{"id":140182,"date":"2021-02-20T10:49:32","date_gmt":"2021-02-20T10:49:32","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=140182"},"modified":"2023-09-15T05:27:59","modified_gmt":"2023-09-15T05:27:59","slug":"ncert-solutions-for-7th-class-maths-chapter-4-simple-equations","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-4-simple-equations\/","title":{"rendered":"NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations"},"content":{"rendered":"\n

Class 7: Maths Chapter 4 solutions. Complete Class 7 Maths Chapter 4 Notes.<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations<\/strong><\/h2>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

Page No: 81<\/p>\n\n\n\n

Exercise 4.1<\/h4>\n\n\n\n

1. Complete the last column of the table:<\/strong><\/p>\n\n\n\n

S. No.<\/strong><\/td>Equation<\/strong><\/td>Value<\/strong><\/td>Say, whether the Equation is satisfied. (Yes \/ No)<\/strong><\/td><\/tr>
1<\/td>x+3=0<\/td>x=3<\/td><\/tr>
2<\/td>x+3=0<\/td>x=0<\/td><\/tr>
3<\/td>x+3=0<\/td>x=\u22123<\/td><\/tr>
4<\/td>x\u22127=1<\/td>x=7<\/td><\/tr>
5<\/td>x\u22127=1<\/td>x=8<\/td><\/tr>
6<\/td>5x=25<\/td>x=0<\/td><\/tr>
7<\/td>5x=25<\/td>x=5<\/td><\/tr>
8<\/td>5x=25<\/td>x=\u22125<\/td><\/tr>
9<\/td>m\/3=2<\/td>m=\u22126<\/td><\/tr>
10<\/td>m\/3=2<\/td>m=0<\/td><\/tr>
11<\/td>m\/3=2<\/td>m=6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

S. No.<\/strong><\/td>Equation<\/strong><\/td>Value<\/strong><\/td>Say, whether the Equation is satisfied. (Yes \/ No)<\/strong><\/td><\/tr>
1<\/td>x+3=0<\/td>x=3<\/td>No<\/td><\/tr>
2<\/td>x+3=0<\/td>x=0<\/td>No<\/td><\/tr>
3<\/td>x+3=0<\/td>x=\u22123<\/td>Yes<\/td><\/tr>
4<\/td>x\u22127=1<\/td>x=7<\/td>No<\/td><\/tr>
5<\/td>x\u22127=1<\/td>x=8<\/td>Yes<\/td><\/tr>
6<\/td>5x=25<\/td>x=0<\/td>No<\/td><\/tr>
7<\/td>5x=25<\/td>x=5<\/td>Yes<\/td><\/tr>
8<\/td>5x=25<\/td>x=\u22125<\/td>No<\/td><\/tr>
9<\/td>m\/3=2<\/td>m=\u22126<\/td>No<\/td><\/tr>
10<\/td>m\/3=2<\/td>m=0<\/td>No<\/td><\/tr>
11<\/td>m\/3=2<\/td>m=6<\/td>Yes<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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=\u22122)
(c) 7n+5=19(n=2)
(d) 4p\u22123=13(p=1)
(e) 4p\u22123=13(p=\u22124)
(f) 4p\u22123=13(p=0)<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) n+5=19(n=1)
Putting n=1 in L.H.S.,
1 + 5 = 6
\u2235 L.H.S. \u2260 R.H.S.,
\u2234 n=1 is not the solution of given equation.<\/p>\n\n\n\n

(b) 7n+5=19(n=\u22122)
Putting n=\u22122 in L.H.S.,
7(\u22122)+5=\u221214+5=\u22129
\u2235 L.H.S. \u2260 R.H.S.,
\u2234 n=\u22122 is not the solution of given equation.<\/p>\n\n\n\n

(c) 7n+5=19(n=2)
Putting n=2 in L.H.S.,
7(2)+5=14+5=19
\u2235 L.H.S. = R.H.S.,
\u2234 n=2is the solution of given equation.<\/p>\n\n\n\n

(d) 4p\u22123=13(p=1)
Putting p=1 in L.H.S.,
4(1)\u22123=4\u22123=1
\u2235 L.H.S. \u2260 R.H.S.,
\u2234 p=1 is not the solution of given equation.<\/p>\n\n\n\n

(e) 4p\u22123=13(p=\u22124)
Putting p=\u22124 in L.H.S.,
4(\u22124)\u22123=\u221216\u22123=\u221219
\u2235 L.H.S. \u2260 R.H.S.,
\u2234 p=\u22124 is not the solution of given equation.<\/p>\n\n\n\n

(f) 4p\u22123=13(p=0)
Putting p=0 in L.H.S.,
4(0)\u22123=0\u22123=\u22123
\u2235 L.H.S. \u2260 R.H.S.,
\u2234 p=0 is not the solution of given equation.<\/p>\n\n\n\n

3. Solve the following equations by trial and error method:
(i) 5p+2=17
(ii) 3m\u221214=4<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 5p+2=17
Putting p=\u22123 in L.H.S. 5(\u22123)+2 = \u221215+2=\u221213
\u2235\u221213\u226017
Therefore, p=\u22123 is not the solution.<\/p>\n\n\n\n

Putting p=\u22122 in L.H.S. 5(\u22122)+2=\u221210+2=\u22128
\u2235\u22128\u226017
Therefore, p=\u22122 is not the solution.<\/p>\n\n\n\n

Putting p=\u22121 in L.H.S. 5(\u22121)+2=\u22125+2=\u22123
\u2235\u22123\u226017
Therefore, p=\u22121 is not the solution.<\/p>\n\n\n\n

Putting p=0 in L.H.S. 5(0)+2=0+2=2
\u2235 2\u226017
Therefore, p=0 is not the solution.<\/p>\n\n\n\n

Putting p=1 in L.H.S. 5(1)+2=5+2=7
\u22357\u226017
Therefore, p=1 is not the solution.<\/p>\n\n\n\n

Putting p=2 in L.H.S. 5(2)+2=10+2=12
\u223512\u226017
Therefore, p=2 is not the solution.<\/p>\n\n\n\n

Putting p=3 in L.H.S. 5(3)+2=15+2=17
\u223517=17
Therefore, p=3 is the solution.<\/p>\n\n\n\n

(ii) 3m\u221214=4
Putting m=\u22122 in L.H.S. 3(\u22122)\u221214=\u22126\u221214=\u221220
\u2235 \u221220\u22604
Therefore, m=\u22122 is not the solution.<\/p>\n\n\n\n

Putting m=\u22121 in L.H.S. 3(\u22121)\u221214=\u22123\u221214=\u221217
\u2235\u221217\u22604
Therefore, m=\u22121 is not the solution.<\/p>\n\n\n\n

Putting m=0 in L.H.S. 3(0)\u221214=0\u221214=\u221214
\u2235\u221214\u22604
Therefore, m=0 is not the solution.<\/p>\n\n\n\n

Putting m=1 in L.H.S. 3(1)\u221214=3\u221214=\u221211
\u2235\u221211\u22604
Therefore, m=1 is not the solution.<\/p>\n\n\n\n

Putting m=2 in L.H.S. 3(2)\u221214=6\u221214=\u22128
\u2235\u22128\u22604
Therefore, m=2is not the solution.<\/p>\n\n\n\n

Putting m=3 in L.H.S. 3(3)\u221214=9\u221214=\u22125
\u2235\u22125\u22604
Therefore, m=3 is not the solution.<\/p>\n\n\n\n

Putting m=4 in L.H.S. 3(4)\u221214=12\u221214=\u22122
\u2235\u22122\u22604
Therefore, m=4 is not the solution.<\/p>\n\n\n\n

Putting m=5 in L.H.S. 3(5)\u221214=15\u221214=1
\u22351\u22604
Therefore, m=5 is not the solution.<\/p>\n\n\n\n

Putting m=6 in L.H.S. 3(6)\u221214=18\u221214=4
\u22354=4
Therefore, m = 6m =6 is the solution.<\/p>\n\n\n\n

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 mm plus 7 gets you 7.
(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.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) x+4=9
(ii) y\u22122=8
(iii) 10a=70
(iv) b\/5=6
(v) 3\/4.t=15
(vi) 7m+7=77
(vii) x\/4\u22124=4
(viii) 6y\u22126=60
(ix) z\/3+3=30<\/p>\n\n\n\n

5. Write the following equations in statement form:
(i) p+4=15
(ii) m\u22127=3
(iii) 2m=7
(iv) m\/5=3
(v) 3m\/5=6
(vi) 3p+4=25
(vii) 4p\u22122=18
(viii) p\/2+2=8<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(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.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

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. (Tale mm to be the number of Parmit\u2019s marbles.)
(ii) Laxmi\u2019s father is 49 years old. He is 4 years older than three times Laxmi\u2019s age. (Take Laxmi\u2019s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class are 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\u2218.)<\/strong>
Answer<\/strong><\/p>\n\n\n\n

(i) Let m be the number of Parmit\u2019s marbles.
\u2234 5m+7=37<\/p>\n\n\n\n

(ii) Let the age of Laxmi be y years.
\u2234 3y+4=49<\/p>\n\n\n\n

(iii) Let the lowest score be l.
\u2234 2l+7=87<\/p>\n\n\n\n

(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b
\u2234 2b+b+b=180\u2218
\u21d2 4b=180\u2218 [Angle sum property of a \u0394]<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

Page No. 86

Exercise 4.2<\/strong><\/p>\n\n\n\n

1. Give first the step you will use to separate the variable and then solve the equations:<\/strong><\/p>\n\n\n\n

(a) x\u22121=0<\/strong><\/p>\n\n\n\n

(b) x+1=0<\/strong><\/p>\n\n\n\n

(c) x\u22121=5<\/strong><\/p>\n\n\n\n

(d) x+6=2<\/strong><\/p>\n\n\n\n

(e) y\u22124=\u22127<\/strong><\/p>\n\n\n\n

(f) y\u22124=4<\/strong><\/p>\n\n\n\n

(g) y+4=4<\/strong><\/p>\n\n\n\n

(h) y+4=\u22124<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) x\u22121=0<\/p>\n\n\n\n

\u21d2 x\u22121+1=0+1 [Adding 1 both sides]<\/p>\n\n\n\n

\u21d2 x=1<\/p>\n\n\n\n

\u21d2 x+1\u22121=0\u22121 [Subtracting 1 both sides]<\/p>\n\n\n\n

\u21d2 x=\u22121<\/p>\n\n\n\n

\u21d2 x\u22121+1=5+1 [Adding 1 both sides]<\/p>\n\n\n\n

\u21d2 x=6<\/p>\n\n\n\n

\u21d2 x+6\u22126=2\u22126 [Subtracting 6 both sides]<\/p>\n\n\n\n

\u21d2 x=\u22124<\/p>\n\n\n\n

\u21d2 y\u22124+4=\u22127+4 [Adding 4 both sides]<\/p>\n\n\n\n

\u21d2 y=\u22123<\/p>\n\n\n\n

\u21d2 y\u22124+4=4+4 [Adding 4 both sides]<\/p>\n\n\n\n

\u21d2 y=8<\/p>\n\n\n\n

\u21d2 y+4\u22124=4\u22124 [Subtracting 4 both sides]<\/p>\n\n\n\n

\u21d2 y=0<\/p>\n\n\n\n

\u21d2 y+4\u22124=\u22124\u22124 [Subtracting 4 both sides]<\/p>\n\n\n\n

\u21d2 y=\u22128<\/p>\n\n\n\n

2. Give first the step you will use to separate the variable and then solve the equations<\/strong><\/p>\n\n\n\n

(a) 3l = 42<\/p>\n\n\n\n

(b) b\/2 = 6<\/p>\n\n\n\n

(c) p\/7 = 4<\/p>\n\n\n\n

(d) 4x = 25<\/p>\n\n\n\n

(e) 8y = 36<\/p>\n\n\n\n

(f) z\/3 = 54<\/p>\n\n\n\n

(g) a\/5 = 7\/15<\/p>\n\n\n\n

(h) 20t = \u221210<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 3l=42<\/p>\n\n\n\n

\u21d2 3l\/3=42\/3 [Dividing both sides by 3]<\/p>\n\n\n\n

\u21d2 l=14<\/p>\n\n\n\n

\u21d2 b\/2\u00d72=6\u00d72 [Multiplying both sides by 2]<\/p>\n\n\n\n

\u21d2 b=12<\/p>\n\n\n\n

\u21d2 p\/7\u00d77=4\u00d77[Multiplying both sides by 7]<\/p>\n\n\n\n

\u21d2 p=28<\/p>\n\n\n\n

\u21d2 4×4=25\/4 [Dividing both sides by 4]<\/p>\n\n\n\n

\u21d2 x=25\/4<\/p>\n\n\n\n

\u21d2 8y\/8=36\/8 [Dividing both sides by 8]<\/p>\n\n\n\n

\u21d2 y=9\/2<\/p>\n\n\n\n

\u21d2 z\/3\u00d73=5\/4\u00d73 [Multiplying both sides by 3]<\/p>\n\n\n\n

\u21d2 z=15\/4<\/p>\n\n\n\n

\u21d2 a\/5\u00d75=7\/15\u00d75 [Multiplying both sides by 5]<\/p>\n\n\n\n

\u21d2 a=73<\/p>\n\n\n\n

\u21d2 20t\/20=\u221210\/20 [Dividing both sides by 20]<\/p>\n\n\n\n

\u21d2 t=\u22121\/2<\/p>\n\n\n\n

3. Give first the step you will use to separate the variable and then solve the equations<\/strong><\/p>\n\n\n\n

(a) 3n\u22122=46<\/p>\n\n\n\n

(b) 5m+7=17<\/p>\n\n\n\n

(c) 20p\/3=40<\/p>\n\n\n\n

(d) 3p\/10=6<\/p>\n\n\n\n

Step I: 3n\u22122+2=46+2=48<\/p>\n\n\n\n

[Adding 2 both sides]<\/p>\n\n\n\n

Step II: 3n\/3=48\/3<\/p>\n\n\n\n

\u21d2 n=16 [Dividing both sides by 3]<\/p>\n\n\n\n

Step I: 5m+7\u22127=17\u22127=17\u22127 <\/p>\n\n\n\n

\u21d2 5m=10 [Subtracting 7 both sides]<\/p>\n\n\n\n

Step II: 5m\/5=10\/5<\/p>\n\n\n\n

\u21d2m=2 [Dividing both sides by 5]<\/p>\n\n\n\n

Step I: 20p\/3\u00d73 = 40\u00d73 <\/p>\n\n\n\n

\u21d2 20p = 120 [Multiplying both sides by 3]<\/p>\n\n\n\n

Step II: 20p\/20=120\/20<\/p>\n\n\n\n

\u21d2 p=6 [Dividing both sides by 20]<\/p>\n\n\n\n

Step I: 3p\/10\u00d710 = 6\u00d710<\/p>\n\n\n\n

\u21d2 3p=60 [Multiplying both sides by 10]<\/p>\n\n\n\n

Step II: 3p\/3 = 60\/3<\/p>\n\n\n\n

\u21d2 p=20 [Dividing both sides by 3]<\/p>\n\n\n\n

4. Solve the following equation:<\/strong><\/p>\n\n\n\n

(a) 10 p = 100<\/p>\n\n\n\n

(b) 10p + 10 =100<\/p>\n\n\n\n

(c) p\/4 = 5<\/p>\n\n\n\n

(d) \u2212p\/3 = 5<\/p>\n\n\n\n

(e) 3p\/4 = 6<\/p>\n\n\n\n

(f) 3s = \u22129<\/p>\n\n\n\n

(g) 3s + 12 = 0<\/p>\n\n\n\n

(h) 3s = 0<\/p>\n\n\n\n

(i) 2q = 6<\/p>\n\n\n\n

(j) 2q\u22126 = 0<\/p>\n\n\n\n

(k) 2q+6 = 0<\/p>\n\n\n\n

(l) 2q+6 = 12<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 10p=100<\/p>\n\n\n\n

\u21d2 10p\/10=100\/10 [Dividing both sides by 10]<\/p>\n\n\n\n

\u21d2 p=10<\/p>\n\n\n\n

\u21d2 10p+10\u221210 = 100\u221210 [Subtracting both sides 10]<\/p>\n\n\n\n

\u21d2 10p=90
\u21d2 10p\/10=90\/10 [Dividing both sides by 10]<\/p>\n\n\n\n

\u21d2 p=9<\/p>\n\n\n\n

\u21d2 p\/4 \u00d7 4 = 5\u00d74 [Multiplying both sides by 4]<\/p>\n\n\n\n

\u21d2 p = 20<\/p>\n\n\n\n

\u21d2 \u2212p\/3\u00d7(\u22123) = 5\u00d7(\u22123) [Multiplying both sides by \u2013 3]<\/p>\n\n\n\n

\u21d2 p = \u221215<\/p>\n\n\n\n

\u21d2 3p\/4 \u00d7 4 = 6\u00d74 [Multiplying both sides by 4]<\/p>\n\n\n\n

\u21d2 3p = 24
\u21d2 3p\/3 = 24\/3 [Dividing both sides by 3]<\/p>\n\n\n\n

\u21d2 p=8<\/p>\n\n\n\n

\u21d2 3s\/3 = \u22129\/3 [Dividing both sides by 3]<\/p>\n\n\n\n

\u21d2 s = \u22123<\/p>\n\n\n\n

\u21d2 3s+12\u221212=0\u221212 [Subtracting both sides 10]<\/p>\n\n\n\n

\u21d2 3s=\u221212
\u21d2 3s\/3 = \u221212\/3 [Dividing both sides by 3]<\/p>\n\n\n\n

\u21d2 s=\u22124<\/p>\n\n\n\n

\u21d2 3s\/3=0\/3 [Dividing both sides by 3]<\/p>\n\n\n\n

\u21d2 s=0<\/p>\n\n\n\n

\u21d2 2q\/2 = 6\/2 [Dividing both sides by 2]<\/p>\n\n\n\n

\u21d2 q = 3<\/p>\n\n\n\n

\u21d2 2q\u22126+6 = 0+6 [Adding both sides 6]<\/p>\n\n\n\n

\u21d2 2q = 6
\u21d2 2q\/2 = 6\/2 [Dividing both sides by 2]<\/p>\n\n\n\n

\u21d2 q=3<\/p>\n\n\n\n

\u21d2 2q+6\u22126 = 0\u22126 [Subtracting both sides 6]<\/p>\n\n\n\n

\u21d2 2q = \u22126
\u21d2 2q\/2 = \u22126\/2 [Dividing both sides by 2]<\/p>\n\n\n\n

\u21d2 q = \u22123<\/p>\n\n\n\n

\u21d2 2q+6\u22126 = 12\u22126 [Subtracting both sides 6]<\/p>\n\n\n\n

\u21d2 2q = 6
\u21d2 2q\/2 = 6\/2 [Dividing both sides by 2]<\/p>\n\n\n\n

\u21d2 q = 3<\/p>\n\n\n\n

Page No. 89<\/p>\n\n\n\n

Exercise 4.3<\/strong><\/p>\n\n\n\n

1. Solve the following equations:<\/strong><\/p>\n\n\n\n

(a) 2y + 5\/2 = 37\/2<\/p>\n\n\n\n

(b) 5t + 28 = 10<\/p>\n\n\n\n

(c) a\/5 + 3 = 2<\/p>\n\n\n\n

(d) q\/4 + 7=5<\/p>\n\n\n\n

(e) 5\/2 x = 10<\/p>\n\n\n\n

(f) 5\/2 x = 25\/4<\/p>\n\n\n\n

(g) 7m + 19\/2 =13<\/p>\n\n\n\n

(h) 6z +10 = \u22122<\/p>\n\n\n\n

(i) 3l\/2 = 2\/3<\/p>\n\n\n\n

(j) 2b\/3 \u2212 5 = 3<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 2y + 5\/2 = 37\/2<\/p>\n\n\n\n

\u21d2 2y = 37\/2\u22125\/2
\u21d2 2y =37\u22125\/2<\/p>\n\n\n\n

\u21d2 2y =32\/2
\u21d2 2y =16\/2<\/p>\n\n\n\n

\u21d2 y = 8<\/p>\n\n\n\n

\u21d2 5t =10\u221228
\u21d2 5t = \u221218<\/p>\n\n\n\n

\u21d2 t =\u221218\/5<\/p>\n\n\n\n

\u21d2 a\/5=2\u22123\u21d2 a\/5=\u22121<\/p>\n\n\n\n

\u21d2 a=\u22121\u00d75<\/p>\n\n\n\n

\u21d2 a=\u22125<\/p>\n\n\n\n

\u21d2 q\/4 = 5\u22127
\u21d2 q\/4 = \u22122<\/p>\n\n\n\n

\u21d2 q = \u22122\u00d74<\/p>\n\n\n\n

\u21d2 q = \u22128<\/p>\n\n\n\n

\u21d2 5x = 10\u00d72<\/p>\n\n\n\n

\u21d2 5x = 20<\/p>\n\n\n\n

\u21d2 x = 20\/5<\/p>\n\n\n\n

\u21d2 x = 4<\/p>\n\n\n\n

\u21d2 5x = 25\/4 \u00d72 [Multiplying both sides by 2]<\/p>\n\n\n\n

\u21d2 5x = 25\/2<\/p>\n\n\n\n

\u21d2 x = 25\/(2\u00d75)<\/p>\n\n\n\n

\u21d2 x = 5\/2<\/p>\n\n\n\n

\u21d2 7m = 13 \u2212 19\/2<\/p>\n\n\n\n

\u21d2 7m = 26\u2212 19\/2<\/p>\n\n\n\n

\u21d2 7m = 7\/2
\u21d2 m = 7\/(2\u00d77)
\u21d2 m = 12<\/p>\n\n\n\n

(h) 6z + 10 = \u22122<\/p>\n\n\n\n

\u21d2 6z = \u22122\u221210<\/p>\n\n\n\n

\u21d2 6z = \u221212<\/p>\n\n\n\n

\u21d2 z = \u221212\/6<\/p>\n\n\n\n

\u21d2 z = \u22122<\/p>\n\n\n\n

\u21d2 3l = 2\/3 \u00d72
\u21d2 3l = 4\/3<\/p>\n\n\n\n

\u21d2 l = 4\/3 \u00d73<\/p>\n\n\n\n

\u21d2l = 49<\/p>\n\n\n\n

\u21d2 2b\/3 = 3+5<\/p>\n\n\n\n

\u21d2 2b\/3 = 8<\/p>\n\n\n\n

\u21d2 2b = 8\u00d73<\/p>\n\n\n\n

\u21d2 2b = 24<\/p>\n\n\n\n

\u21d2 b = 24\/2<\/p>\n\n\n\n

\u21d2 b = 12<\/p>\n\n\n\n

2. Solve the following equations:<\/strong><\/p>\n\n\n\n

(a) 2(x+4)=12<\/p>\n\n\n\n

(b) 3(n\u22125)=21<\/p>\n\n\n\n

(c) 3(n\u22125)=\u221221<\/p>\n\n\n\n

(d) 3\u22122(2\u2212y)=7<\/p>\n\n\n\n

(e) \u22124(2\u2212x)=9<\/p>\n\n\n\n

(f) 4(2\u2212x)=9<\/p>\n\n\n\n

(g) 4+5(p\u22121)=34<\/p>\n\n\n\n

(h) 34\u22125(p\u22121)=4<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 2(x+4) = 12<\/p>\n\n\n\n

\u21d2 x+4 = 12\/2<\/p>\n\n\n\n

\u21d2 x+4 = 6<\/p>\n\n\n\n

\u21d2 x = 6\u22124<\/p>\n\n\n\n

\u21d2 x = 2<\/p>\n\n\n\n

\u21d2 n\u22125 = 21\/3<\/p>\n\n\n\n

\u21d2 n\u22125 = 7<\/p>\n\n\n\n

\u21d2 n =7+5<\/p>\n\n\n\n

\u21d2 n =12<\/p>\n\n\n\n

\u21d2 n\u22125 = \u221221\/3<\/p>\n\n\n\n

\u21d2 n\u22125 = \u22127<\/p>\n\n\n\n

\u21d2 n = \u22127+5<\/p>\n\n\n\n

\u21d2 n = \u22122<\/p>\n\n\n\n

\u21d2 \u22122(2\u2212y) = 7\u22123<\/p>\n\n\n\n

\u21d2 \u22122(2\u2212y) = 4<\/p>\n\n\n\n

\u21d2 2\u2212y = 4\/\u22122<\/p>\n\n\n\n

\u21d2 2\u2212y = \u22122<\/p>\n\n\n\n

\u21d2 \u2212y = \u22122\u22122<\/p>\n\n\n\n

\u21d2 \u2212y = \u22124<\/p>\n\n\n\n

\u21d2 y = 4<\/p>\n\n\n\n

\u21d2 \u22124\u00d72 \u2212 x\u00d7(\u22124) = 9<\/p>\n\n\n\n

\u21d2 \u22128 + 4x = 9<\/p>\n\n\n\n

\u21d2 4x = 9+8<\/p>\n\n\n\n

\u21d2 4x = 17<\/p>\n\n\n\n

\u21d2 x = 17\/4<\/p>\n\n\n\n

\u21d2 4\u00d72 \u2212 x\u00d7(4) = 9<\/p>\n\n\n\n

\u21d2 8\u22124x =9<\/p>\n\n\n\n

\u21d2 \u22124x = 9\u22128<\/p>\n\n\n\n

\u21d2 \u22124x = 1<\/p>\n\n\n\n

\u21d2 x = \u22121\/4<\/p>\n\n\n\n

\u21d2 5(p\u22121) = 34\u22124<\/p>\n\n\n\n

\u21d2 5(p\u22121) = 30<\/p>\n\n\n\n

\u21d2 p\u22121 = 30\/5<\/p>\n\n\n\n

\u21d2 p\u22121 = 6<\/p>\n\n\n\n

\u21d2 p = 6+1<\/p>\n\n\n\n

\u21d2 p = 7p
\u21d2 p =7<\/p>\n\n\n\n

\u21d2 \u22125(p\u22121) = 4\u221234<\/p>\n\n\n\n

\u21d2 \u22125(p\u22121) = \u221230<\/p>\n\n\n\n

\u21d2 p\u22121 = \u221230\/\u22125<\/p>\n\n\n\n

\u21d2 p\u22121 = 6<\/p>\n\n\n\n

\u21d2 p =6+1<\/p>\n\n\n\n

\u21d2 p =7<\/p>\n\n\n\n


3. Solve the following equations:<\/strong><\/p>\n\n\n\n

(a) 4=5(p\u22122)<\/p>\n\n\n\n

(b) \u22124=5(p\u22122)<\/p>\n\n\n\n

(c) \u221216=\u22125(2\u2212p)<\/p>\n\n\n\n

(d) 10 = 4 +3(t+2)<\/p>\n\n\n\n

(e) 28 = 4+ 3(t+5)<\/p>\n\n\n\n

(f) 0 = 16 +4(m\u22126)<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 4 = 5(p-2)
\u21d2 4 = 5p – 10
\u21d2 5p – 10 = 4
\u21d2 5p = 4+10
\u21d2 p = 14\/5<\/p>\n\n\n\n

(b) -4 = 5(p-2)
\u21d2 -4 = 5p – 10
\u21d2 5p – 10 = -4
\u21d2 5p = -4 + 10
\u21d2 p = 6\/5<\/p>\n\n\n\n

(c) \u221216=\u22125(2\u2212p)
-16 = -10 + 5p
\u21d2 -10 + 5p = -16
\u21d2 5p = -16 + 10
\u21d2 p = -6\/5<\/p>\n\n\n\n

(d) 10 = 4 +3(t+2)
\u21d2 10 = 4 + 3t + 6
\u21d2 10 = 10 + 3t
\u21d2 10 + 3t = 10
\u21d2 3t = 10 – 10
\u21d2 t = 0<\/p>\n\n\n\n

(e) 28 = 4 +3(t+5)
\u21d2 28 = 4 + 3t + 15
\u21d2 28 = 18 + 3t
\u21d2 18 + 3t = 28
\u21d2 3t = 28 -18
\u21d2 t = 10\/3<\/p>\n\n\n\n

(f) 0 = 16 +4(m\u22126)
\u21d2 0 = 16 + 4m – 24
\u21d2 -8 + 4m = 0
\u21d2 4m = 8
\u21d2 m = 8\/4
\u21d2 m = 2<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

4. (a) Construct 3 equations starting with x=2.<\/strong><\/p>\n\n\n\n

(b) Construct 3 equations starting with x=\u22122<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) 3 equations starting with x=2.<\/p>\n\n\n\n

(i) x=2<\/p>\n\n\n\n

Multiplying both sides by 10, 10x=20<\/p>\n\n\n\n

Adding 2 both sides 10x+2 =20+2 = 10x + 2 = 22<\/p>\n\n\n\n

(ii) x=2<\/p>\n\n\n\n

Multiplying both sides by 5 5x=10<\/p>\n\n\n\n

Subtracting 3 from both sides 5x\u22123=10\u22123 = 5x\u22123=7<\/p>\n\n\n\n

(iii) x=2<\/p>\n\n\n\n

Dividing both sides by 5 x 5=2\/5<\/p>\n\n\n\n

(b) 3 equations starting with x=\u22122.<\/p>\n\n\n\n

(i) x=\u22122<\/p>\n\n\n\n

Multiplying both sides by 3 3x=\u22126<\/p>\n\n\n\n

(ii) x=\u22122<\/p>\n\n\n\n

Multiplying both sides by 3 3x=\u22126<\/p>\n\n\n\n

Adding 7 to both sides 3x+7=\u22126+7 = 3x+7=1<\/p>\n\n\n\n

(iii) x=\u22122<\/p>\n\n\n\n

Multiplying both sides by 3 3x=\u22126<\/p>\n\n\n\n

Adding 10 to both sides 3x+10=\u22126+10= 3x+10=4<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

Page No. 91<\/p>\n\n\n\n

Exercise 4.4<\/strong><\/p>\n\n\n\n

1. Set up equations and solve them to find the unknown numbers in the following cases:<\/strong><\/p>\n\n\n\n

(a) Add 4 to eight times a number; you get 60.<\/strong><\/p>\n\n\n\n

(b) One-fifth of a number minus 4 gives 3.<\/strong><\/p>\n\n\n\n

(c) If I take three-fourth of a number and add 3 to it, I get 21.<\/strong><\/p>\n\n\n\n

(d) When I subtracted 11 from twice a number, the result was 15.<\/strong><\/p>\n\n\n\n

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.<\/strong><\/p>\n\n\n\n

(f) Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.<\/strong><\/p>\n\n\n\n

(g) Answer thinks of a number. If he takes away 7 from 5\/2 of the number, the result is 11\/2.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) Let the number be x.<\/p>\n\n\n\n

According to the question, 8x+4=60<\/p>\n\n\n\n

\u21d2 8x = 60\u22124<\/p>\n\n\n\n

\u21d2 8x = 56<\/p>\n\n\n\n

\u21d2 x = 56\/8<\/p>\n\n\n\n

(b) Let the number be y.<\/p>\n\n\n\n

According to the question, y\/5\u22124=3<\/p>\n\n\n\n

\u21d2 y\/5 = 3+4<\/p>\n\n\n\n

\u21d2 y\/5 = 7<\/p>\n\n\n\n

\u21d2 y = 7\u00d75<\/p>\n\n\n\n

\u21d2 y = 35<\/p>\n\n\n\n

According to the question, 3\/4.z+3=21<\/p>\n\n\n\n

\u21d2 3\/4 z = 21\u22123<\/p>\n\n\n\n

\u21d2 3\/4 z = 18<\/p>\n\n\n\n

\u21d2 3z = 18\u00d74<\/p>\n\n\n\n

\u21d2 3z = 72<\/p>\n\n\n\n

\u21d2 z = 72\/3<\/p>\n\n\n\n

\u21d2 z = 24<\/p>\n\n\n\n

According to the question, 2x\u221211=15<\/p>\n\n\n\n

\u21d2 2x = 15+11<\/p>\n\n\n\n

\u21d2 2x = 26<\/p>\n\n\n\n

\u21d2 x = 26\/2<\/p>\n\n\n\n

\u21d2 x = 13<\/p>\n\n\n\n

According to the question, 50\u22123m=8<\/p>\n\n\n\n

\u21d2 \u22123m = 8\u221250<\/p>\n\n\n\n

\u21d2 \u22123m = \u221242<\/p>\n\n\n\n

\u21d2 m = \u221242\/\u22123<\/p>\n\n\n\n

\u21d2 m = 14<\/p>\n\n\n\n

According to the question, (n+190\/5=8<\/p>\n\n\n\n

\u21d2 n+19 = 8\u00d75<\/p>\n\n\n\n

\u21d2 n+19 = 40<\/p>\n\n\n\n

\u21d2 n = 40\u221219<\/p>\n\n\n\n

\u21d2 n = 21<\/p>\n\n\n\n

According to the question, 5\/2 x\u22127=11\/2<\/p>\n\n\n\n

\u21d2 5\/2 x = 11\/2+7<\/p>\n\n\n\n

\u21d2 5\/2 x = (11+14)\/2<\/p>\n\n\n\n

\u21d2 5\/2 x = 25\/2<\/p>\n\n\n\n

\u21d2 5x = (25\u00d72)\/2<\/p>\n\n\n\n

\u21d2 5x = 25<\/p>\n\n\n\n

\u21d2 x = 25\/5<\/p>\n\n\n\n

\u21d2 x = 5<\/p>\n\n\n\n

2. Solve the following:<\/strong><\/p>\n\n\n\n

(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?<\/strong><\/p>\n\n\n\n

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40\u2218. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180\u2218.)<\/strong><\/p>\n\n\n\n

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) Let the lowest marks be y.<\/p>\n\n\n\n

According to the question, 2y+7=87<\/p>\n\n\n\n

\u21d2 2y = 87\u22127<\/p>\n\n\n\n

\u21d2 2y = 80<\/p>\n\n\n\n

\u21d2 y = 80\/2<\/p>\n\n\n\n

\u21d2 y = 40<\/p>\n\n\n\n

Thus, the lowest score is 40.<\/p>\n\n\n\n

(b) Let the base angle of the triangle be b.<\/p>\n\n\n\n

Given, a=40\u00b0,b=c<\/p>\n\n\n\n

Since, a+b+c=180\u00b0 [Angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2 40\u00b0+b+b=180\u00b0<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

\u21d2 40\u00b0 +2b = 180\u00b0<\/p>\n\n\n\n

\u21d2 2b= 180\u00b0 \u2212 40\u00b0<\/p>\n\n\n\n

\u21d2 2b = 140\u2218<\/p>\n\n\n\n

\u21d2 b = 140\u00b0\/2<\/p>\n\n\n\n

\u21d2 b = 70\u00b0<\/p>\n\n\n\n

Thus, the base angles of the isosceles triangle are 70\u2218 each.<\/p>\n\n\n\n

(c) Let the score of Rahul be x runs and Sachin\u2019s score is 2x.<\/p>\n\n\n\n

According to the question, x+2x =198<\/p>\n\n\n\n

\u21d2 3x = 198<\/p>\n\n\n\n

\u21d2 x = 198\/3<\/p>\n\n\n\n

\u21d2 x = 66<\/p>\n\n\n\n

Thus, Rahul\u2019s score = 66 runs<\/p>\n\n\n\n

And Sachin\u2019s score = 2 x 66 = 132 runs.<\/p>\n\n\n\n

3. Solve the following:<\/strong><\/p>\n\n\n\n

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?<\/strong><\/p>\n\n\n\n

(ii) Laxmi\u2019s father is 49 years old. He is 4 years older than three times Laxmi\u2019s age. What is Laxmi\u2019s age?<\/strong><\/p>\n\n\n\n

(iii) People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Let the number of marbles Parmit has be m.<\/p>\n\n\n\n

According to the question, 5m+7=37<\/p>\n\n\n\n

\u21d2 5m=37\u22127<\/p>\n\n\n\n

\u21d2 5m=30<\/p>\n\n\n\n

\u21d2 m=30\/5<\/p>\n\n\n\n

\u21d2 m=6<\/p>\n\n\n\n

Thus, Parmit has 6 marbles.<\/p>\n\n\n\n

(ii) Let the age of Laxmi be y years.<\/p>\n\n\n\n

Then her father\u2019s age = (3y+4) years<\/p>\n\n\n\n

According to question, 3y+4=49<\/p>\n\n\n\n

\u21d2 3y=49\u22124<\/p>\n\n\n\n

\u21d2 3y=45<\/p>\n\n\n\n

\u21d2 y=45\/3<\/p>\n\n\n\n

\u21d2 y=15<\/p>\n\n\n\n

Thus, the age of Laxmi is 15 years.<\/p>\n\n\n\n

(iii) Let the number of fruit trees be t.<\/p>\n\n\n\n

Then the number of non-fruits tree = 3t+2<\/p>\n\n\n\n

According to the question, t+3t+2 = 102<\/p>\n\n\n\n

\u21d2 4t+2 =102<\/p>\n\n\n\n

\u21d2 4t = 102\u22122<\/p>\n\n\n\n

\u21d2 4t = 100<\/p>\n\n\n\n

\u21d2 t = 100\/4<\/p>\n\n\n\n

\u21d2 t = 25<\/p>\n\n\n\n

Thus, the number of fruit trees are 25.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

4. Solve the following riddle:<\/strong><\/p>\n\n\n\n

I am a number, Tell my identity!<\/strong><\/em><\/p>\n\n\n\n

Take me seven times over, And add a fifty!<\/strong><\/em><\/p>\n\n\n\n

To reach a triple century, You still need forty!<\/strong><\/em><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let the number be n.<\/p>\n\n\n\n

According to the question, 7n+50+40 = 300<\/p>\n\n\n\n

\u21d2 7n+90 = 300<\/p>\n\n\n\n

\u21d2 7n = 300\u221290<\/p>\n\n\n\n

\u21d2 7n = 210<\/p>\n\n\n\n

\u21d2 n = 210\/7<\/p>\n\n\n\n

\u21d2 n = 30<\/p>\n\n\n\n

Thus, the required number is 30.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 4: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations PDF<\/a><\/p>\n\n\n\n

Chapter-wise NCERT Solutions Class 7 Maths<\/h2>\n\n\n\n
Chapter 1 Integers<\/a><\/td><\/tr>
Chapter 2 Fractions and Decimals<\/a><\/td><\/tr>
Chapter 3 Data Handling<\/a><\/td><\/tr>
Chapter 4 Simple Equations<\/a><\/td><\/tr>
Chapter 5 Lines and Angles<\/a><\/td><\/tr>
Chapter 6 The Triangle and its Properties<\/a><\/td><\/tr>
Chapter 7 Congruence of Triangles<\/a><\/td><\/tr>
Chapter 8 Comparing Quantities<\/a><\/td><\/tr>
Chapter 9 Rational Numbers<\/a><\/td><\/tr>
Chapter 10 Practical Geometry<\/a><\/td><\/tr>
Chapter 11 Perimeter and Area<\/a><\/td><\/tr>
Chapter 12 Algebraic Expressions<\/a><\/td><\/tr>
Chapter 13 Exponents and Powers<\/a><\/td><\/tr>
Chapter 14 Symmetry<\/a><\/td><\/tr>
Chapter 15 Visualising Solid Shapes<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Class 7: Maths Chapter 4 solutions. Complete Class 7 Maths Chapter 4 Notes. NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations NCERT 7th Maths Chapter 4, class 7 Maths Chapter 4 solutions Page No: 81 Exercise 4.1 1. Complete the last column of the table: S. No. Equation Value Say, whether the Equation is […]<\/p>\n","protected":false},"author":302,"featured_media":627313,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1550],"boards":[1180],"class_list":["post-140182","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-ncert-maths-class-7","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions Class 7, Maths Chapter 4- IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations | Browse all Class 7 Maths Solutions of NCERT books- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-4-simple-equations\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 7th Class Maths: Chapter 4-Simple Equations\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 4 solutions. 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