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 B
Solution:
To maintain the same expenditure when the price of a commodity is increased by 50%, the consumption must be reduced by a fraction.

Let’s break it down:

1. Let the original price be \( P \) and the original quantity consumed be \( Q \).
– Original expenditure = \( P \times Q \).

2. The new price after a 50% increase is \( 1.5P \).
– Let the new quantity consumed be \( Q_{\text{new}} \).

3. To keep the same expenditure, the new expenditure should be equal to the original expenditure:
\[
1.5P \times Q_{\text{new}} = P \times Q
\]

4. Now, divide both sides of the equation by \( P \):
\[
1.5 \times Q_{\text{new}} = Q
\]

5. Solving for \( Q_{\text{new}} \):
\[
Q_{\text{new}} = \frac{Q}{1.5} = \frac{2Q}{3}
\]

This means that the new quantity consumed must be \( \frac{2}{3} \) of the original quantity.

6. To find the fraction by which consumption is reduced:
The reduction in consumption is:
\[
Q – Q_{\text{new}} = Q – \frac{2Q}{3} = \frac{Q}{3}
\]

Thus, the consumption must be reduced by \( \frac{1}{3} \) of its original amount.

Answer: The consumption must be reduced by \( \frac{1}{3} \) to keep the expenditure the same.

Answer: Option C
Solution:
To calculate the population after 2 years with an annual increase of 4%, you can use the formula for compound growth:

\[
P_{\text{new}} = P_{\text{initial}} \times (1 + r)^t
\]

Where:
– \( P_{\text{new}} \) is the population after time \( t \),
– \( P_{\text{initial}} \) is the initial population,
– \( r \) is the annual growth rate (as a decimal),
– \( t \) is the time in years.

Given:
– \( P_{\text{initial}} = 50,000 \),
– \( r = 4\% = 0.04 \),
– \( t = 2 \) years.

Substitute the values into the formula:
\[
P_{\text{new}} = 50,000 \times (1 + 0.04)^2
\] \[
P_{\text{new}} = 50,000 \times (1.04)^2
\] \[
P_{\text{new}} = 50,000 \times 1.0816
\] \[
P_{\text{new}} = 54,080
\]

So, after 2 years, the population will be 54,080.

Answer: Option D
Solution:
We are given that:
– The distance between points A and B is 5 cm.
– The distance from A to C is 3 cm.
– The distance from C to B (CB) is \( 5 – 3 = 2 \) cm.
– The length of AC is increased by 6%.

Now, let’s calculate the effect of the 6% increase in AC.

1. New length of AC:
– The new length of AC is increased by 6%, so:
\[
\text{New AC} = AC \times (1 + 0.06) = 3 \times 1.06 = 3.18 \text{ cm}.
\] 2. New length of CB:
– The total distance AB remains 5 cm.
– The new length of CB will be the total distance (5 cm) minus the new length of AC (3.18 cm):
\[
\text{New CB} = 5 – 3.18 = 1.82 \text{ cm}.
\] 3. Change in the length of CB:
– The original length of CB was 2 cm, and now it is 1.82 cm.
– The decrease in the length of CB is:
\[
2 – 1.82 = 0.18 \text{ cm}.
\]

Thus, the length of CB is decreased by 0.18 cm.

Answer: Option A
Solution:
Let’s break down the changes step by step:

Initial cost of the article = Rs. 75.

1. First increase by 20%:
– The increased cost after 20% increase is calculated as:
\[
\text{Increased cost} = 75 \times \left(1 + \frac{20}{100}\right) = 75 \times 1.20 = 90 \, \text{Rs.}
\] 2. Then reduce by 20%:
– The new cost after a 20% reduction is calculated on the increased cost of Rs. 90:
\[
\text{Reduced cost} = 90 \times \left(1 – \frac{20}{100}\right) = 90 \times 0.80 = 72 \, \text{Rs.}
\]

Thus, the present cost of the article is Rs. 72.

Answer: Option B
Solution:
To maintain the same expenses when the price of sugar rises by 25%, the family will need to decrease its consumption of sugar.

Let’s break it down:

1. **Let the original price of sugar be \( P \) and the original quantity consumed be \( Q \).**
– Original expenditure = \( P \times Q \).

2. The new price of sugar after a 25% increase is \( 1.25P \).
– Let the new quantity consumed be \( Q_{\text{new}} \).

3. To keep the expenditure the same, the new expenditure should be equal to the original expenditure:**
\[
1.25P \times Q_{\text{new}} = P \times Q
\]

4. Divide both sides of the equation by \( P \):
\[
1.25 \times Q_{\text{new}} = Q
\]

5. Solving for \( Q_{\text{new}} \):
\[
Q_{\text{new}} = \frac{Q}{1.25} = \frac{4Q}{5}
\]

This means the new quantity consumed must be \( \frac{4}{5} \) of the original quantity.

6. To find the fraction by which consumption is decreased:
The reduction in consumption is:
\[
Q – Q_{\text{new}} = Q – \frac{4Q}{5} = \frac{Q}{5}
\]

Thus, the family will have to decrease its consumption of sugar by \( \frac{1}{5} \) (or 20%) to maintain the same expenditure.

Answer: The family will have to decrease its consumption of sugar by 20%.

Answer: Option B
Solution:
When each side of a rectangular field is diminished by 40%, the area of the field will also decrease. Let’s break it down:

1. Let the original length and breadth of the field be \( L \) and \( B \), respectively.
– The original area of the field is \( \text{Area}_{\text{original}} = L \times B \).

2. When the length and breadth are both decreased by 40%, the new length and breadth become:
– New length = \( 0.60L \) (since 40% decrease means keeping 60% of the original length),
– New breadth = \( 0.60B \).

3. The new area of the field is:
\[
\text{Area}_{\text{new}} = 0.60L \times 0.60B = 0.36L \times B.
\]

4. The decrease in area is:
\[
\text{Area decrease} = \text{Area}_{\text{original}} – \text{Area}_{\text{new}} = L \times B – 0.36L \times B = (1 – 0.36) L \times B = 0.64L \times B.
\]

This means the area has decreased by 64%.
Thus, the area of the field is diminished by 64%.

Answer: Option B
Solution:
We need to determine the overall effect on expenditure when:
– The price of a commodity decreases by 20%.
– The consumption of the commodity increases by 20%.

Step 1: Define Variables
Let:
– The original price per unit of the commodity be **\( P \)**.
– The original quantity consumed be **\( Q \)**.
– The original expenditure = **\( P \times Q \)**.

Step 2: New Price and New Quantity
– Since the price decreases by **20%**, the new price = **\( 0.80P \)**.
– Since the quantity increases by **20%**, the new quantity = **\( 1.20Q \)**.

Step 3: New Expenditure
New expenditure = **New Price × New Quantity**
\[
\text{New Expenditure} = (0.80P) \times (1.20Q)
\] \[
= 0.96 (P \times Q)
\]

Step 4: Percentage Change in Expenditure
The percentage change in expenditure is:
\[
\left( \frac{\text{New Expenditure} – \text{Original Expenditure}}{\text{Original Expenditure}} \right) \times 100
\] \[
= \left( \frac{0.96PQ – PQ}{PQ} \right) \times 100
\] \[
= (-0.04) \times 100
\] \[
= -4\%
\]

Conclusion:
The expenditure decreases by 4%.

Answer: Option B
Solution:
The population of a town increases by **2.5% annually**, but at the same time, it decreases by **0.5% due to migration**. The net annual growth rate is:

\[
\text{Net Growth Rate} = 2.5\% – 0.5\% = 2\% \text{ per year}
\]

We need to determine the percentage increase in **2 years**. Since the growth is **compounded**, we use the compound growth formula:

\[
\text{Total Percentage Increase} = \left(1 + \frac{r}{100}\right)^t – 1
\]

where:
– \( r = 2\% \) (net growth rate),
– \( t = 2 \) years.

Substituting the values:

\[
\text{Total Percentage Increase} = \left(1 + \frac{2}{100}\right)^2 – 1
\]

\[
= (1.02)^2 – 1
\]

\[
= 1.0404 – 1
\]

\[
= 0.0404 \text{ or } 4.04\%
\]

Final Answer:
The population will increase by 4.04% in 2 years.

Answer: Option D
Solution:
Let the total number of votes cast be **\( x \)**.

One candidate got **65%** of the total votes, while the other candidate got **35%** of the total votes.

The majority (the difference between the two candidates’ votes) is given as **2748 votes**.

Step 1: Express the Majority in Terms of \( x \)
\[
\text{Majority} = (\text{Votes received by winner}) – (\text{Votes received by loser})
\] \[
= 65\% \text{ of } x – 35\% \text{ of } x
\] \[
= (0.65x – 0.35x) = 0.30x
\]

Step 2: Solve for \( x \)
\[
0.30x = 2748
\] \[
x = \frac{2748}{0.30}
\] \[
x = 9160
\]

Final Answer:
The total number of votes cast is 9160.

Answer: Option A
Solution:
Let Narayan’s monthly income be **\( x \)**.

Step 1: Calculate Amount Spent on Education**
Narayan spends **30%** of his income on education:
\[
\text{Education Expense} = \frac{30}{100} \times x = 0.30x
\] \[
\text{Remaining Income} = x – 0.30x = 0.70x
\]

Step 2: Calculate Amount Spent on Food
He spends 50% of the remaining income on food:
\[
\text{Food Expense} = \frac{50}{100} \times 0.70x = 0.35x
\] \[
\text{Remaining Income after Food} = 0.70x – 0.35x = 0.35x
\]

Step 3: Account for Rent and Remaining Balance
After paying Rs. 1000*as rent, he is left with Rs. 1800:
\[
0.35x – 1000 = 1800
\]

Step 4: Solve for \( x \)
\[
0.35x = 2800
\] \[
x = \frac{2800}{0.35}
\] \[
x = 8000
\] Final Answer:
Narayan’s monthly income is Rs. 8000.

Answer: Option B
Solution:
The statement **”P is 6 times greater than Q”** means that **P is 6 times more than Q**, or mathematically:

\[
P = Q + 6Q = 7Q
\]

Now, we need to determine by what **percentage Q is smaller than P**.

Step 1: Use the Percentage Decrease Formula
The percentage decrease is calculated as:

\[
\text{Percentage decrease} = \left(\frac{\text{Difference}}{\text{Original Value}}\right) \times 100
\]

Here:
– The **difference** between P and Q is:
\[
P – Q = 7Q – Q = 6Q
\] – The **original value** in this case is **P**, because we are finding how much smaller Q is compared to P.

Thus,

\[
\text{Percentage decrease} = \left(\frac{6Q}{7Q}\right) \times 100
\]

Step 2: Simplify
\[
= \left(\frac{6}{7}\right) \times 100
\]

\[
= 85.71\%
\]

Final Answer:
Q is 85.71% smaller than P.

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 \( x \):

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

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

\[
= \left( 0.9286 \right) \times 100
\]

\[
= 92.86\%
\]

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

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

Given:
\[
m + w = 35000
\] After a **6% increase** in men and a **4% increase** in women, the total population becomes **36760**. Mathematically, we express this as:

\[
1.06m + 1.04w = 36760
\]

Step 1: Solve the Two Equations**
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.