Mathematics Part i Solutions Solutions for Class 6 Math Chapter 2 Joining Parts are provided here with simple step-by-step explanations. These solutions for Joining Parts are extremely popular among class 6 students for Math Joining Parts Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 6 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 6 Math are prepared by experts and are 100% accurate.
Page No 24:
Question 1:
What part of a metre is got by dividing 1 metre into 15 equal parts? To get this, into how many equal parts should we divide every 
metre?
Answer:
Number of equal parts in which 1 metre is to be divided = 15
∴ Part of a metre got by dividing 1 metre into 15 equal parts =
Number of equal parts in which metre must be divided to get part of a metre = 5
Page No 24:
Question 2:
What form of metre and of metre do we get from the above problem?
Answer:
From the above problem:
Number of equal parts in which metre must be divided to get part of a metre = 5
Number of equal parts in which metre must be divided to get part of a metre = 10
Page No 24:
Question 3:
1 metre is divided into 10 equal parts. Can we put together some of these to get metre? How about metre?
Answer:
Number of equal parts in which 1 metre is to be divided = 10
∴ Part of a metre got by dividing 1 metre into 10 equal parts =
Number of parts to be put together to get metre = 5
It is not possible to get metre from part of a metre.
Page No 26:
Question 1:
Write as a fraction with denominator 21.
Answer:
means 1 part out of 3 equal parts.
Each of these 3 parts is again divided into 7 equal parts.
∴ Total number of parts = 7 × 3 = 21
Number of parts taken = 7 × 1 = 7
⇒
Page No 26:
Question 2:
Can you write as a fraction with denominators 10? With denominator 100?
Answer:
cannot be written as a fraction with denominators 10 and 100 because 10 and 100 are not the multiples of 3.
Page No 26:
Question 3:
Write as a fraction with denominator 100
Answer:
means 1 part out of 4 equal parts.
Each of these 4 parts is again divided into 25 equal parts.
∴ Total number of parts = 25 × 4 = 100
Number of parts taken = 25 × 1 = 25
⇒
Page No 26:
Question 4:
Write each pair of fractions given below as fractions with the same denominator.
(i)
(ii)
(iii)
Answer:
(i)
means 1 part out of 3 equal parts.
Each of these 3 parts is again divided into 4 equal parts.
∴ Total number of parts = 4 × 3 = 12
Number of parts taken = 4 × 1 = 4
⇒
means 1 part out of 4 equal parts.
Each of these 4 parts is again divided into 3 equal parts.
∴ Total number of parts = 3 × 4 = 12
Number of parts taken = 3 × 1 = 3
⇒
(ii)
means 1 part out of 4 equal parts.
Each of these 4 parts is again divided into 6 equal parts.
∴ Total number of parts = 6 × 4 = 24
Number of parts taken = 6 × 1 = 6
⇒
means 1 part out of 6 equal parts.
Each of these 6 parts is again divided into 4 equal parts.
∴ Total number of parts = 4 × 6 = 24
Number of parts taken = 4 × 1 = 4
⇒
(iii)
means 1 part out of 3 equal parts.
Each of these 3 parts is again divided into 6 equal parts.
∴ Total number of parts = 6 × 3 = 18
Number of parts taken = 6 × 1 = 6
⇒
means 1 part out of 6 equal parts.
Each of these 6 parts is again divided into 3 equal parts.
∴ Total number of parts = 3 × 6 = 18
Number of parts taken = 3 × 1 = 3
⇒
Page No 28:
Question 1:
Can you now mentally reduce the fraction below to their lowest terms?
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i)
10 and 30 are multiples of 5.
It can be again simplified because 2 and 6 are multiples of 2.
(ii)
20 and 40 are multiples of 5.
It can be again simplified because 4 and 8 are multiples of 4.
(iii)
15 and 20 are multiples of 5.
(iv)
12 and 16 are multiples of 4.
(v)
16 and 24 are multiples of 4.
It can be again simplified because 4 and 6 are multiples of 2.
(vi)
18 and 45 are multiples of 3.
It can be again simplified because 6 and 15 are multiples of 3.
Page No 30:
Question 1:
Ravi completed of a job the first day and of it on the second. How much of the job did he do in these two days together? How much remains to be done?
Answer:
Part of the job done on the first day =
Part of the job done on the second day =
Whole work = 1 =
If we add part to the part of the job done in two days, i.e., we get:
∴ Part of the job left to be done =
Page No 30:
Question 2:
Reema and Suma have pots of the same size. of Rema’s pot are filled with water. Suma pours water from her pot to Rema’s pot till it is full. How much water remains in Suma’s pot now?
Answer:
Part of water the pot can hold = 1 =
Part of the water filled in Rema’s pot =
Part of the water filled in Suma’s pot =
If we add part to part, then we get
Therefore, Suma pours part of the water from her pot to Rema’s pot, such that Rema’s pot is completely filled.
Now, if we add part to part, then we get
∴ Part of the water left in Suma’s pot =
Page No 33:
Question 1:
A new road is being constructed in Vellimala panchayat. During the first week, of the work was done and during the second week. What part of the work was finished by the end of second week? The whole work was finished by the end of third week. What part was done during the third week?
Answer:
Part of the work done in the first week =
Part of the work done in the second week =
∴ Part of the work finished at the end of the second week =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 4 and 5 = 4 × 5 = 20.
and
∴
∴ Part of the work finished at the end of the second week =
Whole work = 1 =
If we add part of the work to the part of the work done in two weeks, i.e., we get:
∴ Part of the work done in the third week =
Page No 33:
Question 2:
A child sleeps for of the day and spends of the day at school. What part of the day do these together make?
Answer:
Part of the day child sleeps =
Part of the day child spends in school =
Total part of the day spend by the child in sleeping and in school =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 3 and 4 = 3 × 4 = 12.
and
∴
∴ Part of the day spend together =
Page No 33:
Question 3:
Abdu and Raghavan take up a contract to till a piece of farm land. Abdu tills of the land a day and Raghavan tills of the land a day. What part of the land do they together till in a day?
Answer:
Part of the land tilled by Abdu in a day =
Part of the land tilled by Raghavan in a day =
Part of land they together till in a day =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator we need should be a multiple of 7 and 5 = 7 × 5 = 35.
and
∴
∴ Part of land they together till in a day =
Page No 36:
Question 1:
Athira took some money from her Sanchayika deposit. She spent of this to buy books and to buy uniform. The remaining, she used to buy uniform for her brother.
(i) How much of the total amount did she use to buy books and uniform for herself?
(ii) How much did she use to buy uniform for her brother?
Answer:
(i)
Part of the money spent on books =
Part of the money spent on uniform for herself =
∴ Part of the money used to buy books and uniform for herself =
(ii)
Total money took by Athira from her Sanchayika deposit = 1 =
Part of the money spent by Athira on herself =
∴ Part of the money used by her to buy uniform for her brother =
Page No 36:
Question 2:
Beans, cucumber and Amaranth are grown in the school vegetable garden. Beans are grown in
of the garden, cucumber in of the garden and amaranth in the remaining part.
(i) How much of the garden is used for beans and cucumber?
(ii) How much is used to amaranth?
Answer:
(i)
Part of the vegetable garden in which beans are grown =
Part of the vegetable garden in which cucumber are grown =
∴ Part of the vegetable garden used for beans and cucumber =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 7 and 5 = 7 × 5 = 35.
and
∴
∴ Part of the vegetable garden used for beans and cucumber =
(ii)
Total part of the vegetable garden = 1 =
Part of the vegetable garden used for beans and cucumber =
∴ Part of vegetable garden used for amaranth =
Page No 37:
Question 1:
Shameem rides of the distance of his school on his bicycle and travels by bus. He walks the rest of the distance.
(i) How much of the total distance does he go by bicycle and bus?
(ii) How much does he walk?
Answer:
(i) Part of the total distance covered by Shameem on his bicycle =
Part of the total distance covered by Shameem by bus =
∴ Part of the total distance covered by Shameem on his bicycle and by bus =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 10 and 5 = 10 × 5 = 50.
and
∴
∴ Part of the total distance covered by Shameem on his bicycle and by bus =
(ii) Total part of distance to Shameem’s school = 1 =
Part of the total distance covered by Shameem on his bicycle and by bus =
∴ Part of the total distance covered by him by walking =
Page No 37:
Question 2:
Raju decides of fence the compound around his home. The total length of the fence is 75 metre. On first day, metres of the fence was built and on the second, metres
(i) What length of the fence was completed during these two days?
(ii) What is the remaining length to be built?
Answer:
(i) Total length of the fence = 75 m
Fence built on the first day =
Fence built on the second day =
∴ Fence built in these two days =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 1, 2 and 4 = 1 × 2 × 4 = 8
, and
∴
∴ Fence built in the two days =
(ii) Remaining length to be built =
Page No 37:
Question 3:
A milk society got litres of milk in the morning and litres in the evening. The society sold litres that day.
(i) How many litres of milk did the society get that day?
(ii) Find the remaining quantity after the sales.
Answer:
(i)
Quantity of milk got by the milk society in the morning =
Quantity of milk got by the milk society in the evening =
Quantity of milk got by the society in the day =
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 1, 4 and 2 = 1 × 4 × 2 = 8
, and
∴
∴ Quantity of milk got by the society in that day =
(ii) Quantity of milk sold =
Now,
Remaining quantity of milk after sales =
Page No 37:
Question 4:
Find the larger in each pair of these numbers. Subtract the smaller from the larger of these pairs.
(i)
(ii)
(iii)
(iv)
Answer:
(i)
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 3 and 2 = 3 × 2 = 6.
and
Now, is made up of 4 equal parts out of 6 equal parts and is made up of 3 such parts.
∴
Now,
(ii)
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 5 and 6 = 5 × 6 = 30.
and
Now, is made up of 24 equal parts out of 30 equal parts and is made up of 25 such parts.
∴
Now,
(iii)
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 3 and 5 = 3 × 5 = 15.
and
Now, is made up of 10 equal parts out of 15 equal parts and is made up of 9 such parts.
∴
Now,
(iv)
First, we change these fractions to fractions with the same denominator.
Thus, the denominator required should be a multiple of 3 and 10 = 3 × 10 = 30.
and
Now, is made up of 10 equal parts out of 30 equal parts and is made up of 9 such parts.
∴
Now,
Page No 38:
Question 1:
Murali bought a 20 metre long string. He cut out a piece metres long and another metres long. What is the length of the string that remains?
Answer:
Length of the string bought by Murali = 20 m
Length of the first piece cut by him =
Length of the second piece cut by him =
Length of the remaining string =
Now,
Length of the remaining string =
Page No 38:
Question 2:
Amarnath needs two iron wires each of the length metres to hang up rubber sheets to dry. How much wire does he need in all?
Answer:
Length of each wire needed by Amarnath =
Now,
Total length of the wire needed by him =
∴ Total length of the wire needed by him = m
View NCERT Solutions for all chapters of Class 6