Question 1

Two-part linear fare model (fixed + per-km): An auto-rickshaw fare consists of a fixed charge plus a per-kilometer charge. A 10 km trip costs ₹85 and a 15 km trip costs ₹120. What will be the fare for a 25 km trip at the same rates?

Accepted Answer

Introduction / Context: Many fare systems combine a base (fixed) fee and a distance-based fee. With two trip-cost data points, we can solve for both unknowns and then predict the cost for any distance under the same pricing scheme. Given Data / Assumptions: Let fixed fee = F and per-km rate = k F + 10k = 85 F + 15k = 120 Concept / Approach: Subtract the two equations to find k, then back-substitute to find F. Use the linear model to compute the 25 km fare. Step-by-Step Solution: (F + 15k) − (F + 10k) = 120 − 85 → 5k = 35 → k = 7F = 85 − 10k = 85 − 70 = 15Fare(25 km) = F + 25k = 15 + 25*7 = ₹190 Verification / Alternative check: Check the 15 km cost: 15 + 15*7 = 15 + 105 = ₹120 (consistent). Why Other Options Are Wrong: ₹175, ₹180, ₹200, and ₹225 do not match F + 25k with F = 15 and k = 7. Common Pitfalls: Assuming direct proportion from 10 km to 25 km without including the fixed charge, which skews the estimate. Final Answer: ₹190

Question 2

Age ratio now, ratio after 10 years: The present ages of Vikas and Vishal are in the ratio 15:8. After 10 years, their ages will be in the ratio 5:3. Find their current ages in years.

Accepted Answer

Introduction / Context: Age-ratio problems often require setting up variables from the initial ratio and then applying a time shift to match a future ratio. Solving the resulting equation yields the scale factor and thus the present ages. Given Data / Assumptions: Now: Vikas : Vishal = 15 : 8 → ages = 15x and 8x After 10 years: (15x + 10) : (8x + 10) = 5 : 3 Concept / Approach: Cross-multiply the future ratio to solve for x. Then compute actual ages by multiplying the ratio terms by x. Step-by-Step Solution: 3(15x + 10) = 5(8x + 10)45x + 30 = 40x + 50 → 5x = 20 → x = 4Ages now: 15x = 60 years; 8x = 32 years Verification / Alternative check: After 10 years: 70 and 42, ratio 70:42 = 5:3, so the condition holds. Why Other Options Are Wrong: Reversed orders or unreduced ratios do not satisfy the future ratio condition when checked. Common Pitfalls: Forgetting to add 10 to both ages or cross-multiplying incorrectly in the ratio equation. Final Answer: 60 years, 32 years

Question 3

Consecutive multiples of 3, fixed sum: The sum of three consecutive multiples of 3 is 72. What is the largest of these three numbers?

Accepted Answer

Introduction / Context: Consecutive multiples introduce a constant step equal to the base multiple. Represent the three numbers with a single parameter and solve the linear sum equation to recover the sequence. Given Data / Assumptions: Let the numbers be 3k, 3k + 3, 3k + 6 Sum = 72 Concept / Approach: Set up the sum: 9k + 9 = 72. Solve for k, then compute the largest term (3k + 6). Step-by-Step Solution: 9k + 9 = 72 → 9k = 63 → k = 7Numbers: 21, 24, 27; largest = 27 Verification / Alternative check: The sum 21 + 24 + 27 = 72 confirms the calculation. Why Other Options Are Wrong: 21 and 24 are smaller members of the set; 36 and 30 are not consistent with the given sum for consecutive triples. Common Pitfalls: Using 3k, 3k + 1, 3k + 2 instead of the correct step size of 3 for multiples of 3. Final Answer: 27

Question 4

Father–son ages with average and multiple relation: The sum of the ages of a father and his son equals 4 times the son’s age. If their average age is 28 years, what is the son’s current age?

Accepted Answer

Introduction / Context: This age problem provides two linked pieces of information: a multiple relationship and an average. Together they determine the son’s age uniquely. Given Data / Assumptions: Father + Son = 4 * Son → Father = 3 * Son Average age = 28 → (Father + Son)/2 = 28 Concept / Approach: Use the average to get the total, then apply the multiple relation to solve for the son’s age directly. Step-by-Step Solution: Total = 2 * 28 = 56But Total = Father + Son = 4 * SonSo, 4 * Son = 56 → Son = 14 years Verification / Alternative check: Father = 42; average = (42 + 14)/2 = 28; sum 56 equals 4 times the son’s age, consistent. Why Other Options Are Wrong: 16 and 12 do not satisfy both the multiple and average conditions; “Data inadequate” is false because the two conditions are sufficient. Common Pitfalls: Interpreting “sum is 4 times the son’s age” incorrectly as the father being 4 times the son’s age—watch the exact wording. Final Answer: 14 years

Question 5

Product and sum of two numbers (quadratic recovery): The product of two numbers is 192 and their sum is 28. What is the smaller of the two numbers?

Accepted Answer

Introduction / Context: When sum and product of two numbers are known, the numbers are roots of the quadratic t^2 − (sum)t + (product) = 0. Solving the quadratic gives both values; choose the smaller as required. Given Data / Assumptions: x + y = 28 xy = 192 Concept / Approach: Construct the quadratic: t^2 − 28t + 192 = 0 and solve via factoring or the quadratic formula. Step-by-Step Solution: t^2 − 28t + 192 = 0Discriminant = 28^2 − 4*192 = 784 − 768 = 16Roots = (28 ± 4)/2 → 16 and 12Smaller number = 12 Verification / Alternative check: 12 + 16 = 28 and 12 * 16 = 192, both conditions satisfied. Why Other Options Are Wrong: 14, 18, and 10 do not form a pair with another number to meet both the sum and product simultaneously; 16 is the larger root. Common Pitfalls: Arithmetic slips with the discriminant or mixing up which root is smaller. Final Answer: 12

Question 6

Three consecutive even numbers, sum equals a fraction of 176 minus 14: The sum of three consecutive even numbers is 14 less than one-fourth of 176. What is the middle number?

Accepted Answer

Introduction / Context: This problem involves setting up consecutive even integers and equating their sum to a numerical expression derived from 176. Careful handling of the phrase “14 less than one-fourth of 176” is key. Given Data / Assumptions: Numbers: n, n + 2, n + 4 (all even) One-fourth of 176 = 44; sum = 44 − 14 = 30 Concept / Approach: Form the sum equation and solve for n. The middle number is n + 2. Step-by-Step Solution: n + (n + 2) + (n + 4) = 303n + 6 = 30 → 3n = 24 → n = 8Middle number = n + 2 = 10 Verification / Alternative check: Numbers are 8, 10, 12 with sum 30. One-fourth of 176 is 44; 44 − 14 = 30. Condition holds. Why Other Options Are Wrong: 6 and 8 are not the middle term; 12 is the largest; “Data inadequate” is invalid as the information is sufficient. Common Pitfalls: Interpreting “14 less than one-fourth of 176” incorrectly as 14 less than 176/4 after other operations; compute 176/4 first, then subtract 14. Final Answer: 10

Question 7

Three-number chain ratios totaling 300: The sum of three numbers is 300. If the ratio between the first and second is 5:9 and the ratio between the second and third is 9:11, find the second number.

Accepted Answer

Introduction / Context: When two ratios share a common term (here the second number), they can be chained to produce a three-number proportion. With the total sum given, we can scale the base ratio to match the sum and extract the requested term. Given Data / Assumptions: A:B = 5:9 B:C = 9:11 A + B + C = 300 Concept / Approach: Because B is common, align the ratios to 5:9 and 9:11 → the triplet is A:B:C = 5:9:11. Sum the parts, set equal to 300, solve for the scale factor, and compute B. Step-by-Step Solution: A:B:C = 5:9:11 (since the 9s align)Total parts = 5 + 9 + 11 = 25Each part = 300 / 25 = 12Second number B = 9 * 12 = 108 Verification / Alternative check: Check totals: A = 60, B = 108, C = 132; sum = 300. Ratios A:B = 5:9 and B:C = 9:11 hold. Why Other Options Are Wrong: 12 and 60 are single-part or A values; 132 is C; 90 does not satisfy the chained ratios when scaled to 300. Common Pitfalls: Mismatching the shared term when chaining ratios or adding parts incorrectly. Final Answer: 108

Question 8

Linear equation from a percent-like fraction of a number: When 20 is subtracted from a number, the result equals seven-twelfths of that number. What is the sum of the digits of the number?

Accepted Answer

Introduction / Context: This algebra problem expresses a number reduced by 20 as a fixed fraction of itself. Solving the resulting linear equation gives the number, and then its digit sum is requested. Given Data / Assumptions: N − 20 = (7/12) * N N is a whole number (implied by digit-sum question) Concept / Approach: Bring all N terms to one side and constants to the other. Solve for N and compute its digit sum. Step-by-Step Solution: N − (7/12)N = 20(5/12)N = 20 → N = 20 * (12/5) = 48Sum of digits of 48 = 4 + 8 = 12 Verification / Alternative check: Check: 48 − 20 = 28 and (7/12)*48 = 28, consistent. Why Other Options Are Wrong: 44 and 48 are the number itself or irrelevant; 16 and 10 are incorrect digit sums for N = 48. Common Pitfalls: Sign errors or misinterpreting “seven-twelfths of the number” as something like N − 7/12 rather than (7/12) * N. Final Answer: 12

Question 9

Scaling a combined purchase (pens and pencils): The total cost of 36 pens and 42 pencils is ₹460. What is the cost of 18 pens and 21 pencils at the same unit prices?

Accepted Answer

Introduction / Context: Because the basket (18 pens, 21 pencils) is exactly half of (36 pens, 42 pencils), the total cost scales linearly by the same factor when unit prices remain constant. Given Data / Assumptions: Cost(36 pens + 42 pencils) = ₹460 Unit prices fixed; no discounts or bulk changes Target: 18 pens + 21 pencils (half of both counts) Concept / Approach: Linear scaling applies: if quantities are halved, the total cost is halved. Step-by-Step Solution: Target basket = 1/2 of the original basketCost = 1/2 * ₹460 = ₹230 Verification / Alternative check: Let pen price be p and pencil price be q. Then 36p + 42q = 460. For 18p + 21q, exactly half: (36p + 42q)/2 = 230. Why Other Options Are Wrong: ₹203, ₹302, ₹320, ₹240 do not match the linear scaling by 1/2 of the original total cost. Common Pitfalls: Overthinking and trying to solve for p and q individually; not needed when the target combination is a fixed scalar multiple. Final Answer: ₹230/-

Question 10

Linear scaling of a mixed basket (pens and pencils): The cost of 5 pens and 8 pencils is ₹31 at fixed unit prices. What is the cost of 15 pens and 24 pencils at the same rates?

Accepted Answer

Introduction / Context: Here, the target basket (15 pens, 24 pencils) is exactly 3 times the original basket (5 pens, 8 pencils). With constant unit prices, the total cost scales by the same factor. Given Data / Assumptions: Cost(5 pens + 8 pencils) = ₹31 Target: 15 pens + 24 pencils = 3 * (5 pens + 8 pencils) Unit prices are constant. Concept / Approach: Apply linear scaling: multiply the original cost by 3. Step-by-Step Solution: Cost(target) = 3 * ₹31 = ₹93 Verification / Alternative check: Let pen be p and pencil be q. Then 5p + 8q = 31. The target cost is 3*(5p + 8q) = 93. Why Other Options Are Wrong: ₹99 and ₹96 are not integral multiples of ₹31; “Cannot be determined” is incorrect because the scaling is exact; ₹90 is also not 3 times ₹31. Common Pitfalls: Trying to compute p and q separately, which is unnecessary when the target is a simple multiple. Final Answer: ₹93

Question 11

Linear system substitution: If 2x + 3y = 29 and y = x + 3, find the value of x.

Accepted Answer

Introduction / Context: This is a straightforward two-equation linear system where one equation already expresses y in terms of x. Substitution directly yields x. Given Data / Assumptions: 2x + 3y = 29 y = x + 3 Concept / Approach: Substitute y into the first equation and solve for x. Then y can be found if desired. Step-by-Step Solution: 2x + 3(x + 3) = 292x + 3x + 9 = 29 → 5x = 20 → x = 4 Verification / Alternative check: Then y = x + 3 = 7; check: 2*4 + 3*7 = 8 + 21 = 29, consistent. Why Other Options Are Wrong: 5, 6, 7, 8 do not satisfy the substituted equation 2x + 3(x + 3) = 29. Common Pitfalls: Arithmetic slips in combining like terms or mis-substituting y. Final Answer: 4

Question 12

Solve a 2×2 linear system, then sum: If 3x + 7y = 75 and 5x − 5y = 25, find the value of x + y.

Accepted Answer

Introduction / Context: We are given a pair of linear equations and asked for x + y directly. Solving the system gives x and y, which we can then add. Alternatively, manipulation can sometimes find x + y without individual values. Given Data / Assumptions: 3x + 7y = 75 5x − 5y = 25 Concept / Approach: First simplify the second equation to x − y = 5. Substitute y = x − 5 into the first and solve for x. Then compute x + y. Step-by-Step Solution: From 5x − 5y = 25 → x − y = 5 → y = x − 53x + 7(x − 5) = 75 → 3x + 7x − 35 = 7510x = 110 → x = 11 → y = 6x + y = 11 + 6 = 17 Verification / Alternative check: Plug back: 3*11 + 7*6 = 33 + 42 = 75 and 5*11 − 5*6 = 55 − 30 = 25, both hold. Why Other Options Are Wrong: 14, 15, 16, and 18 are not equal to the verified sum 17 for the unique solution. Common Pitfalls: Missing the simple reduction of the second equation by dividing by 5, or sign errors when substituting y. Final Answer: 17

Question 13

Evaluate an expression after solving a 2×2 system: If x + y − 7 = 0 and 3x + y − 13 = 0, then compute the value of 4x^2 + y^2 + 4xy.

Accepted Answer

Introduction / Context: This problem couples a small linear system with an algebraic expression. Note that 4x^2 + y^2 + 4xy equals (2x + y)^2, which suggests a quick path once 2x + y is found from the system. Given Data / Assumptions: x + y = 7 3x + y = 13 Concept / Approach: Subtract equations to get x, find y, then compute 2x + y and square it. This avoids expanding 4x^2 + y^2 + 4xy directly. Step-by-Step Solution: (3x + y) − (x + y) = 13 − 7 → 2x = 6 → x = 3Then y = 7 − x = 4Compute 2x + y = 2*3 + 4 = 10Value = (2x + y)^2 = 10^2 = 100 Verification / Alternative check: Direct expansion: 4x^2 + y^2 + 4xy with x = 3, y = 4 gives 36 + 16 + 48 = 100, confirming. Why Other Options Are Wrong: 75, 85, 91, 110 do not match the computed value from the uniquely determined solution. Common Pitfalls: Expanding incorrectly or missing the identity (2x + y)^2, which simplifies the calculation. Final Answer: 100

Question 14

Solve a linear pair to find x: If 3x + y = 81 and 81x − y = 3, determine the value of x.

Accepted Answer

Introduction / Context: We are given two linear equations in x and y. Adding them immediately eliminates y, giving x directly without the need for substitution or more complex algebra. Given Data / Assumptions: 3x + y = 81 81x − y = 3 Concept / Approach: Use elimination by addition: adding the two equations cancels y. Solve for x and then (optionally) back-solve for y to verify. Step-by-Step Solution: (3x + y) + (81x − y) = 81 + 384x = 84 → x = 1(Optional) y = 81 − 3x = 81 − 3 = 78 Verification / Alternative check: Check the second equation: 81*1 − 78 = 3, which holds. Why Other Options Are Wrong: 17/16, 17/8, 17/4, and 15/4 do not satisfy both equations; only x = 1 does. Common Pitfalls: Attempting substitution and making arithmetic slips with large coefficients. Elimination here is faster and reduces error risk. Final Answer: 1

Question 15

Solve the simultaneous equations in real numbers: 6x + 3y = 7xy and 3x + 9y = 11xy. Find the ordered pair (x, y) that satisfies both equations (choose the non-trivial solution if more than one arises).

Accepted Answer

Introduction: We are given two equations with product terms (xy), which suggests rearranging to isolate either x or y, or converting to a form that allows elimination. The task is to find the ordered pair (x, y) that satisfies both equations simultaneously. Non-trivial solutions are preferred if multiple solutions arise. Given Data / Assumptions: Equations: 6x + 3y = 7xy and 3x + 9y = 11xy. Real-number solutions; trivial (x, y) = (0, 0) may appear but the meaningful solution has nonzero variables. Standard algebraic operations are valid. Concept / Approach: Treat each equation as linear in one variable after moving the xy-term to one side. Factor where possible, then eliminate using substitution or simultaneous methods. Because xy multiplies both variables, dividing by a variable is permitted only if that variable is nonzero; we keep track of this carefully. Step-by-Step Solution: Equation (1): 6x + 3y = 7xy → bring terms together.Equation (2): 3x + 9y = 11xy → bring terms together.From (1): 7xy − 6x − 3y = 0 → factor by grouping is difficult; switch to elimination.From (2): 11xy − 3x − 9y = 0.Treat (1) as linear in x: 7xy − 6x = 3y → x(7y − 6) = 3y → x = (3y)/(7y − 6), provided 7y − 6 ≠ 0.Substitute into (2): 3x + 9y = 11xy.3*(3y/(7y − 6)) + 9y = 11*(3y/(7y − 6))*y.Multiply throughout by (7y − 6) to clear denominators and solve; this yields y = 3/2 (nonzero) and then x = 1.A trivial solution (0, 0) also satisfies both, but the non-trivial solution is (1, 3/2). Verification / Alternative check: Substitute x = 1, y = 3/2 into both equations: (1) 6*1 + 3*(3/2) = 6 + 4.5 = 10.5; 7*1*(3/2) = 10.5. (2) 3*1 + 9*(3/2) = 3 + 13.5 = 16.5; 11*1*(3/2) = 16.5. Both hold exactly. Why Other Options Are Wrong: Each alternative fails one (or both) equations upon substitution; signs and magnitudes do not satisfy both identities simultaneously. Common Pitfalls: Dividing by zero inadvertently or discarding valid solutions. Always verify candidates in both equations. Final Answer: x = 1, y = 3/2

Question 16

Deepak keeps hens (2 legs each) and goats (4 legs each). If the total number of animal heads is 90 and the total number of legs is 248, how many goats does Deepak have?

Accepted Answer

Introduction: Classic heads-and-legs problems are solved using linear equations. Each animal contributes one head; hens contribute 2 legs and goats contribute 4 legs. Two equations quickly reveal the counts of each type. Given Data / Assumptions: Total heads = 90 (hens + goats). Total legs = 248 (2 per hen, 4 per goat). Let h = hens, g = goats; both are nonnegative integers. Concept / Approach: Use the system: h + g = 90 and 2h + 4g = 248. Divide the second by 2 to simplify, then eliminate h to solve for g. Step-by-Step Solution: h + g = 90 … (1)2h + 4g = 248 → h + 2g = 124 … (2)Subtract (1) from (2): (h + 2g) − (h + g) = 124 − 90 → g = 34.From (1): h = 90 − 34 = 56 (sanity check). Verification / Alternative check: Legs: 2*56 + 4*34 = 112 + 136 = 248, matches perfectly. Why Other Options Are Wrong: 32, 36, 38 lead to leg counts not equal to 248; “Cannot be determined” is incorrect because the system has a unique integer solution. Common Pitfalls: Forgetting each animal has one head; mixing leg coefficients; or not simplifying the equations. Final Answer: 34

Question 17

Two digits have a sum of 15 and their difference is 3 (order unspecified). What is the product of the two digits?

Accepted Answer

Introduction: This is a simple linear system about two digits. We know the sum and the absolute difference, which uniquely determines the pair (up to order). Once the digits are found, compute their product. Given Data / Assumptions: Let the digits be a and b with 0 ≤ a,b ≤ 9. a + b = 15. |a − b| = 3. Concept / Approach: Solve the system using simultaneous equations. Consider both possibilities for the difference: a − b = 3 or b − a = 3. Either way, the product is the same. Step-by-Step Solution: Case 1: a − b = 3. Then adding with a + b = 15 gives 2a = 18 → a = 9, b = 6 → product = 54.Case 2: b − a = 3. Then adding with a + b = 15 gives 2b = 18 → b = 9, a = 6 → product = 54.Thus, regardless of order, product is 54. Verification / Alternative check: Try both pairs (9, 6) and (6, 9). Sum 15, difference 3, product 54 in both cases. Why Other Options Are Wrong: 56, 63, 42 do not correspond to the product of any pair of digits whose sum is 15 and difference is 3; “Cannot be determined” is incorrect because the product is fixed. Common Pitfalls: Thinking order affects product (it does not) or miscalculating the difference constraint. Final Answer: 54

Question 18

Consider the fraction with numerator 6x + 1 and denominator 7 − 4x. For x in the closed interval [-2, 2], determine all values of x for which the numerator is greater than the denominator (i.e., 6x + 1 > 7 − 4x).

Accepted Answer

Introduction: We must compare the linear expressions 6x + 1 and 7 − 4x over the interval [-2, 2]. The task reduces to solving a first-degree inequality and then intersecting with the given domain. Given Data / Assumptions: Inequality: 6x + 1 > 7 − 4x. Domain: −2 ≤ x ≤ 2. Strict inequality is required (“greater than”). Concept / Approach: Rearrange to gather x-terms on one side and constants on the other. Solve for x and then apply the domain restriction to express the final answer in interval language. Step-by-Step Solution: 6x + 1 > 7 − 4x6x + 4x > 7 − 1 → 10x > 6x > 6/10 = 3/5.Intersect with domain −2 ≤ x ≤ 2 gives (3/5, 2]. Verification / Alternative check: Test a point: x = 1 (allowed). 6(1)+1 = 7; 7 − 4(1) = 3 → 7 > 3 holds. Test x = 0.5 (just below 3/5): 6(0.5)+1 = 4; 7 − 4(0.5) = 5 → 4 > 5 is false; boundary logic confirmed. Why Other Options Are Wrong: Options including “≤” at 3/5 treat the strict inequality incorrectly; other ranges either over- or under-include values outside the solution set. Common Pitfalls: Using ≥ instead of > at the boundary or forgetting to intersect with the domain [-2, 2]. Final Answer: 3/5 < x ≤ 2

Question 19

Solve the system (x/4) + (y/3) = 10/24 and (x/2) + y = 1, and find the value of x + y.

Accepted Answer

Introduction: Two linear equations in two variables can be solved by substitution or elimination. After finding x and y, compute x + y. The fractions reduce cleanly, making arithmetic straightforward. Given Data / Assumptions: (x/4) + (y/3) = 10/24 = 5/12. (x/2) + y = 1. x, y are real numbers. Concept / Approach: Convert the first equation to a common denominator, or express one variable from the second equation and substitute into the first. Solve precisely to avoid rounding. Step-by-Step Solution: From (x/2) + y = 1 → y = 1 − x/2.Substitute into (x/4) + (y/3) = 5/12:x/4 + (1 − x/2)/3 = 5/12 → x/4 + 1/3 − x/6 = 5/12.Take common denominator 12: 3x − 2x + 4 = 5 → x = 1.Then y = 1 − 1/2 = 1/2 → x + y = 3/2. Verification / Alternative check: Plug x = 1, y = 1/2 back: LHS of first = 1/4 + 1/6 = 5/12; second = 1/2 + 1/2 = 1. Both satisfy the system. Why Other Options Are Wrong: 1, 2, 4, and 5/2 do not match the computed sum x + y = 3/2. Common Pitfalls: Mishandling fractional arithmetic or failing to reduce 10/24 to 5/12 correctly. Final Answer: 3/2

Question 20

The combined cost of 21 pencils and 9 clippers is ₹ 819 (uniform pricing per item). What is the cost of 7 pencils and 3 clippers together?

Accepted Answer

Introduction: When pricing is linear and uniform per item, any proportional subset of a given combination costs the same proportion of the total. Here, (7 pencils + 3 clippers) is exactly one-third of (21 pencils + 9 clippers). Given Data / Assumptions: 21 pencils + 9 clippers cost ₹ 819 in total. Uniform price per pencil and per clipper (no bulk discounts). We need cost of 7 pencils + 3 clippers. Concept / Approach: Note that (7, 3) is exactly one-third of (21, 9). Therefore, cost scales by 1/3. Step-by-Step Solution: Total combo (21, 9) → ₹ 819.Desired combo (7, 3) = (1/3) * (21, 9).Cost = (1/3) * 819 = ₹ 273. Verification / Alternative check: Let pencil = p, clipper = c. Then 21p + 9c = 819. Divide by 3: 7p + 3c = 273. Same result. Why Other Options Are Wrong: Other figures do not maintain the linear proportionality to the original combination cost. Common Pitfalls: Attempting to find individual prices unnecessarily; proportional reasoning suffices. Final Answer: ₹ 273

Linear Equation Questions