Question 1

Rohan went to the post-office to buy five-rupee, two-rupee and one-rupee stamps. He paid the clerk ₹ $20$, and since he had no change, he gave Rohan three more one-rupee stamps. If the number of stamps of each type that he had ordered initially was …

Accepted Answer

Concept & Logic This is a Linear Diophantine Equation problem with boundary conditions. We need to form an equation for the total cost based on the number of stamps, adjust for the "change" given as extra stamps, and test integer values that satisfy the condition (more than one of each type). Step-by-Step Solution • Understanding the Cost: Rohan gave ₹ 20. The clerk had no change, so instead of returning money, the clerk gave 3 extra one-rupee stamps. This implies the clerk owed Rohan exactly ₹ 3 in change. Therefore, the true cost of Rohan's *initial* order was ₹ 20 - ₹ 3 = ₹ 17. • Setting up the Equation: Let the initial number of ₹ 5 stamps be x. Let the initial number of ₹ 2 stamps be y. Let the initial number of ₹ 1 stamps be z. Total cost: 5x + 2y + z = 17 Condition: He ordered *more than one* of each type, so x \ge 2, y \ge 2, z \ge 2. • Testing Values: Since 5x grows fastest, test x first. If x = 3: Cost is 15. Remaining for y and z is 17 - 15 = 2. But y \ge 2 alone costs ₹ 4. So x cannot be 3. Therefore, x MUST be 2. If x = 2: 5(2) + 2y + z = 17 \Rightarrow 10 + 2y + z = 17 \Rightarrow 2y + z = 7. Now test valid values for y (where y \ge 2): If y = 2: 2(2) + z = 7 \Rightarrow 4 + z = 7 \Rightarrow z = 3. (Valid, as z \ge 2) If y = 3: 2(3) + z = 7 \Rightarrow 6 + z = 7 \Rightarrow z = 1. (Invalid, as z must be \ge 2) So the initial order was exactly: x = 2, y = 2, z = 3. Initial stamps = 2 + 2 + 3 = 7 stamps. • Final Calculation: The question asks for the total number of stamps he *bought* (walked away with). He got his initial 7 stamps, plus the 3 extra one-rupee stamps given as change. Total = 7 + 3 = 10 stamps. Exam Strategy & Shortcut Deduct the change logically first: he bought ₹17 worth of stamps. Since he bought at least 2 of each, subtract the minimum baseline: 2 × ₹5 + 2 × ₹2 + 2 × ₹1 = ₹10 + ₹4 + ₹2 = ₹16. He has ₹1 left to spend. He can only spend it on one more ₹1 stamp. So his base order is 2, 2, 3. That's 7 stamps. Add the 3 given as change to immediately get 10. Common Pitfall Failing to read the final question carefully. Many students will correctly deduce the initial order (7 stamps) and look for it in the options, or they might calculate the total cost instead of the count. Always verify what the final sentence is asking for. Final Answer Therefore, the correct answer is 10.

Question 2

Study the following information carefully to answer these questions:  A young girl Roopa leaves home with $x$ flowers and goes to the bank of a nearby river. On the bank of the river, there are four places of worship, standing in a row. She dips all…

Accepted Answer

Concept & Logic: This problem is a classic iterative sequence puzzle that translates into a linear equation. The core logic involves tracking the number of flowers through repeated operations of multiplying by 2 (doubling) and subtracting a constant y (offering). By establishing an equation for the final state (zero flowers remaining), we can find the exact relationship between the initial number of flowers x and the offered amount y. Step-by-Step Solution: • Given: • Initial flowers = x = 30 • Flowers offered at each temple = y • The number of flowers doubles before each offering. • There are 4 temples, and she ends with 0 flowers. • Calculation: 1. After 1st river & temple: She dips x flowers \rightarrow becomes 2x. She offers y flowers \rightarrow remaining is 2x - y. 2. After 2nd river & temple: Dips remaining \rightarrow becomes 2(2x - y) = 4x - 2y. Offers y \rightarrow remaining is 4x - 2y - y = 4x - 3y. 3. After 3rd river & temple: Dips remaining \rightarrow becomes 2(4x - 3y) = 8x - 6y. Offers y \rightarrow remaining is 8x - 6y - y = 8x - 7y. 4. After 4th river & temple: Dips remaining \rightarrow becomes 2(8x - 7y) = 16x - 14y. Offers y \rightarrow remaining is 16x - 14y - y = 16x - 15y. 5. Equate to final state: We know she is left with 0 flowers. 16x - 15y = 0 16x = 15y 6. Solve for y given x = 30: 16(30) = 15y 480 = 15y y = 32 Exam Strategy & Shortcut: Instead of writing out all 4 steps, you can use a generalized formula for this specific type of puzzle. If an initial quantity x doubles n times, and a constant amount y is subtracted after each doubling, the final amount is given by the formula: Final = 2^n(x) - (2^n - 1)y For n = 4 temples: Final = 2^4(x) - (2^4 - 1)y 0 = 16x - 15y Substitute x = 30, and you immediately get 16(30) = 15y \implies y = 32. Common Pitfall: A very common mistake is trying to calculate the numbers sequentially with actual values instead of algebra, getting lost in the arithmetic, or forgetting to distribute the factor of 2 when doubling the remaining amounts (e.g., writing 2x - y \rightarrow 4x - y instead of the correct 4x - 2y). Always establish the algebraic relationship first. Final Answer: Therefore, the correct answer is 32.

Question 3

$200 \div 25 	imes 4 + 12 - 3 = x$

Accepted Answer

Concept & Rule: This is a direct application of the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction). When mathematical operators appear in a sequence, they must be resolved in this strict hierarchical order to arrive at the correct result, regardless of how they are arranged from left to right. Step-by-Step Solution: Given: 200 \div 25 × 4 + 12 - 3 Calculation: Step 1: Perform Division (D). (200 \div 25) × 4 + 12 - 3 8 × 4 + 12 - 3 Step 2: Perform Multiplication (M). (8 × 4) + 12 - 3 32 + 12 - 3 Step 3: Perform Addition (A). (32 + 12) - 3 44 - 3 Step 4: Perform Subtraction (S). 44 - 3 = 41 Exam Strategy & Shortcut: Chunking: To solve this rapidly under pressure, mentally group the multiplication and division into one "chunk" and the addition/subtraction into another. Visually process 200 \div 25 × 4 as (200/25) × 4 = 8 × 4 = 32. Then append the simple arithmetic: 32 + 9 = 41. You should not need paper for this question. Common Pitfall: The most frequent error is reading left-to-right but accidentally prioritizing multiplication over division based on visual grouping. A student might incorrectly evaluate 25 × 4 = 100 first, and then do 200 \div 100 = 2, resulting in a final wrong answer of 2 + 12 - 3 = 11. Always divide before you multiply if division comes first sequentially in a chunk! Final Answer: Therefore, the correct answer is 41.

Question 4

Solve $1\frac{4}{5} + 20 - 280 \div 25 = x$

Accepted Answer

Concept & Rule: This problem is a direct test of the BODMAS rule mixed with Fraction and Decimal Conversions. To solve this efficiently, it is often easier to convert simple mixed fractions and division operations into decimals, perform the linear addition/subtraction, and convert back to a fraction if required by the options. Step-by-Step Solution: Calculation: Step 1: Convert the mixed fraction into a decimal. 1(4/5) = 1 + 0.8 = 1.8 Step 2: Process the division component. 280 \div 25 To divide by 25 easily, multiply the number by 4 and divide by 100. 280 × 4 = 1120 \rightarrow (1120/100) = 11.2 Step 3: Substitute the decimal values back into the equation. 1.8 + 20 - 11.2 = x Step 4: Perform addition and subtraction from left to right. 21.8 - 11.2 = x x = 10.6 Step 5: Convert the final decimal back to a mixed fraction. 10.6 = 10 + (6/10) = 10(3/5) Exam Strategy & Shortcut: Fraction Isolation: Keep whole numbers and fractions separate to avoid creating large improper fractions. Equation: 1 + (4/5) + 20 - (280/25) Simplify (280/25) by dividing by 5: (56/5) Convert (56/5) to a mixed fraction: 11(1/5) Now evaluate: (1 + 20 - 11) + ((4/5) - (1/5)) Whole numbers: 21 - 11 = 10 Fractions: (3/5) Combine: 10(3/5). This skips decimals entirely! Common Pitfall: Students often convert everything into a massive single fraction, calculating (9/5) + (100/5) - (56/5). While this works here because the denominators happen to align easily to 5, it forms a bad habit that wastes time on harder questions. Treat whole numbers and fractional parts independently when adding or subtracting. Final Answer: Therefore, the correct answer is 10(3/5).

Question 5

Directions: What approximate value should come in place of question mark($x$) in the following questions? (NOTE: You are not expected to calculate the exact value).  $421 \div 35 	imes 299.97 \div 25.05 = x^2$

Accepted Answer

Concept & Strategy: This is an Approximation problem involving a chain of multiplication and division. Because multiplication and division have the same precedence in BODMAS (evaluated left-to-right), rounding the numbers to highly compatible multiples ensures smooth cancellation of fractions without needing to compute large intermediary decimals. Step-by-Step Solution: Calculation: Step 1: Approximate the values to nearest compatible integers. • 421 \approx 420 (Since it is easily divisible by 35) • 35 remains 35 • 299.97 \approx 300 • 25.05 \approx 25 Step 2: Rewrite the equation with approximations. 420 \div 35 × 300 \div 25 = x^2 Step 3: Convert the chain of operations into a single fraction for easier visual cancellation. ((420/35)) × ((300/25)) = x^2 Step 4: Simplify the individual parts. (420/35) = 12 (300/25) = 12 Step 5: Multiply the simplified results. 12 × 12 = x^2 144 = x^2 Step 6: Solve for x. x = 12 Exam Strategy & Shortcut: Fraction Scaling: If (420/35) isn't instantly obvious, scale it up or down. Divide both by 7 to get (60/5) = 12. Similarly for (300/25), think of 100 having four 25s. Therefore, 300 has 3 × 4 = 12 twenty-fives. Realizing both blocks equal 12 means you are looking at 12 × 12 = x^2, which immediately gives x=12 without even calculating 144. Common Pitfall: A severe mistake here is evaluating strictly left to right without converting to fractions, like trying to do (420 \div 35) = 12, then 12 × 300 = 3600, and then 3600 \div 25. While it gives the same answer, dealing with larger numbers like 3600 increases the mental load and risk of simple errors compared to keeping the numbers small and modular. Final Answer: Therefore, the correct answer is 12.

Question 6

Evaluate : $123 	imes 999 + 123$

Accepted Answer

Concept & Formula: This problem can be efficiently solved using the Distributive Law of Multiplication over addition: a × b + a × c = a × (b + c) Step-by-Step Solution: • Step 1: Rewrite the expression to make the common factor explicit: 123 × 999 + 123 × 1 • Step 2: Factor out the common term 123: 123 × (999 + 1) • Step 3: Simplify inside the parentheses and multiply: 123 × 1000 = 123000 Exam Strategy & Shortcut: Distributive Property: Never multiply 123 × 999 directly. Recognize that 999 is 1 less than 1000. Adding 123 to 123 × 999 simply completes the group to make exactly 123 × 1000, which instantly gives 123000. Common Pitfall: • Brute Force Multiplication: Spending valuable time performing long multiplication for 123 × 999 increases the risk of calculation errors. Always look for algebraic simplifications when numbers like 99, 999, or 9999 appear. Final Answer: Therefore, the correct answer is 123000.

Question 7

Solve 4376 + 3209 - 1784 + 97 = 3125 + x

Accepted Answer

Concept & Formula: To find the value of x, group all constants on one side of the equation using the standard rules of arithmetic operations (BODMAS). Step-by-Step Solution: • Step 1: Simplify the left-hand side (LHS) of the equation: Add positive numbers first: 4376 + 3209 + 97 = 7682 Subtract the negative number: 7682 - 1784 = 5898 • Step 2: Equate the simplified LHS to the right-hand side (RHS): 5898 = 3125 + x • Step 3: Isolate x by subtracting 3125 from 5898: x = 5898 - 3125 x = 2773 Exam Strategy & Shortcut: Unit Digit Trick: Calculate the unit digit of both sides: LHS unit digit: 6 + 9 - 4 + 7 = 18 - 4 = 14 \rightarrow 4 RHS expression: 5 + x \rightarrow must end in 4. Therefore, the unit digit of x must be 9 (since 5 + 9 = 14). Options (b) and (c) end in 3, option (d) ends in 93. Let's check the tens digits: LHS last two digits: 76 + 09 - 84 + 97 = 98 RHS last two digits: 25 + x \rightarrow 98 - 25 = 73. Thus, the answer must end in 73, which leaves 2773 as the correct choice. Common Pitfall: • Premature Transposition Error: When shifting 3125 to the other side, make sure its sign flips to negative. Accidentally adding it instead of subtracting is a common point of failure under exam stress. Final Answer: Therefore, the correct answer is 2773.

Question 8

Solve $14 	imes 627 \div \sqrt{1089} = x^3 + 141$

Accepted Answer

Concept & Formula: This problem evaluates operational hierarchy (BODMAS/VBODMAS) along with finding perfect square roots and solving basic cubic equations. • Square root calculation: \sqrt{1089} = 33 Step-by-Step Solution: • Step 1: Find the square root of 1089: \sqrt{1089} = 33 \quad (since 30^2 = 900 and 40^2 = 1600) • Step 2: Substitute back into the equation and apply division first: 14 × (627 \div 33) = x^3 + 141 627 \div 33 = 19 • Step 3: Perform the multiplication: 14 × 19 = 266 • Step 4: Equate and solve for x^3: 266 = x^3 + 141 x^3 = 266 - 141 = 125 • Step 5: Compute the cube root: x = \sqrt[3]{125} = 5 Exam Strategy & Shortcut: Quick Estimation: Knowing basic squares and cubes speeds up this process tremendously. Knowing \sqrt{1089} = 33 simplifies 627/33 to just under 20 (specifically 19). 14 × 19 is roughly 266. Subtracting 141 yields exactly 125. Recognizing 125 as a perfect cube (5^3) immediately reveals x = 5. Common Pitfall: • Wrong Order of Operations: Executing 14 × 627 before dividing by \sqrt{1089} creates massive numbers that are difficult to manage. Always perform division before multiplication according to BODMAS rules. Final Answer: Therefore, the correct answer is 5.

Question 9

If $x + y + z = 0$, then $x^3 + y^3 + z^3 + 3xyz$ is equal to

Accepted Answer

Concept & Formula: This problem is built upon a well-known algebraic identity regarding cubic sums: x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) If the condition x + y + z = 0 is fulfilled, the entire right side becomes zero, leading to: x^3 + y^3 + z^3 = 3xyz Step-by-Step Solution: • Given: x + y + z = 0 • Step 1: From the standard cubic identity, we know that if x+y+z=0, then: x^3 + y^3 + z^3 = 3xyz • Step 2: The question asks for the value of the expression x^3 + y^3 + z^3 + 3xyz. Substitute 3xyz in place of x^3 + y^3 + z^3: (3xyz) + 3xyz • Step 3: Combine the like terms: 6xyz Exam Strategy & Shortcut: Value Substitution Method: Choose simple values for x, y, z that satisfy the condition x + y + z = 0. Let x = 1, y = -1, z = 0. Substitute these test values into the required expression: 1^3 + (-1)^3 + 0^3 + 3(1)(-1)(0) = 1 - 1 + 0 + 0 = 0 Now check the options using our selected values (6xyz = 6(1)(-1)(0) = 0). While option (a) is also 0, trying a second set like 1, 1, -2 quickly eliminates option (a) because 1 + 1 - 8 + 3(1)(1)(-2) = -6 - 6 = -12, matching 6xyz = 6(1)(1)(-2) = -12. Common Pitfall: • Misreading the Expression: Students frequently mix up x^3 + y^3 + z^3 - 3xyz = 0 with the required expression. They assume the answer is 0 automatically without noticing that the question contains a plus sign (+ 3xyz) instead of a minus sign. Final Answer: Therefore, the correct answer is 6 xyz.

Question 10

Direction: In the question given below, if the given mathematical symbols are changed from '+' to '\div', '-' to '	imes', '\div' to '-' and from '	imes' to '+', then choose your answers form the following options.  Solve $67 	imes 119 + 17 - 27 \…

Accepted Answer

Concept & Formula: This problem requires evaluating an expression after performing character-based operator substitutions. Once the operators are replaced, evaluation must follow the standard BODMAS rule hierarchy: 1. Brackets 2. Orders (Powers/Roots) 3. Division / Multiplication (left-to-right) 4. Addition / Subtraction (left-to-right) Step-by-Step Solution: • Given: The original expression is 67 × 119 + 17 - 27 \div 259. • Step 1: Apply the symbol transformations: • × \rightarrow + • + \rightarrow \div • - \rightarrow × • \div \rightarrow - The new expression becomes: 67 + 119 \div 17 × 27 - 259 • Step 2: Perform division according to BODMAS (119 \div 17 = 7): 67 + 7 × 27 - 259 • Step 3: Perform multiplication (7 × 27 = 189): 67 + 189 - 259 • Step 4: Perform addition (67 + 189 = 256): 256 - 259 • Step 5: Complete the subtraction: -3 Exam Strategy & Shortcut: Rapid Substitution & Mental Division: Write the operational signs directly above or below the original question to save time rewriting the digits. Immediately calculate 119 \div 17 = 7. Next, evaluate 7 × 27 = 189. Finally, use units-digit tracking or quick rounding: 67 + 189 = 256, and 256 - 259 = -3. Common Pitfall: • Ignoring Order of Operations: A frequent mistake is performing the calculations linearly from left to right (e.g., executing 67 + 119 first). Always follow the mandatory BODMAS precedence table. Final Answer: Therefore, the correct answer is -3.

Question 11

Solve $\frac{(0.73)^3 + (0.27)^3}{(0.73)^2 + (0.27)^2 - 0.73 	imes 0.27} = x$

Accepted Answer

Concept & Formula: This expression matches the structure of a standard cubic algebraic identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2) Rearranging this identity to match our rational expression format yields: (a^3 + b^3/a^2 - ab + b^2) = a + b Step-by-Step Solution: • Step 1: Assign variable mappings for the decimal figures: Let a = 0.73 and b = 0.27. • Step 2: Substitute these variables into the given expression structural template: (a^3 + b^3/a^2 + b^2 - ab) • Step 3: Expand the cubic terms in the numerator using the identity: ((a + b)(a^2 - ab + b^2)/(a^2 - ab + b^2)) • Step 4: Cancel the identical algebraic expressions from both the numerator and the denominator: a + b • Step 5: Compute the final numeric sum: 0.73 + 0.27 = 1.00 = 1 Exam Strategy & Shortcut: Algebraic Form Matching: Identify the template (a^3 + b^3/a^2 - ab + b^2) immediately upon looking at the structure. Recognize that this template always simplifies directly down to (a + b). The calculation becomes a simple addition exercise: 0.73 + 0.27 = 1. Common Pitfall: • Long Multiplication of Decimals: Attempting to manually calculate (0.73)^3 or (0.27)^3 leads to multi-digit fractional decimals, wasting critical exam runtime and introducing immediate transcription or precision errors. Final Answer: Therefore, the correct answer is 1.

Question 12

For what value of $x$ the statement $\left( \frac{x}{15} \right) \left( \frac{x}{135} \right) = 1$ is true?

Accepted Answer

Concept & Logic When multiplying fractions involving the same unknown variable, combine the numerators and denominators to form a quadratic equation, then isolate the variable. Step-by-Step Solution • Given: ( (x/15) ) ( (x/135) ) = 1 • Calculation: Multiply the fractions on the left side: (x^2/15 × 135) = 1 • Cross-multiply to isolate x^2: x^2 = 15 × 135 • To avoid large multiplication, factor 135 into smaller, manageable pieces to easily find the square root: x^2 = 15 × (15 × 9) x^2 = 15^2 × 3^2 • Take the square root of both sides: x = 15 × 3 • x = 45 Exam Strategy & Shortcut Instead of multiplying 15 × 135 to get 2025 and then struggling to find its square root, break 135 down immediately into 15 × 9. Taking the square root of (15 × 15 × 9) visually reduces straight to 15 × 3 = 45. Common Pitfall Multiplying the denominators completely (15 × 135 = 2025) and then spending excessive time trying to extract the square root of a large number without a calculator. Final Answer Therefore, the correct answer is 45.

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