Maths-

General

Easy

Question

# What is the number that is to be subtracted from 𝑡𝑜 𝑔𝑒𝑡

Hint:

### Frame the equation according to the question and then get the number to be

subtracted

## The correct answer is: 5/24

### Given rational number = and let the other rational number to be subtracted from

= x

On subtraction, we get the result to be

That is, we have

On rearranging the equation, we get

On taking LCM of 12 and 8 , we have 24 as the LCM.

### Related Questions to study

Maths-

### The circle touch each other internally. The sum of their area is 116 pie cm^{2} and the distance between their centres is 6 cm. Find the radii of the circles.

It is given that the distance between their centres is 6 cm

r

The sum of their area of two circles = 116 cm

(r

[ (r

(r

2 (r

2 (r

Using splitting the middle term,

We get, (r

(r

Since radius is always positive, r

r

r

_{2}- r_{1 }= 6 r_{2}= 6 + r_{1}The sum of their area of two circles = 116 cm

^{2}(r

_{1})^{2}+ (r_{2})^{2}= 116[ (r

_{1})^{2}+ (6 + r_{1})^{2}] = 116(r

_{1})^{2}+ 36 + (r_{1})^{2}+ 12 r = 1162 (r

_{1})^{2}+ 12 r + 36 – 116 = 02 (r

_{1})^{2}+ 12 r – 80 = 0 (r_{1})^{2}+ 6 r – 40 = 0Using splitting the middle term,

We get, (r

_{1})^{2}+ 10 r – 4r – 40 = 0(r

_{1 }+ 10 )( r_{1}– 4) = 0 r_{1}= 4 , - 10Since radius is always positive, r

_{1}= 4 cmr

_{2}= 4 + 6 = 10 cm### The circle touch each other internally. The sum of their area is 116 pie cm^{2} and the distance between their centres is 6 cm. Find the radii of the circles.

Maths-General

It is given that the distance between their centres is 6 cm

r

The sum of their area of two circles = 116 cm

(r

[ (r

(r

2 (r

2 (r

Using splitting the middle term,

We get, (r

(r

Since radius is always positive, r

r

r

_{2}- r_{1 }= 6 r_{2}= 6 + r_{1}The sum of their area of two circles = 116 cm

^{2}(r

_{1})^{2}+ (r_{2})^{2}= 116[ (r

_{1})^{2}+ (6 + r_{1})^{2}] = 116(r

_{1})^{2}+ 36 + (r_{1})^{2}+ 12 r = 1162 (r

_{1})^{2}+ 12 r + 36 – 116 = 02 (r

_{1})^{2}+ 12 r – 80 = 0 (r_{1})^{2}+ 6 r – 40 = 0Using splitting the middle term,

We get, (r

_{1})^{2}+ 10 r – 4r – 40 = 0(r

_{1 }+ 10 )( r_{1}– 4) = 0 r_{1}= 4 , - 10Since radius is always positive, r

_{1}= 4 cmr

_{2}= 4 + 6 = 10 cmMaths-

### A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

It is given that diameter of the wheel = 77 cm

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

### A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

Maths-General

It is given that diameter of the wheel = 77 cm

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

Radius = = 38.5

Circumference of the wheel = 2 r

= 2 × × 38.5 = 242 cm

It is given that Speed of the bucket = 1.1 m/s

Time taken = 1 min 28 sec = 88 sec

Distance covered by the wheel = speed × time

= 1.1 × 88 = 96.8 m

= 96.8 × 100 cm

= 9680 cm

No of revolutions =

= = 40 revolutions

Maths-

### A rectangle with one side 4 cm inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

It is given that radius of the circle = 2.5 cm

AC = 2 × radius = 5 cm

Length of the rectangle, AB = 4 cm ( given)

We need to find breadth of the rectangle i.e. BC

Using Pythagoras Theorem

AC

5

25 – 16 = BC

9 = BC

Area of rectangle = length breadth

= 4 × 3 = 12 cm

AC = 2 × radius = 5 cm

Length of the rectangle, AB = 4 cm ( given)

We need to find breadth of the rectangle i.e. BC

Using Pythagoras Theorem

AC

^{2}= AB^{2}+ BC^{2}5

^{2}= 4^{2}+ BC^{2}25 – 16 = BC

^{2}9 = BC

^{2 }BC = 3 cmArea of rectangle = length breadth

= 4 × 3 = 12 cm

^{2}### A rectangle with one side 4 cm inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

Maths-General

It is given that radius of the circle = 2.5 cm

AC = 2 × radius = 5 cm

Length of the rectangle, AB = 4 cm ( given)

We need to find breadth of the rectangle i.e. BC

Using Pythagoras Theorem

AC

5

25 – 16 = BC

9 = BC

Area of rectangle = length breadth

= 4 × 3 = 12 cm

AC = 2 × radius = 5 cm

Length of the rectangle, AB = 4 cm ( given)

We need to find breadth of the rectangle i.e. BC

Using Pythagoras Theorem

AC

^{2}= AB^{2}+ BC^{2}5

^{2}= 4^{2}+ BC^{2}25 – 16 = BC

^{2}9 = BC

^{2 }BC = 3 cmArea of rectangle = length breadth

= 4 × 3 = 12 cm

^{2}Maths-

### The area of a circular ring enclosed between two concentric circles is 286 cm^{2}. Find the radii of the two circles, given that their difference is 7 cm.

Let radius of inner circle be r

It is given that difference between the radii of both the circles is 7 cm.

Radius of outer circle – Radius of inner circle = 7 cm

Radius of outer circle = 7 + r

Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm

We know that area enclosed between two concentric circles is the difference of the areas of two circles

Area of outer circle – Area of inner circle = 286 cm

(r + 7)

× ((r + 7)

r

49 + 14 r = 91

14 r = 91 – 49 = 42

r = 3 cm

So, radius of inner circle = 3 cm

Radius of outer circle = 3 + 7 = 10 cm

It is given that difference between the radii of both the circles is 7 cm.

Radius of outer circle – Radius of inner circle = 7 cm

Radius of outer circle = 7 + r

Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm

^{2}We know that area enclosed between two concentric circles is the difference of the areas of two circles

Area of outer circle – Area of inner circle = 286 cm

^{2}.(r + 7)

^{2}- p r^{2 }= 286× ((r + 7)

^{2}– r^{2 }) = 286r

^{2}+ 7^{2}+ 14 r – r^{2}= 9149 + 14 r = 91

14 r = 91 – 49 = 42

r = 3 cm

So, radius of inner circle = 3 cm

Radius of outer circle = 3 + 7 = 10 cm

### The area of a circular ring enclosed between two concentric circles is 286 cm^{2}. Find the radii of the two circles, given that their difference is 7 cm.

Maths-General

Let radius of inner circle be r

It is given that difference between the radii of both the circles is 7 cm.

Radius of outer circle – Radius of inner circle = 7 cm

Radius of outer circle = 7 + r

Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm

We know that area enclosed between two concentric circles is the difference of the areas of two circles

Area of outer circle – Area of inner circle = 286 cm

(r + 7)

× ((r + 7)

r

49 + 14 r = 91

14 r = 91 – 49 = 42

r = 3 cm

So, radius of inner circle = 3 cm

Radius of outer circle = 3 + 7 = 10 cm

It is given that difference between the radii of both the circles is 7 cm.

Radius of outer circle – Radius of inner circle = 7 cm

Radius of outer circle = 7 + r

Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm

^{2}We know that area enclosed between two concentric circles is the difference of the areas of two circles

Area of outer circle – Area of inner circle = 286 cm

^{2}.(r + 7)

^{2}- p r^{2 }= 286× ((r + 7)

^{2}– r^{2 }) = 286r

^{2}+ 7^{2}+ 14 r – r^{2}= 9149 + 14 r = 91

14 r = 91 – 49 = 42

r = 3 cm

So, radius of inner circle = 3 cm

Radius of outer circle = 3 + 7 = 10 cm

Maths-

### A circular track has an inside circumference of 1320 m. If the width of the track is 7m, what is the outside circumference

It is given that Circumference of inner circle = 1320 m

2 r = 1320

2 × × r = 1320

r = 210 m

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

= 210 + 7 = 217 m

Circumference of outer circle = 2 r

= 2 × × 217

= 1364 m

2 r = 1320

2 × × r = 1320

r = 210 m

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

= 210 + 7 = 217 m

Circumference of outer circle = 2 r

= 2 × × 217

= 1364 m

### A circular track has an inside circumference of 1320 m. If the width of the track is 7m, what is the outside circumference

Maths-General

It is given that Circumference of inner circle = 1320 m

2 r = 1320

2 × × r = 1320

r = 210 m

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

= 210 + 7 = 217 m

Circumference of outer circle = 2 r

= 2 × × 217

= 1364 m

2 r = 1320

2 × × r = 1320

r = 210 m

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

= 210 + 7 = 217 m

Circumference of outer circle = 2 r

= 2 × × 217

= 1364 m

Maths-

### In 2015 the populations of City X and City Y were equal. From 2010 to 2015 , the population of City X increased by 20% and the population of City decreased by 10%. If the population of City was 120,000 in 2010 , what was the population of City Y in 2010?

Explanation:

1: population of city x in 2010 = 120000

it increases by 20 % in 5 years . so in 2015 it will be

Thus population of city x in 2015 is 144000.

2: the population of city x and y are same in year 2015.

3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010

4: let population of city y in year 2015 be 100.

it gets increased by 10% =90

thus-

X =160000

population of city x is .

Hence, Option C is correct.

- We have given In 2015 the populations of City X and City Y were equal. From 2010 to 2015, the population of City X increased by 20% and the population of City Y decreased by 10%
- We have to find what was the population of City Y in 2010.

1: population of city x in 2010 = 120000

it increases by 20 % in 5 years . so in 2015 it will be

Thus population of city x in 2015 is 144000.

2: the population of city x and y are same in year 2015.

3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010

4: let population of city y in year 2015 be 100.

it gets increased by 10% =90

thus-

X =160000

population of city x is .

Hence, Option C is correct.

### In 2015 the populations of City X and City Y were equal. From 2010 to 2015 , the population of City X increased by 20% and the population of City decreased by 10%. If the population of City was 120,000 in 2010 , what was the population of City Y in 2010?

Maths-General

Explanation:

1: population of city x in 2010 = 120000

it increases by 20 % in 5 years . so in 2015 it will be

Thus population of city x in 2015 is 144000.

2: the population of city x and y are same in year 2015.

3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010

4: let population of city y in year 2015 be 100.

it gets increased by 10% =90

thus-

X =160000

population of city x is .

Hence, Option C is correct.

- We have given In 2015 the populations of City X and City Y were equal. From 2010 to 2015, the population of City X increased by 20% and the population of City Y decreased by 10%
- We have to find what was the population of City Y in 2010.

1: population of city x in 2010 = 120000

it increases by 20 % in 5 years . so in 2015 it will be

Thus population of city x in 2015 is 144000.

2: the population of city x and y are same in year 2015.

3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010

4: let population of city y in year 2015 be 100.

it gets increased by 10% =90

thus-

X =160000

population of city x is .

Hence, Option C is correct.

Maths-

### The area enclosed between two concentric circles is 770 cm^{2}. If the radius of the outer circle is 21cm, calculate the radius of inner circle

It is given that radius of outer circle, r

Let radius of inner circle = r

Area enclosed between two concentric circles = 770 cm

Area of outer circle – Area of inner circle = 770 cm

(r

× (21)

× (21

441 – r

r

r = 14 cm

Hence, radius of inner circle is 14 cm.

_{1}= 21 cmLet radius of inner circle = r

Area enclosed between two concentric circles = 770 cm

^{2}.Area of outer circle – Area of inner circle = 770 cm

^{2}.(r

_{1})^{2}- r^{2 }= 770× (21)

^{2}- × (r)^{2}= 770× (21

^{2}– r^{2 }) = 770441 – r

^{2}= 245r

^{2}= 441 – 245 = 196r = 14 cm

Hence, radius of inner circle is 14 cm.

### The area enclosed between two concentric circles is 770 cm^{2}. If the radius of the outer circle is 21cm, calculate the radius of inner circle

Maths-General

It is given that radius of outer circle, r

Let radius of inner circle = r

Area enclosed between two concentric circles = 770 cm

Area of outer circle – Area of inner circle = 770 cm

(r

× (21)

× (21

441 – r

r

r = 14 cm

Hence, radius of inner circle is 14 cm.

_{1}= 21 cmLet radius of inner circle = r

Area enclosed between two concentric circles = 770 cm

^{2}.Area of outer circle – Area of inner circle = 770 cm

^{2}.(r

_{1})^{2}- r^{2 }= 770× (21)

^{2}- × (r)^{2}= 770× (21

^{2}– r^{2 }) = 770441 – r

^{2}= 245r

^{2}= 441 – 245 = 196r = 14 cm

Hence, radius of inner circle is 14 cm.

Maths-

A researcher surveyed a random sample of students from a large university about how often they see movies. Using the sample data, the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is . Which of the following is the most appropriate conclusion about all students at the university, based on the given estimate and margin of error?

- We have given the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4%.
- We have to find which of the option is the most appropriate conclusion about all students at the university.
- Step 1 of 1:

There was a 4% error estimation error. But it didn’t specify whether it was caused by over estimation or under estimation.

So, the researcher is between (23 - 24) % and (23+24)% - Therefore, It is plausible that the percentage of students who see a movie at least once per month is between 19% and 27%.
- Hence, Option D is correct.

A researcher surveyed a random sample of students from a large university about how often they see movies. Using the sample data, the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is . Which of the following is the most appropriate conclusion about all students at the university, based on the given estimate and margin of error?

Maths-General

- We have given the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4%.
- We have to find which of the option is the most appropriate conclusion about all students at the university.
- Step 1 of 1:

There was a 4% error estimation error. But it didn’t specify whether it was caused by over estimation or under estimation.

So, the researcher is between (23 - 24) % and (23+24)% - Therefore, It is plausible that the percentage of students who see a movie at least once per month is between 19% and 27%.
- Hence, Option D is correct.

Maths-

### If , then find the value of

Given

Step 1 of 2:

Find the value of x

Step 2 of 2:

Put the value of x

Rationalizing in the upper equation

Final Answer:

Hence, the value of is 98.

Step 1 of 2:

Find the value of x

^{2}Step 2 of 2:

Put the value of x

^{2}inRationalizing in the upper equation

Final Answer:

Hence, the value of is 98.

### If , then find the value of

Maths-General

Given

Step 1 of 2:

Find the value of x

Step 2 of 2:

Put the value of x

Rationalizing in the upper equation

Final Answer:

Hence, the value of is 98.

Step 1 of 2:

Find the value of x

^{2}Step 2 of 2:

Put the value of x

^{2}inRationalizing in the upper equation

Final Answer:

Hence, the value of is 98.

Maths-

### A right circular cone has a volume of cubic inches. If the height of the cone is 2 inches, what is the radius, in inches, of the base of the cone?

Explanation:

We know that the volume of a cone is given by .

Height is given 2 inches

So,

= 36

r = 6

Hence, Option B is correct.

- We have given a right circular cone with volume , if the height of the cone is .
- We have to find the radius of the cone
- We know that the volume of a cone is given by .

We know that the volume of a cone is given by .

Height is given 2 inches

So,

= 36

r = 6

Hence, Option B is correct.

### A right circular cone has a volume of cubic inches. If the height of the cone is 2 inches, what is the radius, in inches, of the base of the cone?

Maths-General

Explanation:

We know that the volume of a cone is given by .

Height is given 2 inches

So,

= 36

r = 6

Hence, Option B is correct.

- We have given a right circular cone with volume , if the height of the cone is .
- We have to find the radius of the cone
- We know that the volume of a cone is given by .

We know that the volume of a cone is given by .

Height is given 2 inches

So,

= 36

r = 6

Hence, Option B is correct.

Maths-

### In the given figure OACB is a quadrant of a circle. The radius OA = 3.5 cm, OD = 2cm. Calculate the area of shaded portion

It is given that radius of the circle = 3.5 cm

Area of the circle = r

= × (3.5)

Area of quadrant of a circle = = = 9.625 cm

Now, Area of triangle OAD = × base × height

= × 3.5 × 2 = 3.5 cm

Area of shaded region

= Area of quadrant of the circle – Area of triangle

= 9.625 – 3.5 = 6.125 cm

Area of the circle = r

^{2}= × (3.5)

^{2}= 38.5 cm^{2}Area of quadrant of a circle = = = 9.625 cm

^{2}Now, Area of triangle OAD = × base × height

= × 3.5 × 2 = 3.5 cm

^{2}Area of shaded region

= Area of quadrant of the circle – Area of triangle

= 9.625 – 3.5 = 6.125 cm

^{2}### In the given figure OACB is a quadrant of a circle. The radius OA = 3.5 cm, OD = 2cm. Calculate the area of shaded portion

Maths-General

It is given that radius of the circle = 3.5 cm

Area of the circle = r

= × (3.5)

Area of quadrant of a circle = = = 9.625 cm

Now, Area of triangle OAD = × base × height

= × 3.5 × 2 = 3.5 cm

Area of shaded region

= Area of quadrant of the circle – Area of triangle

= 9.625 – 3.5 = 6.125 cm

Area of the circle = r

^{2}= × (3.5)

^{2}= 38.5 cm^{2}Area of quadrant of a circle = = = 9.625 cm

^{2}Now, Area of triangle OAD = × base × height

= × 3.5 × 2 = 3.5 cm

^{2}Area of shaded region

= Area of quadrant of the circle – Area of triangle

= 9.625 – 3.5 = 6.125 cm

^{2}Maths-

### The inner circumference of a circular track is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the circumference of outer circle at the rate of Rs 2 per meter

It is given that Circumference of inner circle = 220 m

2 r = 220

2 × × r = 220

r = 35 m

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

= 35 + 7 = 42 m

Circumference of outer circle = 2 r

= 2 × × 42

= 264 m

Cost of fencing per meter = Rs. 2

Cost of fencing outer circle = Rs. 2 × 264

= Rs. 528

2 r = 220

2 × × r = 220

r = 35 m

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

= 35 + 7 = 42 m

Circumference of outer circle = 2 r

= 2 × × 42

= 264 m

Cost of fencing per meter = Rs. 2

Cost of fencing outer circle = Rs. 2 × 264

= Rs. 528

### The inner circumference of a circular track is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the circumference of outer circle at the rate of Rs 2 per meter

Maths-General

It is given that Circumference of inner circle = 220 m

2 r = 220

2 × × r = 220

r = 35 m

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

= 35 + 7 = 42 m

Circumference of outer circle = 2 r

= 2 × × 42

= 264 m

Cost of fencing per meter = Rs. 2

Cost of fencing outer circle = Rs. 2 × 264

= Rs. 528

2 r = 220

2 × × r = 220

r = 35 m

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

= 35 + 7 = 42 m

Circumference of outer circle = 2 r

= 2 × × 42

= 264 m

Cost of fencing per meter = Rs. 2

Cost of fencing outer circle = Rs. 2 × 264

= Rs. 528

Maths-

### Is rational or irrational? Explain.

is a rational number because it is in the form of p/q with p = 5 and q = 7 and both p and q are integers

Final Answer:

Hence is a rational number.

Final Answer:

Hence is a rational number.

### Is rational or irrational? Explain.

Maths-General

is a rational number because it is in the form of p/q with p = 5 and q = 7 and both p and q are integers

Final Answer:

Hence is a rational number.

Final Answer:

Hence is a rational number.

Maths-

### How many times will the wheel of a car rotate in a journey of 88 km if its known that the diameter of the wheel is 56 cm?

It is given that diameter of the wheel = 56 cm

Radius = = = 28 cm

Circumference of the circle = 2 r

= 2 × × 28

= 176 cm

Total distance = 88 km = 8800000 cm

No of rotations = = = 50000 rotations

Radius = = = 28 cm

Circumference of the circle = 2 r

= 2 × × 28

= 176 cm

Total distance = 88 km = 8800000 cm

No of rotations = = = 50000 rotations

### How many times will the wheel of a car rotate in a journey of 88 km if its known that the diameter of the wheel is 56 cm?

Maths-General

It is given that diameter of the wheel = 56 cm

Radius = = = 28 cm

Circumference of the circle = 2 r

= 2 × × 28

= 176 cm

Total distance = 88 km = 8800000 cm

No of rotations = = = 50000 rotations

Radius = = = 28 cm

Circumference of the circle = 2 r

= 2 × × 28

= 176 cm

Total distance = 88 km = 8800000 cm

No of rotations = = = 50000 rotations

Maths-

### Express the following recurring decimal as a vulgar fraction

1.3454545454545…………

Hint:

A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.

Solution

The given number is 1.3454545454545…… which can be written as

Let’s say X=

Multiplying both sides by 10

................(1)

Step 2 of 3:

Subtracting equation (2) by equation (1)

Simplifying the fraction:

Final Answer:

Hence, the p/q form of 1.3454545454545…… is .

Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.

A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.

Solution

The given number is 1.3454545454545…… which can be written as

Let’s say X=

Multiplying both sides by 10

................(1)

Step 2 of 3:

Subtracting equation (2) by equation (1)

Simplifying the fraction:

Final Answer:

Hence, the p/q form of 1.3454545454545…… is .

Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.

### Express the following recurring decimal as a vulgar fraction

1.3454545454545…………

Maths-General

Hint:

A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.

Solution

The given number is 1.3454545454545…… which can be written as

Let’s say X=

Multiplying both sides by 10

................(1)

Step 2 of 3:

Subtracting equation (2) by equation (1)

Simplifying the fraction:

Final Answer:

Hence, the p/q form of 1.3454545454545…… is .

Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.

A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.

Solution

The given number is 1.3454545454545…… which can be written as

Let’s say X=

Multiplying both sides by 10

................(1)

Step 2 of 3:

Subtracting equation (2) by equation (1)

Simplifying the fraction:

Final Answer:

Hence, the p/q form of 1.3454545454545…… is .

Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.