Question 1

Evaluate (469 + 174)^2 − (469 − 174)^2.

Accepted Answer

Given data Expression: (a + b)^2 − (a − b)^2 with a = 469, b = 174. Concept / Approach Use the identity: (a + b)^2 − (a − b)^2 = 4ab. Step-by-step calculation 4ab = 4 × 469 × 174First compute 469 × 174 = 469×100 + 469×70 + 469×4 = 46900 + 32830 + 1876 = 81606Multiply by 4: 4 × 81606 = 326424 Alternative verification Use difference of squares: (A^2 − B^2) = (A − B)(A + B)A = 469 + 174, B = 469 − 174 ⇒ (2×174)(2×469) = 4×174×469 (same result). Common pitfalls Expanding both squares fully; identity is faster and less error-prone. Final Answer Value = 326424.

Question 2

A man has Rs. 480 in one-rupee, five-rupee, and ten-rupee notes, with an equal number of each. Find the total number of notes.

Accepted Answer

Given data Total amount = Rs. 480 Three denominations: Rs. 1, Rs. 5, Rs. 10 Counts equal ⇒ let each count be x. Concept / Approach Total value = 1·x + 5·x + 10·x = 16x. Step-by-step calculation 16x = 480 ⇒ x = 480 ÷ 16 = 30Total number of notes = 3x = 90 Verification Value check: 30×1 + 30×5 + 30×10 = 30 + 150 + 300 = 480. Common pitfalls Taking 480 ÷ 3 to get x (ignores denomination values). Final Answer Total notes = 90.

Question 3

Two examination rooms A and B have unknown student counts. If 10 students move from A to B, the rooms then have the same number. If instead 20 students move from B to A, room A then has twice as many students as room B. How many students are in room…

Accepted Answer

Given data Let A = x, B = y. x − 10 = y + 10 (after sending 10 from A to B, both equal). x + 20 = 2(y − 20) (after sending 20 from B to A, A is double B). Concept / Approach Form two linear equations in x and y and solve simultaneously. Step-by-step calculation From x − 10 = y + 10 ⇒ x − y = 20 … (1)From x + 20 = 2(y − 20) ⇒ x − 2y = −60 … (2)Subtract (2) from (1): (x − y) − (x − 2y) = 20 − (−60)⇒ y = 80Then x = y + 20 = 100 Verification Move 10: (100 − 10, 80 + 10) = (90, 90) ✔Move 20: (100 + 20, 80 − 20) = (120, 60), and 120 = 2 × 60 ✔ Common pitfalls Adding or averaging student counts instead of setting up equations. Sign mistakes when moving students between rooms. Final Answer 100

Question 4

David starts at the 11th floor and rides up at 57 floors/min; Albert starts at the 51st floor and rides down at 63 floors/min. At which floor do they meet?

Accepted Answer

Given data David: position D(t) = 11 + 57t Albert: position A(t) = 51 − 63t t in minutes, both start at t = 0. Concept / Approach They meet where floors are equal: set D(t) = A(t) and solve for t, then compute the floor. Step-by-step calculation 11 + 57t = 51 − 63t120t = 40 ⇒ t = 1/3 min = 20 sMeeting floor = 11 + 57 × (1/3) = 11 + 19 = 30 Verification Albert: 51 − 63 × (1/3) = 51 − 21 = 30 ✔ Common pitfalls Adding speeds but forgetting to equate positions. Arithmetic slips when handling fractions of a minute. Final Answer 30

Question 5

A fires 5 shots in the time B fires 3 shots. A hits (kills) once in 3 shots; B hits once in 2 shots. When B has missed 27 times, how many has A killed?

Accepted Answer

Given data Rate of fire: A : B = 5 : 3 (same time interval). Accuracy per shot: A hits with probability 1/3; B hits with probability 1/2 (miss prob = 1/2). B has missed 27 times. Concept / Approach From B's miss probability (1/2), infer total shots fired by B; then scale to A's shots using the rate 5:3; finally compute A's kills using his hit probability (1/3). Step-by-step calculation B's miss prob = 1/2 ⇒ expected misses = (1/2) × (B shots)27 = (1/2) × (B shots) ⇒ B shots = 54By firing-rate ratio, A shots = 54 × (5/3) = 90A's kills = (1/3) × 90 = 30 Verification B's expected hits = (1/2) × 54 = 27; misses = 27 (as given). A's kills align with his accuracy. Common pitfalls Using 5:3 as accuracy instead of rate of fire. Confusing B's misses with B's shots. Final Answer 30

Question 6

One-third of Rahul's National Savings Certificate equals one-half of his Public Provident Fund. If his total savings are Rs. 1,50,000, how much is in the Public Provident Fund?

Accepted Answer

Given data NSC = N, PPF = P N/3 = P/2 ⇒ 2N = 3P ⇒ N = (3/2)P N + P = 1,50,000 Concept / Approach Express both amounts in terms of a single variable and use the total. Step-by-step calculation (3/2)P + P = 1,50,000(5/2)P = 1,50,000 ⇒ P = 1,50,000 × (2/5) = 60,000 Verification Then N = 90,000; check: N/3 = 30,000 and P/2 = 30,000 ✔ Final Answer Rs. 60,000

Question 7

If a − b = 3 and a^2 + b^2 = 29, find ab.

Accepted Answer

Concept / Approach Use the identity (a − b)^2 = a^2 + b^2 − 2ab. Step-by-step calculation (a − b)^2 = 3^2 = 9a^2 + b^2 − 2ab = 29 − 2ab = 92ab = 29 − 9 = 20 ⇒ ab = 10 Common pitfalls Using (a + b)^2 identity by mistake; here we need (a − b)^2. Final Answer 10

Question 8

The price of 10 chairs equals that of 4 tables. If 15 chairs and 2 tables cost Rs. 4,000, find the total cost of 12 chairs and 3 tables.

Accepted Answer

Given data Let chair price = c, table price = t. 10c = 4t ⇒ t = (5/2)c = 2.5c. 15c + 2t = 4000. Concept / Approach Express t in terms of c; solve for c from the total, then compute the required combination. Step-by-step calculation 15c + 2(2.5c) = 4000 ⇒ 15c + 5c = 20c = 4000c = 200; then t = 2.5 × 200 = 500Required: 12c + 3t = 12×200 + 3×500 = 2400 + 1500 = 3900 Verification 10c = 2000 and 4t = 2000, consistent with 10 chairs = 4 tables. Final Answer Rs. 3,900

Question 9

Eight people plan to share equally the cost of a rental car. If one withdraws and the remaining seven share the entire cost equally, by how much does each share increase?

Accepted Answer

Given data Total cost = C. Original share = C/8; new share (after one withdraws) = C/7. Concept / Approach The increase is the difference between the new and old per-person shares. Step-by-step calculation Increase = C/7 − C/8 = C(1/7 − 1/8)= C((8 − 7)/56) = C/56 Interpretation Each remaining person pays an extra one fifty-sixth of the total cost. Common pitfalls Reporting the increase relative to the old share instead of the total cost; the question asks for the absolute increase in terms of total cost. Final Answer 1/56 of the total cost

Question 10

Free notebooks were distributed equally among n children. Each child received (n/8) notebooks. If the number of children had been half, each would have received 16 notebooks. How many notebooks were distributed in total?

Accepted Answer

Given data Number of children = n; each got n/8 notebooks. Total notebooks T = n × (n/8) = n^2/8. If children were n/2, each would get 16 ⇒ T ÷ (n/2) = 16. Concept / Approach Equate the two expressions for the same total T to solve for n. Step-by-step calculation T/(n/2) = 16 ⇒ 2T/n = 16 ⇒ T = 8nBut also T = n^2/8 ⇒ n^2/8 = 8nn^2 = 64n ⇒ n = 64 (n > 0)T = 8n = 8 × 64 = 512 Verification Original per-child = n/8 = 64/8 = 8; 64 children × 8 = 512 ✔With n/2 = 32 children: 512/32 = 16 each ✔ Final Answer 512

Question 11

A week has 5 working days of 8 hours each. Regular pay is Rs. 2.40 per hour; overtime pay is Rs. 3.20 per hour. If a man earns Rs. 432 in 4 weeks, how many total hours did he work?

Accepted Answer

Given data Regular hours per week = 5 × 8 = 40 ⇒ in 4 weeks, up to 160 hours at regular rate. Rates: Rs. 2.40/hr (regular), Rs. 3.20/hr (overtime). Total pay = Rs. 432. Concept / Approach Assume he works all 160 regular hours (common in such problems) and the rest, if any, is overtime. Step-by-step calculation Pay for 160 regular hours = 160 × 2.40 = Rs. 384Overtime pay needed = 432 − 384 = Rs. 48Overtime hours = 48 / 3.20 = 15 hoursTotal hours = 160 + 15 = 175 Verification Wage = 160 × 2.40 + 15 × 3.20 = 384 + 48 = 432 ✔ Common pitfalls Calculating all hours at the regular rate or mixing rates; overtime applies beyond 40 hours/week. Final Answer 175 hours

Question 12

A man has hens and cows. If the total number of heads is 48 and the total number of feet is 140, how many hens are there?

Accepted Answer

Given data Let hens = h (2 feet), cows = c (4 feet). Heads: h + c = 48 Feet: 2h + 4c = 140 Concept / Approach Solve the linear system (or use the substitution/elimination method). Step-by-step calculation Divide feet equation by 2: h + 2c = 70Subtract heads: (h + 2c) − (h + c) = 70 − 48 ⇒ c = 22Then h = 48 − 22 = 26 Verification Feet: 2×26 + 4×22 = 52 + 88 = 140 ✔ Final Answer 26

Question 13

A tank takes 25 buckets to fill. If bucket capacity is reduced to two-fifths of the present, how many buckets are needed to fill the tank?

Accepted Answer

Given data Current requirement = 25 buckets. New bucket capacity = two-fifths of the current capacity. Concept / Approach For a fixed tank volume, the number of buckets required is inversely proportional to the capacity per bucket. Step-by-step calculation Let each current bucket hold C units. Tank volume V = 25C.New bucket capacity = (2/5)C.Required buckets N = V / [(2/5)C] = (25C) / ((2/5)C) = 25 × (5/2) = 62.5. Verification Halving capacity would double buckets to 50; reducing to two-fifths is even smaller capacity, so count should be 25 × 2.5 = 62.5 (consistent). Common pitfalls Multiplying by 2/5 instead of dividing by 2/5 (remember inverse proportionality). Forcing an integer answer; the arithmetic result is 62.5. Final Answer 62.5 buckets.

Question 14

The price of 2 sarees and 4 shirts is Rs. 1600. With the same money, one can buy 1 saree and 6 shirts. How much is payable for 12 shirts?

Accepted Answer

Given data 2 sarees + 4 shirts = Rs. 1600. 1 saree + 6 shirts = Rs. 1600. Concept / Approach Let saree price = a, shirt price = b. Form two linear equations and solve. Step-by-step calculation 2a + 4b = 1600 … (1)a + 6b = 1600 … (2)From (2): a = 1600 − 6b.Substitute in (1): 2(1600 − 6b) + 4b = 1600 ⇒ 3200 − 12b + 4b = 1600 ⇒ −8b = −1600 ⇒ b = 200.Then a = 1600 − 6×200 = 400.Cost of 12 shirts = 12 × 200 = Rs. 2400. Verification Check (1): 2×400 + 4×200 = 800 + 800 = 1600 ✔; (2): 400 + 1200 = 1600 ✔. Common pitfalls Mis-eliminating variables or arithmetic slips while substituting. Final Answer Rs. 2400.

Question 15

Rs. 1360 is divided among A, B, and C so that A gets two-thirds of what B gets and B gets one-fourth of what C gets. What is B's share?

Accepted Answer

Given data A = (2/3)B. B = (1/4)C ⇒ C = 4B. Total A + B + C = 1360. Concept / Approach Express all shares in terms of B and solve for B. Step-by-step calculation Sum = (2/3)B + B + 4B = (2/3 + 5)B = (17/3)B.(17/3)B = 1360 ⇒ B = 1360 × 3 / 17 = 80 × 3 = Rs. 240. Verification A = (2/3)×240 = 160; C = 4×240 = 960; total = 160 + 240 + 960 = 1360 ✔. Common pitfalls Inverting the relation B = (1/4)C incorrectly. Final Answer Rs. 240.

Question 16

Apply order of operations (BODMAS/PEMDAS) Compute the value of 42060 / 15 + 5.

Accepted Answer

Introduction / Context: This is a direct arithmetic evaluation using order of operations: division before addition. Such questions test speed and accuracy under basic BODMAS/PEMDAS rules. Given Data / Assumptions: Expression: 42060 / 15 + 5. Operations are division and addition. No parentheses alter the default precedence. Concept / Approach: Evaluate 42060 / 15 first, then add 5 to the quotient. Keeping computations clean avoids transcription errors with larger numbers. Step-by-Step Solution: Divide: 42060 / 15 = 2804 (since 15 * 2804 = 42060).Add: 2804 + 5 = 2809. Verification / Alternative check: Break division: 42060 / 3 = 14020; 14020 / 5 = 2804. Then add 5 → 2809. Same result. Why Other Options Are Wrong: 2804 omits the final addition; 2103, 289, and 2814 result from arithmetic slips or wrong operation order. Common Pitfalls: Adding before dividing; dividing incorrectly by 10 or 5; copying a partial result as final. Final Answer: 2809

Question 17

Simple equation: solve for the missing addend If 9873 + x = 13200, find the value of x.

Accepted Answer

Introduction / Context: This is a straightforward linear equation with one unknown addend. Such problems appear frequently as warm-ups for algebraic manipulation and mental math. Given Data / Assumptions: Equation: 9873 + x = 13200. We need x. Concept / Approach: Isolate x by subtracting 9873 from both sides. This is the inverse operation to addition and yields the unknown directly. Step-by-Step Solution: x = 13200 − 9873.Compute: 13200 − 9800 = 3400; subtract remaining 73 → 3400 − 73 = 3327. Verification / Alternative check: Add back to confirm: 9873 + 3327 = 13200. Correct. Why Other Options Are Wrong: 3237 and 3337 reflect digit transpositions; “None of these” is incorrect because a valid option (3327) exists; 3427 overshoots by 100. Common Pitfalls: Borrowing mistakes in subtraction; rushing and misplacing tens or hundreds. Final Answer: 3327

Question 18

Evaluate using precedence rules Compute the value of (7 + 7 + 7 ÷ 7) / (5 + 5 + 5 ÷ 5).

Accepted Answer

Introduction / Context: This problem demonstrates strict adherence to operator precedence (division before addition) inside both numerator and denominator. Mastery of BODMAS/PEMDAS prevents common mistakes. Given Data / Assumptions: Expression: (7 + 7 + 7 ÷ 7) / (5 + 5 + 5 ÷ 5). No hidden parentheses beyond those shown. Concept / Approach: Resolve the division terms first in both numerator and denominator. Then perform the additions, and finally divide the simplified numerator by the simplified denominator. Step-by-Step Solution: Numerator: 7 + 7 + (7 ÷ 7) = 7 + 7 + 1 = 15.Denominator: 5 + 5 + (5 ÷ 5) = 5 + 5 + 1 = 11.Overall value = 15 / 11. Verification / Alternative check: Convert to mixed numbers if desired: 15/11 = 1 and 4/11, but the fractional form is exact and preferred. Why Other Options Are Wrong: 1 would require numerator = denominator; 1/5 and 3/11 result from misplacing the division; 11/15 is the reciprocal, not the value. Common Pitfalls: Adding before dividing; treating 7 ÷ 7 as 7; forgetting to compute the 5 ÷ 5 term in the denominator. Final Answer: 15 / 11

Question 19

Multi-step simplification with factoring Evaluate [(272 − 32)(124 + 176)] / (17 × 15 − 15).

Accepted Answer

Introduction / Context: This expression features grouped additions and subtractions that should be simplified before multiplication, followed by a final division. It assesses orderly reduction of terms and clean arithmetic. Given Data / Assumptions: Numerator: (272 − 32)(124 + 176). Denominator: (17 × 15 − 15). Standard precedence rules apply inside each bracket. Concept / Approach: First compute each bracket. Then multiply the bracket results for the numerator. Compute the denominator next. Finally, divide to obtain the simplified value. Look for convenient numbers that help mental arithmetic. Step-by-Step Solution: 272 − 32 = 240.124 + 176 = 300.Numerator = 240 × 300 = 72,000.Denominator = 17 × 15 − 15 = 255 − 15 = 240.Value = 72,000 / 240 = 300. Verification / Alternative check: Divide numerator and denominator by 240 directly: (240 × 300) / 240 = 300. This quick cancellation confirms the result. Why Other Options Are Wrong: 0 and 2.25 are far from the computed value; “None of these” is invalid since 300 is available; 150 would be obtained if one mistakenly halved the denominator twice. Common Pitfalls: Multiplying before finishing bracket simplifications; arithmetic slips with 17 × 15; forgetting to subtract 15 after the product. Final Answer: 300

Question 20

Order of operations with implied square Evaluate (69 − 14 × 3 + 2) / [9 × 5 − (5)^2].

Accepted Answer

Introduction / Context: This expression combines multiplication, subtraction, addition, and an implied square. Correct evaluation depends on strict application of precedence: powers, then multiplication/division, then addition/subtraction, all left to right. Given Data / Assumptions: Numerator: 69 − 14 × 3 + 2. Denominator: 9 × 5 − (5)^2. Assume (5)^2 denotes 25. Concept / Approach: Compute products and powers first, then combine with additions/subtractions. After reducing numerator and denominator, perform the final division, simplifying to decimal if appropriate. Step-by-Step Solution: Numerator: 14 × 3 = 42; then 69 − 42 + 2 = 27 + 2 = 29.Denominator: 9 × 5 = 45; (5)^2 = 25; then 45 − 25 = 20.Value = 29 / 20 = 1.45. Verification / Alternative check: Convert to fraction and mixed number: 29/20 = 1 and 9/20 = 1.45; arithmetic is consistent. Why Other Options Are Wrong: 2.75, 26.5, and 265 are orders of magnitude off; 1.2 reflects miscomputing the numerator or denominator. Common Pitfalls: Treating (5)^2 as 10; adding before doing the product 14 × 3; forgetting the final +2 in the numerator. Final Answer: 1.45

Simplification Questions