NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 tenths
Question: 1 Can you now write the following as decimals

Hundreds (100)	Tens (10)	Ones (1)	Tenths $\small \left ( \frac{1}{10} \right )$
5	3	8	1
2	7	3	4
3	5	4	6

Answer: Yes, we can write to them in Decimal form. The Numbers in the decimal forms are:

i) 538.1
ii) 273.4
iii) 354.6

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 subtopic fractions as decimals

Question: 1 Write $\small \frac{3}{2},\frac{4}{5},\frac{8}{5}$ in decimal notation.

Answer: The number in decimal forms are:

$\frac{3}{2}=1.5$
$\frac{4}{5}=0.8$
$\frac{8}{5}=1.6$

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.1

Question: 1 Write the following as numbers in the given table.

Answer: Observing from the figure, we get

Hundreds

(100)

Tens

(10)

Ones

(1)

Tenths

$\small (\frac{1}{10})$

a) 0

b) 1

Question: 2 Write the following decimals in the place value table.

(a) $\small 19.4$ (b) $\small 0.3$ (c) $\small 10.6$ (d) $\small 205.9$

Answer: The Value table of the numbers are:

Number	Hundreds	Tens	Ones	Tenths
19.3	0	1	9	3
0.3	0	0	0	3
10.6	0	1	0	6
205.9	2	0	5	9

Question: 3 Write each of the following as decimals :

(a) Seven-tenths (b) Two tens and nine-tenths

(c) Fourteen point six (d) One hundred and two ones

(e) Six hundred point eight

Answer:

(a) Seven-tenths :

$7\times\frac{1}{10}=\frac{7}{10}=0.7$

(b) Two tens and nine-tenths:

$2\times10+9\times\frac{1}{10}=20+\frac{9}{10}=20+0.9=20.9$

(d) One hundred and two ones

$1\times100+2\times1=100+2=102.0$

(e) Six hundred point eight

Question: 4 Write each of the following as decimals:

(a) $\small \frac{5}{10}$ (b) $\small 3+\frac{7}{10}$ (c) $\small 200+60+5+\frac{1}{10}$ (d) $\small 70+\frac{8}{10}$
(e) $\small \frac{88}{10}$ (f) $\small 4\frac{2}{10}$ (g) $\small \frac{3}{2}$ (h) $\small \frac{2}{5}$ (i) $\small \frac{12}{5}$ (j) $\small 3\frac{3}{5}$ (k) $\small 4\frac{1}{2}$

Answer: As we know when we divide a number by 10, the result is that number with a decimal after one digit,(we count the digit from right to left). So Keeping that in mind,

(a) $\small \frac{5}{10}=0.5$
(b) $\small 3+\frac{7}{10}=3+0.7=3.7$
(c) $\small 200+60+5+\frac{1}{10}=265+0.1=265.1$
(d) $\small 70+\frac{8}{10}=70+0.8=70.8$
(e) $\small \frac{88}{10}=8.8$
(f) $\small 4\frac{2}{10}=4+\frac{2}{10}=4+0.2=4.2$
(g) $\small \frac{3}{2}=\frac{3}{2}\times\frac{5}{5}=\frac{15}{10}=1.5$
(h) $\small \frac{2}{5}=\frac{2}{5}\times\frac{2}{2}=\frac{4}{10}=0.4$
(i) $\small \frac{12}{5}=\frac{12}{5}\times\frac{2}{2}=\frac{24}{10}=2.4$
(j) $\small 3\frac{3}{5}=3+\frac{3}{5}=3+\frac{3}{5}\times\frac{2}{2}=3+\frac{6}{10}=3+0.6=3.6$
(k) $\small 4\frac{1}{2}=4+\frac{1}{2}=4+\frac{1}{2}\times\frac{5}{5}=4+\frac{5}{10}=4+0.5=4.5$

Question: 5 Write the following decimals as fractions. Reduce the fractions to lowest form.

(a) $\small 0.6$ (b) $\small 2.5$ (c) $\small 1.0$ (d) $\small 3.8$ (e) $\small 13.7$ (f) $\small 21.2$ (g) $\small 6.4$

Answer: Converting Decimals into Fractions. we get

(a) $\small 0.6=\frac{6}{10}=\frac{3}{5}$
(b) $\small 2.5=\frac{25}{10}=\frac{5}{2}$
(c) $\small 1.0=1$
(d) $\small 3.8=\frac{38}{10}=\frac{19}{5}$
(e) $\small 13.7=\frac{137}{10}$
(f) $\small 21.2=\frac{212}{10}=\frac{106}{5}$
(g) $\small 6.4=\frac{64}{10}=\frac{32}{5}$

Question: 6 Express the following as cm using decimals.

(a) 2 mm (b) 30 mm (c) 116 mm (d) 4 cm 2 mm (e) 162 mm (f) 83 mm

Answer: As we know
1 cm = 10 mm
1 mm = 0.1 cm
So,

(a) 2 mm $=\frac{2}{10}\:cm=0.2\:cm$

(b) 30 mm $=\frac{30}{10}\:cm=3.0\: cm$

(d) 4 cm 2 mm $=4\:cm+\frac{2}{10}\:cm=4+0.2=4.2\: cm$

(e) 162 mm $=\frac{162}{10}\:cm=16.2\: cm$

(f) 83 mm $=\frac{83}{10}\:cm=8.3\: cm$

Question: 7 Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?

(a) $\small 0.8$ (b) $\small 5.1$ (c) $\small 2.6$ (d) $\small 6.4$ (e) $\small 9.1$ (f) $\small 4.9$

Answer:

(a) $\small 0.8$ Lies Between 0 and 1. and 1 is the closest whole number to it.
(b) $\small 5.1$ Lies between 5 and 6 and 5 is the closest whole number to it.
(c) $\small 2.6$ Lies between 2 and 3 and 3 is the closest whole number to it.
(d) $\small 6.4$ Lies between 6 and 7 and 6 is the closest whole number to it.
(e) $\small 9.1$ Lies between 9 and 10 and 9 is the closest whole number to it.
(f) $\small 4.9$ Lies between 4 and 5. and 5 is the closest whole number to it.

Question: 8 Show the following numbers on the number line.

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

Answer: The numbers on the number line are:

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

Question: 9 Write the decimal number represented by the points A, B, C, D on the given number line.

Answer: As we can see the points lies in the number line,

$A=\frac{8}{10}=0.8$

$B=1+\frac{3}{10}=1+0.3=1.3$

$C=2+\frac{2}{10}=2+0.2=2.2$

$D=2+\frac{9}{10}=2+0.9=2.9$

Question: 10 (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?

Answer: length of Ramesh’s notebook = 9 cm 5 mm

As we know,

$1mm=\frac{1}{10}cm$

So Length in the unit of cm :

$=9cm+5\times\frac{1}{10}cm=9cm+0.5cm=9.5cm$

Hence the Length of Ramesh's notebook is 9.5 cm.

Question: 10 (b) The length of a young gram plant is 65 mm. Express its length in cm.

Answer: length of a young gram plant is 65 mm

As we know,

$1mm=\frac{1}{10}cm$

So,

$65mm=65\times\frac{1}{10}cm=\frac{65}{10}cm=6.5cm$

So Length of young gram plant is 6.5 cm.

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.2

Question:1 Complete the table with the help of these boxes and use decimals to write the number.

Ones

Tenths

Hundredsths

Number

(a)

(b)

(c)

Answer:

Ones

Tenths

Hundredths

Number

(a)

0

2

6

0.26

(b)

1

3

8

1.38

(c)

1

2

8

1.28

Question:2 Write the numbers given in the following place value table in decimal form.

Hundreds

100

Tens

10

Ones

1

Tenths

$\small \frac{1}{10}$

Hundredths

$\small \frac{1}{100}$

Thousaandths

$\small \frac{1}{1000}$

(a)

(b)

(c)

(d)

(e)

0

1

0

2

0

0

0

3

1

1

3

2

0

1

2

2

6

0

9

2

5

3

2

0

4

0

0

5

2

1

Answer: As we can see from the table, the numbers are:

(a) $3+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25$

(b) $100+2+\frac{6}{10}+\frac{3}{100}=100+2+0.6+0.03=102.63$

(c) $30+\frac{2}{100}+\frac{5}{1000}=30+0.02+0.005=30.025$

(d) $200+10+1+\frac{9}{10}+\frac{2}{1000}=211+0.9+0.002=211.902$

(e) $10+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12+0.2+0.04+0.001=12.241$

Question: 3 Write the following decimals in the place value table.

(a) $\small 0.29$ (b) $\small 2.08$ (c) $\small 19.60$ (d) $\small 148.32$ (e) $\small 200.812$

Answer:

(a)

$\small 0.29=0.2+0.09=\frac{2}{10}+\frac{9}{100}$

(b)

$\small 2.08=2+0.08=2+\frac{8}{100}$

(c)

$\small 19.60=19+0.6=10+9+\frac{6}{10}$

(d)

$\small 148.32=148+0.3+0.02=100+40+8+\frac{3}{10}+\frac{2}{100}$

(e)

$\small 200.812=200+0.8+0.01+0.002=200+\frac{8}{10}+\frac{1}{100}+\frac{2}{1000}$

So, the table becomes:

Hundreds

Tens

Ones

Tenths

Hundredths

Thousands

0

0

0

2

9

0

0

0

2

0

8

0

0

1

9

6

0

0

1

4

8

3

2

0

2

0

0

8

1

2

Question: 4 Write each of the following as decimals.

(a) $\small 20+9+\frac{4}{10}+\frac{1}{100}$ (b) $\small 137+\frac{5}{100}$ (c) $\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}$
(d) $\small 23+\frac{2}{10}+\frac{6}{1000}$ (e) $\small 700+20+5+\frac{9}{100}$

Answer:

Writing the number in a decimal form:

(a)

$\small 20+9+\frac{4}{10}+\frac{1}{100}=29+0.4+0.01=29.41$

(b)

$\small 137+\frac{5}{100}=137+0.05=137.05$

(c)

$\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}=0.7+0.06+0.004=0.764$
(d)

$\small 23+\frac{2}{10}+\frac{6}{1000}=23+0.2+0.006=23.206$

(e)

$\small 700+20+5+\frac{9}{100}=725+0.09=725.09$

Question: 5 Write each of the following decimals in words.

(a) $\small 0.03$ (b) $\small 1.20$ (c) $\small 108.56$ (d) $\small 10.07$ (e) $\small 0.032$ (f) $\small 5.008$

Answer: As we know After decimal, we call one digit at a time.

(a) $\small 0.03$ = Zero point zero three

(b) $\small 1.20$ = One point two zero

(c) $\small 108.56$ = One hundred eight point five six

(d) $\small 10.07$ = Ten point zero seven

(e) $\small 0.032$ = Zero point zero three two

(f) $\small 5.008$ = Five point zero zero eight.

Question:6 Between which two numbers in tenths place on the number line does each of the given number lies?

(a) $\small 0.06$ (b) $\small 0.45$ (c) $\small 0.19$ (d) $\small 0.66$ (e) $\small 0.92$ (f) $\small 0.57$

Answer: Here, we have a zoomed version of the number line in which the interval between two number is 0.1 instead of 1 which we use normally. So,

(a) $\small 0.06$ lies between 0.0 and 0.1

(b) $\small 0.45$ lies between 0.4 and 0.5

(c) $\small 0.19$ lies between 0.1 and 0.2

(d) $\small 0.66$ lies between 0.6 and 0.7

(e) $\small 0.92$ lies between 0.9 and 1.0

(f) $\small 0.57$ lies between 0.5 and 0.6.

Question:7 Write as fractions in lowest terms.

(a) $\small 0.60$ (b) $\small 0.05$ (c) $\small 0.75$ (d) $\small 0.18$ (e) $\small 0.25$ (f) $\small 0.125$ (g) $\small 0.066$

Answer: The Numbers in the lowest form of the fraction are:

(a)

$\small 0.60=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

(b)

$\small 0.05=\frac{5}{100}=\frac{1}{20}$

(c)

$\small 0.75=\frac{75}{100}=\frac{3}{4}$

(d)

$\small 0.18=\frac{18}{100}=\frac{9}{50}$

(e)

$\small 0.25=\frac{25}{100}=\frac{1}{4}$

(f)

$\small 0.125=\frac{125}{1000}=\frac{5}{40}=\frac{1}{8}$

(g)

$\small 0.066=\frac{66}{1000}=\frac{33}{500}$

CBSE NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.3

Question: 1 Which is greater? (a) $\small 0.3$ or $\small 0.4$ (b) $\small 0.07$ or $\small 0.02$ (c) $\small 3$ or $\small 0.8$ (d) $\small 0.5$ or $\small 0.05$ (e) $\small 1.23$ or $\small 1.2$ (f) $\small 0.099$ or $\small 0.19$

(g) $\small 1.5$ or $\small 1.50$ (h) $\small 1.431$ or $\small 1.490$ (i) $\small 3.3$ or $\small 3.300$ (j) $\small 5.64$ or $\small 5.603$

Answer: We compare Decimals just like the way we compare normal numbers. First, we see the leftmost digit of the numbers and if they are same then we move toward the right digits. So,

(a) $\small 0.3$ or $\small 0.4$

0.4 is greater as while number part (the number before the decimal) is the same.

(b) $\small 0.07$ or $\small 0.02$

0.07 is greater

(c) $\small 3$ or $\small 0.8$

3 is greater as the whole part of the numbers are 3 and 0.so obviously 3 is greater than 0.

(d) $\small 0.5$ or $\small 0.05$

0.5 is greater

(e) $\small 1.23$ or $\small 1.2$

1.23 is greater

(f) $\small 0.099$ or $\small 0.19$

0.19 is greater

(g) $\small 1.5$ or $\small 1.50$

As we know that after the decimal, we can add as many zeros as we want in the rightmost to the numbers just like we can add zeroes to the left side of any whole number. this we can do because those zeroes do not carry any value so they don't affect the numbers.

So. They both are the same number.

(h) $\small 1.431$ or $\small 1.490$

1.490 is greater

(i) $\small 3.3$ or $\small 3.300$

As we know in decimal the RIGHTMOST zeros don't carry any value.so,

They both are the same number

(j) $\small 5.64$ or $\small 5.603$

5.64 is greater.

Question:2 Make five more examples and find the greater number from them.

Answer: we can take any two number from the number line and compare them. For example,

1) 9.9 > 3.3

2) 6.6 < 9.9

3) 3.6 > 3.3

4) 9.6 < 9.9

5) 6.9 > 3.9

Solutions for NCERT class 6 maths chapter 8 decimals topic 8.5 using decimals

Question:(i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.

Answer:As we know that

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

2 rupees 5 paise :

$=2+5\times\frac{1}{100}=2+\frac{5}{100}=2+0.05=2.05\:rupees$

and 2 rupees 50 paise:

$=2+50\times\frac{1}{100}=2+\frac{50}{100}=2+0.5=2.5\:rupees$

Question:(ii) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals?

Answer: As we know that

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

20 rupees 7 paise:

$=20+7\times\frac{1}{100}=20+\frac{7}{100}=20+0.07=20.07\:rupees$

and 21 rupees 75 paise:

$=21+75\times\frac{1}{100}=21+\frac{75}{100}=21+0.75=21.75\:rupees$

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.5.2 length

Question:1 Can you write 4 mm in ‘cm’ using decimals?

Answer:As we know,

1 cm = 10 mm

$1 mm = \frac{1}{10}\:cm$

So,

$4 mm = 4\times\frac{1}{10}\:cm=\frac{4}{10}\:cm=0.4cm$ .

Question:2 How will you write 7cm 5 mm in ‘cm’ using decimals?

Answer: As we know,

1 cm = 10 mm

$1mm=\frac{1}{10}cm$

So,

7cm 5 mm :

$=7 + 5\times\frac{1}{10}=7+\frac{5}{10}=7+0.5=7.5cm$ .

Question:3 Can you now write 52 m as ‘km’ using decimals? How will you write 340 m as ‘km’ using decimals? How will you write 2008 m in ‘km’?

Answer: As we know,

1 km = 1000 m

$1 m =\frac{1}{1000}km$

So,

52 m in km :

$52 m =52\times\frac{1}{1000}km=\frac{52}{1000}km=0.052km$
340 m in km :

$340m =340\times\frac{1}{1000}km=\frac{340}{1000}km=0.34km$

2008 m in km :

$2008m =2008\times\frac{1}{1000}km=\frac{2008}{1000}km=2.008km$

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.5.3 Weight

Question:1 Can you now write 456 g as ‘kg’ using decimals?

Answer: As we know,

1 kg = 1000 g

$1\:g=\frac{1}{1000}\:kg$

So,

456 g in kg:

$456\:g=456\times\frac{1}{1000}\:kg=\frac{456}{1000}\:kg=0.456\:kg$

Question:2 How will you write 2 kg 9 g in ‘kg’ using decimals?

Answer: As we know,

1 kg = 1000 g

$1\:g = \frac{1}{1000}\:kg$

So,

2 kg 9 g in kg:

$2\:kg+9\:g = 2\:kg+\frac{9}{1000}\:kg=2\:kg+0.009\:kg=2.009\:kg$

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.4

Question:1 Express as rupees using decimals.

(a) 5 paise (b) 75 paise (c) 20 paise (d) 50 rupees 90 paise (e) 725 paise

Answer: As we know,

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

(a) 5 paise

$=5\times\frac{1}{100}=\frac{5}{100}=0.05\:rupees$

(b) 75 paise

$=75\times\frac{1}{100}=\frac{75}{100}=0.75\:rupees$

(c) 20 paise

$=20\times\frac{1}{100}=\frac{20}{100}=0.20\:rupees$

(d) 50 rupees 90 paise

$=50+90\times\frac{1}{100}=50+\frac{90}{100}=50+0.9=50.9\:rupees$

(e) 725 paise

$=725\times\frac{1}{100}=\frac{725}{100}=7.25\:rupees$

Question:2 Express as metres using decimals.

(a) 15 cm (b) 6 cm (c) 2 m 45 cm (d) 9 m 7 cm (e) 419 cm

Answer: As we know,

1 m = 100 cm

Which means

$1\:cm=\frac{1}{100}\:m$ .

So,

(a) 15 cm

$15\:cm=15\times\frac{1}{100}=\frac{15}{100}=0.15\:m$

(b) 6 cm

$6\:cm=6\times\frac{1}{100}=\frac{6}{100}=0.06\:m$

(c) 2 m 45 cm

$2+45\:cm=2+45\times\frac{1}{100}=2+\frac{45}{100}=2+0.45=2.45\:m$

(d) 9 m 7 cm

$9+7\:cm=9+7\times\frac{1}{100}=9+\frac{7}{100}=9+0.07=9.07\:m$

(e) 419 cm

$419\:cm=419\times\frac{1}{100}=\frac{419}{100}=4.19\:m$

Question:3 Express as cm using decimals.

(a) 5 mm (b) 60 mm (c) 164 mm (d) 9 cm 8 mm (e) 93 mm

Answer: As we know,

1 cm = 10 mm

which means

$1\:mm=\frac{1}{10}\:cm$

So,

(a) 5 mm :

$5\:mm=5\times\frac{1}{10}=\frac{5}{10}=0.5\:cm$

(b) 60 mm

$60\:mm=60\times\frac{1}{10}=\frac{60}{10}=6.0\:cm$

(c) 164 mm

$164\:mm=164\times\frac{1}{10}=\frac{164}{10}=16.4\:cm$

(d) 9 cm 8 mm

$9\:cm+8\:mm=9+8\times\frac{1}{10}=9+\frac{8}{10}=9+0.8=9.8\:cm$

(e) 93 mm

$93\:mm=93\times\frac{1}{10}=\frac{93}{10}=9.3\:cm$

Question:4 Express as km using decimals.

(a) 8 m (b) 88 m (c) 8888 m (d) 70 km 5 m

Answer: As we know,

1 km = 1000 m

which means

$1\:m=\frac{1}{1000}\:km$

So,

(a) 8 m

$8\:m=8\times\frac{1}{1000}=\frac{8}{1000}=0.008\:km$

(b) 88 m

$88\:m=88\times\frac{1}{1000}=\frac{88}{1000}=0.088\:km$

(c) 8888 m

$8888\:m=8888\times\frac{1}{1000}=\frac{8888}{1000}=8.888\:km$

(d) 70 km 5 m

$70\:km+5\:m=70+5\times\frac{1}{1000}=70+\frac{5}{1000}=70+0.005=70.005\:km$

Question:5 Express as kg using decimals.

(a) 2 g (b) 100 g (c) 3750 g
(d) 5 kg 8 g (e) 26 kg 50 g

Answer: As we know,

1 kg = 1000 g

which means

$1 g = \frac{1}{1000}kg$

So,

(a) 2 g

$2 g = \frac{2}{1000}kg=0.002kg$

(b) 100 g

$100 g = \frac{100}{1000}kg=0.1kg$

(c) 3750 g

$3750 g = \frac{3750}{1000}kg=3.75kg$

(d) 5 kg 8 g

$5kg\:\:8g=5+8g =5kg+ \frac{8}{1000}kg=5kg+0.008kg=5.008kg$

(e) 26 kg 50 g

$26kg\:\:50g=26+50g =26kg+ \frac{50}{1000}kg=26kg+0.05kg=26.05kg$

Solutions for NCERT class 6 maths chapter 8 decimals topic 8.6 addition of numbers with decimals

Question:(i) Find

$\small 0.29 + 0.36$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$\small \\0.29$

$\small +$ $\small \\0.36$

$\small \\0.65$

Hence

$\small 0.29 + 0.36=0.65$

Question:(ii) Find

$\small 0.7 + 0.08$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$+\:0.08$

Hence,

$\small 0.7 + 0.08=0.78$

Question:(iii) Find

$\small 1.54 + 1.80$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$+\:1.80$

.

Question:(iv) Find

$\small 2.66 + 1.85$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$+\:1.85$

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.5

Question:1 Find the sum in each of the following :

(a) $\small 0.007 + 8.5 + 30.08$

(b) $\small 15 + 0.632 + 13.8$

(c) $\small 27.076 + 0.55 + 0.004$

(d) $\small 25.65 + 9.005 + 3.7$

(e) $\small 0.75 + 10.425 + 2$

(f) $\small 280.69 + 25.2 + 38$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

(a) $\small 0.007 + 8.5 + 30.08$

$+\:30.080$

(b) $\small 15 + 0.632 + 13.8$

$\small 15.000$

$\small 0.632$

$\small +\:13.800$

$\small 29.432$

(c) $\small 27.076 + 0.55 + 0.004$

$+\:0.004$

(d) $\small 25.65 + 9.005 + 3.7$

$+\:3.700$

(e) $\small 0.75 + 10.425 + 2$

$+\:2.000$

(f) $\small 280.69 + 25.2 + 38$

$+\:38.00$

Question: 2 Rashid spent Rs $\small 35.75$ for Maths book and Rs $\small 32.60$ for Science book. Find the total amount spent by Rashid.

Answer:

Amount spend on maths book = Rs $\small 35.75$

Amount spend on science book = Rs $\small 32.60$

Total amount spend = Rs $\small 35.75$ Rs $\small 32.60$

= Rs

Hence the total amount spends by Rashid is Rs 68.35.

Question:3 Radhika’s mother gave her Rs $\small 10.50$ and her father gave her Rs $\small 15.80$ , find the total amount given to Radhika by the parents.

Answer:

The amount Radhika got from mother = Rs $\small 10.50$

The amount Radhika got from father = Rs $\small 15.80$

Total amount Radhika have = Rs $\small 10.50$ Rs $\small 15.80$

= Rs

Hence the total amount of money Radhika got from her parents is Rs 26.30.

Question:4 Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.

Answer:

Length of cloth for shirt = 3 m 20 cm

Length of cloth for trouser = 2 m 5 cm

As we know.

For adding two numbers their unit must be the same.

So making length in meter:

$3m+20cm=3\:m+\frac{20}{100}m$

$=3\:m+0.2m$

and

$2m+5cm=2\:m+\frac{5}{100}m$

$=2\:m+0.05m$

Now,

Total Length of the cloth = 3.2 m + 2.05 m

= 5.25 m

Hence Nasreen bought clothe with a length of 5.25 m.

Question:5 Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?

Answer:

The distance covered in the morning = 2 km 35 m

The distance covered in the evening = 1 km 7 m

As we know.

For adding two numbers their unit must be the same.

So Lengths in the Kilometer :

$2 km+35m=2km+\frac{35}{1000}km$

And

$1km+7m=1km+\frac{7}{1000}km$

$=1.007\:km$

Now,

Total Distance covered by Naresh = 2.035 km + 1.007 km

= 3.042 km

Hence the distance covered by Naresh is 3.042 km.

Question:6 Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?

Answer:

The distance travelled by bus = 15 km 268 m

The distance travelled by car = 7 km 7 m

The distance travelled by foot = 500 m

As we know.

For adding two numbers their unit must be the same.

So,

$15 km +268 m = 15 km + \frac{268}{1000}km$

Also

$7km+7m=7km+\frac{7}{1000}km$

And,

$500m=\frac{500}{1000}km$

So,

The total distance Sunita traveled = 15.268 km + 7.007 km + 0.5 km

= 22.775 km

Hence the distance between Sunita's residence and school is 22.775 km.

Question:7 Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850g flour. Find the total weight of his purchases.

Answer:

amount of rice purchased by Ravi = 5 kg 400 g.

amount of sugar purchased by Ravi = 2 kg 20 g.

amount of flour purchased by Ravi = 10 kg 850 g.

As we know.

For adding two numbers their unit must be the same.

So,

$5\:kg+400g=5kg+\frac{400}{1000}kg$

Also,

$2\:kg+20g=2kg+\frac{20}{1000}kg$

And,

$10\:kg+850g=10kg+\frac{850}{1000}kg$

So,

The total weight of the grain Ravi purchased = 5.4 kg + 2.002 kg + 10.85 kg

= 18.270 kg

Hence Ravi purchased total grain having the weight of 18.270 kg.

Solutions for NCERT class 7 maths chapter 8 decimals topic 8.7 subtraction of decimals

Question:1 Subtract $\small 1.85$ from $\small 5.46$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

Question:2 Subtract $\small 5.25$ from $\small 8.28$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

Question:3 Subtract $\small 0.95$ from $\small 2.29$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

Question: 4 Subtract $\small 2.25$ from

$\small 5.68$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

CBSE NCERT solutions for class 7 maths chapter 8 decimals-Exercise: 8.6

Question: 1 Subtract :

(a) Rs $\small 18.25$ from Rs $\small 20.75$

(b) $\small 202.54$ m from 250 m

(c) Rs $\small 5.36$ from Rs $\small 8.40$

(d) $\small 2.051$ km from $\small 5.206$ km

(e) $\small 0.314$ kg from $\small 2.107$ kg

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

(a) Rs $\small 18.25$ from Rs $\small 20.75$

Rs

(b) $\small 202.54$ m from 250 m\

m

(c)Rs $\small 5.36$ from Rs $\small 8.40$

Rs

(d ) $\small 2.051$ km from $\small 5.206$ km

km

(e) $\small 0.314$ kg from $\small 2.107$ kg

kg

Question: 2 Find the value of :

(a) $\small 9.756 - 6.28$

(b) $\small 21.05 -15.27$

(c) $\small 18.5 - 6.79$

(d) $\small 11.6 - 9.847$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

(a) $\small 9.756 - 6.28$

(b) $\small 21.05 -15.27$

(c) $\small 18.5 - 6.79$

(d) $\small 11.6 - 9.847$

Question: 3 Raju bought a book for Rs $\small 35.65$ . He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Answer: Total money Raju gave to shopkeeper = Rs 50

The price of the book = Rs 35.65

Money shopkeeper will return Raju = Rs 50.00 - Rs 35.65

= Rs 14.35

Hence Raju got 14.35 Rs from the shopkeeper.

Question: 4 Rani had Rs $\small 18.50$ . She bought one ice-cream for Rs $\small 11.75$ . How much money does she have now?

Answer: Total Money Rani had = Rs 18.50

The money she spent on ice-cream = 11.75

The Remaining money Rani have = Rs 18.50 - Rs 11.75

= Rs 6.75

Hence Rani has 6.75 Rs now.

Question:5 Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?

Answer: Length of cloth initially = 20 m 5 cm = 20.05 m

Length of cloth Tina cuts = 4 m 50 cm = 4.50 m

The length of remaining cloth = 20.05 m - 4.50 m

= 15.55 m

= 15 m 55 cm

Hence the length of remaining cloth Tina have is 15 m 55 cm.

Question:6 Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?

Answer: Total distance travelled by Namita = 20 km 50 m = 20.05 km

The distance travelled by bus = 10 km 200 m = 10.20 km

The remaining distance which is traveled by auto = 20.05 km - 10.20 km

= 9.850 km.

Hence the distance travelled by Namita in auto is 9.850 km.

Question:7 Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?

Answer: Total weight of the vegetables = 10 kg

weight of onion = 3 kg 500 g = 3.5 kg

weight of tomato = 2 kg 75 g = 2.075 kg

Total weight of onion and potato = 3.5 kg + 2.075 kg

= 5.575 kg

So, The rest weight of potato = 10 kg - 5.575 kg

= 4.425 kg

Hence, the weight of the potato is 4.425 kg.

Chapter No. Chapter Name

Chapter 1 NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers

Chapter 2 NCERT solutions for class 6 maths chapter 2 Whole Numbers

Chapter 3 NCERT solutions for class 6 maths chapter 3 Playing with Numbers

Chapter 4 NCERT Solutions for class 6 maths chapter 4 Basic Geometrical Ideas

Chapter 5 NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes

Chapter 6 NCERT solutions for class 6 maths chapter 6 Integers

Chapter 7 NCERT Solutions for class 6 maths chapter 7 Fractions

Chapter 8 NCERT solutions for class 6 maths chapter 8 Decimals

Chapter 9 NCERT solutions for class 6 maths chapter 9 Data Handling

Chapter 10 NCERT solutions for class 6 maths chapter 10 Mensuration

Chapter 11 NCERT Solutions for class 6 maths chapter 11 Algebra

Chapter 12 NCERT solutions for class 6 maths chapter 12 Ratio and Proportion

Chapter 13 NCERT solutions for class 6 maths chapter 13 Symmetry

Chapter 14 NCERT Solutions for class 6 maths chapter 14 Practical Geometry

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Chapter No.	Chapter Name
Chapter 1	NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers
Chapter 2	NCERT solutions for class 6 maths chapter 2 Whole Numbers
Chapter 3	NCERT solutions for class 6 maths chapter 3 Playing with Numbers
Chapter 4	NCERT Solutions for class 6 maths chapter 4 Basic Geometrical Ideas
Chapter 5	NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes
Chapter 6	NCERT solutions for class 6 maths chapter 6 Integers
Chapter 7	NCERT Solutions for class 6 maths chapter 7 Fractions
Chapter 8	NCERT solutions for class 6 maths chapter 8 Decimals
Chapter 9	NCERT solutions for class 6 maths chapter 9 Data Handling
Chapter 10	NCERT solutions for class 6 maths chapter 10 Mensuration
Chapter 11	NCERT Solutions for class 6 maths chapter 11 Algebra
Chapter 12	NCERT solutions for class 6 maths chapter 12 Ratio and Proportion
Chapter 13	NCERT solutions for class 6 maths chapter 13 Symmetry
Chapter 14	NCERT Solutions for class 6 maths chapter 14 Practical Geometry

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals NCERT Solutions for Class 6 Maths Chapter 8 Decimals Solutions of NCERT for class ...

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 tenths
Question: 1 Can you now write the following as decimals

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.1

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.2

CBSE NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.3

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.5.2 length

Solutions for NCERT class 6 maths chapter 8 decimals topic 8.6 addition of numbers with decimals

Solutions for NCERT class 7 maths chapter 8 decimals topic 8.7 subtraction of decimals

Question:1 Subtract $\small 1.85$ from $\small 5.46$

CBSE NCERT solutions for class 7 maths chapter 8 decimals-Exercise: 8.6

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 tenthsQuestion: 1 Can you now write the following as decimals

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.1

NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.2

CBSE NCERT solutions for class 6 maths chapter 8 decimals-Exercise: 8.3

Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.5.2 length

Solutions for NCERT class 6 maths chapter 8 decimals topic 8.6 addition of numbers with decimals

Solutions for NCERT class 7 maths chapter 8 decimals topic 8.7 subtraction of decimals

Question:1 Subtract from

CBSE NCERT solutions for class 7 maths chapter 8 decimals-Exercise: 8.6

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Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 tenths
Question: 1 Can you now write the following as decimals

Question:1 Subtract $\small 1.85$ from $\small 5.46$