Maths-

General

Easy

Question

# Venu has as many sisters as he has brothers. If Karuna, Venu’s sister has thrice as many brothers as she has sisters, then Venu has how many sisters?

Hint:

### let no. of sisters for venu be x

By condition no. of brothers of venu is also x

If Karuna, Venu’s sister has thrice as many brothers as she has sisters

Find no. of karuna’s brothers and no. of karuna’s sisters and equate them as per above condition.

## The correct answer is: 2 sisters

### Ans :- Venu has 2 sisters

Explanation :-

let no. of sisters for venu be x

By condition no. of brothers of venu is also x

Step 1:-Find no. of karuna’s brothers and no. of karuna’s sisters

No. of sisters of karuna is no. of sister of venu except herself

No. of karunas sisters = no. of sister for venu - 1 = x -1

No. of brothers of karuna are no. of brothers of venu including venu.

No. of brothers of karuna = x + 1

Step 2:-equate the condition

If Karuna, Venu’s sister has thrice as many brothers as she has sisters

Now ,

∴venu has 2 sisters.

### Related Questions to study

Maths-

### A pit 10 m long and 2 m wide is filled with 4.8 cubic m. of sand. What is the depth of the pit?

Hint:

A pit is a rectangular cylinder. So, we use as the formula for volume of the pit where are the length, width and depth of the pit, respectively and solve the problem.

Explanations:

Step 1 of 1:

The parameters of the pit is given by, = 10; b = 2 and volume = 4.8

Therefore,

Final Answer:

The depth of the pit is 0.24 m

A pit is a rectangular cylinder. So, we use as the formula for volume of the pit where are the length, width and depth of the pit, respectively and solve the problem.

Explanations:

Step 1 of 1:

The parameters of the pit is given by, = 10; b = 2 and volume = 4.8

Therefore,

Final Answer:

The depth of the pit is 0.24 m

### A pit 10 m long and 2 m wide is filled with 4.8 cubic m. of sand. What is the depth of the pit?

Maths-General

Hint:

A pit is a rectangular cylinder. So, we use as the formula for volume of the pit where are the length, width and depth of the pit, respectively and solve the problem.

Explanations:

Step 1 of 1:

The parameters of the pit is given by, = 10; b = 2 and volume = 4.8

Therefore,

Final Answer:

The depth of the pit is 0.24 m

A pit is a rectangular cylinder. So, we use as the formula for volume of the pit where are the length, width and depth of the pit, respectively and solve the problem.

Explanations:

Step 1 of 1:

The parameters of the pit is given by, = 10; b = 2 and volume = 4.8

Therefore,

Final Answer:

The depth of the pit is 0.24 m

Maths-

### The circumference of Earth is estimated to be 40,030 kilometers at the equator. Which of the following best approximates the diameter, in miles, of Earth's equator? (1 kilometer miles)

Given,

Circumference of the earth at the equator = 40,030 kilomteres

The equator can be considered as a circle,

And the circumference of a circle = , where r is the radius of the circle.

As, diameter of a circle is twice the radius, that is, diameter = 2r ,

So we get

cicum ference diameter

Thus,

Putting the value of the circumference in the above equation

Using the calculator, we get

Thus, we get

Diameter of the earth 12741.94 kilometres

It is given that,

This is the unit conversion factor.

Thus, we have

= 7917.45

Thus, the correct option is C)

Circumference of the earth at the equator = 40,030 kilomteres

The equator can be considered as a circle,

And the circumference of a circle = , where r is the radius of the circle.

As, diameter of a circle is twice the radius, that is, diameter = 2r ,

So we get

cicum ference diameter

Thus,

Putting the value of the circumference in the above equation

Using the calculator, we get

Thus, we get

Diameter of the earth 12741.94 kilometres

It is given that,

This is the unit conversion factor.

Thus, we have

= 7917.45

Thus, the correct option is C)

### The circumference of Earth is estimated to be 40,030 kilometers at the equator. Which of the following best approximates the diameter, in miles, of Earth's equator? (1 kilometer miles)

Maths-General

Given,

Circumference of the earth at the equator = 40,030 kilomteres

The equator can be considered as a circle,

And the circumference of a circle = , where r is the radius of the circle.

As, diameter of a circle is twice the radius, that is, diameter = 2r ,

So we get

cicum ference diameter

Thus,

Putting the value of the circumference in the above equation

Using the calculator, we get

Thus, we get

Diameter of the earth 12741.94 kilometres

It is given that,

This is the unit conversion factor.

Thus, we have

= 7917.45

Thus, the correct option is C)

Circumference of the earth at the equator = 40,030 kilomteres

The equator can be considered as a circle,

And the circumference of a circle = , where r is the radius of the circle.

As, diameter of a circle is twice the radius, that is, diameter = 2r ,

So we get

cicum ference diameter

Thus,

Putting the value of the circumference in the above equation

Using the calculator, we get

Thus, we get

Diameter of the earth 12741.94 kilometres

It is given that,

This is the unit conversion factor.

Thus, we have

= 7917.45

Thus, the correct option is C)

Maths-

### Length of the fence of a trapezium shaped field ABCD is 120m. If BC = 48m , CD = 17m and AD = 40m , Find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans :-660 m

Explanation :-

Step 1:- Find the length of AB from perimeter and other given side lengths.

Perimeter of trapezium = 120 m

Given BC = 48 m ;CD = 17 m and AD = 40 m.

Side AB is perpendicular to the parallel sides AD and BC

Perimeter of trapezium = AB+BC+CD+DA

Step 2:- find the area of trapezium

= 660 m

Therefore, Area of trapezium shaped field ABCD = 660 m

^{2}.Explanation :-

Step 1:- Find the length of AB from perimeter and other given side lengths.

Perimeter of trapezium = 120 m

Given BC = 48 m ;CD = 17 m and AD = 40 m.

Side AB is perpendicular to the parallel sides AD and BC

Perimeter of trapezium = AB+BC+CD+DA

Step 2:- find the area of trapezium

= 660 m

^{2}Therefore, Area of trapezium shaped field ABCD = 660 m

^{2}.### Length of the fence of a trapezium shaped field ABCD is 120m. If BC = 48m , CD = 17m and AD = 40m , Find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Maths-General

Ans :-660 m

Explanation :-

Step 1:- Find the length of AB from perimeter and other given side lengths.

Perimeter of trapezium = 120 m

Given BC = 48 m ;CD = 17 m and AD = 40 m.

Side AB is perpendicular to the parallel sides AD and BC

Perimeter of trapezium = AB+BC+CD+DA

Step 2:- find the area of trapezium

= 660 m

Therefore, Area of trapezium shaped field ABCD = 660 m

^{2}.Explanation :-

Step 1:- Find the length of AB from perimeter and other given side lengths.

Perimeter of trapezium = 120 m

Given BC = 48 m ;CD = 17 m and AD = 40 m.

Side AB is perpendicular to the parallel sides AD and BC

Perimeter of trapezium = AB+BC+CD+DA

Step 2:- find the area of trapezium

= 660 m

^{2}Therefore, Area of trapezium shaped field ABCD = 660 m

^{2}.Maths-

### Mukesh has some goats and hens in his shed. Upon counting, Mukesh found that the total number of legs is 112 and the total number of heads is 40. Find the number of hens in his shed.

Ans :- There are 24 hens in the shed

Explanation :-

let no. of hens be x and no. of goats be y .

No. of legs for each hen = 2 ;No. of legs for each goat = 4

No. of heads for each hen = 1 ;No. of heads for each goat = 1

Step 1:- Form the equation based on given conditions

Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112

x No. of legs for each hen + y No. of legs for each goat = 112

— Eq1

Total no. of heads = No. of heads for x hens + No. of head for y goats = 40

x No. of heads for each hen + y No. of heads for each goat = 40

—Eq2

Step 2:- eliminate y to find x

Doing 2(Eq2) -(Eq1)

∴There are 24 hens in the shed

Explanation :-

let no. of hens be x and no. of goats be y .

No. of legs for each hen = 2 ;No. of legs for each goat = 4

No. of heads for each hen = 1 ;No. of heads for each goat = 1

Step 1:- Form the equation based on given conditions

Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112

x No. of legs for each hen + y No. of legs for each goat = 112

— Eq1

Total no. of heads = No. of heads for x hens + No. of head for y goats = 40

x No. of heads for each hen + y No. of heads for each goat = 40

—Eq2

Step 2:- eliminate y to find x

Doing 2(Eq2) -(Eq1)

∴x = 24 |

### Mukesh has some goats and hens in his shed. Upon counting, Mukesh found that the total number of legs is 112 and the total number of heads is 40. Find the number of hens in his shed.

Maths-General

Ans :- There are 24 hens in the shed

Explanation :-

let no. of hens be x and no. of goats be y .

No. of legs for each hen = 2 ;No. of legs for each goat = 4

No. of heads for each hen = 1 ;No. of heads for each goat = 1

Step 1:- Form the equation based on given conditions

Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112

x No. of legs for each hen + y No. of legs for each goat = 112

— Eq1

Total no. of heads = No. of heads for x hens + No. of head for y goats = 40

x No. of heads for each hen + y No. of heads for each goat = 40

—Eq2

Step 2:- eliminate y to find x

Doing 2(Eq2) -(Eq1)

∴There are 24 hens in the shed

Explanation :-

let no. of hens be x and no. of goats be y .

No. of legs for each hen = 2 ;No. of legs for each goat = 4

No. of heads for each hen = 1 ;No. of heads for each goat = 1

Step 1:- Form the equation based on given conditions

Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112

x No. of legs for each hen + y No. of legs for each goat = 112

— Eq1

Total no. of heads = No. of heads for x hens + No. of head for y goats = 40

x No. of heads for each hen + y No. of heads for each goat = 40

—Eq2

Step 2:- eliminate y to find x

Doing 2(Eq2) -(Eq1)

∴x = 24 |

Maths-

### The capacity of a cup is 300 ml. The capacity of a jug is twice that of a cup. What is the capacity of 3 cups and 3 jugs?

Hint:

We simply find out the capacity of a jug and solve the problem using multiplication.

Explanations:

Step 1 of 3:

The capacity of a cup is 300 ml.

The capacity of a jug is twice that of a cup, hence the capacity of a jug is ml.

Step 2 of 3:

The capacity of 3 jugs is

The capacity of 3 jugs is

Step 3 of 3:

The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)

Final Answer:

The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr

We simply find out the capacity of a jug and solve the problem using multiplication.

Explanations:

Step 1 of 3:

The capacity of a cup is 300 ml.

The capacity of a jug is twice that of a cup, hence the capacity of a jug is ml.

Step 2 of 3:

The capacity of 3 jugs is

The capacity of 3 jugs is

Step 3 of 3:

The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)

Final Answer:

The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr

### The capacity of a cup is 300 ml. The capacity of a jug is twice that of a cup. What is the capacity of 3 cups and 3 jugs?

Maths-General

Hint:

We simply find out the capacity of a jug and solve the problem using multiplication.

Explanations:

Step 1 of 3:

The capacity of a cup is 300 ml.

The capacity of a jug is twice that of a cup, hence the capacity of a jug is ml.

Step 2 of 3:

The capacity of 3 jugs is

The capacity of 3 jugs is

Step 3 of 3:

The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)

Final Answer:

The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr

We simply find out the capacity of a jug and solve the problem using multiplication.

Explanations:

Step 1 of 3:

The capacity of a cup is 300 ml.

The capacity of a jug is twice that of a cup, hence the capacity of a jug is ml.

Step 2 of 3:

The capacity of 3 jugs is

The capacity of 3 jugs is

Step 3 of 3:

The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)

Final Answer:

The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr

Maths-

### A mother said to her son, ‘the sum of our present ages is twice my age 12 years ago and nine years hence, the sum of our ages will be thrice my age 14 years ago’. What is her son’s present age?

Ans :- present age of son is 12 years

Explanation :-

Step 1:- find the equation by 1st condition.

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

The sum of our present ages is twice my age 12 years ago.

x + y =2( y-12) [ the age of 12 years ago is present age -12 years]

— Eq1

Step 2:- find the equation by applying second condition

After 9 years the sons age = x+9

After 9 years the mothers age = y+9

Mother age 14 years ago = present age - 14 years = y - 14

After 9 years ,the sum of our ages will be thrice my age 14 years ago

— Eq2

Step 3:- find value of x by eliminating y

Doing 2(Eq1)-Eq2 to eliminate y

∴ present age of son = x= 12 years.

Explanation :-

Step 1:- find the equation by 1st condition.

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

The sum of our present ages is twice my age 12 years ago.

x + y =2( y-12) [ the age of 12 years ago is present age -12 years]

— Eq1

Step 2:- find the equation by applying second condition

After 9 years the sons age = x+9

After 9 years the mothers age = y+9

Mother age 14 years ago = present age - 14 years = y - 14

After 9 years ,the sum of our ages will be thrice my age 14 years ago

— Eq2

Step 3:- find value of x by eliminating y

Doing 2(Eq1)-Eq2 to eliminate y

### A mother said to her son, ‘the sum of our present ages is twice my age 12 years ago and nine years hence, the sum of our ages will be thrice my age 14 years ago’. What is her son’s present age?

Maths-General

Ans :- present age of son is 12 years

Explanation :-

Step 1:- find the equation by 1st condition.

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

The sum of our present ages is twice my age 12 years ago.

x + y =2( y-12) [ the age of 12 years ago is present age -12 years]

— Eq1

Step 2:- find the equation by applying second condition

After 9 years the sons age = x+9

After 9 years the mothers age = y+9

Mother age 14 years ago = present age - 14 years = y - 14

After 9 years ,the sum of our ages will be thrice my age 14 years ago

— Eq2

Step 3:- find value of x by eliminating y

Doing 2(Eq1)-Eq2 to eliminate y

∴ present age of son = x= 12 years.

Explanation :-

Step 1:- find the equation by 1st condition.

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

The sum of our present ages is twice my age 12 years ago.

x + y =2( y-12) [ the age of 12 years ago is present age -12 years]

— Eq1

Step 2:- find the equation by applying second condition

After 9 years the sons age = x+9

After 9 years the mothers age = y+9

Mother age 14 years ago = present age - 14 years = y - 14

After 9 years ,the sum of our ages will be thrice my age 14 years ago

— Eq2

Step 3:- find value of x by eliminating y

Doing 2(Eq1)-Eq2 to eliminate y

Maths-

### The cross section of a canal is a trapezium is shape. If the canal is 10m wide at the top and 6m wide at the bottom and the area of cross section is 840sq.m. Find the depth of the canal?

Ans :- 105 m.

Explanation :-

Given the parallel sides of trapezium are 10 cm and 6 cm.

Sum of lengths of parallel sides =10+6 =16 cm

Area of trapezium = 840 sq. m

Let height or depth of cross section be h m

Therefore ,the depth of the canal is 105 m .

Explanation :-

Given the parallel sides of trapezium are 10 cm and 6 cm.

Sum of lengths of parallel sides =10+6 =16 cm

Area of trapezium = 840 sq. m

Let height or depth of cross section be h m

Therefore ,the depth of the canal is 105 m .

### The cross section of a canal is a trapezium is shape. If the canal is 10m wide at the top and 6m wide at the bottom and the area of cross section is 840sq.m. Find the depth of the canal?

Maths-General

Ans :- 105 m.

Explanation :-

Given the parallel sides of trapezium are 10 cm and 6 cm.

Sum of lengths of parallel sides =10+6 =16 cm

Area of trapezium = 840 sq. m

Let height or depth of cross section be h m

Therefore ,the depth of the canal is 105 m .

Explanation :-

Given the parallel sides of trapezium are 10 cm and 6 cm.

Sum of lengths of parallel sides =10+6 =16 cm

Area of trapezium = 840 sq. m

Let height or depth of cross section be h m

Therefore ,the depth of the canal is 105 m .

Maths-

### A certain colony of bacteria began with one cell, and the population doubled every 20 minutes. What was the population of the colony after 2 hours?

The relation between hours and minutes is given by

1 hour = 60 minutes

This can be written as

1 hour =

So, we get

2 hours = 2.

That is , 2 hours have 6 sessions of 20 minutes.

Given, the initial population is 1.

Also, the population doubles every 20 minutes.

And this happens 6 times,

So the population of the colony after 2 hours

64

Thus, the correct option is D)

1 hour = 60 minutes

This can be written as

1 hour =

So, we get

2 hours = 2.

That is , 2 hours have 6 sessions of 20 minutes.

Given, the initial population is 1.

Also, the population doubles every 20 minutes.

And this happens 6 times,

So the population of the colony after 2 hours

64

Thus, the correct option is D)

### A certain colony of bacteria began with one cell, and the population doubled every 20 minutes. What was the population of the colony after 2 hours?

Maths-General

The relation between hours and minutes is given by

1 hour = 60 minutes

This can be written as

1 hour =

So, we get

2 hours = 2.

That is , 2 hours have 6 sessions of 20 minutes.

Given, the initial population is 1.

Also, the population doubles every 20 minutes.

And this happens 6 times,

So the population of the colony after 2 hours

64

Thus, the correct option is D)

1 hour = 60 minutes

This can be written as

1 hour =

So, we get

2 hours = 2.

That is , 2 hours have 6 sessions of 20 minutes.

Given, the initial population is 1.

Also, the population doubles every 20 minutes.

And this happens 6 times,

So the population of the colony after 2 hours

64

Thus, the correct option is D)

Maths-

### A Rectangular sheet of paper, 36 cm x 22 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed?

Hint:

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(=

Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 5.73 cm(=

Step 3 of 3:

The volume of the formed cylinder is cm

Final Answer:

The volume of the cylinder so formed is 2770.16 cm

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(=

*h*), and the circumference of the cylinder base would be 36 cm.Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 5.73 cm(=

*r*).Step 3 of 3:

The volume of the formed cylinder is cm

^{3}Final Answer:

The volume of the cylinder so formed is 2770.16 cm

^{3}.### A Rectangular sheet of paper, 36 cm x 22 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed?

Maths-General

Hint:

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(=

Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 5.73 cm(=

Step 3 of 3:

The volume of the formed cylinder is cm

Final Answer:

The volume of the cylinder so formed is 2770.16 cm

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(=

*h*), and the circumference of the cylinder base would be 36 cm.Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 5.73 cm(=

*r*).Step 3 of 3:

The volume of the formed cylinder is cm

^{3}Final Answer:

The volume of the cylinder so formed is 2770.16 cm

^{3}.Maths-

### The parallel sides of a trapezium are 20cm and 10cm. Its non parallel sides are both equal. Each being 13cm. Find the area of the trapezium ?

Ans :- 180 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 20 cm .

Length of non parallel sides is 13 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 20 - 10 = 10 cm .

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

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 180 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 20 cm .

Length of non parallel sides is 13 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 20 - 10 = 10 cm .

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

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 180 cm

^{2}.### The parallel sides of a trapezium are 20cm and 10cm. Its non parallel sides are both equal. Each being 13cm. Find the area of the trapezium ?

Maths-General

Ans :- 180 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 20 cm .

Length of non parallel sides is 13 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 20 - 10 = 10 cm .

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

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 180 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 20 cm .

Length of non parallel sides is 13 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 20 - 10 = 10 cm .

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

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 180 cm

^{2}.Maths-

### In a fraction, if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1. Instead, if the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8. Find the fraction.

Ans :- 8/13 is the value of fraction .

Explanation :-

Step 1:- Frame the equation based on given conditions.

let the numerator of fraction be x and denominator of fraction is y

if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1

We get — Eq1

If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.

We get

— Eq2

Step 2:- eliminate the y and find x

Doing Eq2 -3(Eq1) to eliminate y .

Step 3:- find y by substituting the value of x in eq1.

We know the fraction is x/y =

Explanation :-

Step 1:- Frame the equation based on given conditions.

let the numerator of fraction be x and denominator of fraction is y

if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1

We get — Eq1

If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.

We get

— Eq2

Step 2:- eliminate the y and find x

Doing Eq2 -3(Eq1) to eliminate y .

∴x = 8 |

∴y = 13 |

### In a fraction, if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1. Instead, if the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8. Find the fraction.

Maths-General

Ans :- 8/13 is the value of fraction .

Explanation :-

Step 1:- Frame the equation based on given conditions.

let the numerator of fraction be x and denominator of fraction is y

if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1

We get — Eq1

If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.

We get

— Eq2

Step 2:- eliminate the y and find x

Doing Eq2 -3(Eq1) to eliminate y .

Step 3:- find y by substituting the value of x in eq1.

We know the fraction is x/y =

Explanation :-

Step 1:- Frame the equation based on given conditions.

let the numerator of fraction be x and denominator of fraction is y

if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1

We get — Eq1

If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.

We get

— Eq2

Step 2:- eliminate the y and find x

Doing Eq2 -3(Eq1) to eliminate y .

∴x = 8 |

∴y = 13 |

Maths-

### What is the volume of a cylindrical container with base radius 5 cm and height 20 cm?

Hint:

The volume of a cylindrical container with base radius r and height h, is cubic units.

Explanations:

Step 1 of 1:

Given, r = 5 and h = 20.

Therefore, the volume of the cylindrical container is given by

Final Answer:

The required volume of the cylindrical container is 1571.43 cm3

The volume of a cylindrical container with base radius r and height h, is cubic units.

Explanations:

Step 1 of 1:

Given, r = 5 and h = 20.

Therefore, the volume of the cylindrical container is given by

Final Answer:

The required volume of the cylindrical container is 1571.43 cm3

### What is the volume of a cylindrical container with base radius 5 cm and height 20 cm?

Maths-General

Hint:

The volume of a cylindrical container with base radius r and height h, is cubic units.

Explanations:

Step 1 of 1:

Given, r = 5 and h = 20.

Therefore, the volume of the cylindrical container is given by

Final Answer:

The required volume of the cylindrical container is 1571.43 cm3

The volume of a cylindrical container with base radius r and height h, is cubic units.

Explanations:

Step 1 of 1:

Given, r = 5 and h = 20.

Therefore, the volume of the cylindrical container is given by

Final Answer:

The required volume of the cylindrical container is 1571.43 cm3

Maths-

### In the x - y plane, the graph of line has slope 3 . Line k is parallel to line and contains the point ( 3, 10 ). Which of the following is an equation of line k ?

Given,

Slope of line = 3

As, line is parallel to , their slopes are equal.

So, slope of line k (m) = 3

Also, passes through the point (3, 10)

The equation of a line in point-slope form is given by

where, m is the slope of the line and

( a, b ) is any point lying on the line.

Using the above formula, the equation of line is given by

Simplifying the above equation, we have

Adding 10 both sides, we get

Finally, we have

Hence, the equation of line k is y = 3x + 1

Thus, the correct option is D)

Slope of line = 3

As, line is parallel to , their slopes are equal.

So, slope of line k (m) = 3

Also, passes through the point (3, 10)

The equation of a line in point-slope form is given by

where, m is the slope of the line and

( a, b ) is any point lying on the line.

Using the above formula, the equation of line is given by

Simplifying the above equation, we have

Adding 10 both sides, we get

Finally, we have

Hence, the equation of line k is y = 3x + 1

Thus, the correct option is D)

### In the x - y plane, the graph of line has slope 3 . Line k is parallel to line and contains the point ( 3, 10 ). Which of the following is an equation of line k ?

Maths-General

Given,

Slope of line = 3

As, line is parallel to , their slopes are equal.

So, slope of line k (m) = 3

Also, passes through the point (3, 10)

The equation of a line in point-slope form is given by

where, m is the slope of the line and

( a, b ) is any point lying on the line.

Using the above formula, the equation of line is given by

Simplifying the above equation, we have

Adding 10 both sides, we get

Finally, we have

Hence, the equation of line k is y = 3x + 1

Thus, the correct option is D)

Slope of line = 3

As, line is parallel to , their slopes are equal.

So, slope of line k (m) = 3

Also, passes through the point (3, 10)

The equation of a line in point-slope form is given by

where, m is the slope of the line and

( a, b ) is any point lying on the line.

Using the above formula, the equation of line is given by

Simplifying the above equation, we have

Adding 10 both sides, we get

Finally, we have

Hence, the equation of line k is y = 3x + 1

Thus, the correct option is D)

Maths-

### The ratio of the lengths of the parallel sides of a trapezium is 3:2. The shortest distance between them is 15cm. If the area of the trapezium is 450 sq. cm. Find the sum of the lengths of the parallel sides ?

Ans :- 60 cm

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x>y).

Given ratio of x:y = 3:2

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 450 sq.cm

Altitude (or) height = 15 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We know Sum of lengths of parallel sides =

So, Sum of lengths of parallel sides = 60 cm.

Therefore, Sum of lengths of parallel sides is 60 cm.

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x>y).

Given ratio of x:y = 3:2

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 450 sq.cm

Altitude (or) height = 15 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We know Sum of lengths of parallel sides =

So, Sum of lengths of parallel sides = 60 cm.

Therefore, Sum of lengths of parallel sides is 60 cm.

### The ratio of the lengths of the parallel sides of a trapezium is 3:2. The shortest distance between them is 15cm. If the area of the trapezium is 450 sq. cm. Find the sum of the lengths of the parallel sides ?

Maths-General

Ans :- 60 cm

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x>y).

Given ratio of x:y = 3:2

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 450 sq.cm

Altitude (or) height = 15 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We know Sum of lengths of parallel sides =

So, Sum of lengths of parallel sides = 60 cm.

Therefore, Sum of lengths of parallel sides is 60 cm.

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x>y).

Given ratio of x:y = 3:2

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 450 sq.cm

Altitude (or) height = 15 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We know Sum of lengths of parallel sides =

So, Sum of lengths of parallel sides = 60 cm.

Therefore, Sum of lengths of parallel sides is 60 cm.

Maths-

### The sum of two numbers is 52 and their difference is 20. find the numbers?

Ans :- 16 and 36 are the numbers

Explanation :-

Step 1:-Frame the system of linear equations based on the

let the two number be x and y where y > x

Given sum of = 52

x + y =52 — Eq1

Given difference = 20

Difference is bigger number y - smaller number x = 20

y = x = 20 — Eq2

Step 2:- eliminate the x and find y

Doing Eq2 +Eq1 to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴16 and 36 are the numbers which satisfy the given conditions.

Explanation :-

Step 1:-Frame the system of linear equations based on the

let the two number be x and y where y > x

Given sum of = 52

x + y =52 — Eq1

Given difference = 20

Difference is bigger number y - smaller number x = 20

y = x = 20 — Eq2

Step 2:- eliminate the x and find y

Doing Eq2 +Eq1 to eliminate x .

∴y = 36 |

∴x = 16 |

### The sum of two numbers is 52 and their difference is 20. find the numbers?

Maths-General

Ans :- 16 and 36 are the numbers

Explanation :-

Step 1:-Frame the system of linear equations based on the

let the two number be x and y where y > x

Given sum of = 52

x + y =52 — Eq1

Given difference = 20

Difference is bigger number y - smaller number x = 20

y = x = 20 — Eq2

Step 2:- eliminate the x and find y

Doing Eq2 +Eq1 to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴16 and 36 are the numbers which satisfy the given conditions.

Explanation :-

Step 1:-Frame the system of linear equations based on the

let the two number be x and y where y > x

Given sum of = 52

x + y =52 — Eq1

Given difference = 20

Difference is bigger number y - smaller number x = 20

y = x = 20 — Eq2

Step 2:- eliminate the x and find y

Doing Eq2 +Eq1 to eliminate x .

∴y = 36 |

∴x = 16 |