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
More Solutions :
👉 Angles
👉 Integers
👉 HCF-LCM
