QUESTION 1
Problem: Find the HCF of $4052$ and $12576$ using Euclid’s Division Algorithm.
Options:
A. $2$
B. $4$
C. $6$
D. $8$
Solution:
1. Apply Euclid’s Division Lemma on $12576$ and $4052$:
$12576 = 4052 \times 3 + 420$
2. Since remainder $420 \neq 0$, apply division lemma on $4052$ and $420$:
$4052 = 420 \times 9 + 272$
3. Apply division lemma on $420$ and $272$:
$420 = 272 \times 1 + 148$
4. Apply division lemma on $272$ and $148$:
$272 = 148 \times 1 + 124$
5. Apply division lemma on $148$ and $124$:
$148 = 124 \times 1 + 24$
6. Apply division lemma on $124$ and $24$:
$124 = 24 \times 5 + 4$
7. Apply division lemma on $24$ and $4$:
$24 = 4 \times 6 + 0$
8. Since the remainder is now $0$, the last divisor is $4$.
Correct Answer: B. $4$
QUESTION 2
Problem: Express $3825$ as a product of its prime factors.
Options:
A. $3^2 \times 5^2 \times 17$
B. $3 \times 5^3 \times 17$
C. $3^2 \times 5 \times 17^2$
D. $3^3 \times 5^2 \times 17$
Solution:
1. Divide $3825$ by $3$:
$3825 \div 3 = 1275$
2. Divide $1275$ by $3$:
$1275 \div 3 = 425$
3. Divide $425$ by $5$:
$425 \div 5 = 85$
4. Divide $85$ by $5$:
$85 \div 5 = 17$
5. $17$ is a prime number.
6. Thus, $3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$.
Correct Answer: A. $3^2 \times 5^2 \times 17$
QUESTION 3
Problem: Find the LCM of $96$ and $404$ by prime factorization method, and hence find their HCF.
Options:
A. $\text{LCM} = 9696$, $\text{HCF} = 4$
B. $\text{LCM} = 4848$, $\text{HCF} = 8$
C. $\text{LCM} = 9696$, $\text{HCF} = 2$
D. $\text{LCM} = 19392$, $\text{HCF} = 4$
Solution:
1. Prime factorization of $96$ and $404$:
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
2. $\text{HCF}(96, 404) = 2^2 = 4$
3. Using $\text{LCM} = \frac{a \times b}{\text{HCF}}$:
$\text{LCM}(96, 404) = \frac{96 \times 404}{4} = 96 \times 101 = 9696$
Correct Answer: A. $\text{LCM} = 9696$, $\text{HCF} = 4$
QUESTION 4
Problem: If two positive integers $p$ and $q$ can be expressed as $p = a b^2$ and $q = a^3 b$, where $a, b$ are prime numbers, then find $\text{LCM}(p, q)$.
Options:
A. $a b$
B. $a^2 b^2$
C. $a^3 b^2$
D. $a^3 b^3$
Solution:
1. $p = a^1 b^2$
2. $q = a^3 b^1$
3. $\text{LCM}$ takes the highest power of each prime factor present:
$\text{LCM}(p, q) = a^{\max(1, 3)} \times b^{\max(2, 1)} = a^3 b^2$
Correct Answer: C. $a^3 b^2$
Question 5
———–
Problem: $\sqrt{5} + \sqrt{2}$ is __________ number:
A. a rational
B. an irrational
C. a whole
D. a natural
Solution:
1. Both $\sqrt{5}$ and $\sqrt{2}$ are irrational numbers because 5 and 2 are prime integers that are not perfect squares.
2. The sum of two distinct positive square roots of prime integers ($\sqrt{p} + \sqrt{q}$) is always an irrational number.
3. Therefore, $\sqrt{5} + \sqrt{2}$ is an irrational number.
Correct Answer: B. an irrational
Question 6
Problem: A rectangular field has a perimeter of 350 kilometers. Three cyclists start together and can cycle 300 km, 262.5 km, and 175 km a day, around the field. How many days will they meet again?
A. 20 days
B. 24 days
C. 28 days
D. 30 days
Solution:
1. Calculate time taken by each cyclist to complete 1 round:
– First cyclist: $\text{Time}_1 = \frac{350}{300} = \frac{7}{6} \text{ days}$
– Second cyclist: $\text{Time}_2 = \frac{350}{262.5} = \frac{350}{\frac{525}{2}} = \frac{700}{525} = \frac{4}{3} \text{ days}$
– Third cyclist: $\text{Time}_3 = \frac{350}{175} = 2 = \frac{2}{1} \text{ days}$
2. Find the LCM of the fractional times:
$$\text{LCM}\left(\frac{7}{6}, \frac{4}{3}, \frac{2}{1}\right) = \frac{\text{LCM}(\text{numerators})}{\text{HCF}(\text{denominators})}$$
– LCM of numerators $(7, 4, 2) = 28$
– HCF of denominators $(6, 3, 1) = 1$
– $\text{LCM} = \frac{28}{1} = 28 \text{ days}$
3. Therefore, they will meet again after 28 days.
Correct Answer: C. 28 days
QUESTION 7
Problem: Find the HCF of $a = 2^3 \times 3^2 \times 5$ and $b = 2^2 \times 3^3 \times 5^2 \times 7$.
Options:
A. $180$
B. $540$
C. $360$
D. $90$
Solution:
1. $\text{HCF}$ takes the lowest power of each common prime factor:
$\text{HCF}(a, b) = 2^{\min(3, 2)} \times 3^{\min(2, 3)} \times 5^{\min(1, 2)}$
2. $\text{HCF}(a, b) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180$
Correct Answer: A. $180$
QUESTION 8
Problem: What is the exponent of $2$ in the prime factorization of $144$?
Options:
A. $2$
B. $3$
C. $4$
D. $5$
Solution:
1. Prime factorize $144$:
$144 = 2 \times 72 = 2^2 \times 36 = 2^3 \times 18 = 2^4 \times 9 = 2^4 \times 3^2$
2. The power of $2$ is $4$.
Correct Answer: C. $4$
QUESTION 9
Problem: Find the total number of prime factors of $2^3 \times 3^4 \times 5^2 \times 7^1$.
Options:
A. $4$
B. $10$
C. $24$
D. $12$
Solution:
1. Distinct prime factors involved are $2, 3, 5, 7$.
2. The total count of distinct prime factors is $4$.
3. (If total prime factors with multiplicity are asked: $3 + 4 + 2 + 1 = 10$). Here distinct prime factors = $4$.
Correct Answer: A. $4$
QUESTION 10
Problem: The total number of factors of a prime number is:
Options:
A. $1$
B. $2$
C. $3$
D. Infinitely many
Solution:
1. By definition, a prime number has exactly two distinct positive factors: $1$ and the number itself.
Correct Answer: B. $2$
QUESTION 11
Problem: The HCF of two numbers is $27$ and their LCM is $162$. If one of the numbers is $54$, find the other number.
Options:
A. $36$
B. $81$
C. $108$
D. $72$
Solution:
1. Using the property $a \times b = \text{HCF} \times \text{LCM}$:
$54 \times b = 27 \times 162$
2. Solve for $b$:
$b = \frac{27 \times 162}{54} = \frac{162}{2} = 81$
Correct Answer: B. $81$
QUESTION 12
Problem: The ratio of two numbers is $3 : 4$ and their HCF is $4$. Find their LCM.
Options:
A. $12$
B. $24$
C. $36$
D. $48$
Solution:
1. Let the two numbers be $3x$ and $4x$, where $x = \text{HCF} = 4$.
2. The numbers are $a = 3 \times 4 = 12$ and $b = 4 \times 4 = 16$.
3. $\text{LCM}(12, 16) = \frac{12 \times 16}{4} = 48$.
Alternatively, $\text{LCM} = 3 \times 4 \times x = 12 \times 4 = 48$.
Correct Answer: D. $48$
Question 13
Problem: The HCF of 376 and 168 is:
A. 9
B. 8
C. 7
D. None of above
Solution:
Using Euclid’s Division Algorithm:
1. $376 = 168 \times 2 + 40$
2. $168 = 40 \times 4 + 8$
3. $40 = 8 \times 5 + 0$
Since the remainder is 0, the last non-zero divisor is 8.
Correct Answer: B. 8
Question 14
Problem: If the sum of LCM and HCF of two numbers is 1332 and their LCM is 1308 more than their HCF, then the product of two numbers is:
A. 13200
B. 19800
C. 15840
D. 6600
Solution:
1. Let $\text{LCM} = L$ and $\text{HCF} = H$.
2. Given equations:
– $L + H = 1332$
– $L – H = 1308$
3. Add both equations to solve for $L$:
$$2L = 2640 \implies L = 1320$$
4. Find $H$:
$$H = 1332 – 1320 = 12$$
5. Using $\text{Product of two numbers} = \text{LCM} \times \text{HCF}$:
$$\text{Product} = 1320 \times 12 = 15840$$
Correct Answer: C. 15840
Question 15
Problem: The HCF of two numbers is 58 and their LCM is 580. If one number is 116, then find the sum of the two numbers.
A. 402
B. 404
C. 406
D. 408
Solution:
1. Let the two numbers be $a = 116$ and $b$.
2. Using $a \times b = \text{HCF} \times \text{LCM}$:
$$116 \times b = 58 \times 580$$
$$b = \frac{58 \times 580}{116} = \frac{580}{2} = 290$$
3. Find the sum of the two numbers ($a + b$):
$$\text{Sum} = 116 + 290 = 406$$
Correct Answer: C. 406
QUESTION 16
Problem: Can two numbers have $18$ as their HCF and $380$ as their LCM?
Options:
A. Yes, always
B. No, because HCF must divide LCM completely
C. Yes, depending on the numbers
D. Cannot be determined
Solution:
1. The HCF of two numbers must always divide their LCM completely without any remainder.
2. Check if $380$ is divisible by $18$:
$380 \div 18 = 21.111…$ ($380 = 18 \times 21 + 2$)
3. Since $18$ does not divide $380$ completely, no two numbers can have $18$ as HCF and $380$ as LCM.
Correct Answer: B. No, because HCF must divide LCM completely
QUESTION 17
Problem: If the sum of LCM and HCF of two numbers is $1260$ and their LCM is $900$ more than their HCF, find the product of the two numbers.
Options:
A. $194400$
B. $19440$
C. $205200$
D. $180000$
Solution:
1. Let $\text{LCM} = L$ and $\text{HCF} = H$.
$L + H = 1260$
$L – H = 900$
2. Adding both equations:
$2L = 2160 \implies L = 1080$
3. Find $H$:
$H = 1260 – 1080 = 180$
4. Product of numbers $= L \times H = 1080 \times 180 = 194400$.
Correct Answer: A. $194400$
QUESTION 18
Problem: Find the smallest prime number which divides the sum of $a$ and $b$, where $a$ is the smallest composite number and $b$ is the smallest prime number.
Options:
A. $2$
B. $3$
C. $5$
D. $7$
Solution:
1. Smallest prime number $b = 2$.
2. Smallest composite number $a = 4$.
3. Sum $a + b = 4 + 2 = 6$.
4. Prime factorization of $6 = 2 \times 3$.
5. The smallest prime number dividing $6$ is $2$.
Correct Answer: A. $2$
QUESTION 19
Problem: Find the largest number that divides $2053$ and $967$ leaving a remainder of $5$ and $7$ respectively.
Options:
A. $64$
B. $128$
C. $32$
D. $96$
Solution:
1. Subtract the respective remainders:
$2053 – 5 = 2048$
$967 – 7 = 960$
2. Find $\text{HCF}(2048, 960)$:
$2048 = 2^{11}$
$960 = 2^6 \times 3 \times 5 = 64 \times 15$
3. $\text{HCF}(2048, 960) = 2^6 = 64$.
Correct Answer: A. $64$
Question 20
Problem: If the sum of LCM and HCF of two numbers is 1705 and their LCM is 1695 more than their HCF, then the product of two numbers is ____.
A. 4500
B. 5500
C. 7500
D. 8500
Solution:
1. Let $\text{LCM} = L$ and $\text{HCF} = H$.
– $L + H = 1705$
– $L – H = 1695$
2. Add both equations:
$$2L = 3400 \implies L = 1700$$
3. Find $H$:
$$H = 1705 – 1700 = 5$$
4. Calculate Product:
$$\text{Product} = \text{LCM} \times \text{HCF} = 1700 \times 5 = 8500$$
Correct Answer: D. 8500
Question 21
Problem: Find the smallest number which when divided by 36 and 52 leaves remainder 10 and 26 respectively.
A. 442
B. 468
C. 452
D. 480
Solution:
1. Find the difference between divisors and their remainders:
– $36 – 10 = 26$
– $52 – 26 = 26$
The common difference $k = 26$.
2. Calculate $\text{LCM}(36, 52)$:
– $36 = 2^2 \times 3^2$
– $52 = 2^2 \times 13$
– $\text{LCM}(36, 52) = 2^2 \times 3^2 \times 13 = 468$
3. Subtract the common difference $k$:
$$\text{Required Number} = 468 – 26 = 442$$
Correct Answer: A. 442
Question 22
Problem: Find the smallest number which when increased by 6 is exactly divisible by both 190 and 140.
A. 2708
B. 2676
C. 2654
D. 2650
Solution:
1. Let the required number be $x$.
2. According to the condition, $x + 6 = \text{LCM}(190, 140)$.
3. Calculate $\text{LCM}(190, 140)$:
– $190 = 2 \times 5 \times 19$
– $140 = 2^2 \times 5 \times 7$
– $\text{LCM}(190, 140) = 2^2 \times 5 \times 7 \times 19 = 2660$
4. Solve for $x$:
$$x + 6 = 2660 \implies x = 2660 – 6 = 2654$$
Correct Answer: C. 2654
Question 23
Problem: Show that the given expression is a composite number: $29 \times 5 \times 14 + 14$
Solution:
1. Factor out 14 from the expression:
$$29 \times 5 \times 14 + 14 = 14 \times (29 \times 5 + 1)$$
2. Simplify inside the brackets:
$$14 \times (145 + 1) = 14 \times 146$$
3. Express into prime factors:
$$14 = 2 \times 7$$
$$146 = 2 \times 73$$
$$\text{Expression} = 2^2 \times 7 \times 73 = 2044$$
4. Since the expression can be written as a product of prime factors other than 1 and itself, it is a composite number.
Question 24
Problem: Write all the prime factors of 22547140.
Solution:
1. Perform prime factorization on 22547140:
– $22547140 \div 2 = 11273570$
– $11273570 \div 2 = 5636785$
– $5636785 \div 5 = 1127357$
– $1127357 \div 7 = 161051$
– $161051 \div 11 = 14641$
– $14641 \div 11 = 1331$
– $1331 \div 11 = 121$
– $121 \div 11 = 11$
– $11 \div 11 = 1$
2. Full prime factorization:
$$22547140 = 2^2 \times 5 \times 7 \times 11^5$$
3. Distinct prime factors:
The prime factors of 22547140 are 2, 5, 7, and 11.
QUESTION 25
Problem: Find the greatest number of 4 digits which is exactly divisible by $15, 24,$ and $36$.
Options:
A. $9720$
B. $9360$
C. $9840$
D. $9960$
Solution:
1. Find $\text{LCM}(15, 24, 36)$:
$15 = 3 \times 5$
$24 = 2^3 \times 3$
$36 = 2^2 \times 3^2$
$\text{LCM}(15, 24, 36) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$
2. The largest 4-digit number is $9999$.
3. Divide $9999$ by $360$:
$9999 = 360 \times 27 + 279$
4. Subtract remainder $279$ from $9999$:
$9999 – 279 = 9720$.
Correct Answer: A. $9720$
QUESTION 26
Problem: Find the least number which when divided by $12, 16, 24,$ and $36$ leaves a remainder of $7$ in each case.
Options:
A. $151$
B. $144$
C. $137$
D. $155$
Solution:
1. Calculate $\text{LCM}(12, 16, 24, 36)$:
$12 = 2^2 \times 3$
$16 = 2^4$
$24 = 2^3 \times 3$
$36 = 2^2 \times 3^2$
$\text{LCM} = 2^4 \times 3^2 = 16 \times 9 = 144$
2. Required number $= \text{LCM} + \text{Remainder} = 144 + 7 = 151$.
Correct Answer: A. $151$
QUESTION 27
Problem: Find the smallest 3-digit number which is divisible by $6, 8,$ and $12$.
Options:
A. $108$
B. $120$
C. $112$
D. $124$
Solution:
1. Find $\text{LCM}(6, 8, 12)$:
$6 = 2 \times 3$, $8 = 2^3$, $12 = 2^2 \times 3$
$\text{LCM}(6, 8, 12) = 2^3 \times 3 = 24$
2. Smallest 3-digit number is $100$.
3. Divide $100$ by $24$:
$100 = 24 \times 4 + 4$
4. To make it divisible, add $(24 – 4) = 20$ to $100$:
$100 + 20 = 120$.
Correct Answer: B. $120$
QUESTION 28
Problem: Find the smallest number which when diminished by $7$ is divisible by $21, 28, 36,$ and $45$.
Options:
A. $1267$
B. $1253$
C. $1260$
D. $1274$
Solution:
1. Let the number be $x$. Then $x – 7 = \text{LCM}(21, 28, 36, 45)$.
2. Calculate $\text{LCM}$:
$21 = 3 \times 7$
$28 = 2^2 \times 7$
$36 = 2^2 \times 3^2$
$45 = 3^2 \times 5$
$\text{LCM} = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1260$
3. $x – 7 = 1260 \implies x = 1260 + 7 = 1267$.
Correct Answer: A. $1267$
QUESTION 29
Problem: Determine the largest positive integer that divides $398, 436,$ and $542$ leaving remainders $7, 11,$ and $15$ respectively.
Options:
A. $17$
B. $19$
C. $23$
D. $31$
Solution:
1. Subtract respective remainders:
$398 – 7 = 391$
$436 – 11 = 425$
$542 – 15 = 527$
2. Find $\text{HCF}(391, 425, 527)$:
$391 = 17 \times 23$
$425 = 17 \times 25 = 17 \times 5^2$
$527 = 17 \times 31$
3. Common factor is $17$. So $\text{HCF} = 17$.
Correct Answer: A. $17$
QUESTION 30
Problem: Three bells toll together at intervals of $9, 12, 15$ minutes respectively. If they start tolling together, after what time will they next toll together?
Options:
A. $180 \text{ minutes}$ ($3 \text{ hours}$)
B. $90 \text{ minutes}$ ($1.5 \text{ hours}$)
C. $360 \text{ minutes}$ ($6 \text{ hours}$)
D. $120 \text{ minutes}$ ($2 \text{ hours}$)
Solution:
1. The bells will toll together after a time interval equal to $\text{LCM}(9, 12, 15)$.
2. Prime factorization:
$9 = 3^2$
$12 = 2^2 \times 3$
$15 = 3 \times 5$
3. $\text{LCM}(9, 12, 15) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \text{ minutes}$.
4. $180 \text{ minutes} = 3 \text{ hours}$.
Correct Answer: A. $180 \text{ minutes}$ ($3 \text{ hours}$)
QUESTION 31
Problem: A sweet seller has $420$ Kaju barfis and $130$ Badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
Options:
A. $10$
B. $20$
C. $15$
D. $30$
Solution:
1. To minimize tray area, the number of barfis in each stack must be maximum, which is $\text{HCF}(420, 130)$.
2. Using Euclid’s division algorithm:
$420 = 130 \times 3 + 30$
$130 = 30 \times 4 + 10$
$30 = 10 \times 3 + 0$
3. $\text{HCF}(420, 130) = 10$.
Correct Answer: A. $10$
QUESTION 32
Problem: In a school, there are two sections – section A and section B of class X. There are $32$ students in section A and $36$ students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.
Options:
A. $288$
B. $144$
C. $576$
D. $1152$
Solution:
1. Minimum number of books required $= \text{LCM}(32, 36)$.
2. Prime factorizations:
$32 = 2^5$
$36 = 2^2 \times 3^2$
3. $\text{LCM}(32, 36) = 2^5 \times 3^2 = 32 \times 9 = 288$.
Correct Answer: A. $288$
QUESTION 33
Problem: A merchant has $120$ liters of oil of one kind, $180$ liters of another kind, and $240$ liters of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Options:
A. $60 \text{ liters}$
B. $30 \text{ liters}$
C. $40 \text{ liters}$
D. $120 \text{ liters}$
Solution:
1. The greatest capacity of the tin $= \text{HCF}(120, 180, 240)$.
2. Factorizations:
$120 = 60 \times 2$
$180 = 60 \times 3$
$240 = 60 \times 4$
3. $\text{HCF}(120, 180, 240) = 60 \text{ liters}$.
Correct Answer: A. $60 \text{ liters}$
QUESTION 34
Problem: Show that $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.
Solution:
1. First expression: $7 \times 11 \times 13 + 13$
$= 13 \times (7 \times 11 + 1)$
$= 13 \times (77 + 1) = 13 \times 78 = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2$.
Since it has prime factors other than $1$ and itself, it is a composite number.
2. Second expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
$= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$
$= 5 \times (1008 + 1) = 5 \times 1009$.
Since $1009$ is a prime number, the expression is a product of primes $5$ and $1009$, making it a composite number.
QUESTION 35
Problem: Prove that $3 + 2\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is irrational.
Solution:
1. Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.
2. Then there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$3 + 2\sqrt{5} = \frac{a}{b}$
3. Rearranging the terms:
$2\sqrt{5} = \frac{a}{b} – 3 = \frac{a – 3b}{b}$
$\sqrt{5} = \frac{a – 3b}{2b}$
4. Since $a, b$ are integers, $\frac{a – 3b}{2b}$ is a rational number.
5. This implies that $\sqrt{5}$ is rational, which contradicts the given fact that $\sqrt{5}$ is irrational.
6. Hence, our assumption was wrong, and $3 + 2\sqrt{5}$ is irrational.
QUESTION 36
Problem: Check whether $6^n$ can end with the digit $0$ for any natural number $n$.
Solution:
1. If any number ends with the digit $0$, its prime factorization must contain both $2$ and $5$ as factors (since $10 = 2 \times 5$).
2. Prime factorization of $6^n$:
$6^n = (2 \times 3)^n = 2^n \times 3^n$
3. By the Fundamental Theorem of Arithmetic, the prime factorization of $6^n$ is unique and contains only the primes $2$ and $3$.
4. Since $5$ is not a prime factor of $6^n$, $6^n$ cannot end with the digit $0$ for any natural number $n \in \mathbb{N}$.
QUESTION 37
Problem: Prove that $\sqrt{3}$ is an irrational number.
Solution:
1. Assume to the contrary that $\sqrt{3}$ is rational.
2. Then $\sqrt{3} = \frac{a}{b}$, where $a, b$ are co-prime integers and $b \neq 0$.
3. Squaring both sides:
$3 = \frac{a^2}{b^2} \implies a^2 = 3b^2 \quad \text{— (Equation 1)}$
4. Thus $3$ divides $a^2$, which implies $3$ divides $a$ (since $3$ is prime).
5. So, let $a = 3c$ for some integer $c$.
6. Substituting $a = 3c$ into Equation 1:
$(3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies b^2 = 3c^2$
7. This means $3$ divides $b^2$, so $3$ divides $b$.
8. Therefore, $3$ is a common factor of both $a$ and $b$.
9. This contradicts our initial assumption that $a$ and $b$ are co-prime (having no common factor other than $1$).
10. Hence, $\sqrt{3}$ is irrational.
QUESTION 38
Problem: Without actually performing long division, state whether $\frac{13}{3125}$ will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
Options:
A. Terminating
B. Non-terminating repeating
C. Non-terminating non-repeating
D. Cannot be determined
Solution:
1. Prime factorize the denominator $3125$:
$3125 = 5 \times 625 = 5^5 = 2^0 \times 5^5$
2. The denominator is of the form $2^n \times 5^m$, where $n = 0$ and $m = 5$ are non-negative integers.
3. Therefore, $\frac{13}{3125}$ has a **terminating** decimal expansion.
Correct Answer: A. Terminating
QUESTION 39
Problem: What is the maximum number of digits in the repeating block of digits in the decimal expansion of $\frac{1}{17}$?
Options:
A. $16$
B. $17$
C. $15$
D. $8$
Solution:
1. For any rational number $\frac{p}{q}$ with a non-terminating repeating decimal expansion, the maximum number of digits in the repeating period is less than $q$, i.e., $q – 1$.
2. Here $q = 17$, so the maximum number of digits in the repeating block is $17 – 1 = 16$.
(Actual expansion: $\frac{1}{17} = 0.\overline{0588235294117647}$, which has $16$ repeating digits).
Correct Answer: A. $16$
QUESTION 40
Problem: The product of two numbers is $1600$ and their HCF is $5$. Find the LCM of the numbers.
Options:
A. $320$
B. $160$
C. $640$
D. $800$
Solution:
1. Formula: $\text{LCM} \times \text{HCF} = \text{Product of numbers}$
2. $\text{LCM} \times 5 = 1600 \implies \text{LCM} = \frac{1600}{5} = 320$.
Correct Answer: A. $320$