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Exercise 4.1 - Chapter 4 Simple Equations class 7 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Complete the last column of the table.

S. 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)

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m1efdba87.gif

m = − 6

(x)

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m1efdba87.gif

m = 0

(xi)

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m1efdba87.gif

m = 6

Answer:

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

L.H.S. = x + 3

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

∴ No, the equation is not satisfied.

(iii) x + 3 = 0

L.H.S. = x + 3

By putting x = −3,

L.H.S. = − 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisfied.

(iv) x − 7 = 1

L.H.S. = x − 7

By putting x = 7,

L.H.S. = 7 − 7 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(v) x − 7 = 1

L.H.S. = x − 7

By putting x = 8,

L.H.S. = 8 − 7 = 1 = R.H.S.

∴ Yes, the equation is satisfied.

(vi) 5= 25

L.H.S. = 5x

By putting x = 0,

L.H.S. = 5 × 0 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(vii) 5= 25

L.H.S. = 5x

By putting x = 5,

L.H.S. = 5 × 5 = 25 = R.H.S.

∴ Yes, the equation is satisfied.

(viii) 5= 25

L.H.S. = 5x

By putting x = −5,

L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ix) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif = 2

L.H.S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif

By putting m = −6,

L. H. S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_647bb9f2.gif  ≠ R.H.S.

∴No, the equation is not satisfied.

(x) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif = 2

L.H.S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif

By putting m = 0,

L.H.S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m4f182c3c.gif  ≠ R.H.S.

∴No, the equation is not satisfied.

(xi) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif = 2

L.H.S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m2939255a.gif

By putting m = 6,

L.H.S. = https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/8772/Chapter%204_html_m15b4b892.gif  = R.H.S.

∴ Yes, the equation is satisfied.

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)

Answer:

(a) n + 5 = 19 (n = 1)

Putting n = 1 in L.H.S.,

n + 5 = 1 + 5 = 6 ≠ 19

As L.H.S. ≠ R.H.S.,

Therefore, n = 1 is not a solution of the given equation, n + 5 = 19.

(b) 7n + 5 = 19 (n = −2)

Putting n = −2 in L.H.S.,

7n + 5 = 7 × (−2) + 5 = −14 + 5 = −9 ≠ 19

As L.H.S. ≠ R.H.S.,

Therefore, n = −2 is not a solution of the given equation, 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)

Putting n = 2 in L.H.S.,

7n + 5 = 7 × (2) + 5 = 14 + 5 = 19 = R.H.S.

As L.H.S. = R.H.S.,

Therefore, n = 2 is a solution of the given equation, 7n + 5 = 19.

(d) 4− 3 = 13 (p = 1)

Putting p = 1 in L.H.S.,

4− 3 = (4 × 1) − 3 = 1 ≠ 13

As L.H.S ≠ R.H.S.,

Therefore, p = 1 is not a solution of the given equation, 4− 3 = 13.

(e) 4p − 3 = 13 (p = −4)

Putting p = −4 in L.H.S.,

4p − 3 = 4 × (−4) − 3 = − 16 − 3 = −19 ≠ 13

As L.H.S. ≠ R.H.S.,

Therefore, p = −4 is not a solution of the given equation, 4− 3 = 13.

(f) 4− 3 = 13 (p = 0)

Putting p = 0 in L.H.S.,

4p − 3 = (4 × 0) − 3 = −3 ≠ 13

As L.H.S. ≠ R.H.S.,

Therefore, p = 0 is not a solution of the given equation, 4− 3 = 13.

Question 3:

Solve the following equations by trial and error method:

(i) 5p + 2 = 17 (ii) 3m − 14 = 4

Answer:

(i) 5+ 2 = 17

Putting p = 1 in L.H.S.,

(5 × 1) + 2 = 7 ≠ R.H.S.

Putting p = 2 in L.H.S.,

(5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.

Putting p = 3 in L.H.S.,

(5 × 3) + 2 = 17 = R.H.S.

Hence, p = 3 is a solution of the given equation.

(ii) 3m − 14 = 4

Putting m = 4,

(3 × 4) − 14 = −2 ≠ R.H.S.

Putting m = 5,

(3 × 5) − 14 = 1 ≠ R.H.S.

Putting m = 6,

(3 × 6) − 14 = 18 − 14 = 4 = R.H.S.

Hence, m = 6 is a solution of the given equation.

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) − 2 = 8

(iii) 10a = 70

(iv) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_m3faab1a8.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_309ceac7.gif

(vi) Seven times of m is 7m.

7m + 7 = 77

(vii) One-fourth of a number x is https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_6376ca82.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_33e3878c.gif

(viii) Six times of y is 6y.

6y − 6 = 60

(ix) One-third of z is https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_m1d2ac133.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1861/Chapter%204_html_m779f774a.gif

Question 5:

Write the following equations in statement forms:

(i) p + 4 = 15 (ii) m − 7 = 3

(iii) 2m = 7 (iv) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1865/Chapter%204_html_m19425f99.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1865/Chapter%204_html_m2bf78223.gif  (vi) 3p + 4 = 25

(vii) 4p − 2 = 18 (viii) https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/34/1865/Chapter%204_html_64a41d4.gif

Answer:

(i) The sum of p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Twice of a number m is 7.

(iv) One-fifth of m is 3.

(v) Three-fifth of is 6.

(vi) Three times of a number p, when added to 4, gives 25.

(vii) When 2 is subtracted from four times of a number p, it gives 18.

(viii) When 2 is added to half of a number p, it gives 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.)

Answer:

(i) Let Parmit has m marbles.

5 × Number of marbles Parmit has + 7 = Number of marbles Irfan has

5 × m + 7 = 37

5m + 7 = 37

(ii) Let Laxmi be y years old.

3 × Laxmi’s age + 4 = Laxmi’s father’s age

3 × y + 4 = 49

3y + 4 = 49

(iii) Let the lowest marks be l.

2 × Lowest marks + 7 = Highest marks

2 × l + 7 = 87

l + 7 = 87

(iv) An isosceles triangle has two of its angles of equal measure.

Let base angle be b.

Vertex angle = 2 × Base angle = 2b

Sum of all interior angles of a Δ = 180°

b + b + 2b = 180°

4b = 180°

Also Read : Exercise-4.2-Chapter-4-Simple-Equations-class-7-ncert-solutions-Maths

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