Test post Ziarul

Answer: Option D
Solution:
Step 1: Represent the odd and even numbers
– The four consecutive odd numbers can be expressed as \( x, x+2, x+4, x+6 \).
Their sum is:
\[
\text{Sum of odd numbers} = x + (x+2) + (x+4) + (x+6) = 4x + 12.
\] Their average is:
\[
\text{Average of odd numbers} = \frac{4x + 12}{4} = x + 3.
\]

– The three consecutive even numbers can be expressed as \( y, y+2, y+4 \).
Their sum is:
\[
\text{Sum of even numbers} = y + (y+2) + (y+4) = 3y + 6.
\] Their average is:
\[
\text{Average of even numbers} = \frac{3y + 6}{3} = y + 2.
\]

—

Step 2: Use the given conditions
1. The average of the odd numbers is 6 less than the average of the even numbers:
\[
x + 3 = (y + 2) – 6.
\] Simplify:
\[
x + 3 = y – 4 \implies x = y – 7.
\]

2. The sum of the three even numbers is 16 less than the sum of the four odd numbers:
\[
3y + 6 = (4x + 12) – 16.
\] Simplify:
\[
3y + 6 = 4x – 4.
\] Rearrange:
\[
3y = 4x – 10.
\]

—

Step 3: Substitute \( x = y – 7 \) into the second condition
Substitute \( x = y – 7 \) into \( 3y = 4x – 10 \):
\[
3y = 4(y – 7) – 10.
\] Simplify:
\[
3y = 4y – 28 – 10.
\] \[
3y = 4y – 38.
\] \[
y = 38.
\]

—

Step 4: Solve for \( x \)
Using \( x = y – 7 \):
\[
x = 38 – 7 = 31.
\]

—

Step 5: Find the average of \( x, x+2, x+4, x+6 \)
The average of the four odd numbers is:
\[
\text{Average of odd numbers} = x + 3 = 31 + 3 = 34.
\]

—

Final Answer:
The average of \( x, x+2, x+4, x+6 \) is 34.

Answer: Option C
Solution:
To express 1.14 as a percentage of 1.9, we use the formula:
\[
\text{Percentage} = \left( \frac{1.14}{1.9} \right) \times 100
\] Let’s calculate that.

1.14 expressed as a percentage of 1.9 is 60.%.

Answer: Option A
Solution:
To find the percentage of candidates who failed in both English and Mathematics, we can use the principle of inclusion-exclusion.

Given:
– 80% passed in English.
– 85% passed in Mathematics.
– 73% passed in both English and Mathematics.

Here’s how we can calculate it:
1. Find the percentage of candidates who passed in at least one of the subjects:
\[
\text{Passed in either subject} = \text{Passed in English} + \text{Passed in Mathematics} – \text{Passed in both}
\] \[
= 80\% + 85\% – 73\%
\] \[
= 92\%
\] 2. Find the percentage of candidates who failed in both subjects:
\[
\text{Failed in both subjects} = 100\% – \text{Passed in either subject}
\] \[
= 100\% – 92\% = 8\%
\]

So, 8% of candidates failed in both subjects.

Answer: Option A
Solution:
Let the third number be \( x \).

Step 1: Express the First and Second Numbers in Terms of \( x \)
– The first number is 30% more than \( x \):
\[
\text{First number} = x + 0.30x = 1.30x
\] – The second number is **40% more than \( x \):
\[
\text{Second number} = x + 0.40x = 1.40x
\]

Step 2: Find What Percent the First is of the Second
We need to find what percent the first number (1.30x) is of the second number (1.40x):

\[
\text{Required percentage} = \left( \frac{\text{First number}}{\text{Second number}} \right) \times 100
\]

\[
= \left( \frac{1.30x}{1.40x} \right) \times 100
\]

Cancel \( x \):

\[
= \left( \frac{1.30}{1.40} \right) \times 100
\]

\[
= \left( \frac{130}{140} \right) \times 100
\]

\[
= 92.86\%
\]

Final Answer:
The first number is 92.86% of the second number.

Answer: Option C
Solution:
Let the number of men in the city be **\( m \)** and the number of women be **\( w \)**.

We are given that the total population is:

\[
m + w = 35000
\]

After an increase of **6% in the number of men** and **4% in the number of women**, the new population becomes 36760. This can be written as:

\[
1.06m + 1.04w = 36760
\]

Step 1: Solve for \( w \)
We have the system of equations:

1. \( m + w = 35000 \)
2. \( 1.06m + 1.04w = 36760 \)

Express \( m \) in terms of \( w \)
From equation (1):

\[
m = 35000 – w
\]

Substitute into equation (2)
\[
1.06(35000 – w) + 1.04w = 36760
\]

\[
37100 – 1.06w + 1.04w = 36760
\]

\[
37100 – 36760 = 1.06w – 1.04w
\]

\[
340 = 0.02w
\]

\[
w = \frac{340}{0.02} = 17000
\]

Final Answer:
The initial number of women in the city was 17,000.

Answer: Option D
Solution
To solve this, we first need to recall how the volume of a cuboid is calculated. The volume \( V \) of a cuboid is given by:

\[
V = \text{length} \times \text{breadth} \times \text{height}
\]

Let the initial dimensions of the cuboid be:
– Length = \( l \)
– Breadth = \( b \)
– Height = \( h \)

The initial volume \( V_0 \) is:

\[
V_0 = l \times b \times h
\]

Now, the new dimensions are increased by the following percentages:
– Length increases by 10% → new length = \( l \times 1.10 \)
– Breadth increases by 20% → new breadth = \( b \times 1.20 \)
– Height increases by 50% → new height = \( h \times 1.50 \)

So, the new volume \( V_1 \) is:

\[
V_1 = (l \times 1.10) \times (b \times 1.20) \times (h \times 1.50)
\]

\[
V_1 = l \times b \times h \times 1.10 \times 1.20 \times 1.50
\]

\[
V_1 = V_0 \times (1.10 \times 1.20 \times 1.50)
\]

First, calculate the product of the factors:

\[
1.10 \times 1.20 = 1.32
\] \[
1.32 \times 1.50 = 1.98
\]

Thus:

\[
V_1 = V_0 \times 1.98
\]

The volume has increased by a factor of 1.98, so the percentage change in volume is:

\[
\text{Percentage change in volume} = (1.98 – 1) \times 100 = 0.98 \times 100 = 98\%
\]

Therefore, the volume of the cuboid has increased by 98%.

Answer: Option C
Solution:
If the price of rice falls by 20%, this means the new price is 80% of the previous price.

Let the original price of rice per kilogram be \( P \). The total cost for 20 kg of rice at the original price would be:

\[
\text{Total cost for 20 kg} = 20 \times P
\]

Now, since the price falls by 20%, the new price per kilogram of rice is:

\[
\text{New price per kg} = 0.80 \times P
\]

With the same amount of money, the quantity of rice that can be bought at the new price is:

\[
\text{Quantity of rice} = \frac{\text{Total cost for 20 kg}}{\text{New price per kg}} = \frac{20 \times P}{0.80 \times P}
\]

The \( P \) terms cancel out, and we get:

\[
\text{Quantity of rice} = \frac{20}{0.80} = 25 \, \text{kg}
\]

So, with the same amount of money, you can now buy 25 kg of rice.

Answer: Option D
Solution:
Let the unknown number be \( x \).

According to the problem, 30% of \( x \) when subtracted from 91 gives the number itself. This can be written as the equation:

\[
91 – 0.30x = x
\]

Now, let’s solve for \( x \).

1. First, add \( 0.30x \) to both sides to move the terms involving \( x \) to one side:

\[
91 = x + 0.30x
\]

2. Combine the terms on the right-hand side:

\[
91 = 1.30x
\]

3. Now, divide both sides by 1.30 to solve for \( x \):

\[
x = \frac{91}{1.30} = 70
\]

So, the number is 70.

Answer: Option B
Solution:
To find the population at the start of the third year, we need to apply the formula for compound interest since the population is increasing at a rate of 10% per annum.

The formula for population after \( t \) years is:

\[
P = P_0 \times (1 + r)^t
\]

Where:
– \( P_0 \) is the initial population,
– \( r \) is the rate of increase (in decimal form),
– \( t \) is the number of years.

Given:
– Initial population, \( P_0 = 100,000 \),
– Rate of increase, \( r = 10\% = 0.10 \),
– We need to find the population at the start of the third year, which means after 2 full years of increase, so \( t = 2 \).

First, let’s calculate the population at the start of the third year (after 2 years):

\[
P = 100,000 \times (1 + 0.10)^2
\]

\[
P = 100,000 \times (1.10)^2
\]

\[
P = 100,000 \times 1.21 = 121,000
\]

So, the population at the start of the third year is 121,000.