Maths-

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Easy

Question

# Ram's mother is four times as old as Ram now. After sixteen years, she will be twice as old as Ram. Find their present ages.

Hint:

### let the rams present age be x ; find the ram’s mothers age now

Calculate ram’s and ram’s mothers age after 16 yr by adding 16 to their present ages . After sixteen years, she will be twice as old as Ram.

Apply the condition and calculate the present ages .

## The correct answer is: 32 years

### Ans :- The present age of ram is 8 and the present age of ram’s mother 32 (in years) .

Explanation :-

Step 1:- calculate present age ram’s mother

let the rams present age be x

Ram's mother is four times as old as Ram now

Now , Ram's mother = 4x

Step 2:-After 16 year calculate the ram’s and ram’s mother age

After 16 yr ram’s age = x+16

After 16 yr ram’s mother age = 4x+16

Step 3:-Apply the given condition and find their respective ages

But After sixteen years, she will be twice as old as Ram

So, 4x+16 = 2(x+16)

4x +16 = 2x + 32

4x-2x = 32 -16

2x = 16

x =8

∴The present age of ram = x = 8 yrs and the present age of ram’s mother = 4x = 32 yrs.

### Related Questions to study

Maths-

### Tanay obtained 98 marks in a mathematics test. His score is the highest in the class and it is also 8 more than three times the lowest score. Create the equation and calculate the lowest score.

Ans :- 30 is the lowest score of the class

Explanation :-

let the lowest score be x,

Given tanay's score is 8 more than three times the lowest score.

Tanay’s score = 8 + 3(x) .

Where tanay’s score = 98 (given)

So, 3x +8 = 98 3x = 98 -8

3x = 90

30 is the lowest score in the class.

Explanation :-

let the lowest score be x,

Given tanay's score is 8 more than three times the lowest score.

Tanay’s score = 8 + 3(x) .

Where tanay’s score = 98 (given)

So, 3x +8 = 98 3x = 98 -8

3x = 90

### Tanay obtained 98 marks in a mathematics test. His score is the highest in the class and it is also 8 more than three times the lowest score. Create the equation and calculate the lowest score.

Maths-General

Ans :- 30 is the lowest score of the class

Explanation :-

let the lowest score be x,

Given tanay's score is 8 more than three times the lowest score.

Tanay’s score = 8 + 3(x) .

Where tanay’s score = 98 (given)

So, 3x +8 = 98 3x = 98 -8

3x = 90

30 is the lowest score in the class.

Explanation :-

let the lowest score be x,

Given tanay's score is 8 more than three times the lowest score.

Tanay’s score = 8 + 3(x) .

Where tanay’s score = 98 (given)

So, 3x +8 = 98 3x = 98 -8

3x = 90

Maths-

### Find three consecutive odd numbers whose sum is 159.

Ans :- The three consecutive odd numbers which satisfy the given condition are 51,53 and 55

Explanation :-

let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .

Sum of three consecutive odd numbers = 159.

(2n+1) + (2n+3) +(2n+5) = 159

2n + 2n + 2n +1+3+5 =159

6n + 9 =159

6n = 150

3(2n) = 150

2n = 50

We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55

∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.

Explanation :-

let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .

Sum of three consecutive odd numbers = 159.

(2n+1) + (2n+3) +(2n+5) = 159

2n + 2n + 2n +1+3+5 =159

6n + 9 =159

6n = 150

3(2n) = 150

2n = 50

We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55

∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.

### Find three consecutive odd numbers whose sum is 159.

Maths-General

Ans :- The three consecutive odd numbers which satisfy the given condition are 51,53 and 55

Explanation :-

let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .

Sum of three consecutive odd numbers = 159.

(2n+1) + (2n+3) +(2n+5) = 159

2n + 2n + 2n +1+3+5 =159

6n + 9 =159

6n = 150

3(2n) = 150

2n = 50

We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55

∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.

Explanation :-

let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .

Sum of three consecutive odd numbers = 159.

(2n+1) + (2n+3) +(2n+5) = 159

2n + 2n + 2n +1+3+5 =159

6n + 9 =159

6n = 150

3(2n) = 150

2n = 50

We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55

∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.

Maths-

### Solve 3x - 2y = 6 and

Ans :- x = 0 ; y = -3

Explanation :-

Step 1 :- find x by substituting in eq 2.

Step 2 :- substitute value of y and find x

x = 0 and y = -3 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by substituting in eq 2.

### Solve 3x - 2y = 6 and

Maths-General

Ans :- x = 0 ; y = -3

Explanation :-

Step 1 :- find x by substituting in eq 2.

Step 2 :- substitute value of y and find x

x = 0 and y = -3 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by substituting in eq 2.

Maths-

### Solve the following by substitution method 2x - 3y = 7 and x + 6y = 11.

Ans :- x = 5 ; y = 1

Explanation :-

Step 1 :- find x by substituting x = 11-6y in eq 2.

Step 2 :- substitute value of y and find x

x = 5 and y = 1 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by substituting x = 11-6y in eq 2.

### Solve the following by substitution method 2x - 3y = 7 and x + 6y = 11.

Maths-General

Ans :- x = 5 ; y = 1

Explanation :-

Step 1 :- find x by substituting x = 11-6y in eq 2.

Step 2 :- substitute value of y and find x

x = 5 and y = 1 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by substituting x = 11-6y in eq 2.

Maths-

### Solve the following by using elimination method: 2x + y = 6 , 3y = 8 + 4x

Ans :- x = 1 ; y = 4

Explanation :-

Step 1 :- find y by eliminating x

Eliminating x by doing eq 2- 2(eq1) then

Step 2 :- substitute value of y and find x

x = 1 and y = 4 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find y by eliminating x

Eliminating x by doing eq 2- 2(eq1) then

### Solve the following by using elimination method: 2x + y = 6 , 3y = 8 + 4x

Maths-General

Ans :- x = 1 ; y = 4

Explanation :-

Step 1 :- find y by eliminating x

Eliminating x by doing eq 2- 2(eq1) then

Step 2 :- substitute value of y and find x

x = 1 and y = 4 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find y by eliminating x

Eliminating x by doing eq 2- 2(eq1) then

Maths-

### Solve 2a – 3/b = 12 and 5a – 7/b = 1

Ans :- a = -81 ; b = -1/58

Explanation :-

Step 1:- find the value of 1/b by eliminating a

By 5eq1 - 2eq2 we get

Step 2:- find the value of a by substitution -1/b = 58 in eq 1

a = -81 and b = -1/58 are the solution of the given equation.

Explanation :-

Step 1:- find the value of 1/b by eliminating a

By 5eq1 - 2eq2 we get

### Solve 2a – 3/b = 12 and 5a – 7/b = 1

Maths-General

Ans :- a = -81 ; b = -1/58

Explanation :-

Step 1:- find the value of 1/b by eliminating a

By 5eq1 - 2eq2 we get

Step 2:- find the value of a by substitution -1/b = 58 in eq 1

a = -81 and b = -1/58 are the solution of the given equation.

Explanation :-

Step 1:- find the value of 1/b by eliminating a

By 5eq1 - 2eq2 we get

Maths-

### Solve the following equations by using the elimination method: x - y = 1 , 3x - y = 9

Ans :- x = 4 ; y = 3

Explanation :-

Step 1 :- find x by eliminating y

Eliminating x by subtracting eq 2 - eq1 then

Step 2 :- substitute value of x and find y

x = 4 and y = 3 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by eliminating y

Eliminating x by subtracting eq 2 - eq1 then

### Solve the following equations by using the elimination method: x - y = 1 , 3x - y = 9

Maths-General

Ans :- x = 4 ; y = 3

Explanation :-

Step 1 :- find x by eliminating y

Eliminating x by subtracting eq 2 - eq1 then

Step 2 :- substitute value of x and find y

x = 4 and y = 3 is the solution of the given pair of equations.

Explanation :-

Step 1 :- find x by eliminating y

Eliminating x by subtracting eq 2 - eq1 then

Maths-

### Solve the following system of linear equations: 2x - 4y = 6 , x - 3y = 12

Ans :- x = -15 ; y = -9

Explanation :-

2x - 4y = 6 — eq 1

x - 3y = 12 — eq 2

Step 1 :- find y by eliminating x

Eliminating x by subtracting e2q 2- eq1 then

Step 2 :- substitute value of y and find x

x = -15 and y = -9 is the solution of the given pair of equations.

Explanation :-

2x - 4y = 6 — eq 1

x - 3y = 12 — eq 2

Step 1 :- find y by eliminating x

Eliminating x by subtracting e2q 2- eq1 then

### Solve the following system of linear equations: 2x - 4y = 6 , x - 3y = 12

Maths-General

Ans :- x = -15 ; y = -9

Explanation :-

2x - 4y = 6 — eq 1

x - 3y = 12 — eq 2

Step 1 :- find y by eliminating x

Eliminating x by subtracting e2q 2- eq1 then

Step 2 :- substitute value of y and find x

x = -15 and y = -9 is the solution of the given pair of equations.

Explanation :-

2x - 4y = 6 — eq 1

x - 3y = 12 — eq 2

Step 1 :- find y by eliminating x

Eliminating x by subtracting e2q 2- eq1 then

Maths-

### Solve: x - 2y = 8 , 4x + 2y = 7 by using elimination method.

Ans :- x = 3 ; y = -5/2

Explanation :-

x - 2y = 8 — eq 1

4x = 2y = 7 — eq 2

Step 1 :- find x by eliminating y

Eliminating x by adding eq 2 and eq1 then

Step 2 :- substitute value of x and find y

and y = -5/2 is the solution of the given pair of equations.

Explanation :-

x - 2y = 8 — eq 1

4x = 2y = 7 — eq 2

Step 1 :- find x by eliminating y

Eliminating x by adding eq 2 and eq1 then

### Solve: x - 2y = 8 , 4x + 2y = 7 by using elimination method.

Maths-General

Ans :- x = 3 ; y = -5/2

Explanation :-

x - 2y = 8 — eq 1

4x = 2y = 7 — eq 2

Step 1 :- find x by eliminating y

Eliminating x by adding eq 2 and eq1 then

Step 2 :- substitute value of x and find y

and y = -5/2 is the solution of the given pair of equations.

Explanation :-

x - 2y = 8 — eq 1

4x = 2y = 7 — eq 2

Step 1 :- find x by eliminating y

Eliminating x by adding eq 2 and eq1 then

Maths-

For the solution of the system of equations above, what is the value of ?

The given equations are

First, we eliminate y by adding the equations together

Simplifying, we get

We have,

Thus, we get

As , we must have x =0 .

Thus, the correct option is C).

First, we eliminate y by adding the equations together

Simplifying, we get

We have,

Thus, we get

As , we must have x =0 .

Thus, the correct option is C).

For the solution of the system of equations above, what is the value of ?

Maths-General

The given equations are

First, we eliminate y by adding the equations together

Simplifying, we get

We have,

Thus, we get

As , we must have x =0 .

Thus, the correct option is C).

First, we eliminate y by adding the equations together

Simplifying, we get

We have,

Thus, we get

As , we must have x =0 .

Thus, the correct option is C).

Maths-

### If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD and ar(EFGH) = 40 sq.cm , find the ar (parallelogram ABCD).

Ans :- 80 cm

Explanation :-

Step 1:- Find area of parallelogram ABCD

Let height Dk = h and base AB = b

We get area of parallelogram ABCD = base × height = bh

Step 2:- Find height GI and EJ

In parallelogram line joining midpoints of opposites sides is parallel to the base and

equal to the base length i.e HF = AB = b

Let us consider ,

Here HF // AB and passing through mid point of side AD of

So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.

As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB

As we know perpendicular length between DC and HF is h/2 then GI = h/2

As we know perpendicular length between AB and HF is h/2 then EJ = h/2

Step 3:- Find the areas of EFH and HGF

As the area of triangle = ½ base × height

Area of = ½ base length of hf × height EJ =

Area of = ½ base length of hf × height GI =

Area of EFGH = Area of + Area of =

Step 4:- Equate the areas of EFGH

We Know that area of parallelogram ABCD = bh = 80 sq.cm.

^{2}Explanation :-

Step 1:- Find area of parallelogram ABCD

Let height Dk = h and base AB = b

We get area of parallelogram ABCD = base × height = bh

Step 2:- Find height GI and EJ

In parallelogram line joining midpoints of opposites sides is parallel to the base and

equal to the base length i.e HF = AB = b

Let us consider ,

Here HF // AB and passing through mid point of side AD of

So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.

As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB

As we know perpendicular length between DC and HF is h/2 then GI = h/2

As we know perpendicular length between AB and HF is h/2 then EJ = h/2

Step 3:- Find the areas of EFH and HGF

As the area of triangle = ½ base × height

Area of = ½ base length of hf × height EJ =

Area of = ½ base length of hf × height GI =

Area of EFGH = Area of + Area of =

Step 4:- Equate the areas of EFGH

We Know that area of parallelogram ABCD = bh = 80 sq.cm.

### If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD and ar(EFGH) = 40 sq.cm , find the ar (parallelogram ABCD).

Maths-General

Ans :- 80 cm

Explanation :-

Step 1:- Find area of parallelogram ABCD

Let height Dk = h and base AB = b

We get area of parallelogram ABCD = base × height = bh

Step 2:- Find height GI and EJ

In parallelogram line joining midpoints of opposites sides is parallel to the base and

equal to the base length i.e HF = AB = b

Let us consider ,

Here HF // AB and passing through mid point of side AD of

So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.

As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB

As we know perpendicular length between DC and HF is h/2 then GI = h/2

As we know perpendicular length between AB and HF is h/2 then EJ = h/2

Step 3:- Find the areas of EFH and HGF

As the area of triangle = ½ base × height

Area of = ½ base length of hf × height EJ =

Area of = ½ base length of hf × height GI =

Area of EFGH = Area of + Area of =

Step 4:- Equate the areas of EFGH

We Know that area of parallelogram ABCD = bh = 80 sq.cm.

^{2}Explanation :-

Step 1:- Find area of parallelogram ABCD

Let height Dk = h and base AB = b

We get area of parallelogram ABCD = base × height = bh

Step 2:- Find height GI and EJ

In parallelogram line joining midpoints of opposites sides is parallel to the base and

equal to the base length i.e HF = AB = b

Let us consider ,

Here HF // AB and passing through mid point of side AD of

So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.

As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB

As we know perpendicular length between DC and HF is h/2 then GI = h/2

As we know perpendicular length between AB and HF is h/2 then EJ = h/2

Step 3:- Find the areas of EFH and HGF

As the area of triangle = ½ base × height

Area of = ½ base length of hf × height EJ =

Area of = ½ base length of hf × height GI =

Area of EFGH = Area of + Area of =

Step 4:- Equate the areas of EFGH

We Know that area of parallelogram ABCD = bh = 80 sq.cm.

Maths-

### A soft drink is available in two packs. i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Hint:

The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.

Explanations:

Step 1 of 3:

The volume of a cylinder, with a rectangular base of length and b width and height of the cylinder be h , is , (since the area of a rectangle is length x width).

So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,

Step 2 of 3:

We know that, if the radius of a cylinder is r and height be h , then the volume is .

So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,

cm

Step 3 of 3:

The tin can have a capacity of 300 cm

So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm

Final Answer:

The plastic cylinder has greater capacity by 85 cm

The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.

Explanations:

Step 1 of 3:

The volume of a cylinder, with a rectangular base of length and b width and height of the cylinder be h , is , (since the area of a rectangle is length x width).

So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,

Step 2 of 3:

We know that, if the radius of a cylinder is r and height be h , then the volume is .

So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,

cm

^{3}Step 3 of 3:

The tin can have a capacity of 300 cm

^{3 }and the plastic cylinder has of 385 cm^{3}So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm

^{3}more.Final Answer:

The plastic cylinder has greater capacity by 85 cm

^{3 }more than the tin can.### A soft drink is available in two packs. i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Maths-General

Hint:

The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.

Explanations:

Step 1 of 3:

The volume of a cylinder, with a rectangular base of length and b width and height of the cylinder be h , is , (since the area of a rectangle is length x width).

So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,

Step 2 of 3:

We know that, if the radius of a cylinder is r and height be h , then the volume is .

So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,

cm

Step 3 of 3:

The tin can have a capacity of 300 cm

So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm

Final Answer:

The plastic cylinder has greater capacity by 85 cm

The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.

Explanations:

Step 1 of 3:

The volume of a cylinder, with a rectangular base of length and b width and height of the cylinder be h , is , (since the area of a rectangle is length x width).

So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,

Step 2 of 3:

We know that, if the radius of a cylinder is r and height be h , then the volume is .

So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,

cm

^{3}Step 3 of 3:

The tin can have a capacity of 300 cm

^{3 }and the plastic cylinder has of 385 cm^{3}So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm

^{3}more.Final Answer:

The plastic cylinder has greater capacity by 85 cm

^{3 }more than the tin can.Maths-

### Two transversals are cutting two parallel lines as shown in the diagram. Find the missing values: x, y and

ANS :- x = 90 ,the triangle is an obtuse angled triangle.

Explanation :- Given the two lines are parallel and two are transversals .

As alternate interior angles are equal we get, z° = 20°

∴z = 20

Using sum of angles in triangle is 180° we get ,

20°+ 30°+ 2x°=180° (sum of angles in a triangle)

2x°+50° = 180°

2x° = 130° x° = 65°

∴ x = 65

2x° = y° ( vertically opposite angles are equal)

130° = y°

∴ y = 130

Explanation :- Given the two lines are parallel and two are transversals .

As alternate interior angles are equal we get, z° = 20°

∴z = 20

Using sum of angles in triangle is 180° we get ,

20°+ 30°+ 2x°=180° (sum of angles in a triangle)

2x°+50° = 180°

2x° = 130° x° = 65°

∴ x = 65

2x° = y° ( vertically opposite angles are equal)

130° = y°

∴ y = 130

### Two transversals are cutting two parallel lines as shown in the diagram. Find the missing values: x, y and

Maths-General

ANS :- x = 90 ,the triangle is an obtuse angled triangle.

Explanation :- Given the two lines are parallel and two are transversals .

As alternate interior angles are equal we get, z° = 20°

∴z = 20

Using sum of angles in triangle is 180° we get ,

20°+ 30°+ 2x°=180° (sum of angles in a triangle)

2x°+50° = 180°

2x° = 130° x° = 65°

∴ x = 65

2x° = y° ( vertically opposite angles are equal)

130° = y°

∴ y = 130

Explanation :- Given the two lines are parallel and two are transversals .

As alternate interior angles are equal we get, z° = 20°

∴z = 20

Using sum of angles in triangle is 180° we get ,

20°+ 30°+ 2x°=180° (sum of angles in a triangle)

2x°+50° = 180°

2x° = 130° x° = 65°

∴ x = 65

2x° = y° ( vertically opposite angles are equal)

130° = y°

∴ y = 130

Maths-

### Find the value of x and classify the triangle by its angles.

ANS :- x = 90 ,the triangle is an obtuse angled triangle.

Explanation :-

Given , the angles of the triangle are 50°,30°and (x+10)°.

50°+ 30°+ x+10°=180° (sum of angles in a triangle)

x°+90° = 180°

x° = 90°

∴ x = 90

We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)

One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle

Explanation :-

Given , the angles of the triangle are 50°,30°and (x+10)°.

50°+ 30°+ x+10°=180° (sum of angles in a triangle)

x°+90° = 180°

x° = 90°

∴ x = 90

We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)

One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle

### Find the value of x and classify the triangle by its angles.

Maths-General

ANS :- x = 90 ,the triangle is an obtuse angled triangle.

Explanation :-

Given , the angles of the triangle are 50°,30°and (x+10)°.

50°+ 30°+ x+10°=180° (sum of angles in a triangle)

x°+90° = 180°

x° = 90°

∴ x = 90

We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)

One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle

Explanation :-

Given , the angles of the triangle are 50°,30°and (x+10)°.

50°+ 30°+ x+10°=180° (sum of angles in a triangle)

x°+90° = 180°

x° = 90°

∴ x = 90

We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)

One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle

Maths-

Let A be the event that the house selected at random is a single storey house with slate roof.

From the table given, it can be seen that

The number of houses which are single storey having slate roof =4.

Total number of houses =48

So,

The number of outcomes favourable to A (m)=4

Total number of outcomes (n)= 48

Hence, the probability of selecting a single storey house with slate roof,

Thus, the correct option is A)

From the table given, it can be seen that

The number of houses which are single storey having slate roof =4.

Total number of houses =48

So,

The number of outcomes favourable to A (m)=4

Total number of outcomes (n)= 48

Hence, the probability of selecting a single storey house with slate roof,

Thus, the correct option is A)

Maths-General

Let A be the event that the house selected at random is a single storey house with slate roof.

From the table given, it can be seen that

The number of houses which are single storey having slate roof =4.

Total number of houses =48

So,

The number of outcomes favourable to A (m)=4

Total number of outcomes (n)= 48

Hence, the probability of selecting a single storey house with slate roof,

Thus, the correct option is A)

From the table given, it can be seen that

The number of houses which are single storey having slate roof =4.

Total number of houses =48

So,

The number of outcomes favourable to A (m)=4

Total number of outcomes (n)= 48

Hence, the probability of selecting a single storey house with slate roof,

Thus, the correct option is A)