Std 6 th Mathematics- 5. Decimal Fractions page 2

Question 1:

Write the proper number in the empty boxes.
(1) $\frac{3}{5}=\frac{3\times x}{5\times x}=\frac{x}{10}=x$

(2) $\frac{25}{8}=\frac{25\times x}{8\times 125}=\frac{x}{1000}=3.125$

(3) $\frac{21}{2}=\frac{21\times x}{2\times x}=\frac{x}{10}=x$

(4) $\frac{22}{40}=\frac{11}{20}=\frac{11\times x}{20\times 5}=\frac{x}{100}=x$

Answer 1:

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

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

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

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

Question 2:

Convert the common fractions into decimal fractions.
(1) $\frac{3}{4}$

(2) $\frac{4}{5}$

(3) $\frac{9}{8}$

(4) $\frac{17}{20}$

(5) $\frac{36}{40}$

(6) $\frac{7}{25}$

(7) $\frac{19}{200}$

Answer 2:

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

Question 3:

Convert the decimal fractions into common fractions.
(1) 27.5
(2) 0.007
(3) 90.8
(4) 39.15
(5) 3.12
(6) 70.400

Answer 3:

(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 = $\frac{704}{10}$

Question 1:

If, 317 × 45 = 14265, then 3.17 × 4.5 = ?

Answer 1:

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

Question 2:

If, 503 × 217 = 109151, then 5.03 × 2.17 = ?

Answer 2:

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

Question 3:

Multiply.
(1) 2.7 × 1.4
(2) 6.17 × 3.9
(3) 0.57 × 2
(4) 5.04 × 0.7

Answer 3:

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

Question 4:

Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs 42 rupees per kg, how much did he pay for it?

Answer 4:

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 = ₹ 42/kg

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

= 94.5 kg × ₹ 42/kg

= $\frac{945}{10}\times \frac{42}{1}$

= $\frac{945\times 42}{10\times 1}$

= $\frac{39690}{10}$

= ₹ 3,969

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

Question 5:

Vedika has 23.50 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?

Answer 5:

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

$\begin{array}{cccccc}& 2& 3& .& 45& 100\\ -& 2& 1& .& 2& 5\\ & & 2& .& 2& 5\\ & & & & & \end{array}$
Thus, the length of the cloth left over is 2.25 m.

Question 1:

Carry out the following divisions.
(1) 4.8 ÷ 2
(2) 17.5 ÷ 5
(3) 20.6 ÷ 2
(4) 32.5 ÷ 25

Answer 1:

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

Question 2:

A road is 4 km 800 m long. If trees are planted on both its sides at intervals of 9.6 m, how many trees were planted?

Answer 2:

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.

Question 3:

Pradnya exercises regularly by walking along a circular path on a field. If she walks a distance of 3.825 km in 9 rounds of the path, how much does she walk in one round?

Answer 3:

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

= $\frac{3825}{1000}÷\frac{9}{1}$

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

= $\frac{425}{1000}$

= 0.425 km

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

Question 4:

A pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for 9500 rupees. What is the cost per quintal of hirada ? (1 quintal = 100 kg)

Answer 4:

Amount paid for 0.25 quintal of hirada = ₹ 9,500

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

= ₹ 9,500 ÷ 0.25

= $\frac{9500}{1}÷\frac{25}{100}$

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

= 380 × 100

= ₹ 38,000

Thus, the cost per quintal of hirada is ₹ 38,000.