Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.
Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.4
Question 1.
Set up equations and solve them to find the undnown numbers in the following cases :
(a) Add 4 to eight times a number; you get 60.
(b) One fifth of a number minus 4 gives 3.
(c) If 1 take three fourths of a number and count up 3 more, 1 get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the rusult is \(\frac{7}{11}\).
Solution:
(a) Let a number be x.
According to question, + 4 = 60
⇒ 8x = 60 – 4
⇒ x = \(\frac{56}{8}\) =7
Required number is 7.
(b) \(\frac{x}{5}\) – 4 = 3
⇒ \(\frac{x-20}{5}\) = 3
⇒ x – 20 = 15
⇒ x = 15 + 20
⇒ x = 35
Required number is 35.
(c) \(\frac{3 x}{4}\) + 3 = 21
\(\frac{3 x+12}{4}\) = 21
⇒ 3x + 12 = 84
⇒ 3x = 84-12
⇒ x = \(\frac{72}{3}\)
∴ x = 24
Required number is 24.
(d) 2x -11 = 15
⇒ 2x = 15 + 11
⇒ x = \(\frac{26}{2}\)
∴ x = 13
Required number is 13.
(e) 50 – 3x = 8
⇒ 50 – 8 = 3x
⇒ 3x = 42
x = \(\frac{42}{3}\)
∴ x = 14
Required number is 14.
(f) \(\frac{x+19}{5}\) = 8
⇒ x + 19 = 8 x 5
⇒ x = 40 – 19
⇒ x = 21
Required number is 21.
(g) \(\frac{5 x}{2}-7=\frac{11}{2}\)
⇒ \(\frac{5 x+14}{2}=\frac{11}{2}\)
⇒ 2(5x -14) = 11 x 2
⇒ 10x – 28 = 22
⇒ 10x = 22 + 28
⇒ x = \(\)
∴ x = 5
Required number is 5.
Question 2.
Solve the following :
(a) 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. What is the lowest score ?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°).
(c) Smita’s mother is 34 years old. Two years from now mother’s age will be 4 times Smita’s present age. ?What is Smita’s present age ?
(d) 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?
Solution:
(a) Let lowest score be x
According to question, 2x + 7 = 87
⇒ 2x = 87 – 7
⇒ x = \(\frac{80}{2}\)
∴ x = 40
Required lowest score is 40.
(b) Let base angle be x now base angle are equal We know that the sum of three angles of a triangle is 180°.
x = x
hence, 40° + x + x = 180°
⇒ 2x = 180°-40°
⇒ x = \(\frac{140}{2}\)
∴ x = 70°
Required base angles are 70° and 70°.
(c) Let Smita presentage be x.
According to question 4x – 2 = 34
⇒ 4x = 34 + 2
⇒ x = \(\frac{36}{4}\)
∴ x = 9
Smita’s present age is 9 years.
(d) Let Rahul’s scores be
hence sachin’s scores = 2x
A.t.Q 2x + x = 200-2
⇒ 3x = 198
⇒ x = \(\frac{198}{3}\)
∴ x = 66
Hence, required Rahul’s scores be 99 runs.
Sachin’s scores be 2 x 66 = 132 runs.
Question 3.
Solve the following :
(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 ?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age ?
(iii) Maya, Madhra and Mohsina are friends studying in the same class. In a call test in geography, Maya got 16 out of 25. Madhura got 20. Their average score was 19. How much did Mohsina score ?
(iv) People ofSutidargram 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 ?
Solution:
(i) Let pannit’s marbles be m.
A.t.Q, 5m + 7 = 37
⇒ 5m = 37 – 7
⇒ x = \(\frac{30}{5}\)
∴ x = 6
Hence, parmit have 6 marbles. (ii) Let Laxmi’s age be years.
A.t.Q, 3y + 4 = 49
3y = 49 – 4
⇒ y = \(\frac{45}{3}\)
⇒ y = 15
Hence, Laxmi’s age is 15 years.
(iii) Let Mohsina score be x.
According to question, \(\frac{16+20+x}{3}\) = 19
⇒ 36 + x = 19 x 3
⇒ x = 57 — 36
∴ x = 21
Hence, Mohsina score is 21.
(iv) Let the number of fruit trees planted be x.
According to questions^: + (3x + 2) = 102
⇒ 4x = 102-2
⇒ x = \(\frac{100}{4}\)
∴ x = 25
Hence, the number of fruit trees planted be 25.
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!
Solution:
Let I be denoted by x.
According to question, (7x + 50) + 40 = 300
⇒ 7x + 90 = 300
⇒ 7x = 300-90
⇒ x = \(\frac{210}{7}\)
∴ x = 30
Thus I am 30.