Question

# A plate of 3 cm thick, 9 cm broad, and 27 m long is melted into a cube. find the difference in surface area of the two solids correct to the nearest whole number.

- 37
- 29
- 39
- 27

Hint:

### TSA of cuboid = 2[lb + bh + hl]

TSA of cube = 6x2

Volume of cube = x3

Volume of cuboid = l x b x h

Where , l is length, b is breadth , h is height

x is side of cube

## The correct answer is: 27

### We are given the dimensions of the plate

Thickness h = 3 cm

Breadth b= 9cm

Length l = 27cm

This plate is melted into cube , let x be the length of side of cube

So, as the plate is melted into cube the volume of both the solids will be same

Therefore, l x b x h = x^{3}

27 x 9 x 3 = x^{3}

729 = x^{3}

As we know, 729 is the cube of 9

Therefore, x = 9

Surface area of the plate = 2[lb + bh + hl]

= 2[(27)(9) + (9)(3) + (3)(27)]

= 2[243 + 27 + 81]

= 2(351)

= 702

Surface area of the cube = 6x^{2}

= 6(9)^{2}

= 6 (81)

= 486

Difference between both the areas= 702 – 486

= 216

Therefore the difference between the volumes is surface area .

^{3}

^{3}

As we know, 729 is the cube of 9

Therefore, x = 9

Surface area of the plate = 2[lb + bh + hl]

^{2}

^{2}

Difference between both the areas= 702 – 486

Therefore the difference between the volumes is surface area .

### Related Questions to study

### Which number is next in the sequence?

35, 85, 135, 185, __

a

_{n}= a + (n – 1) × d

Solution :- We have given the pattern of numbers

35, 85, 135, 185, __

We have to find the next number.

We can see clearly that it is an Arithmetic progression With Common difference = 85 - 35 = 50

= 135 - 85 = 50

= 185 - 135 = 50

Therefore next term will be

185 + Common difference

185 + 50 = 235

Therefore, the pattern will be

35, 85, 135, 185, 235

### Which number is next in the sequence?

35, 85, 135, 185, __

a

_{n}= a + (n – 1) × d

Solution :- We have given the pattern of numbers

35, 85, 135, 185, __

We have to find the next number.

We can see clearly that it is an Arithmetic progression With Common difference = 85 - 35 = 50

= 135 - 85 = 50

= 185 - 135 = 50

Therefore next term will be

185 + Common difference

185 + 50 = 235

Therefore, the pattern will be

35, 85, 135, 185, 235

### A cuboidal water tank is 6m long , 5m wide and 4.5 m deep.

How many litres of water can it hold ?

The given cuboidal water tank has-

- Length = 6m,
- Width = 5m and
- Height = 4.5m

Volume of a cuboid = Length * Width * Height

∴ Volume of the cuboidal tank = 6 * 5 * 4.5

∴ Volume of the cuboidal tank = 135 m3.

Now, we know that-

1 = 1,000 L

∴ 135 = 1,000 * 135

∴ 135 = 1,35,000 L.

### A cuboidal water tank is 6m long , 5m wide and 4.5 m deep.

How many litres of water can it hold ?

The given cuboidal water tank has-

- Length = 6m,
- Width = 5m and
- Height = 4.5m

Volume of a cuboid = Length * Width * Height

∴ Volume of the cuboidal tank = 6 * 5 * 4.5

∴ Volume of the cuboidal tank = 135 m3.

Now, we know that-

1 = 1,000 L

∴ 135 = 1,000 * 135

∴ 135 = 1,35,000 L.

### Show the conjecture is false by finding a counterexample.

Two supplementary angles form a linear pair.

Given, two supplementary angles always form a liner pair

We have to determine if the given statement is true or false.

When the sum of measures of two angles is 180 degrees, then the angles are called supplementary angles.

When two lines intersect each other at a single point, linear pairs of angles are formed.

From the figure,

∠1 + ∠2 = 180°

Linear pair of angles occurs in a straight line.

So, two supplementary angles do not always form a linear pair.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

Two supplementary angles form a linear pair.

Given, two supplementary angles always form a liner pair

We have to determine if the given statement is true or false.

When the sum of measures of two angles is 180 degrees, then the angles are called supplementary angles.

When two lines intersect each other at a single point, linear pairs of angles are formed.

From the figure,

∠1 + ∠2 = 180°

Linear pair of angles occurs in a straight line.

So, two supplementary angles do not always form a linear pair.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

Two adjacent angles always form a linear pair

Two adjacent angles always form a linear pair.

We have to prove the statement is false.

The linear pair of angles is a pair of angles which is created by the intersection of two straight lines and the sum of a linear pair of angles is 180°.

Now, if we take an angle which is less than 180° and bisect that angle with the help of a straight line, then we will get two adjacent angles. And, the sum of those two adjacent angles isn't 180°.

In this way, we can come to a conclusion that every pair of adjacent angles will be not considered as the linear pair of angles.

So, the given conjecture is false.

### Show the conjecture is false by finding a counterexample.

Two adjacent angles always form a linear pair

Two adjacent angles always form a linear pair.

We have to prove the statement is false.

The linear pair of angles is a pair of angles which is created by the intersection of two straight lines and the sum of a linear pair of angles is 180°.

Now, if we take an angle which is less than 180° and bisect that angle with the help of a straight line, then we will get two adjacent angles. And, the sum of those two adjacent angles isn't 180°.

In this way, we can come to a conclusion that every pair of adjacent angles will be not considered as the linear pair of angles.

So, the given conjecture is false.

### A cubical tank 50 cm in length and 36 cm in breadth contain water. A cube of x cm edge is dropped into it and fully immersed. If the rise in water level is 15 cm, solve for x. and hence find the T.S.A of the cubical tank

Length l = 50cm

Breadth b= 36cm

The dimensions of cube dropped are

Side = x cm

We have to find the value of x and TSA of cubical tank.

It is given that rise in water level is 15 cm , so it will be the height of the water in the cubical tank

h = 15 cm

Therefore the volume of the cube dropped will be

Volume = l b h

x^{3} = 50 36 15

= 27000

As 27000 is a cube of 30

x^{3 }= (30)^{3}

therefore, x = 30

The TSA of the cubical tank = 2[lb + bh + hl]

= 2[(50)(36) + (36)(15) + (15)(50)l]

= 2[1800 + 540 + 750]

= 2[3090]

= 6180

Which is approximately equal to 6200.

Therefore, the correct option is a) 30 , 6200.

### A cubical tank 50 cm in length and 36 cm in breadth contain water. A cube of x cm edge is dropped into it and fully immersed. If the rise in water level is 15 cm, solve for x. and hence find the T.S.A of the cubical tank

Length l = 50cm

Breadth b= 36cm

The dimensions of cube dropped are

Side = x cm

We have to find the value of x and TSA of cubical tank.

It is given that rise in water level is 15 cm , so it will be the height of the water in the cubical tank

h = 15 cm

Therefore the volume of the cube dropped will be

Volume = l b h

x^{3} = 50 36 15

= 27000

As 27000 is a cube of 30

x^{3 }= (30)^{3}

therefore, x = 30

The TSA of the cubical tank = 2[lb + bh + hl]

= 2[(50)(36) + (36)(15) + (15)(50)l]

= 2[1800 + 540 + 750]

= 2[3090]

= 6180

Which is approximately equal to 6200.

Therefore, the correct option is a) 30 , 6200.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is even, then the two numbers must be even.

We have given a statement

If the product of two numbers is even, then the two numbers must be even.

We have to show the given conjecture is false.

Let us take two even numbers 6 and 8

Product of considered numbers = 6 x 8 = 48

Product we get is an even number , so the given statement is true for considered values.

If we consider one odd and one even number , 5 and 6

Product = 5 x 6 = 30

So, the product obtained is an even which contradicts the given statement

So, We can say that if the product of two numbers is even, then the two numbers need be even.

So, the given conjecture is false.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is even, then the two numbers must be even.

We have given a statement

If the product of two numbers is even, then the two numbers must be even.

We have to show the given conjecture is false.

Let us take two even numbers 6 and 8

Product of considered numbers = 6 x 8 = 48

Product we get is an even number , so the given statement is true for considered values.

If we consider one odd and one even number , 5 and 6

Product = 5 x 6 = 30

So, the product obtained is an even which contradicts the given statement

So, We can say that if the product of two numbers is even, then the two numbers need be even.

So, the given conjecture is false.

### Show the conjecture is false by finding a counterexample.

If a + b = 0, then a = b = 0.

We have given a statement

If a + b = 0, then a = b = 0

We have to show the given conjecture is false.

Let us take a = 0 and b = 0

Then a + b = 0 + 0 = 0

Given statement is true for considered values of a and b.

Let us take a = 3 and b = - 3

Then a + b = 3 + (- 3) = 3 – 3 = 0

From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .

Therefore, the given conjecture is false.

### Show the conjecture is false by finding a counterexample.

If a + b = 0, then a = b = 0.

We have given a statement

If a + b = 0, then a = b = 0

We have to show the given conjecture is false.

Let us take a = 0 and b = 0

Then a + b = 0 + 0 = 0

Given statement is true for considered values of a and b.

Let us take a = 3 and b = - 3

Then a + b = 3 + (- 3) = 3 – 3 = 0

From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .

Therefore, the given conjecture is false.

### calculate the surface area of a cube and volume of a cube with the given diagonal length of 15cm

D= 15 cm

We have to find the Surface area and volume of the cube.

We know that Diagonal of the cube = a

Where a is side of the cube

Comparing with the given value we get

a = 15 cm

Therefore , Surface area of cube = 6a^{2}

= 6(15)^{2}

= 6(225)

= 1350

And, volume of the given cube = a^{3}

= (15)^{3}

= 3375

Therefore the correct option is c)1350, 3375.

### calculate the surface area of a cube and volume of a cube with the given diagonal length of 15cm

D= 15 cm

We have to find the Surface area and volume of the cube.

We know that Diagonal of the cube = a

Where a is side of the cube

Comparing with the given value we get

a = 15 cm

Therefore , Surface area of cube = 6a^{2}

= 6(15)^{2}

= 6(225)

= 1350

And, volume of the given cube = a^{3}

= (15)^{3}

= 3375

Therefore the correct option is c)1350, 3375.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is positive, then the two numbers must both be positive.

Solution:-

We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.

If we consider two positive numbers 7 and 9

Then , their product, 7 x 9 = 63

Hence, the statement is correct for positive integers

Suppose we take two negative integers - 7 and - 9

Then, their product, - 7 x - 9 = 63

So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is positive, then the two numbers must both be positive.

Solution:-

We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.

If we consider two positive numbers 7 and 9

Then , their product, 7 x 9 = 63

Hence, the statement is correct for positive integers

Suppose we take two negative integers - 7 and - 9

Then, their product, - 7 x - 9 = 63

So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

The square root of any positive integer x is always less than x.

The square root of any positive integer x is always less than

Let us take any positive integer x

We consider that the square root of any interger x is always less than x

But if we consider integer 1 then,

The square root of 1 is 1.

Which is not less than 1

So, the given statement is false

### Show the conjecture is false by finding a counterexample.

The square root of any positive integer x is always less than x.

The square root of any positive integer x is always less than

Let us take any positive integer x

We consider that the square root of any interger x is always less than x

But if we consider integer 1 then,

The square root of 1 is 1.

Which is not less than 1

So, the given statement is false

### Show the conjecture is false by finding a counterexample.

All prime numbers are odd.

All prime numbers are odd

Before deciding we will know the definition of prime numbers

Definition - a prime number has only 2 factors - itself and 1.

Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.

Therefore, it shows that All prime numbers are not odd.

### Show the conjecture is false by finding a counterexample.

All prime numbers are odd.

All prime numbers are odd

Before deciding we will know the definition of prime numbers

Definition - a prime number has only 2 factors - itself and 1.

Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.

Therefore, it shows that All prime numbers are not odd.

### Complete the conjecture.

The sum of the first n odd positive integers is ____.

a

_{n}= a + (n – 1) × d

To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:

S = n/2[2a + (n − 1) × d]

To find: Sum of first n odd natural numbers

The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.

- a = 1
- d = 2
- t
_{n}= (2n – 1)

S = n/2[2a + (n − 1) × d]

S = n /2 [2 + 2n – 2]

S = n /2 [2n]

S = n

^{2}

The sum of first n odd natural numbers is n

^{2}.

### Complete the conjecture.

The sum of the first n odd positive integers is ____.

a

_{n}= a + (n – 1) × d

To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:

S = n/2[2a + (n − 1) × d]

To find: Sum of first n odd natural numbers

The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.

- a = 1
- d = 2
- t
_{n}= (2n – 1)

S = n/2[2a + (n − 1) × d]

S = n /2 [2 + 2n – 2]

S = n /2 [2n]

S = n

^{2}

The sum of first n odd natural numbers is n

^{2}.

### A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm^{2}, then find the side of the cube?

Area of square=side × side=a²

Diagonal of cube= unit

Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.

- We are given that,

Side of square=Diagonal of cube

*Let the edge of the cube be 'a' cm.*

As we know,

length of the diagonal of cube= ×edge of the cube=a cm

- The length of the side of the square is equal to the length of the diagonal of the cube.

- Area of square(drawn on diagonal of cube)

- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²

Therefore , the side of the cube is 155 cm.

### A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm^{2}, then find the side of the cube?

Area of square=side × side=a²

Diagonal of cube= unit

Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.

- We are given that,

Side of square=Diagonal of cube

*Let the edge of the cube be 'a' cm.*

As we know,

length of the diagonal of cube= ×edge of the cube=a cm

- The length of the side of the square is equal to the length of the diagonal of the cube.

- Area of square(drawn on diagonal of cube)

- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²

Therefore , the side of the cube is 155 cm.

### What are all the possible values of b for which is it factorable using only integer coefficients and constant?

Given, is it factorable so be must be sum of factors of (-8)(3)

Factors of -24 are

(1,-24) b can be -24 + 1 = -23

(2,-12) b can be -12 + 2 = -10

(3,-8) b can be -8 + 3 = -5

(4,-6) b can be 4 - 6 = -2

(6,-4) b can be 6 - 4 = 2

(8,-3) b can be 8 - 3 = 5

(12,-2) b can be 12 - 2 = 10

(24,-1) b can be 24 - 1 = 23

So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}

### What are all the possible values of b for which is it factorable using only integer coefficients and constant?

Given, is it factorable so be must be sum of factors of (-8)(3)

Factors of -24 are

(1,-24) b can be -24 + 1 = -23

(2,-12) b can be -12 + 2 = -10

(3,-8) b can be -8 + 3 = -5

(4,-6) b can be 4 - 6 = -2

(6,-4) b can be 6 - 4 = 2

(8,-3) b can be 8 - 3 = 5

(12,-2) b can be 12 - 2 = 10

(24,-1) b can be 24 - 1 = 23

So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}

### Find a counterexample to show that the following conjecture is false.

Conjecture: The square of any integer is always greater than the integer.

We have to prove that given statement false.

Number obtained when a number is multiplied by itself three times is called a cube number.

If m = n², then m is a perfect square where m and n are natural numbers.

Example: consider 1

Square of 1 = 1

Example: consider 2

Square of 2 = 4

Therefore, the square of every natural number is not always greater than the number itself.

Therefore, given statement is false.

### Find a counterexample to show that the following conjecture is false.

Conjecture: The square of any integer is always greater than the integer.

We have to prove that given statement false.

Number obtained when a number is multiplied by itself three times is called a cube number.

If m = n², then m is a perfect square where m and n are natural numbers.

Example: consider 1

Square of 1 = 1

Example: consider 2

Square of 2 = 4

Therefore, the square of every natural number is not always greater than the number itself.

Therefore, given statement is false.