Maths-

General

Easy

Question

# A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine is needed to fill this capsule ?

Hint:

### The volume of sphere =

## The correct answer is: 22.46mm3 ( approx )

### Explanation:

- We have given a capsule with diameter 3.5mm
- We have to find the how much medicine require to fill the capsule.

Step 1 of 1:

Since the capsule is spherical in shape, the amount of medicine needed to fill the capsule is the volume of a sphere.

The volume of sphere =

Diameter of the spherical capsule, d = 3.5mm

Radius of the spherical capsule, r = 3.5 / 2mm = 1.75 mm

Now, the medicine is needed to fill this capsule = The volume of the spherical capsule

The medicine needed to fill this capsule =

= 4/3 22/7 1.75mm 1.75mm 1.75mm

22 .46mm^{3} (approx)^{}

22 .46mm

^{3}(approx)

^{}

### Related Questions to study

Maths-

### Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

Solution :-

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

### Prove that the sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of its sides.

Maths-General

Solution :-

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Aim :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. where , and are lengths of diagonals

Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.

Explanation(proof ) :-

As we know opposite sides of parallelogram are equal i.e AD = BC = L

AX and BY are perpendicular to line DC and intersect at X and Y respectively.

Consider ΔADX and ΔBCY ,

AX = BY (distance between two parallel lines is always equal)

AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)

Both have right angles at X and Y .

By RHS congruence rule,

ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

Let DX = CY = a and AC = ; BD =

Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a

Applying pythagoras theorem in ΔACX ,We

Applying pythagoras theorem in ΔBYD, we get

— Eq2

Adding equation 2 and 1 we get ,

— Eq3

Applying pythagoras theorem in ΔADX,

—- Eq4

Substitute Eq4 in Eq3

We get

Maths-

### Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

Explanation:

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

- We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
- We have to find the Radius of new sphere and the ratio of S and S’

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

### Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the i) Radius r’ of the new sphere ii) ratio of S and S’

Maths-General

Explanation:

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

- We have given twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’.
- We have to find the Radius of new sphere and the ratio of S and S’

We have given 27 solid spheres each of radius r and surface area s

So,

(i)

Volume of each sphere will be

Volume of big sphere

So,

Step 2 of 2:

Surface area is

and

Therefore,

=

Or

1 : 9

Maths-

### Bloom’s category: Application

Level: Medium

Identify points, lines and planes in the given image of a floor.

HINT :- Identify points, lines and planes in the given image using the definitions of them .

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

### Bloom’s category: Application

Level: Medium

Identify points, lines and planes in the given image of a floor.

Maths-General

HINT :- Identify points, lines and planes in the given image using the definitions of them .

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

ANS :- 15 points(intersections points) ,8 lines and 1 plane

Explanation :-

A plane is considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions.

A point is the fundamental object in geometry. It is represented by a dot and point represents position only.

A line (straight line) can be thought of as a connected set of infinitely many points. It extends infinitely far in two opposite directions. A line has infinite length, zero width, and zero height.

We have 3 horizontal and 5 vertical lines which have 15 (= 5 x 3 ) intersection points and

8 lines (= 5+3)

Here all these lie on the plane (surface of floor)

∴ 15 points(intersections points) ,8 lines and 1 plane

Maths-

### A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

Explanation:

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

Step 2 of 2:

Now inside surface area is 246.48m

r

r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

- We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
- We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

^{2}Step 2 of 2:

Now inside surface area is 246.48m

^{2}r

^{2}= 39.22r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

^{2}and the inside volume is 82.07m^{3}### A dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of RS 4989.60. If the cost of white washing is Rs 20 per square metre , find the i) inside surface area of the dome ii) Volume of the air inside the dome

Maths-General

Explanation:

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

Step 2 of 2:

Now inside surface area is 246.48m

r

r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

- We have given a dome of a building is in the form of a hemisphere. From inside it was white washed at the cost of . If the cost of white washing is per square metre.
- We have to find the (i) Inside surface area of dome. (ii) Volume of the air inside the dome

It is given that the total cost of white washing is Rs 4989.60.

The cost of white washing per unit square meter is Rs 20.

So,

Total inside surface area will be

=

=

= 246.48m

^{2}Step 2 of 2:

Now inside surface area is 246.48m

^{2}r

^{2}= 39.22r = 6.26

So, The volume will be

= 82.07

Hence, The inside surface area is 246.48m

^{2}and the inside volume is 82.07m^{3}Maths-

### Find the volume of a sphere whose surface area is 154 sq.cm

Explanation:

We have given surface area 154cm

So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

- We have given surface area .
- We have to find the volume of the sphere.

We have given surface area 154cm

^{2}.So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

^{3}.### Find the volume of a sphere whose surface area is 154 sq.cm

Maths-General

Explanation:

We have given surface area 154cm

So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

- We have given surface area .
- We have to find the volume of the sphere.

We have given surface area 154cm

^{2}.So,

r = 3.5

Volume of sphere

Thus, the volume of the sphere is 179.67cm

^{3}.Maths-

### A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

Explanation:

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

- We have given a hemisphere with thickness 1cm.
- We have to find the volume of iron used to make the tank.

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

^{3 }of iron used to make the tank### A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

Maths-General

Explanation:

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

- We have given a hemisphere with thickness 1cm.
- We have to find the volume of iron used to make the tank.

Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.

The Volume of hemisphere of base radius r is equal to

The inner radius of the tank r = 1m

Thickness of iron = 1cm = 1/100 m =0.01m

Outer radius of the tank, R = 1m + 0.01m = 1.01m

The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.

Volume of the iron used to make the tank

0.06348 m

^{3 }of iron used to make the tankMaths-

### How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Explanation:

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

- We have given a hemispherical bowl of diameter 10.5cm
- We have to find how many litre of milk this bowl can hold.

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

### How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Maths-General

Explanation:

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

- We have given a hemispherical bowl of diameter 10.5cm
- We have to find how many litre of milk this bowl can hold.

The diameter of the hemispherical bowl is

Radius will be

So, The volume of the hemispherical bowl will be

Also we know that

So,

Maths-

### of an isosceles triangle ABC is acute in which and . Prove that

Solution :-

Aim :- prove

Hint :- Using pythagoras theorem, find the length of side AB and BC .

Substitute the RHS of in terms of AC and CD Explanation(proof

Using pythagoras theorem,in triangle BDC

— Eq1

Using pythagoras theorem,in triangle BDA — Eq2

Substitute Eq2 in Eq1

Substitute AD = AC - CD (from diagram)

Hence proved

Aim :- prove

Hint :- Using pythagoras theorem, find the length of side AB and BC .

Substitute the RHS of in terms of AC and CD Explanation(proof

Using pythagoras theorem,in triangle BDC

— Eq1

Using pythagoras theorem,in triangle BDA — Eq2

Substitute Eq2 in Eq1

Substitute AD = AC - CD (from diagram)

Hence proved

### of an isosceles triangle ABC is acute in which and . Prove that

Maths-General

Solution :-

Aim :- prove

Hint :- Using pythagoras theorem, find the length of side AB and BC .

Substitute the RHS of in terms of AC and CD Explanation(proof

Using pythagoras theorem,in triangle BDC

— Eq1

Using pythagoras theorem,in triangle BDA — Eq2

Substitute Eq2 in Eq1

Substitute AD = AC - CD (from diagram)

Hence proved

Aim :- prove

Hint :- Using pythagoras theorem, find the length of side AB and BC .

Substitute the RHS of in terms of AC and CD Explanation(proof

Using pythagoras theorem,in triangle BDC

— Eq1

Using pythagoras theorem,in triangle BDA — Eq2

Substitute Eq2 in Eq1

Substitute AD = AC - CD (from diagram)

Hence proved

Maths-

### The Diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Explanation:

Let the diameter of moon be ,Then the diameter of earth will be 4d

So, Radius of moon will be . And the radius of earth will be

Since, both earth and moon are sphere

We can use formula of volume of sphere .

And volume of earth will be

Therefore, the fraction will be

- We have given the diameter of moon is approximately one fourth of the diameter of the earth.
- We have to find the fraction of the volume of the earth is the volume of the moon

Let the diameter of moon be ,Then the diameter of earth will be 4d

So, Radius of moon will be . And the radius of earth will be

Since, both earth and moon are sphere

We can use formula of volume of sphere .

And volume of earth will be

Therefore, the fraction will be

### The Diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Maths-General

Explanation:

Let the diameter of moon be ,Then the diameter of earth will be 4d

So, Radius of moon will be . And the radius of earth will be

Since, both earth and moon are sphere

We can use formula of volume of sphere .

And volume of earth will be

Therefore, the fraction will be

- We have given the diameter of moon is approximately one fourth of the diameter of the earth.
- We have to find the fraction of the volume of the earth is the volume of the moon

Let the diameter of moon be ,Then the diameter of earth will be 4d

So, Radius of moon will be . And the radius of earth will be

Since, both earth and moon are sphere

We can use formula of volume of sphere .

And volume of earth will be

Therefore, the fraction will be

Maths-

### The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

Explanation:

We will first find the volume of the metallic ball

Now we know that

Density =

Put the value of density and volume in formula, we will get mass

Density =

8.9 =

Mass = 8.9 310.399gm

= 2762.5511gm

- We have given diameter of metallic ball .and the density of the metal is
- We have to find the mass of the ball.

We will first find the volume of the metallic ball

Now we know that

Density =

Put the value of density and volume in formula, we will get mass

Density =

8.9 =

Mass = 8.9 310.399gm

= 2762.5511gm

### The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

Maths-General

Explanation:

We will first find the volume of the metallic ball

Now we know that

Density =

Put the value of density and volume in formula, we will get mass

Density =

8.9 =

Mass = 8.9 310.399gm

= 2762.5511gm

- We have given diameter of metallic ball .and the density of the metal is
- We have to find the mass of the ball.

We will first find the volume of the metallic ball

Now we know that

Density =

Put the value of density and volume in formula, we will get mass

Density =

8.9 =

Mass = 8.9 310.399gm

= 2762.5511gm

Maths-

### Find the amount of water displaced by a solid spherical ball of diameter 0.21m

Explanation:

We know that the volume of water displaced by the given sphere will be equal to volume of sphere.

So,

Volume of sphere will be

- We have given a solid spherical ball with diameter
- We have to find the volume of water displaced by the given sphere.

We know that the volume of water displaced by the given sphere will be equal to volume of sphere.

So,

Volume of sphere will be

### Find the amount of water displaced by a solid spherical ball of diameter 0.21m

Maths-General

Explanation:

We know that the volume of water displaced by the given sphere will be equal to volume of sphere.

So,

Volume of sphere will be

- We have given a solid spherical ball with diameter
- We have to find the volume of water displaced by the given sphere.

We know that the volume of water displaced by the given sphere will be equal to volume of sphere.

So,

Volume of sphere will be

Maths-

### Write truth table for the converse of p → q.

Hint:

If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p q is

q p

Solution

The converse of the conditional statement p q is q p

Final Answer:

Hence, we have drawn the truth table above.

If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p q is

q p

Solution

The converse of the conditional statement p q is q p

Final Answer:

Hence, we have drawn the truth table above.

### Write truth table for the converse of p → q.

Maths-General

Hint:

If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p q is

q p

Solution

The converse of the conditional statement p q is q p

Final Answer:

Hence, we have drawn the truth table above.

If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p q is

q p

Solution

The converse of the conditional statement p q is q p

Final Answer:

Hence, we have drawn the truth table above.

Maths-

### Find the volume of a sphere whose radius is 0.63 m

Explanation:

We know that the volume of sphere with radius is .

Now put the value of

So,

- We have given a sphere with radius .
- We have to find the volume of sphere.

We know that the volume of sphere with radius is .

Now put the value of

So,

### Find the volume of a sphere whose radius is 0.63 m

Maths-General

Explanation:

We know that the volume of sphere with radius is .

Now put the value of

So,

- We have given a sphere with radius .
- We have to find the volume of sphere.

We know that the volume of sphere with radius is .

Now put the value of

So,

Maths-

### Make a valid conclusion in the situation.

If two points lie in a plane, then the line containing them lies in the plane.

Points A and B lie in plane PQR.

Hint:

Law of Detachment states that if p q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.

Solution

Consider the statement into two separate statements

p: Two points lie in a plane

q: Line containing the two points is in the plane

So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as

p q

We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B lie in the plane PQR.

Final Answer:

Hence we can conclude that the line containing points A and B lie in the plane PQR.

Law of Detachment states that if p q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.

Solution

Consider the statement into two separate statements

p: Two points lie in a plane

q: Line containing the two points is in the plane

So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as

p q

We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B lie in the plane PQR.

Final Answer:

Hence we can conclude that the line containing points A and B lie in the plane PQR.

### Make a valid conclusion in the situation.

If two points lie in a plane, then the line containing them lies in the plane.

Points A and B lie in plane PQR.

Maths-General

Hint:

Law of Detachment states that if p q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.

Solution

Consider the statement into two separate statements

p: Two points lie in a plane

q: Line containing the two points is in the plane

So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as

p q

We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B lie in the plane PQR.

Final Answer:

Hence we can conclude that the line containing points A and B lie in the plane PQR.

Law of Detachment states that if p q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.

Solution

Consider the statement into two separate statements

p: Two points lie in a plane

q: Line containing the two points is in the plane

So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as

p q

We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B lie in the plane PQR.

Final Answer:

Hence we can conclude that the line containing points A and B lie in the plane PQR.

Maths-

### The conditional p → q is only false when

a. p = T, q = T

b. p = T, q = F

c. p = F, q = F

d. p = F, q = T

Hint:

p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true

Solution

The truth table of p → q is given as

Hence, we can see that p → q is only false when p is True and q is False.

Final Answer:

So, p → q is only false when p = T and q = F. Hence option b is correct.

p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true

Solution

The truth table of p → q is given as

Hence, we can see that p → q is only false when p is True and q is False.

Final Answer:

So, p → q is only false when p = T and q = F. Hence option b is correct.

### The conditional p → q is only false when

a. p = T, q = T

b. p = T, q = F

c. p = F, q = F

d. p = F, q = T

Maths-General

Hint:

p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true

Solution

The truth table of p → q is given as

Hence, we can see that p → q is only false when p is True and q is False.

Final Answer:

So, p → q is only false when p = T and q = F. Hence option b is correct.

p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true

Solution

The truth table of p → q is given as

Hence, we can see that p → q is only false when p is True and q is False.

Final Answer:

So, p → q is only false when p = T and q = F. Hence option b is correct.