Maths-

General

Easy

Question

# In a school, 𝑡ℎ of the students are boys, if there are 240 girls, find the number of boys in the school?

Hint:

### Find the total number of students and then subtract number of girls from it to

get the total number of boys.

## The correct answer is: 400

### Let the total number of students = x

Number of girls = 240

So, number of boys = x - 240…(i)

Given that of the total students are boys. That is, …(ii)

On equating (i) and (ii) , we get

Multiply by 8 on both sides.

We have

Hence total number of students = 640

From (ii), we get total number of boys

So, number of boys in the school = 400

### Related Questions to study

Maths-

### If 24 pairs of trousers of equal size can be prepared with 54 m of cloth, what length of cloth is required for each pair of trousers?

Let the cloth required for one trouser = x meters

Then the cloth required for 24 equal size trousers = 54 m

That is, 24x = 54

On dividing both the sides by 24, we get

Hence, cloth required for one trouser

Then the cloth required for 24 equal size trousers = 54 m

That is, 24x = 54

On dividing both the sides by 24, we get

Hence, cloth required for one trouser

### If 24 pairs of trousers of equal size can be prepared with 54 m of cloth, what length of cloth is required for each pair of trousers?

Maths-General

Let the cloth required for one trouser = x meters

Then the cloth required for 24 equal size trousers = 54 m

That is, 24x = 54

On dividing both the sides by 24, we get

Hence, cloth required for one trouser

Then the cloth required for 24 equal size trousers = 54 m

That is, 24x = 54

On dividing both the sides by 24, we get

Hence, cloth required for one trouser

Maths-

### Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:

Let’s say x =

Upon dividing 24 by 11 we got

The resultant decimal is a non-terminating and repeating decimal.

Final Answer:

As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of is and it is a non-terminating and repeating decimal.

Upon dividing 24 by 11 we got

The resultant decimal is a non-terminating and repeating decimal.

Final Answer:

As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of is and it is a non-terminating and repeating decimal.

### Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:

Maths-General

Let’s say x =

Upon dividing 24 by 11 we got

The resultant decimal is a non-terminating and repeating decimal.

Final Answer:

As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of is and it is a non-terminating and repeating decimal.

Upon dividing 24 by 11 we got

The resultant decimal is a non-terminating and repeating decimal.

Final Answer:

As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of is and it is a non-terminating and repeating decimal.

Maths-

### By what rational number should be multiplied to obtain ?

**Complete step by step solution:**

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

we get

So, must be multiplied with to get .

**By what rational number should be multiplied to obtain ?**

**Maths-General**

**Complete step by step solution:**

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

we get

So, must be multiplied with to get .

Maths-

### By what rational number should be multiplied to obtain ?

**Complete step by step solution:**

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

we get

So, must be multiplied with to get .

**By what rational number should be multiplied to obtain ?**

**Maths-General**

**Complete step by step solution:**

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

Given rational number = and let the other rational number be x

On multiplication, we get the result to be

That is, we have

On dividing both sides by ,

we get

So, must be multiplied with to get .

### Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:

### Find the cost of 3 m of cloth at Rs 36. per meter.

### Find the cost of 3 m of cloth at Rs 36. per meter.

### The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm

### The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm

### The area enclosed by the circumference of two concentric circles is 364.5 cm

### The area enclosed by the circumference of two concentric circles is 364.5 cm

### The area of a circular ring is 352 m

### The area of a circular ring is 352 m

Maths-

### Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:

Let’s say x =

Upon dividing 59 by 8 we got

The resultant decimal is a terminating decimal

Final Answer:

As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of is 7.375 and it is a terminating decimal.

Upon dividing 59 by 8 we got

The resultant decimal is a terminating decimal

Final Answer:

As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of is 7.375 and it is a terminating decimal.

Maths-General

Let’s say x =

Upon dividing 59 by 8 we got

The resultant decimal is a terminating decimal

Final Answer:

As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of is 7.375 and it is a terminating decimal.

Upon dividing 59 by 8 we got

The resultant decimal is a terminating decimal

Final Answer:

As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of is 7.375 and it is a terminating decimal.

Maths-

### A sim earns Rs. 40000 per month. He spends of his income on food, of the remaining on education of children and of the remainder on the house rent. How much money is still left with him?

Total money earned by Asim = 40000 Rs

Money spent on food from his income = Rs

Amount left with him = 40000 - 16000 = 24000 Rs

Money spent on education of children from the remaining money = Rs

Amount left with him = 24000 - 10000 = 14000 Rs

Money spent on house rent from the remaining money =

Amount left with him = 14000 - 6000 = 8000 Rs

Money still left with him = 8000 Rs

Money spent on food from his income = Rs

Amount left with him = 40000 - 16000 = 24000 Rs

Money spent on education of children from the remaining money = Rs

Amount left with him = 24000 - 10000 = 14000 Rs

Money spent on house rent from the remaining money =

Amount left with him = 14000 - 6000 = 8000 Rs

Money still left with him = 8000 Rs

### A sim earns Rs. 40000 per month. He spends of his income on food, of the remaining on education of children and of the remainder on the house rent. How much money is still left with him?

Maths-General

Total money earned by Asim = 40000 Rs

Money spent on food from his income = Rs

Amount left with him = 40000 - 16000 = 24000 Rs

Money spent on education of children from the remaining money = Rs

Amount left with him = 24000 - 10000 = 14000 Rs

Money spent on house rent from the remaining money =

Amount left with him = 14000 - 6000 = 8000 Rs

Money still left with him = 8000 Rs

Money spent on food from his income = Rs

Amount left with him = 40000 - 16000 = 24000 Rs

Money spent on education of children from the remaining money = Rs

Amount left with him = 24000 - 10000 = 14000 Rs

Money spent on house rent from the remaining money =

Amount left with him = 14000 - 6000 = 8000 Rs

Money still left with him = 8000 Rs

Maths-

### Find the cost of 3 m of cloth at Rs 36. per meter.

Given that the cost of 1 m of cloth

∴ Cost of

= 124.93 Rs

Hence cost of meter = Rs

∴ Cost of

= 124.93 Rs

Hence cost of meter = Rs

### Find the cost of 3 m of cloth at Rs 36. per meter.

Maths-General

Given that the cost of 1 m of cloth

∴ Cost of

= 124.93 Rs

Hence cost of meter = Rs

∴ Cost of

= 124.93 Rs

Hence cost of meter = Rs

Maths-

### Find the area of circular ring whose external and internal radii are 10cm and 3 cm respectively

It is given that External radii, r

Internal radii, r

Area of circular ring = Area of outer ring – Area of inner ring

= (r

= × (r

= × (10

= × 91 = 286 cm

_{1}= 10 cmInternal radii, r

_{2}= 3 cmArea of circular ring = Area of outer ring – Area of inner ring

= (r

_{1})^{2}- r^{2 }= × (r

_{1})^{2}- × (r_{2})^{2}= × (10

^{2}– 3^{2})= × 91 = 286 cm

^{2}### Find the area of circular ring whose external and internal radii are 10cm and 3 cm respectively

Maths-General

It is given that External radii, r

Internal radii, r

Area of circular ring = Area of outer ring – Area of inner ring

= (r

= × (r

= × (10

= × 91 = 286 cm

_{1}= 10 cmInternal radii, r

_{2}= 3 cmArea of circular ring = Area of outer ring – Area of inner ring

= (r

_{1})^{2}- r^{2 }= × (r

_{1})^{2}- × (r_{2})^{2}= × (10

^{2}– 3^{2})= × 91 = 286 cm

^{2}Maths-

### The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm^{2}

It is given that Circumference of the circle = 123.2 cm

(i). Circumference of the circle = 2 r

123.2 = 2 × × r

19.6 = r

(ii) Area of circle = r

= × (19.6)

= 1207.36 cm

(i). Circumference of the circle = 2 r

123.2 = 2 × × r

19.6 = r

(ii) Area of circle = r

^{2}= × (19.6)

^{2}= 1207.36 cm

^{2}### The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm^{2}

Maths-General

It is given that Circumference of the circle = 123.2 cm

(i). Circumference of the circle = 2 r

123.2 = 2 × × r

19.6 = r

(ii) Area of circle = r

= × (19.6)

= 1207.36 cm

(i). Circumference of the circle = 2 r

123.2 = 2 × × r

19.6 = r

(ii) Area of circle = r

^{2}= × (19.6)

^{2}= 1207.36 cm

^{2}Maths-

### The area enclosed by the circumference of two concentric circles is 364.5 cm^{2}. If the circumference of the inner circle is 88cm, calculate the radius of the outer circle.

It is given that Circumference of inner circle = 88 cm

2 r = 88

2 × × r = 8

r = 14 cm

Let radius of outer circle = r

Area enclosed by the circumference of two concentric circles = 364.5 cm

i.e. Area of outer circle – Area of inner circle = 364.5 cm

(r

× (r

((r

((r

(r

r

2 r = 88

2 × × r = 8

r = 14 cm

Let radius of outer circle = r

_{1}Area enclosed by the circumference of two concentric circles = 364.5 cm

^{2}i.e. Area of outer circle – Area of inner circle = 364.5 cm

^{2}(r

_{1})^{2}- r^{2 }= 364.5× (r

_{1})^{2}- × (14)^{2}= 364.5((r

_{1})^{2}– 14^{2}) = 364.5((r

_{1})^{2}– 14^{2}=(r

_{1})^{2}= 311.977r

_{1 }= 17.66 cm### The area enclosed by the circumference of two concentric circles is 364.5 cm^{2}. If the circumference of the inner circle is 88cm, calculate the radius of the outer circle.

Maths-General

It is given that Circumference of inner circle = 88 cm

2 r = 88

2 × × r = 8

r = 14 cm

Let radius of outer circle = r

Area enclosed by the circumference of two concentric circles = 364.5 cm

i.e. Area of outer circle – Area of inner circle = 364.5 cm

(r

× (r

((r

((r

(r

r

2 r = 88

2 × × r = 8

r = 14 cm

Let radius of outer circle = r

_{1}Area enclosed by the circumference of two concentric circles = 364.5 cm

^{2}i.e. Area of outer circle – Area of inner circle = 364.5 cm

^{2}(r

_{1})^{2}- r^{2 }= 364.5× (r

_{1})^{2}- × (14)^{2}= 364.5((r

_{1})^{2}– 14^{2}) = 364.5((r

_{1})^{2}– 14^{2}=(r

_{1})^{2}= 311.977r

_{1 }= 17.66 cmMaths-

### A track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track

It is given that Circumference of inner circle = 352 m

2 r = 352

2 r = 352

r = 56 m

Circumference of outer circle = 396 m

2 r = 396

2 × r = 3

r = 63 m

Width of the track

= Radius of outer circle – Radius of inner circle

= 63 – 56

= 7 m

2 r = 352

2 r = 352

r = 56 m

Circumference of outer circle = 396 m

2 r = 396

2 × r = 3

r = 63 m

Width of the track

= Radius of outer circle – Radius of inner circle

= 63 – 56

= 7 m

### A track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track

Maths-General

It is given that Circumference of inner circle = 352 m

2 r = 352

2 r = 352

r = 56 m

Circumference of outer circle = 396 m

2 r = 396

2 × r = 3

r = 63 m

Width of the track

= Radius of outer circle – Radius of inner circle

= 63 – 56

= 7 m

2 r = 352

2 r = 352

r = 56 m

Circumference of outer circle = 396 m

2 r = 396

2 × r = 3

r = 63 m

Width of the track

= Radius of outer circle – Radius of inner circle

= 63 – 56

= 7 m

Maths-

### The area of a circular ring is 352 m^{2}. If the diameter of its outer circle is 32 m, find the diameter of its inner circle.

It is given that Area of circular ring = 352 m

Diameter of outer circle = 32 m

Radius of outer circle, r

Now, we know that area of the ring is difference of the areas of two circle

Let radius of outer circle be r

i.e. Area of ring = Area of outer circle – Area of inner circle

352 = (r

352 = × (16)

352 = × (16

112 = 256 – r

r

r =

r = 12

Diameter = 2 radius = 2 12 = 24 m

^{2}Diameter of outer circle = 32 m

Radius of outer circle, r

_{1}= = = 16 mNow, we know that area of the ring is difference of the areas of two circle

Let radius of outer circle be r

i.e. Area of ring = Area of outer circle – Area of inner circle

352 = (r

_{1})^{2}- r^{2}352 = × (16)

^{2}- × (r)^{2}352 = × (16

^{2}– r^{2})112 = 256 – r

^{2}r

^{2}= 256 – 112 = 144r =

r = 12

Diameter = 2 radius = 2 12 = 24 m

### The area of a circular ring is 352 m^{2}. If the diameter of its outer circle is 32 m, find the diameter of its inner circle.

Maths-General

It is given that Area of circular ring = 352 m

Diameter of outer circle = 32 m

Radius of outer circle, r

Now, we know that area of the ring is difference of the areas of two circle

Let radius of outer circle be r

i.e. Area of ring = Area of outer circle – Area of inner circle

352 = (r

352 = × (16)

352 = × (16

112 = 256 – r

r

r =

r = 12

Diameter = 2 radius = 2 12 = 24 m

^{2}Diameter of outer circle = 32 m

Radius of outer circle, r

_{1}= = = 16 mNow, we know that area of the ring is difference of the areas of two circle

Let radius of outer circle be r

i.e. Area of ring = Area of outer circle – Area of inner circle

352 = (r

_{1})^{2}- r^{2}352 = × (16)

^{2}- × (r)^{2}352 = × (16

^{2}– r^{2})112 = 256 – r

^{2}r

^{2}= 256 – 112 = 144r =

r = 12

Diameter = 2 radius = 2 12 = 24 m

Maths-

### The diameter of a circular park is 94.4 m. A 2.8 m wide road is constructed outside this circular park, i.e., around it. Find the cost of paving this road at the rate of Rs. 50 square meter

Since the road of width 2.8 m is on outside , it means

Diameter of inner circle = 94.4 m

Radius = = = 47.2 m

We know that area of a circle = p r

Radius of outer circle = Radius of inner circle + Width of road

= 47.2 + 2.8 = 50 m

Area of the road = Area of outer circle – Area of inner circle

= × (50)

= × ( 50 – 47.2)(50 + 47.2)

( a

= 2.8 × 97.2

= 855.36 m

Now, Cost of constructing 1 m

Cost of constructing 855.36 m

= Rs. 42768

Diameter of inner circle = 94.4 m

Radius = = = 47.2 m

We know that area of a circle = p r

^{2}Radius of outer circle = Radius of inner circle + Width of road

= 47.2 + 2.8 = 50 m

Area of the road = Area of outer circle – Area of inner circle

= × (50)

^{2}- × (47.2)^{2}= × ( 50 – 47.2)(50 + 47.2)

( a

^{2}– b^{2}) = (a – b)(a + b)= 2.8 × 97.2

= 855.36 m

^{2}Now, Cost of constructing 1 m

^{2}road = Rs. 5Cost of constructing 855.36 m

^{2}road = Rs. 50 855.36= Rs. 42768

### The diameter of a circular park is 94.4 m. A 2.8 m wide road is constructed outside this circular park, i.e., around it. Find the cost of paving this road at the rate of Rs. 50 square meter

Maths-General

Since the road of width 2.8 m is on outside , it means

Diameter of inner circle = 94.4 m

Radius = = = 47.2 m

We know that area of a circle = p r

Radius of outer circle = Radius of inner circle + Width of road

= 47.2 + 2.8 = 50 m

Area of the road = Area of outer circle – Area of inner circle

= × (50)

= × ( 50 – 47.2)(50 + 47.2)

( a

= 2.8 × 97.2

= 855.36 m

Now, Cost of constructing 1 m

Cost of constructing 855.36 m

= Rs. 42768

Diameter of inner circle = 94.4 m

Radius = = = 47.2 m

We know that area of a circle = p r

^{2}Radius of outer circle = Radius of inner circle + Width of road

= 47.2 + 2.8 = 50 m

Area of the road = Area of outer circle – Area of inner circle

= × (50)

^{2}- × (47.2)^{2}= × ( 50 – 47.2)(50 + 47.2)

( a

^{2}– b^{2}) = (a – b)(a + b)= 2.8 × 97.2

= 855.36 m

^{2}Now, Cost of constructing 1 m

^{2}road = Rs. 5Cost of constructing 855.36 m

^{2}road = Rs. 50 855.36= Rs. 42768

Maths-

### The diameter of a circular park is 104 m. A road of width 3.5 m is on the outside around it. Find the cost of constructing the road at Rs. 15 per square meter.

Since the road of width 3.5 m is on outside , it means

Diameter of inner circle = 104 m

Radius = = = 52 m

Radius of outer circle = Radius of inner circle + Width of road

= 52 + 3.5 = 55.5 m

We know that area of a circle = r

Area of the road = Area of outer circle – Area of inner circle

= × (55.5)

= × ( 55.5 – 52)(55.5 + 52)

( a

= 3.5 × 107.5

= 1182.5 m

Now, Cost of constructing 1 m

Cost of constructing 1182.5 m

= Rs. 17737.5

Diameter of inner circle = 104 m

Radius = = = 52 m

Radius of outer circle = Radius of inner circle + Width of road

= 52 + 3.5 = 55.5 m

We know that area of a circle = r

^{2}Area of the road = Area of outer circle – Area of inner circle

= × (55.5)

^{2}- × 52= × ( 55.5 – 52)(55.5 + 52)

( a

^{2}– b^{2}) = (a – b)(a + b)= 3.5 × 107.5

= 1182.5 m

^{2}Now, Cost of constructing 1 m

^{2}road = Rs. 15Cost of constructing 1182.5 m

^{2}road = Rs. 15 × 1182.5= Rs. 17737.5

### The diameter of a circular park is 104 m. A road of width 3.5 m is on the outside around it. Find the cost of constructing the road at Rs. 15 per square meter.

Maths-General

Since the road of width 3.5 m is on outside , it means

Diameter of inner circle = 104 m

Radius = = = 52 m

Radius of outer circle = Radius of inner circle + Width of road

= 52 + 3.5 = 55.5 m

We know that area of a circle = r

Area of the road = Area of outer circle – Area of inner circle

= × (55.5)

= × ( 55.5 – 52)(55.5 + 52)

( a

= 3.5 × 107.5

= 1182.5 m

Now, Cost of constructing 1 m

Cost of constructing 1182.5 m

= Rs. 17737.5

Diameter of inner circle = 104 m

Radius = = = 52 m

Radius of outer circle = Radius of inner circle + Width of road

= 52 + 3.5 = 55.5 m

We know that area of a circle = r

^{2}Area of the road = Area of outer circle – Area of inner circle

= × (55.5)

^{2}- × 52= × ( 55.5 – 52)(55.5 + 52)

( a

^{2}– b^{2}) = (a – b)(a + b)= 3.5 × 107.5

= 1182.5 m

^{2}Now, Cost of constructing 1 m

^{2}road = Rs. 15Cost of constructing 1182.5 m

^{2}road = Rs. 15 × 1182.5= Rs. 17737.5

Maths-

### A wire is in a circular shape of radius 35 cm. if it is bent in the form of square, find the area of square so formed.

It is given that radius of the circle = 35 cm

Circumference of the circle = 2 r

= 2 × ×35

= 220 cm

Since same wire is used, circumference/perimeter of both the shapes will be equal.

Perimeter of square = Circumference of the circle

4 × (side) = 220 cm

Side = 55 cm

Now, Area of square = (side)

= 55

Circumference of the circle = 2 r

= 2 × ×35

= 220 cm

Since same wire is used, circumference/perimeter of both the shapes will be equal.

Perimeter of square = Circumference of the circle

4 × (side) = 220 cm

Side = 55 cm

Now, Area of square = (side)

^{2}= 55

^{2}= 3025 cm^{2}### A wire is in a circular shape of radius 35 cm. if it is bent in the form of square, find the area of square so formed.

Maths-General

It is given that radius of the circle = 35 cm

Circumference of the circle = 2 r

= 2 × ×35

= 220 cm

Since same wire is used, circumference/perimeter of both the shapes will be equal.

Perimeter of square = Circumference of the circle

4 × (side) = 220 cm

Side = 55 cm

Now, Area of square = (side)

= 55

Circumference of the circle = 2 r

= 2 × ×35

= 220 cm

Since same wire is used, circumference/perimeter of both the shapes will be equal.

Perimeter of square = Circumference of the circle

4 × (side) = 220 cm

Side = 55 cm

Now, Area of square = (side)

^{2}= 55

^{2}= 3025 cm^{2}

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