SEBA Class 6 Mathematics Chapter 10 Algebra Solutions, SEBA Class 6 Maths Notes in English Medium, SEBA Class 6 Mathematics Chapter 10 Algebra Notes to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 6 Mathematics Chapter 10 Algebra Notes and select needs one.
SEBA Class 6 Mathematics Chapter 10 Algebra
Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 6 General Mathematics Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 6 Mathematics Chapter 10 Algebra Solutions for All Subject, You can practice these here.
Algebra
Chapter – 10
| Exercise – 10 (A) |
1. Fill in the blanks with variables Suppose x is a number.
(a) If we add 10 to the number we will get ____________.
Ans: If we add 10 to the number we will get x + 10
(b) If we subtract 17 from the number we will get ____________.
Ans: If we subtract 17 from the number we will get x – 17
(c) If we multiply the number by 9 we will get ____________.
Ans: If we multiply the number by 9 we will get 9x
(d) If we divide the number by 15 we will get ____________.
Ans: If we divide the number by 15 we will get x/15
2. Choose the correct answer:
If 5 is added to a number x, the sum is-
(a) 5x
(b) x + 5
(c) x/5
(d) 5 – x
Ans: (b) x + 5
3. 3 times a number is-
(a) m + 3
(b) m – 3
(c) 3m
(d) m/3
Ans: (c) 3m
4. Riyan scored 76 marks in Mathematics. We do not know his marks in science. If we take the mark scored in science as x, then what is the total marks scored by Riyan in these two subjects?
Ans: Riyan scored 76 marks in Mathematics.
Let his marks in Science = x
∴ The total marks scored by Riyan in these two subjects = x + 76
5. Ankita had some chocolates with her. Mridusmita has 5 chocolates more than Ankita. Use an appropriate variable to write the algebraic expression for the number of chocolates Mridusmita has.
Ans: Let Ankita had x chocolates.
∴ The number of chocolates Mridusmita has (x + 5)
6. Some patterns were made with matchsticks. Look at the patterns and complete the following table.
Ans:
Ans:
Try these: Write ten algebraic expressions and write how they are formed. Express the following in algebraic terms:
(i) Sum of a and 10.
Ans: a + 10
(ii) Product of a and 10.
Ans: a × 10 = 10a
(iii) -m is multiplied by 8.
Ans: -m × 8 = – 8m
(iv) Multiply x by 7 and subtract 10 from the product.
Ans: 7 × x – 10 = 7x – 10
(v) Subtract the product of 9 and y from 18.
Ans: 18 – 9 × y = 18 – 9y
(vi) Subtract 8 from -m.
Ans: – m – 8
(vii) Multiply – p by 5.
Ans: – p × 5 = – 5p
(viii) Subtract 11 from 2m.
Ans: 2m – 11
(ix) Divide y by 5.
Ans: y ÷ 5 = y/5
(x) Divide m by 7 and subtract 12 from the quotient.
Ans: m / 7 – 12
| Exercise – 10 (B) |
1. State which of the following are equations. Give reasons for your answer.
(a) x – 19 = 10
Ans: x – 19 = 10
⇒ x = 10 + 19
⇒ x = 29
∴ x – 19 = 10 is an equation.
(b) a + 9 = -9
Ans: a + 9 = – 9
⇒ a = – 9 – 9
⇒ a = – 18
∴ a + 9 = -18 is an equation.
(c) 3b = 15
Ans: 3b = 15
⇒ b = 15/3 = 5
∴ 3b = 15 is an equation.
(d) 21 – 1 > 5
Ans: 21 – 1 > 5
⇒ 20 > 5
∴ 21 – 1 > 5 is not an equation.
(e) 2n + 6 < 18
Ans: 2n + 6 < 18
⇒ 2n < 18 – 6
⇒ 2n < 12
⇒ n < 12 / 2
⇒ n < 6
∴ 2n + 6 < 18 is not an equation.
2. Identify the variables in the following equations.
(i) 3p – 7 = 5
Ans: Variable of is p.
(ii) (5q) / 3 = 8
Ans: Variable of is q.
(iii) 3 × 9 – 11 = r
Ans: Variable of is r.
3. Choose the correct answer.
(i) Value of m in the equation m + 4 = 7 is:
(a) 0
(b) 4
(c) 3
(d) 7
Ans: (c) 3
Sol:
m + 4 = 7
⇒ m = 7 – 4 = 3
(ii) Solution of the equation 4p = 20 is.
(a) p = 4
(b) p = 5
(c) p = 20
(d) p = 0
Ans: (b)
Sol:
4p = 20
⇒ p = 20 / 4 = 5
4. (i) Complete the following table.
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2x + 3 |
Ans:
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2x + 3 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 |
(ii) From the table above find the solution to the equation 2x + 3 = 11.
Ans: If x = 4, 3x + 3 = 11 satisfy.
∴ x = 4
5. State whether the value of the variable shown in the table against the equation satisfy the corresponding equation.
| Equation | Value of variable | Equation satisfied Yes/No |
| l + 9 = 101 | 1 = 92 | |
| m – 9 = 12 | m = 15 | |
| 2p = 18 | p = 3 | |
| h – 7 = 0 | h = 4 | |
| 3a = 2a + 2 | a = 1 | |
| b + 4 = 1 | b = – 3 | |
| 3q + 4 = 7 | q = 2 |
Ans: (i) l + 9 = 101
⇒ 92 + 9 = 101
⇒ 101 = 101
∴ Equation is satisfied
(ii) m – 9 = 12
⇒ 15 – 9 = 12
6 = 12
∴ 6 ≠ 12
∴ Equation is not satisfied
(iii) 2p = 18
⇒ 2 × 3 = 18
⇒ 6 = 18
∴ 6 ≠ 18
∴ Equation is not satisfied
(iv) h – 7 = 0
⇒ 4 – 7 = 0
⇒ -3 = 0
∴ -3 ≠ 0
∴ Equation is not satisfied
(v) 3a = 2a + 2
⇒ 3 × 1 = 2 × 1 + 2
⇒ 3
∴ 3 ≠ 4
∴ Equation is not satisfied
(vi) b + 4 = 1
⇒ -3 + 4 = 1
⇒ 1 = 1
∴ Equation is satisfied
(vii) 3q + 4 = 7
⇒ 3 × 2 + 4 = 7
⇒ 6 + 4 = 7
⇒ 10 = 7
∴ 10 ≠ 7
∴ Equation is not satisfied
6. Solve the following equations by trial and error method.
(i) 2r + 1 = 7
Ans: 2r + 1 = 7
r = 1, 2 × 1 + 1 = 2 + 1 = 3, L.H.S.= R.H.S
r = 2, 2 × 2 + 1 = 4 + 1 = 5, L.H.S.= R.H.S
r = 3, 2 × 3 + 1 = 7, L.H.S.= R.H.S
(ii) 4m = 24
Ans: 4m = 24
m = 2, 4 × 2 = 8, L.H.S. = R.H.S
m = 4, 4 × 4 = 16, L.H.S.= R.H.S
m = 6, 4 × 6 = 24, L.H.S. = R.H.S
7. State the following statements in equations.
(i) When Rs. 10 is subtracted from the cost of Rs. 3 kg of rice, Rs. 80 is left.
Ans: Let one kilogram of rice = Rs. x
∴ According to question, 3x – 10 = 80
(ii) In a cricket game when 7 runs are added to the runs scored in 5 overs, the total run is 39. If the number of runs per over is same, express the above statement in an equation.
Ans: Let the number of runs per over = x
∴ According to question, 5x + 7 = 39
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