# Frank Learning Maths Class 6 Solutions Chapter 6

## Frank Learning Maths Class 6 Solutions Chapter 6 Decimals

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 6 Decimals. Here students can easily find step by step solutions of all the problems for Decimals. Here students will find solutions for Exercise 6.1, 6.2, 6.3, 6.4, 6.5 and 6.6. Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Decimals Exercise 6.1 Solution

Question no – (1)

Solution :

(a) 9851250

Here underline digit is 9

So, the place value of 9 is,

= 9851250 → 9000000

(b) 25728

Here underline digit is 7

So, the place value of 7 is,

= 25728 → 700

(c) 24.325

Here underline digit is 3

So, the place value of 3 is,

= 24.325 → 0.3

(d) 625.004

Here underline digit is 4

So, the place value of 4 is,

= 625.004 → 0.004

(e) 672.89

Here underline digit is 9

So, the place value of 9 is,

= 672.89 → 0.09

(f) 1232.760

Here underline digit is 2

So, the place value of 2 is,

= 1232.760 → 2

Question no – (2)

Solution :

(a) 9 ones and 3 hundredths 4 thousandths

(9 × 1) + (3 × 1/100) + (4 × 1/1000)

= 9 + 0.03 + 0.004

= 9.034

So, the decimal form is 9.034

(b) 5 hundreds 2 ones and 5 hundredths

= (5 × 100) + (2 × 1) + (5 × 1/100)

= 500 + 2 + 0.05

= 502.05

Thus, the decimal form of 5 hundreds 2 ones and 5 hundredths is 502.05.

(c) Decimal form of 1 one and 1 hundredth,

= (1 × 1) + (1 × 1/100)

= 1 + 0.01

= 1.01

Thus, decimal form of 1 one and 1 hundredth is 1.01

(d) Decimal form of 4 tens 1 one and 2 tenths,

= (4 × 10) + (1 × 1) + (2 × 1/10)

= 40 + 1 + 0.2

= 41.2

So, the decimal form of 4 tens 1 one and 2 tenths is 41.2

(e) 4 thousands 5 hundreds 2 tens and 2 tenths 3 hundredths.

(4 × 1000) + (5 × 100) + (2 × 10) + (2 × 1/10) + (3 × 1/100)

= 4000 + 500 + 20 + 0.2 + 0.03

= 4520.23

So, the decimal form of 4 thousands 5 hundreds 2 tens and 2 tenths 3 hundredths is 4520.23.

Question no – (3)

Solution :

(a) (5 × 1) + (0 × 1/10) + (0 × 1/100) + (7 × 1/1000)

= 5 + 0 + 0 + 0.007

= 5.007

The corresponding decimal number is = 5.007

(b) (3 × 100) + (0 × 10) + (3 × 1) + (5 × 1/10) + (4 × 1/100)

= 300 + 0 + 3 + 0.5 + 0.04

= 303.54

The corresponding decimal number is = 303.54

(c) (1 × 10) + (2 × 1) + (0 × 1/10) + (1 × 1/100) + (0 × 1/1000)

= 10 + 2 + 0 + 0.01 + 0

= 12.01

The corresponding decimal number is = 12.01

(d) (9 × 100) + (8 × 10) + (2 × 1) + (4 × 0.1) + (6 × 0.01) + (3 × 0.001)

= 900 + 80 + 2 + 0.4 + 0.6 + 0.003

= 982.463

The corresponding decimal number is = 982.463

(e) (7 × 100) + (2 × 10) + (0 × 1) + (0 × 1/10) + (3 × 1/100) + (4 × 1/1000)

= 700 + 20 + 0 + 0 + 0.03 + 0.004

= 720.034

The corresponding decimal number is = 720.034

Question no – (4)

Solution :

(a) Given decimal number, 425.325

In expanded form,

= (4 × 100) + (2 × 10) + (5 × 1) + (3 × 1/10) + (2 × 1/100) + (5 × 1/1000)

In Word,

= 4 hundreds 2 tens 5 ones, 3 tenths, 2 hundredths, 5 thousandths

(b) Given decimal number, 425.325

In expanded form,

= (4 × 100) + (2 × 10) + (5 × 1) + (3 × 1/10) + (2 × 1/100) + (5 × 1/1000)

In Word,

= 4 hundreds 2 tens 5 ones, 3 tenths, 2 hundredths, 5 thousandths

(c) Given decimal number, 100.001

In expanded form,

= (1 × 100) + (1 × 1/1000)

In Word,

= 1 hundred and 1 thousandths.

(d) Given decimal number, 308.79

In expanded form,

= (3 × 100) + (8 × 1) + (7 × 1/10) + (9 × 1/100)

In Word form,

= 3 hundreds 8 ones, 7 tenths, 9 hundredths.

(e) Given decimal number, 560.035

In expanded form,

= (5 × 100) + (6 × 10) + (3 × 1/100) + (5 × 1/1000)

In Word form,

= 5 hundreds, 6 tens, 3 hundredths, 5 thousandths

Question no – (5)

Solution :

 60 hundredths 60/100 0.6 0.6 6/10 70 hundredths 70/100 0.7 0.7 7/10 90 hundredths 90/100 0.9 0.9 9/10 20 hundredths 20/100 0.2 0.2 2/10 30 hundredths 30/100 0.3 0.3 3/10

Question no – (7)

Solution :

(a) 30 + 4 + 5/10 + 7/100

= 34+ 0.5 + 0.07

= 34.57

(b) 400 + 40 + 0/10 + 4/100 + 5/1000

= 440 + 0.04 + 0.005

= 440.045

(c) 5 + 0/10 + 3/100 + 7/1000

= 5 + 0.03 + 0.007

= 5037

(d) 700 + 30 + 2 + 5/10 + 3/100 + 9/1000

= 732 + 0.5 + 0.03 + 0.009

= 732.539

(e) 900 + 20 + 7 + 5/10 + 2/100

= 927 + 0.5 + 0.02

= 927.57

(f) 159 + 2/10 + 3/100 + 7/1000

= 159 + 0.2 + 0.03 + 0.007

= 159.237

Question no – (9)

Solution :

 Hundreds (100) Tens (10) ones (1) Tenth (1/10) Hundredth (1/100) Thousandth (1/1000) (a) 2 5 4 (b) 5 7 (c) 7 4 5 (d) 2 0 0 0 0 2 (e) 3 2

Question no – (10)

Solution :

Let, the five be, 0.5, 0.3, 0.4, 0.2

Now,

Decimals Exercise 6.2 Solution

Question no – (1)

Solution :

(a) 4.46

= 46

The decimal part = 46

(b) 25.426

= 426

The decimal part = 426

(c) 432.85

= 85

The decimal part = 85

(d) 100.050

= 050

The decimal part = 050

Question no – (2)

Solution :

(a) 44.002

= 44

Whole number part = 44

(b) 106.005

= 106

Whole number part = 106

(c) 625.250

= 625

Whole number part = 625

(d) 1045.246

= 1045

Whole number part = 1045

Question no – (3)

Solution :

(a) 5.46

= 2

The number of decimal places is 2

(b) 9.0450

= 4

The number of decimal places is 4

(c) 25.01

= 2

The number of decimal places is 2

(d) 550.02

= 2

The number of decimal places is 2

Question no – (4)

Solution :

(a) 0.7

= 7/10

0.7 in fraction = 7/10

(b) 0.85

= 85/100

= 17/20

0.85 in fraction = 17/20

(c) 13.5

= 135/10

= 27/2

13.5 in fraction = 27/2

(d) 6.4

= 64/10

= 32/5

6.4 in fraction = 32/5

(e) 0.08

= 8/100

= 2/25

0.08 in fraction = 2/25

(f) 0.125

= 125/1000

= 1/8

0.125 in fraction = 1/8

(g) 8.625

= 8625/1000

= 37/4

8.625 in fraction = 37/4

Question no – (5)

Solution :

(a) 10/25

10/25 = 0.4

(b) 15/20

15/20 = 0.75

(c) 7/8

7/8 = 0.875

(d) 8/125

8/125 = 0.064

(e) 5 4/10

= 54/10

= 5.4

Question no – (6)

Solution :

(a) 0.75

= 75/100

= 3/4

0.75 in the lowest form = 3/4

(b) 0.20

= 20/100

= 1/5

0.20 in the lowest form = 1/5

(c) 0.150

= 15/100

= 3/20

0.150 in the lowest form = 3/20

(d) 0.055

= 55/1000

= 11/200

0.055 in the lowest form = 11/200

(e) 0.125

= 125/1000

= 1/8

0.125 in the lowest form = 1/8

Question no – (7)

Solution :

Given numbers are, 0.5, 2.5, 3.5, 1.5

Now on the number line,

Question no – (8)

Solution :

(a) Given, 0.5

= 0.5 lies between 0 and 1

(b) Given, 5.5

= 5.5 lies between 5 and 6

(c) Given, 7.7

= 7.7 lies between 7 and 8

(d) Given, 6.3

= 6.3 lies between 6 and 7

(e) Given, 9.1

= 9.1 lies between 9 and 10

Question no – (9)

Solution :

(a) Given, 0.45

= 0.45 lies between 0.4 and 0.5

(b) Given, 0.77

= 0.77 lies between 0.7 and 0.8

(c) Given, 0.03

= 0.03 lies between 0 and 0.1

(d) Given, 0.29

= 0.29 lies between 0.2 and 0.3

Question no – (10)

Solution :

A = 8/10 = 0.8

B = 13/10 = 1.3

C = 22/10 = 2.2

D = 29/10 = 2.9

Try These 6 (i) :

Question no – (1)

Solution :

 Tens (10) Ones (1) Tenths (1/10) Hundredths(1/100) Decimal form (a) 3 ones and 5 tenths 3 5 3.5 (b) Five tens six ones and two hundredths 5 6 0 2 56.02

Question no – (2)

Solution :

(a) Decimal notation of 2 4/10,

= 2 + 4/10

= 2 + 0.4

= 2.4

So, the decimal notation of 2 4/10 is 2.4

(b) Decimal notation of 30 6/100,

= 30 + 6/100

= 30 + 0.06

= 30.06

So, the decimal notation of 30 6/100 is 30.06

Question no – (3)

Solution :

The place value of two 4s in 34.574,

The place value of first 4 is 40

The place value of second 4 is .004

For more better understanding :

Try These 6 (ii) :

Question no – (1)

Solution :

Given decimals, 3.007, 3.07, 3.7, 3.77

Here, 3.07 and 3.77 are Like decimals.

Question no – (2)

Solution :

Given, (a) 5.5 5.05 < 5.005 < 5.555

(b) 5.0055.05 < 5.5 < 5.555

(c) 5.05 < 5.5 < 5.005 < 5.555

(d) 5.555 < 5.05 <5.005 < 5.5

Here, only option – (b) is in the correct order.

(b) 5.0055.05 < 5.5 < 5.555 is in the correct order.

Question no – (3)

Solution :

(a) 2 1/2

= 2 + 1/2

= 1/2 + 0.5

= 2.5

(b) 2.3

= 2.3

(c) 2

= 2

(d) 1/4

= 0.25

Therefore,

= 1/4 < 2 < 2.3 < 2 1/2

Question no – (4)

Solution :

(a) 0.5 __ 0.9

= 0.5 < 0.9

(b) 16.0 __ 16

= 16.0 = 16

(c) 0.4 __ 1/2

= 0.4 < 1/2

(d) 108 __ 99.9

= 108 > 99.9

Question no – (5)

Solution :

(a) Given, unlike decimals, 945, 39.118, 72.8

Now, Into like decimals,

= 945.000, 39,118, 72.800

(b) Given, unlike decimals, 846.57, 239.7, 80.628

Now, into Like decimals,

= 846.570, 239.700, 80,628

Question no – (6)

Solution :

(a) 7.2

Two more equivalent decimals,

= 7.2, 9.5, .4

(b) 25.90

Two more equivalent decimals,

= 25.90, 30,14 70, 69

(c) 83.500

Two more equivalent decimals,

= 83, 500, 83,696, 25,969

Decimals Exercise 6.3 Solution

Question no – (1)

Solution :

(a) 7.05 __ 7.50

= 7.05 < 7.50

(b) 1.7 __ 1 7/10

= 1.7 = 1 7/10

(c) 5.85 __ 5.058

= 5.85 > 5.058

(d) 12.001 __ 12.10

= 12.001 < 12.10

(e) 24. 36 __ 20.72

= 24. 36 > 20.72

Question no – (2)

Solution :

(a) 0.5 or 0.3

= 0.5 > 0.3

= 0.5 is greater.

(b) 4 or 0.5

= 4 > 0.5

= 4 is greater.

(c) 5.46 or 5.406

= 5.46 > 5.406

= 5.46 is greater.

(d) 0.99 or 0.19

= 0.99 > 0.19

= 0.99 is greater.

(e) 140.65 or 37.999

= 140.65 > 37.999

= 140.65 is greater.

(f) 18.3457 or 18.3467

= 18.3457 < 18.3467

= 18.3467 is greater.

Question no – (3)

Solution :

Let, the five numbers be, 0.0099, 0.969, 0.198969, 2.969 and 10.01

Now, (i) 0.01 > 0.0099

(ii) 2.3 > 0.969

(iii) 0.2 > 0.198969

(iv) 3 > 2.969

(v) 100 > 10.01

Question no – (4)

Solution :

(a) Given decimal numbers = 0.011, 1.001, 0.101, 0.110

Now, in ascending order = 0.11, 0.101, 0110, 1.001

(b) Given decimal numbers = 5.65, 6.56, 5.56, 5.05, 5.66

Now, in ascending order = 5.05, 5.56, 5.65, 5.66, 6.56

Question no – (5)

Solution :

(a) Given, decimal numbers = 8.3, 8.73, 7.83, 7.38, 8.043, 8.04

Now, in descending order = 8.73, 8.3, 8.043, 8.04, 7.83, 7.38

(b) Given, decimal numbers = 7.77, 7.077, 77.7, 77.07, 7.007

Now, in descending order = 77.7, 77.07, 7.77, 7.077, 7.007

Decimals Exercise 6.4 Solution

Question no – (1)

Solution :

(a) 72 mm

As we know that,

1 centimetre = 10 millimeters

Now, 72/10 cm

= 7.2 cm

(b) 540 mm

= 450/10 cm

= 45 cm

(c) 20 cm 20 mm

= 20 cm 20 mm

= 20 + 20/10

= 20 + 2

= 22 cm

(d) 5 mm

= 5 mm = 5/10 cm

= 0.5 cm

Question no – (2)

Solution :

As we know, 1 metre = 100 Centimetre

So now,

(a) 550 cm

= 550/100 m

= 5.5 m

(b) 4 m 40 cm

= 4 + 40/100

= 4 + 0.4

= 44 m

(c) 130 cm

= 130/100

= 1.3 m

(d) 1255 cm

= 1255/100

= 12.55 m

Question no – (3)

Solution :

As we know that,

1 kilometre = 1000 metres

Now, (a) 5445 m

= 5445/1000

= 5.445 km

(b) 50 km 345 m

= 50 + 345/1000

= 54.345 km

(c) 8 m

= 8/1000

= 0.008 km

(d) 225 m

= 225/1000

= 0.225 km

Question no – (4)

Solution :

As we know that, 1 kg = 1000 g

(a) 25 g

= 25/1000

= 0.025 kg

(b) 25 kg 750 g

= 25 + 750/1000

= 25 + 0.75

= 20.75 kg

(c) 5 g

= 5/1000 kg

= 09.005 kg

(d) 5005 g

= 5005/1000

= 5.005 kg

Question no – (5)

Solution :

As we know that, 1 rupee = 100 paise

(a) 50 paise

= 50/100

= 0.5 rupee

50 paise = 0.5 rupee

(b) 2 paise

= 2/100

= 0.02 rupee

2 paise = 0.02 rupee

(c) 420 paise

= 420/100

= 4.2 rupee

420 paise = 4.2 rupees

(d) 10 rupees 75 paise

= 10 + 75/100

= 10.75 rupee

10 rupees 75 paise = 10.75 rupees

(e) 2020 paise

= 2020/100

= 20.2 rupee

2020 paise = 20.2 rupees

Question no – (6)

Solution :

(a) Rs 10.50

= 10.50 × 100 paise

= 1050 paise

(b) Rs 100.100

= 100.1000 × 100 paise

= 10010 paise

(c) 10.10 m

= 10.10 × 100 cm

= 10.0 cm

(d) 3.05 km

= 3.05 × 1000

= 3050 m

(e) 50.210 kg

= 50.210 × 1000

= 50210 g

(f) 5.2 cm

= 5.2 × 10

= 52 mm

(g) 20.300 km

= 20.300 × 1000

= 20300 m

Question no – (7)

Solution :

(a) 10 centimetres in metres and kilometres.

First in metres

= 10 cm

= 10/100 m

= 0.1 m

Now in kilometres,

= 1/10 × 1000

= 0.0001 km

(b) 25 mm into cm, m and km

First in cm.

= 25 mm = 25/10

= 2.5 cm

Second in m,

= 25/10 × 100

= 0.025 m

Now in km,

= 25/10 × 100 × 1000

= 0.000025 km

(c) 5000 mL to L.

As we know, 1 litre = 1000 mL

∴ 5000 mL

= 5000/1000

= 5L

Question no – (8)

Solution :

As per the given question,

Mr. Sharma bought spinach = 500 g,

onions = 750 g

garlic = 100 g

ginger = 150 g

potatoes = 750 g

The total weight of all the vegetables in kg,

= (500 + 750 + 100 + 150 + 750) g 2250 g

= 2250/1000

= 2.25 kg

Therefore, the total weight of all the vegetables will be 2.25 kg.

Question no – (9)

Solution :

According to the question,

The length of the top of Mahima’s study table is = 125 cm.

Its length in metres = ?

Now length in metres,

= 125 cm

= 125/100 m

= 1.25 m

Thus, the length of the top of Mahima’s study table is 1.25 m.

Question no – (10)

Solution :

As per the given question,

Isha has 2000 paise in her piggy bank.

Isha’s money in rupees = ?

We know, 1 rupee = 100 paise

2000 paise

= 2000/100

= 20 Rupees

Therefore, Isha has 20 rupee in her piggy bank.

Decimals Exercise 6.5 Solution

Question no – (1)

Solution :

(a) 9.839, 5.69, 3.03

9.839

5.690

+ 3.030
——————-
18.559

(b) 25.050, 40.550

25.05

+ 40.55
——————-
65.60

(c) 28.35, 250, 8.20, 45.375

28.350

250.000

8.200

+ 45.375
——————-
731.925

(d) 65.75, 2.35, 8.735, 0.456, 7.6

65.750

2.350

8.735

0.456

+ 7.600
——————-
76.891

(e) 15.5, 0.046, 220.660, 2.750

15.500

0.046

220.660

+ 2.750
——————-
238.956

Question no – (2)

Solution :

As per the question,

Vikas purchased a pair of shoes for = Rs 636.50,

A pair of socks for = Rs 49.90

A shirt for = Rs 955.55

Total amount,

= (636.50 + 955.55 + 49.90)

= 1641.95 rupees

Thus, Vikas spend total 1641.95 rupees.

Question no – (3)

Solution :

According to the question,

Vidushi bought mangoes = 2.25 kg

Apples = 2 kg

Litchis = 3.03 kg

Bananas = 1.632 kg

Total fruits,

= (2.250 + 2.000 + 3.030 + 1.632)

= 8.912 kg fruits

Therefore, Vidushi bought total 8.912 kg fruits.

Question no – (4)

Solution :

In a city, rain poured for three consecutive days.

On the 1st day, it was 2.35 cm;

on the 2nd day, it was 5.36 cm;

on the 3rd day, it was 7.10 cm

Total rainfall in three days,

= (2.35 + 5.36 + 7.10)

= 14.81 cm

Therefore, the total rainfall for three days is 14.81 cm.

Question no – (5)

Solution :

Mona and Sonu have ribbons of length = 2 m 30 cm and 4 m 20 cm.

The sum of the lengths of both the ribbons,

= 2 m 30 cm = 2.03 m

= 4 m 20 cm = 4.02 m

Now, (2.03 + 4.02)

= 6.05 m

Therefore, the sum of the lengths of both the ribbons is 6.05 m.

Question no – (6)

Solution :

According to the question,

A bank account holder earns an interest of

Rs 5512.72 in the 1st month

Rs 5421.62 in the 2nd month

Rs 4460.02 in the 3rd month

Quarterly interest paid by bank,

= (5512.72 + 5421.62 + 4460.02)

= 15394.36 Rs

Thus, the quarterly interest paid by bank to the account holder] is 15394.36 Rs.

Question no – (7)

Solution :

As per the given question,

A labourer’s day-wise earnings,

Monday – Rs 220.30,

Tuesday – Rs 150.75

Wednesday – Rs 360.65

Thursday – Rs 275.70

Friday – Rs 175.25

Saturday – Rs 300.30

Sunday – No earnings

Total earnings of the labourer,

= (220.30 + 150.75 + 360.65 + 275.25 + 175.25 + 300.30)

= 1482.95 Rs.

Therefore, the total earnings of the labourer in a week is 1482.95 Rs.

Question no – (8)

Solution :

According to the question,

Krishna walks 2 km 10 m in the morning

In the evening = 1 km 20 m

Total distance he walks in a day,

= 2 km 10 m = 2.01 km

= 1 km 20 = 1.02 km

Now, (2.01 + 1.02)

= 3.03 km

Hence, the total distance Krishna walks in a day is 3.03 km.

Question no – (9)

Solution :

According to the question,

Lakshman travel by train = 520 km 230 m

Travel by car = 220 km 360 m

Travel by aeroplane = 1025 km

The total distance covered by Lakshman,

520 km 230 m = 520.23

220 km 360 m = 220.36

Now, (520.23 + 220.36 + 1025.00)

= 1765.59

Therefore, the total distance covered by Lakshman is 1765.5 km

Try These 6 (iii) :

Question no – (1)

Solution :

(a) 0.03 + 0.09

0.03

+ 0.09
————–
0.12

(b) 0.25 + 0.25

0.25

+ 0.25
————–
0.50

(c) 0.11 + 0.19

0.11

+ 0.19
————–
0.30

(d) 0.01 + 0.10

0.01

+ 0.10
————–
0.11

Question no – (2)

Solution :

Option – (c) 0.12 + 23/10 will have sum as 0.40

0.28

+ 0.12
————
0.40

Question no – (3)

Solution :

The series by adding 0.02 to 0.91 and so on,

= (0.91), (0.91 + 0.02), (0.91 + 0.02 + 0.02)

= 0.91, 0.93, 0.95

Question no – (4)

Solution :

(a) 0.47

= With 0.47, I have to add 3 hundredths.

(b) 0.88

= With 0.88, I have to add 2 hundredths.

Decimals Exercise 6.6 Solution

Question no – (1)

Solution :

(a) 61.2 – 32.3

61.2

– 32,3
————-
28.9

(b) 8.08 – 37

8.08

– 3.70
————-
32.97

(c) 55 – 32.03

55.00

– 32.03
————-
12.97

(d) 20.389 – 0.4

20.389

0.400
————-
19.989

(e) 8.937 – 0.848

8.937

– 0,848
————-
8.089

(f) 435.819 – 100.008

435.819

– 100.008
————-
335.811

Question no – (2)

Solution :

(a) 24.50 + 34.68 – 12.76

= 59.18 – 12.76

= 46.42   …(Simplified)

(b) 200 + 100.23 – 28.72

= 300.23 – 28.72

= 271.28   …(Simplified)

(c) 66.30 – 7.333 + 6.666

= 72.966 – 7.333

= 65.633   …(Simplified)

(d) 185.30 – 150.605 + 156.490 – 23.35

= (185.30 + 156.490) – (105.605 + 23.35)

= 241.79 – 128.955

= 121.835   …(Simplified)

Question no – (3)

Solution :

150.00 – 65.20

= 84.80

So, 84.80 should be added to 65.20 to get 150

Question no – (4)

Solution :

= 100.00 – 33.33

= 66.67

Therefore, 66.67 should be subtracted from 100 to get 33.33.

Question no – (5)

Solution :

As per the question we get,

The school bag of a Class VI student weighs = 5 kg 500 g

The school bag of a Class VII student weighs = 4 kg 225 g

Now, 5.500 – 4.225

= 1.225 kg

Therefore, Bag of the class vi is heavier by 1.225 kg.

Question no – (6)

Solution :

According to the question,

A godown contains 175.255 kg of fruits,

Out of which 35.023 kg of fruits are rotten.

Fresh fruits,

= 175.225 – 35.023

= 140.232 kg

Hence, there are 140.232 kg of fruits are fresh.

Question no – (7)

Solution :

As per the given question,

Muskan needs 23.25 m cloth of red colour.

She could get only 15.75 m from one shop

Cloth she still need to buy,

= 23.25 – 15.75

= 7.50 m

Thus, she need to buy more 7.50 m cloth.

Question no – (8)

Solution :

According to the question,

Sudarshan goes nearby town, which is 10.25 km far from his place.

He first takes a bus and covers 7.45 km

Remaining distance covers by walking only

Sudarshan covers by walking,

= 10,25 – 7.45

= 2.70 km

Thus, Sudarshan walk 2.70 km.

Question no – (9)

Solution :

Given in the question,

Pankaj buys 10 kg of grocery items for his house.

Out of this 2 kg 250 g is rice,

3 kg 330 g is sugar,

4 kg is wheat

Remaining is toiletries = ?

The weight of toiletries,

= {10 – (2.250 + 3.330 + 4)}

= (10 – 9.580) kg

= 0.420 kg

Therefore, the weight of toiletries is 0.420 kg

Question no – (10)

Solution :

= 142.5 – 95.0

= 47.5 km

Therefore, 95 km is 47.5 km less than 142.5 km.

Question no – (11)

Solution :

= 65.5 km – 37.0

= 28.5 km

Therefore, 65.5 km is 28.5 km more than 37 km.

Question no – (12)

Solution :

(a) Rs 56.25 from Rs 98.75

(98.75 – 56.25)

= 42.50

(b) 4.051 km from 16.207 km

16.207 – 4.051

= 12.156

(c) 0.514 kg from 12.75 kg

12.750 – 0.514

= 12.236

(d) 407.5 m from 600 m.

600.0 – 407.5

= 192.5 m

Question no – (13)

Solution :

(a) Given temperature, 41°F

We know to convert in Celsius,

First we Subtract it by 32, then Multiply it by 5 and lastly Divide with 9.

Now, (41°F − 32) × 5/9

= 5°C

So, 41° Fahrenheit in Celsius = 5°C

(b) Given temperature, 100°C

We know to convert in Fahrenheit, First we multiply it by 9, then divide it by 5 and lastly Add with 32,

Now, (100°C × 9/5) + 32

= 212°F

So, 100°Celsius in Fahrenheit = 212°F

(c) Given temperature, 101.3°F

We know to convert in Celsius, First we Subtract it by 32, then Multiply it by 5 and lastly Divide with 9.

Now, (101.3°F − 32) × 5/9

= 38.5°C

So, 101.3° Fahrenheit in Celsius = 38.5°C

(d) Given temperature, 27.5°C

We know to convert in Fahrenheit, First we multiply it by 9, then divide it by 5 and lastly Add with 32,

Now, (27.5°C × 9/5) + 32

= 99.5 °F

Thus, 27.5° Celsius in Fahrenheit = 99.5°F

Question no – (14)

Solution :

According to the question,

Reema is suffering from fever.

Temperature in the evening is 101.8°F and

Decreases to 99.2°F next morning

Decrease in temperature,

= 101.8 – 992

= 2.6° Fahrenheit

Decimal Chapter Check-up Solution :

Question no – (1)

Solution :

(a) Given Like decimals are = 1.500, 2.750, 3.006, 4.10

Now, into unlike decimals = 1.5, 2.75, 3.006, 4.1

(b) Given Unlike decimals are = 2.3, 7.09, 8.33, 10.375

Now, into like decimals = 2.300, 7.090, 8,330, 10,375

Question no – (2)

Solution :

(a) 16.007 in expanded form

= (1 × 10) + (6 × 1) + (7 × 1/1000)

(b) 356.089 in expanded form

= (3 × 1000) + (5 × 10) + (6 × 1) + (8 × 1/1000) + (9 × 1/1000)

Question no – (3)

Solution :

(a) 0.15

= 15/100

= 3/20

So, 0.15 into fraction = 3/20

(b) 3.165

= 3165/100

= 633/200

So, 3.165 into fraction = 633/200

(c) 8.73

= 873/100

So, 8.73 into fraction = 873/100

Question no – (4)

Solution :

(a) 10 3/4

= 10 3/4 = 43/4

= 43 × 25/4 × 25

= 1125/100

= 11.25

Hence, 10 3/4 into decimal = 11.25

(b) 217/25

= 217/25 = 217 × 4/25 × 4

= 868/100

= 8.86

So, 217/25 into decimal = 8.86

(c) 15/12

Thus, 15/12 into decimal = 1.25

(d) 1 2/50

= 1 2/50 = 52/50

= 52 × 2/50 × 2

= 104/100

= 1.04

Therefore, 1 2/50 into decimal = 1.04

Question no – (5)

Solution :

(a) Given numbers = 7.025, 7.520, 7.052, 7.500, 7.502

Now in ascending order = 7.025, 7.05, 2, 7.500, 7.502, 7.520

(b) Given decimal numbers = 14.67, 14.678, 14.876, 14.768, 14.867

Now in ascending order = 14,67, 14,678, 14.768, 14,867, 14,876

Question no – (6)

Solution :

(a) 41.8 + 39.52 + 5.01

= 41.80 + 39.52 + 5.01

= 86.32    …(Simplified)

(b) 85.43 + 24.956 – 36.256

= 85.3 + 24.956 – 36.256

= 110.256 – 36.256

= 74   …(Simplified)

(c) 0.008 + 0.156 + 3.098

= 0.008 + 0.156 + 3.098

= 3.262   …(Simplified)

(d) 280.69 + 25.2 – 38.56

= 280.69 + 25.2 – 38.56

= 305.71 – 38.56

= 297.15   …(Simplified)

(e) 389.68 – 199.96

= 389.68 – 199.96

= 189.72   …(Simplified)

Question no – (7)

Solution :

1000.00

– 765.38
——————-
= 234.62

Thus, 234.62 should be added to 765.38 to get 1000.

Question no – (8)

Solution :

= 156.75 – (98.75 – 39.045) ….(according to the question)

= 156.75 – 59.705

= 97.045

Question no – (9)

Solution :

As per the given question,

Radhika’s mother gave her = Rs 10.50

Father gave her = Rs 15.80

Total amount,

= 10.50 + 15.80

= 26.30 rupees

Therefore, the total amount given to Radhika by her parents is 26.30 rupees.

Question no – (10)

Solution :

Nasreen bought,

for her shirt and = 3 m 20 cm

for her trouser = 2 m 5 cm

Now, 3 m 20 cm = 3.2 m,

= 2 m 5 cm = 2.05 m

Total length of cloth bought by her,

= (3.2 + 2.05)

= 5.25 m

Therefore, the total length of cloth bought by her is 5.25 m.

Question no – (11)

Solution :

According to the question,

Ravi purchased 5 kg 400 g rice,

2 kg 20 g sugar

10 kg 850 g flour.

Now, 5 kg 400 g = 5.400 kg,

2 kg 20 g = 2.02g,

10 kg 850 g = 10.850 kg

Total weight of his purchases,

= (5.400 + 2.020 + 10.850) kg

= 18.270 kg

Hence, the total weight of his purchases is 18.270 kg

Question no – (12)

Solution :

As per the given question,

Tina had 20 m 5 cm long cloth.

She cuts 4 m 50 cm length of cloth from this for making a curtain.

Cloth is left with her,

20 m 5 cm = 20.05 m,

4 m 50 cm = 4.5 m

Now, (20.05 – 4.5)m

= 15.45 m

Hence, 15.45 m cloth is left with her.

Question no – (13)

Solution :

Given in the question,

The thickness of a book is = 3.95 cm

whereas thickness of a copy is = 2.86 cm

Total thickness,

= (3.95 + 2.86) cm

= 6.81 cm

Therefore, the total thickness of the two is 6.81 cm.

Question no – (14)

Solution :

From the question we get,

A drum of oil has a capacity of 750 litres.

Two containers of capacity 180.75 litres and 249.85 litres

The quantity of oil left in the drum,

= {750 – (180.75 + 249.85)}

= 750 – 430.60

= 319.40 L

Therefore, the quantity of oil left in the drum is 319.40 Litres

Question no – (15)

Solution :

Given, pea plant measured 8.5 cm and it grows 0.85 cm everyday,

Initial height = 8.5 cm

Growth per day = 0.85 cm

Height after 2 days

= 8.5 + (2 × 0.85)

= 8.5 + 1.7

= 10.2 cm

Thus, the height of the plant after 2 days will be 10.2 cm.

Question no – (16)

Solution :

According to the given table ,

Milk provides least energy and Rice provides maximum.

Fraction,

= 3/5.3

= 30/53

Question no – (17)

Solution :

Magic Square

The sum of each row, column and diagonal is 9.0.

So now,

 1.2 5.4 2.4 4.2 3 1.8 3.6 0.6 4.8

Question no – (18)

Solution :

(a) I am the sum of 6 tenths and 78 hundredths.

= (6 × 1/10) + (78 × 1/100) …(according to the question)

= 0.6 + 0.78

= 1.38

Thus, here I am 1.38

(b) I am the difference of 4 tenths and 6 thousandths

= (4 × 1/10) – (6 × 1/1000)

= 0.4 – 0.006

= 0.394

Therefore, here I am = 0.394

(c) I separate the whole number part of the decimal number from the part that is less than 1.

0 separate the whole number part of the decimal number from the part that is less than 1.

I am = 0

(e) I represent the temperature of a healthy human being in °F.

Fahrenheit thermometer represent the temperature of a healthy human being in °F.

So, here I am = Fahrenheit thermometer.

Previous Chapter Solution :

Updated: June 6, 2023 — 5:13 am