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Simple Equations Ex-4.4

NCERT solutions for Maths Simple Equations 

NCERT Solutions for Class 7 Maths Exercise 4.4

NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.4)
Question 1.Set up equations and solve them to find the unknown numbers in the following cases:
  1. Add 4 to eight times a number; you get 60.
  2. One-fifth of a number minus 4 gives 3.
  3. If I take three-fourth of a number and add 3 to it, I get 21.
  4. When I subtracted 11 from twice a number, the result was 15.
  5. Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
  6. Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.
  7. Answer thinks of a number. If he takes away 7 from 52\frac{5}{2} of the number, the result is 112.\frac{{11}}{2}.

Answer:

(a) Let the number be x.x.

According to the question, 8x+4=608x + 4 = 60

\Rightarrow 8x=6048x = 60 – 4 \Rightarrow 8x=568x = 56

\Rightarrow x=568x = \frac{{56}}{8} \Rightarrow x=7x = 7

(b) Let the number be y.y.

According to the question, y54=3\frac{y}{5} – 4 = 3

\Rightarrow y5=3+4\frac{y}{5} = 3 + 4 \Rightarrow y5=7\frac{y}{5} = 7

\Rightarrow y=7×5y = 7 \times 5 \Rightarrow y=35y = 35

(c) Let the number be z.z.

According to the question, 34z+3=21\frac{3}{4}z + 3 = 21

\Rightarrow 34z=213\frac{3}{4}z = 21 – 3 \Rightarrow 34z=18\frac{3}{4}z = 18 \Rightarrow 3z=18×43z = 18 \times 4

\Rightarrow 3z=723z = 72 \Rightarrow z=723z = \frac{{72}}{3} \Rightarrow z=24z = 24

(d) Let the number be x.x.

According to the question, 2x11=152x – 11 = 15

\Rightarrow 2x=15+112x = 15 + 11 \Rightarrow 2x=262x = 26

\Rightarrow x=262x = \frac{{26}}{2} \Rightarrow x=13x = 13

(e) Let the number be m.m.

According to the question, 503m=850 – 3m = 8

\Rightarrow 3m=850– 3m = 8 – 50 \Rightarrow 3m=42– 3m = – 42

\Rightarrow m=423m = \frac{{ – 42}}{{ – 3}} \Rightarrow m=14m = 14

(f) Let the number be n.n.

According to the question, n+195=8\frac{{n + 19}}{5} = 8

\Rightarrow n+19=8×5n + 19 = 8 \times 5 \Rightarrow n+19=40n + 19 = 40

\Rightarrow n=4019n = 40 – 19 \Rightarrow n=21n = 21

(g) Let the number be x.x.

According to the question, 52x7=112\frac{5}{2}x – 7 = \frac{{11}}{2}

\Rightarrow 52x=112+7\frac{5}{2}x = \frac{{11}}{2} + 7 \Rightarrow 52x=11+142\frac{5}{2}x = \frac{{11 + 14}}{2}

\Rightarrow 52x=252\frac{5}{2}x = \frac{{25}}{2} \Rightarrow 5x=25×225x = \frac{{25 \times 2}}{2} \Rightarrow 5x=255x = 25

\Rightarrow x=255x = \frac{{25}}{5} \Rightarrow x=5x = 5


NCERT Solutions for Class 7 Maths Exercise 4.4

Question 2.Solve the following:
  1. 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?
  2. In an isosceles triangle, the base angles are equal. The vertex angle is 40.40^\circ . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180.180^\circ .)
  3. 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?

Answer:

(a) Let the lowest marks be y.y.

According to the question, 2y+7=872y + 7 = 87

\Rightarrow 2y=8772y = 87 – 7 \Rightarrow 2y=802y = 80 \Rightarrow y=802y = \frac{{80}}{2}

\Rightarrow y=40y = 40

Thus, the lowest score is 40.

(b) Let the base angle of the triangle be b.b.

Given, a=40,b=ca = 40^\circ ,b = c

Since, a+b+c=180a + b + c = 180^\circ [Angle sum property of a triangle]

\Rightarrow 40+b+b=18040^\circ + b + b = 180^\circ

\Rightarrow 40+2b=18040^\circ + 2b = 180^\circ

\Rightarrow 2b=180402b = 180^\circ – 40^\circ \Rightarrow 2b=1402b = 140^\circ

\Rightarrow b=1402b = \frac{{140^\circ }}{2} \Rightarrow b=70b = 70^\circ

Thus, the base angles of the isosceles triangle are 7070^\circ each.

(c) Let the score of Rahul be xx runs and Sachin’s score is 2x.2x.

According to the question, x+2x=198x + 2x = 198

\Rightarrow 3x=1983x = 198 \Rightarrow x=1983x = \frac{{198}}{3}

\Rightarrow x=66x = 66

Thus, Rahul’s score = 66 runs

And Sachin’s score = 2 x 66 = 132 runs.


NCERT Solutions for Class 7 Maths Exercise 4.4

Question 3.Solve the following:
  1. 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?
  2. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
  3. 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?

Answer:

(i) Let the number of marbles Parmit has be m.m.

According to the question, 5m+7=375m + 7 = 37

\Rightarrow 5m=3775m = 37 – 7 \Rightarrow 5m=305m = 30

\Rightarrow m=305m = \frac{{30}}{5} \Rightarrow m=6m = 6

Thus, Parmit has 6 marbles.

(ii) Let the age of Laxmi be yy years.

Then her father’s age = (3y+4)\left( {3y + 4} \right) years

According to question, 3y+4=493y + 4 = 49

\Rightarrow 3y=4943y = 49 – 4 \Rightarrow 3y=453y = 45

\Rightarrow y=453y = \frac{{45}}{3} \Rightarrow y=15y = 15

Thus, the age of Laxmi is 15 years.

(iii) Let the number of fruit trees be t.t.

Then the number of non-fruits tree = 3t+23t + 2

According to the question, t+3t+2=102t + 3t + 2 = 102

\Rightarrow 4t+2=1024t + 2 = 102 \Rightarrow 4t=10224t = 102 – 2

\Rightarrow 4t=1004t = 100 \Rightarrow t=1004t = \frac{{100}}{4}

\Rightarrow t=25t = 25

Thus, the number of fruit trees are 25.


NCERT Solutions for Class 7 Maths Exercise 4.4

Question 4.Solve the following riddle:

I am a number, Tell my identity!

Take me seven times over, And add a fifty!

To reach a triple century, You still need forty!

Answer:

Let the number be n.n.

According to the question, 7n+50+40=3007n + 50 + 40 = 300

\Rightarrow 7n+90=3007n + 90 = 300 \Rightarrow 7n=300907n = 300 – 90

\Rightarrow 7n=2107n = 210 \Rightarrow n=2107n = \frac{{210}}{7}

\Rightarrow n=30n = 30

Thus, the required number is 30.


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