Balbharati Solutions Class 6 Mathematics Equations
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 26 and 27, Equations. Here students can easily find step by step solutions of all the problems for Equations. 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 Equations solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.
Equations all Question Solutions :
Practice Set 26 :
Question no – (1)
Solution :
(1) 16 ÷ 2
= 16 ÷ 2
= 16 – 8
(2) 5 × 2
= 5 × 2
= 12 -2
(3) 8 × 3
= 8 × 3
= 20 +4
(4) 19 – 10
= 19 – 10
= 3 × 3
(5) 9 + 4
= 9 + 4
= 15 -2
(6) 10 – 2
= 10 – 2
= 20 – 12
(7) 72 ÷ 3
= 72 ÷ 3
= 12 × 2
(8) 37 – 2
= 37 – 2
= 7 × 5
(9) 4 + 5
= 4 + 5
= 10 – 1
(10) 6 + 7
= 6 + 7
= 10 + 3
Practice Set 26 :
Question no – (1)
Solution :
(1) The sun of a certain number and 3
= Let a be a certain number.
= Therefore, a + 3
(2) The difference obtained by subtracting 11 from another number
= Let another number is b.
= Thus, b – 11
(3) The product of 15 and another number
= Let another number is c.
= Thus, 15 × c
(4) Four times a number is 24
= Let a number is d.
= Thus, 4 × d = 24
Question no – (2)
Solution :
(1) x + 9 = 11
Given, x + 9 = 11
Thus, x + 9 – 9 = 11 – 9
x = 2
Therefore, the operation is subtracted 9 from both sides.
(2) x – 4 = 9
x – 4 = 9
Thus, x – 4 + 4 = 9 + 4
x = 13
Therefore, the operation is adding 4 from both sides.
(3) 8x = 24
8x = 24
Thus, 8x/8 = 24 /8
x = 3
Therefore, the operation is dividing 8 from both sides.
(4) x/6 = 3
x/6 = 3
Thus, x/6 * 6 = 3*6
x = 18
Therefore, the operation is multiplying 6 on both sides.
Question no – (4)
Solution :
(1) y – 5 = 1
Adding 5 on both sides,
We get, y – 5 +5 = 1 + 5
Therefore, y = 6
(2) 8 = t + 5
Subtracting 5 from both sides,
We get, 8 – 5 = t + 5 – 5
= 3 = t
= t = 3
(4) 19 = m – 4
Adding both sides by 4,
We get, 19 + 4 = m – 4 + 4
Therefore, 23 = m
(5) P/4 = 9
Multiplying 4 on both sides,
We get, 4 × P/4 = 4 × 9
Therefore, p = 3
(6) x + 10 = 5
Subtracting 10 from both sides
We get, x + 10 – 10 = 5 – 10
Therefore, x = -5
(7) m – 5 = – 12
Adding 5 on both sides,
We get, m – 5 + 5 = – 12 + 5
Therefore, m = -7
(8) p + 4 = – 1
Subtracting both sides by 4,
We get, p + 4 – 4 = -1 – 4
Therefore, p = – 5
Question no – (5)
Solution :
(1) Supposed, Haraba owns x sheep.
After he selling 34 of them in the market, he still has 176 sheep.
Thus, x – 34 = 176
Now, adding both side by 34 ,
We get, x – 34 + 34 = 176 + 34
Therefore, x = 210
Hence, 210 sheep did Haraba have at first.
(2) Let Sakshi made x jam bottles
Thus, x – 7 = 12
x – 7 + 7 = 12 + 7 ….. ( Since, adding both sides by 7)
We get, x = 19
Since, for 1 bottle = 250 g jam
Thus, for 19 bottle,
= 19 × 250
= 4750 gm of jam
Hence, Sakshi made 19 bottles and it’s weight is 4750 gm.
(3) Let x kilograms of wheat bought Archana.
Since, She requires 12 kg per month.
Thus, For 3 month = 3 * 12 = 36 kg.
Also, She got enough wheat milled for 3 months. After that, she had 14 kg left.
Therefore, x – 36 = 14
Adding both sides by 36 ,
We get x – 36 + 36
= 14 + 36
= x = 50
Therefore, Archana bought 50 kg wheat altogether.
More Solutions :
👉 Angles
👉 Integers
👉 HCF-LCM
