Question 1

Equal count of three note types: A man has ₹480 in ₹1, ₹5, and ₹10 notes. The count of notes of each denomination is equal. What is the total number of notes he has?

Accepted Answer

Introduction / Context:This is a neat arithmetic application of equal counts across denominations. The total value is a simple multiple of the sum of the denominations, scaled by the common count. We then multiply by three to get the total number of notes.Given Data / Assumptions:Total money = ₹480.Denominations: ₹1, ₹5, ₹10.Let n be the equal count for each denomination.Concept / Approach:Total value = n*(1 + 5 + 10) = 16n. Solve 16n = 480, then compute total notes = 3n. This linear setup avoids any need for trial and error.Step-by-Step Solution:16n = 480 ⇒ n = 480/16 = 30Total number of notes = 3n = 90Verification / Alternative check:Compute value by denomination: 30*₹1 + 30*₹5 + 30*₹10 = ₹30 + ₹150 + ₹300 = ₹480, exactly the given total.Why Other Options Are Wrong:45, 60, 75, 120: Do not correspond to 3n with n an integer satisfying 16n = 480.Common Pitfalls:Confusing equal value with equal count, or summing denominations incorrectly (1 + 5 + 10 = 16, not 15).Final Answer:90

Question 2

Adult fare is six times child fare: Between Rajpur and Mirpur, one adult ticket costs ₹114, which is six times a child ticket. What is the total fare for 4 adults and 5 children together for the same trip?

Accepted Answer

Introduction / Context:Typical fare-ratio problems use a direct proportional relation between adult and child fares. Given the adult price, we find the child price, and then scale to the specified group counts to compute the total bill.Given Data / Assumptions:Adult fare A = ₹114.A = 6 * child fare (C) ⇒ C = A/6.Group: 4 adults and 5 children.Concept / Approach:Compute the unit child fare, then sum 4*A + 5*C. This is a single-step arithmetic calculation once C is identified from the ratio.Step-by-Step Solution:C = 114 / 6 = ₹19Total = 4*114 + 5*19 = 456 + 95 = ₹551Verification / Alternative check:Check dimensions: rupees across, and integer arithmetic throughout. No rounding is needed as values are exact.Why Other Options Are Wrong:₹505/-, ₹545/-, ₹560/-, ₹572/-: Do not match the exact linear sum 456 + 95.Common Pitfalls:Using 5 children * ₹18 (rounding) or misreading “six times” as “six more.” “Six times” means a multiplier, not an additive increase.Final Answer:₹ 551/-

Question 3

Ages ratio in the future: Two people are currently 36 years and 50 years old. After n years, their ages will be in the ratio 3:4. Find the value of n.

Accepted Answer

Introduction / Context:Age-ratio problems are solved by forming an equation on future ages. Linear growth by n years preserves differences but changes ratios; we equate the target ratio and solve for n directly.Given Data / Assumptions:Current ages: 36 and 50 years.After n years, ratio becomes 3:4.n is nonnegative and integral in typical age problems.Concept / Approach:Write (36 + n)/(50 + n) = 3/4 and cross-multiply to obtain a simple linear equation in n. Ensure the solution is feasible (age cannot be negative, but here it is clearly positive).Step-by-Step Solution:(36 + n)/(50 + n) = 3/44(36 + n) = 3(50 + n)144 + 4n = 150 + 3n ⇒ n = 6Verification / Alternative check:After 6 years: ages are 42 and 56. Ratio 42:56 simplifies to 3:4, confirming the result.Why Other Options Are Wrong:3, 4, 7, 8: Substituting these values does not yield the target 3:4 ratio.Common Pitfalls:Accidentally using (50 − n) or mixing up who is older. Always add n to both ages for “after n years.”Final Answer:6

Question 4

Billing error with sock pairs: A customer ordered 4 pairs of black socks and some pairs of brown socks. A black pair costs twice a brown pair. The clerk mistakenly swapped the counts of black and brown pairs on the bill, increasing the bill by 50%. …

Accepted Answer

Introduction / Context:This question blends unit pricing with a mistaken interchange of quantities. It uses proportional reasoning to compare the correct bill with the incorrect one. Translating prices and quantities into algebra makes the comparison precise.Given Data / Assumptions:Original order: 4 black pairs, x brown pairs.Black pair price = 2p; Brown pair price = p.Clerk swapped counts: billed 4 brown and x black.Mistake increased the bill by 50%.Concept / Approach:Compute the correct bill B and mistaken bill B′, then use B′ = 1.5B to solve for x. Finally reduce the ratio 4 : x to simplest form.Step-by-Step Solution:Correct bill B = 4*(2p) + x*(p) = 8p + xpMistaken bill B′ = 4*(p) + x*(2p) = 4p + 2xpGiven B′ = 1.5B ⇒ 4p + 2xp = 1.5(8p + xp) = 12p + 1.5xp2xp − 1.5xp = 12p − 4p ⇒ 0.5xp = 8p ⇒ x = 16Original ratio (black : brown) = 4 : 16 = 1 : 4Verification / Alternative check:Plug back x = 16: B = 8p + 16p = 24p; B′ = 4p + 32p = 36p = 1.5 * 24p. Works exactly.Why Other Options Are Wrong:2:1, 1:2, 4:1, 3:2: Do not satisfy B′ = 1.5B when substituted.Common Pitfalls:Using “twice” as 2p more instead of 2p total, or interchanging prices rather than quantities.Final Answer:1:4

Question 5

Equal-split bill with one shortfall: Eleven friends share a hotel bill equally. Ten paid ₹60 each; the 11th paid ₹50 more than his equal share to settle the bill. How much did he pay?

Accepted Answer

Introduction / Context:Bill-splitting questions hinge on identifying the equal share and then incorporating stated deviations (someone paying extra) to balance to the total. Here, the last friend covers not just his share but also a shortfall caused by others paying less than their equal shares.Given Data / Assumptions:Total bill = 11 * (equal share s).Ten friends each paid ₹60.Eleventh paid (s + ₹50).Concept / Approach:Express the total paid both as 11s and as the sum of individual payments, then equate to solve for s. Finally, compute the 11th friend’s payment s + 50.Step-by-Step Solution:Total by sum = 10*60 + (s + 50) = 600 + s + 50 = 650 + sBut total must also equal 11s.11s = 650 + s ⇒ 10s = 650 ⇒ s = 65Amount paid by 11th = s + 50 = 65 + 50 = ₹115Verification / Alternative check:Compute total: 10*60 + 115 = 600 + 115 = 715. Equal-share total: 11*65 = 715. Both match.Why Other Options Are Wrong:₹105, ₹110, ₹120, ₹125: Do not satisfy the equality 11s = 650 + s.Common Pitfalls:Assuming the 11th paid ₹110 by adding ₹50 to ₹60; however his equal share is ₹65, so he pays ₹115.Final Answer:₹115

Question 6

Two-digit number with digit relation and reversal effect: The units digit is 1 less than twice the tens digit. After interchanging the digits, the difference (new − original) is 20 less than the original number. Find the original number.

Accepted Answer

Introduction / Context:This problem combines a digit relationship with a condition on the difference between the reversed number and the original. Setting up algebra for digits and using the reversal difference formula make it tractable.Given Data / Assumptions:Tens digit = t; Units digit = u.u = 2t − 1.After swapping digits, difference (new − original) is less than the original by 20 ⇒ (new − original) = original − 20.Concept / Approach:Original = 10t + u; New = 10u + t. Then (new − original) = 9(u − t). Combine this with the linear relation u = 2t − 1 and the “less by 20” equation to solve for t.Step-by-Step Solution:new − original = 9(u − t)Given 9(u − t) = (10t + u) − 20Substitute u = 2t − 1 ⇒ 9((2t − 1) − t) = 10t + (2t − 1) − 209(t − 1) = 12t − 21 ⇒ 9t − 9 = 12t − 21 ⇒ 3t = 12 ⇒ t = 4u = 2*4 − 1 = 7 ⇒ Number = 47Verification / Alternative check:Reversed is 74. new − original = 74 − 47 = 27; original − 20 = 47 − 20 = 27. Condition holds exactly.Why Other Options Are Wrong:59, 23, 35, 41: They fail either the digit relation u = 2t − 1 or the “difference” condition.Common Pitfalls:Misinterpreting “less than the original by 20” as subtracting 20 from the new number. Carefully equate (new − original) to (original − 20).Final Answer:47

Question 7

Direction: In the following question, three quantities I, II and III are given. Compare quantity I, quantity II and quantity III on its basis. (Only quantity is to be considered)  Two equations (a) and (b) are given below: (a) 3x - 4y = 2 (b) x - 2y…

Accepted Answer

Concept & Algebraic Comparison: To compare multiple quantities, solve each independent mathematical problem first. For linear equations, use standard substitution or elimination methods. For profit and loss, use the fundamental formulas linking Cost Price (CP), Marked Price (MP), Markup Percentage, and Discount Percentage. Step-by-Step Solution: • Calculate Quantity I & II: • Equation (a): 3x - 4y = 2 • Equation (b): x - 2y = 8 \Rightarrow 2x - 4y = 16 (multiplying the entire equation by 2) • Subtract the modified equation (b) from (a): (3x - 2x) - (4y - 4y) = 2 - 16 \Rightarrow x = -14. • Substitute x = -14 into (b): -14 - 2y = 8 \Rightarrow -2y = 22 \Rightarrow y = -11. • Therefore, Quantity I = -14 and Quantity II = -11. • Calculate Quantity III: • Given: Marked Price (MP) = 1000, Discount = 200, Markup amount = 250. • Cost Price (CP) = MP - Markup amount = 1000 - 250 = 750. • Markup percent (P) = (\frac{Markup amount}{CP}) × 100 = ((250/750)) × 100 = 33.33%. • Discount percent (Q) = (\frac{Discount}{MP}) × 100 = ((200/1000)) × 100 = 20%. • Value of (Q - P) = 20 - 33.33 = -13.33. • Therefore, Quantity III = -13.33. • Comparison: • We have: I = -14, II = -11, III = -13.33. • Arranging in ascending order: -14 Quantity III > Quantity I. Exam Strategy & Shortcut: When dealing with simple linear equations, coefficient elimination is typically faster than substitution. For the profit/loss section, remember that markup percentage is strictly calculated on the Cost Price (CP), whereas discount percentage is calculated on the Marked Price (MP). Common Pitfall: A very common mistake is calculating the markup percentage (P%) on the Marked Price instead of the Cost Price, which would yield an incorrect 25% instead of 33.33%, altering the final comparison. Final Answer: Therefore, the correct answer is Quantity II > Quantity III > Quantity I.

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