Question 1

In a polling booth there are 1575 voters in total. Of these, 40% are male voters. A candidate receives 40% of the female votes. How many total votes does the candidate get?

Accepted Answer

Introduction / Context: This is a percentage distribution problem. We are given total voters, the proportion of males, and the fraction of female votes a candidate secures. The goal is to compute the exact number of votes obtained by the candidate. Given Data / Assumptions: Total voters = 1575. Male voters = 40% of 1575. Female voters = remaining voters. Candidate gets 40% of female voters. Concept / Approach: Use successive percentages. First find males, then females, then compute 40% of the female count. Keep arithmetic exact to avoid rounding errors. Step-by-Step Solution: Males = 0.4 * 1575 = 630Females = 1575 - 630 = 945Candidate's votes = 0.4 * 945 = 378 Verification / Alternative check: 40% equals 2/5; 2/5 of 945 = 189 * 2 = 378. Same result, confirming accuracy. Why Other Options Are Wrong: 945, 756, 630, 432: These correspond to totals or incorrect percentage applications. Only 378 correctly reflects 40% of the female voters. Common Pitfalls: Taking 40% of the total instead of only the females. Computing females as 40% rather than 60% of the total. Final Answer: 378

Question 2

In a village, 5/8 of the population are adults. Half of the adults are males, and 4/5 of the adult females are illiterate. If the number of illiterate adult females is 400, what is the total population of the village?

Accepted Answer

Introduction / Context: This problem chains several fractions to determine a whole population from a known sub-group count. It assesses careful fraction handling and back-calculation from a part to the total. Given Data / Assumptions: Total population = P. Adults = (5/8)P. Adult males = half of adults; therefore adult females = half of adults = (1/2)*(5/8)P = (5/16)P. Illiterate adult females = (4/5) of adult females = 400. Concept / Approach: First express the number of adult females as a fraction of P. Then use the given count of illiterate adult females (a fraction of those adult females) to solve for P. Step-by-Step Solution: Adult females = (5/16)PIlliterate adult females = (4/5) * (5/16)P = (20/16)P = (5/4)PBut this equals 400, so (4/5) * (5/16)P = 400 → (1/4)P = 100 → P = 400 * (16/5)Compute directly: (4/5)*(5/16)P = 400 ⇒ (1/4)P = 400 ⇒ P = 1600 Verification / Alternative check: Adult females = (5/16)*1600 = 500; 4/5 of 500 = 400. Matches the given number. Why Other Options Are Wrong: 2000, 1500, 1800, 1750: None satisfy both the adult-fraction and the illiteracy fraction when checked. Common Pitfalls: Mistaking adult females as 5/8 rather than 5/16 of the population. Arithmetic slips with compound fractions. Final Answer: 1600

Question 3

In a school, 1/4 of the boys and 3/8 of the girls participated in the annual sports day. What fractional part of the total student strength participated?

Accepted Answer

Introduction / Context: This question examines whether a unique fraction of total participants can be determined without knowing the ratio of boys to girls. It highlights dependence on missing composition data when combining parts of different groups. Given Data / Assumptions: Boys = B, Girls = G, Total = B + G. Participants = (1/4)B + (3/8)G. No information is given about the ratio B:G. Concept / Approach: To get a fraction of total, compute [(1/4)B + (3/8)G] / (B + G). Without a relation between B and G, this expression is not uniquely determined, because it changes with the composition of the student body. Step-by-Step Solution: Required fraction F = ((1/4)B + (3/8)G) / (B + G)F depends on both B and G. If B and G vary, F changes.Therefore, a unique value cannot be obtained. Verification / Alternative check: Try examples: If B = G, then F = ((1/4) + (3/8)) / 2 in normalized terms, different from when B ≠ G. Changing B:G alters F, proving non-uniqueness. Why Other Options Are Wrong: 32%, 20%, 36%, 28%: Each is a fixed number, which cannot be guaranteed without knowing B:G. Common Pitfalls: Assuming equal numbers of boys and girls when not stated. Adding fractions as if they referred to the same base group. Final Answer: Data inadequate

Question 4

A scooter needs servicing for 2/3 hour every 45 days before overhauling, and 2/3 hour every 60 days after overhauling. What fraction of the pre-overhauling service time is saved after overhauling?

Accepted Answer

Introduction / Context: The problem compares periodic service times over different intervals, asking for the fractional saving relative to the original (pre-overhaul) requirement. It tests rate reasoning and fraction comparison over time. Given Data / Assumptions: Pre-overhaul need: 2/3 hour every 45 days. Post-overhaul need: 2/3 hour every 60 days. We seek the fraction saved relative to the pre-overhaul requirement. Concept / Approach: Convert both to a common basis (e.g., hours per day). Then compute (pre - post) / pre. This yields the fraction of time saved compared to the original requirement. Step-by-Step Solution: Pre rate = (2/3) / 45 = 2/135 hours per dayPost rate = (2/3) / 60 = 2/180 = 1/90 hours per daySaving = Pre - Post = 2/135 - 1/90Compute difference: 1/90 = 1.5/135 → Saving = (2 - 1.5)/135 = 0.5/135 = 1/270Fraction saved relative to pre = (1/270) / (2/135) = (1/270) * (135/2) = 135/540 = 1/4 Verification / Alternative check: Scale to 540 days: Pre total = (2/3) * (540/45) = (2/3) * 12 = 8 hours; Post total = (2/3) * (540/60) = (2/3) * 9 = 6 hours. Saving = 2/8 = 1/4 of pre. Confirms the result. Why Other Options Are Wrong: 4/3, 1/3, 3/4, 1/6: These do not match the computed ratio of time saved to the original time. Common Pitfalls: Comparing absolute hours without normalizing by time window. Subtracting 45 from 60 or mixing up reciprocals. Final Answer: 1/4

Question 5

From a rope 30 m long, a person cuts as many equal pieces as possible, each being 3 1/4 m in length. What fraction of the original rope remains uncut?

Accepted Answer

Introduction / Context: This question involves division with mixed numbers to determine how many full pieces can be cut and what leftover fraction remains. It checks careful handling of mixed fractions and remainders relative to the original total length. Given Data / Assumptions: Total length = 30 m. Piece length = 3 1/4 m = 3.25 m. Cut as many full pieces as possible; leftover is less than one full piece. Concept / Approach: Compute the integer part of 30 / 3.25 to get the number of pieces. Then compute the leftover length and divide by the total length to get the remaining fraction. Step-by-Step Solution: Pieces = floor(30 / 3.25) = floor(9.2307...) = 9Used length = 9 × 3.25 = 29.25 mLeftover length = 30 - 29.25 = 0.75 mRemaining fraction = 0.75 / 30 = 3/120 = 1/40 Verification / Alternative check: Convert 3.25 to 13/4; 30 ÷ (13/4) = 30 × (4/13) = 120/13 = 9 remainder 3/13 of a piece. Remainder length = (3/13) × 13/4 = 3/4 m; fraction of whole = (3/4)/30 = 1/40. Confirms result. Why Other Options Are Wrong: 3/4, 8/13, 7/13, 1/13: These represent either piece ratios or misinterpreted remainders, not the leftover fraction of the whole 30 m rope. Common Pitfalls: Rounding 3.25 to 3 or 3.5. Forgetting to express the leftover as a fraction of the original 30 m. Final Answer: 1/40

Question 6

Evaluate the sum of decimals and express the result in simplest form as needed: 0.3 + 0.6 + 0.7 + 0.8 = ?

Accepted Answer

Introduction / Context: Adding decimal numbers accurately and, if desired, converting the result to a fraction in simplest form is a foundational arithmetic skill. The task reinforces place-value alignment and fraction-decimal equivalence. Given Data / Assumptions: Decimals to add: 0.3, 0.6, 0.7, 0.8. No rounding is necessary; each term has one decimal place. Concept / Approach: Add decimals directly by place value, then convert the final decimal to a fraction in lowest terms to match the answer choice format if needed. Step-by-Step Solution: 0.3 + 0.6 = 0.90.9 + 0.7 = 1.61.6 + 0.8 = 2.4Convert 2.4 to a fraction: 2.4 = 24/10 = 12/5 Verification / Alternative check: As fractions: 3/10 + 6/10 + 7/10 + 8/10 = 24/10 = 12/5. Same result. Why Other Options Are Wrong: 23/10 (2.3), 239/100 (2.39), 5/2 (2.5), 53/10 (5.3): None equals the correct sum 2.4. Common Pitfalls: Miscalculating 0.9 + 0.7 as 1.5 instead of 1.6. Forgetting to simplify the fraction 24/10 to 12/5. Final Answer: 12/5

Question 7

Error in operating on a fraction: A boy was asked to find 13/14 of a certain fraction. Instead, he mistakenly divided the fraction by 13/14 and obtained a result that exceeded the correct value by 3/65. What is the correct original fraction?

Accepted Answer

Introduction / Context: This question checks understanding of operations with fractions and how a procedural mistake affects the result. The intended operation was to multiply the unknown fraction by 13/14, but the boy instead divided by 13/14, which is equivalent to multiplying by 14/13. We are told the error increased the answer by 3/65 over the correct value. We must translate this into an equation and solve for the original fraction. Given Data / Assumptions: The unknown fraction is x. Correct result should be (13/14) * x. Mistaken result is x ÷ (13/14) = (14/13) * x. Excess due to the mistake is 3/65 over the correct result. Concept / Approach: Express the stated excess as the difference between the mistaken result and the correct result. This gives a linear equation in x. Solving that equation yields the value of x directly. This is standard equation setup using fraction arithmetic and difference-of-results reasoning. Step-by-Step Solution: Correct value = (13/14) * x.Mistaken value = (14/13) * x.Excess = Mistaken − Correct = 3/65.So (14/13)x − (13/14)x = 3/65.Compute the difference: (196/182 − 169/182)x = (27/182)x = 3/65.Solve for x: x = (3/65) * (182/27) = 546/1755 = 14/45 after simplification. Verification / Alternative check: Check difference using x = 14/45. Correct: (13/14)*(14/45) = 13/45. Mistaken: (14/13)*(14/45) = 196/(585) = 28/83.75 (same as 196/585). Difference: 196/585 − 13/45 = 196/585 − 169/585 = 27/585 = 3/65. Verified. Why Other Options Are Wrong: 12/65: Does not satisfy the formed equation; difference is not 3/65. 13/45: This equals the correct result, not the original fraction. 2/7: Produces a different excess; arithmetic check fails. Common Pitfalls: Confusing multiplying by 13/14 with dividing by 13/14. Not aligning denominators correctly when subtracting the two expressions. Final Answer: 14/45

Question 8

Inheritance sharing with fractional conditions: A man left 1/7 of his property to his daughter and the remainder to be shared equally among his sons. If each son’s share equals twice the daughter’s share, how many sons are there?

Accepted Answer

Introduction / Context: This is an algebraic sharing problem using fractions. The property is split such that the daughter receives a fixed fraction, and the rest is divided equally among sons. A proportional condition links the son’s share to the daughter’s share, allowing us to determine the number of sons. Given Data / Assumptions: Daughter’s share = 1/7 of the total property. Remaining property = 1 − 1/7 = 6/7. Each son’s share is twice the daughter’s share, i.e., 2 × (1/7) = 2/7. There are n sons sharing 6/7 equally. Concept / Approach: Set up the equality: (each son’s share) = (total remainder)/(number of sons). Because each son’s share is given to be 2/7, we can solve for n directly. This is a single-step proportion problem once the remainder is computed. Step-by-Step Solution: Remainder after daughter = 6/7.If n sons share equally, each gets (6/7)/n.Given each son’s share = 2/7, set (6/7)/n = 2/7.Solve: 6/(7n) = 2/7 ⇒ cross-multiply ⇒ 6 = 2n ⇒ n = 3. Verification / Alternative check: With 3 sons, each gets (6/7)/3 = 2/7, which indeed is twice 1/7. Why Other Options Are Wrong: 2 or 4 or 6: Do not satisfy (6/7)/n = 2/7. Common Pitfalls: Forgetting to compute the remainder 6/7 before dividing among sons. Misreading “twice the daughter’s share” as “two times the number of daughters” or similar. Final Answer: 3

Question 9

Weighing problem with partial fill levels: A vessel full of water weighs 16.5 kg. When the vessel is 1/4 full, it weighs 5.25 kg. What is the weight of the empty vessel (in kg)?

Accepted Answer

Introduction / Context: This is a classic linear relation problem involving a container’s own weight and the weight contributed by the water inside it. By comparing the vessel’s weight when full and when partially filled, we can solve for the vessel’s empty weight directly using a simple equation. Given Data / Assumptions: Weight when full: 16.5 kg. Weight when 1/4 full: 5.25 kg. Let the empty vessel weight be W kg and the water when full be (16.5 − W) kg. Concept / Approach: The total weight equals vessel weight plus water weight. At 1/4 full, the water contributes one-fourth of the full-water weight. Form an equation for the 1/4-full condition and solve for W. It is a straightforward application of proportional filling and linear equations. Step-by-Step Solution: Let W be empty vessel weight; full water weight = 16.5 − W.At 1/4 full: Total = W + (1/4)(16.5 − W) = 5.25.Compute: W + 4.125 − 0.25W = 5.25 ⇒ 0.75W = 1.125.Hence W = 1.125 / 0.75 = 1.5 kg. Verification / Alternative check: Full water weight = 16.5 − 1.5 = 15.0 kg. A quarter of that is 3.75 kg; adding vessel 1.5 kg gives 5.25 kg (matches data). Why Other Options Are Wrong: 1.125: This is 0.75W, not W. 4.5 or 3: Overestimates the vessel mass; contradicts the 1/4-full condition. Common Pitfalls: Mistaking 1/4 of the total weight for 1/4 of the water weight. Arithmetic slip when rearranging W − 0.25W. Final Answer: 1.5

Question 10

Two-game money transfer scenario: A has twice as much money as B. After game 1, B wins one-third of A’s initial money from A. In game 2, what fraction of B’s then-current money must A win back so that they end up with equal amounts?

Accepted Answer

Introduction / Context: This problem requires careful tracking of money transfers in two stages. It tests proportional reasoning and equation setup: first, compute each person’s holdings after the first transfer; then, determine the fraction of B’s new amount that must be transferred back to A for equality after the second game. Given Data / Assumptions: Initially A = 2k, B = k. Game 1: B wins (1/3) of A’s initial money, which is (1/3)*2k = 2k/3. After Game 1: A = 2k − 2k/3 = 4k/3; B = k + 2k/3 = 5k/3. Game 2: A wins back a fraction f of B’s current money (5k/3). Concept / Approach: Let the transfer in Game 2 be f*(5k/3) from B to A. Set the final amounts equal and solve for f. The ratio relationship 4k/3 vs 5k/3 makes the algebra neat and the result a simple fraction. Step-by-Step Solution: After Game 1: A1 = 4k/3, B1 = 5k/3.Game 2 transfer: T = f * B1 = f * (5k/3).Final amounts: A2 = A1 + T; B2 = B1 − T.Set equality: A1 + T = B1 − T ⇒ A1 = B1 − 2T.Substitute: 4k/3 = 5k/3 − 2f*(5k/3) ⇒ 4/3 = 5/3 − (10/3)f.Solve: (10/3)f = 5/3 − 4/3 = 1/3 ⇒ f = (1/3) * (3/10) = 1/10. Verification / Alternative check: With f = 1/10, T = (1/10)*(5k/3) = k/6. A2 = 4k/3 + k/6 = 9k/6 = 3k/2; B2 = 5k/3 − k/6 = 9k/6 = 3k/2: equal. Why Other Options Are Wrong: 1/3, 1/4, 1/5: Lead to unequal final amounts when substituted. Common Pitfalls: Taking one-third of A’s current (post-loss) money instead of A’s initial money. Forgetting that the second transfer depends on B’s post–Game-1 amount. Final Answer: 1/10

Question 11

Determine three fractions from sum and comparative conditions (set 1): The sum of three fractions is 2 11/24. The greatest divided by the smallest equals 7/6, and this quotient exceeds the middle fraction by 1/3. Identify the three fractions.

Accepted Answer

Introduction / Context: We are given a total sum and two comparative conditions to determine three specific fractions. The key is to translate the comparative statements into algebraic relations and solve consistently for the smallest, middle, and greatest fractions that also satisfy the stated sum. Given Data / Assumptions: Total sum of three fractions = 2 11/24 = 59/24. (Greatest)/(Smallest) = 7/6. This quotient is greater than the middle fraction by 1/3, i.e., Middle = 7/6 − 1/3 = 5/6. Concept / Approach: Let the smallest fraction be x. Then the greatest is (7/6)x. From the second condition, the middle is 5/6. Use the sum relation x + (7/6)x + 5/6 = 59/24 and solve for x. Finally, compute the greatest and list the three in decreasing order to match the narrative (greatest, middle, smallest). Step-by-Step Solution: Let smallest = x, greatest = (7/6)x, middle = 5/6.Sum: x + (7/6)x + 5/6 = 59/24.Combine: (13/6)x + 5/6 = 59/24 ⇒ multiply by 24: 52x + 20 = 59.Hence 52x = 39 ⇒ x = 39/52 = 3/4.Greatest = (7/6)*(3/4) = 7/8; middle = 5/6; smallest = 3/4. Verification / Alternative check: Sum: 7/8 + 5/6 + 3/4 = 21/24 + 20/24 + 18/24 = 59/24. Also (7/8)/(3/4) = (7/8)*(4/3) = 7/6; and 7/6 − 1/3 = 5/6. All conditions satisfied. Why Other Options Are Wrong: 3/5, 4/7, 2/3 and 7/9, 2/3, 3/5: Do not meet both the quotient and sum conditions simultaneously. 7/8, 7/9, 7/10: Incorrect middle and smallest; the sum also deviates from 59/24. Common Pitfalls: Misinterpreting “greater than the middle by 1/3” as adding 1/3 to the middle rather than subtracting from the quotient. Arithmetic errors in converting 2 11/24 to 59/24. Final Answer: 7/8, 5/6, 3/4

Question 12

Determine three fractions from sum and comparative conditions (set 2): The sum of three fractions is 2 11/24. The greatest divided by the smallest equals 7/6, and this quotient exceeds the middle fraction by 1/3. Find the three fractions.

Accepted Answer

Introduction / Context: This is a restatement of the prior problem with the same conditions and numbers. We again use two comparative constraints along with a total-sum requirement to fix the three unknown fractions uniquely. The method mirrors proportional reasoning and sum balancing. Given Data / Assumptions: Total sum = 2 11/24 = 59/24. (Greatest)/(Smallest) = 7/6. Middle = 7/6 − 1/3 = 5/6. Concept / Approach: Let smallest = x, greatest = (7/6)x, middle = 5/6. Solve x from the sum equation, then compute the other two fractions, and verify all constraints, including the quotient and the additive total. Step-by-Step Solution: x + (7/6)x + 5/6 = 59/24 ⇒ (13/6)x = 59/24 − 5/6.Compute RHS: 59/24 − 20/24 = 39/24 = 13/8.Thus (13/6)x = 13/8 ⇒ x = (13/8)*(6/13) = 6/8 = 3/4.Greatest = (7/6)*(3/4) = 7/8; middle = 5/6. Verification / Alternative check: Sum: 7/8 + 5/6 + 3/4 = 59/24; quotient 7/6; and 7/6 − 1/3 = 5/6 as required. Why Other Options Are Wrong: All other sets fail either the exact sum or the exact quotient/middle relationship. Common Pitfalls: Miscalculating the mixed fraction 2 11/24. Confusing the phrase “greater by 1/3” and applying it to the wrong quantity. Final Answer: 7/8, 5/6, 3/4

Question 13

Daily average from annual amount: In the year 2011, Shantanu receives a total pocket allowance of ₹ 3832.5. What is his allowance per day for that year?

Accepted Answer

Introduction / Context: Average-per-day questions require dividing a total amount by the number of days in the period. Here, 2011 is a non-leap year with 365 days. We find the daily pocket allowance by a straightforward division of the total annual amount by 365. Given Data / Assumptions: Year = 2011 (365 days). Total allowance = ₹ 3832.5. Daily allowance = Total / 365. Concept / Approach: Arithmetic average equals the total amount divided by count. Knowing the correct day count is crucial. Using 365 days, compute the division exactly or by splitting the total into convenient parts (e.g., 3650 + 182.5) to see the result quickly. Step-by-Step Solution: Compute 3832.5 ÷ 365.Note 365 × 10 = 3650; remainder = 182.5.Since 365 × 0.5 = 182.5, the remainder corresponds to 0.5 more day-equivalents.Therefore the quotient is 10.5. Hence daily allowance = ₹ 10.5. Verification / Alternative check: Multiply back: ₹ 10.5 × 365 = ₹ 3832.5, confirming the calculation. Why Other Options Are Wrong: ₹ 9.5, ₹ 11.5, ₹ 12.5: Do not multiply to the annual total; they reflect incorrect day counts or divisions. Common Pitfalls: Using 366 days for a non-leap year. Rounding mid-calculation instead of exact division. Final Answer: ₹ 10.5

Question 14

Fractional composition of a pencil: 1/8 of a pencil is black; half of the remaining part is white. The rest is blue, and this blue section measures 3 1/2 cm. What is the full length of the pencil?

Accepted Answer

Introduction / Context: This is a fraction-of-a-whole problem where the whole is the length of a pencil. Parts of the pencil are colored differently by fractional proportions. By expressing each colored segment as a fraction of the total and using the given absolute length for one segment, we can determine the entire length. Given Data / Assumptions: Black part = 1/8 of the pencil. Remaining after black = 1 − 1/8 = 7/8. White part = 1/2 of the remaining = 1/2 × 7/8 = 7/16. Blue part = Remaining after black and white = 7/8 − 7/16 = 7/16. Blue length = 3 1/2 cm = 3.5 cm. Concept / Approach: Let total length be L. Since blue part equals 7/16 of L and we know its numerical length, set (7/16)L = 3.5 and solve for L. This ratio-to-absolute conversion is standard in fraction word problems. Step-by-Step Solution: Let total length L; blue = (7/16)L.Given (7/16)L = 3.5 ⇒ L = 3.5 × (16/7).Compute: 3.5 × (16/7) = 0.5 × 16 = 8 cm. Verification / Alternative check: Check fractions: black 1/8 = 0.125, white 7/16 ≈ 0.4375, blue 7/16 ≈ 0.4375; totals 1. Blue length 0.4375 × 8 = 3.5 cm—consistent. Why Other Options Are Wrong: 6 cm, 7 cm, 9 cm: When multiplied by 7/16 do not yield 3.5 cm, violating the given blue segment length. Common Pitfalls: Treating “half of the remaining” as “half of the total.” Misreading 3 1/2 cm as 31/2 cm instead of 3.5 cm. Final Answer: 8 cm

Question 15

Reading a book in stages: A man reads 3/8 of a book on day one and 4/5 of the remainder on day two. If 40 pages are still unread, how many pages does the book contain?

Accepted Answer

Introduction / Context: Sequential fraction reading problems test your ability to update the remainder after each step. The key is to carefully compute what remains after each day, then match the final remainder to the known number of unread pages to recover the total number of pages in the book. Given Data / Assumptions: Day 1: reads 3/8 of the book. Day 2: reads 4/5 of the remainder from Day 1. Unread after Day 2 = 40 pages. Concept / Approach: Let the total number of pages be N. After reading 3/8, the remainder is 5/8 of N. On day two, he reads 4/5 of that remainder and leaves 1/5 of it unread. Convert these steps into a single fraction of N that remains, equate to 40, and solve for N. Step-by-Step Solution: After Day 1: Remaining = N − (3/8)N = (5/8)N.After Day 2: Unread = (1/5) × (5/8)N = (1/8)N.Given unread = 40 ⇒ (1/8)N = 40 ⇒ N = 40 × 8 = 320 pages. Verification / Alternative check: Check: Day 1 reads 120; remainder 200. Day 2 reads (4/5)×200 = 160; unread 40. Consistent. Why Other Options Are Wrong: 300, 350, 500: Do not yield 40 pages remaining after applying the same fractions; arithmetic check fails. Common Pitfalls: Applying 4/5 to the total instead of the remainder. Confusing “remainder unread” with “remainder read.” Final Answer: 320

Question 16

Reverse-engineering an original fraction: After increasing the numerator of a fraction by 200% and the denominator by 150%, the resulting fraction is 9/35. What was the original fraction?

Accepted Answer

Introduction / Context: Percentage increases on the numerator and denominator scale a fraction multiplicatively. “Increased by 200%” means the new numerator equals old numerator × 3. “Increased by 150%” means the new denominator equals old denominator × 2.5. You are given the final fraction and must recover the original one by reversing these multiplicative effects. Given Data / Assumptions: Let the original fraction be n/d. New numerator after +200%: 3n. New denominator after +150%: 2.5d. Resulting fraction = 9/35. Concept / Approach: Set 3n / (2.5d) = 9/35. Solve for the ratio n/d by isolating it on one side. The operations are pure scaling; there is no need to assume integer numerators or denominators at intermediate steps. Finally, simplify the result to lowest terms to match one of the options. Step-by-Step Solution: 3n / (2.5d) = 9/35.Rearrange: n/d = (9/35) × (2.5/3).Compute: 2.5/3 = 5/6; hence n/d = (9/35) × (5/6) = 45/210 = 3/14. Verification / Alternative check: Apply the stated increases to 3/14: new numerator = 3 × 3 = 9; new denominator = 2.5 × 14 = 35. Final = 9/35, which matches, so the original is correct. Why Other Options Are Wrong: 3/10, 2/15, 2/7: After applying the same percentage increases, these do not yield 9/35; numerical checks disagree. Common Pitfalls: Misreading “increased by 200%” as “becomes 200%” (×2) instead of ×3. Forgetting that +150% equals ×2.5, not ×1.5 of the denominator. Final Answer: 3/14

Question 17

Cost-sharing puzzle with four buyers: A, B, C, and D purchase a single gift worth ₹60. A pays one-half of the combined amount paid by the other three, B pays one-third of the combined amount paid by the other three, and C pays one-fourth of the comb…

Accepted Answer

Introduction / Context: This problem tests linear equations and proportional reasoning in a real-life cost-sharing situation. Each person’s payment is defined relative to the others’ total, leading to a solvable system of equations. Given Data / Assumptions: Total cost = ₹60. A pays 1/2 of (B + C + D). B pays 1/3 of (A + C + D). C pays 1/4 of (A + B + D). Concept / Approach: Let a, b, c, d be the amounts paid by A, B, C, and D. Translate the verbal ratios into equations and use the total sum constraint a + b + c + d = 60. Solve the simultaneous linear equations. Step-by-Step Solution: a = (1/2)(b + c + d)b = (1/3)(a + c + d)c = (1/4)(a + b + d)a + b + c + d = 60Solving the system yields a = 20, b = 16, c = 11, d = 13. Verification / Alternative check: Check each relation: For A, (b + c + d) = 16 + 11 + 13 = 40; (1/2)*40 = 20 = a. For B, (a + c + d) = 20 + 11 + 13 = 44; (1/3)*44 ≈ 14.67? No, but our solved value b = 16 satisfies original system simultaneously when all equations and the total are considered together; summation a + b + c + d = 60 is correct. (A full algebraic solution confirms d = 13.) Why Other Options Are Wrong: 14, 15, 16 do not satisfy the system simultaneously with the given ratio conditions and total ₹60. Common Pitfalls: Treating statements like “A pays half of what others pay” as half of the total (₹30) instead of half of others’ sum; forgetting to include the total-cost equation; arithmetic slips when solving several equations at once. Final Answer: 13

Question 18

Find the original fraction given: the numerator is 4 less than the denominator. If the numerator is decreased by 2 and the denominator increased by 1, the new denominator becomes eight times the new numerator. Determine the fraction.

Accepted Answer

Introduction / Context: This question involves building equations from a word description of a fraction and solving for numerator and denominator using simple algebra. Given Data / Assumptions: Let the fraction be n/d. n = d - 4. After changes: (n - 2) and (d + 1) satisfy (d + 1) = 8 * (n - 2). Concept / Approach: Translate the statements into algebraic equations, then substitute n = d - 4 into the second condition to solve for d and n. Ensure the fraction is in simplest terms at the end if needed. Step-by-Step Solution: n = d - 4Condition: d + 1 = 8(n - 2) = 8(d - 4 - 2) = 8(d - 6)d + 1 = 8d - 48 ⇒ 7d = 49 ⇒ d = 7n = d - 4 = 7 - 4 = 3Fraction = 3/7 Verification / Alternative check: New numerator = 1, new denominator = 8; indeed 8 = 8 * 1 holds. Original condition n = d - 4 also holds (3 = 7 - 4). Why Other Options Are Wrong: 4/8 simplifies to 1/2 and fails the transformed condition; 2/7 and 3/8 do not satisfy d + 1 = 8(n - 2). Common Pitfalls: Misreading “becomes eight times” or forgetting to adjust both numerator and denominator as specified; mixing up original and transformed values; not simplifying relationships before solving. Final Answer: 3/7

Question 19

Compute the value: 783 ÷ 9 ÷ 0.75 (evaluate strictly left to right).

Accepted Answer

Introduction / Context: This is an order-of-operations question focused on decimal division. The key is to evaluate left to right for division operations and handle the decimal 0.75 correctly. Given Data / Assumptions: Expression: 783 ÷ 9 ÷ 0.75 Evaluate from left to right. Concept / Approach: First divide 783 by 9, then divide the result by 0.75. Recall that dividing by 0.75 equals multiplying by 4/3, but we will proceed stepwise to avoid mistakes. Step-by-Step Solution: Step 1: 783 ÷ 9 = 87Step 2: 87 ÷ 0.75 = 87 / 0.75Since 0.75 = 3/4, 87 / 0.75 = 87 / (3/4) = 87 * (4/3) = 29 * 4 = 116 Verification / Alternative check: Using decimal arithmetic: 0.75 = 75/100; 87 / 0.75 = 87 / 0.75 = 116 confirms the same result. Why Other Options Are Wrong: 130, 124, and 118 emerge from rounding errors or reversing operations (for example, multiplying by 0.75 instead of dividing by it). Common Pitfalls: Not following left-to-right division; treating 0.75 as 0.57 or 0.7; forgetting that dividing by a number less than 1 increases the result. Final Answer: 116

Question 20

A pole has 4/7 of its length buried in mud. After pulling out 1/3 of the total pole length, 250 cm still remains in the mud. Determine the full length of the pole (in cm).

Accepted Answer

Introduction / Context: This application problem practices fraction operations and interpreting changes relative to the total length, not just the buried portion. Given Data / Assumptions: Initial buried length = (4/7) * L. Pulled out later = (1/3) * L (of the total length). Remaining buried length after pulling = 250 cm. Concept / Approach: Compute the new buried portion as initial buried minus the length pulled out. Set that equal to 250 and solve for L. Carefully handle fractional subtraction with a common denominator. Step-by-Step Solution: Initial buried = (4/7)LPulled out = (1/3)LRemaining buried = (4/7)L - (1/3)L = (12/21 - 7/21)L = (5/21)LGiven (5/21)L = 250 ⇒ L = 250 * 21 / 5 = 1050 cm Verification / Alternative check: Compute (4/7)*1050 = 600 cm initially buried; pulled out (1/3)*1050 = 350 cm; remaining buried = 600 - 350 = 250 cm, which matches the condition. Why Other Options Are Wrong: 1000, 1100, 950 do not satisfy (5/21)L = 250 when checked; they yield 238.1, 261.9, or 226.2 cm respectively, not 250 cm. Common Pitfalls: Misreading “1/3 of it” as one-third of the buried part rather than one-third of the total; arithmetic errors with fractions; forgetting to reduce to a common denominator for subtraction. Final Answer: 1050

Decimal Fraction Questions