Important Questions for CBSE Class 6 Maths Chapter 8

CBSE Class 6 Maths Important Questions Chapter 8 - Decimals - Free PDF Download

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Ch-8 Decimals

Write the decimal number represented by the points S on the given number line.
1. 1.2
2. 1.6
3. 1.5
4. 1.4
3 g = ________ kg
1. 0.03
2. 0.3
3. None of these
4. 0.003
Find: 0.29 + 0.36
1. 0.065
2. 0.65
3. None of these
4. 65
31.24 + 7.03 = _______
1. zero
2. 38.07
3. 38.2
4. 38.27
1.24 + 7.23 = _______
1. 8.45
2. 8.07
3. 8.4
4. 8.47
Match the following-
Column I Column II
a. 2m 5cm i. 20 Rupees 15 paise
b. 14.035 km ii. Rs. 0.50
c. Rs 20.15 iii. 2.05 m
d. 50 paise iv. 14km 35m
Fill in the blanks.
1. $25.135 = 25 + \frac{1}{__{}} + \frac{3}{__{}} + \frac{5}{__{}}$
2. $66.606 = 60 +__{} + \frac{6}{__{}} + 0 + \frac{6}{__{}}$
3. $14 + \frac{5}{1000} + \frac{6}{10} = 14.__{} 0__{}$
4. $\frac{4}{100} + 5 =__.0__$
State True or False:-
1. Rs. 60.75 can also expressed as 60Rs 75p.
2. Adding 70.25 and 14.6 gives 84.85.
3. 1 kg is equals to 100 grams.
4. Expanded form of 23.05 is $20 + 3 + \frac{0}{10} + \frac{5}{100}$ .
Arrange the following decimal numbers in descending order.
8.51, 8.06, 9.10, 8.57
Express $\frac{6}{100}$ as decimal.
Express the following as cm using decimals : 30 mm
Show 1.1 on the number line.
Express the following as cm using decimals
116 mm.
Subtract 0.314 kg from 2.107 kg.
Subtract the sum of 114.753 and 70.14 from the sum of 93.12 and 212.15.

Column I	Column II
a. 2m 5cm	i. 20 Rupees 15 paise
b. 14.035 km	ii. Rs. 0.50
c. Rs 20.15	iii. 2.05 m
d. 50 paise	iv. 14km 35m

Answer

1. 1.5
  Explanation: S is lies after 1 and fifth part toward right. So the decimal number represented by the points S on the given number line is 1.5.
1. 0.003
  Explanation: 1000 g = 1 kg3 g = $\frac{3}{1000}$ kg= 0.003 kg
1. 0.65
  Explanation: 0.29 + 0.36 = 0.65
1. 38.27
  Explanation: 31.24 + 7.03 = 38.27
1. 8.47
  Explanation: 1.24 + 7.23 = 8.47
1. $\to$ iii
  100 cm = 1 m
  so, 5 cm = $\frac{5}{100}$ m
  Therefore, 2 m 5 cm = 2 m + $\frac{5}{100}$ m
  = 2 m + 0.05 m
  = 2.05 m
2. $\to$ iv
  1 km = 1000 m
  14.035 km = 14.035 $\times$ 1000 m
  = 14035 m
  = 14000 m + 35 m
  = 14 km + 35 m = 14 km 35 m.
3. $\to$ i
  1 Rupee = 100 paise
  Rs. 20.15 = 20.15 $\times$ 100 p
  = 2015 paise
  = 2000 paise + 15 paise
  = 20 Rupees + 15 paise = 20 Rupees 15 paise.
4. $\to$ ii
  100 paise = 1 Rupee
  50 paise = Rs. $\frac{50}{100}$ = Rs. 0.50.
1. 10, 100, 1000.
  In the decimal number, 25.135, 1 is in tenths place. So its place value is $\frac{1}{10}$ , 3 is in hundredths place. So, its place value is $\frac{3}{100}$ .
  6 is in thousandths place. So, its place value is $\frac{6}{1000}$ .
  Therefore, $25.135 = 25 + \frac{1}{10} + \frac{3}{100} + \frac{5}{1000}$
2. 6, 10, 1000
  In the decimal number, 66..606, the place values of each 6 from left to right are: 60, 6, $\frac{6}{10}$ , $\frac{0}{100}, \frac{6}{1000}$
  Therefore, 66.606 = 60 + 6 + $\frac{6}{10}$ + 0 + $\frac{6}{1000}$
3. 6, 5
  The given expanded form 14 + $\frac{5}{1000} + \frac{6}{10}$ can be written in the correct order as, 14 + $\frac{6}{10} + \frac{0}{100} + \frac{5}{1000}$
  i.e., 6 is in tenths place, 0 is in hundredths place and 5 is in thousandths place.
  So, the decimal number can be written as 14.605.
4. 5, 4
  The given expanded form $\frac{4}{100} + 5$ can be written in the correct order as, 5 + $\frac{0}{10} + \frac{4}{100}$ .
  i.e., 0 is in tenths place and 4 is in hundredths place. So, the decimal number can be written as 5.04.
1. True(Rs.60.75 = Rs 60 + Rs 0.75 = Rs 60 + Rs $\frac{75}{100}$ = 60Rs 75p)
2. True ( By changing both numbers in to like decimals and adding, 70.25 + 14.60 = 84.85)
3. False (We know that 1Kg = 1000g)
4. True ( The expanded form of 23.05 = 2 Tens + 3 Ones + 0 Tenths + 5 Hundredths = 20 + 3 + $\frac{0}{10}$ + $\frac{5}{100}$ )
The given decimal numbers are like decimals. By checking the digits before decimal, we observe that 9.10 is largest of given numbers. Among the remaining numbers, by checking the tenths place, we observe that 8.06 is the least. Out of remaining numbers, by checking the hundredths place, we observe that 8.57 is bigger than 8.51. So, by arranging the given numbers from big to small, we have,
$\Rightarrow$ 9.10, 8.57, 8.51, 8.06
$\frac{6}{100}$ = 0.06
$\begin{array}{l} 1 m m = \frac{1}{10} c m \\ s o 30 m m = \frac{30}{10} c m \\ = 3 c m = 3.0 c m \end{array}$
We know that 1.1 is more than one but less than two. There is one-tenth in it. Divide the unit length between 1 and 2 into 10 equal parts and take 1 part as shown below :
116 mm = $\frac{116}{10}$ cm [ $∵$ 1 mm $= \frac{1}{10}$ cm] $= \frac{110 + 6}{10}$ cm $= (\frac{110}{10} + \frac{6}{10})$ cm $= (11 + \frac{6}{10})$ cm
= 11.6 cm
0.314 kg from 2.107 kg = 1.793 kg
Sum of 114.753 and 70.14:- $\frac{+ \begin{matrix} 114.753 \\ 70.140 \end{matrix}}{184.893}$ Sum of 212.15 and 93.12:- $\frac{+ \begin{matrix} 212.15 \\ 93.12 \end{matrix}}{305.27}$
Difference in their sums = $\frac{- \begin{matrix} 305.270 \\ 184.893 \end{matrix}}{120.377}$

Important Questions for CBSE Class 6 Maths Chapter 8 – Decimals

Important Questions for CBSE Class 6 Maths Chapter 8 - Decimals

CBSE Class 6 Maths Important Questions Chapter 8 - Decimals - Free PDF Download