Number System, LCM/HCF & Divisibility: previous-year questions

Answer: (C) 115

With $c$ correct, $u$ unattempted and the rest wrong ($c + u \le 30$), the score is $4c + u$. Scores of 113 ($c = 28$, $u = 1$), 114 ($c = 28$, $u = 2$) and 117 ($c = 29$, $u = 1$) are all possible. For 115 we would need $c = 28$ with $u = 3$ (31 questions) or $c = 29$ with $u = -1$, both impossible, so 115 cannot be scored. In fact, near the maximum of 120 the unattainable scores are 119, 118 and 115. Method for such questions: fix the number of correct answers at the highest value that does not overshoot, then see whether the remaining marks fit into the remaining questions.

Answer: (C) $h^2 + L^2$

For two numbers, $ab = hL$, and here $a + b = h + L$. So $a^2 + b^2 = (a + b)^2 - 2ab = (h + L)^2 - 2hL = h^2 + L^2$. Option B is $a + b$ itself, not the sum of squares.

Answer: (B) 57463822

With boys : girls = 7 : 4, the total must be a multiple of 11. Using the alternating-digit test, 29435417, 28444625 and 29434526 each give 11, so they are divisible by 11. 57463822 gives 9, so it is not a multiple of 11 and cannot be the total.

Answer: (D) 40

Let the number of correct answers be $x$; then wrong answers $=75-x$, since all questions were attempted. Marks: $2x-(75-x)=45$, so $3x=120$ and $x=40$. Check: $40\times2-35\times1=80-35=45$. Shortcut: if all 75 were right the score would be 150; each wrong answer costs 3 marks (2 lost + 1 deducted), so the number of wrong answers is $(150-45)\div3=35$, leaving 40 right.

Answer: (D) 9

Write the two numbers around their mean, 8795: $8798\times8792=(8795+3)(8795-3)=8795^2-3^2=8795^2-9$. Adding 9 gives $8795^2$, a perfect square, and 9 is the least such number, because the product lies just below $8795^2$ and far above the previous square $8794^2$ (which is $2\times8795-1=17589$ less than $8795^2$). In general, for two numbers that differ by $2k$, the product falls short of the square of their mean by $k^2$.

Answer: (D) 69

Let the numbers be $x, x+2, x+4, x+6, x+8, x+10$. Their average is $x + 5 = 64$, so $x = 59$ and the greatest is $x + 10 = 69$. Shortcut: for an even count of consecutive odd numbers the average lies midway between the two middle terms (63 and 65), giving 59, 61, 63, 65, 67, 69.

Answer: (A) 7

Let there be $c$ cows and $h$ hens. Heads $= c + h$ and legs $= 4c + 2h$. Given $4c + 2h = 2(c + h) + 14$, the hen terms cancel and $2c = 14$, so $c = 7$. The number of hens cannot be found and is not needed. Shortcut: every cow has 2 legs more than 'twice its head', so cows $= \frac{14}{2} = 7$.

Answer: (C) $\frac{4}{5}$

Convert to decimals: $\frac{3}{4} = 0.75$ and $\frac{5}{6} \approx 0.833$. Among the options $\frac{2}{3} \approx 0.667$, $\frac{1}{2} = 0.5$, $\frac{4}{5} = 0.8$ and $\frac{9}{10} = 0.9$, so only $\frac{4}{5}$ lies between the two. With LCM 60: $\frac{45}{60} < \frac{48}{60} < \frac{50}{60}$.

Answer: (C) 100.101

Convert the integer and fractional parts separately. Integer part: $4=(100)_2$. Fractional part: $0.625=\frac{1}{2}+\frac{1}{8}=(0.101)_2$ - or multiply repeatedly by 2: $0.625\times2=1.25$ (1), $0.25\times2=0.5$ (0), $0.5\times2=1.0$ (1). Hence $4.625=(100.101)_2$. Check the distractors: $100.001=4.125$, $100.110=4.75$, $100.011=4.375$. The place values after the binary point are $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$

Answer: (B) 6, 8

Multiplying 6789 by 5 gives 33945, so the missing digits are 6 (thousands place) and 8 (units place).

Answer: (C) 4

A composite number must have more than two factors; 1 isn't composite (or prime), and 2, 3 are prime, so 4 (factors 1, 2, 4) is the smallest composite number.

Answer: (B) 19

Counting occurrences of the digit 7 in the units place (17, 27, ..., 97 — 9 times) and the tens place (70-79 — 10 times) gives a total of 19.

Answer: (B) 24

For fractions, $\text{LCM} = \dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$. Here, LCM of numerators $(4,8,3) = 24$, and HCF of denominators $(3,9,5) = 1$, so the LCM of the three fractions is $\dfrac{24}{1}=24$.

Answer: (B) 301

The number of boxes must leave a remainder of 1 when divided by the LCM of 3, 4, 5 and 6 (which is 60), so it has the form $60k+1$, and must also be exactly divisible by 7. Testing $60k+1$ for divisibility by 7 gives the smallest valid value at $k=5$: $60\times5+1=301$, and $301 \div 7 = 43$ exactly.

Answer: (B) 2.002 < 2.02 < 2.2 < 2.222

Write all the numbers to three decimal places: 2.002, 2.020, 2.200, 2.222. Now the order is clear: $2.002 < 2.02 < 2.2 < 2.222$. Option (A) wrongly starts with 2.2 and option (C) places 2.002 after 2.22.

Answer: (C) 12

$12 \otimes 16 = \text{GCD}(12,16) = 4$, and $24 \otimes 36 = \text{GCD}(24,36) = 12$. Combining these with $\oplus$ (LCM): $\text{LCM}(4,12) = 12$.

Answer: (C) 2

The unit digit of powers of 232 depends on the unit digit of 2: $2^1=2, 2^2=4, 2^3=8, 2^4=16$ (unit 6), then cycle repeats every 4. $33 \div 4$ remainder 1 → unit digit same as $2^1 = 2$.

Answer: (C) 312

Total strikes in 12 hours = $1+2+3+...+12 = \frac{12 \times 13}{2} = 78$. In 24 hours = $78 \times 2 = 156$. In 2 days = $156 \times 2 = 312$.

Answer: (B) 99856

The square root of 99999 ≈ 316.2. $316^2 = 99856$. So 99856 is the greatest five-digit perfect square.

Answer: (D) 1.

$17 \equiv -1 \pmod{18}$. So $17^{200} \equiv (-1)^{200} = 1 \pmod{18}$. Remainder = 1.