Number system

Answer: (c) 1049

Pages 1 to 999 use 9 + 180 + 2700 = 2889 digits. The remaining 200 digits give 50 four-digit page numbers, 1000 to 1049.

1. Pages 1–9: 9 digits; 10–99: 90 × 2 = 180; 100–999: 900 × 3 = 2700. Total 2889.
2. Digits left = 3089 − 2889 = 200, and 200 ÷ 4 = 50 four-digit pages.
3. These are pages 1000 to 1049, so the book has 1049 pages.

Remember · Use the digit blocks 9, 180, 2700; divide what remains by the next digit length.

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Answer: (b) S2 alone is sufficient to answer the question

Two whole numbers have an odd sum only when one is even and the other odd, and then their product is even. So S2 alone gives a definite 'No'. S1 alone fails: 3 × 5 = 15 is odd but 2 × 3 = 6 is even.

1. S1 alone: p = 3, q = 5 gives pq = 15 (odd); p = 2, q = 3 gives pq = 6 (even) — not sufficient.
2. S2 alone: for whole numbers, an odd sum means one number is even and the other odd.
3. Then pq has an even factor, so pq is even — a definite 'No'. S2 is sufficient.
4. Since S2 works alone, S1 is not needed, so (c) and (d) fail.
5. Like the question itself, this reading takes p and q to be whole numbers.

Remember · In data sufficiency a definite 'No' is an answer; odd + even = odd, and an even factor makes a product even.

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Answer: (d) 8

The units give 7 + 6 = 13 (carry 1), and the hundreds rise from 5 + 1 to 7, so the tens column must also carry: B + 4 = A + 10, i.e. A = B − 6. For 7A3 to be divisible by 3, 10 + A must be a multiple of 3, so A = 2 and B = 8.

1. Units: 7 + 6 = 13 → write 3, carry 1.
2. Hundreds: 5 + 1 + carry = 7, so the tens column must carry 1.
3. Tens: B + 3 + 1 = A + 10 → A = B − 6; so B is 6, 7, 8 or 9 and A is 0, 1, 2 or 3.
4. 7A3 divisible by 3 → 7 + A + 3 = 10 + A is a multiple of 3 → A = 2.
5. B = 8. Check: 587 + 136 = 723, and 7 + 2 + 3 = 12.

Remember · In digit-sum puzzles, go column by column with carries, then use the divisibility rule to fix the digit.

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Answer: (d) 9

Decoded, the expression is 10 − 5 × 1 ÷ 5. Multiplication and division come before subtraction, so 5 × 1 ÷ 5 = 1 and the value is 10 − 1 = 9.

1. 10#5@1$5 = 10 − 5 × 1 ÷ 5.
2. 5 × 1 = 5, then 5 ÷ 5 = 1.
3. 10 − 1 = 9.
4. Check: working strictly left to right would give (10 − 5) × 1 ÷ 5 = 1, which ignores BODMAS.

Remember · After decoding the symbols, apply BODMAS strictly: division and multiplication before subtraction.

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Answer: (c) 4

The known digits add to 30, already a multiple of 3, so B itself must be a multiple of 3: 0, 3, 6 or 9 — four values.

1. 4 + 2 + 5 + 2 + 7 + 4 + 6 = 30.
2. 30 + B is divisible by 3 exactly when B is divisible by 3.
3. B can be 0, 3, 6 or 9 → 4 values.

Remember · Divisibility by 3 depends only on the digit sum; if the known digits already give a multiple of 3, the unknown digit must too.

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Answer: (d) 8

Treat • as one digit x and write every term by place value. The equation 23x + 34 = 210 + x gives x = 8.

1. Let • = x. Then 1• = 10 + x, 2• = 20 + x, •3 = 10x + 3, •1 = 10x + 1 and 21• = 210 + x.
2. Left side: x + (10 + x) + (20 + x) + (10x + 3) + (10x + 1) = 23x + 34.
3. 23x + 34 = 210 + x, so 22x = 176 and x = 8.
4. Check: 8 + 18 + 28 + 83 + 81 = 218, which is 21• with • = 8.

Remember · Write each symbol-number by place value (tens digit × 10 + units digit), then solve one linear equation.

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Answer: (d) 0 and 9

Squaring removes the signs: X² lies between 1 and 9, and Y² between 0 and 1. So X² − Y² is more than 1 − 1 = 0 and less than 9 − 0 = 9.

1. −3 < X < −1 gives 1 < X² < 9.
2. −1 < Y < 1 gives 0 ≤ Y² < 1.
3. Smallest value of X² − Y²: just above 1 − 1 = 0. Largest: just below 9 − 0 = 9.
4. So X² − Y² lies between 0 and 9.
5. Check: X = −2.9 and Y = 0 give 8.41, more than 8, so (c) is too narrow.

Remember · For the range of a difference, pair the largest first term with the smallest second term and vice versa; square before combining.

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Answer: (a) XY

With X at least 2, the product XY is at least 2Y, while Y/X is at most Y/2, X/Y is below 1 and (X + Y)/XY is below 1. XY is always the largest.

1. Try the smallest values X = 2, Y = 3: XY = 6, X/Y ≈ 0.67, Y/X = 1.5, (X + Y)/XY = 5/6.
2. In general X ≥ 2, so XY ≥ 2Y > Y > Y/X.
3. X/Y < 1 because Y > X, and (X + Y)/XY = 1/X + 1/Y ≤ 1/2 + 1/3 < 1.
4. So XY is the largest.

Remember · For 'which is largest' under conditions, test the smallest allowed values first, then confirm with a general inequality.

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Answer: (b) 2

Swapping the first and last digits of a three-digit number changes it by 99 × (difference of those digits). 198 ÷ 99 = 2, so the digits differ by 2 — the number is 103, which becomes 301.

1. Let the number be 100a + c (middle digit 0); interchanged, it is 100c + a.
2. (100c + a) − (100a + c) = 99(c − a) = 198, so c − a = 2.
3. With a + c = 4: a = 1 and c = 3, so the number is 103.
4. Check: 301 − 103 = 198.

Remember · Interchanging the end digits of a three-digit number changes it by 99 × their difference; divide the change by 99.

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Answer: (d) Neither 1 nor 2 nor 3

y is always 8 less than x. When x is positive, y can be positive or negative (x = 10, y = 2; x = 2, y = −6), and when x is negative, y is even more negative. None of the three statements must hold.

1. Rewrite the condition as y = x − 8.
2. x = 2 gives y = −6: both need not be positive.
3. x = 10 gives y = 2: y need not be negative when x is positive.
4. If x is negative, y = x − 8 is below −8, so y can never be positive.

• ✗ 1. x = 2, y = −6 satisfies x − y = 8 with y negative.
• ✗ 2. x = 10, y = 2 satisfies the equation with both positive.
• ✗ 3. A negative x makes y = x − 8 even more negative, never positive.

Remember · For 'must be true' statements, one counter-example disproves each; express one variable in terms of the other first.

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Answer: (a) 47

The mean of five consecutive odd numbers is the middle (third) one, so the third number is 39 and the list starts at 35. The mean of thirteen consecutive odd numbers is the 7th number: 35 + 6 × 2 = 47.

1. Mean of 5 consecutive odd numbers = the 3rd number = 39, so the first number is 39 − 4 = 35.
2. The 13 numbers are 35, 37, …, 59 (35 + 12 × 2 = 59), all 2-digit.
3. Mean of an evenly spaced list = middle (7th) term = 35 + 6 × 2 = 47.
4. Check: (35 + 59) ÷ 2 = 47.

Remember · For an evenly spaced list, the mean equals the middle term, or the average of the first and last terms.

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Answer: (b) 4

If a number and its reverse are both multiples of 7, so is their difference, 99 × (first digit − last digit). Since 99 is not a multiple of 7, the first and last digits must differ by 7 — giving just 168, 861, 259 and 952.

1. Let the number be 100a + 10b + c; its reverse is 100c + 10b + a.
2. Their difference is 99(a − c). Both are multiples of 7 and 99 is not, so a − c must be a multiple of 7.
3. The digits are all different, so a − c = ±7: (a, c) = (1, 8), (8, 1), (2, 9), (9, 2) or (7, 0). The last fails: 7b0 is divisible by 7 only if b = 0 or 7, which repeats a digit.
4. Find the middle digit in each case: 1b8 → 168 = 7 × 24; 8b1 → 861 = 7 × 123; 2b9 → 259 = 7 × 37; 9b2 → 952 = 7 × 136.
5. Each case has exactly one suitable middle digit, so there are 4 such numbers.
6. Check: 861 and 952 are the reverses of 168 and 259, and all four have three different digits.

Remember · A 3-digit number and its reverse differ by 99 × (first − last digit); this turns 'both divisible by 7' into an end-digit condition.

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Answer: (c) 90

Numbers between 99 and 1000 are the 3-digit numbers 100 to 999. With 8 fixed in the units place, the hundreds digit has 9 choices and the tens digit 10 choices, giving 9 × 10 = 90.

1. Numbers strictly between 99 and 1000: 100 to 999.
2. Units digit fixed as 8.
3. Hundreds digit: 1–9 (9 ways); tens digit: 0–9 (10 ways).
4. Count = 9 × 10 = 90.
5. Check: 108, 118, …, 998 rise in steps of 10, so the count is (998 − 108) ÷ 10 + 1 = 90.

Remember · Fix the given digit, then multiply the free choices for the other places — the leading digit cannot be 0.

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Answer: (b) 38

Next year's age is 2 more than last year's, so we need a square that is 2 less than a cube: 25 + 2 = 27. Mr. X was 25, is 26 now and turns 27 next year. The next cube is 64, which is 38 years away from his present age.

1. If last year's age is n² and next year's is m³, then m³ − n² = 2.
2. Try cubes: 8 − 2 = 6 (not a square), 27 − 2 = 25 = 5² ✓, 64 − 2 = 62, 125 − 2 = 123, 216 − 2 = 214 — only 27 works for a human age.
3. So he was 25 last year, is 26 now and will be 27 next year.
4. After 27, the next cube is 4³ = 64.
5. Years to wait from his present age: 64 − 26 = 38.

Remember · 'Square last year, cube next year' means the two numbers differ by 2 — list squares and cubes side by side.

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Answer: (d) 45

A number and its reverse differ by 9 × (difference of the digits). When the larger is less than twice the smaller, the quotient is 1 and the remainder is that difference; the biggest such case is 94 ÷ 49, which leaves 45.

1. Let the larger number be 10a + b and the smaller 10b + a, with a > b.
2. Their difference is 9(a − b). If the larger is less than twice the smaller, the quotient is 1 and the remainder is exactly 9(a − b).
3. Quotient 1 needs 10a + b < 2(10b + a), i.e. 8a < 19b. With a = 9 this needs b ≥ 4, so the best pair is 94 and 49: remainder 9 × 5 = 45.
4. If the quotient is 2 or more, then 8a ≥ 19b forces b ≤ 3, so the smaller number is at most 39 and the remainder is less than 39.
5. Largest possible remainder = 45 (94 = 1 × 49 + 45).

Remember · A number and its reverse differ by 9 × (difference of digits); when the quotient is 1, that difference is the remainder.

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Answer: (d) None of the above

A 2-digit number and its reverse differ by 9 × (difference of the digits), so the digits must differ by 3. That gives 41, 52, 63, 74, 85, 96 and 14, 25, 36, 47, 58, 69 (and 30, if 03 is allowed) — at least 12 numbers, far more than 3, 4 or 5.

1. (10a + b) − (10b + a) = 9(a − b), which is 27 in size when the digits differ by 3.
2. Tens digit 3 more than units: 41, 52, 63, 74, 85, 96 (and 30, if its reverse 03 = 3 is accepted).
3. Tens digit 3 less than units: 14, 25, 36, 47, 58, 69.
4. That is 12 numbers (13 counting 30); even one direction alone gives 6. None of 3, 4 or 5 fits.

Remember · Number − reverse = 9 × (digit difference). Divide the given difference by 9 to get the digit gap, then list the pairs.

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Answer: (b) 342

Pages 1–9 use 9 digits, pages 10–99 use 90 × 2 = 180 digits, and pages 100–150 use 51 × 3 = 153 digits. The total is 342.

1. Pages 1–9: 9 pages × 1 digit = 9.
2. Pages 10–99: 90 pages × 2 digits = 180.
3. Pages 100–150: 51 pages × 3 digits = 153.
4. Total = 9 + 180 + 153 = 342.

Remember · Count digits band by band: pages in a band = last − first + 1, times the digits on each page.

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Answer: (d) 3

All five clubs meet together on days that are common multiples of 2, 3, 4, 5 and 6, that is every LCM = 60 days. Within 180 days that happens on days 60, 120 and 180 — three times.

1. All five meet on a day that is a multiple of 2, 3, 4, 5 and 6.
2. LCM(2, 3, 4, 5, 6) = 60.
3. Within 180 days: days 60, 120 and 180 — 3 times.

Remember · Events repeating every a, b, c … days coincide every LCM(a, b, c …) days.

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Answer: (b) 21

If there are n students, each gives Rs. n, so the collection is n². With the extra Rs. 2 it is 443, so n² = 441 and n = 21.

1. n students each give Rs. n: total = n².
2. n² + 2 = 443, so n² = 441 and n = 21.
3. Check: 21 × 21 = 441; 441 + 2 = 443.

Remember · 'As many rupees as the number of people' makes the total a perfect square — look for the square near the figure.

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Answer: (b) R + S is divisible by 10

Write R = 5a and S = 5b. The difference, product and sum of squares always keep the factor 5 (or 25), but R + S = 5(a + b) is a multiple of 10 only when a + b is even — 5 + 10 = 15 shows it can fail.

1. Let R = 5a and S = 5b, where a and b are different integers.
2. R − S = 5(a − b), R × S = 25ab and R² + S² = 25(a² + b²): always true.
3. R + S = 5(a + b) is divisible by 10 only if a + b is even. Counter-example: R = 5, S = 10 gives 15.

• ✗ (a) R − S = 5(a − b) is always a multiple of 5.
• ✓ (b) Fails for R = 5, S = 10: the sum 15 is not divisible by 10.
• ✗ (c) R × S = 25ab is always a multiple of 25.
• ✗ (d) R² + S² = 25(a² + b²) is always a multiple of 5.

Remember · For 'not necessarily true', try one quick counter-example with the smallest numbers allowed.

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