Maths-

General

Easy

Question

# Given a rectangular room whose width is 4/7 of its length y and perimeter z. Write down an equation connecting y and z. Calculate the length of the rectangular room when the given perimeter is 4400 cm.

Hint:

### let the length of the rectangular room is y and width is 4/7 of length(i.e4y/7)

Perimeter of rectangular room z = 4400cm

Find perimeter in terms of lengths and width and solve to get the equation

Now substitute the value of z in the equation to get y.

## The correct answer is: 1400 cm.

### Ans :- length of rectangle is 1400 cm .

Explanation :-

Step 1:- frame the equation from the given set of conditions .

let the length of the rectangular room is y

width is 4/7 of length width = 4y/7

Perimeter of rectangle = 2(length + width) = 2(y + )

Z =

The equation connecting y and z is 22y = 7z

Step 2:- find the y by substituting z=4400cm(given)

∴ y = 1400 cm.

∴ length of the rectangle is 1400 cm .

### Related Questions to study

Maths-

### A drainage tile is a cylindrical shell 21 cm long. The inside and outside diameters are 4.5 cm and 5.1 cm respectively. What is the volume of the clay required for the tile?

Hint:

Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.

Explanations:

Step 1 of 2:

Given, h = 21cm; r = 4.5/2 cm (inner radius) and R = 5.1/2 cm (outer radius)

The volume with inner radius cm

The volume with outer radius = cm

Step 2 of 2:

The volume of the clay required for the tile

Final Answer:

The volume of the clay required for the tile is 95.04 cm

Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.

Explanations:

Step 1 of 2:

Given, h = 21cm; r = 4.5/2 cm (inner radius) and R = 5.1/2 cm (outer radius)

The volume with inner radius cm

^{3}The volume with outer radius = cm

^{3}Step 2 of 2:

The volume of the clay required for the tile

Final Answer:

The volume of the clay required for the tile is 95.04 cm

^{3}### A drainage tile is a cylindrical shell 21 cm long. The inside and outside diameters are 4.5 cm and 5.1 cm respectively. What is the volume of the clay required for the tile?

Maths-General

Hint:

Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.

Explanations:

Step 1 of 2:

Given, h = 21cm; r = 4.5/2 cm (inner radius) and R = 5.1/2 cm (outer radius)

The volume with inner radius cm

The volume with outer radius = cm

Step 2 of 2:

The volume of the clay required for the tile

Final Answer:

The volume of the clay required for the tile is 95.04 cm

Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.

Explanations:

Step 1 of 2:

Given, h = 21cm; r = 4.5/2 cm (inner radius) and R = 5.1/2 cm (outer radius)

The volume with inner radius cm

^{3}The volume with outer radius = cm

^{3}Step 2 of 2:

The volume of the clay required for the tile

Final Answer:

The volume of the clay required for the tile is 95.04 cm

^{3}Maths-

### Find the area of the given figure.

Ans :- 40.5cm

Explanation :-

Step 1:- Find the area of square

Given side of square = 3 cm .

Area of square =

Step 2:- Find the area of trapezium

We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .

Height (or) distance between parallel lines = 10 - 3 = 7 cm

Step 3:- Find the Area of Figure

Area of polygon = area of square + area of trapezium

Area of polygon = 9 + 31.5 = 40.5 cm

Therefore, the area of the given figure is 40.5cm

^{2}.Explanation :-

Step 1:- Find the area of square

Given side of square = 3 cm .

Area of square =

Step 2:- Find the area of trapezium

We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .

Height (or) distance between parallel lines = 10 - 3 = 7 cm

Step 3:- Find the Area of Figure

Area of polygon = area of square + area of trapezium

Area of polygon = 9 + 31.5 = 40.5 cm

^{2}.Therefore, the area of the given figure is 40.5cm

^{2}.### Find the area of the given figure.

Maths-General

Ans :- 40.5cm

Explanation :-

Step 1:- Find the area of square

Given side of square = 3 cm .

Area of square =

Step 2:- Find the area of trapezium

We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .

Height (or) distance between parallel lines = 10 - 3 = 7 cm

Step 3:- Find the Area of Figure

Area of polygon = area of square + area of trapezium

Area of polygon = 9 + 31.5 = 40.5 cm

Therefore, the area of the given figure is 40.5cm

^{2}.Explanation :-

Step 1:- Find the area of square

Given side of square = 3 cm .

Area of square =

Step 2:- Find the area of trapezium

We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .

Height (or) distance between parallel lines = 10 - 3 = 7 cm

Step 3:- Find the Area of Figure

Area of polygon = area of square + area of trapezium

Area of polygon = 9 + 31.5 = 40.5 cm

^{2}.Therefore, the area of the given figure is 40.5cm

^{2}.Maths-

### Four years back in time, a father was 3 times as old as his son then was. 8 years later father will be 2 times as old as his son will then be. Find the ages of son and father.

Ans :- Age of Son is 16 years and Age of father is 40 years .

Explanation :-

Step 1:- frame the equations from the given set of conditions .

let present ages of son is x years and present age of father is y years

Fours years back, age of son = x-4

Fours years back, age of father = y-4

Four years back in time, a father was 3 times as old as his son then was

—Eq1

After 8 years, age of son = x+8

After 8 years, age of father = y+8

8 years later father will be 2 times as old as his son will then be

— Eq2

Step 2:- find x by eliminating y

Doing Eq1-Eq2 to eliminate x

Step 3:- substitute x=16 in Eq2.

∴ The present age of son =x=16 years and present age of father = 40 years

Explanation :-

Step 1:- frame the equations from the given set of conditions .

let present ages of son is x years and present age of father is y years

Fours years back, age of son = x-4

Fours years back, age of father = y-4

Four years back in time, a father was 3 times as old as his son then was

—Eq1

After 8 years, age of son = x+8

After 8 years, age of father = y+8

8 years later father will be 2 times as old as his son will then be

— Eq2

Step 2:- find x by eliminating y

Doing Eq1-Eq2 to eliminate x

∴ x= 16 |

∴ y = 40 |

### Four years back in time, a father was 3 times as old as his son then was. 8 years later father will be 2 times as old as his son will then be. Find the ages of son and father.

Maths-General

Ans :- Age of Son is 16 years and Age of father is 40 years .

Explanation :-

Step 1:- frame the equations from the given set of conditions .

let present ages of son is x years and present age of father is y years

Fours years back, age of son = x-4

Fours years back, age of father = y-4

Four years back in time, a father was 3 times as old as his son then was

—Eq1

After 8 years, age of son = x+8

After 8 years, age of father = y+8

8 years later father will be 2 times as old as his son will then be

— Eq2

Step 2:- find x by eliminating y

Doing Eq1-Eq2 to eliminate x

Step 3:- substitute x=16 in Eq2.

∴ The present age of son =x=16 years and present age of father = 40 years

Explanation :-

Step 1:- frame the equations from the given set of conditions .

let present ages of son is x years and present age of father is y years

Fours years back, age of son = x-4

Fours years back, age of father = y-4

Four years back in time, a father was 3 times as old as his son then was

—Eq1

After 8 years, age of son = x+8

After 8 years, age of father = y+8

8 years later father will be 2 times as old as his son will then be

— Eq2

Step 2:- find x by eliminating y

Doing Eq1-Eq2 to eliminate x

∴ x= 16 |

∴ y = 40 |

Maths-

### A number which is formed by two digits is 3 times the sum of the digits. While interchanging the digits it is 45 more than the actual number. Find the number.

Ans :- 27 is the number which satisfies the given condition.

Explanation :-

Step 1:- frame the equations from the given set of conditions .

Given The number formed is 3 times the sum of digits

— Eq1

If digits are reversed the number is reduced by 45

—- Eq2

Step 2:- Substitute Eq1 in Eq2

We get

Step 3:- find the value of y by substituting x=2 in Eq2

∴ 27 is the number which satisfy the given conditions.

Explanation :-

Step 1:- frame the equations from the given set of conditions .

Given The number formed is 3 times the sum of digits

— Eq1

If digits are reversed the number is reduced by 45

—- Eq2

Step 2:- Substitute Eq1 in Eq2

We get

∴ x = 2 |

∴ y = 7 |

### A number which is formed by two digits is 3 times the sum of the digits. While interchanging the digits it is 45 more than the actual number. Find the number.

Maths-General

Ans :- 27 is the number which satisfies the given condition.

Explanation :-

Step 1:- frame the equations from the given set of conditions .

Given The number formed is 3 times the sum of digits

— Eq1

If digits are reversed the number is reduced by 45

—- Eq2

Step 2:- Substitute Eq1 in Eq2

We get

Step 3:- find the value of y by substituting x=2 in Eq2

∴ 27 is the number which satisfy the given conditions.

Explanation :-

Step 1:- frame the equations from the given set of conditions .

Given The number formed is 3 times the sum of digits

— Eq1

If digits are reversed the number is reduced by 45

—- Eq2

Step 2:- Substitute Eq1 in Eq2

We get

∴ x = 2 |

∴ y = 7 |

Maths-

### The gas mileage M (s), in miles per gallon, of a car traveling s miles per hour is modeled by the function below, where .

According to the model, at what speed, in miles per hour, does the car obtain its greatest gas mileage?

Given,

Gas mileage is denoted by M(s), in miles per gallon,

Speed of the car is denoted by s miles per hour.

The relation between gas mileage and speed is given by

We need to find the value of for which M(s) is maximum.

We use the second derivative test.

First, we find

That is

TO find critical points,

Which gives

Solving for , we have

The only critical point, we get is s = 48 . So, we check if M (s) has a maximum at s = 48

Now

So

Thus, M (s) has a maximum at s = 48

The correct option is B)

Gas mileage is denoted by M(s), in miles per gallon,

Speed of the car is denoted by s miles per hour.

The relation between gas mileage and speed is given by

We need to find the value of for which M(s) is maximum.

We use the second derivative test.

First, we find

That is

TO find critical points,

Which gives

Solving for , we have

The only critical point, we get is s = 48 . So, we check if M (s) has a maximum at s = 48

Now

So

Thus, M (s) has a maximum at s = 48

The correct option is B)

### The gas mileage M (s), in miles per gallon, of a car traveling s miles per hour is modeled by the function below, where .

According to the model, at what speed, in miles per hour, does the car obtain its greatest gas mileage?

Maths-General

Given,

Gas mileage is denoted by M(s), in miles per gallon,

Speed of the car is denoted by s miles per hour.

The relation between gas mileage and speed is given by

We need to find the value of for which M(s) is maximum.

We use the second derivative test.

First, we find

That is

TO find critical points,

Which gives

Solving for , we have

The only critical point, we get is s = 48 . So, we check if M (s) has a maximum at s = 48

Now

So

Thus, M (s) has a maximum at s = 48

The correct option is B)

Gas mileage is denoted by M(s), in miles per gallon,

Speed of the car is denoted by s miles per hour.

The relation between gas mileage and speed is given by

We need to find the value of for which M(s) is maximum.

We use the second derivative test.

First, we find

That is

TO find critical points,

Which gives

Solving for , we have

The only critical point, we get is s = 48 . So, we check if M (s) has a maximum at s = 48

Now

So

Thus, M (s) has a maximum at s = 48

The correct option is B)

Maths-

### Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm. Find y and z

Ans :- y = 12 and z = 8

Explanation :-

Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.

Step 1:- form the system of linear equations using the given condition

AB = CD (opposite sides of the rectangle are equal.)

— Eq1

BC = DA(opposite sides of the rectangle are equal.)

— Eq2

Step 2:- substitute the Eq1 in Eq2.

Step 3:- the value of z in Eq1.

Explanation :-

Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.

Step 1:- form the system of linear equations using the given condition

AB = CD (opposite sides of the rectangle are equal.)

— Eq1

BC = DA(opposite sides of the rectangle are equal.)

— Eq2

Step 2:- substitute the Eq1 in Eq2.

∴ z = 8 |

∴ y = 12 |

### Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm. Find y and z

Maths-General

Ans :- y = 12 and z = 8

Explanation :-

Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.

Step 1:- form the system of linear equations using the given condition

AB = CD (opposite sides of the rectangle are equal.)

— Eq1

BC = DA(opposite sides of the rectangle are equal.)

— Eq2

Step 2:- substitute the Eq1 in Eq2.

Step 3:- the value of z in Eq1.

Explanation :-

Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.

Step 1:- form the system of linear equations using the given condition

AB = CD (opposite sides of the rectangle are equal.)

— Eq1

BC = DA(opposite sides of the rectangle are equal.)

— Eq2

Step 2:- substitute the Eq1 in Eq2.

∴ z = 8 |

∴ y = 12 |

Maths-

### Find the volume of a right circular cylinder of length 80 cm and diameter of the base 14 cm.

Hint:

A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and height h, is cubic units.

Therefore, the volume of

Explanations:

Step 1 of 1:

The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius

r = 14/2 = 7cm.

The height is given by, h = 80 cm

the given right circular cylinder

cm3

Final Answer:

The volume of the given right circular cylinder is 12320 cm3.

A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and height h, is cubic units.

Therefore, the volume of

Explanations:

Step 1 of 1:

The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius

r = 14/2 = 7cm.

The height is given by, h = 80 cm

the given right circular cylinder

cm3

Final Answer:

The volume of the given right circular cylinder is 12320 cm3.

### Find the volume of a right circular cylinder of length 80 cm and diameter of the base 14 cm.

Maths-General

Hint:

A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and height h, is cubic units.

Therefore, the volume of

Explanations:

Step 1 of 1:

The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius

r = 14/2 = 7cm.

The height is given by, h = 80 cm

the given right circular cylinder

cm3

Final Answer:

The volume of the given right circular cylinder is 12320 cm3.

A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and height h, is cubic units.

Therefore, the volume of

Explanations:

Step 1 of 1:

The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius

r = 14/2 = 7cm.

The height is given by, h = 80 cm

the given right circular cylinder

cm3

Final Answer:

The volume of the given right circular cylinder is 12320 cm3.

Maths-

### Find the area of the polygon in the given picture, cm,

Ans :- 31.75cm

Explanation :-

Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.

OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm

Step 1:- Find the area of ΔMAN

Consider ΔMAN ,

we have MA = 2 cm and NA = 2.5 cm with right angle at A

Area of triangle ΔMAN = ½ × b × h = ½ × 2 × 2.5 = 2.5 cm

Step 2:- Find the area of ANOD

Consider ANOD,

As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to

both. Here ANOD is a Trapezium .parallel sides are DO and AD

Here AD is height ; AD = MD - MA = 7-2 = 5 cm

Area of trapezium ANOD

Step 3 :- Find the area of ΔDOP

Consider ΔDOP ,

We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

Step 4 :- Find the area of ΔMBR

Consider ΔMBR ,

we have MB = 4 cm and RB = 2.5 cm with right angle at B

Area of triangle ΔMAN = ½ × b × h = ½ × 4 × 2.5 = 5 cm

Step 5 :- Find the area of RBCQ

Consider RBCQ,

As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to

both. Here RBCQ is a Trapezium .parallel sides are RB and CQ

Here BC is height ; BC = MC - MB = 6 -4 = 2 cm

Area of trapezium ANOD

Step 6 :- Find the area of ΔQCP

Consider ΔQCP ,

We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

Step 7:-Find the area of the polygon.

Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +

area of ΔMBR + area of RBCQ + area of ΔQCP

Area of polygon = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm

Therefore, area of polygon is 31.75 cm

^{2}.Explanation :-

Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.

OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm

Step 1:- Find the area of ΔMAN

Consider ΔMAN ,

we have MA = 2 cm and NA = 2.5 cm with right angle at A

Area of triangle ΔMAN = ½ × b × h = ½ × 2 × 2.5 = 2.5 cm

^{2}Step 2:- Find the area of ANOD

Consider ANOD,

As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to

both. Here ANOD is a Trapezium .parallel sides are DO and AD

Here AD is height ; AD = MD - MA = 7-2 = 5 cm

Area of trapezium ANOD

Step 3 :- Find the area of ΔDOP

Consider ΔDOP ,

We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

^{2}Step 4 :- Find the area of ΔMBR

Consider ΔMBR ,

we have MB = 4 cm and RB = 2.5 cm with right angle at B

Area of triangle ΔMAN = ½ × b × h = ½ × 4 × 2.5 = 5 cm

^{2}Step 5 :- Find the area of RBCQ

Consider RBCQ,

As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to

both. Here RBCQ is a Trapezium .parallel sides are RB and CQ

Here BC is height ; BC = MC - MB = 6 -4 = 2 cm

Area of trapezium ANOD

Step 6 :- Find the area of ΔQCP

Consider ΔQCP ,

We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

^{2}Step 7:-Find the area of the polygon.

Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +

area of ΔMBR + area of RBCQ + area of ΔQCP

Area of polygon = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm

^{2}Therefore, area of polygon is 31.75 cm

^{2}### Find the area of the polygon in the given picture, cm,

Maths-General

Ans :- 31.75cm

Explanation :-

Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.

OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm

Step 1:- Find the area of ΔMAN

Consider ΔMAN ,

we have MA = 2 cm and NA = 2.5 cm with right angle at A

Area of triangle ΔMAN = ½ × b × h = ½ × 2 × 2.5 = 2.5 cm

Step 2:- Find the area of ANOD

Consider ANOD,

As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to

both. Here ANOD is a Trapezium .parallel sides are DO and AD

Here AD is height ; AD = MD - MA = 7-2 = 5 cm

Area of trapezium ANOD

Step 3 :- Find the area of ΔDOP

Consider ΔDOP ,

We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

Step 4 :- Find the area of ΔMBR

Consider ΔMBR ,

we have MB = 4 cm and RB = 2.5 cm with right angle at B

Area of triangle ΔMAN = ½ × b × h = ½ × 4 × 2.5 = 5 cm

Step 5 :- Find the area of RBCQ

Consider RBCQ,

As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to

both. Here RBCQ is a Trapezium .parallel sides are RB and CQ

Here BC is height ; BC = MC - MB = 6 -4 = 2 cm

Area of trapezium ANOD

Step 6 :- Find the area of ΔQCP

Consider ΔQCP ,

We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

Step 7:-Find the area of the polygon.

Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +

area of ΔMBR + area of RBCQ + area of ΔQCP

Area of polygon = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm

Therefore, area of polygon is 31.75 cm

^{2}.Explanation :-

Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.

OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm

Step 1:- Find the area of ΔMAN

Consider ΔMAN ,

we have MA = 2 cm and NA = 2.5 cm with right angle at A

Area of triangle ΔMAN = ½ × b × h = ½ × 2 × 2.5 = 2.5 cm

^{2}Step 2:- Find the area of ANOD

Consider ANOD,

As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to

both. Here ANOD is a Trapezium .parallel sides are DO and AD

Here AD is height ; AD = MD - MA = 7-2 = 5 cm

Area of trapezium ANOD

Step 3 :- Find the area of ΔDOP

Consider ΔDOP ,

We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

^{2}Step 4 :- Find the area of ΔMBR

Consider ΔMBR ,

we have MB = 4 cm and RB = 2.5 cm with right angle at B

Area of triangle ΔMAN = ½ × b × h = ½ × 4 × 2.5 = 5 cm

^{2}Step 5 :- Find the area of RBCQ

Consider RBCQ,

As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to

both. Here RBCQ is a Trapezium .parallel sides are RB and CQ

Here BC is height ; BC = MC - MB = 6 -4 = 2 cm

Area of trapezium ANOD

Step 6 :- Find the area of ΔQCP

Consider ΔQCP ,

We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C

Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm

^{2}Step 7:-Find the area of the polygon.

Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +

area of ΔMBR + area of RBCQ + area of ΔQCP

Area of polygon = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm

^{2}Therefore, area of polygon is 31.75 cm

^{2}Maths-

### A man bought 3 paise and 5 paise denomination stamps for Rs 1.00. In total he bought 22 stamps. Find out total number of 3 paise stamps bought by him.

Ans :- There are 5 stamps of 3 paise.

Explanation :-

let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y

Step 1:-Frame the equations from the given conditions.

Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22

— Eq1

Total cost of stamps = 3 no. of 3 paise stamps + 5 no. of 5 paise stamps = 100 paise

— Eq2

Step 2:- Eliminate y to find x

Doing 5(Eq1) -(Eq2) to eliminate y

∴ The no. of 3 paise stamps = x = 5 stamps

Explanation :-

let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y

Step 1:-Frame the equations from the given conditions.

Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22

— Eq1

Total cost of stamps = 3 no. of 3 paise stamps + 5 no. of 5 paise stamps = 100 paise

— Eq2

Step 2:- Eliminate y to find x

Doing 5(Eq1) -(Eq2) to eliminate y

∴ x = 5 |

### A man bought 3 paise and 5 paise denomination stamps for Rs 1.00. In total he bought 22 stamps. Find out total number of 3 paise stamps bought by him.

Maths-General

Ans :- There are 5 stamps of 3 paise.

Explanation :-

let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y

Step 1:-Frame the equations from the given conditions.

Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22

— Eq1

Total cost of stamps = 3 no. of 3 paise stamps + 5 no. of 5 paise stamps = 100 paise

— Eq2

Step 2:- Eliminate y to find x

Doing 5(Eq1) -(Eq2) to eliminate y

∴ The no. of 3 paise stamps = x = 5 stamps

Explanation :-

let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y

Step 1:-Frame the equations from the given conditions.

Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22

— Eq1

Total cost of stamps = 3 no. of 3 paise stamps + 5 no. of 5 paise stamps = 100 paise

— Eq2

Step 2:- Eliminate y to find x

Doing 5(Eq1) -(Eq2) to eliminate y

∴ x = 5 |

Maths-

### A ball pen costs Rs 3.50 more than pencil. Adding 3 ball pens and 2 pencils sum up to Rs 13. Taking y and z as costs of ball pen and pencil respectively. Write down 2 simultaneous equations in terms of y and z which satisfy the above statement. Find out the values of y and z.

Ans :- cost ball pen = 4.00 rs and cost of pencil = 0.50 rs

Explanation :-

Taking y and z as costs of ball pen and pencil respectively

L.e Cost of ball pen = y and cost of pencil = z

Step 1:- form the system of linear equations using the given condition

A ball pen costs Rs 3.50 more than pencil

so, y = z + 3.50 —- Eq1

Adding 3 ball pens and 2 pencils sum up to Rs 13.

So, 3y + 2z = 13 —- Eq2

We got Eq1 and Eq2 are simultaneous equations .

Step 2:- substitute the Eq1 in Eq2.

Step 3:-substitute the value of z in Eq1 to get the value of y

∴ z = 0.50 rs is the cost of pencil and y = 4.00 rs is the cost of ball pen.

Explanation :-

Taking y and z as costs of ball pen and pencil respectively

L.e Cost of ball pen = y and cost of pencil = z

Step 1:- form the system of linear equations using the given condition

A ball pen costs Rs 3.50 more than pencil

so, y = z + 3.50 —- Eq1

Adding 3 ball pens and 2 pencils sum up to Rs 13.

So, 3y + 2z = 13 —- Eq2

We got Eq1 and Eq2 are simultaneous equations .

Step 2:- substitute the Eq1 in Eq2.

∴ z = 0.50 |

∴ y = 4.00 |

### A ball pen costs Rs 3.50 more than pencil. Adding 3 ball pens and 2 pencils sum up to Rs 13. Taking y and z as costs of ball pen and pencil respectively. Write down 2 simultaneous equations in terms of y and z which satisfy the above statement. Find out the values of y and z.

Maths-General

Ans :- cost ball pen = 4.00 rs and cost of pencil = 0.50 rs

Explanation :-

Taking y and z as costs of ball pen and pencil respectively

L.e Cost of ball pen = y and cost of pencil = z

Step 1:- form the system of linear equations using the given condition

A ball pen costs Rs 3.50 more than pencil

so, y = z + 3.50 —- Eq1

Adding 3 ball pens and 2 pencils sum up to Rs 13.

So, 3y + 2z = 13 —- Eq2

We got Eq1 and Eq2 are simultaneous equations .

Step 2:- substitute the Eq1 in Eq2.

Step 3:-substitute the value of z in Eq1 to get the value of y

∴ z = 0.50 rs is the cost of pencil and y = 4.00 rs is the cost of ball pen.

Explanation :-

Taking y and z as costs of ball pen and pencil respectively

L.e Cost of ball pen = y and cost of pencil = z

Step 1:- form the system of linear equations using the given condition

A ball pen costs Rs 3.50 more than pencil

so, y = z + 3.50 —- Eq1

Adding 3 ball pens and 2 pencils sum up to Rs 13.

So, 3y + 2z = 13 —- Eq2

We got Eq1 and Eq2 are simultaneous equations .

Step 2:- substitute the Eq1 in Eq2.

∴ z = 0.50 |

∴ y = 4.00 |

Maths-

### In a town there are 2 brothers namely Anand and Suresh. Adding 1/3 rd anand’s age and suresh age would sum up to 10. Also adding Anand’s age together with 1⁄2 of suresh age would sum up to 10. Find Anand and Suresh ages.

Ans :- age of anand = 6 yrs ; age of suresh = 8 yrs

Explanation :-

let the age of Anand be x years and age of suresh be y years

Step 1:- Frame the equations using the given conditions.

Adding 1/3 rd anand’s age and suresh age would sum up to 10.

So, — Eq1

Adding Anand’s age together with 1⁄2 of suresh age would sum up to 10

So, — Eq2

Step 2:- Eliminate x and find y

Doing, 3(Eq1)-Eq2 to eliminate x.

Step 3:- Find x by substituting y= 8 in Eq2

∴ age of Anand is 6 years and age of suresh is 8 years .

Explanation :-

let the age of Anand be x years and age of suresh be y years

Step 1:- Frame the equations using the given conditions.

Adding 1/3 rd anand’s age and suresh age would sum up to 10.

So, — Eq1

Adding Anand’s age together with 1⁄2 of suresh age would sum up to 10

So, — Eq2

Step 2:- Eliminate x and find y

Doing, 3(Eq1)-Eq2 to eliminate x.

∴ y = 8 |

∴ x = 6 |

### In a town there are 2 brothers namely Anand and Suresh. Adding 1/3 rd anand’s age and suresh age would sum up to 10. Also adding Anand’s age together with 1⁄2 of suresh age would sum up to 10. Find Anand and Suresh ages.

Maths-General

Ans :- age of anand = 6 yrs ; age of suresh = 8 yrs

Explanation :-

let the age of Anand be x years and age of suresh be y years

Step 1:- Frame the equations using the given conditions.

Adding 1/3 rd anand’s age and suresh age would sum up to 10.

So, — Eq1

Adding Anand’s age together with 1⁄2 of suresh age would sum up to 10

So, — Eq2

Step 2:- Eliminate x and find y

Doing, 3(Eq1)-Eq2 to eliminate x.

Step 3:- Find x by substituting y= 8 in Eq2

∴ age of Anand is 6 years and age of suresh is 8 years .

Explanation :-

let the age of Anand be x years and age of suresh be y years

Step 1:- Frame the equations using the given conditions.

Adding 1/3 rd anand’s age and suresh age would sum up to 10.

So, — Eq1

Adding Anand’s age together with 1⁄2 of suresh age would sum up to 10

So, — Eq2

Step 2:- Eliminate x and find y

Doing, 3(Eq1)-Eq2 to eliminate x.

∴ y = 8 |

∴ x = 6 |

Maths-

### From a cylindrical wooden log of length 30 cm and base radius 72 cm, biggest cuboid of square base is made. Find the volume of wood wasted?

Hint:

Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find

Explanations:

Step 1 of 3:

In ,

AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm

Step 2 of 3:

For wooden log, base radius r = 72cm and height h = 30cm

So, the volume of the wooden log = cm

For square cuboid, side a = cm

Then, the volume of the cuboid = cm

Step 3 of 3:

The

is cm

Log volume – cuboid volume

Final Answer:

The volume of wood wasted

Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find

*a*and subtract the volume of square cuboid from the volume of log to solve the problem.Explanations:

Step 1 of 3:

In ,

AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm

Step 2 of 3:

For wooden log, base radius r = 72cm and height h = 30cm

So, the volume of the wooden log = cm

^{3}For square cuboid, side a = cm

Then, the volume of the cuboid = cm

^{3}Step 3 of 3:

The

is cm

^{3}. volume of the wood wasted is given by,Log volume – cuboid volume

Final Answer:

The volume of wood wasted

### From a cylindrical wooden log of length 30 cm and base radius 72 cm, biggest cuboid of square base is made. Find the volume of wood wasted?

Maths-General

Hint:

Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find

Explanations:

Step 1 of 3:

In ,

AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm

Step 2 of 3:

For wooden log, base radius r = 72cm and height h = 30cm

So, the volume of the wooden log = cm

For square cuboid, side a = cm

Then, the volume of the cuboid = cm

Step 3 of 3:

The

is cm

Log volume – cuboid volume

Final Answer:

The volume of wood wasted

Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find

*a*and subtract the volume of square cuboid from the volume of log to solve the problem.Explanations:

Step 1 of 3:

In ,

AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm

Step 2 of 3:

For wooden log, base radius r = 72cm and height h = 30cm

So, the volume of the wooden log = cm

^{3}For square cuboid, side a = cm

Then, the volume of the cuboid = cm

^{3}Step 3 of 3:

The

is cm

^{3}. volume of the wood wasted is given by,Log volume – cuboid volume

Final Answer:

The volume of wood wasted

Maths-

### Find the area of the figure.

.

Ans :- 340 cm

Explanation :-

By observing the diagram EFGH is a quadrilateral where all sides are equal so

EFGH is a rhombus .

ABCD is quadrilateral with AD // BC .

so, ABCD is trapezium with AD and BC as parallel sides.

STEP 1:- Find the area of rhombus EFGH

GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)

Area of rhombus =

Area of rhombus =

We get area of rhombus EFGH is 70cm

STEP 2:- Find the area of trapezium ABCD

AD = 24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm

= 270 cm

Area of trapezium ABCD is 270 cm

Step 3:- Find the area of figure

Area of figure = area of rhombus EFGH + area of trapezium ABCD .

Area of figure = 70+270 = 340 cm

Therefore , the area of the entire figure is 340cm

^{2}.Explanation :-

By observing the diagram EFGH is a quadrilateral where all sides are equal so

EFGH is a rhombus .

ABCD is quadrilateral with AD // BC .

so, ABCD is trapezium with AD and BC as parallel sides.

STEP 1:- Find the area of rhombus EFGH

GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)

Area of rhombus =

Area of rhombus =

We get area of rhombus EFGH is 70cm

^{2}.STEP 2:- Find the area of trapezium ABCD

AD = 24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm

= 270 cm

^{2}Area of trapezium ABCD is 270 cm

^{2}Step 3:- Find the area of figure

Area of figure = area of rhombus EFGH + area of trapezium ABCD .

Area of figure = 70+270 = 340 cm

^{2}.Therefore , the area of the entire figure is 340cm

^{2}.### Find the area of the figure.

.

Maths-General

Ans :- 340 cm

Explanation :-

By observing the diagram EFGH is a quadrilateral where all sides are equal so

EFGH is a rhombus .

ABCD is quadrilateral with AD // BC .

so, ABCD is trapezium with AD and BC as parallel sides.

STEP 1:- Find the area of rhombus EFGH

GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)

Area of rhombus =

Area of rhombus =

We get area of rhombus EFGH is 70cm

STEP 2:- Find the area of trapezium ABCD

AD = 24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm

= 270 cm

Area of trapezium ABCD is 270 cm

Step 3:- Find the area of figure

Area of figure = area of rhombus EFGH + area of trapezium ABCD .

Area of figure = 70+270 = 340 cm

Therefore , the area of the entire figure is 340cm

^{2}.Explanation :-

By observing the diagram EFGH is a quadrilateral where all sides are equal so

EFGH is a rhombus .

ABCD is quadrilateral with AD // BC .

so, ABCD is trapezium with AD and BC as parallel sides.

STEP 1:- Find the area of rhombus EFGH

GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)

Area of rhombus =

Area of rhombus =

We get area of rhombus EFGH is 70cm

^{2}.STEP 2:- Find the area of trapezium ABCD

AD = 24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm

= 270 cm

^{2}Area of trapezium ABCD is 270 cm

^{2}Step 3:- Find the area of figure

Area of figure = area of rhombus EFGH + area of trapezium ABCD .

Area of figure = 70+270 = 340 cm

^{2}.Therefore , the area of the entire figure is 340cm

^{2}.Maths-

The system of equations above is graphed in the xy -plane. At which of the following points do the graphs of the equations intersect?

The given equations are:

We will use the method of substitution to solve these equations.

From the first equation, we use the value of in the second equation,

We get an equation in one variable.

Simplifying this, we get

Taking all the terms on one side, we have

Thus, we get a quadratic equation,

factorization to solve We use middle term the above equation.

factorization to solve We use middle term the above equation.

Now, taking common from the first two terms and 2 from the last two terms, we have

So, we get two solutions for given by

Using these values of x in the first equation, we get two values of y

Thus, we get two solutions for the given system of equation

We can see that ( 5, 14 ) is in the options.

The correct option is D).

We will use the method of substitution to solve these equations.

From the first equation, we use the value of in the second equation,

We get an equation in one variable.

Simplifying this, we get

Taking all the terms on one side, we have

Thus, we get a quadratic equation,

factorization to solve We use middle term the above equation.

factorization to solve We use middle term the above equation.

Now, taking common from the first two terms and 2 from the last two terms, we have

So, we get two solutions for given by

Using these values of x in the first equation, we get two values of y

Thus, we get two solutions for the given system of equation

We can see that ( 5, 14 ) is in the options.

The correct option is D).

The system of equations above is graphed in the xy -plane. At which of the following points do the graphs of the equations intersect?

Maths-General

The given equations are:

We will use the method of substitution to solve these equations.

From the first equation, we use the value of in the second equation,

We get an equation in one variable.

Simplifying this, we get

Taking all the terms on one side, we have

Thus, we get a quadratic equation,

factorization to solve We use middle term the above equation.

factorization to solve We use middle term the above equation.

Now, taking common from the first two terms and 2 from the last two terms, we have

So, we get two solutions for given by

Using these values of x in the first equation, we get two values of y

Thus, we get two solutions for the given system of equation

We can see that ( 5, 14 ) is in the options.

The correct option is D).

We will use the method of substitution to solve these equations.

From the first equation, we use the value of in the second equation,

We get an equation in one variable.

Simplifying this, we get

Taking all the terms on one side, we have

Thus, we get a quadratic equation,

factorization to solve We use middle term the above equation.

factorization to solve We use middle term the above equation.

Now, taking common from the first two terms and 2 from the last two terms, we have

So, we get two solutions for given by

Using these values of x in the first equation, we get two values of y

Thus, we get two solutions for the given system of equation

We can see that ( 5, 14 ) is in the options.

The correct option is D).

Maths-

### Find the area of an isosceles trapezium ABCD, if the parallel sides are 6cm and 14 cm, its non parallel sides is 5cm each.

Ans :- 30 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 6 and 14 cm .

Length of non parallel sides is 5 and 5 cm

We get a triangle and parallelogram ABED

we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )

BE = BA-AE = 14- 6 = 8 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 30 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 6 and 14 cm .

Length of non parallel sides is 5 and 5 cm

We get a triangle and parallelogram ABED

we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )

BE = BA-AE = 14- 6 = 8 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 30 cm

^{2}.### Find the area of an isosceles trapezium ABCD, if the parallel sides are 6cm and 14 cm, its non parallel sides is 5cm each.

Maths-General

Ans :- 30 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 6 and 14 cm .

Length of non parallel sides is 5 and 5 cm

We get a triangle and parallelogram ABED

we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )

BE = BA-AE = 14- 6 = 8 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 30 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 6 and 14 cm .

Length of non parallel sides is 5 and 5 cm

We get a triangle and parallelogram ABED

we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )

BE = BA-AE = 14- 6 = 8 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 30 cm

^{2}.