# NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals: If you go to a shop and ask for the price of pencils, the shop keeper tells you a pencil costs rupees 3.5. If you need 4 pencils, how much do you have to pay to the shopkeeper? In such situations, you need calculations using decimals. You have to add 3.5 four times which will give 14 as the answer, so you have to give 14 rupees to the shopkeeper. Calculations using decimals are necessary for our daily life. In this article, you will get NCERT solutions for class 6 maths chapter 8 decimals. In this chapter, you will study about decimal numbers, addition and subtraction of decimal numbers. Also, you will study how to convert fractions to decimals and also decimals to fractions. There are many questions in the solutions of NCERT for class 6 maths chapter 8 decimals which will give more clarity to the concepts. You will also find that how to read the decimal number which is different from the whole number. You will read 24.365 as twenty-four point three six five, not twenty-four point three sixty-five. The number 5 can be written in decimal form as 5.000, 3.5 as 3.500000000, There are 38 questions in 6 exercises of this chapter. You will get detailed explanations of these questions in the CBSE NCERT solutions for class 6 maths chapter 8 decimals. It will help you in understanding the chapter in a more easy way. You can get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.

Topic-wise and exercise questions are solved in the NCERT solutions for class 6 maths chapter 8 decimals. To go to respective topics or exercise click the links below.

Topic 8.2

Topic 8.5

Topic 8.5.2

Topic 8.5.3

Topic 8.6

Topic 8.7

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

Exercise 8.5

Exercise 8.6

## 8.1 Introduction

8.2 Tenths

8.3 Hundredths

8.4 Comparing Decimals

8.5 Using Decimals

8.5.1 Money

8.5.2 Length

8.5.3 Weight

8.6 Addition of Numbers with Decimals

8.7 Subtraction of Decimals

## Solutions of NCERT for class 6 maths chapter 8 decimals topic 8.2 tenthsQuestion: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

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

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

Observing from the figure, we get

 Hundreds (100) Tens (10) Ones (1) Tenths $\small (\frac{1}{10})$ a)      0 3 1 2 b)    1 1 0 4

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

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

(e) Six hundred point eight

(a) Seven-tenths :

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

(b) Two tens and nine-tenths:

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

(c) Fourteen point six :

$=14.6$

(d) One hundred and two ones

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

(e) Six hundred point eight

$=600.8$

Question: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)  $\small 4\frac{1}{2}$

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)

$\small 4\frac{1}{2}=4+\frac{1}{2}=4+\frac{1}{2}\times\frac{5}{5}=4+\frac{5}{10}=4+0.5=4.5$

(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) $\small 6.4$

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)

$\small 6.4=\frac{64}{10}=\frac{32}{5}$

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

$=\frac{30}{10}\:cm=3.0\: cm$

(c) 116 mm

$=\frac{116}{10}\:cm=11.6\: cm$

(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

$=\frac{83}{10}\:cm=8.3\: cm$

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

(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.

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

The numbers on the number line are:

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

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

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

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

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

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

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.

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

 Ones Tenths Hundredsths Number (a) (b) (c)

 Ones Tenths Hundredths Number (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28

 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

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)

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

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

(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: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)   $\small 700+20+5+\frac{9}{100}$

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)

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

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

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.

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

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) $\small 0.066$

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

(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 $\small 5.603$

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.

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

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:

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

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

As we know,

1 cm = 10 mm

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

So,

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

As we know,

1 cm = 10 mm

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

So,

7cm 5 mm :

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

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

As we know,

1 kg = 1000 g

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

So,

456 g in kg:

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

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

As we know that

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

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

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

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

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

$\small 0.29 + 0.36$

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

$\small 0.7 + 0.08$

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,

$0.70$

$+\:0.08$

$0.78$

Hence,

$\small 0.7 + 0.08=0.78$

Question:(iii) Find

$\small 1.54 + 1.80$

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,

$1.54$

$+\:1.80$

$3.34$.

Question:(iv) Find

$\small 2.66 + 1.85$

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,

$2.66$

$+\:1.85$

$4.51$

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)    $\small 280.69 + 25.2 + 38$

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)    $\small 0.007 + 8.5 + 30.08$

$0.007$

$8.500$

$+\:30.080$

$38.587$

(b)    $\small 15 + 0.632 + 13.8$

$\small 15.000$

$\small 0.632$

$\small +\:13.800$

$\small 29.432$

(c)    $\small 27.076 + 0.55 + 0.004$

$27.076$

$0.550$

$+\:0.004$

$27.630$

(d)    $\small 25.65 + 9.005 + 3.7$

$25.650$

$9.005$

$+\:3.700$

$38.355$

(e)    $\small 0.75 + 10.425 + 2$

$0.750$

$10.425$

$+\:2.000$

$13.175$

(f)    $\small 280.69 + 25.2 + 38$

$280.69$

$25.20$

$+\:38.00$

$343.89$

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  $68.35$

Hence the total amount spends by Rashid is Rs 68.35.

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 $26.30$

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

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:

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

$=3\:m+0.2m$

$=3.2m$

and

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

$=2\:m+0.05m$

$=2.05m$

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.

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 :

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

$=2km+0.035km$

$=2.035km$

And

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

$=1km+0.007km$

$=1.007\:km$

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.

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,

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

$= 15 km + 0.268km$

$= 15.268km$

Also

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

$=7km+0.007km$

$=7.007km$

And,

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

$=0.5km$

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.

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,

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

$=5kg+0.4kg$

$=5.4kg$

Also,

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

$=2kg+0.02kg$

$=2.002kg$

And,

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

$=10kg+0.85kg$

$=10.85kg$

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.

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

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

So

$5.46$

$-1.85$

$3.61$

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

So

$8.28$

$-5.25$

$3.03$

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

So

$2.29$

$-0.95$

$1.34$

$\small 5.68$

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

So

$5.68$

$-2.25$

$3.43$

## 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

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 $\small 20.75$

$20.75$

$-18.25$

$2.50$ Rs

(b)  $\small 202.54$ m from 250 m\

$250.00$

$-202.54$

$47.46$ m

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

$8.40$

$-5.36$

$3.04$ Rs

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

$5.206$

$-2.051$

$3.155$ km

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

$2.107$

$-0.314$

$1.793$ 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) $\small 11.6 - 9.847$

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

So

(a) $\small 9.756 - 6.28$

$9.756$

$-6.280$

$3.476$

(b) $\small 21.05 -15.27$

$21.05$

$-15.27$

$5.78$

(c) $\small 18.5 - 6.79$

$18.50$

$-6.79$

$11.71$

(d) $\small 11.6 - 9.847$

$11.600$

$-9.847$

$1.753$

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.

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.

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.

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.

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.

## NCERT Solutions for Class 6 Mathematics - Chapter-wise

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

## NCERT Solutions for Class 6 - Subject wise

Some important findings from the NCERT solutions for class 6 maths chapter 8 decimals which will be useful in real life too are listed below.

Converting a quantity from one unit to another-

• Millimeter to centimeter-

$1mm=\frac{1}{10}cm=0.1cm$

• Centimeter to meter-

$1cm=\frac{1}{100}m=0.01m$

• Paise to rupee

$1\ paise=\frac{1}{100}rupees=0.01 rupees$

$250\ paise=\frac{250}{100}rupees=2.5rupees$

• Grams to kg

$1gm=\frac{1}{1000}kg=0.001kg$

Tip- You don't have to remember these units conversion. If you know the value of the quantity in one unit and other units, you can convert it very easily. In CBSE NCERT solutions for class 6 maths chapter 8 decimals, you will find many unit conversion questions that will further strengthen your fundamentals.