Arithmetic & Number Systems

Question 1 (12)

(a) Euclidean algorithm on $(1260, 882)$:

• $1260 = 1 \times 882 + 378$ (1)
• $882 = 2 \times 378 + 126$ (1)
• $378 = 3 \times 126 + 0$ (1)
• Last non-zero remainder $= 126$, so $\gcd = 126$. (1)

(b) $\text{LCM} = \dfrac{1260 \times 882}{126} = \dfrac{1{,}111{,}320}{126} = 8820$. (3) (Award 1 for correct identity, 1 for product, 1 for answer.)

(c) Each step replaces the pair by a smaller remainder pair; remainders are non-negative and strictly decreasing, so the process must reach 0 in finitely many steps (termination). At each step $\gcd(a,b)=\gcd(b, a \bmod b)$ because any common divisor of $a,b$ divides the remainder and vice-versa, so the gcd is preserved; hence the last non-zero remainder is the gcd. (3)

(d) $1260 = 2^2 \cdot 3^2 \cdot 5 \cdot 7$; $882 = 2 \cdot 3^2 \cdot 7^2$. Common $= 2^1 \cdot 3^2 \cdot 7^1 = 126$. ✓ (2)

Question 2 (10)

(a) (5 marks — 1 array init, 1 loop to √n, 1 inner marking from p², 1 correctness, 1 return)

def sieve(n):
    is_prime = [True]*(n+1)
    is_prime[0] = is_prime[1] = False
    p = 2
    while p*p <= n:
        if is_prime[p]:
            for m in range(p*p, n+1, p):
                is_prime[m] = False
        p += 1
    return [i for i in range(2, n+1) if is_prime[i]]

(b) Any composite $m$ has a prime factor $\le \sqrt{m}$, so it is already crossed out by a prime $p \le \sqrt{m}$; thus once $p^2 > n$ every remaining number is prime and we may stop. We start at $p^2$ because all smaller multiples $p\cdot k$ ($k<p$) were already crossed by the smaller factor $k$. (3)

(c) Primes between 40 and 60: 41, 43, 47, 53, 59. (2)

Question 3 (10)

(a) Convert to improper: $3\tfrac14=\tfrac{13}{4}$, $2\tfrac56=\tfrac{17}{6}$, $1\tfrac78=\tfrac{15}{8}$. (1) LCD of 4,6,8 = 24. (1) $\tfrac{13}{4}=\tfrac{78}{24}$, $\tfrac{17}{6}=\tfrac{68}{24}$, $\tfrac{15}{8}=\tfrac{45}{24}$. (2) $\tfrac{78+68-45}{24}=\tfrac{101}{24}$. (1) $=4\tfrac{5}{24}$ (already lowest terms). (1)

(b) Addition combines quantities of the same unit; a common denominator makes the pieces the same size so the numerators can be counted together. Multiplication of $\tfrac ab \times \tfrac cd$ means "$\tfrac ab$ of $\tfrac cd$" — it scales, so numerators and denominators multiply directly ($\tfrac{ac}{bd}$) with no need for equal-sized pieces. (4)

Question 4 (12)

(a) Marked price $= 800 \times 1.25 = 1000$. (1) Discount $10\%$: SP $= 1000 \times 0.90 = 900$. (2) Profit $= 900 - 800 = 100$; profit% $= \tfrac{100}{800}\times100 = 12.5\%$. (2)

(b) $T = 2\tfrac{4}{12} = 2\tfrac13 = \tfrac73$ years. (1) $I = \dfrac{12000 \times 7.5 \times \tfrac73}{100} = \dfrac{12000 \times 7.5 \times 7}{300} = \dfrac{630000}{300} = 2100$. (3) So $I = ₹2100$.

(c) The 10% decrease applies to the increased base (125%), not the original: net factor $=1.25\times0.90=1.125$, i.e. +12.5%, not +15%. Percentages compound on different bases, so they cannot simply be added/subtracted. (3)

Question 5 (10)

(a) Innermost: $7-3=4$. (1) $2\times4=8$; $4+8=12$. (1) $120\div12=10$. (1) $10-3=7$. (1)

(b) Total parts $=2+3+4=9$; one part $=5400/9=600$. (1) A $=1200$, B $=1800$, C $=2400$. (2)

(c) Total work $=15\times8=120$ worker-days. Days for 20 workers $=120/20=6$ days. (2) Inverse proportion: more workers ⇒ fewer days. (1)

Question 6 (6)

(a) $2{,}05{,}30{,}070$ (2,05,30,070). (2) The digit 5 is in the lakhs place → place value $= 5{,}00{,}000$ (five lakh). (1)

(b) $|x-3|\le 2 \Rightarrow -2 \le x-3 \le 2 \Rightarrow 1 \le x \le 5$. (1) Integers: $\{1,2,3,4,5\}$. (1) On the number line these are all points within distance 2 of 3. (1)

[
  {"claim":"gcd(1260,882)=126 and lcm=8820", "code":"g=gcd(1260,882); l=lcm(1260,882); result=(g==126 and l==8820)"},
  {"claim":"3 1/4 + 2 5/6 - 1 7/8 = 101/24", "code":"result=(Rational(13,4)+Rational(17,6)-Rational(15,8)==Rational(101,24))"},
  {"claim":"SP=900 and profit%=12.5 on cost 800", "code":"sp=800*Rational(125,100)*Rational(90,100); pp=(sp-800)/800*100; result=(sp==900 and pp==Rational(25,2))"},
  {"claim":"Simple interest = 2100", "code":"I=Rational(12000*Rational(15,2)*Rational(7,3),100); result=(I==2100)"},
  {"claim":"BODMAS expression equals 7", "code":"result=(120/(4+2*(7-3))-3==7)"}
]