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MathematicsClass 6CBSE

Form expressions using y, 2 and 7. Every expression must have y in it. use only two number operations. These should be different.

The different expressions that can formed are: 2y + 7, 2y – 7, 7y + 2, 7y-2, (y/2) – 7, (y/7)-2, y – (7/2), y + (7/2)

MathematicsClass 6CBSE

The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students ? (Use s for number of students.)

Number of pencils to be distributed to each student= 5And, let the number of students in class be ‘s’. As per the logic, Number of pencils needed =(Number of students in the class) x. (Number of pencils to be distributed to one student )So, Number of pencils needed = 5 x s = 5s.

MathematicsClass 6CBSE

Rearrange the terms of the following expressions in ascending order of powers of x:5x2, 2x, 4x4, 3x3, 7x5

If the given terms are arranged in the ascending order of powers of x, we get, 2x, 5x2, 3x3, 4x4, 7x5.

MathematicsClass 6CBSE

Write which letters give us the same rule as that given by L.

The other letters which give us the same rule as L are T, V and X because the number of matchsticks required to make each of them is 2.

MathematicsClass 6CBSE

The no. of terms in the expression 3x2y-4x2y2+12xy2-5x is

The no. of terms in the expression 3x2 y-4 x2 y2 +1/2 xy2 - 5x is 4.

MathematicsClass 6CBSE

The numerical coefficient of the terms 12xy2+12xy2 is ________.

The numerical coefficient of the terms 1/2 xy2 + 1/2 xy2 is 1/2.

MathematicsClass 6CBSE

The product of 2 and x is being added to the product of 3 and y is expressed as ________.

The product of 2 and x is being added to the product of 3 and y is expressed as 2x + 3y.

MathematicsClass 6CBSE

The value of 2x – 12 is zero, when x = ________.

The value of 2x – 12 is zero, when x = 6.

MathematicsClass 6CBSE

Fencing the compound of a house costs ₹5452. If the rate is ₹94 per metre, find the perimeter of the compound. If the breadth is 10 m, find its length.

Cost of fencing the compound = ₹5452and the rate of fencing = ₹94 per metre∴ Perimeter of the compound = 5452 ÷ 94 = 58 metresNow breadth of the compound = 10 m.2 [length + breadth] = 58 m∴ length + breadth = 58 + 2 m = 29 m∴ Length of the compound = 29 m – 10 m = 19 m.

MathematicsClass 6CBSE

A rectangle and a square have the same perimeter 100 cm. Find the side of the square. If the rectangle has a breadth 2 cm less than that of the square. Find the breadth, length and area of the rectangle.

Perimeter of the square = 100 cmPerimeter 100Side of the square = Perimeter/4 = 100/425 cm.∴ Breadth of the rectangle = 25 cm – 2 cm = 23 cmNow perimeter of the rectangle = 100 cm∴ 2 [length + breadth] = 100length + breadth = 100 ÷ 2 = 50 cmBut breadth = 23 cm∴ Length = 50 cm – 23 cm = 27 cmNow, Area of the rectangle= length × breadth = 27 cm × 23 cm= 621 sq cm.

MathematicsClass 6CBSE

A rectangular park is 30 metres long and 20 metres broad. A steel wire fence is put up all around it. Find the cost of putting the fence at the rate of ₹15 per metre.

Length of the rectangular park = 30 mBreadth = 20 m∴ Perimeter of the rectangular park = 2(length + breadth)= 2 [30 + 20] = 2 × 50 m = 100 m∴ Cost of fencing all around the park = ₹15 × 100 = ₹1500

MathematicsClass 6CBSE

How many trees can be planted at a distance of 6 metres each around a rectangular plot whose length is 120 m and breadth is 90 m?

Length of the rectangular plot = 120 mBreadth = 90 m∴ Perimeter of the rectangular plot= 2 [length + breadth]= 2 [120 m + 90 m]= 2 × 210 m = 420 mNow distance between two trees = 6 m∴ Number of trees around the rectangular plot = 420 m ÷ 6 m = 70

MathematicsClass 6CBSE

Find the length of a rectangle given that its perimeter is 880 m and breadth is 88 m.

Perimeter of the rectangle = 2 [length + breadth]∴ 2 [length + breadth] = 880length + breadth = 880 ÷ 2 = 440∵ Breadth = 88 m∴ Length = 440 m – 88 m = 352 mHence, the required length = 352 m.

MathematicsClass 6CBSE

Length and breadth of a rectangular paper are 22 cm and 10 cm respectively. Find the area of the paper.

Length of the rectangular paper = 22 cmBreadth = 10 cm∴ Area of the rectangular paper = length × breadth= 22 cm × 10 cm= 220 sq cm

MathematicsClass 6CBSE

Find the area of a square field whose each side is 150 m.

Side of the square field = 150 m∴ Area of the square field = Side × Side= 150 m × 150 m= 22500 sq m.

MathematicsClass 6CBSE

Find the cost of fencing a rectangular park 300 m long and 200 m wide at the rate of ₹4 per metre.

Length of the park = 300 mBreadth = 200 m∴ Perimeter of the park = 2 [length + breadth]= 2 [300 m + 200 m]= 2 × 500 m = 1000 m.Cost of fencing the rectangular park = 1000 × 4 = ₹ 4000

MathematicsClass 6CBSE

Find the perimeter of a square whose side is 15 cm.

Side of the square = 15 cm∴ Perimeter of the square = 15 cm × 4 = 60 cm

MathematicsClass 6CBSE

Which of the following figure has greater perimeter?

∴ Perimeter of the table-top = 2 [length + breadth]= 2 [36 cm + 24 cm]= 2 × 60 cm = 120 cm.Fig. (i) Perimeter of the square = 4 × side= 4 × 4 cm = 16 cmFig. (ii) Perimeter of the rectangle= 2 [length + breadth]= 2[8 cm + 3 cm]= 2 × 11 cm = 22 cmSince 22 cm > 16 cm∴ Rectangle has greater perimeter than the square.

MathematicsClass 6CBSE

Length and breadth of a rectangular table-top are 36 cm and 24 cm respectively. Find its perimeter.

Length of the rectangular table-top = 36 cmand its breadth = 24 cm.

MathematicsClass 6CBSE

The perimeter of a square is 64 cm. Find the length of each side.

Perimeter of the square = 64 cm∴ Length of its side = Perimeter/Number of sides = 64/4= 16 cm.

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