NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1
Question 1.
Complete the last column of the table.
| Sr. No. | Equation | Value | Say, whether the Equation is Satisfied. (Yes/ No) |
|---|---|---|---|
| (i) | x+3 = 0 | x=3 | |
| (ii) | x+3 = 0 | x=0 | |
| (iii) | x+3 = 0 | x=-3 | |
| (iv) | x-7 = 1 | x=7 | |
| (v) | x-7 = 1 | x=8 | |
| (vi) | 5x = 25 | x= 0 | |
| (vii) | 5x = 25 | x= 5 | |
| (viii) | 5x = 25 | x= -5 | |
| (ix) | m/3= 2 | m=-6 | |
| (x) | m/3 = 2 | m=0 | |
| (xi) | m/3= 2 | m=6 |
| Sr. No. | Equation | Value | Say, whether the Equation is Satisfied. (Yes/ No) |
|---|---|---|---|
| (i) | x+3 = 0 | x=3 | No |
| (ii) | x+3 = 0 | x=0 | No |
| (iii) | x+3 = 0 | x=-3 | Yes |
| (iv) | x-7 = 1 | x=7 | No |
| (v) | x-7 = 1 | x=8 | Yes |
| (vi) | 5x = 25 | x= 0 | No |
| (vii) | 5x = 25 | x= 5 | Yes |
| (viii) | 5x = 25 | x= -5 | No |
| (ix) | m/3= 2 | m=-6 | No |
| (x) | m/3 = 2 | m=0 | No |
| (xi) | m/3= 2 | m=6 | Yes |
Question 2.
Check whether the value given in the brackets is a solution to the given equation
or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0)
Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4
Solution:
(i) 5p + 2=17
Putting p=–3 in L.H.S. 5(–3)+2 = –15+2=–13
∵–13≠17
Therefore, p=–3 is not the solution.
Putting p=–2 in L.H.S. 5(–2)+2=–10+2=–8
∵–8≠17
Therefore, p=–2 is not the solution.
Putting p=–1 in L.H.S. 5(–1)+2=–5+2=–3
∵–3≠17
Therefore, p=–1 is not the solution.
Putting p=0 in L.H.S. 5(0)+2=0+2=2
∵2≠17
Therefore, p=0 is not the solution.
Putting p=1 in L.H.S. 5(1)+2=5+2=7
∵7≠17
Therefore, p=1 is not the solution.
Putting p=2 in L.H.S. 5(2)+2=10+2=12
∵12≠17
Therefore, p=2 is not the solution.
Putting p=3 in L.H.S. 5(3)+2=15+2=17
∵17=17
Therefore, p=3 is the solution.
(ii) 3m–14=4
Putting m=–2 in L.H.S. 3(–2)–14=–6–14=–20
∵–20≠4
Therefore, m=–2 is not the solution.
Putting m=–1 in L.H.S. 3(–1)–14=–3–14=–17
∵–17≠4
Therefore, m=–1 is not the solution.
Putting m=0 in L.H.S. 3(0)–14=0–14=–14
∵–14≠4 Therefore, m=0 is not the solution.
Putting m=1 in L.H.S. 3(1)–14=3–14=–11
∵–11≠4
Therefore, m=1 is not the solution.
Putting m=2 in L.H.S. 3(2)–14=6–14=–8
∵–8≠4
Therefore, m=2 is not the solution.
Putting m=3 in L.H.S. 3(3)–14=9–14=–5
∵–5≠4
Therefore, m=3 is not the solution.
Putting m=4 in L.H.S. 3(4)–14=12–14=–2
∵–2≠4
Therefore, m=4 is not the solution.
Putting m=5 in L.H.S. 3(5)–14=15–14=1
∵1≠4
Therefore, m=5 is not the solution.
Putting m=6 in L.H.S. 3(6)–14=18–14=4
∵4=4
Therefore, m=6 is the solution.
Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Answer:
(i) x+4 = 9
(ii) y–2 = 8
(iii) 10a = 70
(iv) b/5= 6
(v) 34t = 15
(vi) 7m + 7 = 77
(vii) x/4 – 4 = 4
(viii) 6y – 6 = 60
(ix) z/3 + 3 = 30
Question 5.
Write the following equations in statement form:
(i) p+4=15
(ii) m–7=3
(iii) 2m=7
(iv) m/5=3
(v) 3m/5 = 6
(vi) 3p+4=25
(vii) 4p–2=18
(viii) p/2 + 2 = 8
Answer:
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8
Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.(Take Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
(i) Let m be the number of Parmit’s marbles.
∴ 5m+7=37
(ii) Let the age of Laxmi be y years.
∴ 3y+4=49
(iii) Let the lowest score be l.
∴ 2l+7=87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.
∴ 2b+b+b=180∘
⇒ 4b=180∘ [Angle sum property of a Δ]
(iv) In an isosceles triangle, the vertex angle is twice either base angle.
(Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
