Fractions And Decimals

Length of ribbon= m

Length of each piece= m

Number of piece

7 pieces

First we need to find the L.C.M of denominator

L.C.M.= 21

We have to make the denominator equal to the L.C.M for all the fractions by multiplying same number in numerator and denominator.

Ascending arrangement:

The reciprocal is obtained by interchanging the number at numerator and denominator position.

Reciprocal is

× 4

3 × 4

means 3 pictures in which two-third of each picture is selected

We find only option B has such pictorial representation

3×9=27

Since the decimal point has shifted two places left of 3 and one place left of 9 so we have to move three places towards left in case of 27

0.027

÷6

The word ‘of’ means multiplication between the two fractions

Flour Required = cups

Sugar required = cups

Total quantity of ingredient for one packet

⇒ Total quantity of ingredient for one packet cups

⇒ Total quantity of ingredient for 10 packet cups

It requires 40 to 50 cups of ingredients

7 × 6

7÷

÷ 5

Amount of apple used on Monday= of 5

Remaining apple =(5-4)= 1 kg

Amount of apple used the next day = of 1

Weight of apples left now

Shaded region of each picture on the left side

Total no. of pictures on the left = 3

Total value

The amount of cake at the beginning =1

Part of cake Rani ate

Part of the cake left after Rani ate

Part of the remaining cake Ravi ate

Part of the cake left after both ate

The reciprocal is obtained by interchanging the number at numerator and denominator position.

Reciprocal is

of 27

⇒ 2×9=18

18

of 45

⇒ 4×9=36

36

4 × 6

of 25

⇒ 2×5=10

Multiplication (×)

Division of two fractions is same as multiplying the first fraction with the reciprocal of second fraction.

32

We have to shift the decimal place to the right by one place since 10 has only one zero

25400

We have to shift the decimal place to the right by three places since 1000 has only three zeroes

9350

We have to shift the decimal place to the right by two places since 100 has only two zeroes

0.47

We have to shift the decimal place to the left by one place since 10 has only one zero

0.047

We have to shift the decimal place to the left by two places since 100 has only two zeroes

0.0047

We have to shift the decimal place to the left by three places since 1000 has only three zeroes

less

The value of a proper fraction is always less than 1. So when they are multiplied the value is less than each of the two fractions.

Multiply, Reciprocal

Division of two fractions is same as multiplying the first fraction with the reciprocal of second fraction.

4

The divisor can be found by dividing the dividend by quotient

100

The Decimal has shifted two place towards left. So the number is divided by 100.

0.5× 0.7=0.35

× (Multiplication)

÷(Division)

667

It is a false statement

The value of a reciprocal of a proper fraction is always greater than 1 and hence it is an improper fraction.

It is a false statement

The value of a reciprocal of an improper fraction is always less than 1 and hence it is a proper fraction.

It is a true statement

Product of more than one fraction is found my multiplying all the numerator and all the denominators.

It is a false statement

The value of an improper fraction is always greater than 1 and hence its product with another improper fraction is always greater than the two fractions.

It is a true statement

The reciprocal is obtained by interchanging the number at numerator and denominator position.

It is a true statement

Since multiplication by 1000 is similar to multiplying by 10 three times, so we have to move the decimal place towards right by three places.

To divide a decimal number by 10, 100 or 1000, we shift the decimal point in the number to the left by as many places as there are zeroes over 1, to get the quotient.

∴ This statement is True.

Since 1 can be written as , and we know that reciprocal of a fraction is obtained by interchanging the numerator and denominator of fraction, therefore reciprocal of is 1 only.

∴ This statement is True.

As we know that, ‘of’ operator in fraction is used for its multiplication,

When dividing a whole number by a fraction, we multiply the whole number by the reciprocal of that fraction.

∴ This statement is false.

As we know that reciprocal of a fraction is obtained by interchanging the numerator and denominator of fraction.

∴ this statement is False.

If 5 is added to both the numerator and the denominator of the fraction , the value would become and as .

∴ The value of fraction changes and it gets increased.

Let the original fraction be .

Now, if denominator is decreased (say by 1) while numerator is kept same, the fraction becomes .

As the value of fraction is inversely proportional to the value of denominator, therefore, by decreasing denominator the value of fraction increases.

Since the number of letters from A to J are 10, therefore,

⇒ 4^th letter from A comes 2/5 of way, which is letter D.

∴ Letter D is the required answer.

Let the number be A.

It is given that,

Also, 1.75 of A = 1.75 × 15 = 26.25

∴ Required answer is 26.25

Total number of students in class = 40

Let the number of students that like to eat rice only be x.

Let the number of students that like to eat chapatti only be y.

Students that like both rice and chapatti are = 40 – x – y

⇒ 40 – 8 – 16

⇒ 40 – 24

⇒ 16

∴ 16 students like both rice and chapatti.

Part of homework completed in 2 hours =

Part of homework completed in 1 hour would be =

As we know , therefore,

Part of homework completed in hours would be =

∴ Renu must have completed 1/3 of her homework in hours.

Let the total number of pages in the book be x.

Pages read by Reemu =

If she reads further 40 pages, she would have readth pages of the book,

On cross-multiplying, we get,

⇒ 5x = 400

⇒ x = 80

∴ Pages read by Reema = Total pages – Pages read

⇒ Required answer =

∴ 24 pages are left to be read.

Let the number in box be A.

We know, When dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

Let the quotient be A.

We know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

Since,, therefore the quotient of given fraction is greater than 1.5

Method 1:

Convert the into a decimal value, which comes out to be 0.76.

This means is smaller than 0.82

Method 2: Convert 0.82 to a fraction form =

Making denominator of both fractions to its LCM,

LCM of 50 and 17 = 850

Also,

So, , therefore,

13/17 is smaller than 0.82

(a)

∴ A 72kg of adult takes 0.64gm of pain relievers.

(b) An adult of 42 kg of weight should take tablets.

As the weight of each tablet = 4/25 grams

⇒ Weight of tablets =

∴ 0.20 grams of pain reliever is the recommended dose for an adult weighing 46 kg.

a) As the dosage listed for adult dog is 1/2 tsp per 9 kg dog weight.

⇒ Dosage needed for 18 kg adult dog will be =

An 18 kg adult dog will have 1 tsp of dosage.

b) As the dosage listed for cat is 1/4 tsp per 1 kg dog weight.

⇒ Dosage needed for 6 kg cat will be = 1/4 × 6

An 18 kg adult dog will have 1.5 tsp of dosage.

c) As the dosage listed for pregnant dog is 1/2 tsp per 4.5 kg dog weight.

⇒ Dosage needed for 18 kg adult dog will be =

An 18 kg adult dog will have 2 tsp of dosage.

Weight of each box = kg

Total weight of chocolates = kg

Number of boxes of chocolates that can be made

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

⇒ 3 × 8 = 24 boxes

∴ 24 chocolate boxes can be made.

Length of each bookmarker

Total length of ribbon = 15m

Number of bookmarker

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

∴ Anvi can make 142 complete bookmarkers.

a) Length of side of square = 8.3 cm

Length of diagonal = Length of square × 1.414

⇒ Length of side = 8.3 × 1.414 = 11.74 cm

b) Length of side of square = 7.875 cm

Length of diagonal = Length of square × 1.414

⇒ Length of side = 7.875 × 1.414 = 11.14 cm

a) Diameter of the circle = 14.35cm

Length of side of square = 0.707 × 14.35 = 10.14 cm

b) Diameter of the circle = 8.63cm

Length of side of square = 0.707 × 8.63 = 6.101 cm

a) The diameter of the disc = 18.7 cm

Distance around the disc = 18.7 × 3.14 = 58.72 cm

b) The diameter of the disc = 6.45 cm

Distance around the disc = 6.45 × 3.14 = 20.25 cm

Rate of cloth = Rs. 53.50 per m

Cost of 27.5 m cloth = Rate × 27.5

⇒ 53.50 × 27.5 = 1471.25

∴ Required cost of cloth is Rs. 1471.25

Since when Nidhi has completed 2/6 of the race, She is at hurdle B, this means the hurdles are placed equidistant from each other and as there are six hurdles, the race is divided into six parts.

a) So, when she is of the way through the race, she will be at hurdle D.

b) When she is of the way through the race, she will be at hurdle E.

c) When Nidhi is over hurdle C, then she would have completed of the way.

Or, as she is half way through the race, she would have completed 1/2 of the race.

∴ Two fractions to tell what part of the race Nidhi has finished when she is over hurdle C will be and .

As 1000m = 1km,

Diameter of earth, D = 12756000m = 12756km

Diameter of planet = d

⇒ 0.058 × 12756 km

⇒ 739.848 km

∴ Diameter of this planet is 739.848 km

As we know that reciprocal of a fraction is obtained by interchanging the numerator and denominator of fraction.

Product of given number and its reciprocal,

∴ Product of a number and its reciprocal is always equal to 1.

Converting the mixed fractions to simple fractions.

Solving the numerator,

Now, Solving Denominator, we get,

Putting the numerator and denominator in the fraction, we get,

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

On solving Numerator of given fraction, we get,

On solving the denominator, we get,

Putting the value of numerator and denominator in given fraction, we get,

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

On simplifying the brackets, we get,

So, the question reduces to

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

Let the number be x.

Solving the equals part first, we get,

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

Now, according to question, it is given that,

⇒ x = 8 × 8 = 64

∴ the number is 64.

Total money given by Heena’s dad = Rs. 500

Amount of electricity bill = Rs 385.70

Money returned to her father = 114.30

Change he received should be 30 paise.

Normal body temperature = 98.6° F

Savitri temperature = 103.1° F

Temperature above normal =

∴ Rise in temperature was 4.5° F

a) In year 1978, the temperature was,

Multiplying both numerator and denominator by, we get

As we know that, to divide a decimal number by 10, 100 or 1000, we shift the decimal point in the number to the left by as many places as there are zeroes over 1, to get the quotient.

So, the temperature in year 1978 was 0.02° C.

Order of five years from coldest to warmest is:

1964 < 1965 < 1978 < 1958 < 2002

b) Since the temperature in 1946 was -0.03° C

Therefore, it lies between the year 1965 and 1978

-0.10 < -0.03 < 0.02

Atomic Weight of Helium, He = 4.0030

Atomic Weight of Hydrogen, H = 1.0080

Atomic Weight of Oxygen, O = 16.0000

a) Difference between Oxygen and Hydrogen = O – H

⇒ Difference between Oxygen and Hydrogen is 14.9920

b) Difference between Oxygen and Helium = O – He

⇒ Difference between Oxygen and Helium is 11.9970

c) Difference between Helium and Hydrogen = He – H

⇒ Difference between Helium and Hydrogen is 2.9950

Correct length of rod, C = 19.33 cm

Length reported by Ravi, R = 19.34 cm

Length reported by Kamal, K = 19.25 cm

Length reported by Tabish, T = 19.27 cm

Error made by Ravi = R – C

Error made by Kamal = K – C

Error made by Tabish = T – C

∴ Error made by Ravi, Kamal and Tabish is 0.01cm, -0.08cm and -0.06 cm respectively.

We can re-write the given decimals into fractions as:

Also, when they are divided,

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

∴ the quotient will be 74.135

Let the number be x, so according to the question,

⇒ x = 520 × 136

⇒ x = 70720

∴ the number is 70720

Area of the entire floor, A_F = 4.5 × 3.6 sq.m

Area of square tile, A_T = 6 × 6 = 36 sq.cm

Number of tiles needed = A_F / A_T

Rate of tile = Rs 23.25

Total cost = Rate × tiles needed

Total cost of flooring is Rs 10462.5

Cloth Sunita has

She gave 1/3 of that to Rehana,

∴ Rehana have m of cloth.

Length of flower garden, L = 22.50 m

Length of brick, b = 0.25 m

As we know, when dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

∴ 90 bricks are needed to make the border.

Cloth required for making one shirt = R metre

Cloth needed for 6 shirts = (R × 6) metres

∴ 14.25 m of cloth is required for making 6 shirts.

Given:

Seats in picture hall = 820

Persons reported in hall = 648

Guess made by 1^st user = full

Guess made by 2^nd user = full

Formula Used\Theory:

Less the difference between guessed value and real value

More it will be a better guess.

⇒ The guess made by 1^st user:-

full = × 820

= 3 × 205

= 615

Difference in reality and guessed = 648 – 615

= 33

⇒ The guess made by 2^nd user:-

full = × 820

= 2 × 273.33

= 546.66

Difference in reality and guessed = 648 – 546.66

= 101.33

∴ The difference of guessed from reality is lower in 1^st user

Conclusion:1^st user made a better guessed

Given:

Number of students in class = 35.

Price of sweets = Rs.740.25

Price of cold drink = Rs.70

Formula Used\Theory:

Dividing total amount by number of students gives

Contribution of each student

⇒ Total amount = Amount of cold drink + Amount of sweets

Total amount = Rs.740.25 + Rs.70

= Rs.810.25

⇒ Amount paid by 35 students = Rs.810.25

Amount paid by 1 student =

= Rs.23.15

Conclusion: Each student contributes Rs.23.15

Given:

The time taken in 5 races are:

3.20 minutes, 3.37 minutes, 3.29 minutes, 3.17 minutes

3.32 minutes

Formula Used\Theory:

Average =

Number of timings of races = 5

Sum of timings of races =

(3.20 + 3.37 + 3.29 + 3.17 + 3.32) minutes

= 16.35

Average =

Average =

Average = 3.27minutes

Conclusion: Average time taken by Rohan is 3.27 minutes

Given:

Length of road = m

Length of work labourer does in 1 day = 7.5m

Formula Used\Theory:

Dividing total length by length of work done in 1 day gives

Number of days to complete the work

7.5m Length of work done in⇒ 1 day

1m Length of work done in⇒ day

m Length of work done in⇒ × day

= ×

= ×

=

10.7days

Conclusion:10.7days are required to complete the work

Given:

The weight of an object on moon is its weight on Earth.

Weight of object is kg on Earth

Formula Used\Theory:

Multiplying the ratio of both with value of one object gives

Value of 2^nd object

If the weight of an object on moon is its weight on Earth.

And weight of object is kg on Earth

Weight of object on earth = kg

= kg

∴ Weight on moon = kg

= kg

= 0.933kg

Conclusion: Weight of object on moon is 0.933kg

Given:

Number of students = 200

Formula Used\Theory:

Multiplying of ratio of each part and total value gives out

Value of each part

(a) The number of students influenced by radio is =

Ratio × total number of students

= × 200

= 9 × 10

= 90students

(b) Number of students influence more by radio than music video channel

= difference of ratio × number of students

= () × 200

= () × 200

=

= 37 × 2

= 74 students

(c) Number of students influence by friends and relatives

= Ratio × total number of students

= × 200

= 3 × 10

= 30

Number of students heard CD in a shop = Ratio × total number of students

= × 200

= 20 students

Given:

Total milk milkman had in his can = L

Milk given to Renu, Kamla and Renuka is L each

Milk given to Shadma L

Milk given to Jassi L

Formula Used\Theory:

Difference of sold item from total gives

Value of remaining part

Renu, Kamla and Renuka is L each

Shadma L

Jassi L = L

Sum of milk sold is

L + L + L + L + L

[]L

L

L

= L

Milk left in his can = total milk – sold milk

= []L

= []L

= []L

=

=

= L milk

Conclusion: L of milk left in his can.

Given:

Anuradha can do a piece of work in 6 hours

Formula Used\Theory:

Division of specific time with total time gives

The amount of work done in that time interval

Let X be a piece of work

If the work X is done in 6 hours

⇒ Then work done in 1 hour is

∴ Part of work X is done in 1 hours

If work done in 1 hour is

⇒ Then work done in 5 hours is

∴ Part of work X is done 5 hours

If work done in 1 hour is

⇒ Then work done in 6 hours is

∴ Whole Part of work X will be done 6 hours

Given:

Rehana requires 5 hours to paint the whole saree

Formula Used\Theory:

Division of specific time with total time gives

The amount of work done in that time interval

Let X be work of painting a saree

If the work X is done in 5 hours

⇒ Then work done in 1 hour is

∴ Part of work X is done in 1 hours

→ Hours = Hours = Hours

If work done in 1 hour is

⇒ Then work done in hours is

=

∴ Part of work X is done hours

→Hours = Hours = Hours

If work done in 1 hour is

⇒ Then work done in hours is

=

∴ Part of work X will be done hours

Given:

Rama has kg of cotton wool

Cotton wool taken by 1 pillow is kg

Formula Used\Theory:

Dividing total weight by weight of 1 item gives

Number of items

Cotton wool Rama had kg

= kg = kg

Cotton wool taken by 1 pillow is kg

= kg = kg

∴ Number of pillows = = = 5

Conclusion: 5 pillows can be constructed

Given:

Radhika has m cloth

Cloth require to make skirt is m

Formula Used\Theory:

Dividing total length by length of 1 skirt gives

Number of skirt

Length of cloth Radhika had kg

= kg = kg

Length of cloth require to make skirt is kg

= kg = kg

∴ Number of skirts = = = 4

Conclusion: 4 skirts can be constructed

Given:

Ravi can walk km in one hour

Distance to walk to his office from home is 10km

Formula Used\Theory:

Speed =

Time =

Speed = km in one hour

= = km in one hour

Distance = 10km

∴ Time =

Time =

Time = 3 Hours

Conclusion: Ravi take 3 hours to each office while walking

Given:

360 km can be traveled by ^th of petrol tank

Formula Used\Theory:

Multiplying opposite ratio with given specific value gives

Total amount value

Let the km driven from full tank be X

Then;

If of petrol tank travels 360km

And full tank travel X km

⇒ of X km is 360km

∴ × X = 360km

X = 360 × km

X = 600km

Conclusion: Full tank of petrol drives 600 km

Given:

of amount earned is Rs.75

Formula Used\Theory:

Multiplying opposite ratio with given specific value gives

Total amount value

Let the total amount earned by Kajol be X

Then;

If of her earnings is Rs.75

And the complete earning is Rs.X

⇒ of Rs.X is Rs.75

∴ × X = Rs.75

X = Rs.75 ×

X = Rs.200

Conclusion: Kajol earns Rs.200

Given:

Total 221 trees in forest

17 trees require to make 1 tonne of paper

Formula Used\Theory:

Dividing total amount by amount of 1 item gives

Number of items

(i) (A) 5 tonnes of paper

17 trees require to make 1 tonne of paper

Then;

5 tonnes of paper require

17 × 5 trees

= 85trees

Fraction of forest =

Fraction of forest =

(B) 10 tonnes of paper

17 trees require to make 1 tonne of paper

Then;

10 tonnes of paper require

17 × 10 trees

= 170trees

Fraction of forest =

Fraction of forest =

(ii) To save part of forest

Means;

× 221 trees to be saved

= 7 × 17

= 119trees to be saved

If 17 trees require to make 1 tonne of paper

Then;

Number of tonnes of paper to be saved =

= 7 tonnes

∴ 7 tonnes of paper are require to save part of forest

Given:

Formula Used\Theory:

Calculation follows BODMAS rule

Bracket Of Division Multiplication Addition Subtraction

In the given calculation

Moving to the 1^st bracket

⇒

=

Moving to the 2^nd bracket

⇒

= ⇒

=

Moving to the 3^rd bracket

⇒

= ⇒

=

∴

= = 5.18

Conclusion: The value of given calculation is 5.18

(1) 2 ×

=

Figure D supports above equation

(2) 2 ×

=

Figure F supports above equation

(3) 2 ×

=

Figure C supports above equation

(4) 4 × = 1

=

Figure B supports above equation

(5) 3 ×

Figure A supports above equation

(6) 3 ×

Figure E supports above equation

Given:

(0.3) × (0.3) – (0.2) × (0.2)

Formula Used\Theory:

Calculation follows BODMAS rule

Bracket Of Division Multiplication Addition Subtraction

⇒ (0.3) × (0.3) – (0.2) × (0.2)

⇒ (0.3 × 0.3)-(0.2 × 0.2)

⇒ 0.09 – 0.04

⇒ 0.05

Conclusion: Result of above calculation is 0.05

Given:

Formula Used\Theory:

Calculation follows BODMAS rule

Bracket Of Division Multiplication Addition Subtraction

As we move to 1^st part

⇒ As we remove decimal from numerator and denominator

We have to multiply 10 both on numerator and denominator

= 2

As we move to other part

⇒ As we remove decimal from numerator and denominator

We have to multiply 10 on numerator and 100 on denominator

= = 0.4

Now as per equation above

2 + 0.4

2.4

Conclusion: Result of above calculation is 2.4

Given:

Formula Used\Theory:

Calculation follows BODMAS rule

Bracket Of Division Multiplication Addition Subtraction

Moving on to 1^st bracket

(0.2 × 0.14)

= 0.028

Moving on to 2^nd bracket

(0.5 × 0.91)

= 0.455

Moving on to the 3^rd bracket

(0.1 × 0.2)

= 0.02

Now as per equation above

=

⇒ As we remove decimal from numerator and denominator

We have to multiply 100 in numerator and 1000 in denominator

= =

= 24.15

Conclusion: The result of above calculation is 24.15

Given:

Square BCDE and an equilateral triangle ABC have a common side BC

Side of triangle is cm

Formula Used\Theory:

All sides of square are equal

All sides of equilateral triangle are equal

If side of triangle is cm

Then;

AB = BC = CA = cm

∴ In square BCDE

If BC = cm ∵ BC is common in both triangle and square

As all sides of square are equal

BC = CD = DE = EB = cm

∴ All sides of equilateral triangle and square are equal

By joining triangle and square as in figure above

It forms a pentagon ABDEC

With sides AB = BD = DE = EC = CA = cm

⇒ Perimeter of ABDEC = AB + BD + DE + EC + CA

=

= 5 × cm

= cm

Conclusion: Perimeter of above figure is cm

Given:

Size of carpet 4 m × m

Room size is

Formula Used\Theory:

Fraction of specific part means Division of specific part with

Total amount

Area of carpet = 4m × m

= 4m × m

= 4m × m

= m²

Area of room =

=

=

= m²

∴ Area need to be cut out is = Area of carpet – Area of room

= m²

= m²

= m²

= m²

Fraction of area cut out is

=

=

Conclusion: part of carpet should be cut out to fit into the room

Given:

Family photograph has length cm and breadth cm

Framed border of uniform width cm.

Formula Used\Theory:

Border is added twice in both length and breadth

In order to require length and breadth of whole frame

Length of frame = length of picture + 2 × width of border

= + 2 × cm

= cm + 2 × cm

= cm + cm

= cm = cm

Breadth of frame = breadth of picture + 2 × width of border

= + 2 × cm

= cm + 2 × cm

= cm + cm

= cm = cm

Area of framed picture = cm × cm

= cm²

= 305.76cm²

Conclusion: Area of framed picture is 305.76cm²

Given:

Cost of burger = Rs.

Cost of Macpuff = Rs.

Formula Used\Theory:

Cost of items = number of items × Cost of 1 item

Cost of 4 burgers =

= 4 × Rs.

= Rs.4 ×

= Rs.4 ×

= Rs.83

Cost of 14 macpuffs =

= 14 × Rs.

= Rs.14 ×

= Rs.14 ×

= Rs.217

Conclusion: Cost of 4 burgers is Rs.83 and 14 macpuffs is Rs.217

Given:

A hill, m in height

^th of its height is under water

Formula Used\Theory:

Difference of fraction from 1 gives value of other part

If ^th of its height is under water

Then;

Height of above water =

1 -

=

= ^th part of hill is above water

If height of hill is m

Then height of hill above water = ^th ×

=

=

= = 76

Conclusion: Height of hill above water is 76m

Given:

Lauryn Williams had a reaction time of 0.214 second

Her total race of 100m time was 11.03 seconds

Formula Used\Theory:

Subtracting reaction time from total time gives

Actual time to cover the distance

Total race time of Lauryn Williams was 11.03seconds

The reaction time of Lauryn Williams was 0.214 seconds

Actual time taken to run 100m = 11.03-0.214seconds

= 10.816 seconds

Conclusion: Actual time taken to run 100m by Lauryn Williams is

10.816 seconds

(i)

= = 1.33

(ii)

= = 0.33

(iii)

= = 24

(iv)

= = 0.4

(v)

=

=

= = 1.857

(vi)

= = 5.66

Given:

Height of smallest container is 10.5cm

Height of largest container is Xcm

Relation between smallest and largest height ⇒

× X = 10.5cm

As in the given relation × X = 10.5cm

7 × X = 10.5cm × 25

X =

To remove the decimal multiply 10 in denominator

X =

X = = 37.5cm

Conclusion: Height of largest container is 37.5cm

As we move from 1^st term to 2^nd the ratio comes to be

=

As we move from 2^nd term to 3^rd the ratio comes to be

=

As we move from 3^rd term to 4^th the ratio comes to be

=

∴ The next upcoming term will be in same ratio as above

X =

=

As we move from 1^st term to 2^nd the ratio comes to be

= 2

As we move from 2^nd term to 3^rd the ratio comes to be

= 2

As we move from 3^rd term to 4^th the ratio comes to be

= 2

∴ The next upcoming term will be in same ratio as above

X = 2 ×

=

As we move from 1^st term to 2^nd the ratio comes to be

= 10

As we move from 2^nd term to 3^rd the ratio comes to be

= 10

As we move from 3^rd term to 4^th the ratio comes to be

= 10

∴ The next upcoming term will be in same ratio as above

X = 10 × 50

= 500

As we move from 1^st term to 2^nd the ratio comes to be

=

As we move from 2^nd term to 3^rd the ratio comes to be

=

As we move from 3^rd term to 4^th the ratio comes to be

=

∴ The next upcoming term will be in same ratio as above

X = × 0.0001

= 0.00001

As we can see that

0.3 value is greater than 0.25

∵ They are positive values

As we move to negative numeral system

Negative sign makes value smaller

Moving from 0 to -∞ we gets decreasing order of values

∴ more the value to the negative sign less it counts

And less the value to negative sign more it counts

As 0.3 value is greater than 0.25

But due to negative sign

-0.3 gets smaller than -0.25

∴ Student error was –0.3 is greater than –0.25

Instead he has to write –0.3 is smaller than –0.25

The student multiplication was:-

=

= (3 × 2)

= 6() =

The actual multiplication is :-

We have to 1^st open full term only then we have to multiply

=

=

=

=

∴ The student error was multiplying terms simply with opening the complete terms.

By making sum of 1^st 3 terms

We get;

=

Which is not greater than 1

By making sum of 1^st 4 terms

We get;

=

Which is not greater than 1

By making sum of 1^st 5 terms

=

=

The fraction comes to greater than 1

∴ the fraction firstly makes the sum greater than 1