Question 1

The sum of seven consecutive numbers is $175$. What is the difference between twice the largest number and thrice the smallest number?

Accepted Answer

### Concept & Strategy The sum of an odd number of consecutive terms (which form an Arithmetic Progression) yields the middle term when divided by the number of terms. This is because the average of an AP is its perfectly centered value. $$	ext{Middle Term} = \frac{	ext{Sum of Terms}}{	ext{Number of Terms}}$$ ### Step-by-Step Solution * **Given:** The sum of $7$ consecutive numbers is $175$. * Calculate the middle term directly by finding the average: $$	ext{Middle Term} = \frac{175}{7} = 25$$ * Since $25$ is the middle term of $7$ consecutive integers, there are exactly $3$ terms before it and $3$ terms after it. * The sequence is: $22, 23, 24, 	extbf{25}, 26, 27, 28$. * Identify the smallest and largest numbers: * Smallest Number = $22$ * Largest Number = $28$ * The question asks for the difference between twice the largest and thrice the smallest: * Twice the largest = $2 	imes 28 = 56$ * Thrice the smallest = $3 	imes 22 = 66$ * Calculate the absolute difference: $$|56 - 66| = 10$$ ### Exam Strategy & Shortcut Never use the standard algebraic setup $x + (x+1) + ... + (x+6) = 175$, which simplifies to $7x + 21 = 175$. While it works, it is mathematically heavier. Identifying the middle term via division ($175 / 7 = 25$) allows you to immediately step back by $3$ to find the smallest ($22$) and step forward by $3$ to find the largest ($28$). ### Common Pitfall The phrasing "difference between A and B" in aptitude exams implies the absolute difference (larger value minus smaller value) unless a specific order of subtraction is stated. Calculating $56 - 66 = -10$ and assuming the answer isn't present (leading to "None of these") is a frequent trap. ### Final Answer **Therefore, the correct answer is 10.**

Question 2

If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?

Accepted Answer

### Concept & Logic Manipulating place values algebraically. We must carefully construct the expression for the "new" number based on the fractional and multiplicative changes to its individual digits. ### Step-by-Step Solution Let the ten's digit be $t$ and the unit's digit be $u$. The original number is $10t + u$. Condition 1: Unit's digit is halved ($u/2$) and ten's digit is doubled ($2t$). The new number formed is: $10(2t) + (u/2) = 20t + 0.5u$. Condition 2: This new number is equal to the number obtained by interchanging the original digits ($10u + t$). $20t + 0.5u = 10u + t$ Rearrange to find the relationship between $t$ and $u$: $20t - t = 10u - 0.5u$ $19t = 9.5u$ Divide both sides by 9.5: $2t = u$ This equation directly translates to: The unit's digit ($u$) is twice the ten's digit ($t$). ### Exam Strategy & Shortcut Plug in options to create a valid number. For (d), if unit is twice ten's, let's pick $t=2, u=4$. Original number = 24. Halve unit (2), double ten's (4). New digits: tens=4, units=2. New number = 42. Interchanged original number (24) is also 42. $42 = 42$. The condition holds perfectly. ### Common Pitfall Constructing the new number incorrectly. A common mistake is writing the new number as $2t + u/2$ instead of properly multiplying the new ten's digit ($2t$) by its place value of 10. ### Final Answer Therefore, the correct answer is **Digit in the unit's place is twice the digit in the ten's place.**.

Question 3

The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from the denominator, it becomes $\frac{2}{3}$. The fraction is

Accepted Answer

### Concept & Strategy This problem translates simple arithmetic operations on the components of a fraction into algebraic equations. However, because we are looking for a specific starting fraction, checking the provided multiple-choice options against the conditions is the absolute fastest and most reliable method. Let the fraction be $\frac{n}{d}$. ### Step-by-Step Solution * **Given:** 1. $n + d = 11$ 2. $\frac{n + 1}{d - 2} = \frac{2}{3}$ * **Calculation (Option Verification):** Let's check the options to see which satisfies both conditions. * **Check Option A ($\frac{5}{6}$):** * Sum: $5 + 6 = 11$ (Condition 1 satisfied). * New fraction: $\frac{5 + 1}{6 - 2} = \frac{6}{4} = \frac{3}{2}$. This is not $\frac{2}{3}$. * **Check Option C ($\frac{3}{8}$):** (Note: Checking C before B because $\frac{3}{8}$ looks more standard as a proper fraction, though any order works). * Sum: $3 + 8 = 11$ (Condition 1 satisfied). * New fraction: $\frac{3 + 1}{8 - 2} = \frac{4}{6} = \frac{2}{3}$. * This perfectly satisfies both conditions. *(For completion, if we solved algebraically: From $n+d=11$, we get $d = 11-n$. Substitute into the second equation: $\frac{n+1}{11-n-2} = \frac{2}{3} \implies \frac{n+1}{9-n} = \frac{2}{3} \implies 3(n+1) = 2(9-n) \implies 3n+3 = 18-2n \implies 5n = 15 \implies n=3$. If $n=3$, then $d=11-3=8$. The fraction is $\frac{3}{8}$.)* ### Exam Strategy & Shortcut Use the "Sum" condition to eliminate options instantly. Option B ($\frac{6}{5}$) has sum $11$. Option D ($\frac{8}{3}$) has sum $11$. Wait, all options sum to $11$! So the first condition doesn't help eliminate. Proceed directly to the second condition mentally: "Add $1$ to top, subtract $2$ from bottom." Apply to A: $\rightarrow \frac{6}{4}$ (No) Apply to C: $\rightarrow \frac{4}{6}$ (Yes, simplifies to $\frac{2}{3}$). Done in 10 seconds. ### Common Pitfall When doing algebra, students might write the second equation incorrectly, such as $\frac{n+1}{d} - 2 = \frac{2}{3}$, misinterpreting "subtracted from the denominator" as subtracted from the whole fraction. Always ensure operations on numerators/denominators are properly grouped in the fraction. ### Final Answer **Therefore, the correct answer is $\frac{3}{8}$.**

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