Numbers - Quiz

Let's say the number we are trying to find is x. We know that the number is multiplied by one-third of itself to get 192, so we can write the equation:

x * (1/3) * x = 192
Simplifying the left side of the equation, we get:

(1/3) * x^2 = 192
Multiplying both sides of the equation by 3, we get:

x^2 = 576
Taking the square root of both sides of the equation, we get:

x = 24
Therefore, the number we are looking for is 24.

Let H be the number of hens and P be the number of pigs. The first equation is:

H + P = 84 (1)

The second equation is:

2H + 4P = 282 (2)

We can solve for P by dividing equation (2) by 2, which gives us:

H + 2P = 141 (3)

Subtracting equation (1) from equation (3) gives us:

H + 2P - H - P = 141 - 84

P = 57

Substituting this value for P into equation (1) gives us:

H + 57 = 84

H = 84 - 57

H = 27

Therefore, there are 27 hens in the poultry farm.

I hope this rephrased version makes the solution more clear. Let me know if you have any further questions.

To maintain the same expenditure on rice when the price increases from Rs. 6 per kg to Rs. 8 per kg, a person must reduce their consumption of sugar by 25%.

To see this, suppose that a person was originally spending X rupees on rice, and consuming Y kg of sugar. Then, the original cost of the sugar can be expressed as 6X rupees/Y kg.

When the price of rice increases to Rs. 8 per kg, the person must reduce their consumption of sugar in order to maintain the same expenditure on rice. Let's say they reduce their consumption of sugar to Y - Z kg, where Z is the amount of sugar they need to reduce.

The cost of the sugar after the reduction is (8X rupees)/(Y - Z kg). Setting this equal to the original cost of the sugar, we have:

(8X rupees)/(Y - Z kg) = 6X rupees/Y kg

Solving for Z, we find that Z = (3/4)Y. In other words, the person must reduce their consumption of sugar by 25% in order to maintain the same expenditure on rice when the price increases from Rs. 6 per kg to Rs. 8 per kg 

 The number x is such that 3(2x + 5) = 105.

This equation can be rewritten as 6x + 15 = 105.

 Subtracting 15 from both sides of the equation gives us 6x = 90, 
which means that x = 15.

 To solve this problem, we can use the fact that the total number of 
marks a student receives is equal to the number of correct answers multiplied by 3, minus the
 number of wrong answers. We can represent this relationship with the equation:

3x - y = 38

Where x is the number of correct answers and y is the number of wrong answers.

We are also given that the total number of questions is 70, so we can set up
 a second equation to represent this relationship:

x + y = 70

We can solve this system of equations by first solving for y in the first equation:

y = 3x - 38

Substituting this expression for y into the second equation gives:

x + (3x - 38) = 70

This simplifies to:

4x - 38 = 70

Adding 38 to both sides gives:

4x = 108

Dividing both sides by 4 gives:

x = 27

Therefore, the student answered 27 questions correctly.   

We can represent the given information as follows:
52 % 6 = 1
53 % 6 = 5
54 % 6 = 1
55 % 6 = 5

For any positive integer x, we have:
5(2x) % 6 = 1
5(2x - 1) % 6 = 5
Therefore, for x = 285, we have:
5(2 * 285) % 6 = 570 % 6 = 1
This can be written as an equation as follows:
5(2 * 285) % 6 = 1
570 % 6 = 1
Solving this equation will confirm that the given information is consistent

To find the number that is as much greater than 36 as it is less than 86, we can use the following steps:
Let the number be represented by the variable x.
Set up the equation x - 36 = 86 - x.
Solve the equation to obtain 2x = 122, which simplifies to x = 61.
The required number is x, which is equal to 61.

Given:

x + y = 42
xy = 437
Find:

|x - y|
Substitute the first equation into the expression for |x - y| to get:

|x - y| = √((x + y)^2 - 4xy)
Substitute the given values for x + y and xy into this expression to get:

|x - y| = √((42^2) - 4(437)) = √(1764 - 1748) = √(16) = 4
Therefore, the absolute difference between the numbers is 4.

Given:

x + (15 - x) = 15
x^2 + (15 - x)^2 = 113
Find:

x and (15 - x)
Substitute the first equation into the expression for (15 - x) to get:

x^2 + (15 - x)^2 = 113
x^2 + 225 + x^2 - 30x + x^2 = 113
3x^2 - 30x + 112 = 0
Factor the quadratic equation:

(x - 7)(x - 8) = 0
Solve for x:

x = 7 or x = 8
Therefore, the two numbers are x = 7 and (15 - x) = 8.

Given:

The ratio between a two-digit number and the sum of the digits of that number is 4:1
The digit in the unit's place is 3 more than the digit in the ten's place
Let the digit in the ten's place be x.
Then the digit in the unit's place is x + 3 and the sum of the digits is 2x + 3. 
The two-digit number can be expressed as:

10x + (x + 3) = 11x + 3
The ratio of the number to the sum of its digits is given by:

(11x + 3) / (2x + 3) = 4/1
Solve for x:

3x = 9
x = 3
Therefore, the two-digit number is 11x + 3 = 11(3) + 3 = 36.