RD Sharma Solutions Class 6 Maths Chapter 7 Decimals

Exercise 7.1

 

Question 1:  Write the following decimals in the place value table:

(i) 52.5

(ii) 12.57

(iii) 15.05

(iv) 74.059

(v) 0.503 

Solution 1:         

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals


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

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-

Solution 2:

(i) Decimal Expiration 307.12

(ii) Decimal Expiration 9543.025

(iii) Decimal Expiration 12.503

 

Question 3:  Write each of the following decimals in words:

(i) 175.04

(ii) 0.21

(iii) 9.004

(iv) 0.459 

Solution 3:

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-1


RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-2


Question 5:  Write each of the following as decimals:
(i) Five and four tenths
(ii) Twelve and four hundredths
(iii) Nine and Seven hundred five thousandths
(iv) Zero point five two six
(v) Forty seven and six thousandths
(vi) Eight thousandths
(vii) Nineteen and nineteen hundredths.
 
Solution 5:

(i) Five and four tenths
= 5 + 4/10   
= 5.4
 
(ii) Twelve and four hundredths
= 12 + 4/100   
= 12.04
 
(iii) Nine and Seven hundred five thousandths
= 9 + 705/1000   
= 9.705
 
(iv) Zero point five two six
= 0.526
 
(v) Forty seven and six thousandths
= 47 + 6/1000   
= 47.006
 
(vi) Eight thousandths
=8/1000   
= 0.008
 
(vii) Nineteen and nineteen hundredths
= 19 + 19/100  
= 19.19
Exercise 7.2  
Question 1:   Write each of the following as decimals:
(i) Three tenths
(ii) Two ones and five tenths
(iii) Thirty and one tenths
(iv) Twenty two and six tenths
(v) One hundred, two ones and three tenths
 
Solution 1:
 
(i) Three tenths
= 3/10 
= 0.3
 
(ii) Two ones and five tenths
= 2 + 5/10  
= 2.5
 
(iii) Thirty and one tenths
= 30 + 1/10  
= 30.1
 
(iv) Twenty two and six tenths
= 22 + 6/10 
= 22.6
 
(v) One hundred, two ones and three tenths
= 100 + 2 + 3/10  
= 102.3
 
Question 2:  Write each of the following as decimals:
(i) 30 + 6 + 2/10
(ii) 700 + 5 + 7/10
(iii) 200 + 60 + 5 + 1/10
(iv) 200 + 70 + 9 + 5/10
 
Solution 2:
(i) 30 + 6 + 2/10
3 tens + 6 ones + 2 tenths
Thus, the decimal is 36.2.
 
(ii) 700 + 5 + 7/10
7 hundreds + 5 ones + 7 tenths
Thus, the decimal is 705.7.
 
(iii) 200 + 60 + 5 + 1/10
2 hundreds + 6 tens + 5 ones + 1 tenths
Thus, the decimal is 265.1.
 
(iv) 200 + 70 + 9 + 5/10
2 hundreds + 7 tens + 9 ones + 5 tenths
Thus, the decimal is 279.5.
 
Question 3:  Write each of the following as decimals:
(i) 22/10 
(ii) 3/2 
(iii) 2/5 
Solution 3:
(i) 22/10
The decimal expiration of  22/10 is 2.2
Thus, the decimal expiration is 2.2
 
(ii) 3/2
 
= 3/2  ×  5/5  (by multiply 5 with numerator and denominator) 
= 15/10  
= 1.5
Thus, the decimal expiration is 1.5
 
(iii) 2/5
 
= 2/5  ×  2/2  (by multiply 2 with numerator and denominator) 
= 4/10  
= 0.4
Thus, the decimal expiration is 0.4


RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-3

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-4

Question 5:  Write the following decimals as fractions. Reduce the fractions to lowest form:
(i) 3.8
(ii) 21.2
(iii) 6.4
(iv) 1.0
 
Solution 5:
(i) 3.8
= 3 + 8 tenths
= 3 + 8/10
By taking LCM of denominators is 10.
= (30 + 8)/10
= 38/10
Converting in standard form
= 19/5
 
(ii) 21.2
= 21 + 2 tenths
= 21 + 2/10
By taking LCM of denominators is 10.
= (210 + 2)/10
= 212/10
Converting in standard form
= 106/5
 
(iii) 6.4
= 6 + 4 tenths
= 6 + 4/10
By taking LCM of denominators is 10.
= (60 + 4)/10
= 64/10
Converting in standard form
= 32/5
(iv) 1.0
= 1 + 0 tenths
=1 
 
Question 6:  Represent the following decimal numbers on the number line:
(i) 0.2
(ii) 1.9
(iii) 1.1
(iv) 2.5
 
Solution 6:   
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-5
 
 
Question 7:  Between which two whole numbers on the number line are the given numbers? Which one is nearer the number?
(i) 0.8
(ii) 5.1
(iii) 2.6
(iv) 6.4
(v) 9.0
(vi) 4.9
 
Solution 7:
(i) 0.8 is in between 0 and 1. 
Hence, it is nearer to 1.
 
(ii) 5.1 is in between 5 and 6.
Hence, it is nearer to 5.
 
(iii) 2.6 is in between 2 and 3.
Hence, it is nearer to 3.
 
(iv) 6.4 is in between 6 and 7.
Hence, it is nearer to 6.
 
(v) 9.0 is already a whole number.
Hence, it is nearer to 9.
 
(vi) 4.9 is in between 4 and 5.
Hence, it is nearer to 5.
 
 
 
Question 8:  Write the decimal number represented by the points on the given number line: A, B, C, D.
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-6

Solution 8:

(A) Location of Point A on number line is 0.8.

(B) Location of Point B on number line is 1.3.

(C) Location of Point C on number line is 1.9.

(D) Location of Point C on number line is 2.96. 

Exercise 7.3
 
Question 1:   Write each of the following as decimals:
(i) Five hundred twenty five and forty hundredths.
(ii) Twelve and thirty five thousandths
(iii) Fifteen and seventeen thousandths
(iv) Eighty eight and forty eight hundredths 
Solution 1:

(i) Five hundred twenty five and forty hundredths.
= 525 + 40/100 
= 525.40
Hence, the decimal form is 525.40.
 
(ii) Twelve and thirty five thousandths
= 12 + 35/1000 
= 12.035
Hence, the decimal form is 12.035.
 
(iii) Fifteen and seventeen thousandths
= 15 + 17/1000 
= 15.017
Hence, the decimal form is 15.017.
 
(iv) Eighty eight and forty eight hundredths
= 88 + 48/100 
= 88.48
Hence, the decimal form is 88.48.
 
Question 2:  Write each of the following as decimals:
(i) 137 + 5/100 
(ii) 20 + 9 + 4/100 
Solution 2:

(i) 137 + 5/100
1 hundred + 3 tens + 7 ones + 5 hundredths
Thus, the decimal is 137.05.
 
(ii) 20 + 9 + 4/100
2 tens + 9 ones + 4 hundredths
Thus, the decimal is 29.04.
 
Question 3:  Write each of the following as decimals:
(i) 8/100
(ii) 300/1000
(iii)  18/1000 
(iv) 208/100
(v) 888/1000
 
Solution 3:

(i) 8/100 = 0.08
Thus, the decimal form of 8/100is 0.08.
 
(ii) 300/1000 = 0.300
Thus, the decimal form of 300/1000 is 0.3.
 
(iii) 18/1000 = 0.018
Thus, the decimal form of 18/1000 is 0.018.
 
(iv) 208/100 = 2.08
Thus, the decimal form of 208/100 is 2.08.
 
(v) 888/1000 = 0.888
Thus, the decimal form of 888/1000 is 0.888.
 
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-7
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-8
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-9
 

Exercise 7.4


 
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-10
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-11
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-13
 
Question 2:  Express the following decimals as fractions in the lowest form:
(i) 0.5
(ii) 2.5
(iii) 0.60
(iv) 0.18
(v) 5.25
(vi) 7.125
(vii) 15.004
(viii) 20.375
(ix) 600.75
(x) 59.48
 
Solution 2:

(i) 0.5
= 0 + 0.5
= 0 + 5 tenths
Change the term in fractions we get,
= 0 + 5/10
= 5/10
Change in lowest form we get,
= 1/2
Thus, the fractional form of 0.5 is  1/2
 
(ii) 2.5
= 2 + 0.5
= 2 + 5 tenths
Change the term in fractions we get,
= 2 + 5/10
By taking LCM of denominators 10 we get,
= (20 + 5)/10
= 25/10
Change in lowest form we get,
= 5/2
Thus, the fractional form of 2.5 is  5/2
 
(iii) 0.60
= 0 + 0.60
= 0 + 60 hundredths
Change the term in fractions we get,
= 0 + 60/100 
= 60/100 
Change in lowest form we get,
= 3/5
Thus, the fractional form of 0.60 is  3/5
 
(iv) 0.18
= 0 + 0.18
= 0 + 18 hundredths
Change the term in fractions we get,
= 0 + 18/100
= 18/100
Change in lowest form we get,
= 9/50
Thus, the fractional form of 0.18 is  9/50
 
(v) 5.25
= 5 + 0.25
= 5 + 25 hundredths
Change the term in fractions we get,
=  5 + 25/100
By taking LCM of denominators 100 we get,
= (500 + 25)/100
= 525/100
Change in lowest form we get,
= 21/4
Thus, the fractional form of 5.25 is  21/4
 
(vi) 7.125
= 7 + 0.125
= 7 + 125 thousandths
Change the term in fractions we get,
= 7+ 125/1000 
By taking LCM of denominators 100 we get,
= (7000 + 125)/1000 
= 7125/1000 
Change in lowest form we get,
=57/8 
Thus, the fractional form of 7.125 is 57/8
 
(vii) 15.004
= 15 + 0.004
= 15 + 4 thousandths
Change the term in fractions we get,
= 15+ 4/1000 
By taking LCM of denominators 1000 we get,
=(15000 + 4)/1000 
=15004/1000 
Change in lowest form we get,
=3751/250 
Thus, the fractional form of 15.004 is 3751/250
 
(viii) 20.375
= 20 + 0.375
= 20 + 375 thousandths
Change the term in fractions we get,
= 20+ 375/1000 
By taking LCM of denominators 1000 we get,
=(20000 + 375)/1000 
=20375/1000 
Change in lowest form we get,
=163/8 
Thus, the fractional form of 20.375 is 163/8
 
(ix) 600.75
= 600 + 0.75
= 600 + 75 hundredths
Change the term in fractions we get,
= 60+ 75/100 
By taking LCM of denominators 100 we get,
=(6000 + 75)/100 
=6075/100 
Change in lowest form we get,
=2403/4 
Thus, the fractional form of 600.75 is 6075/4
 
(x) 59.48
= 59 + 0.48
= 59 + 48 hundredths
Change the term in fractions we get,
= 59+ 48/100 
By taking LCM of denominators 100 we get,
=(5900 + 48)/100 
=5948/100 
Change in lowest form we get,
=1487/25 
Thus, the fractional form of 59.48 is 1487/25 
 

 Exercise 7.5 

Question 1:  Fill in the blanks by using > or < to complete the following:

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-14


Question 2:  Which is greater? Give reason for your answer?

(i) 1.008 or 1.800

(ii) 3.3 or 3.300

(iii) 5.64 or 5.603

(iv) 1.5 or 1.50

(v) 1.431 or 1.439

(vi) 0.5 or 0.05 

Solution 2:

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-15

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-16

 

Exercise 7.6

 
Question 1:  Express as Rupees (Rs.) using decimals:
(i) 15 paisa
(ii) 5 paisa
(iii) 350 paisa
(iv) 2 rupees 60 paisa
 
Solution 1:

(i) 15 paisa
1 paisa = Rs. 1/100
15 paisa = 15/100
15 paisa = Rs. 0.15
Thus, 15 paisa = Rs. 0.15
 
(ii) 5 paisa
1 paisa = Rs. 1/100
5 paisa = 5/100
5 paisa = Rs 0.05
Thus, 5 paisa = Rs. 0.5
 
(iii) 350 paisa
100 paisa = Rs. 1
1 paisa = Rs. 1/100
350 paisa = 350/100
350 paisa = Rs 3.50
Thus, 350 paisa = Rs. 3.50
 
(iv) 2 rupees 60 paisa
1 paisa = Rs. 1/100
2 rupees 60 paisa = 2 + 60/100
2 rupees 60 paisa = Rs 2.60
Thus, 2 rupees 60 paisa = Rs. 2.60
 
Question 2: Express as metres (m) using decimals:
(i) 15 cm
(ii) 8 cm
(iii) 135 cm
(iv) 3 m 65 cm
 
Solution 2:
(i) 15 cm
1 cm = 1/100 m
15 cm = 15/100
15 cm = 0.15 m
Thus, 15 cm = 0.15m
 
(ii) 8 cm
1 cm = 1/100 m
8 cm = 8/100
8 cm = 0.08 m
Thus, 8 cm = 0.08m
 
(iii) 135 cm
1 cm = 1/100 m
135 cm = 135/100 
135 cm = 1.35 m
Thus, 135 cm = 1.35m
 
(iv) 3 m 65 cm
1 cm = 1/100 m
3 m 65 cm = 3 + 65/100
3 m 65 cm = 3.65 m
Thus, 3m 6cm = 3.65m
 
Question 3:   Express as centimetre (cm) using decimals:
(i) 5 mm
(ii) 60 mm
(iii) 175 mm
(iv) 4 cm 5 mm 
Solution 3:

(i) 5 mm
1 mm = 1/10cm
5 mm = 5/10
5 mm = 0.5 cm
Thus, 5mm = 0.5cm.
 
(ii) 60 mm
1 mm = 1/10cm
60 mm = 60/10
60 mm = 6 cm
Thus, 60mm = 6cm.
 
(iii) 175 mm
1 mm = 1/10cm
175 mm = 175/10
175 mm = 17.5 cm
Thus, 175mm = 17.5cm.
 
(iv) 4 cm 5 mm
1 mm = 1/10cm
4 cm 5 mm = 4 + 5/10
4 cm 5 mm = 4.5 cm
Thus, 4cm 5mm = 4.5cm.
 
Question 4:  Express as kilometre (km) using decimals:
(i) 5 m
(ii) 55 m
(iii) 555 m
(iv) 5555 m
(v) 15 km 35 m
 
Solution 4:
(i) 5 m
1 m = 1/1000 km
5 m = 5/1000 km
5 m = 0.005 km
Thus, 5m = 0.005 km
 
(ii) 55 m
1 m = 1/1000 km
55 m = 55/1000  km
55 m = 0.055 km
Thus, 55m = 0.055 km
 
(iii) 555 m
1 m = 1/1000 km
555 m = 555/1000 km
555 m = 0.555 km
Thus, 555m = 0.555 km
 
(iv) 5555 m
1 m = 1/1000 km
5555 m = 555/1000 km
5555 m = 5.555 km
Thus, 5555 m = 5.555 km
 
(v) 15 km 35 m
1 m = 1/1000 km
15 km + 35 m = 15 35/1000km
15 km 35 m = 15.035 km
Thus, 15 km 35m = 15.035 km
 
Question 5:  Express as kilogram (kg) using decimals:
(i) 8 g
(ii) 150 g
(iii) 2750 g
(iv) 5 kg 750 g
(v) 36 kg 50 g
 
Solution 5:
(i) 8 g
1 g = 1/1000 kg
8 g = 8/1000kg
8 g = 0.008 kg
Thus, 8 g = 0.008 kg
 
(ii) 150 g
1 g = 1/1000 kg
150 g = 150/1000
150 g = 0.150 kg
Thus, 150 g = 0.150 kg
 
(iii) 2750 g
1 g = 1/1000 kg
2750 g = 2750/1000
2750 g = 2.750 kg
Thus, 2750 g = 2.750 kg
 
(iv) 5 kg 750 g
1 g = 1/1000 kg
5 kg + 750 g = 5 + 750/1000
5 kg 750 g = 5.750 kg
Thus, 5kg 750 g = 5.750 kg
 
(v) 36 kg 50 g
1 g = 1/1000 kg
36 kg + 50 g = 36 + 50/1000
36 kg 50 g = 36.050 kg
Thus, 36 kg 50 g = 36.050 kg
 
 
Question 6:  Express each of the following without using decimals:
(i) Rs. 5.25
(ii) 8.354 kg
(iii) 3.5 cm
(iv) 3.05 km
(v) 7.54 m
(vi) 15.005 kg
(vii) 12.05 m
(viii) 0.2 cm
 
Solution 6:
(i) Rs 5.25
1 paisa = Rs. 1/100
Rs 5.25 = 5 + 25/100
Rs 5.25 = Rs. 5 + 1/4
By taking LCM of Denominator we get
Rs 5.25 = Rs. (20 + 1)/4 
= Rs. 21/4
Thus, the non-decimal form of Rs. 5.25 is Rs. 21/4
 
(ii) 8.354 kg
1 g = 1/1000 kg
8.354 kg = 8 + 354/1000 kg
By taking LCM of Denominator we get
8.354 kg = (8000 + 354)/1000 kg
8.354 kg = 8354/1000 kg
=8354/1000 kg 
Thus, the non-decimal form of 8.354 kg is 8354/1000  kg 
 
(iii) 3.5 cm
1 mm = 1/10 cm
3.5 kg = 3 + 5/10 cm
By taking LCM of Denominator we get
3.5 kg = (30 + 5)/10 cm
8.354 kg = 35/10 cm
=35/10 cm 
=7/2 cm 
Thus, the non-decimal form of 3.5cm is 7/2 cm
 
(iv) 3.05 km
1 m = 1/1000 km
3.05 m = 3 + 5/1000 km
By taking LCM of Denominator we get
3.05 m = (3000 + 5)/1000 km
3.5 kg = 3005/1000 km
=3005/1000 km 
=601/200 m 
Thus, the non-decimal form of 3.05km is 601/200 m
 
(v) 7.54 m
1 cm = 1/100 m
7.54 m = 7 + 54/100 m
By taking LCM of Denominator we get
7.54 m = (700 + 54)/100 m
7.54 m = 754/100 m
=754/100  m 
=377/50 m 
Thus, the non-decimal form of 7.54m is 377/50 m
 
(vi) 15.005 kg
1 g = 1/1000 kg
15.005 g = 15 + 5/1000 kg
By taking LCM of Denominator we get
15.005 g = (15000 + 5)/1000 kg
15.005 g = 15005/1000 kg
=15005/1000  kg 
=3001/200 kg 
Thus, the non-decimal form of 15.005 kg is 3001/200 kg
 
(vii) 12.05 m
1 cm = 1/100 m
12.05 cm = 12 + 5/100 m
By taking LCM of Denominator we get
12.05 cm = (1200 + 5)/100  m
12.05 cm = 1205/100  m
=1205/100  m 
=241/20 m 
Thus, the non-decimal form of 12.05m is 241/20 m
 
(viii) 0.2 cm
1 mm = 1/10 cm
0.2 mm = 0 + 2/10 cm
By taking LCM of Denominator we get
0.2 mm = (0 + 2)/10  cm
0.2 cm = 2/10  cm
=2/10  cm 
=1/5  cm 
Thus, the non-decimal form of 0.2 m is 1/5  cm

Exercise 7.7

 

Question 1:  Choose the decimal (s) from the brackets which is (are) not equivalent to the given decimals:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500) 

Solution 1:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

0.8 = 0.80, 0.800

0.8  0.85, 0.08

 

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

25.1 = 25.10, 25.100

25.1  25.01, 25.001

 

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500)

45.05 = 45.050, 45.0500

45.05  45.005, 45.500

 

Question 2:  Which of the following are like decimals:

(i) 0.34, 0.07, 5.35, 24.70

(ii) 45.05, 4.505, 20.55, 20.5

(iii) 8.80, 17.08, 8.94, 0.27

(iv) 4.50, 16.80, 0.700, 7.08 

Solution 2:

(i) 0.34, 0.07, 5.35, 24.70

In the given expression all the values are like decimals because the numbers of digits after decimal are equal.

 

(ii) 45.05, 4.505, 20.55, 20.5

In the given expression all the values are unlike decimals as the numbers of digits after decimal are not equal.

 

(iii) 8.80, 17.08, 8.94, 0.27

In the given expression all the values are like decimals because the numbers of digits after decimal are equal.

 

(iv) 4.50, 16.80, 0.700, 7.08

In the given expression all the values are unlike decimals as the numbers of digits after decimal are not equal.

 

Question 3:  Which of the following statements are correct?

(i) 8.05 and 7.95 are like decimals.

(ii) 0.95, 0.306, 7.10 are unlike decimals.

(iii) 3.70 and 3.7 are like decimals.

(iv) 13.59, 1.359, 135.9 are like decimals.

(v) 5.60, 3.04, 0.45 are like decimals. 

Solution 3:

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-17


Question 4:   Convert each of the following sets of unlike decimals to like decimal:

(i) 7.8, 7.85

(ii) 2.02, 3.2

(iii) 0.6, 5.8, 12.765

(iv) 5.296, 5.2, 5.29

(v) 4.3294, 43.29, 432.94 

Solution 4:

(i) 7.8, 7.85

In 7.8 there is 1 digit after decimal.

In 7.85 there are 2 digits after decimal.

7.85 contains 2 digits after decimal point so by changing 7.8 as 7.80

Hence, 7.80 and 7.85 are like decimals.

 

(ii) 2.02, 3.2

In 2.02 there are 2 digits after decimal.

In 3.2 there is 1 digit after decimal.

2.02 contains 2 digits after decimal point so by changing 3.2 as 3.20

Hence, 2.02 and 3.20 are like decimals.

 

(iii) 0.6, 5.8, 12.765

In 0.6 there is 1 digit after decimal.

In 5.8 there is 1 digit after decimal.

In 12.765 there are 3 digits after decimal.

12.765 contains 3 digits after decimal point so by changing 0.6 as 0.600 and 5.8 as 5.800

Hence, 0.600, 5.800 and 12.765 are like decimals.

 

(iv) 5.296, 5.2, 5.29

In 5.296 there are 3 digits after decimal.

In 5.2 there is 1 digit after decimal.

In 5.29 there are 2 digits after decimal.

5.296 contains 3 digits after decimal point so by changing 5.2 as 5.200 and 5.29 as 5.290

Hence, 5.296, 5.200 and 5.290 are like decimals.

 

(v) 4.3294, 43.29, 432.94

In 4.3294 there are 4 digits after decimal.

In 43.29 there are 2 digits after decimal.

In 432.94 there are 2 digits after decimal.

4.3294 contains 4 digits after decimal point so by changing 43.29 as 43.2900 and 432.94 as 432.9400

Hence, 4.3294, 43.2900 and 432.9400 are like decimals.

 

Exercise 7.8

 

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

RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-18


Question 2:  Add the following:

(i) 41.8, 39.24, 5.01 and 62.6

(ii) 4.702, 4.2, 6.02 and 1.27

(iii) 18.03, 146.3, 0.829 and 5.324 

Solution 2:

(i) 41.8, 39.24, 5.01 and 62.6

Change the given values in like fractions.

41.80 + 39.24 + 5.01 + 62.60 = 148.65

Hence, the addition of 41.8, 39.24, 5.01 and 62.6 is 148.65

 

(ii) 4.702, 4.2, 6.02 and 1.27

Change the given values in like fractions.

4.702 + 4.200 + 6.020 + 1.270 = 16.192

Hence, the addition of 4.702, 4.2, 6.02 and 1.27 is 16.192

 

(iii) 18.03, 146.3, 0.829 and 5.324

Change the given values in like fractions.

18.030 + 146.300 + 0.829 + 5.324 = 170.483

Hence, the addition of 18.03, 146.3, 0.829 and 5.324 is 170.483

 

Question 3:  Find the sum in each of the following:

(i) 0.007 + 8.5 + 30.08

(ii) 280.69 + 25.2 + 38

(iii) 25.65 + 9.005 + 3.7

(iv) 27.076 + 0.55 + 0.004 

Solution 3:

(i) 0.007 + 8.5 + 30.08

Change the given values in like fractions.

0.007 + 8.500 + 30.080 = 38.587

Hence, the sum of 0.007 + 8.5 + 30.08 is 170.483

 

(ii) 280.69 + 25.2 + 38

Change the given values in like fractions.

280.69 + 25.20 + 38.00 = 343.89

Hence, the sum of 280.69 + 25.2 + 38 is 343.89

 

(iii) 25.65 + 9.005 + 3.7

Change the given values in like fractions.

25.650 + 9.005 + 3.700 = 38.355

Hence, the sum of 25.650 + 9.005 + 3.700 is 38.355

 

(iv) 27.076 + 0.55 + 0.004

Change the given values in like fractions.

27.076 + 0.550 + 0.004 = 27.630

Hence, the sum of 27.076 + 0.550 + 0.004 is 27.630

 

Question 4:  Radhika’s mother gave her Rs. 10.50 and her father gave her Rs. 15.80, find the total amount given to Radhika by her parents. 

Solution 4:

Radhika’s mother gave her = Rs 10.50

Radhika’s father gave her = Rs 15.80

Total amount given by her parents = Rs 10.50 + Rs 15.80 = Rs 26.30

Thus, the total amount given by her parents is Rs 26.30.

 

Question 5:  Rahul bought 4 kg 90 g apples, 2 kg 60 g of grapes and 5 kg 300 g of mangoes. Find the weight of the fruits he bought in all. 

Solution 5:

Apple’s weight bought by Rahul = 4 kg 90 g = 4.090 kg

Grapes weight bought by Rahul = 2 kg 60 g = 2.060 kg

Mangoes weight bought by Rahul = 5 kg 300 g = 5.300 kg

The weight of all the fruits = 4.090 + 2.060 + 5.300 = 11.450 kg

Hence, the total weight of the fruits is 11.450 kg.

 

Question 6:  Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for skirt. Find the total cloth bought by her. 

Solution 6:

Cloth bought for shirt = 3 m 20 cm = 3.20 m

Cloth bought for skirt = 2 m 50 cm = 2.05 m

Total cloth bought by her = 3.20 + 2.05 = 5.25 m = 5 m 25 cm

Thus, the total cloth bought by Nasreen is 5 m 25 cm.

 

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

Solution 7:

Sunita travelled by bus = 15 km 268 m = 15.268 km

Sunita travelled by car = 7 km 7 m = 7.007 km

Sunita travelled by foot = 500 m = 0.500 km

The total distance from residence to school = 15.268 + 7.007 + 0.500 = 22.775 km

Thus, the distance from her residence to school is 22.775 km.

 

Exercise 7.9

 RD Sharma Solutions Class 6 Maths Chapter 7 Decimals-19


Question 2:  Find the value of:

(i) 9.756 – 6.28

(ii) 21.05 – 15.27

(iii) 18.5 – 6.79

(iv) 48.1 – 0.37

(v) 108.032 – 86.8

(vi) 91.001 – 72.9

(vii) 32.7 – 25.86

(viii) 100  – 26.32 

Solution 2:

(i) 9.756 – 6.28

Change the given values in like fractions.

9.756 – 6.280 = 3.476

Hence, the value of given expiration is 3.476. 

 

(ii) 21.05 – 15.27

Change the given values in like fractions.

21.05 – 15.27 = 5.78

Hence, the value of given expiration is 5.78.

 

(iii) 18.5 – 6.79

Change the given values in like fractions.

18.50 – 6.79 = 11.71

Hence, the value of given expiration is 11.71.

 

(iv) 48.1 – 0.37

Change the given values in like fractions.

48.10 – 0.37 = 47.73

Hence, the value of given expiration is 47.73.

 

(v) 108.032 – 86.8

Change the given values in like fractions.

108.032 – 86.800 = 21.232

Hence, the value of given expiration is 21.232.

 

(vi) 91.001 – 72.9

Change the given values in like fractions.

91.001 – 72.900 = 18.101

Hence, the value of given expiration is 18.101.

 

(vii) 32.7 – 25.86

Change the given values in like fractions.

32.70 – 25.86 = 6.84

Hence, the value of given expiration is 6.84.

 

(viii) 100 – 26.32

Change the given values in like fractions.

100 – 26.32 = 73.68

Hence, the value of given expiration is 73.68.

 

Question 3:  The sum of two numbers is 100. If one of them is 78.01, find the other. 

Solution 3:

One of the number = 78.01
Let the other number will be x
Sum of two numbers = 100
78.01 + x = 100 
x = 100 – 78.01 
x = 21.99 
Thus, the other number is 21.99.

 

Question 4:  Waheeda’s school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the rest she travels by bus. How much distance does she travel by bus? 

Solution 4:

Let the distance travelled by bus is x
Total Distance of school from house = 5 km 350 m = 5.350 km
Total Distance travelled by foot = 1 km 70 m = 1.070 km
1.070 + x = 5.350 
x = 5.350 – 1.070 
x = 4.280km 
Thus, the distance travelled by bus is 4.280 km.
 

Question 5:  Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Solution 5:

Total Amount given to shopkeeper = Rs. 50

Cost of book = Rs. 35.65 

The remaining amount returned by shopkeeper = 50 – 35.65 = Rs 14.35

Thus, the remaining amount returned by the shopkeeper is Rs 14.35.

 

Question 6:  Ruby bought a watermelon weighing 5 kg 200 g. Out of this she gave 2 kg 750 g to her neighbour. What is the weight of the watermelon left with Ruby? 

Solution 6:

Watermelon weight bought by Ruby is = 5 kg 200 g = 5.200 kg

Watermelon weight Ruby gave to neighbour is = 2 kg 750 g = 2.750 kg

Weight of watermelon left with Ruby = 5.200 – 2.750 = 2.450 kg

Thus, the weight of watermelon left with Ruby is 2.450 kg.

 

Question 7:  Victor drove 89.050 km on Saturday and 73.9 km on Sunday. How many kilometres more did he drive on Sunday? 

Solution 7:

Distance travelled on Saturday = 89.050 km

Distance travelled on Sunday = 73.9 km

Distance travelled by Victor more on Saturday = 89.050 – 73.9 = 15.15 km

Thus, 15.15km more he drive on Sunday Victor

 

Question 8:  Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Solution 8:
Total Amount given to shopkeeper = Rs. 50
Cost of the book = Rs 35.65
The balance amount retuned by shopkeeper = 50 – 35.65 = Rs 14.35
Thus, the balance amount retuned by shopkeeper is Rs 14.35.
 
Question 9:  Gopal travelled 125.5 km by bus, 14.25 km by pony and the rest of distance to Kedarnath on foot. If he covered a total distance of 150 km, how much did he travel on foot?
 
Solution 9:
Let Distance travelled by Gopal by foot x
Total Distance travelled by Gopal = 150 km
Distance travelled by Gopal by bus = 125.5 km
Distance travelled by Gopal by pony = 14.25km
150 = 125.5 + 14.25 + x
x = 150 – 125.5 – 14.25
x = 10.25 km
Thus, Distance travelled by Gopal on foot is 10.25 km.
 
Question 10:  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?
 
Solution 10:
Tina had cloth = 20 m 5 cm = 20.05 m
Cloth cut to make curtain = 4m 50 cm = 4.50 m
Cloth left with her = 20.05 – 4.50 = 15.55 m
Thus, the length of cloth left with Tina is 15.55 m.
 
Question 11:   Vineeta bought a book for Rs 18.90, a pen for Rs 8.50 and some papers for Rs 5.05. She gave fifty rupee to the shopkeeper. How much balance did she get back?
 
Solution 11:
Cost of a book = Rs 18.90
Cost of a pen = Rs 8.50
Cost of a papers = Rs 5.05
Total cost = 18.90 + 8.50 + 5.05 = Rs 32.45
Amount give to shopkeeper = Rs 50
The remaining amount returned = 50 – 32.45 = Rs 17.55
Thus, the shopkeeper returned back Rs 17.55.
 
Question 12:  Tanuj walked 8.62 km on Monday, 7.05 km on Tuesday and some distance on Wednesday. If he walked 21.01 km in three days, how much distance did he walk on Wednesday?
 
Solution 12:
Let the walked by Tanuj on Wednesday be x 
Total distance walked by Tanuj in three days = 21.01 km 
Distance walked by Tanuj on Monday = 8.62 km
Distance walked by Tanuj on Tuesday = 7.05 km
21.01 = 8.62 + 7.05 + x
21.01 = 15.67 + 7.05 + x
21.01 – 15.67=x
5.34 km=x
Thus, the distance walked by Tanuj on Wednesday is 5.34 km.
 
Objective Type Questions ::->

Mark the correct alternative in each of the following:
 
Question 1:  3/10 is equal to
(a) 3.1
(b) 1.3
(c) 0.3
(d) 0.03
 
Solution 1: (c)
3/10 = 0.3
Hence, 3/10 = 0.3
 
Question 2:  7/100 is equal to
(a) 7.1
(b) 7.01
(c) 0.7
(d) 0.07
 
Solution 2: (d) 
7/100 = 0.07
Hence, 7/100 = 0.07
 
Question 3:  4/1000 is equal to
(a) 0.004
(b) 0.04
(c) 0.4
(d) 4.001
 
Solution 2: (a) 
4/1000 = 0.004
Hence, 4/1000 = 0.004
 
Question 4:  The value of 37/10000 is
(a) 0.0370
(b) 0.0037
(c) 0.00037
(d) 0.000037
 
Solution 4: (b)
37/10000 = 0.0037 
Hence, 37/10000 = 0.0037 
 
Question 5:  The place value of 5 in 0.04532 is
(a) 5
(b) 5/100
(c) 5/1000
(d) 5/10000
 
Solution 5: (c) 
5 is in the thousandth place.
Hence, place value of 5 in 0.04532 is 5/1000 
 
 
Question 6:   The value of 231/1000 is
(a) 0.231
(b) 2.31
(c) 23.1
(d) 0.0231
 
Solution 6: (a) 

Question 7:  The value of 3 5/1000 is
(a) 3.5
(b) 3.05
(c) 3.005
(d) 3.0005
 
Solution 7: (c) 
3 5/1000
= 3 + 5/1000 
= 3 + 0.005 
= 3.005
 
Question 8:   The value of 3/25 is
(a) 1.2
(b) 0.12
(c) 0.012
(d) None of these
 
Solution 8: (b) 
3/25  
Multiply numerator and denominator by 4.
= (3 × 4)/(25 × 4) 
= 12/100 
= 0.12
 
Question 9:  The value of 2 1/25 is
(a) 2.4
(b) 2.25
(c) 2.04
(d) 2.40
 
Solution 9: (c) 
2 1/25
Multiply numerator and denominator by 4.
= 2 + (1 × 4)/(25 × 4) 
= 2+ 4/100 
= 2 + 0.04
= 2.04
 
Question 10:  4 7/8 is equal to
(a) 4.78
(b) 4.87
(c) 4.875
(d) None of these
 
Solution 10: (c) 
4 7/8
= 4 + 7/8 
Multiply numerator and denominator by 125.
= 4 + (7 × 125)/(8 × 125)
= 4 +875/1000  
= 4 + 0.875 
= 4.875
 
Question 11:   2 + 3/10 + 5/100 is equal to
(a) 2.305
(b) 2.3
(c) 2.35
(d) 0.235
 
Solution 11: (c) 
3/10 + 5/100
By taking LCM of Denominators we get 100.
= (200+30+5)/100 
= 235/100   
= 2.35 
 
Question 12:  3/100 + 5/10000 is equal to
(a) 0.35
(b) 0.305
(c) 0.0305
(d) 0.3005
 
Solution 12: (c) 
3/100 + 5/10000 By taking LCM of Denominators we get 10000.
= 3/100 + 5/10000
= 0.03 + 0.0005 
= 0.0305
 
Question 13:  1 cm is equal is
(a) 0.1 m
(b) 0.01 m
(c) 0.10 m
(d) 0.001 m
 
Solution 13: (b) 
100 cm = 1 m
1 cm = 1/100m 
= 0.01 m
 
Question 14:  1 m is equal to
(a) 0.1 km
(b) 0.01 km
(c) 0.001 km
(d) 0.0001 km
 
Solution 14: (c) 
1000 m = 1 km
1 m = 1/1000m 
= 0.001 km
 
Question 15:   2 kg 5 gm is equal to
(a) 2.5 kg
(b) 2.05 kg
(c) 2.005 kg
(d) 2.6 kg
 
Solution 15: (c) 
1 g = 1/1000  kg 
5 g =  5/1000 kg
= 0.005 kg
Thus, 2 kg 5 gm =2 kg + 0.005 kg = 2.005 kg
 
Question 16:  15 litres and 15 ml is equal to
(a) 15.15 litres
(b) 15.150 litres
(c) 15.0015 litres
(d) 15.015 litres
 
Solution 16: (d) 
1 ml = 1/1000 
15 ml = 15/1000
= 0.015 litre
15 litre + 0.015 litre = 15.015 litres
Hence, 15 litre and 15 ml = 15.015 litres
 
Question 17:   Which of the following are like decimals?
(a) 5.5, 5.05, 5.005, 5.50
(b) 5.5, 0.55, 5.55, 5.555
(c) 5.5, 6.6, 7.7, 8.8
(d) 0.5, 0.56, 0.567, 0.5678
 
Solution 17: (c)
Like decimals are having same number of decimals.
Hence, the like decimals are 5.5. 6.6, 7.7, 8.8. 
 
Question 18:  The value of 0.5 + 0.005 + 5.05 is
(a) 5.55
(b) 5.555
(c) 5.055
(d) 5.550
 
Solution 18: (b)
0.5 + 0.005 + 5.05 
= 5.555
 
Question 19:  0.35 − 0.035 is equal to
(a) 0.3
(b) 0.349
(c) 0.315
(d) 0.353
 
Solution 19: (c) 
0.35 – 0.035 
= 0.315
 
Question 20:  2.5 + 3.05 − 4.005 is equal to
(a) 1.545
(b) 1.455
(c) 1.554
(d) 0.545
 
Solution 20: (a) 
2.5 + 3.05 − 4.005 
= 5.55 − 4.005 
= 1.545
 
Question 21:  Which is greater among 2.3, 2.03, 2.33, 2.05?
(a) 2.3
(b) 2.03
(c) 2.33
(d) 2.05
 
Solution 21: (c) 
When comparing we get 
In 2.33 after decimals have greater number
Hence, 2.33 is greater among all given decimals.



 

NCERT Exemplar Solution Class 6 Mathematics
NCERT Class 6 Mathematics Chapter 1 Number System Exemplar Solutions
NCERT Class 6 Mathematics Chapter 2 Geometry Exemplar Solutions
NCERT Class 6 Mathematics Chapter 3 Integers Exemplar Solutions
NCERT Class 6 Mathematics Chapter 4 Fractions and Decimals Exemplar Solutions
NCERT Class 6 Mathematics Chapter 5 Data Handling Exemplar Solutions
NCERT Class 6 Mathematics Chapter 6 Mensuration Exemplar Solutions
NCERT Class 6 Mathematics Chapter 7 Algebra Exemplar Solutions
NCERT Class 6 Mathematics Chapter 8 Ratio and Proportion Exemplar Solutions
NCERT Class 6 Mathematics Chapter 9 Symmetry and Practical Geometry Exemplar Solutions
RD Sharma Solutions Class 6 Maths
RD Sharma Solutions Class 6 Maths Chapter 1 Knowing our Numbers
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
RD Sharma Solutions Class 6 Maths Chapter 9 Ratio Proportion and Unitary Method
RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geomatrical Concepts
RD Sharma Solutions Class 6 Maths Chapter 11 Angles
RD Sharma Solutions Class 6 Maths Chapter 12 Triangle
RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
RD Sharma Solutions Class 6 Maths Chapter 14 Circles
RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes
RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling Presentation of Data
RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling Pictographs
RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling Bar Graphs