class 7 chapter 4 please answer these questions I will follow u and mark brainlest
Answers
Answer:
1.
Solution:-
Let us assume the required number be x
Eight times a number = 8x
The given above statement can be written in the equation form as,
= 8x + 4 = 60
By transposing 4 from LHS to RHS it becomes – 4
= 8x = 60 – 4
= 8x = 56
Divide both side by 8,
Then we get,
= (8x/8) = 56/8
= x = 7
(b)
Solution:-
Let us assume the required number be x
One-fifth of a number = (1/5) x = x/5
The given above statement can be written in the equation form as,
= (x/5) – 4 = 3
By transposing – 4 from LHS to RHS it becomes 4
= x/5 = 3 + 4
= x/5 = 7
Multiply both side by 5,
Then we get,
= (x/5) × 5 = 7 × 5
= x = 35
(c) .
Solution:-
Let us assume the required number be x
Three-fourths of a number = (3/4) x
The given above statement can be written in the equation form as,
= (3/4) x + 3 = 21
By transposing 3 from LHS to RHS it becomes – 3
= (3/4) x = 21 – 3
= (3/4) x = 18
Multiply both side by 4,
Then we get,
= (3x/4) × 4 = 18 × 4
= 3x = 72
Then,
Divide both side by 3,
= (3x/3) = 72/3
= x = 24
(d) When I subtracted 11 from twice a number, the result was 15.
Solution:-
Let us assume the required number be x
Twice a number = 2x
The given above statement can be written in the equation form as,
= 2x –11 = 15
By transposing -11 from LHS to RHS it becomes 11
= 2x = 15 + 11
= 2x = 26
Then,
Divide both side by 2,
= (2x/2) = 26/2
= x = 13
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Solution:-
Let us assume the required number be x
Thrice the number = 3x
The given above statement can be written in the equation form as,
= 50 – 3x = 8
By transposing 50 from LHS to RHS it becomes – 50
= – 3x = 8 – 50
= -3x = – 42
Then,
Divide both side by -3,
= (-3x/-3) = – 42/-3
= x = 14
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Solution:-
Let us assume the required number be x
The given above statement can be written in the equation form as,
= (x + 19)/5 = 8
Multiply both side by 5,
= ((x + 19)/5) × 5 = 8 × 5
= x + 19 = 40
Then,
By transposing 19 from LHS to RHS it becomes – 19
= x = 40 – 19
= x = 21
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Solution:-
Let us assume the required number be x
5/2 of the number = (5/2) x
The given above statement can be written in the equation form as,
= (5/2) x – 7 = 23
By transposing -7 from LHS to RHS it becomes 7
= (5/2) x = 23 + 7
= (5/2) x = 30
Multiply both side by 2,
= ((5/2) x) × 2 = 30 × 2
= 5x = 60
Then,
Divide both the side by 5
= 5x/5 = 60/5
= x = 12
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?
Solution:-
Let us assume the lowest score be x
From the question it is given that,
The highest score is = 87
Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7
5/2 of the number = (5/2) x
The given above statement can be written in the equation form as,
Then,
= 2x + 7 = Highest score
= 2x + 7 = 87
By transposing 7 from LHS to RHS it becomes -7
= 2x = 87 – 7
= 2x = 80
Now,
Divide both the side by 2
= 2x/2 = 80/2
= x = 40
Hence, the lowest score is 40
(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°).
Solution:-
From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
= b + b + 40o = 180o
= 2b + 40 = 180o
By transposing 40 from LHS to RHS it becomes -40
= 2b = 180 – 40
= 2b = 140
Now,
Divide both the side by 2
= 2b/2 = 140/2
= b = 70o
Hence, 70o is the base angle of an isosceles triangle.
(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?
Solution:-
Let us assume Rahul’s score be x
Then,
Sachin scored twice as many runs as Rahul is 2x
Together, their runs fell two short of a double century,
= Rahul’s score + Sachin’s score = 200 – 2
= x + 2x = 198
= 3x = 198
Divide both the side by 3,
= 3x/3 = 198/3
= x = 66
So, Rahul’s score is 66
And Sachin’s score is 2x = 2 × 66 = 132
3. Solve the following:
Answer:(1). The old couples have a dog as a pet....and they just take care of the dog as their child .......During the meals they used to feed the dog plenty of rice and tibbits of fish........from their own chopstick.......
(2). Because he knew that they were greedy and that's why he just don't take them to the pile of gold..........
(3). The old couple gave feast to their friends and helped their poor neighbours........
(4). The pet dog used to follow the farmer in his fields. A heron walked in the footsteps of the old man to pick up worms.like his master,the dog also never harmed the bird.............
(5).The old farmer can perform the smalll ritual after the muko's death by when the muko came into the farmers dream and says to cut the pine and many more things he said to the farmer and he do...........
Explanation: HOPE THAT U GOT THE ANSWER WHAT U WANTED........