Question 1

It costs ₹ $x$ each to make the first thousand copies of a compact disc and ₹ $y$ to make each subsequent copy. If $z$ is greater than 1000, how much will it cost to make $z$ copies of the compact disc?

Accepted Answer

### Concept & Logic Split the total quantity into two distinct tiers and calculate the cost for each tier separately. Then, combine the costs and factor the algebraic expression to match the given options. $$Total\ Cost = (Cost\ of\ First\ Tier) + (Cost\ of\ Remaining\ Tier)$$ ### Step-by-Step Solution * **Given:** * Cost for first 1000 copies = ₹ $x$ per copy * Cost for copies after 1000 = ₹ $y$ per copy * Total copies = $z$ (where $z > 1000$) * **Calculation:** 1. Calculate the total cost for the first 1000 copies: $Cost_1 = 1000 	imes x = 1000x$ 2. Determine the number of subsequent copies made: $Remaining\ copies = z - 1000$ 3. Calculate the cost for these remaining copies: $Cost_2 = (z - 1000) 	imes y = yz - 1000y$ 4. Add the two costs together for the total cost: $Total\ Cost = 1000x + yz - 1000y$ 5. Rearrange and factor the expression to match the options: Factor out 1000 from the $x$ and $y$ terms: $1000(x - y) + yz$ ### Exam Strategy & Shortcut When algebraic manipulation gets confusing, use the substitution method. Pick simple, concrete numbers. Let $x = 10$, $y = 5$, and $z = 1001$. Cost = $(1000 	imes 10) + (1 	imes 5) = 10000 + 5 = 10005$. Now plug these values into the options: (a) $(1001	imes10) - (1001	imes5) = 10010 - 5005 = 5005$ (Incorrect) (b) $1000(10) + (5	imes1001) = 10000 + 5005 = 15005$ (Incorrect) (c) $1000(10 - 5) + (5	imes1001) = 5000 + 5005 = 10005$ (Matches perfectly) ### Common Pitfall Students often correctly formulate $1000x + y(z - 1000)$ but fail to recognize how it expands and regroups into option (c). Don't assume you made a mistake if your initial expression isn't listed; expand the options to check for equivalence. ### Final Answer **Therefore, the correct answer is $1000(x - y) + yz$.**

Question 2

If $\frac{2}{2 + \frac{2}{2 + \frac{2}{x}}} = 3$, then $x$ is equal to

Accepted Answer

### Concept & Strategy Unlike standard continued fraction problems where you simplify from the bottom up, this is an **Inverse Continued Fraction Equation**. Because the unknown $x$ is buried at the very bottom, but the total value of the expression is known (it equals 3), the fastest strategy is to solve from the **top down**. ### Step-by-Step Solution We will isolate the terms layer by layer, starting from the outermost shell. * **Step 1: Isolate the main denominator** The equation is $\frac{2}{	ext{Denominator}} = 3$. Rearranging this gives: $$	ext{Denominator} = \frac{2}{3}$$ So, $2 + \frac{2}{2 + \frac{2}{x}} = \frac{2}{3}$ * **Step 2: Isolate the next nested fraction** Subtract 2 from both sides: $$\frac{2}{2 + \frac{2}{x}} = \frac{2}{3} - 2$$ $$\frac{2}{2 + \frac{2}{x}} = \frac{2 - 6}{3} = -\frac{4}{3}$$ * **Step 3: Isolate the next denominator** We have $\frac{2}{	ext{Denominator}} = -\frac{4}{3}$. Rearranging gives $	ext{Denominator} = 2 	imes \left(-\frac{3}{4}ight) = -\frac{6}{4} = -\frac{3}{2}$. So, $2 + \frac{2}{x} = -\frac{3}{2}$ * **Step 4: Solve for $x$** Subtract 2 from both sides: $$\frac{2}{x} = -\frac{3}{2} - 2$$ $$\frac{2}{x} = \frac{-3 - 4}{2} = -\frac{7}{2}$$ Invert both sides and multiply to find $x$: $$\frac{x}{2} = -\frac{2}{7}$$ $$x = 2 	imes \left(-\frac{2}{7}ight) = -\frac{4}{7}$$ ### Exam Strategy & Shortcut The top-down approach *is* the shortcut. If you try to simplify this from the bottom up (by finding a common denominator for $2 + 2/x$, then inverting, then adding 2, etc.), you will create massive, algebraically heavy expressions that are incredibly prone to silly mistakes. Working backwards with simple integers is always faster. ### Common Pitfall The biggest danger here is losing track of negative signs. In Step 2, subtracting 2 from $2/3$ yields a negative number ($-4/3$). If you accidentally write positive $4/3$, you will end up with an incorrect positive fraction for $x$. Always double-check your sign arithmetic when moving terms across the equals sign. ### Final Answer Therefore, the correct answer is **-\frac{4}{7}**.

Question 3

If $x = y = 2z$ and $xyz = 256$, then $x$ is equal to

Accepted Answer

### Concept & Strategy This is an **Algebraic Substitution** problem. The strategy is to express all variables ($y$ and $z$) in terms of the target variable ($x$) using the given linear equality. This converts the multi-variable product equation into a single-variable cubic equation. ### Step-by-Step Solution * **Step 1: Express variables in terms of $x$** We are given $x = y = 2z$. From this, we can isolate $y$ and $z$ in terms of $x$: $$y = x$$ $$2z = x \Rightarrow z = \frac{x}{2}$$ * **Step 2: Substitute into the product equation** We are given $xyz = 256$. Substitute $y$ and $z$ with their $x$-equivalents: $$x \cdot (x) \cdot \left(\frac{x}{2}ight) = 256$$ * **Step 3: Solve for $x$** Multiply the terms: $$\frac{x^3}{2} = 256$$ Multiply both sides by 2: $$x^3 = 512$$ Take the cube root of both sides. Since $8 	imes 8 	imes 8 = 512$: $$x = 8$$ ### Exam Strategy & Shortcut For equations with neat integer options, **Option Verification** is incredibly fast. Test the options directly: If $x = 8$ (Option c), then $y = 8$. Since $2z = 8$, $z = 4$. Check the product: $8 	imes 8 	imes 4 = 64 	imes 4 = 256$. It matches perfectly! You can bypass the algebraic solving entirely. ### Common Pitfall A very common mistake is solving for the wrong variable. Students might manipulate the equations to solve for $z$, find $z = 4$, see '4' in the options (Option b), and instantly select it without realizing the question explicitly asked for $x$. Always double-check what the question is requesting. ### Final Answer Therefore, the correct answer is **8**.

Question 4

If $2x^2 + 12x + 18 = 0$, what is the value of $x$?

Accepted Answer

### Concept & Formula To solve a quadratic equation of the form $ax^2 + bx + c = 0$, we can simplify the equation by dividing by the greatest common divisor of the coefficients, and then factorize it or use the perfect square formula. ### Step-by-Step Solution Given equation: $$2x^2 + 12x + 18 = 0$$ Divide the entire equation by 2: $$x^2 + 6x + 9 = 0$$ Recognize the expression as a perfect square $(a + b)^2 = a^2 + 2ab + b^2$: $$(x + 3)^2 = 0$$ Taking the square root on both sides: $$x + 3 = 0 \implies x = -3$$ ### Exam Strategy & Shortcut Instead of solving completely, look at the signs. Since all signs in the equation are positive ($+12x, +18$), the value of $x$ must be negative to cancel them out. This immediately eliminates choices (c) and (d). Test $x = -3$: $$2(-3)^2 + 12(-3) + 18 = 18 - 36 + 18 = 0$$ This confirms the answer efficiently. ### Common Pitfall Students often confuse the sign during factorization, writing $(x - 3)^2 = 0$ instead of $(x + 3)^2 = 0$, leading to an incorrect answer of $3$. Always double-check by expanding your factored expression. ### Final Answer **Therefore, the correct answer is - 3.**

Question 5

The value of $\frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} + \frac{17}{8^2 \cdot 9^2} + \frac{19}{9^2 \cdot 10^…

Accepted Answer

### Concept & Formula This complex-looking sequence is a cleverly disguised telescoping series. Notice the relationship between the numerator and the factors in the denominator: the numerator is exactly the difference between the two squared factors in the denominator. Mathematically, the general term can be written and split as follows: $$T_n = \frac{(n+1)^2 - n^2}{n^2 \cdot (n+1)^2} = \frac{(n+1)^2}{n^2 \cdot (n+1)^2} - \frac{n^2}{n^2 \cdot (n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$$ ### Step-by-Step Solution Apply this splitting logic to every term in the given series: First term: $\frac{3}{1^2 \cdot 2^2} = \frac{4 - 1}{1^2 \cdot 2^2} = \frac{1}{1^2} - \frac{1}{2^2}$ Second term: $\frac{5}{2^2 \cdot 3^2} = \frac{9 - 4}{2^2 \cdot 3^2} = \frac{1}{2^2} - \frac{1}{3^2}$ Third term: $\frac{7}{3^2 \cdot 4^2} = \frac{16 - 9}{3^2 \cdot 4^2} = \frac{1}{3^2} - \frac{1}{4^2}$ $$\vdots$$ Last term: $\frac{19}{9^2 \cdot 10^2} = \frac{100 - 81}{9^2 \cdot 10^2} = \frac{1}{9^2} - \frac{1}{10^2}$ Now, write out the sum of all these rewritten terms: $$	ext{Sum} = \left(\frac{1}{1^2} - \frac{1}{2^2}ight) + \left(\frac{1}{2^2} - \frac{1}{3^2}ight) + \dots + \left(\frac{1}{9^2} - \frac{1}{10^2}ight)$$ Every adjacent fractional pair cancels out ($-\frac{1}{2^2}$ cancels $+\frac{1}{2^2}$, and so on). The only remaining components are the first positive fraction and the last negative fraction: $$	ext{Sum} = \frac{1}{1^2} - \frac{1}{10^2}$$ $$	ext{Sum} = 1 - \frac{1}{100}$$ $$= \frac{99}{100}$$ ### Exam Strategy & Shortcut Whenever you see a series where the numerator equals the difference of the denominator's factors (e.g., $Numerator = b - a$ for a denominator $a \cdot b$), the entire series evaluates to: $$	ext{Sum} = \frac{1}{	ext{First Factor}} - \frac{1}{	ext{Last Factor}}$$ Here, the first factor is $1^2 = 1$ and the last factor is $10^2 = 100$. So, $	ext{Sum} = \frac{1}{1} - \frac{1}{100} = \frac{99}{100}$. You can solve this visually in under 5 seconds. ### Common Pitfall Students frequently overlook the relationship between the numerator and denominator, assuming the problem requires massive brute-force fraction addition, leading them to either skip the question or waste minutes calculating huge LCMs. ### Final Answer **Therefore, the correct answer is 99/100.**

Question 6

Mayank, Mirza, Little and Jaspal bought a motorbike for ₹ 60000. Mayank paid one-half of the sum of the amounts paid by the other boys, Mirza paid one-third of the sum of the amounts paid by the other boys and Little paid one-fourth of the sum of th…

Accepted Answer

### Concept & Strategy The problem tests proportional division of a total sum. The key algebraic insight is that if a person pays a fraction of what the *others* paid, you can easily find their share of the *total* amount. If a person pays $\frac{1}{x}$ of the others, they pay $\frac{1}{x+1}$ of the total sum. ### Step-by-Step Solution * **Given:** Total cost of the motorbike $T = 60000$. * **Calculate Mayank's share:** Mayank pays $\frac{1}{2}$ of the others. Let Mayank's share be $M$. The others paid $(T - M)$. $M = \frac{1}{2}(T - M) \Rightarrow 3M = T \Rightarrow M = \frac{60000}{3} = 20000$. * **Calculate Mirza's share:** Mirza pays $\frac{1}{3}$ of the others. $Mi = \frac{1}{4}T = \frac{60000}{4} = 15000$. * **Calculate Little's share:** Little pays $\frac{1}{4}$ of the others. $L = \frac{1}{5}T = \frac{60000}{5} = 12000$. * **Calculate Jaspal's share:** Jaspal pays the remaining balance. $J = T - (M + Mi + L) = 60000 - (20000 + 15000 + 12000) = 60000 - 47000 = 13000$. ### Exam Strategy & Shortcut Use the ratio-to-fraction shortcut directly: "1/2 of others" $ightarrow \frac{1}{1+2} = \frac{1}{3}$ of Total. "1/3 of others" $ightarrow \frac{1}{1+3} = \frac{1}{4}$ of Total. "1/4 of others" $ightarrow \frac{1}{1+4} = \frac{1}{5}$ of Total. Total fraction paid by the first three = $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{20 + 15 + 12}{60} = \frac{47}{60}$. Jaspal's fraction = $1 - \frac{47}{60} = \frac{13}{60}$. Jaspal's amount = $\frac{13}{60} 	imes 60000 = 13000$. ### Common Pitfall A very common mistake is calculating the fractions based on the total amount directly (e.g., Mayank = 60000/2 = 30000). The prompt explicitly says "sum of the amounts paid by the *other* boys", not the total amount. ### Final Answer **Therefore, the correct answer is ₹ 13000.**

Question 7

If $x = y - \frac{y}{10}$, where $y$ is a positive integer which increases in value, then $x$

Accepted Answer

### Concept & Logic This problem tests the understanding of proportional linear equations. When a variable ($x$) is defined as a constant positive fraction of another variable ($y$), their growth directions will identical. ### Step-by-Step Solution * **Simplify the given equation:** The equation is given as $x = y - \frac{y}{10}$. We can factor out $y$: $$x = y \left(1 - \frac{1}{10}\right)$$ $$x = y \left(\frac{9}{10}\right)$$ $$x = 0.9y$$ * **Analyze the relationship:** The equation $x = 0.9y$ represents a direct, positive linear relationship between $x$ and $y$. The coefficient (0.9) is a positive constant. * **Determine the behavior:** The problem states that $y$ is a positive integer which *increases* in value. Since $x$ is simply $90\%$ of $y$, as $y$ grows larger, $x$ must proportionally grow larger as well. There are no negative signs or exponential constraints that would cause $x$ to reverse direction or decrease. ### Exam Strategy & Shortcut Plug in simple test values to immediately see the trend. Let $y = 10 \Rightarrow x = 10 - 1 = 9$ Let $y = 20 \Rightarrow x = 20 - 2 = 18$ Let $y = 30 \Rightarrow x = 30 - 3 = 27$ The values of $x$ (9, 18, 27) are strictly increasing. ### Common Pitfall Some students overthink the subtraction sign, assuming that subtracting a component ($\frac{y}{10}$) that is also growing will eventually cause the total value to shrink. They fail to realize that the base value ($y$) is growing 10 times faster than the subtracted component. ### Final Answer **Therefore, the correct answer is increases in value.**

Question 8

If $x$ is a positive number, then which of the following fractions has the greatest value?

Accepted Answer

### Concept & Logic This problem analyzes variables inside simple algebraic fractions. A fraction is maximized when its numerator is larger than its denominator (improper fraction value $> 1$). ### Step-by-Step Solution * **Analyze Option (a):** $\frac{x}{x} = 1$ (Always exactly equal to 1 for any positive number). * **Analyze Option (b):** $\frac{x}{x+1}$ (Since the numerator is smaller than the denominator, this value is always $< 1$). * **Analyze Option (c):** $\frac{x+1}{x} = 1 + \frac{1}{x}$ (Since the numerator is strictly greater than the denominator, this value is always $> 1$). * **Analyze Option (d):** $\frac{x+2}{x+3}$ (Since the numerator is smaller than the denominator, this value is always $< 1$). * **Conclusion:** Option (c) is the only fraction expression that produces a value strictly greater than 1, making it the greatest value among all choices. ### Exam Strategy & Shortcut Plug in a basic integer value for $x$, such as $x = 1$: Option (a): $\frac{1}{1} = 1$ Option (b): $\frac{1}{2} = 0.5$ Option (c): $\frac{2}{1} = 2$ Option (d): $\frac{3}{4} = 0.75$ Comparing the absolute values shows that 2 is unambiguously the greatest value. ### Common Pitfall Overcomplicating the variable structure with derivative inequalities can lead to wasted time. Simple classification into proper ($<1$) and improper ($>1$) states resolves the expression relationship immediately. ### Final Answer **Therefore, the correct answer is (x + 1) / x.**

Question 9

Sanket earns twice as much in the month of March as in each of the other months of the year. What part of his entire annual earnings was earned in March?

Accepted Answer

### Concept & Logic There are 12 months in a year. Let the earnings of a normal month be $x$. March is an exception where the earnings are $2x$. $$	ext{Total Annual Earnings} = 	ext{Earnings in March} + 	ext{Earnings in remaining 11 months}$$ $$	ext{Required Fraction} = \frac{	ext{Earnings in March}}{	ext{Total Annual Earnings}}$$ ### Step-by-Step Solution * **Define monthly earnings:** Let the earnings for each of the other 11 months = $x$ Therefore, earnings for March = $2x$ * **Calculate total yearly earnings:** $$	ext{Total Earnings} = 2x + (11 	imes x) = 13x$$ * **Calculate the proportion for March:** $$	ext{Fraction for March} = \frac{2x}{13x} = \frac{2}{13}$$ ### Exam Strategy & Shortcut Assume a convenient value instead of $x$. Let the earnings of a regular month be ₹ 1. * Earnings for 11 months = ₹ 11 * Earnings for March = ₹ 2 * Total annual income = $11 + 2 = 13$ * March fraction = $\frac{2}{13}$ ### Common Pitfall Forgetting that March replaces one of the 12 months, leading to an incorrect total computation like $12x + 2x = 14x$, which gives a wrong fraction of $\frac{2}{14} = \frac{1}{7}$. ### Final Answer **Therefore, the correct answer is \frac{2}{13}.**

Question 10

A tank is $\frac{2}{5}$ full. If 16 litres of water is added to the tank, it becomes $\frac{6}{7}$ full. The capacity of the tank is

Accepted Answer

### Concept & Formula This problem tests the application of linear equations to solve fraction-based word problems. The core insight is that the mathematical difference between the two fractions of the tank's capacity corresponds exactly to the physical volume of water added. Let the total capacity of the tank be represented by $x$. ### Step-by-Step Solution * **Given:** Initial volume = $\frac{2}{5}x$ Final volume = $\frac{6}{7}x$ Volume added = 16 litres. * **Calculation:** Set up the equation based on the change in volume: $$ \frac{6}{7}x - \frac{2}{5}x = 16 $$ * Find a common denominator for 7 and 5, which is 35, to subtract the fractions: $$ \frac{30x}{35} - \frac{14x}{35} = 16 $$ $$ \frac{16x}{35} = 16 $$ * Solve for $x$ by multiplying both sides by $\frac{35}{16}$: $$ x = 16 	imes \frac{35}{16} $$ $$ x = 35 $$ ### Exam Strategy & Shortcut Use the **LCM & Divisibility Method**. The total capacity must be a value that interacts cleanly with the denominators given in the problem (5 and 7) to yield whole numbers. The Least Common Multiple (LCM) of 5 and 7 is 35. Looking at the options (28, 32, 35, 42), only 35 is perfectly divisible by 5. You can immediately identify 35 as the highly probable correct answer without doing the full algebra, saving precious seconds. ### Common Pitfall A frequent mistake is setting up the equation backwards (subtracting the larger fraction from the smaller one) resulting in a negative capacity. Always double-check which fraction represents the "more full" state to ensure your resulting difference is positive. ### Final Answer Therefore, the correct answer is **35 litres**.

Question 11

The fluid contained in a bucket can fill four large bottles or seven small bottles. A full large bottle is used to fill an empty small bottle. What fraction of the fluid is left over in the large bottle when the small one is full?

Accepted Answer

### Concept & Logic This is a comparative volume problem solvable through algebraic substitution or the LCM ratio method. By equating the total volume of the bucket in terms of both large and small bottles, we can establish a direct volume relationship (ratio) between one large bottle and one small bottle. ### Step-by-Step Solution * **Given:** 1 Bucket = 4 Large Bottles ($L$) 1 Bucket = 7 Small Bottles ($S$) Therefore, $4L = 7S$. * **Calculation:** Express the volume of a Large bottle in terms of a Small bottle: $$ L = \frac{7}{4}S $$ * A full large bottle ($L$) is poured into an empty small bottle until the small bottle is full (volume $S$ removed). Find the remaining volume in the large bottle: $$ 	ext{Remaining Volume} = L - S $$ $$ 	ext{Remaining Volume} = \frac{7}{4}S - S = \frac{3}{4}S $$ * Calculate what fraction this remaining fluid is relative to the original large bottle's capacity ($L$): $$ 	ext{Fraction Left} = \frac{	ext{Remaining Volume}}{	ext{Original Volume } L} $$ $$ 	ext{Fraction Left} = \frac{\frac{3}{4}S}{\frac{7}{4}S} $$ $$ 	ext{Fraction Left} = \frac{3}{4} 	imes \frac{4}{7} = \frac{3}{7} $$ ### Exam Strategy & Shortcut Use the **LCM Value Assumption Method**. Let the capacity of the bucket be the LCM of 4 and 7, which is 28 units. Capacity of 1 Large bottle = $\frac{28}{4} = 7$ units. Capacity of 1 Small bottle = $\frac{28}{7} = 4$ units. If you pour from the Large (7 units) to fill the Small (4 units), you are left with $7 - 4 = 3$ units in the Large bottle. The fraction left over in the large bottle is simply $\frac{	ext{Leftover}}{	ext{Total Large}} = \frac{3}{7}$. This avoids fractions entirely! ### Common Pitfall A common misstep is calculating the fraction relative to the small bottle or the whole bucket, rather than the initial volume of the *large bottle*. The question asks: "What fraction of the fluid is left over *in the large bottle*". The denominator must be the large bottle's capacity. ### Final Answer Therefore, the correct answer is **$\frac{3}{7}$**.

Question 12

Peter gave one-fourth of the amount he had to Michael. Michael in turn gave half of what he received from Peter to Sam. If the difference between the remaining amount with Peter and the amount received by Sam is ₹ 500, how much money did Michael rec…

Accepted Answer

### Concept & Logic This problem tracks the distribution of an initial amount through a series of fractional transfers. The strategy is to define Peter's initial money as a variable, track the exact fraction held by each person at the end of the transactions, and use the given absolute difference (₹ 500) to solve for the base amount. ### Step-by-Step Solution * **Given:** Let Peter's initial amount be $x$. Peter gives $\frac{1}{4}x$ to Michael. * **Calculation:** Track the amounts held by everyone: Peter's remaining amount = $x - \frac{1}{4}x = \frac{3}{4}x$ Amount Michael receives = $\frac{1}{4}x$ Amount Sam receives = $\frac{1}{2}$ of Michael's amount = $\frac{1}{2} 	imes \frac{1}{4}x = \frac{1}{8}x$ * Apply the final condition: The difference between Peter's remaining amount and Sam's amount is 500. $$ \frac{3}{4}x - \frac{1}{8}x = 500 $$ * Find a common denominator to subtract the fractions: $$ \frac{6x}{8} - \frac{1x}{8} = 500 $$ $$ \frac{5x}{8} = 500 $$ $$ x = 500 	imes \frac{8}{5} = 800 $$ So, Peter's initial amount was ₹ 800. * The question asks for the amount Michael received from Peter: $$ 	ext{Michael's amount} = \frac{1}{4}x = \frac{1}{4}(800) = 200 $$ ### Exam Strategy & Shortcut Use the **LCM Assumption Method** to avoid working with fractions. Look at the denominators in the problem: one-fourth (4) and half (2). We also know Sam gets half of a fourth, which is an eighth (8). Assume Peter's starting amount is $8$ units. Peter gives Michael 2 units. Peter has $8 - 2 = 6$ units left. Michael gives Sam 1 unit. Sam has 1 unit. Difference between Peter's remaining (6) and Sam's (1) is 5 units. We are told this difference is ₹ 500. So, $5 	ext{ units} = 500 \implies 1 	ext{ unit} = 100$. Michael received 2 units from Peter. $2 	imes 100 = 200$. ### Common Pitfall Many students successfully calculate $x = 800$ (Peter's initial amount) and mistakenly choose 800 if it were an option, or get confused and mark "None of these" if it isn't. Always highlight the exact entity the question is asking for before selecting an option (in this case, what Michael received). ### Final Answer Therefore, the correct answer is **₹ 200**.

Question 13

A man spends one-quarter of his income on food, one-fifth of it on rent and remaining, which is ₹ 231, on other commodities.  Find his total income.

Accepted Answer

### Concept & Formula This is a classic basic fractions problem involving expenditure. Since both the food and rent expenses are expressed as fractions of the *total* income (not the remainder), we can simply add these fractions together to find the total fractional portion spent. The remainder can then be equated to the absolute monetary value given to find the total income. ### Step-by-Step Solution * **Given:** Food expense = $\frac{1}{4}$ of total. Rent expense = $\frac{1}{5}$ of total. Remaining balance = ₹ 231. Let total income be $x$. * **Calculation:** Find the total fraction of income spent: $$ 	ext{Total Spent Fraction} = \frac{1}{4} + \frac{1}{5} $$ $$ 	ext{Total Spent Fraction} = \frac{5}{20} + \frac{4}{20} = \frac{9}{20} $$ * Find the fraction representing the remaining income: $$ 	ext{Remaining Fraction} = 1 - \frac{9}{20} = \frac{11}{20} $$ * Equate this remaining fraction to the actual remaining amount to solve for $x$: $$ \frac{11}{20}x = 231 $$ $$ x = 231 	imes \frac{20}{11} $$ $$ x = 21 	imes 20 = 420 $$ ### Exam Strategy & Shortcut Use the **LCM Value Assumption**. The denominators are 4 and 5, so assume the total income is a multiple of 20 units. Food = $\frac{1}{4}$ of 20 = 5 units. Rent = $\frac{1}{5}$ of 20 = 4 units. Total spent = 9 units. Remaining = $20 - 9 = 11$ units. We are told the remaining amount is 231. Thus, $11 	ext{ units} = 231$, meaning $1 	ext{ unit} = 21$. Total income is 20 units, so $20 	imes 21 = 420$. ### Common Pitfall The main trap is treating the "one-fifth" rent expense as a fraction of the *remaining* income after food, rather than a fraction of the *total* income. The phrasing "one-fifth of **it**" refers back to the original subject, "his income." ### Final Answer Therefore, the correct answer is **₹ 420**.

Question 14

In a class, $\frac{3}{5}$ of the students are girls and rest are boys. If $\frac{2}{9}$ of the girls and $\frac{1}{4}$ of the boys are absent, what part of the total number of students is present?

Accepted Answer

### Concept & Strategy The problem asks for the fraction of the total students present. We can solve this by calculating the individual fractions of girls and boys who are present, and then finding their weighted sum relative to the total population. Alternatively, assuming a convenient plug-in number for the total number of students simplifies the calculation significantly. ### Step-by-Step Solution * **Given:** * Fraction of girls = $\frac{3}{5}$ * Fraction of boys = $1 - \frac{3}{5} = \frac{2}{5}$ * Fraction of absent girls = $\frac{2}{9} \implies$ Fraction of present girls = $1 - \frac{2}{9} = \frac{7}{9}$ * Fraction of absent boys = $\frac{1}{4} \implies$ Fraction of present boys = $1 - \frac{1}{4} = \frac{3}{4}$ * **Calculation:** * Total fraction of present students = (Fraction of girls $	imes$ Present girls) + (Fraction of boys $	imes$ Present boys) $$ 	ext{Present} = \left(\frac{3}{5} 	imes \frac{7}{9}ight) + \left(\frac{2}{5} 	imes \frac{3}{4}ight) $$ $$ 	ext{Present} = \frac{7}{15} + \frac{3}{10} $$ * Find a common denominator for 15 and 10, which is 30: $$ 	ext{Present} = \frac{14}{30} + \frac{9}{30} = \frac{23}{30} $$ ### Exam Strategy & Shortcut **Plug-in Number Method:** Choose a total number of students that is a multiple of the denominators: 5, 9, and 4. Let the total number of students be $5 	imes 9 	imes 4 = 180$. * Number of girls = $\frac{3}{5} 	imes 180 = 108$ * Number of boys = $180 - 108 = 72$ * Present girls = $108 	imes \left(1 - \frac{2}{9}ight) = 108 	imes \frac{7}{9} = 84$ * Present boys = $72 	imes \left(1 - \frac{1}{4}ight) = 72 	imes \frac{3}{4} = 54$ * Total present students = $84 + 54 = 138$ * Required fraction = $\frac{138}{180} = \frac{23}{30}$ ### Common Pitfall Students often calculate the fraction of *absent* students ($\frac{3}{5} 	imes \frac{2}{9} + \frac{2}{5} 	imes \frac{1}{4} = \frac{7}{30}$) and forget to subtract it from 1 to find the present students. Always double-check whether the question asks for the "present" or "absent" share. ### Final Answer **Therefore, the correct answer is 23/30.**

Question 15

Ganeshi's monthly income is twice Jassi's monthly income. Two-third of Jassi's monthly income is equal to Sukhvinder's monthly income. If Sukhvinder's annual income is ₹ 2.34 lakhs, what is Ganeshi's monthly income?

Accepted Answer

### Concept & Strategy The problem links three distinct entities through consecutive multipliers. The critical operational task is maintaining consistency between annual units and monthly units before performing the substitutions. ### Step-by-Step Solution * **Given:** * Ganeshi's monthly income ($G$) = $2 	imes$ Jassi's monthly income ($J$) * Sukhvinder's monthly income ($S$) = $\frac{2}{3} 	imes$ Jassi's monthly income ($J$) * Sukhvinder's annual income = ₹ 2,34,000 * **Calculation:** * Step 1: Find Sukhvinder's monthly income ($S$): $$ S = \frac{2,34,000}{12} = 	ext{₹ } 19,500 $$ * Step 2: Use the relation between $S$ and $J$ to find $J$: $$ 19,500 = \frac{2}{3} 	imes J \implies J = 19,500 	imes \frac{3}{2} = 9,750 	imes 3 = 	ext{₹ } 29,250 $$ * Step 3: Compute Ganeshi's monthly income ($G$): $$ G = 2 	imes J = 2 	imes 29,250 = 	ext{₹ } 58,500 $$ ### Exam Strategy & Shortcut Combine the variable equations down into a direct structural ratio layout: $$ \frac{G}{S} = \frac{2J}{\frac{2}{3}J} = 2 	imes \frac{3}{2} = 3 $$ This directly tells us that Ganeshi's monthly income is exactly 3 times Sukhvinder's monthly income. * Sukhvinder's monthly income = $\frac{2,34,000}{12} = 19,500$. * Ganeshi's monthly income = $19,500 	imes 3 = 58,500$. ### Common Pitfall A common mistake is using the annual figure (2.34 lakhs) directly without converting it to a monthly scale first, leading to massive scale mismatches in the options. ### Final Answer **Therefore, the correct answer is ₹ 58500.**

Question 16

Directions: Each question below consists of a number. Your task is to follow the following steps in order to find the result from the given alternatives.  Step I: Multiply the number by 3 and add the square of the number.  Step II: Divide the result…

Accepted Answer

### Concept & Logic This problem requires following a predefined mathematical algorithm strictly sequentially. When dealing with decimals, converting to fractions often helps maintain precision through intermediate steps. ### Step-by-Step Solution * **Given Input:** $x = 2.5$. It is often easier to use the fraction form $x = \frac{5}{2}$. * **Step I: Multiply the number by 3 and add the square of the number.** $$	ext{Result}_1 = 3(2.5) + (2.5)^2$$ $$	ext{Result}_1 = 7.5 + 6.25 = 13.75$$ * **Step II: Divide the result obtained after Step I by 5 and add 6.** $$	ext{Result}_2 = \frac{13.75}{5} + 6$$ $$	ext{Result}_2 = 2.75 + 6 = 8.75$$ * **Step III: Divide the result obtained after Step II by 3 and then square the resulting number.** $$	ext{Result}_3 = \left( \frac{8.75}{3} ight)^2$$ Convert $8.75$ to a fraction to make squaring easier: $8.75 = \frac{35}{4}$. $$	ext{Result}_3 = \left( \frac{\frac{35}{4}}{3} ight)^2 = \left( \frac{35}{12} ight)^2$$ $$	ext{Result}_3 = \frac{1225}{144}$$ Perform the final division: $$1225 \div 144 \approx 8.5069$$ Rounding to two decimal places gives $8.51$. ### Exam Strategy & Shortcut Avoid heavy fraction squaring by using smart decimal estimation at Step 3. $\frac{8.75}{3}$ is slightly less than 3 (since $\frac{9}{3} = 3$). Let's approximate it: $8.75 \div 3 \approx 2.916$. Since $3^2 = 9$, the answer must be slightly less than 9. Looking at the options, 8.10 is too low (that would require $2.84^2$), making 8.51 the only logical approximation. ### Common Pitfall Struggling with the decimal division in Step II ($13.75 \div 5$). A quick trick for dividing any number by 5 is to double the number and divide by 10. $13.75 	imes 2 = 27.5$. $27.5 \div 10 = 2.75$. ### Final Answer **Therefore, the correct answer is 8.51 (approx.).**

Question 17

A squirrel starts climbing up a tree at the speed of 6 metres a minute but after every 6 metres it slips down 4 metres in the next minute. It will be able to reach the top 120 metres high in

Accepted Answer

### Concept & Strategy This is a classic "cyclic progress" problem (often framed as a monkey climbing a pole). The crucial insight is that during the *last* cycle, the target is reached during the upward phase, ending the journey before any backward slip occurs. ### Step-by-Step Solution * **Given:** * Target height = 120 metres. * Climb phase = 6 metres in 1 minute. * Slip phase = 4 metres in 1 minute. * **Deduction:** Calculate the net progress of one complete cycle: $$	ext{Cycle duration} = 1 + 1 = 2 	ext{ minutes}$$ $$	ext{Net gain per cycle} = 6 - 4 = 2 	ext{ metres}$$ * **Calculation:** Subtract the final unabated climb (6m) from the total height to find the distance that must be covered via the slow, slipping cycles: $$	ext{Cyclic distance} = 120 - 6 = 114 	ext{ metres}$$ Calculate the number of complete cycles required to cover 114 metres: $$	ext{Cycles} = \frac{114}{2} = 57 	ext{ cycles}$$ Calculate the time taken for these 57 cycles: $$57 	imes 2 = 114 	ext{ minutes}$$ Finally, add the 1 minute for the last 6-metre climb to reach the 120m top: $$	ext{Total Time} = 114 + 1 = 115 	ext{ minutes}$$ ### Exam Strategy & Shortcut Use the standardized formula for alternating positive/negative work problems to find the number of pair cycles: $$	ext{Cycles} = \frac{	ext{Total Distance} - 	ext{Final Jump}}{	ext{Net Gain}}$$ $	ext{Cycles} = \frac{120 - 6}{2} = 57$. Total time is $57 	imes 2 = 114$ mins, plus 1 minute for the final jump = 115 mins. ### Common Pitfall Dividing the total height (120) by the net cycle gain (2) directly to get 60 cycles, and thus concluding 120 minutes. This ignores the physical reality that once the top is reached at 115 minutes, the squirrel grabs the edge and won't slip down again. ### Final Answer **Therefore, the correct answer is 115 minutes.**

Question 18

The growth of a medicinal plant of height 1 metre, is 1 cm/day and it is growing straight vertically upward. A man cuts 6 cm from the top, at a regular interval of every 11 days. Its height will be 2 metres after

Accepted Answer

### Concept & Strategy This is a sequential growth and reduction problem. Care must be taken to distinguish between the continuous daily growth and the discrete periodic cuts. We must track the daily height to see exactly when it hits the threshold. ### Step-by-Step Solution * **Given:** * Initial height = 1 m = 100 cm * Target height = 2 m = 200 cm * Growth rate = 1 cm/day * Cut = 6 cm every 11 days. * **Deduction:** In one 11-day period, the plant grows 11 cm and is then cut by 6 cm. $$	ext{Net growth per 11 days} = 11 - 6 = 5 	ext{ cm}$$ * **Calculation:** Find how many 11-day cycles it takes to get near 200 cm. After 19 cycles ($19 	imes 11 = 209$ days): $$	ext{Height} = 100 + 19(5) = 195 	ext{ cm}$$ Now, track the daily growth from day 210 onwards (before the 20th cut happens on day 220): * Day 210: 196 cm * Day 211: 197 cm * Day 212: 198 cm * Day 213: 199 cm * Day 214: 200 cm The plant reaches exactly 2 metres on the 214th day. Since 214 is not listed among the options, the correct choice is "None of these". ### Exam Strategy & Shortcut Evaluate the widely accepted "flawed" logic vs. the strict mathematical reality. Flawed logic: Target growth = 100 cm. Net rate = 5 cm / 11 days. Time = $(100 \div 5) 	imes 11 = 220$ days. Strict math: At day 209, height is 195cm. It grows 1cm/day for 5 days to hit 200cm exactly at day 214. Since 214 is missing, confidently choose "None of these". ### Common Pitfall Blindly assuming the target height is only reached exactly at the end of a full cycle and calculating $(100 \div 5) 	imes 11 = 220$ days (Option a). This ignores the intra-cycle peak where the plant hits 200 cm organically on day 214 before the next cut occurs. ### Final Answer **Therefore, the correct answer is None of these.**

Question 19

If $ x = 5 $, $ y = 3 $, the value of $ \frac{x^3 - y^3}{x^2 - y^2} - \frac{3xy}{x + y} $ will be

Accepted Answer

### Concept & Formula This problem tests your ability to simplify rational algebraic expressions before inserting numerical values. The goal is to combine the two fractions into a much simpler form. The necessary factorization formulas are: $$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $$ $$ x^2 - y^2 = (x - y)(x + y) $$ ### Step-by-Step Solution Let's start with the first fraction in the expression: $$ \frac{x^3 - y^3}{x^2 - y^2} $$ Factor both the numerator and the denominator using the standard formulas: $$ = \frac{(x - y)(x^2 + xy + y^2)}{(x - y)(x + y)} $$ Cancel out the common $ (x - y) $ term from the top and bottom: $$ = \frac{x^2 + xy + y^2}{x + y} $$ Now, bring back the second part of the original expression and subtract it: $$ \frac{x^2 + xy + y^2}{x + y} - \frac{3xy}{x + y} $$ Because both fractions now share a common denominator ($ x + y $), we can combine their numerators: $$ = \frac{x^2 + xy + y^2 - 3xy}{x + y} $$ Combine the like terms ($ +xy - 3xy = -2xy $): $$ = \frac{x^2 - 2xy + y^2}{x + y} $$ Notice that the numerator is now a perfect square trinomial. Factor it: $$ = \frac{(x - y)^2}{x + y} $$ The expression is fully simplified. Now substitute the given values: $ x = 5 $ and $ y = 3 $. $$ = \frac{(5 - 3)^2}{5 + 3} $$ $$ = \frac{(2)^2}{8} $$ $$ = \frac{4}{8} $$ $$ = \frac{1}{2} $$ ### Exam Strategy & Shortcut Instead of algebraic manipulation, you can use direct substitution ("Value Putting") immediately, as the numbers are very small. Fraction 1: $ \frac{5^3 - 3^3}{5^2 - 3^2} = \frac{125 - 27}{25 - 9} = \frac{98}{16} = \frac{49}{8} $ Fraction 2: $ \frac{3(5)(3)}{5 + 3} = \frac{45}{8} $ Difference: $ \frac{49}{8} - \frac{45}{8} = \frac{4}{8} = \frac{1}{2} $. For small integers, basic arithmetic is often faster than performing multi-step polynomial factorization. ### Common Pitfall When simplifying algebraically, students sometimes mistakenly cancel $ (x^2 - y^2) $ out of the denominator without properly factoring the numerator, or miscalculate the combination $ +xy - 3xy $ as $ +2xy $, which leads to an entirely incorrect final value. ### Final Answer **Therefore, the correct answer is $ \frac{1}{2} $.**

Question 20

The value of  $$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}$$ is

Accepted Answer

### Concept & Formula This fraction simplifies using the conditional identity for the sum of cubes: If $A + B + C = 0$, then $A^3 + B^3 + C^3 = 3ABC$. ### Step-by-Step Solution * Let the numerator components be $A = x - y$, $B = y - z$, and $C = z - x$. * Notice that $A + B + C = (x - y) + (y - z) + (z - x) = 0$. * Therefore, the numerator becomes $3(x - y)(y - z)(z - x)$. * Now, let the denominator components be $P = x^2 - y^2$, $Q = y^2 - z^2$, and $R = z^2 - x^2$. * Notice that $P + Q + R = (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0$. * Therefore, the denominator becomes $3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)$. * Expand the denominator using the difference of squares $a^2 - b^2 = (a-b)(a+b)$: $$3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)$$ * Divide the simplified numerator by the simplified denominator: $$\frac{3(x - y)(y - z)(z - x)}{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}$$ * Cancel the common terms $3(x - y)(y - z)(z - x)$ from top and bottom: $$= \frac{1}{(x + y)(y + z)(z + x)}$$ * Rewrite using a negative exponent: $$= [(x + y)(y + z)(z + x)]^{-1}$$ ### Exam Strategy & Shortcut Whenever you see a sum of three cubes in a pattern that looks cyclical (like $x-y, y-z, z-x$), immediately check if they sum to zero. If they do, instantly replace them with 3 times their product. This visual recognition avoids lengthy expansions. ### Common Pitfall A common error is to try expanding $(x - y)^3$ individually. This creates a massive, unmanageable polynomial that guarantees a loss of time and inevitable calculation errors. ### Final Answer Therefore, the correct answer is **$[(x + y) (y + z) (z + x)]^{-1}$**.

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