**Balbharati Solutions Class 6 Mathematics Decimal Fractions**

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Balbharati Class 6 Mathematics Book, Practice Set 14, 15, 16, and 17, Decimal Fractions. Here students can easily find step by step solutions of all the problems for Decimal Fractions. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Decimal Fractions solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.

**Decimal Fractions all Question Solutions :**

**Practice Set (14) : **

**Question no – (1) **

**Solution :**

Place value:

Ten = 70

Unit = 8

Hundredth = 2/10 = 0.02

**Question no – (2) **

**Solution :**

**(1) 905.5 +27. 197**

= 905.500 + 27.197

= 932.697

**(2) 39 + 700.65**

= 039.00 + 700.65

= 739.65

**(3) 40 + 27.7 + 2.451**

= 40.000 + 27.700 + 2.451

= 67.700 + 2.451

= 70.151

**Question no – (3) **

**Solution :**

**(1) 85.96 – 2.345**

= 85.960 + 2.345

= 83.615

**(2) 632.24 – 97.45**

= 632.24 – 97.45

= 534.79

**(3) 200.005 – 17.186**

= 200.005 – 17.186

= 182.819

**Question no – (4) **

**Solution :**

As per the given question,

Avinash travelled 42 km 365 m by bus

12 km 460 m by car and walked 640 m.

Now, By bus = 42 km 365 m

By car = 12 km 460 m

By walked = 640 m

Thus, 42 km 365 m + 12 km 460 m + 640 m

= (42 km + 12 km ) + (365 + 460 + 640) m

= 54 km + (1465 m)

= 54 km + (1 km 465 m)

= 55 km 465 m

**Question no – (5) **

**Solution :**

According to the given question,

Ayesha bought 1.80 m of cloth for her salwaar and 2.25 m for her kurta.

Thus, total cloth,

=1.80 m + 2.25 m

= 4.05 m

Since, for 1 m cloth = ₹120

Thus, for 4.05 m cloth,

= 4.05 * 120

= ₹ 486.00

Hence, she pay ₹ 486 to the shopkeeper.

**Question no – (6) **

**Solution :**

As per the question,

Sujata bought a watermelon weighing 4.25 kg

1 kg 750 g to the children in her neighborhood.

Thus, 4.25 – 1.750 = 2.50 kg

Hence, she have left 2.5 kg watermelon.

**Question no – (7) **

**Solution :**

Anita was driving at a speed of 85.6 km per hour.

Since, road had a speed limit of 55 km per hour.

Thus, 85.6 – 55

= 30.6 km

Therefore,, she reduce 30.6 km per hour speed.

**Practice Set (15) : **

**Question no – (1) **

**Solution :**

(1) 3/5 = 3 × 2/5 × 2 = 6 /10 = 0.6

(2) 25/8 = 25 × 1 25/8 × 125 = 3125/1000 = 3.125

(3) 21/2 = 21 × 5/2 × 5 = 105/10 = 10.5

(4) 22/40 = 11/20 = 11 × 5/20 × 5 = 55/100 = 100

**Question no – (2) **

**Solution :**

**(1) 3/4**

3/4 = 3 × 25 /4 × 25

= 75/100

= 0.75

**(2) 4/5**

= 4 × 2/5 × 2

= 8/10

= 0.8

**(3) 9/8**

= 9 × 125/8 × 125

= 1125/1000

= 1.125

**(4) 17/20**

= 17 × 5/20 × 5

= 85 /100

= 0.85

**(5) 36/40**

= 18/20

= 18 × 5/20 × 5

= 90/100

= 0.9

**(6) 7/25**

= 7 × 4/25 × 4

= 28 /100

= 0.28

**(7) 19/200**

= 19 × 5/200 × 5

= 95 /1000

= 0.095

**Question no – (3) **

**Solution :**

**(1) 27.5**

= 275/10

**(2) 0.007**

= 7/1000

**(3) 90.8**

= 908/10

**(4) 39.15**

= 3915/100

**(5) 3.12**

= 312/100

**(6) 70.400**

= 70400/1000

= 704/10

**Practice Set (16) : **

**Question no – (1) **

**Solution :**

Since, 317 × 45 = 14265

Therefore, 3.17 × 4.5

= 14.265

**Question no – (2) **

**Solution :**

Since, 503 × 217 = 109151

Therefore, 5.03 × 2.17

= 10.9151

**Question no – (3) **

**Solution :**

**(1) 2.7 × 1.4**

= 27/10 × 14/10

= 378/100

= 3.78

**(2) 6.17 × 3.9**

= 617/100 × 39/10

= 24063/1000

= 24.063

**(3) 0.57 × 2**

= 57/100 × 2/1

= 114/100

= 1.14

**(4) 5.04 × 0.7**

= 504/100 × 7/10

= 3528/1000

= 3.528

**Question no – (4) **

**Solution :**

As per the question,

Virendra bought 18 bags of rice, each bag weighing 5.250 kg.

Thus, 18 × 5.250

= 18/1 × 5250/1000

= 94500/1000

= 94.500

Therefore, he buy 94.500 kg rise altogether.

Since, for 1 kg = ₹48

Thus, for 94.5 kg = 48 × 94.5

= 48/1 × 945/10

= 39690/19

= ₹3969

Therefore, he will pay Rs 3969

**Question no – (5) **

**Solution :**

According to the question,

Vedika has 23.50 metres of cloth.

She make 5 curtains of equal size.

Since, for 1 curtains

= 4 m 25 cm

= 4.25 m cloth required

Thus, for 5 curtains

= 5 × 4.25 = 5/1 ×425/100

= 2125/100

= 21.25 m cloth required

Now, total cloth = 23.50 m

Required cloth = 21.25 m

Thus, 23.50 – 21.25 = 2.25 m

Hence, 2.25 m cloth is left.

**Practice Set (17) : **

**Question no – (1) **

**Solution : **

**(1) 4.8 ÷ 2**

= 48/10 ÷ 2/1

= 48/10 × 1/2

= 24/10

= 2.4

**(2) 17.5 ÷ 5**

= 175/10 × 1/5

= 35/10

= 3.5

**(3) 20.6 ÷ 2**

= 206/10 ×1/2

= 103/10

= 10.3

**(4) 32.5 ÷ 25**

= 325/10 × 1/25

= 13/10

= 1.3

**Question no – (2) **

**Solution :**

As per the question,

A road is 4 km 800 m.

Since, trees are planted on both its sides at intervals of 9.6 m

Now, 4 km 800 m

= 4 km + 800 m

= 4800 m

4800 ÷ 9.6 = 4800 × 10/96

= 48000/96

= 500

Thus, 2 × 500 = 1000

Hence, 1000 trees are planted on both it’s sides at intervals of 9.6 m

**Question no – (3) **

**Solution :**

According to the question,

Pradnya walks a distance of 3.825 km in 9 rounds of the path

Thus, 3.825 ÷ 9 = 3825/1000 ÷ 9/1

= 3825/1000 × 1/9

= 425/1000

= 0.425 km

Hence, she walk in one round is 0.425 km

**Question**** no – (4) **

**Solution :**

Since, 1 quintal = 100 kg

Thus, 0.25 quintal

= 0.25 × 100

= 25/100 × 100/1

= 25 kg

We have given that, a pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for 9500 rupees.

Thus for 25 kg = ₹9500

So, for 1 kg

= 9500 /25

= ₹380

Therefore, for 100 kg

= 1 quintal

= 100 × 380

= ₹38,000

Therefore, the cost per quintal of Hirada Rs 8,000

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