Number system

Answer: (d) 14

All three touch 199′, and after that their landing marks coincide every LCM(4, 6, 5) = 60 feet. Counting 199, 259, … up to 979 gives 14 common marks in the range.

1. LCM of 4, 6 and 5 is 60, so common marks are 199 + 60k.
2. 199 − 60 = 139 is below 195, so the first common mark in range is 199′ itself.
3. 199 + 60k ≤ 1000 gives k ≤ 13.35, so k = 0, 1, …, 13.
4. Number of common marks = 14 (199′, 259′, …, 979′).

Remember · Common landing points of different step lengths repeat every LCM of the steps; count terms of the resulting sequence.

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Answer: (b) 111

Break the right side into primes: 2³⁷⁵ × 3²⁵ × 5¹¹⁴. The number of 10s is the smaller of the powers of 2 and 5, i.e., 114, and n must hold none of them, so m + 3 = 114.

1. 75²⁵ = (3 × 5²)²⁵ = 3²⁵ × 5⁵⁰.
2. 25³² = 5⁶⁴.
3. 32⁷⁵ = (2⁵)⁷⁵ = 2³⁷⁵.
4. Right side = 2³⁷⁵ × 3²⁵ × 5¹¹⁴, which contains exactly 10¹¹⁴ (limited by the 5s).
5. n = 2²⁶¹ × 3²⁵ is not divisible by 10, so 10ᵐ × 10³ = 10¹¹⁴ and m = 111.

Remember · Powers of 10 in a product = min(power of 2, power of 5). Count both, take the smaller.

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Answer: (b) II only

t + 1/t falls until t = 1 and rises after it, so Statement I breaks when x and y are on opposite sides of 1. Statement II simplifies to √(1/y² + 1) < √(1/x² + 1), which always holds because y > x > 0.

1. Statement I: try x = 1/2, y = 1. Left = 0·5 + 2 = 2·5; right = 1 + 1 = 2. 2·5 < 2 is false, so I is not always correct.
2. Statement II: divide inside the root by the square of the denominator: √(1 + y²)/y = √(1/y² + 1) and √(1 + x²)/x = √(1/x² + 1).
3. Since 0 < x < y, 1/y² < 1/x², so √(1/y² + 1) < √(1/x² + 1). II always holds.
4. Check II with numbers: x = 1, y = 1·5 gives 1·202 < 1·414.

• ✗ I Fails for x = 1/2, y = 1: 2·5 is not less than 2. The function t + 1/t decreases on (1/3, 1).
• ✓ II Equivalent to 1/y² < 1/x², true whenever 0 < x < y.

Remember · For 'always correct', hunt for one counter-example; for the other statement, simplify until the inequality is obvious.

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Answer: (b) 5

Measuring can remove water as well as add it: take 100 litres three times (300 litres) and take out 1 litre twice. That is 5 measurements, and 4 cannot work because 298 is 2 short of 300 and no single cylinder holds 2 litres.

1. 298 = 300 − 2 = 100 + 100 + 100 − 1 − 1: three fills of 100 litres and two removals of 1 litre = 5 measurements.
2. Could 4 work? With at most two 100-litre fills, the other two measures add at most 50 litres: 250 < 298.
3. So three 100-litre fills are needed, leaving one measure to remove exactly 2 litres — impossible with 1, 6, 25 or 100.
4. Minimum = 5.
5. Check: the greedy pour-in-only count 2 × 100 + 3 × 25 + 3 × 6 + 5 × 1 = 13 measures is the trap option.

Remember · For 'minimum measurements', aim just above the target with the big measure, then remove the excess.

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Answer: (c) 13

The product dominates the sum, so put the two largest values in the product for M and the two smallest for N. M = 5 × 4 + 3 = 23 and N = 2 × 3 + 4 = 10, so M − N = 13.

1. Largest: x × y = 5 × 4 = 20, z = 3 (largest value left): M = 23. (5 × 3 + 4 = 19 is smaller.)
2. Smallest: x × y = 2 × 3 = 6, z = 4 (smallest value left): N = 10. (2 × 4 + 3 = 11 is larger.)
3. M − N = 23 − 10 = 13.

Remember · To maximise or minimise xy + z, decide the product first; the added term matters less.

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Answer: (c) 8

Any power of 6 ends in 6. Powers of 7 end in 7, 9, 3, 1 in a cycle of four; 307 leaves remainder 3 on division by 4, so 7³⁰⁷ ends in 3. The product ends in the last digit of 6 × 3 = 18, i.e., 8.

1. 6¹²⁹ ends in 6.
2. 7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1; the cycle length is 4.
3. 307 = 4 × 76 + 3, so 7³⁰⁷ ends in 3.
4. 6 × 3 = 18, so the unit digit is 8.

Remember · Unit digits: 0, 1, 5, 6 never change; 2, 3, 7, 8 cycle every 4 — use the exponent's remainder on division by 4.

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Answer: (c) 3

Powers of 2 run 64, 128, 256, 512, 1024. Only 2⁷, 2⁸ and 2⁹ — 128, 256 and 512 — have three digits.

1. 2⁶ = 64 (two digits), 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, 2¹⁰ = 1024 (four digits).
2. Three-digit powers of 2: 128, 256, 512 — that is 3.

Remember · Know powers of 2 up to 2¹⁰ = 1024 by heart; they settle many CSAT items instantly.

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Answer: (b) 19

5 is the tens digit in 50–59 (10 times) and the units digit in 15, 25, …, 95 (9 times). Counting appearances, 55 contributes two, and the total is 19.

1. Tens place: 50, 51, …, 59 → 10 appearances.
2. Units place: 15, 25, 35, 45, 55, 65, 75, 85, 95 → 9 appearances.
3. Total = 10 + 9 = 19.

Remember · Count digit appearances place by place (tens, then units); a number like 55 is counted once in each place.

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Answer: (c) Both I and II

HCF × LCM equals the product of the two numbers, so a × b = a³ means b = a². The numbers are a and a²: their difference a(a − 1) is a product of consecutive integers and hence even, and a² is a perfect square.

1. For any two numbers, HCF × LCM = a × b.
2. a × b = a³ gives b = a² (a ≠ 1, as the numbers are distinct).
3. Difference = a² − a = a(a − 1), a product of two consecutive integers, so it is even.
4. a² is a perfect square.
5. Check: 3 and 9 — HCF 3, LCM 9, product 27 = 3³; difference 6, and 9 is a square.

• ✓ I a² − a = a(a − 1) is always even.
• ✓ II The second number is a², a perfect square.

Remember · HCF × LCM = product of the two numbers — the first thing to use in any HCF–LCM item.

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Answer: (d) 7

For divisibility by 11, the alternating digit sums must differ by a multiple of 11. Here that gives 7 − (x + y) ≡ 0 (mod 11), so x + y leaves remainder 7 when divided by 11.

1. Digits from the right: 0, 9, 7, y, 5, x, 4.
2. Odd places (1st, 3rd, 5th, 7th): 0 + 7 + 5 + 4 = 16.
3. Even places (2nd, 4th, 6th): 9 + y + x.
4. 16 − (9 + x + y) = 7 − (x + y) must be a multiple of 11.
5. So x + y = 7 or 18; either way the remainder on division by 11 is 7.
6. Check: x = 0, y = 7 gives 4057790 = 11 × 368890.

Remember · Divisibility by 11: (sum of digits in odd places) − (sum in even places) must be 0 or a multiple of 11.

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Answer: (b) 9

A − B = 9 × (difference of digits), so the digit difference must be a multiple of 3: 3 or 6 (9 would need a 0 as the units digit, making B a one-digit number). That gives 6 + 3 = 9 numbers.

1. A = 10a + b, B = 10b + a, so A − B = 9(a − b), with a > b.
2. 9(a − b) is a multiple of 27 when a − b is 3, 6 or 9.
3. B must be two-digit, so b ≥ 1. a − b = 9 would need b = 0 — not allowed.
4. a − b = 3: 41, 52, 63, 74, 85, 96 → 6 numbers.
5. a − b = 6: 71, 82, 93 → 3 numbers.
6. Total = 9. (Allowing b = 0 would give 12 — the trap option.)

Remember · A two-digit number minus its reverse = 9 × (difference of the digits). Remember that the reverse must stay two-digit.

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Answer: (a) 20

Every tile is 3′ wide, so the tiles lie in one row and their lengths must add up to exactly 100: 3a + 7b + 11c = 100. Using mostly 3′ tiles gives at most 32 tiles; using mostly 11′ tiles gives at least 12, so x − y = 20.

1. The strip is 3′ wide and every tile is 3′ on one side (a 7′ or 11′ side cannot fit across), so 3a + 7b + 11c = 100.
2. Maximum: 100 = 3 × 33 + 1 is not possible with 3′ tiles alone; 3 × 31 + 7 = 100 gives 32 tiles — the most.
3. Minimum: 11 × 8 + 3 × 4 = 100 gives 12 tiles (so does 11 × 4 + 7 × 8).
4. 11 tiles cannot work: with a + b + c = 11, the equation becomes 8c + 4b = 67, and the left side is even. Fewer tiles fail the same way or fall short in length.
5. x − y = 32 − 12 = 20.

Remember · Covering a strip: turn it into an equation of lengths, then push towards the smallest (max count) or largest (min count) pieces.

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