Mathematics Solutions Solutions for Class 6 Math Chapter 5 - Decimal Fractions

Answer:

Answer:

(i)
$\begin{array}{cccccccc}& & \overline{)1}& & & & & \\ & 9& 0& 5& .& 5& 0& 0\\ +& & 2& 7& .& 1& 9& 7\\ & 9& 3& 2& .& 6& 9& 7\\ & & & & & & & \end{array}$
∴ 905.5 + 27.197 = 932.697

(ii)
$\begin{array}{ccccccc}& & 3& 9& .& 0& 0\\ +& 7& 0& 0& .& 6& 5\\ & 7& 3& 9& .& 6& 5\\ & & & & & & \end{array}$
∴ 39 + 700.65 = 739.65

(iii)
$\begin{array}{cccccccc}& & \overline{)1}& \overline{)1}& & & & \\ & & 4& 0& .& 0& 0& 0\\ & & 2& 7& .& 7& 0& 0\\ +& & & 2& .& 4& 5& 1\\ & & 7& 0& .& 1& 5& 1\\ & & & & & & & \end{array}$
∴ 40 + 27.7 + 2.451 = 70.151

Answer:

(i)
$\begin{array}{ccccccc}& 8& 5& .& 9& {}^5\cancel{6}& {}^{10}\cancel{0}\\ -& & 2& .& 3& 4& 5\\ & 8& 2& .& 6& 1& 5\\ & & & & & & \end{array}$
∴ 85.96 − 2.345 = 82.615

(ii)
$\begin{array}{ccccccc}& {}^5\cancel{6}& {}^{12}\cancel{3}& {}^{11}\cancel{2}& .& {}^{11}\cancel{2}& {}^{14}\cancel{4}\\ -& & 9& 7& .& 4& 5\\ & 5& 3& 4& .& 7& 9\\ & & & & & & \end{array}$
∴ 632.24 − 97.45 = 534.79

(iii)
$\begin{array}{cccccccc}& {}^1\cancel{2}& {}^9\cancel{0}& {}^9\cancel{0}& .& {}^9\cancel{0}& {}^9\cancel{0}& {}^{15}\cancel{5}\\ -& & 1& 7& .& 1& 8& 6\\ & 1& 8& 2& .& 8& 1& 9\\ & & & & & & & \end{array}$
∴ 200.005 − 17.186 = 182.819

Answer:

Distance travelled by bus = 42 km 365 m = 42 km + 365 m = 42 km + $\frac{365}{1000}$ km = 42 km + 0.365 km = 42.365 km         (1 km = 1000 m)

Distance travelled by car = 12 km 460 m = 12 km + 460 m = 12 km + $\frac{460}{1000}$ km = 12 km + 0.460 km = 12.460 km

Distance travelled by walking = 640 m = $\frac{640}{1000}$ km = 0.640 km

∴ Total distance travelled altogether

= Distance travelled by bus + Distance travelled by car + Distance travelled by walking

= 42.365 km + 12.460 km + 0.640 km 

= 55.465 km


$\begin{array}{cccccccc}& & & \overline{)1}& & \overline{)1}& & \\ & & 4& 2& .& 3& 6& 5\\ & & 1& 2& .& 4& 6& 0\\ +& & & 0& .& 6& 4& 0\\ & & 5& 5& .& 4& 6& 5\\ & & & & & & & \end{array}$
Thus, the total distance travelled by Avinash altogether is 55.465 km.

Answer:

Length of cloth bought for salwaar = 1.80 m

Length of cloth bought for kurta = 2.25 m

∴ Total length of cloth bought

= Length of cloth bought for salwaar + Length of cloth bought for kurta

= 1.80 m + 2.25 m

= 4.05 m


$\begin{array}{ccccc}& \overline{)1}& & & \\ & 1& .& 8& 0\\ +& 2& .& 2& 5\\ & 4& .& 0& 5\\ & & & & \end{array}$
Rate of cloth = Rs 120/m

∴ Amount paid to the shopkeeper

= Total length of cloth bought × Rate of cloth

=  4.05 m × â‚¹ 120/m

= â‚¹ 486

$\begin{array}{cccccc}& & & 4& 0& 5\\ & & \times & 1& 2& 0\\ & & & 0& 0& 0\\ & & 8& 1& 0& \times \\ & 4& 0& 5& \times & \times \\ & 4& 8& 6& 0& 0\\ & & & & & \end{array}$
Thus, Ayesha paid â‚¹ 486 to the shopkeeper.

Answer:

Total weight of the watermelon = 4.25 kg

Weight of the watermelon given to the children = 1 kg 750 g = 1 kg + 750 g = 1 kg + $\frac{750}{1000}$ kg = 1 kg + 0.75 kg = 1.75 kg

∴ Weight of the watermelon left with her

= Total weight of the watermelon − Weight of the watermelon given to the children

= 4.25 kg − 1.75 kg

= 2.5 kg


$\begin{array}{ccccc}& {}^3\cancel{4}& .& {}^{12}\cancel{2}& 5\\ -& 1& .& 7& 5\\ & 2& .& 5& 0\\ & & & & \end{array}$
Thus, the weight of watermelon left with Sujata is 2.5 kg.

Answer:

Original driving speed of Anita = 85.6 km/h

Speed limit for driving = 55 km/h

∴ Speed reduced by her to be within the speed limit

= Original driving speed of Anita − Speed limit for driving

= 85.6 km/h − 55 km/h

= 30.6 km/h

$\begin{array}{ccccc}& 8& 5& .& 6\\ -& 5& 5& .& 0\\ & 3& 0& .& 6\\ & & & & \end{array}$
Thus, Anita needs to reduce her speed by 30.6 km/h to be within the speed limit.

Answer:

(1) $\frac{3}{5}=\frac{3\times \boxed{2}}{5\times \boxed{2}}=\frac{\boxed{6}}{10}=\boxed{0.6}$

(2) $\frac{{\displaystyle 25}}{{\displaystyle 8}}=\frac{{\displaystyle 25\times \boxed{125}}}{{\displaystyle 8\times 125}}=\frac{{\displaystyle \boxed{3125}}}{{\displaystyle 1000}}=3.125$

(3) $\frac{{\displaystyle 21}}{{\displaystyle 2}}=\frac{{\displaystyle 21\times \boxed{5}}}{{\displaystyle 2\times \boxed{5}}}=\frac{{\displaystyle \boxed{105}}}{{\displaystyle 10}}=\boxed{10.5}$

(4) $\frac{{\displaystyle 22}}{{\displaystyle 40}}=\frac{11}{20}=\frac{{\displaystyle 11\times \boxed{5}}}{{\displaystyle 20\times 5}}=\frac{{\displaystyle \boxed{55}}}{{\displaystyle 100}}=\boxed{0.55}$

Answer:

(1) $\frac{3}{4}=\frac{3\times 25}{4\times 25}=\frac{75}{100}=0.75$

(2) $\frac{4}{5}=\frac{4\times 2}{5\times 2}=\frac{8}{10}=0.8$

(3) $\frac{9}{8}=\frac{9\times 125}{8\times 125}=\frac{1125}{1000}=1.125$

(4) $\frac{17}{20}=\frac{17\times 5}{20\times 5}=\frac{85}{100}=0.85$

(5) $\frac{36}{40}=\frac{36÷4}{40÷4}=\frac{9}{10}=0.9$

(6) $\frac{7}{25}=\frac{7\times 4}{25\times 4}=\frac{28}{100}=0.28$

(7) $\frac{19}{200}=\frac{19\times 5}{200\times 5}=\frac{95}{1000}=0.095$

Answer:

(1) 27.5 = $\frac{275}{10}$

(2) 0.007 = $\frac{7}{1000}$

(3) 90.8 = $\frac{908}{10}$

(4) 39.15 = $\frac{3915}{100}$

(5) 3.12 = $\frac{312}{100}$

(6) 70.400 = 70.4 = $\frac{704}{10}$

Answer:

$$\begin{gathered} 3.17\times 4.5 \\ =\frac{317}{100}\times \frac{45}{10} \\ =\frac{317\times 45}{100\times 10} \\ =\frac{14265}{1000} \\ =14.265 \end{gathered}$$

Answer:

$$\begin{gathered} 5.03\times 2.17 \\ =\frac{503}{100}\times \frac{217}{100} \\ =\frac{503\times 217}{100\times 100} \\ =\frac{109151}{10000} \\ =10.9151 \end{gathered}$$

Answer:

(i) $2.7\times 1.4=\frac{27}{10}\times \frac{14}{10}=\frac{27\times 14}{10\times 10}=\frac{378}{100}=3.78$

(ii) $6.17\times 3.9=\frac{617}{100}\times \frac{39}{10}=\frac{617\times 39}{100\times 10}=\frac{24063}{1000}=24.063$

(iii) $0.57\times 2=\frac{57}{100}\times \frac{2}{1}=\frac{57\times 2}{100\times 1}=\frac{114}{100}=1.14$

(iv) $5.04\times 0.7=\frac{504}{100}\times \frac{7}{10}=\frac{504\times 7}{100\times 10}=\frac{3528}{1000}=3.528$

Answer:

Weight of each bag of rice = 5.250 kg

Number of bags of rice bought = 18

∴ Weight of rice bought altogether = Weight of each bag of rice × Number of bags of rice bought

= 5.250 kg × 18

= $\frac{525}{100}\times \frac{18}{1}$

= $\frac{525\times 18}{100\times 1}$

= $\frac{9450}{100}$

= 94.5 kg

Rate of rice = â‚¹ 42/kg

∴ Total amount paid for the rice = Weight of rice bought altogether × Rate of rice

= 94.5 kg × â‚¹ 42/kg

= $\frac{945}{10}\times \frac{42}{1}$

= $\frac{945\times 42}{10\times 1}$

= $\frac{39690}{10}$

= â‚¹ 3,969

Thus, Virendra bought 94.5 kg of rice altogether and the total amount paid by him is â‚¹ 3,969.

Answer:

Total length of the cloth = 23.50 m

Length of cloth required to make each curtain = 4 m 25 cm = 4 m + 25 cm = 4 m + $\frac{25}{100}$ m = 4 m + 0.25 m = 4.25 m       (1 m = 100 cm)

Number of curtains made of equal size = 5

∴ Length of cloth required to make 5 curtains = Length of cloth required to make each curtain × 5

= 4.25 m × 5

= $\frac{425}{100}\times \frac{5}{1}$

= $\frac{425\times 5}{100\times 1}$

= $\frac{2125}{100}$

= 21.25 m

∴ Amount of cloth left  = Total length of the cloth − Length of cloth required to make 5 curtains 

= 23.50 m − 21.25 m

= 2.25 m

$\begin{array}{cccccc}& 2& 3& .& {}^4\cancel{5}& {}^{10}\cancel{0}\\ -& 2& 1& .& 2& 5\\ & & 2& .& 2& 5\\ & & & & & \end{array}$
Thus, the length of the cloth left over is 2.25 m.

Answer:

(1) 4.8 ÷ 2 = $\frac{48}{10}÷\frac{2}{1}=\frac{48}{10}\times \frac{1}{2}=\frac{24}{10}=2.4$

(2) 17.5 ÷ 5 = $\frac{175}{10}÷\frac{5}{1}=\frac{175}{10}\times \frac{1}{5}=\frac{35}{10}=3.5$

(3) 20.6 ÷ 2 = $\frac{206}{10}÷\frac{2}{1}=\frac{206}{10}\times \frac{1}{2}=\frac{103}{10}=10.3$

(4) 32.5 ÷ 25 = $\frac{325}{10}÷\frac{25}{1}=\frac{325}{10}\times \frac{1}{25}=\frac{13}{10}=1.3$

Answer:

Total length of the road = 4 km 800 m = 4 km + 800 m = 4000 m + 800 m = 4800 m                   (1 km = 1000 m)

Distance between two plants = 9.6 m

∴ Number of trees planted on each side of the road = (Total length of the road ÷ Distance between two plants) + 1

= (4800 m ÷ 9.6 m) + 1

= $\left(\frac{4800}{1}÷\frac{96}{10}\right)$ + 1

= $\left(\frac{4800}{1}\times \frac{10}{96}\right)$ + 1

= 500 + 1

= 501

∴ Total number of trees planted on both sides of the road = Number of trees planted on each side of the road × 2 = 501 × 2 = 1002

Thus, the number of trees planted on both sides of the road is 1002.

Answer:

Total distance walked by Pradnya in 9 rounds of the path = 3.825 km

∴ Distance walked by Pradnya in one round of the path = Total distance walked by Pradnya in 9 rounds of the path ÷ 9

= 3.825 km ÷ 9

= $\frac{3825}{1000}÷\frac{{\displaystyle 9}}{1}$

= $\frac{3825}{1000}\times \frac{{\displaystyle 1}}{9}$

= $\frac{425}{1000}$

= 0.425 km

Thus, the distance walked by Pradnya in one round is 0.425 km.

Answer:

Amount paid for 0.25 quintal of hirada = â‚¹ 9,500

∴ Cost per quintal of hirada = Amount paid for 0.25 quintal of hirada ÷ 0.25

= â‚¹ 9,500 ÷ 0.25

= $\frac{9500}{1}÷\frac{25}{100}$

= $\frac{9500}{1}\times \frac{100}{25}$

= 380 × 100

= â‚¹ 38,000

Thus, the cost per quintal of hirada is â‚¹ 38,000.