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MathematicsClass 6CBSE

_____ is an integer which is neither positive nor negative.

0 is an integer which is neither positive nor negative.

MathematicsClass 6CBSE

When we subtract -10 from 18 we get ______.

When we subtract -10 from 18 we get 28.

MathematicsClass 6CBSE

Verify the ‘Euler’s formula’ V + F = E + 2 for the given figures.(a) A triangular prism having 5 faces, 9 edges and 6 vertices.(b) A rectangular prism with 6 faces, 12 edges and 8 vertices.(c) A pentagonal prism with 7 faces, 15 edges and 10 vertices. (d) A tetrahedron -with 4 faces, 6 edges and 4 vertices.

(a) Here, F = 5, E = 9 and V = 6∴ V + F = E + 2⇒ 6 + 5 = 9 + 2⇒ 11 = 11Hence, verified.(b) Here, F = 6, E = 12 and V = 8∴ V + F = E + 2⇒ 8 + 6 = 12 + 2⇒ 14 = 14Hence, verified.(c) Here, F = 7, E = 15 and V = 10∴ V + F = E + 2⇒ 10 + 7 = 15 + 2⇒ 17 = 17Hence, verified.(d) Here, F = 4, E = 6 and V = 4∴ V + F = E + 2⇒ 4 + 4 = 6 + 2⇒ 8 = 8Hence, verified.

MathematicsClass 6CBSE

In the given figure, find the measure of the angles marked with a, b, c, d, e and f.

∠a = 180° -129° = 51°∠b = 180° – (51° + 92°)= 180° – 143° = 37°∠c = 180° – 88° = 92°∠d = 180° – 152° = 28°∠e = 180° – 143° = 37°∠f= 180° – (∠e + ∠d)= 180° – (37° + 28°)= 180°- 65°= 115°∠g = 180° – ∠f = 180° – 115° = 65°

MathematicsClass 6CBSE

What is the measure of straight angle?

The measure of straight angle is 180°.

MathematicsClass 6CBSE

All equilateral triangle are isosceles, but all isosceles triangle are not equilateral. Justify the statement.

An isosceles triangle is any triangle with 2 sides that are equal in length. So, every equilateral triangle is a special case of an isosceles triangle since not just 2 sides are equal, but all 3 are. But every isosceles triangle is not equilateral, because you can have 2 sides of equal length and a third side that is either longer or shorter than those 2 sides. For example, if the triangle is a right-angle triangle and the two sides that meet to make the right angle are the same length, then the 3rd side would be longer than those two.

MathematicsClass 6CBSE

If B is the mid point of (AC) and C is the point of (BD)   where A, B, C, D lie on a straight line, say why AB = CD?

∵ B is the mid-point of AC∴ AB = BC …(1)∵C is the mid-point of BD∴ BC = CD …(2)In view of (1) and (2), we getAB = CD.

MathematicsClass 6CBSE

Draw a hexagon and write its sides and diagonals?

HexagonSides of hexagon: AB, BC, CD, DE, EF and FA.Diagonals of hexagon: AC, AD, AE, BD, BE, BF, CE, CF, and DF

MathematicsClass 6CBSE

What is the angle name for half a revolution?

Straight Angle (180°)

MathematicsClass 6CBSE

How many faces a tetrahedron have?

In geometry, a tetrahedron is a polyhedron composed of four triangular faces, three of which meet at each corner or vertex.

MathematicsClass 6CBSE

Ill cows, 185 sheep and 296 goats are to be taken across a river. There is only one boat and the boatsman says; he will take the same number and same kind of animals in each trip. Find the largest number of animals in each trip and the number of trips he will have to make.

To arrange the books in the required way,we have to find the greatest number that divides 336, 192 and 144 exactly.So, HCF of 336, 192 and 144 isCommon factors are 2 × 2 × 2 × 2 × 3 = 48∴ HCF = 48, i.e., each stack contains 48 books.∴ Number of stacks of English books = 336 + 48 = 7Number of stacks of Hindi books = 192 4 - 48 = 4Number of stacks of Urdu books = 144 4 - 48 = 3

MathematicsClass 6CBSE

Three sets of English, Hindi, and Urdu books are to be stacked in such a way that the books are stored subject wise and the height of each stack is the same. The numbers of English, Hindi and Urdu books are 336, 192 and 144 respectively. Assuring that the books have that same thickness, determine the number of stacks of English, Hindi and Urdu books.

Using BODMAS Rule, we have40 + [20 – {28 ÷ 7 – 3 + (30 – 5 of 4)}]= 40 + [20 – {28 ÷ 7 – 3 + (30 – 20)}]= 40 + [20 – {28 ÷ 7 – 3 + 10}]= 40 + [20 – [4 – 3 + 10}]= 40 + [20-11] = 40 + 9 = 49.

MathematicsClass 6CBSE

Find the greatest number that will divide 455, 582 and 710 leaving remainders 14, 15 and 17 respectively.

Hence, the required number is 9.Given numbers are 455, 582 and 710 and the respective remainders are 14, 15 and 17.We have 455 – 14 = 441, 582 – 15 = 567 and 710 – 17 = 693.Now let us find their HCF.Common factors are 3 and 7.∴ HCF = 3 × 7 = 21Hence, the required number is 21.

MathematicsClass 6CBSE

Find the greatest number which divides 82 and 132 leaving 1 and 6, respectively as remainders.

Given numbers are 82 and 132 and the remainders are 1 and 6 respectively.We have, 82 – 1 = 81 and 132 – 6 = 126So, we need to find the HCF of 81 and 126Common factor is 3 (occurring twice).∴ HCF = 3 × 3 = 9

MathematicsClass 6CBSE

Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Given numbers are 18, 24 and 32, we haveThus, LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288The smallest 4-digit number = 1000Now, we write multiples of 288, till we get a 4-digit number.288 × 1 = 288, 288 × 2 = 576,288 × 3 = 864, 288 × 4 = 1152Hence, 1152 is the required number.

MathematicsClass 6CBSE

Find the LCM of 12 and 30.

Given numbers are 12 and 3012 = 2 × 2 × 3;30 = 2 × 3 × 5∴ LCM = 2 × 2 × 3 × 5 = 60Hence, the LCM of 12 and 30 = 60.

MathematicsClass 6CBSE

The HCF and LCM of two numbers are 6 and 120 respectively. If one of the numbers is 24, find the other number.

Given that: HCF = 6LCM = 120Let the two numbers be a and b, where a = 24, b = ?We know that: a × b = HCF × LCM⇒ 24 × b = 6 × 120⇒ b = 6 × 12024⇒ b = 30Hence, the other number is 30.

MathematicsClass 6CBSE

Is 80136 divisible by 11?

Sum of the digits at odd places = 6 + 1 + 8 = 15Sum of the digits at even places = 3 + 0 = 3Difference of the two sums = 15 – 3 = 12,which is neither 0 nor the multiple of 11.Hence, 80136 is not divisible by 11.

MathematicsClass 6CBSE

The sum of two numbers is 25 and their product is 144. Find the numbers.

Hence, 11 is a factor of 1,10,011The product of two numbers is 144.∴ The possible factors are 1 × 144, 2 × 72, 3 × 48, 4 × 36, 6 × 24, 8 × 18, 9 × 16, 12 × 12Here, we observe that out of these factors, we take 9 and 16.Product = 9 × 16 = 144 and sum = 9 + 16 = 25Hence, the required numbers are 9 and 16.

MathematicsClass 6CBSE

Without actual division, show that 11 is a factor of 1,10,011.

Here 1,10,011 = 1,10,000 + 11= 11 × 10,000 + 11 × 1= 11 × (10,000 + 1)= 11 × 10,001It is clear that 11 is a factor of 11 × 10,001.

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