2
Breadth of the rectangle = 25 cm
We know that area = L × B
It can be written as
L = $\dfrac{{{\rm{Area}}}}{{\rm{L}}}$
By substituting the values
L = $\dfrac{{400}}{{25}}$ = 16 cm
We know that perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (16 + 25) = 2 × 41 = 82 cm
Hence, the perimeter of the rectangle is 82 cm.
Question 16 of 20

16. Question
1 point(s)
A rectangle with area 740 cm2 has 37 cm length. What is the perimeter of the rectangle?
- 74 cm
- 84 cm
- 120 cm
- 114 cm
Correct

Incorrect

Hint

The dimensions of rectangle are
Length = 37 cm
Area = 740 cm2
We know that
Area of rectangle = L × B
It can be written as
B = $\dfrac{{{\rm{Area}}}}{{\rm{L}}} = \dfrac{{740}}{{37}}$ = 20 cm
So, the perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (20 + 37) = 2 × 57 = 114 cm
Hence, the perimeter of the rectangle is 114 cm.
Question 17 of 20

17. Question
1 point(s)
A steel plate measures 5 cm × 12 cm. How many plates will be required to cover a gate of size 2 m × 3 m? Also, find the total cost of the plates at the rate of Rs 20 per tile
- 2000
- 4000
- 20000
- 40000
Correct

Incorrect

Hint

Measure of steel plate = 5 cm × 12 cm
Size of the gate = 2 m × 3 m = 200 cm × 300 cm
So, the area of steel plate = 5 cm × 12 cm = 60 cm2
Area of gate = 200 cm × 300 cm = 60000 cm2
No. of plates required to cover the gate = $\dfrac{{{\rm{Area of gate}}}}{{{\rm{Area of one plate}}}}$ Substituting the values
No. of tiles required to cover the wall = $\dfrac{{60000}}{{60}}$ =1000 tiles
It is given that
Cost of one plate = Rs 20
So, the cost of 1000 plates = 1000 × 20 = Rs 20000
Hence, 1000 number of plates are required to cover the wall and the cost is Rs 20000.
Question 18 of 20

18. Question
1 point(s)
A triangle has a perimeter of 75 cm. The longest side is of 35 cm and the shortest side is of 15 cm. What is the length of the third side?
- 25 cm
- 20 cm
- 30 cm
- 35 cm
Correct

Incorrect

Hint

It is given that
Longest side of triangle = 35 cm
Shortest side of triangle = 15 cm
In order to find the length of third side
We know that perimeter of a triangle is the sum of all three sides of a triangle
So, the length of third side = perimeter of triangle – sum of length of other two sides
By substituting the values
Length of third side = 75 – (35 + 15) = 25 cm.
Hence, the length of third side is 25 cm.
Question 19 of 20

19. Question
1 point(s)
Split the following shapes into rectangles and find the perimeter of each. (The measures are given in centimeters)
- 23 cm² and 27 cm²
- 23 cm² and 20 cm²
- 25 cm² and 20 cm²
- 25 cm² and 23 cm²
Correct

Incorrect

Hint

The given figure has two rectangles I and II.
For rectangle I,
Length = 10 cm
Breadth = 2 cm
So, the perimeter of rectangle I = 2(L + B) = 2(10 + 2) = 24 cm2
For rectangle II,
Length = 10 cm
Breadth = 1.5 cm
So, perimeter of rectangle II = 2(10 + 1.5) = 23 cm2
Question 20 of 20

20. Question
1 point(s)
How many tiles with dimensions 10 cm and 13 cm will be needed to fit a region whose length and breadth are 100 cm and 169 cm respectively.
- 240 tiles
- 130 tiles
- 150 tiles
- 200 tiles
Correct

Incorrect

Hint

Tile dimensions = 10 cm × 13 cm
Region dimensions = 100 cm × 169 cm
So, the area of tile = 10 cm × 13 cm = 130 cm2
Similarly, area of region = 100 cm × 169 cm = 16900 cm2
No. of tiles which is required to cover the region = $\dfrac{{{\rm{Area of region}}}}{{{\rm{Area of one tile}}}}$
By substituting the values
No. of tiles which is required to cover the region $\dfrac{{16900}}{{130}} = 130$ tiles

Mathematics Class VI

Olympiad Questions-Mensuration-Class VI

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