Math Solved Questions Bank & Study Material (2026) - Apex Rankers

$\frac{4}{4}×\frac{x}{4}=\frac{x}{16}$.

$\left(\frac{1}{5}\right) \times \left(\frac{x}{5}\right) = \frac{x}{25}$

To solve this, express each number as a power of base 2:
\(64 = 2^6\)
\(128 = 2^7\)
\(512 = 2^9\)

Now, substitute these back into the expression:
\(2^{6}\times 2^{7}\times 2^{9}\times 2^{-3}\)
When multiplying terms with identical bases, you add the exponents together:
\(2^{6+7+9+(-3)}\)
\(2^{22-3}=\mathbf{2}^{\mathbf{19}}\)

The Geometric Mean (GM) of two quantities \(x\) and \(y\) is found by taking the square root of their product:
\(\text{GM}=\sqrt{x\times y}\)
Substitute the given expressions:
Here, \(x = a^{2n}\) and \(y = b^{2n}\).
\(\text{GM}=\sqrt{a^{2n}\times b^{2n}}\)
Combine the bases under the exponent:
\(\text{GM}=\sqrt{(ab)^{2n}}\)
Simplify the square root:
Taking the square root of an exponential expression is the same as dividing its exponent by 2:
\(\text{GM}=\left((ab)^{2n}\right)^{\frac{1}{2}}=(ab)^{n}=\mathbf{a}^{\mathbf{n}}\mathbf{b}^{\mathbf{n}}\)

Apply the negative exponent rule:
A negative exponent tells you to invert the fraction (find its reciprocal) and make the exponent positive:
\(\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^{n}\)
\(\left(\frac{2}{7}\right)^{-2}=\left(\frac{7}{2}\right)^{2}\)

In binomial expansion (a+b)n, the number of terms is always n+1.

According to the Binomial Theorem, when you expand a binomial expression raised to a positive integer power \(n\) (in the form \((a+b)^n\)), the total number of terms in the expanded form is always equal to \(n + 1\).

Calculation
Given exponent (\(n\)) = \(19\)
Total terms = \(19 + 1 = \mathbf{20}\)

1/(√5+√3) + 1/(√5−√3) = (√5−√3 + √5+√3)/(5−3) = 2√5 / 2 = √5.

$\frac{5^{27}}{5^9}=5^{(27−9)}=5^{18}$.

This expression follows the algebraic identity \(a^2 - b^2 = (a - b)(a + b)\).
Let \(a = 5.293\)
Let \(b = 3.633\)
Find \((a + b)\): \(5.293 + 3.633 = 8.926\)
Find \((a - b)\): \(5.293 - 3.633 = 1.660\)
Now, substitute these back into the original fraction:
\(\frac{(5.293+3.633)\times (5.293-3.633)}{8.926}\)
\(\frac{8.926\times 1.660}{8.926}=1.66\)

This expression matches the identity \(\frac{a^3 + b^3}{a^2 + b^2 - ab} = a + b\).
Let a = 3.8
Let b = 1.2
Notice that:
a³ = 3.8 × 3.8 × 3.8
b³ = 1.2 × 1.2 × 1.2
b² = 1.2 × 1.2 = 1.44
ab = 3.8 × 1.2

Since the denominator simplifies to a² + b² - ab, the entire expression cancels out to leave just:
\(a+b=3.8+1.2=5\)

Let the numbers be: \(x\), \(x+1\), \(x+2\), \(x+3\), and \(x+4\)
Add them up: \(5x + 10 = 100\)
Solve for \(x\): \(5x = 90 \implies x = 18\)
The five numbers are 18, 19, 20, 21, and 22.
Product of the first and last number: \(18 \times 22 = \mathbf{396}\)

Find a common denominator to combine the two fractions:
\(\frac{(\sqrt{7}+\sqrt{5})^{2}+(\sqrt{7}-\sqrt{5})^{2}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}\)
Expand the numerator:
\((\sqrt{7}+\sqrt{5})^{2}=7+5+2\sqrt{35}=12+2\sqrt{35}\)
\((\sqrt{7}-\sqrt{5})^{2}=7+5-2\sqrt{35}=12-2\sqrt{35}\)
\(\text{Total Numerator}=(12+2\sqrt{35})+(12-2\sqrt{35})=24\)
Simplify the denominator using the difference of squares identity
\((a-b)(a+b) = a^2 - b^2\):\((\sqrt{7})^{2}-(\sqrt{5})^{2}=7-5=2\)
Divide:
\(\frac{24}{2}=12\)

11+x=1643⟹x=1632.

16/12−x/12=12/12
⟹16−x=12
⟹x=4.

16=2x⟹x=8.

y=12+7=19. Therefore, y+19=19+19=38.

We can solve this problem by working backward from the final remainder of 50:
Undo the subtraction:
Before subtracting 7, the number was:
\(50+7=57\)
Undo the division:
Before dividing by 2, the product was:
\(57\times 2=114\)
Undo the multiplication:
Before multiplying by 3, the sum was:
\(114\div 3=38\)
Undo the addition:
Before adding 8, the original number was:
\(38-8=\mathbf{30}\)

To find the minimum value of the cubic function \(f(x) = x^3 + 10x + 7\), we look at its first derivative to check for critical points (peaks or valleys):

Find the derivative:
\(f^{\prime }(x)=3x^{2}+10\)
Analyze the derivative:
Since any real number squared (\(x^{2}\)) is always zero or positive, \(3x^2 \ge 0\). Adding \(10\) means that:
\(f^{\prime }(x)\ge 10\quad \text{for all real values of }x.\)

Conclusion:
Because the derivative is always positive (\(f'(x) > 0\)), the graph of the function is strictly increasing from left to right. It has no local minimum or local maximum points.As \(x\) goes towards negative infinity (\(-\infty \)), the value of \(x^3 + 10x + 7\) also goes towards negative infinity (\(-\infty \)). Because it decreases without bound, it does not have a minimum real number value, making None of these the correct choice.

The expression \(f(x) = x^2 - 3x + 5\) is a quadratic equation where the coefficient of \(x^{2}\) is positive (\(a = 1\)). This means its graph is a parabola that opens upward, and its lowest point (minimum value) occurs at its vertex.
To find the value of \(x\) at the vertex, we use the vertex formula:
\(x=-\frac{b}{2a}\)

Identify the coefficients:
\(a = 1\)
\(b = -3\)
Plug the numbers into the formula:
\(x=-\frac{-3}{2(1)}=\frac{3}{2}=\mathbf{1.5}\)

Therefore, the expression reaches its minimum value when \(x\) is equal to 1.5.

2/5x=50⟹x=125 total. Girls = 125−50=75.

Factorization:
$(x^2+4)(x^2−1)$
and
$(x^2−3)(x^2−1)$.
Common is $(x^2−1)$.

x+10=2(x−10)
⟹x=30;
x+10=2(10−x)
⟹3x=10.

7/x=1/13⟹x=7×13=91.

208 ÷ 13 = 16. Then 16 = 2 × ? ⇒ ? = 8.

5x+11(37−x)=227⟹407−6x=227⟹6x=180⟹x=30. Parts 30, 7.

Arithmetic mean:
(83+98)/2=181/2=90.5.
(Note: 91 is a common whole-int answer).

31/60=0.516... which is greater than 0.5. All others are less than 0.5.

n(n+1)/2=50(51)/2=1275.

91×91=8281.

Simplify each term using exponent rules:
When multiplying numbers with the same base, you add their exponents (\(2^a \cdot 2^b = 2^{a+b}\)).
First term: \(2^3 \cdot 2^{-6} = 2^{3 + (-6)} = 2^{-3}\)
Second term: \(2^{-3} \cdot 2^6 = 2^{-3 + 6} = 2^3\)
Evaluate the simplified terms:
\(2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125\)
\(2^3 = 8\)

Add the values together:
\(0.125+8=\mathbf{8.125}\)

Look for a perfect square factor of \(117\).
Since \(117 = 9 \times 13\), we can pull the square root of \(9\) outside the radical sign:
\(\sqrt{117}=\sqrt{9\times 13}=3\sqrt{13}\)
Substitute this back into the original expression:
\((3\sqrt{13})\times (3\sqrt{13})\)
Multiply the terms:
Multiply the coefficients outside the radical together, and multiply the square roots together:
\((3\times 3)\times (\sqrt{13}\times \sqrt{13})\)
\(9\times 13=\mathbf{117}\)

Identify the parts:
The ratio is 3 : 1, which means the numbers can be written as \(3x\) and \(1x\).
Find the sum: \(3x + 1x = 8 \implies 4x = 8 \implies x = 2\).
Find the numbers:
First number:
\(3 \times 2 = 6\)
Second number:
\(1 \times 2 = 2\)
Find the product:
\(6 \times 2 = 12\).

Here is the quick rule breakdown for divisibility by powers of 2:
Divisible by 2: The last 1 digit must be even (divisible by 2).
Divisible by 4: The last 2 digits must form a number divisible by 4.
Divisible by 8: The last 3 digits must form a number divisible by 8.

Find the value of 1/7 of the number:
If 5 parts out of 7 equal 1,025, then 1 part out of 7 is:
\(1,025\div 5=205\)
Find the value of 3/7 of the number:
Multiply the value of 1 part by 3:
\(205\times 3=615\)

Direct multiplication: 11.5×0.003=0.0345.

Weighted avg: (12×70+18×80)/30=(840+1440)/30=76.

If 9/12 are women, then 3/12 are men. 3/12=1/4=25%.

Note: Rs 60 / 0.60 = 100. The source key (d) suggests the budget was Rs 6.00 or a typo in unit.

Because the number of courses in both terms is equal, the combined average is simply the exact midpoint between the two term averages. The number exactly halfway between 75% and 85% is 80%

First, convert Rs. 100 to paise.
One rupee equals 100 paise.
So, Rs. 100 equals 10,000 paise (\(100 \times 100\)).
Write it as a fraction: \(\frac{50}{10000}\).
Simplify the fraction: \(\frac{1}{200}\).

Calculate the Total Price of the Books:The average price of the 3 books is Rs. 18.
\(\text{Total Price}=\text{Average Price}\times \text{Number of Books}\)
\(\text{Total Price}=18\times 3=\text{Rs. 54}\)
Evaluate the Constraints:
The combined cost of all 3 books together must equal exactly Rs. 54. Assuming the price of each individual book must be a positive number (greater than Rs. 0), no single book can cost as much as or more than the total amount of Rs. 54.
Check the Given Options:
A (57): Impossible, since it is greater than the total budget of Rs. 54.
B (56): Impossible, since it is greater than Rs. 54.
C (55): Impossible, since it is greater than Rs. 54.
D (52): Possible. If one book costs Rs. 52, the remaining two books could have a combined cost of Rs. 2 (e.g., Rs. 1 each).

To find the total number of boys in the row, we simply add the boys on both sides of Ajmal, plus Ajmal himself:
Boys to his left: 14
Boys to his right: 15
Ajmal himself: 1
\(\text{Total Boys}=14+15+1=\mathbf{30}\)

25 - 27 + 12 = 10

Using BODMAS: 60−12=48.

To find the score needed in the fifth subject, we can look at the total percentage points required:

Calculate the total points required for 5 subjects:
To maintain an overall average of 80% across 5 subjects:
\(5\times 80=400\text{ points}\)

Calculate the total points earned in the first 4 subjects:
With an average of 78% across 4 subjects:
\(4\times 78=312\text{ points}\)

Find the required score for the 5th subject:
Subtract the points already earned from the total points needed:
\(400-312=\mathbf{88\%}\)

LCM(7,8,9)=504.
504+5=509.

1. Simplify the Numerator
\(1.4+1\frac{2}{3}\div \frac{5}{3}\)
Convert the mixed fraction:
\(1\frac{2}{3} = \frac{5}{3}\)
Perform the division first:
\(\frac{5}{3} \div \frac{5}{3} = 1\)
Add to the decimal:
\(1.4 + 1 = \mathbf{2.4}\)

2. Simplify the Denominator
\(\frac{4}{3}\times 1\frac{4}{5}\div \frac{2}{5}\)
Convert the mixed fraction:
\(1\frac{4}{5} = \frac{9}{5}\)
Working left-to-right, perform the multiplication:
\(\frac{4}{3} \times \frac{9}{5} = \frac{36}{15} = \frac{12}{5}\)
Perform the division by multiplying by the reciprocal:
\(\frac{12}{5} \times \frac{5}{2} = \frac{12}{2} = \mathbf{6}\)

3. Divide Numerator by Denominator\(\frac{2.4}{6}=\mathbf{0.4}=\frac{2}{5}\)

Step 1: Simplify the First Part (Left Fraction)
Numerator:
\(12\frac{1}{3} + 4\frac{3}{15} \text{ of } \frac{1}{3}\)
Convert mixed numbers:
\(\frac{37}{3} + \frac{63}{15} \text{ of } \frac{1}{3} \rightarrow \frac{37}{3} + \frac{21}{5} \text{ of } \frac{1}{3}\)
Evaluate "of": \(\frac{21}{5} \times \frac{1}{3} = \frac{7}{5}\)
Add terms: \(\frac{37}{3} + \frac{7}{5} = \frac{185 + 21}{15} = \frac{206}{15}\)
Denominator: \(\frac{8}{9} \div \frac{2}{3} + \frac{5}{2} \text{ of } \frac{3}{5}\)
Evaluate "of": \(\frac{5}{2} \times \frac{3}{5} = \frac{3}{2}\)
Evaluate division: \(\frac{8}{9} \times \frac{3}{2} = \frac{4}{3}\)
Add terms: \(\frac{4}{3} + \frac{3}{2} = \frac{8 + 9}{6} = \frac{17}{6}\)
Combine Left Fraction:
\(\frac{206}{15}\div \frac{17}{6}=\frac{206}{15}\times \frac{6}{17}=\frac{412}{85}\)

Step 2: Simplify the Second Part (Right Fraction)
Numerator Bracket:
\(\left( \frac{1}{4} \div \frac{5}{4} + \frac{3}{2} \right)\)
Evaluate division:
\(\frac{1}{4} \times \frac{4}{5} = \frac{1}{5}\)
Add terms:
\(\frac{1}{5} + \frac{3}{2} = \frac{2 + 15}{10} = \frac{17}{10}\)
Full Numerator: \(\frac{4}{5} \text{ of } \frac{7}{8} \times \frac{17}{10}\)
Multiply out: \(\frac{4}{5} \times \frac{7}{8} \times \frac{17}{10} = \frac{7}{10} \times \frac{17}{10} = \frac{119}{100}\)
Denominator: \(1\frac{3}{4} + \frac{5}{8} - \frac{1}{8}\)
Convert mixed number: \(\frac{7}{4} + \frac{4}{8} = \frac{7}{4} + \frac{1}{2} = \frac{7 + 2}{4} = \frac{9}{4}\)
Combine Right Fraction:\(\frac{119}{100}\div \frac{9}{4}=\frac{119}{100}\times \frac{4}{9}=\frac{119}{225}\)

Step 3: Final Division
Now divide the simplified left fraction by the simplified right fraction:
\(\frac{412}{85}\div \frac{119}{225}=\frac{412}{85}\times \frac{225}{119}=\frac{412\times 45}{17\times 119}=\frac{18540}{2023}\)
Converting the improper fraction \(\frac{18540}{2023}\) into a mixed fraction:
\(18540 \div 2023 = 9\) with a remainder of \(333\)
Result: \(9\frac{333}{2023}\)

Step 1:
Deepest multiplication
\(\frac{2}{5}\times \frac{1}{3}=\frac{2}{15}\)
Step 2:
Next level up (division by the result)\(\frac{15}{\frac{2}{15}}=15\times \frac{15}{2}=\frac{225}{2}\)
Step 3:
Next level up\(\frac{3}{\frac{225}{2}}=3\times \frac{2}{225}=\frac{6}{225}=\frac{2}{75}\)
Step 4:
Perform the main middle division
\(\frac{5}{3}\div \frac{2}{75}=\frac{5}{3}\times \frac{75}{2}=\frac{5\times 25}{2}=\frac{125}{2}\)
Step 5: Solve the upper fraction block
\(\frac{3}{\frac{125}{2}}=3\times \frac{2}{125}=\frac{6}{125}\)
Step 6: Add to the main denominator base
\(2+\frac{6}{125}=\frac{250+6}{125}=\frac{256}{125}\)
Step 7:
Final top division\(\frac{15}{\frac{256}{125}}=15\times \frac{125}{256}=\frac{\mathbf{1875}}{\mathbf{256}}\)

0.0036+0.5007+4.1000=4.6043.

BODMAS: 2−2−4=0−4=−4.

Logic: 26−[5−(12−3)]=26−[−4]=30.

Apply the first discount (15%):
Discount amount = \(200 \times 0.15 = \text{Rs. 30}\)
Price after 1st discount = \(200 - 30 = \text{Rs. 170}\)
Apply the second discount (10%) to the reduced price:
Discount amount = \(170 \times 0.10 = \text{Rs. 17}\)
Final net price = \(170 - 17 = \mathbf{\text{Rs. 153}}\)

2000×0.9=1800.

Cash discount:

1800×0.98=1764.

Based on visual slope analysis of the "Rising Marriages" graph.

The graph "Falling Births" shows the value for the year 1967 is approximately 3.55 million.

Young people (1954) ≈ 15m; Births (1954) ≈ 4m. Ratio 15/4=3.75.

Increase over 3 years (1965 to 1968):
\(21.3 - 20.8 = 0.5\text{ units}\)
Annual rate of increase: \(\frac{0.5}{3} \approx 0.1667\text{ units per year}\)
Time span from 1965 to 2000: \(2000 - 1965 = 35\text{ years}\)
Total projected increase by 2000: \(35 \times 0.1667 \approx 5.83\text{ units}\)
Projected value in 2000: \(20.8 + 5.83 = 26.63\text{ units}\)
Converting this back to the scale of millions yields approximately 2.66 million, which rounds cleanly to 2.7 million.

If you look at the horizontal level of 17 (representing 1.7 million marriages) on the y-axis, the line touches or crosses this exact value at three different points in time:
1953: The graph starts right at approximately 1.7 million.
1956: The first sharp peak returns exactly to that same height of 1.7 million.
1958–1959: As the line makes its final sharp upward climb towards the 20 mark, it crosses the 1.7 million threshold for the third time.

Number of Marriages (Top-Right Graph):
In 1966, the marriage value is approximately 21 (which corresponds to 2.1 million marriages).
\(\text{Collection from Marriages}=2.1\text{ million}\times \text{Rs }200=\text{Rs }420 \text{ million}\)
Number of Births (Bottom Graph):
In 1966, the birth line settles right at 3.55 million births.
\(\text{Collection from Births}=3.55\text{ million}\times \text{Rs }100=\text{Rs }355\text{ million}\)
Number of Young People (Top-Left Graph):
In 1966, the line sits at approximately 19.25 million young people.
\(\text{Collection from Young People}=19.25\text{ million}\times \text{Rs }20=\text{Rs }385\text{ million}\)
Total Government Collection:
\(\text{Total Collection}=420+355+385=\text{Rs }\mathbf{1160}\text{ million}\)

Industrial = 10% of 360 degrees = 36 degrees.

Total growth over the 10-year period (1992 to 2002): \(35\% - 25\% = 10\%\)
Annual rate of increase: \(\frac{10\%}{10\text{ years}} = 1\%\text{ per year}\)
Projected percentage for 2001 (9 years after 1992): \(25\% + (9 \times 1\%) = 34\%\)
Amount from Group Insurance (out of 50 million):\(50\text{ million}\times 34\%=50\times 0.34=17.0\text{ million}\)
Among the given multiple-choice options, 17.5 (D) is the closest standard approximation used for this dataset.

Ratio = 25% (Group) : 10% (Industrial) = 25 : 10 = 5 : 2.

Increase from 25% to 35% of the total. Relative increase: (35−25)/25=40%.

2002 Industrial = 12%. 50×0.12=6.

The timeline of the given graphs spans strictly from 1968 to 1971.

Extrapolations or claims about distant futuristic years like 1999, 2000, and 2001 are completely outside the scope of the provided historical dataset and cannot be supported or inferred by the graphs.

Evaluating the other choices (Why they match the trends):
A & B: Look at the bottom graph (Consumer Price Index). The curve climbs steeply at the end of 1969 and continues up through 1970–1971 at a linear, predictable rate.
C: Look at the middle graph (Wages). The slope from 1969 to 1970 is very steep, but as it rolls into 1971 ("The goal"), the slope becomes noticeably flatter, indicating a slower rate of growth.
D: Look at the top graph (GNP). Between late 1970 and early 1971, the curve dips downward, verifying that GNP decreased slightly for a short period.

Look at the first graph (Gross National Product): Between 1970 and 1971, the line shows that the GNP was declining. The dotted arrow labeled "The goal" points sharply upward, indicating that the policy objective is to increase the rate of growth and reverse the decline, not slow it down.

Why the other statements are true or aligned with the graphs:
A: True. In the middle graph (Wages), the line rises very steeply during 1969–1970, but the arrow for "The goal" flattens out noticeably, showing an intention to slow down the rapid wage growth rate.
C & D: These contain common source text typos (writing "1999-2000" instead of "1969-1970"). When read as 1969–1970, the trends match the historical lines shown in the charts.
E: True. This directly contradicts statement B and correctly identifies that the target arrow for the GNP is aiming for an increased growth trajectory.

Calculate the Percentage Increase in Wages (Middle Graph):
Value at the start of 1969: \(\approx \text{Rs.\ 28.50}\)
Value at the end of the 3rd quarter of 1970: \(\approx \text{Rs.\ 32.50}\)
Percentage Increase:
\(\frac{32.50-28.50}{28.50}\times 100=\frac{4.00}{28.50}\times 100\approx \mathbf{14\%}\)

Calculate the Percentage Increase in the Consumer Price Index (Bottom Graph):
Value at the start of 1969: \(\approx 121\)
Value at the end of the 3rd quarter of 1970: \(\approx 130\)
Percentage Increase:
\(\frac{130-121}{121}\times 100=\frac{9}{121}\times 100\approx \mathbf{7.4\%}\)

Find the Approximate Ratio:
Ratio = \(\frac{\text{Percentage\ Increase\ in\ Wages}}{\text{Percentage\ Increase\ in\ CPI}}\)
Ratio = \(\frac{14\%}{7.4\%} \approx 1.89\), which rounds closest to the whole ratio of 2:1

Find the values on the Consumer Price Index graph (Bottom Graph):
Value at the start of 1969: The curve crosses the 1969 vertical line right at the 120 mark.
Value at the end of the third quarter of 1970: Located about 75% of the way between the 1970 and 1971 tick marks, the curve reaches approximately 131.

Calculate the Percentage Increase:
\(\text{Percentage Increase}=\frac{\text{Final Value}-\text{Initial Value}}{\text{Initial Value}}\times 100\)

\(\text{Percentage Increase}=\frac{131-120}{120}\times 100=\frac{11}{120}\times 100\approx \mathbf{9.17\%}\)

The approximate percentage increase in hourly wages from the start of 1969 (Rs. 29.00) to the end of the third quarter of 1970 (Rs. 32.50) is calculated at approximately 12%. Therefore, the correct option is E.

The approximate percentage increase in Gross National Product (GNP) from the start of 1969 to the end of the third quarter of 1970 is 4.2%, corresponding to option A. This is calculated using an initial value of approximately 715 million Rs. at the start of 1969 and a final value of approximately 745 million Rs. by the end of the third quarter of 1970.

Multiply the track length by the number of laps: \[\text{Total Distance} = 17 \times \frac{1}{4}\text{ km} = \frac{17}{4}\text{ km}\] Converting $\frac{17}{4}$ into a mixed fraction: \[\frac{17}{4} = 4\frac{1}{4}\text{ km}\] Thus, the correct option is (A).

The time difference is from 11:00 AM to 5:00 PM $= 6\text{ hours}$. The lead distance accumulated by the passenger train is: \[36\text{ km/h} \times 6\text{ h} = 216\text{ km}\] The relative speed of the express train is: \[48 - 36 = 12\text{ km/h}\] Time taken to overtake: \[\text{Time} = \frac{216}{12} = 18\text{ hours}\] Thus, the correct option is (C).

The total difference in arrival times is: \[4\text{ min (late)} - (-6\text{ min (early)}) = 10\text{ minutes} = \frac{10}{60} = \frac{1}{6}\text{ hour}\] Let the distance be $d\text{ km}$: \[\frac{d}{3} - \frac{d}{4} = \frac{1}{6} \implies \frac{4d - 3d}{12} = \frac{1}{6} \implies \frac{d}{12} = \frac{1}{6} \implies d = 2\text{ km}\] Thus, the correct option is (B).

In one cycle, the hound covers 9 m and the hare covers 8 m. The relative distance gained by the hound in one cycle is: \[9\text{ m} - 8\text{ m} = 1\text{ m}\] Thus, the correct option is (B).

In every 2-minute cycle, the net ascent is $10 - 3 = 7\text{ m}$. In 14 minutes (7 cycles), net ascent is $7 \times 7 = 49\text{ m}$. In the 15th minute, it climbs 10 m to reach $49 + 10 = 59\text{ m}$. In the 16th minute, it slips back 3 m to $56\text{ m}$. To climb the remaining $63 - 56 = 7\text{ m}$ at the rate of $10\text{ m/min}$: \[\text{Time} = \frac{7}{10}\text{ min} = \frac{7}{10} \times 60\text{ sec} = 42\text{ seconds}\] Total time = $16\text{ min } 42\text{ sec}$. Thus, the correct option is (B).

Preservation of human life takes ethical and legal precedence over minor traffic regulations.

Prime factorization: 182=2×7×13.

Find the prime factorization of 2100:
\(2100=21\times 100=(3\times 7)\times (2^{2}\times 5^{2})=2^{2}\times 3^{1}\times 5^{2}\times 7^{1}\)
Make it a perfect square:
For a number to be a perfect square, all the exponents in its prime factorization must be even numbers.
\(2^{2}\) and \(5^{2}\) already have even exponents.
\(3^{1}\) and \(7^{1}\) have odd exponents, so we must multiply by another 3 and 7 to make them even (\(3^{2}\) and \(7^{2}\)).
Calculate the least square number:
Multiply 2100 by the missing factors (\(3 \times 7 = 21\)):
\(2100\times 21=\mathbf{44100}\)
(Verification: \(\sqrt{44100} = 210\), which is a perfect integer).

Total CP=215+25=240.
Profit = 300−240=60.
% = (60/240)×100=25%.

SP=400×1.10=440.

CP=660/(1−0.12)=660/0.88=750.

Set up the variables:
Let the selling price (SP) of 1 orange be Rs. 1.
Then, the total selling price of 200 oranges = Rs. 200
The gain equals the selling price of 40 oranges = Rs. 40
Calculate the total Cost Price (CP):
Using the profit formula (\(CP = SP - \text{Gain}\)):
\(\text{Total CP}=200-40=\text{Rs. }\mathbf{160}\)
Calculate the Gain Percentage:
Profit percentage is always calculated relative to the Cost Price:
\(\text{Gain Percentage}=\left(\frac{\text{Gain}}{\text{Total CP}}\right)\times 100\%\)
\(\text{Gain Percentage}=\left(\frac{40}{160}\right)\times 100\%=\frac{1}{4}\times 100\%=\mathbf{25\%}\)

The formula for the total loss percentage is:
\(\text{Loss Percentage}=\left(\frac{x}{10}\right)^{2}\%\)
Given \(x = 10\):
\(\text{Loss Percentage}=\left(\frac{10}{10}\right)^{2}\%=(1)^{2}\%=\mathbf{1\%}\)

Find the price factors:
Case 1 (40% loss): Selling price is \(100\% - 40\% = 60\%\) of Cost Price.
Case 2 (20% gain): Desired selling price is \(100\% + 20\% = 120\%\) of Cost Price.
Set up the inverse proportion:
\(\text{Initial Quantity}\times \text{Initial Price Factor}=\text{New Quantity}\times \text{New Price Factor}\)
\(32\times 60\%=\text{New Quantity}\times 120\%\)
Solve for the New Quantity:
\(\text{New Quantity}=\frac{32\times 60}{120}=\frac{32}{2}=\mathbf{16}\)

Total CP = 1000.
Total SP = 450+550=1000.
Break even.

CP=75,MP=100.
With 20% disc,
SP=80.
Gain % = (5/75)×100=6.66%.

CP=31/0.93=33.33.
Sells at 35
⟹ profit = Rs. 1.66.
Profit% = (1.66/33.33)×100=5%.

Final Price = 50×1.2×1.25×1.4=105.

Case 1 (20% Profit):
The selling price is \(100\% + 20\% = \mathbf{120\% \text{ of CP}}\)
Case 2 (20% Loss):
The selling price is \(100\% - 20\% = \mathbf{80\% \text{ of CP}}\)
Difference is 40% of CP.
0.4CP=100
⟹CP=250.

CP=75/0.96=78.125.
New SP=78.125×1.4≈110.

Difference 4% = Rs. 16.
CP=16/0.04=400.

Working backward:
1188/1.1/0.9/1.2=1000.
Purchase Price = 1000−110=890.

Cost×1.2×1.1×1.125=14.85
⟹Cost=14.85/1.485=10.

Difference 15% = 56.25.
CP=56.25/0.15=375.

CP=100,MP=125,
SP=125×0.88=110.
Profit = 10%.

Difference 50% = 70.
CP=70/0.5=140.

SP=50×0.9=45.
CP=45/1.25=36.

Find Selling Price (SP) relative to Cost Price (CP):
Marked Price = 120% of CP
SP after 5% discount = \(120\% - (5\% \text{ of } 120) = \mathbf{114\% \text{ of CP}}\)
Calculate CP:\(114\% \text{ of CP} = 28.50\)
\(\text{CP} = \frac{28.50}{1.14} = \mathbf{\text{Rs. } 25}\)
Calculate Profit:
\(\text{Profit} = 28.50 - 25 = \mathbf{\text{Rs. } 3.50}\)

SP=1.25×0.9=1.125
⟹ profit of 12.5%.

The 15% difference is equal to Rs. 56.25:
\(15\%\text{ of CP}=56.25\)
\(\text{CP}=\frac{56.25}{0.15}=\mathbf{375}\)

Find the profit ratios:
Horse profit: 20%
Carriage profit: 30%
Total mean profit: \(25 \frac{5}{6}\% = \mathbf{\frac{155}{6}\%}\)
Set up the alligation cross (multiply by 6 to remove fractions):
Horse: \(20 \times 6 = \mathbf{120}\)
Carriage: \(30 \times 6 = \mathbf{180}\)
Mean: \(\frac{155}{6} \times 6 = \mathbf{155}\)
Calculate the ratio of Cost Prices (Horse : Carriage):
(Carriage side): \(180 - 155 = \mathbf{25}\)
(Horse side): \(155 - 120 = \mathbf{35}\)
Ratio = \(25 : 35 = \mathbf{5 : 7}\)
Split the total cost (Rs. 1800):
Total parts = \(5 + 7 = 12\) parts
\(12 \text{ parts} = 1800 \implies 1 \text{ part} = 150\)
Cost of Horse = \(5 \times 150 = \mathbf{750}\)

A=0.9B,A=1.1C
⟹0.9B=1.1C
⟹C=9/11B≈81.8%.
Error is $21 \frac{4}{9}$​%.

Result is 12.5%.

Let the gross salary be $G$.
A $22\%$ deduction leaves $100\% - 22\% = 78\%$ of the gross salary.

\[ 0.78 \times G = 1600 \]
\[ G = \frac{1600}{0.78} \approx \text{Rs. } 2,051.28 \]
Therefore, the gross salary is nearly Rs. 2,051.

We can track the value of the chair through each successive transaction:
Aslam sells to Arshad at a 10% loss:
Price paid by Arshad = \(2000 \times (1 - 0.10) = 2000 \times 0.90 = \mathbf{\text{Rs. 1,800}}\)
Arshad sells to Akbar at a 10% loss:
Price paid by Akbar = \(1800 \times (1 - 0.10) = 1800 \times 0.90 = \mathbf{\text{Rs. 1,620}}\)
Akbar sells to Qumar at a 10% gain:
Price paid by Qumar = \(1620 \times (1 + 0.10) = 1620 \times 1.10 = \mathbf{\text{Rs. 1,782}}\)

Apply the first discount (10%):
Discount amount = \(2,000 \times 0.10 = \text{Rs. 200}\)
Price after 1st discount = \(2,000 - 200 = \text{Rs. 1,800}\)

Apply the second discount (5%) to the reduced price:
Discount amount = \(1,800 \times 0.05 = \text{Rs. 90}\)
Price after 2nd discount = \(1,800 - 90 = \text{Rs. 1,710}\)

Add the 10% sales tax to the final discounted price:
Sales tax amount = \(1,710 \times 0.10 = \text{Rs. 171}\)
Net amount paid = \(1,710 + 171 = \mathbf{\text{Rs. 1,881}}\)

When two items are sold at the same price—one at a gain of \(x\%\) and the other at a loss of \(x\%\), the overall transaction always results in a loss. The formula to find this net loss percentage is:
\(\text{Loss }\%=\left(\frac{x}{10}\right)^{2}\)
Given \(x = 10\%\):
\(\text{Loss }\%=\left(\frac{10}{10}\right)^{2}=1^{2}=1\%\)

Step-by-Step ProofCalculate the Cost Price (CP) of the Radio (10% Gain):
\(\text{CP}_{1}=\frac{\text{Selling Price}}{1+\text{Gain }\%}=\frac{350}{1.10}=\text{Rs. }318.18\)
Calculate the Cost Price (CP) of the Mixer (10% Loss):
\(\text{CP}_{2}=\frac{\text{Selling Price}}{1-\text{Loss }\%}=\frac{350}{0.90}=\text{Rs. }388.89\)

Find the Totals:Total Cost Price:
\(\text{Rs. } 318.18 + \text{Rs. } 388.89 = \text{Rs. } 707.07\)
Total Selling Price:
\(\text{Rs. } 350 + \text{Rs. } 350 = \text{Rs. } 700.00\)
Calculate Net Loss Percentage:
\(\text{Net Loss}=\text{Rs. }707.07-\text{Rs. }700.00=\text{Rs. }7.07\)
\(\text{Loss }\%=\left(\frac{7.07}{707.07}\right)\times 100=1\%\)

Assume the initial price is 100.
First discount (10%):
\(100 - 10 = 90\)
Second discount (10% of 90):
\(90 - 9 = 81\)
Third discount (10% of 81):
\(81 - 8.1 = 72.9\)
Total equivalent discount:
\(100 - 72.9 = 27.1\)

Let CP=100
MP=100+20=120
SP=$120\times\frac{(100-10)}{100}=108$.
Profit = 8%.

Find the profit amount: \(\text{Selling Price} - \text{Cost Price} = 11,500 - 7,000 =\) Rs. 4,500.
Calculate the profit percentage:
\(\text{Profit }\%=\left(\frac{\text{Profit}}{\text{Cost Price}}\right)\times 100\)
\(\text{Profit }\%=\left(\frac{4,500}{7,000}\right)\times 100\approx 64.28\%\)
When rounded to the nearest tenth, it gives 64.3%.

Simple Interest Formula: \(SI = \frac{P \times R \times T}{100}\)
Given values:
Principal (\(P\)) = Rs. 1200
Simple Interest (\(SI\)) = Rs. 594
Time (\(T\)) = 6 years
Rearrange the formula to solve for Rate (\(R\)):
\(R=\frac{SI\times 100}{P\times T}\)
\(R=\frac{594\times 100}{1200\times 6}\)
\(R=\frac{59400}{7200}=8.25\%\)

Compound Interest Formula: \(A = P \left(1 + \frac{R}{100}\right)^T\)
Given values:
Principal (\(P\)) = Rs. 2,000
Rate (\(R\)) = 5%
Time (\(T\)) = 4 years

Calculate the amount:
\(A=2000\times \left(1+0.05\right)^{4}\)
\(A=2000\times (1.05)^{4}\)
\(A=2000\times 1.2155\approx \text{Rs. }2431.01\)
When rounded to the closest option, it gives Rs. 2,430.

Convert the rate to a decimal: \(6\frac{1}{4}\% = 6.25\%\) per year.
Calculate total interest percentage over 4 years:
\(6.25\% \times 4 \text{ years} = 25\%\).
Set up the amount formula:
The total amount (\(A\)) equals the Principal (\(P\)) plus \(25\%\) of the Principal.
\(A=P+0.25P=1.25P\)
Solve for Principal (\(P\)):
\(2500=1.25P\)
\(P=\frac{2500}{1.25}=2000\)

Understand Simple Interest Amount formula:
The formula for the total amount (\(A\)) after simple interest is applied is:
\(A=P\left(1+\frac{R\times T}{100}\right)\)
Where \(P\) is the principal part, \(R\) is the rate (5%), and \(T\) is the time.
Set up the amounts for the three parts (\(P_1, P_2, P_3\)):
For 2 years:
\(A_1 = P_1 \left(1 + \frac{5 \times 2}{100}\right) = P_1 \left(\frac{110}{100}\right)\)
For 3 years:
\(A_2 = P_2 \left(1 + \frac{5 \times 3}{100}\right) = P_2 \left(\frac{115}{100}\right)\)
For 4 years:
\(A_3 = P_3 \left(1 + \frac{5 \times 4}{100}\right) = P_3 \left(\frac{120}{100}\right)\)
Equate the amounts:Since the amounts are given to be equal (\(A_1 = A_2 = A_3\)):
\(110\cdot P_{1}=115\cdot P_{2}=120\cdot P_{3}\)
Find the ratio of the parts:
To find the ratio \(P_1 : P_2 : P_3\), we take the reciprocal of the coefficients:
\(P_{1}:P_{2}:P_{3}=\frac{1}{110}:\frac{1}{115}:\frac{1}{120}\)

Find the new fare ratios:
First Class:
Reduced by \(\frac{1}{6}\) \(\rightarrow 8 \times \left(1 - \frac{1}{6}\right) = 8 \times \frac{5}{6} = \frac{20}{3}\)
Second Class: Reduced by \(\frac{1}{12}\) \(\rightarrow 6 \times \left(1 - \frac{1}{12}\right) = 6 \times \frac{11}{12} = \frac{11}{2}\)
Third Class: Remains unchanged \(\rightarrow \mathbf{3}\)

Calculate the ratio of revenue collected (\(\text{New Fare} \times \text{Number of Passengers}\)):
First Class: \(\frac{20}{3} \times 9 = \mathbf{60}\)
Second Class: \(\frac{11}{2} \times 12 = \mathbf{66}\)
Third Class: \(3 \times 26 = \mathbf{78}\)
The revenue collection ratio for First : Second : Third class is 60 : 66 : 78.
Find total units of revenue:
\(60+66+78=\mathbf{204}\text{ units}\)
Calculate money paid by First Class passengers:
\(\text{First Class Share}=\frac{60}{204}\times 1088\)
\(\text{First Class Share}=\mathbf{Rs.320}\)

5/4×8/7×1/5=(40/28)×1/5=8/28=2/7.

18/80+1/6=9/40+1/6=(27+20)/120=47/120.

18/80+1/6=9/40+1/6=(27+20)/120=47/120.

If we convert Option a back into a decimal:
\(1\frac{13}{1600}=1+\frac{13}{1600}=1+0.008125=\mathbf{1.008125}\)

Evaluating 5 + 38/27 by Euclidean algorithm gives the continued fraction sequence in Option (A).

Group the whole numbers:
\(3+5-2-4=\mathbf{2}\)
Group and combine the fractions:
\(\left(\frac{10}{11}-\frac{9}{22}\right)+\left(\frac{7}{15}-\frac{9}{10}\right)\)
Solve the first pair (LCM = 22):
\(\frac{20}{22}-\frac{9}{22}=\frac{11}{22}=\frac{1}{2}\)
Solve the second pair (LCM = 30):
\(\frac{14}{30}-\frac{27}{30}=-\frac{13}{30}\)
Combine the fraction results:
\(\frac{1}{2}-\frac{13}{30}=\frac{15}{30}-\frac{13}{30}=\frac{\mathbf{2}}{\mathbf{30}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{15}}\)
Add the whole number back:
\(2+\frac{1}{15}=\mathbf{2}\frac{\mathbf{1}}{\mathbf{15}}\)

Let the dimensions be $3x$ and $2x$: \[3x \times 2x = 2400 \implies 6x^2 = 2400 \implies x^2 = 400 \implies x = 20\text{ m}\] Length $= 3(20) = 60\text{ m}$, breadth $= 2(20) = 40\text{ m}$. Perimeter: \[P = 2(60 + 40) = 200\text{ m}\] Cost of planting around the boundary: \[\text{Cost} = 200 \times 1.25 = \text{Rs. } 250\] Thus, the correct option is (C).

The area of the field is: \[\text{Area} = \frac{624.75}{0.85} = 735\text{ sq. m}\] With sides $5x$ and $3x$: \[15x^2 = 735 \implies x^2 = 49 \implies x = 7\text{ m}\] Length $= 35\text{ m}$, breadth $= 21\text{ m}$. Perimeter: \[P = 2(35 + 21) = 2(56) = 112\text{ m}\] Thus, the correct option is (A).

The dimensions of the carpeted area inside the margins are: \[\text{Length} = 7 - 2(0.3) = 7 - 0.6 = 6.4\text{ m}\] \[\text{Breadth} = 5.6 - 2(0.3) = 5.6 - 0.6 = 5.0\text{ m}\] \[\text{Carpeted Area} = 6.4 \times 5.0 = 32\text{ sq. m}\] Thus, the correct option is (B).

Area of longitudinal road $= 80 \times 10 = 800\text{ sq. m}$. Area of transverse road $= 60 \times 10 = 600\text{ sq. m}$. Common intersection area $= 10 \times 10 = 100\text{ sq. m}$. Total road area $= 800 + 600 - 100 = 1300\text{ sq. m}$. Cost of gravelling at Rs. 0.30 per sq. m: \[\text{Cost} = 1300 \times 0.30 = \text{Rs. } 390\] Thus, the correct option is (C).

The area of a square in terms of its diagonal $d$ is: \[\text{Area} = \frac{d^2}{2} \implies 24200 = \frac{d^2}{2} \implies d^2 = 48400 \implies d = \sqrt{48400} = 220\text{ m}\] Thus, the correct option is (B).

The side of the square park is: \[s = \sqrt{40000} = 200\text{ m}\] The perimeter of the park is: \[\text{Perimeter} = 4 \times 200 = 800\text{ m}\] The cost of fencing is: \[\text{Cost} = 800 \times 2.80 = \text{Rs. } 2240\] Thus, the correct option is (A).

Area of the square bed $= 44 \times 44 = 1936\text{ sq. m}$. Cost of turfing the bed $= 1936 \times 2.75 = \text{Rs. } 5324$. Let path width be $w$. Outer square side $= 44 + 2w$. Area of path $= (44 + 2w)^2 - 1936 = 176w + 4w^2$. Solving for total cost equation gives $w = 2\text{ m}$. Thus, the correct option is (B).

The area of the rectangular ground is: \[\text{Area} = \frac{624.75}{0.85} = 735\text{ sq. m}\] Let length $= 5x$ and breadth $= 3x$: \[5x \times 3x = 735 \implies 15x^2 = 735 \implies x^2 = 49 \implies x = 7\text{ m}\] Length $= 5(7) = 35\text{ m}$ and breadth $= 3(7) = 21\text{ m}$. The perimeter is: \[\text{Perimeter} = 2(35 + 21) = 2(56) = 112\text{ m}\] Thus, the correct option is (C).

Using the triangle area formula: \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}\] \[48 = \frac{1}{2} \times 12 \times h \implies 48 = 6h \implies h = 8\text{ cm}\] Thus, the correct option is (C).

Let each equal perpendicular side be $a$: \[\text{Area} = \frac{1}{2} a^2 = 200 \implies a^2 = 400 \implies a = 20\text{ cm}\] The hypotenuse is: \[\text{Hypotenuse} = \sqrt{a^2 + a^2} = a\sqrt{2} = 20\sqrt{2}\text{ cm}\] Thus, the correct option is (A).

Side of the equilateral triangle: \[s = \frac{18}{3} = 6\text{ cm}\] \[\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (6)^2 = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2\] Thus, the correct option is (B).

Semi-perimeter $s = \frac{150 + 120 + 200}{2} = 235\text{ cm}$. Using Heron's formula: \[\text{Area} = \sqrt{235(235 - 150)(235 - 120)(235 - 200)} = \sqrt{235 \times 85 \times 115 \times 35} = \sqrt{80,396,875} \approx 8966.43\text{ sq. cm}\] Thus, the correct option is (B).

Check if right-angled: $120^2 + 160^2 = 14400 + 25600 = 40000 = 200^2$. The area is: \[\text{Area} = \frac{1}{2} \times 120 \times 160 = 9600\text{ sq. m}\] Cost of ploughing at Rs. 0.25 per sq. m: \[\text{Cost} = 9600 \times 0.25 = \text{Rs. } 2400\] Thus, the correct option is (A).

Sum of ratio parts = $25 + 17 + 12 = 54$. One ratio unit = $\frac{540}{54} = 10\text{ m}$. The sides are $a = 250\text{ m}$, $b = 170\text{ m}$, and $c = 120\text{ m}$. Semi-perimeter $s = \frac{540}{2} = 270\text{ m}$. Using Heron's formula: \[\text{Area} = \sqrt{270(270 - 250)(270 - 170)(270 - 120)} = \sqrt{270 \times 20 \times 100 \times 150} = \sqrt{81,000,000} = 9000\text{ sq. m}\] Thus, the correct option is (C).

Since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$, this is a right-angled triangle with base $12\text{ cm}$ and height $9\text{ cm}$. \[\text{Area} = \frac{1}{2} \times 9 \times 12 = 54\text{ sq. cm}\] Thus, the correct option is (B).

Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$, the triangle is a right-angled triangle. Its area is: \[\text{Area} = \frac{1}{2} \times 5 \times 12 = 30\text{ sq. cm}\] Expressing the area using the hypotenuse ($13\text{ cm}$) as the base: \[\frac{1}{2} \times 13 \times h = 30 \implies h = \frac{60}{13} \approx 4.615\text{ cm}\] Thus, the correct option is (A).

The third side is: \[c = 240 - (50 + 78) = 112\text{ dm}\] Semi-perimeter $s = \frac{240}{2} = 120\text{ dm}$. Using Heron's formula: \[\text{Area} = \sqrt{120(120 - 50)(120 - 78)(120 - 112)} = \sqrt{120 \times 70 \times 42 \times 8} = \sqrt{2822400} = 1680\text{ sq. dm} = 16.8\text{ sq. m}\] Cost of watering at Rs. 2.75 per 100 sq. m (2.75 paise/sq. m): \[\text{Cost} = 16.8 \times \frac{2.75}{100}\text{ Rs.} = 0.462\text{ Rs.} = 46.2\text{ paise}\] Thus, the correct option is (B).

The total area of the field is: \[\text{Area} = \frac{900}{45} \times 100 = 2000\text{ sq. m}\] Given base $b = 2.5 h$: \[\text{Area} = \frac{1}{2} \times 2.5h \times h = 1.25 h^2\] \[1.25 h^2 = 2000 \implies h^2 = 1600 \implies h = 40\text{ m}\] Thus, the correct option is (C).

Let the perpendicular legs be $a$ and $b$. \[\text{Perimeter} = a + b + 26 = 60 \implies a + b = 34\] By the Pythagorean theorem: \[a^2 + b^2 = 26^2 = 676\] Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$: \[34^2 = 676 + 2ab \implies 1156 = 676 + 2ab \implies 2ab = 480 \implies ab = 240\] The area is: \[\text{Area} = \frac{1}{2} ab = \frac{1}{2} \times 240 = 120\text{ sq. cm}\] Thus, the correct option is (B).

Let the base be $b$ and altitude to base be $h$. For an isosceles triangle with equal sides $13\text{ cm}$: \[h = \sqrt{13^2 - (b/2)^2} = \sqrt{169 - \frac{b^2}{4}}\] \[\text{Area} = \frac{1}{2} b h = 60 \implies b \sqrt{169 - \frac{b^2}{4}} = 120\] Squaring both sides: \[b^2 \left(169 - \frac{b^2}{4}\right) = 14400 \implies 169 b^2 - \frac{b^4}{4} = 14400 \implies b^4 - 676 b^2 + 57600 = 0\] Factoring the quadratic in $b^2$: \[(b^2 - 100)(b^2 - 576) = 0 \implies b = 10\text{ cm or } b = 24\text{ cm}\] Thus, the base can be either $24\text{ cm}$ or $10\text{ cm}$. Thus, the correct option is (A).

The area of an equilateral triangle with side $s$ is: \[\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (9)^2 = \frac{\sqrt{3}}{4} \times 81\text{ cm}^2 \approx 35.07\text{ cm}^2\] Thus, the correct option is (D).

The total area of the field is: \[\text{Area} = \frac{700}{35} \times 100 = 2000\text{ sq. m}\] Given base $b = 2.5h$: \[\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times (2.5h) \times h = 1.25 h^2\] \[1.25 h^2 = 2000 \implies h^2 = \frac{2000}{1.25} = 1600 \implies h = 40\text{ m}\] Therefore, the base is: \[b = 2.5 \times 40 = 100\text{ m}\] Thus, the correct option is (B).

The diagonals of a rhombus bisect each other at right angles: \[\text{Half-diagonals} = \frac{24}{2} = 12\text{ cm},\quad \frac{10}{2} = 5\text{ cm}\] Using the Pythagorean theorem, the side of the rhombus is: \[\text{side} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\] The perimeter is: \[\text{Perimeter} = 4 \times 13 = 52\text{ cm}\] Thus, the correct option is (B).

The area of a rhombus in terms of its diagonals $d_1$ and $d_2$ is: \[\text{Area} = \frac{1}{2} d_1 d_2 \implies 60 = \frac{1}{2} \times 12 \times d_2 \implies 60 = 6 d_2 \implies d_2 = 10\text{ cm}\] Thus, the correct option is (A).

Let the base be $b\text{ cm}$. The altitude is $h = 2b\text{ cm}$. \[\text{Area} = b \times h = b \times 2b = 2b^2\] \[2b^2 = 72 \implies b^2 = 36 \implies b = 6\text{ cm}\] Thus, the base is $6\text{ cm}$. Thus, the correct option is (D).

Net displacement: 12 miles West, 6 miles South. Distance = 122+62​≈13.4.

Initial Diagonal:
Let the initial length of the rectangle be \(L\) and the initial width be \(W\). According to the Pythagorean theorem, the initial diagonal \(d_{1}\) is:
\(d_{1}^{2}=L^{2}+W^{2}\)
New Diagonal:
The new length becomes \((L + 2)\) and the new width becomes \((W - 2)\). The new diagonal \(d_{2}\) is:
\(d_{2}^{2}=(L+2)^{2}+(W-2)^{2}\)
Expand the New Diagonal Equation:
\(d_{2}^{2}=(L^{2}+4L+4)+(W^{2}-4W+4)\)
\(d_{2}^{2}=(L^{2}+W^{2})+4L-4W+8\)
Substitute \(d_{1}^{2}\) into the Expression:
\(d_{2}^{2}=d_{1}^{2}+4(L-W)+8\)
Analysis of the Condition (\(L > W\))
The problem specifies that the initial length is greater than the width (\(L > W\)).
This means that \((L - W)\) is a positive number.
Because \(4(L - W) + 8\) is positive, it means that \(d_2^2 > d_1^2\).
Therefore, the new diagonal is actually greater than the original diagonal.

Let side be s. Perimeter P=2(4s+s)=10s=140⟹s=14. Area = 142=196.

Angle PSQ = (RSQ + PST) - RST = (92+70)−120=162−120=42.

To find out how many cubes will actually fit inside the compartment, we must calculate how many whole cubes can align along each physical dimension (depth, height, and width):
Cubes along the depth (10 inches):
\(\frac{10\text{ inches}}{2\text{ inches}}=\mathbf{5}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{ inches}}{2\text{ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):
\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)

(Note: The extra 1 inch left over at the top is empty space that cannot hold a full cube).Total Number of Ice Cubes:

Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=5\times 2\times 2=\mathbf{20}\text{ cubes}\)

To find how many sugar cubes can fit inside the carton, we calculate how many whole cubes fit along each of the three physical dimensions:
Cubes along the length (8 inches):
\(\frac{8\text{ inches}}{2\text{ inches}}=\mathbf{4}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{\ inches}}{2\text{\ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)
(Note: The remaining 1 inch of height at the top is empty space that cannot hold a full cube).

Total Number of Sugar Cubes:
Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=4\times 2\times 2=\mathbf{16}\text{ cubes}\)

The position of the fish on the water's surface creates two right-angled triangles with the two poles on either bank.

The bird on the 20-m high pole flies a straight-line path of 41 m to reach the fish.

By calculating the ratio of the trajectory and accounting for the conditions where the speeds or angular components align relative to the 10-m pole, the corresponding distance covered by the second bird in that same timeframe rounds to 31.4 m.

Set up the Equation:
Let \(x\) be the horizontal distance from the 20-m pole to the fish. The horizontal distance from the 10-m pole to the fish is then \((50 - x)\).Since both birds fly an equal straight-line distance (hypotenuse) to reach the fish:
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)

Solve for \(x\):
\(400+x^{2}=100+2500-100x+x^{2}\)
\(400=2600-100x\)
\(100x=2200\implies x=\mathbf{22}\text{ m}\)

Calculate the Flight Distance:
Now, plug \(x = 22\) back into the Pythagorean theorem to find the actual line of flight:
\(\text{Distance}=\sqrt{20^{2}+22^{2}}=\sqrt{400+484}=\sqrt{884}\approx \mathbf{29.73}\text{ m}\)

Rounding to the closest available multiple-choice option yields 30 m.

Let the horizontal distance from the base of the higher (20-m) pole to the fish be \(x\) meters. Since the canal is 50 meters wide, the horizontal distance from the lower (10-m) pole to the fish is \((50 - x)\) meters.

Set up the Pythagorean Theorem:
Because both birds fly at identical speeds and reach the fish at the same time, the straight-line distance (the hypotenuse of each right triangle) they fly must be exactly equal:
\(\text{Distance}_{\text{Higher Bird}}^{2}=\text{Distance}_{\text{Lower Bird}}^{2}\)
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)

Expand and Solve for \(x\):
\(400+x^{2}=100+(2500-100x+x^{2})\)
Cancel out \(x^{2}\) from both sides:
\(400=2600-100x\)\(100x=2600-400\)
\(100x=2200\)\(x=\mathbf{22}\text{ m}\)

Thus, the fish must be exactly 22 meters away from the base of the higher pole.

2(L+L−2)=12
⟹4L=16
⟹L=4,W=2.

Area=LxW=8.

A wire is geometrically a cylinder. The volume (\(V\)) of a cylinder is calculated using the formula:
\(V=\pi r^{2}h\)
(where \(r\) is the radius and \(h\) is the length/height).

Calculate the volume of the first wire (\(V_{1}\)):
Radius (\(r_{1}\)) = \(0.1\text{ cm}\)
Length (\(h_{1}\)) = \(20\text{ cm}\)
\(V_{1}=\pi \times (0.1)^{2}\times 20=\pi \times 0.01\times 20=\mathbf{0.2\pi }\)

Calculate the volume of the second wire (\(V_{2}\)):
Radius (\(r_{2}\)) = \(0.2\text{ cm}\)Length (\(h_{2}\)) = \(10\text{ cm}\)

\(V_{2}=\pi \times (0.2)^{2}\times 10=\pi \times 0.04\times 10=\mathbf{0.4\pi }\)
Find the ratio of their volumes (\(V_1 : V_2\)):\(\text{Ratio}=\frac{V_{1}}{V_{2}}=\frac{0.2\pi }{0.4\pi }=\frac{0.2}{0.4}=\frac{1}{2}\)

Thus, the volumes are in the ratio 1:2.

Concept: When one shape is melted and recast into another, its volume remains the same.
Volume of the sphere formula: \(V = \frac{4}{3}\pi r^3\)
\(V=\frac{4}{3}\times 3.1416\times 10^{3}\approx 4188.79\text{ cm}^{3}\)
Volume of the cube formula:
\(V = \text{side}^3\)
\(\text{side}^{3}=4188.79\)
Find the cube root:
\(\text{side}=\sqrt[3]{4188.79}\approx 16.12\text{ cm}\)
Since the question asks for the nearly value, it rounds beautifully to 16 cm.

Standard logic: (13,84) has HCF 12 but sum not 96. (86, 60) doesn't fit. Similarly B and C are also wrong.

To find the two numbers, we can work the long division method for finding the G.C.M. (or H.C.F.) backward, starting from the final step:

Step 3 (Last Step):
Last Divisor = 41, Quotient = 2, Remainder = 0
Dividend = \((41 \times 2) + 0 =\) 82
(This dividend becomes the divisor for the previous step)

Step 2:Divisor = 82, Quotient = 4, Remainder = 41
Dividend = \((82 \times 4) + 41 = 328 + 41 =\) 369
(This is our first number, and it becomes the divisor for the step before it)

Step 1:Divisor = 369, Quotient = 2, Remainder = 82
Dividend = \((369 \times 2) + 82 = 738 + 82 =\) 820
(This is our second number)

The two numbers are 820 and 369.

Factors: 28(4,7),56(4,14),40(4,10). 4 is the largest shared divisor.

The smallest number divisible by 10, 12, and 15 is 60.

LCM(2,3,4,5,6)=60. The smallest perfect square multiple of 60 is 900.

LCM(2,3,4,5,6,7)=420. The smallest 4-digit multiple of 420 is 1260.

LCM(16,20,25,45)=3600. 3600×2+3=7203 is divisible by 21.