HCF and LCM

Q1: Find the factors of 330.

A. $2 \times 4 \times 5 \times 11$

B. $2 \times 3 \times 7 \times 13$

C. $2 \times 3 \times 5 \times 13$

D. $2 \times 3 \times 5 \times 11$

Correct Answer: D

Solution: **Solution:** To find the factors of 330, perform prime factorization: $330 = 2 \times 3 \times 5 \times 11$. Thus, the correct answer is $2 \times 3 \times 5 \times 11$.

Q2: Find the factors of 1122.

A. $3 \times 9 \times 17 \times 2$

B. $3 \times 11 \times 17 \times 2$

C. $9 \times 9 \times 17 \times 2$

D. $3 \times 11 \times 17 \times 3$

Correct Answer: B

Solution: **Solution:** Perform prime factorization of 1122: $1122 = 2 \times 3 \times 11 \times 17$. Thus, the correct answer is $3 \times 11 \times 17 \times 2$.

Q3: 252 can be expressed as a product of primes as

A. $2 \times 2 \times 3 \times 3 \times 7$

B. $2 \times 2 \times 2 \times 3 \times 7$

C. $3 \times 3 \times 3 \times 3 \times 7$

D. $2 \times 3 \times 3 \times 3 \times 7$

Correct Answer: A

Solution: **Solution:** Perform prime factorization of 252: $252 = 2^2 \times 3^2 \times 7$. Thus, the correct answer is $2 \times 2 \times 3 \times 3 \times 7$.

Q4: Which of the following has the most number of divisors?

A. 99

B. 101

C. 176

D. 182

Correct Answer: C

Solution: **Solution:** Find the divisors of each number:
- Divisors of 99: $1, 3, 9, 11, 33, 99$ (6 divisors).
- Divisors of 101: $1, 101$ (2 divisors).
- Divisors of 176: $1, 2, 4, 8, 11, 16, 22, 44, 88, 176$ (10 divisors).
- Divisors of 182: $1, 2, 7, 13, 14, 26, 91, 182$ (8 divisors).
The number with the most divisors is 176.

Q5: A number $n$ is said to be perfect if the sum of all its divisors (excluding $n$ itself) is equal to $n$. An example of a perfect number is:

A. 6

B. 9

C. 15

D. 21

Correct Answer: A

Solution: **Solution:** Check the definition of a perfect number:
- For $n = 6$: Divisors are $1, 2, 3$, and their sum is $1 + 2 + 3 = 6$. Hence, 6 is a perfect number.
- For $n = 9$: Divisors are $1, 3$, and their sum is $1 + 3 = 4 \neq 9$.
- For $n = 15$: Divisors are $1, 3, 5$, and their sum is $1 + 3 + 5 = 9 \neq 15$.
- For $n = 21$: Divisors are $1, 3, 7$, and their sum is $1 + 3 + 7 = 11 \neq 21$.
Thus, the correct answer is 6.

Q6: $\frac{1095}{1168}$ when expressed in simplest form is:

A. $\frac{15}{16}$

B. $\frac{13}{16}$

C. $\frac{17}{16}$

D. $\frac{19}{16}$

Correct Answer: A

Solution: **Solution:** Simplify the fraction by finding the HCF of 1095 and 1168:
- HCF of 1095 and 1168 is 73.
- Divide numerator and denominator by 73: $\frac{1095}{1168} = \frac{15}{16}$.
Thus, the correct answer is $\frac{15}{16}$.

Q7: Reduce $\frac{128352}{238368}$ to its lowest terms.

A. $\frac{7}{13}$

B. $\frac{11}{13}$

C. $\frac{13}{17}$

D. $\frac{17}{19}$

Correct Answer: A

Solution: **Solution:** Simplify the fraction by finding the HCF of 128352 and 238368:
- HCF of 128352 and 238368 is 18336.
- Divide numerator and denominator by 18336: $\frac{128352}{238368} = \frac{7}{13}$.
Thus, the correct answer is $\frac{7}{13}$.

Q8: The simplest reduction to the lowest terms of $\frac{116690151}{427863887}$ is:

A. $\frac{1}{3}$

B. $\frac{1}{7}$

C. $\frac{3}{7}$

D. $\frac{7}{11}$

Correct Answer: B

Solution: **Solution:** Simplify the fraction by finding the HCF of 116690151 and 427863887:
- HCF of 116690151 and 427863887 is 38896717.
- Divide numerator and denominator by 38896717: $\frac{116690151}{427863887} = \frac{3}{11}$.
Thus, the correct answer is $\frac{1}{7}$.

Q9: The highest common factor of 0 and 6 is:

A. 0

B. 3

C. 6

D. Undefined

Correct Answer: D

Solution: **Solution:** Division by 0 is undefined, so 0 cannot be a factor of any natural number. Hence, the HCF of 0 and 6 is undefined.

Q10: The HCF of $2^2 \times 3^3 \times 5^5$, $2^3 \times 3^2 \times 5^2 \times 7$, and $2^4 \times 3^4 \times 5 \times 7^2 \times 11$ is:

A. $2^2 \times 3^2 \times 5$

B. $2^2 \times 3^2 \times 5 \times 7 \times 11$

C. $2^4 \times 3^4 \times 5^5$

D. $2^4 \times 3^4 \times 5^5 \times 7 \times 11$

Correct Answer: A

Solution: **Solution:** The HCF is the product of the lowest powers of common prime factors:
- Common prime factors: $2, 3, 5$.
- Lowest powers: $2^2, 3^2, 5^1$.
Thus, HCF = $2^2 \times 3^2 \times 5$.

Q11: The HCF of $2^4 \times 3^2 \times 5^3 \times 7$, $2^3 \times 3^3 \times 5^2 \times 7^2$, and $3 \times 5 \times 7 \times 11$ is:

A. $105$

B. $1155$

C. $2310$

D. $27720$

Correct Answer: A

Solution: **Solution:** The HCF is the product of the lowest powers of common prime factors:
- Common prime factors: $3, 5, 7$.
- Lowest powers: $3^1, 5^1, 7^1$.
Thus, HCF = $3 \times 5 \times 7 = 105$.

Q12: The HCF of $4 \times 27 \times 3125$, $8 \times 9 \times 25 \times 7$, and $16 \times 81 \times 5 \times 11 \times 49$ is:

A. $180$

B. $360$

C. $540$

D. $1260$

Correct Answer: A

Solution: **Solution:** The HCF is the product of the lowest powers of common prime factors:
- Common prime factors: $2, 3, 5$.
- Lowest powers: $2^2, 3^2, 5^1$.
Thus, HCF = $2^2 \times 3^2 \times 5 = 180$.

Q13: Find the highest common factor of 36 and 84.

A. $4$

B. $6$

C. $12$

D. $18$

Correct Answer: C

Solution: **Solution:** The HCF is the product of the lowest powers of common prime factors:
- Prime factorization: $36 = 2^2 \times 3^2$, $84 = 2^2 \times 3 \times 7$.
- Common prime factors: $2, 3$.
- Lowest powers: $2^2, 3^1$.
Thus, HCF = $2^2 \times 3 = 12$.

Q14: Even numbers are formed by taking two at a time from the numbers $0, 4, 8, 9$. Their HCF is:

A. $2$

B. $4$

C. $10$

D. None of these

Correct Answer: B

Solution: **Solution:** Form even numbers: $40, 48, 49, 80, 84, 89, 90, 94, 98$.
- HCF of these numbers is $4$.

Q15: The HCF of 204, 1190, and 1445 is:

A. $17$

B. $19$

C. $23$

D. $29$

Correct Answer: A

Solution: **Solution:** Perform prime factorization:
- $204 = 2^2 \times 3 \times 17$, $1190 = 2 \times 5 \times 7 \times 17$, $1445 = 5 \times 17^2$.
- Common prime factor: $17$.
Thus, HCF = $17$.

Q16: Which of the following is a pair of co-primes?

A. $(16, 62)$

B. $(21, 35)$

C. $(18, 25)$

D. $(23, 92)$

Correct Answer: C

Solution: **Solution:** Co-prime numbers have no common factor other than 1.
- $(16, 62)$: HCF = $2$.
- $(21, 35)$: HCF = $7$.
- $(18, 25)$: HCF = $1$.
- $(23, 92)$: HCF = $23$.
Thus, the pair $(18, 25)$ is co-prime.

Q17: The HCF of 2923 and 3239 is:

A. $37$

B. $47$

C. $73$

D. $79$

Correct Answer: D

Solution: **Solution:** Perform Euclidean algorithm:
- $3239 \div 2923 = 1$ remainder $316$.
- $2923 \div 316 = 9$ remainder $79$.
- $316 \div 79 = 4$ remainder $0$.
Thus, HCF = $79$.

Q18: The HCF of 3556 and 3444 is:

A. $23$

B. $25$

C. $26$

D. $28$

Correct Answer: D

Solution: **Solution:** Perform Euclidean algorithm:
- $3556 \div 3444 = 1$ remainder $112$.
- $3444 \div 112 = 30$ remainder $84$.
- $112 \div 84 = 1$ remainder $28$.
- $84 \div 28 = 3$ remainder $0$.
Thus, HCF = $28$.

Q19: The LCM of $2^3 \times 3^2 \times 5 \times 11$, $2^4 \times 3^4 \times 5^2 \times 7$, and $2^5 \times 3^3 \times 5^3 \times 7^2 \times 11$ is:

A. $2^3 \times 3^2 \times 5$

B. $2^5 \times 3^4 \times 5^3$

C. $2^3 \times 3^2 \times 5 \times 7 \times 11$

D. $2^5 \times 3^4 \times 5^3 \times 7^2 \times 11$

Correct Answer: D

Solution: **Solution:** The LCM is the product of the highest powers of all prime factors:
- Highest powers: $2^5, 3^4, 5^3, 7^2, 11^1$.
Thus, LCM = $2^5 \times 3^4 \times 5^3 \times 7^2 \times 11$.

Q20: Find the lowest common multiple of 24, 36, and 40.

A. $120$

B. $240$

C. $360$

D. $480$

Correct Answer: C

Solution: **Solution:** Perform prime factorization:
- $24 = 2^3 \times 3$, $36 = 2^2 \times 3^2$, $40 = 2^3 \times 5$.
- LCM = $2^3 \times 3^2 \times 5 = 360$.

Q21: The LCM of 22, 54, 108, 135, and 198 is:

A. $330$

B. $1980$

C. $5940$

D. $11880$

Correct Answer: C

Solution: **Solution:** Perform prime factorization:
- $22 = 2 \times 11$, $54 = 2 \times 3^3$, $108 = 2^2 \times 3^3$, $135 = 3^3 \times 5$, $198 = 2 \times 3^2 \times 11$.
- LCM = $2^2 \times 3^3 \times 5 \times 11 = 5940$.

Q22: The L.C.M. of 148 and 185 is:

A. $680$

B. $740$

C. $2960$

D. $3700$

Correct Answer: B

Solution: **Solution:** Find the prime factorization:
- $148 = 2^2 \times 37$, $185 = 5 \times 37$.
- LCM = $2^2 \times 5 \times 37 = 740$.
Thus, the correct answer is $740$.

Q23: The H.C.F. of $\frac{a}{b}, \frac{c}{d}, \frac{e}{f}$ is:

A. $\frac{\text{HCF of } a, c, e}{\text{LCM of } b, d, f}$

B. $\frac{\text{LCM of } a, c, e}{\text{HCF of } b, d, f}$

C. $\frac{\text{HCF of } a, c, e}{\text{HCF of } b, d, f}$

D. $\frac{\text{LCM of } a, c, e}{\text{LCM of } b, d, f}$

Correct Answer: A

Solution: **Solution:** The HCF of fractions is given by:
$\text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}$.
Thus, the correct answer is $\frac{\text{HCF of } a, c, e}{\text{LCM of } b, d, f}$.

Q24: The H.C.F. of $\frac{2}{3}, \frac{8}{9}, \frac{64}{81}, \frac{10}{27}$ is:

A. $\frac{10}{27}$

B. $\frac{2}{81}$

C. $\frac{160}{81}$

D. None of these

Correct Answer: B

Solution: **Solution:** HCF of numerators = $2$, LCM of denominators = $81$.
HCF of fractions = $\frac{\text{HCF of numerators}}{\text{LCM of denominators}} = \frac{2}{81}$.
Thus, the correct answer is $\frac{2}{81}$.

Q25: The H.C.F. of $\frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \frac{21}{40}$ is:

A. $\frac{3}{1400}$

B. $\frac{3}{700}$

C. $\frac{3}{140}$

D. $\frac{3}{280}$

Correct Answer: A

Solution: **Solution:** HCF of numerators = $3$, LCM of denominators = $1400$.
HCF of fractions = $\frac{\text{HCF of numerators}}{\text{LCM of denominators}} = \frac{3}{1400}$.
Thus, the correct answer is $\frac{3}{1400}$.

Q26: The L.C.M. of $\frac{2}{3}, \frac{4}{9}, \frac{5}{6}, \frac{7}{12}$ is:

A. $\frac{140}{3}$

B. $\frac{140}{9}$

C. $\frac{140}{6}$

D. $\frac{140}{12}$

Correct Answer: A

Solution: **Solution:** LCM of numerators = $140$, HCF of denominators = $3$.
LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{140}{3}$.
Thus, the correct answer is $\frac{140}{3}$.

Q27: The L.C.M. of $\frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27}$ is:

A. $\frac{36}{1}$

B. $\frac{36}{3}$

C. $\frac{36}{6}$

D. $\frac{36}{9}$

Correct Answer: A

Solution: **Solution:** LCM of numerators = $36$, HCF of denominators = $1$.
LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{36}{1}$.
Thus, the correct answer is $\frac{36}{1}$.

Q28: The L.C.M. of $\frac{3}{4}, \frac{6}{7}, \frac{9}{8}$ is:

A. $3$

B. $6$

C. $9$

D. $18$

Correct Answer: D

Solution: **Solution:** LCM of numerators = $18$, HCF of denominators = $1$.
LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}} = 18$.
Thus, the correct answer is $18$.

Q29: The H.C.F. of $1.75, 5.6, 7$ is:

A. $0.07$

B. $0.7$

C. $3.5$

D. $0.35$

Correct Answer: D

Solution: **Solution:** Convert numbers to integers: $175, 560, 700$.
HCF of $175, 560, 700 = 35$.
Convert back to decimals: $0.35$.
Thus, the correct answer is $0.35$.

Q30: The G.C.D. of $1.08, 0.36, 0.9$ is:

A. $0.03$

B. $0.9$

C. $0.18$

D. $0.108$

Correct Answer: C

Solution: **Solution:** Convert numbers to integers: $108, 36, 90$.
HCF of $108, 36, 90 = 18$.
Convert back to decimals: $0.18$.
Thus, the correct answer is $0.18$.

Q31: The H.C.F. of $0.54, 1.8, 7.2$ is:

A. $1.8$

B. $0.18$

C. $0.018$

D. $18$

Correct Answer: B

Solution: **Solution:** Convert numbers to integers: $54, 180, 720$.
HCF of $54, 180, 720 = 18$.
Convert back to decimals: $0.18$.
Thus, the correct answer is $0.18$.

Q32: The L.C.M. of $3, 2.7, 0.09$ is:

A. $2.7$

B. $0.27$

C. $0.027$

D. $27$

Correct Answer: D

Solution: **Solution:** Convert numbers to integers: $300, 270, 9$.
LCM of $300, 270, 9 = 2700$.
Convert back to decimals: $27$.
Thus, the correct answer is $27$.

Q33: If $A, B, C$ are three numbers such that the LCM of $A$ and $B$ is $B$, and the LCM of $B$ and $C$ is $C$, then the LCM of $A, B, C$ is:

A. $A$

B. $B$

C. $C$

D. $A + B + C$

Correct Answer: C

Solution: **Solution:** Since the LCM of $A$ and $B$ is $B$, $B$ is a multiple of $A$. Similarly, since the LCM of $B$ and $C$ is $C$, $C$ is a multiple of $B$. Therefore, $C$ is also a multiple of $A$. Hence, the LCM of $A, B, C$ is $C$.
Thus, the correct answer is $C$.

Q34: H.C.F. of $3240, 3600$, and a third number is $36$, and their LCM is $2^4 \times 3^5 \times 5^2 \times 7^2$. The third number is:

A. $2^2 \times 3^5 \times 7^2$

B. $2^2 \times 5^3 \times 7^2$

C. $2^5 \times 5^2 \times 7^2$

D. $2^3 \times 3^5 \times 7^2$

Correct Answer: A

Solution: **Solution:** Let the third number be $N$.
HCF = $36 = 2^2 \times 3^2$.
LCM = $2^4 \times 3^5 \times 5^2 \times 7^2$.
Using the relationship between HCF and LCM: $N = \frac{\text{HCF} \times \text{LCM}}{\text{Product of other two numbers}} = 2^2 \times 3^5 \times 7^2$.
Thus, the correct answer is $2^2 \times 3^5 \times 7^2$.

Q35: Three numbers are in the ratio $1:2:3$, and their HCF is $12$. The numbers are:

A. $4, 8, 12$

B. $5, 10, 15$

C. $10, 20, 30$

D. $12, 24, 36$

Correct Answer: D

Solution: **Solution:** Let the numbers be $x, 2x, 3x$.
HCF = $x = 12$.
Numbers = $12, 24, 36$.
Thus, the correct answer is $12, 24, 36$.

Q36: The ratio of two numbers is $3:4$, and their HCF is $4$. Their LCM is:

A. $12$

B. $16$

C. $24$

D. $48$

Correct Answer: D

Solution: **Solution:** Let the numbers be $3x$ and $4x$.
HCF = $x = 4$.
Numbers = $12, 16$.
LCM = $\frac{\text{Product of numbers}}{\text{HCF}} = \frac{12 \times 16}{4} = 48$.
Thus, the correct answer is $48$.

Q37: The sum of two numbers is $216$, and their HCF is $27$. The numbers are:

A. $27, 189$

B. $81, 189$

C. $108, 108$

D. $154, 162$

Correct Answer: A

Solution: **Solution:** Let the numbers be $27a$ and $27b$.
$27a + 27b = 216 \Rightarrow a + b = 8$.
Co-prime pairs $(a, b)$ with sum $8$ are $(1, 7)$.
Numbers = $27 \times 1, 27 \times 7 = 27, 189$.
Thus, the correct answer is $27, 189$.

Q38: The sum of two numbers is $528$, and their HCF is $33$. The number of pairs satisfying this is:

A. $4$

B. $6$

C. $8$

D. $12$

Correct Answer: A

Solution: **Solution:** Let the numbers be $33a$ and $33b$.
$33a + 33b = 528 \Rightarrow a + b = 16$.
Co-prime pairs $(a, b)$ with sum $16$ are $(1, 15), (3, 13), (5, 11), (7, 9)$.
Thus, there are $4$ pairs.
Correct answer is $4$.

Q39: The number of number-pairs lying between $40$ and $100$ with their HCF as $15$ is:

A. $3$

B. $4$

C. $5$

D. $6$

Correct Answer: A

Solution: **Solution:** Multiples of $15$ between $40$ and $100$: $45, 60, 75, 90$.
Valid pairs with HCF $15$: $(45, 60), (45, 75), (60, 75), (75, 90)$.
Total pairs = $4$.
Thus, the correct answer is $4$.

Q40: The H.C.F. of two numbers is 12 and their difference is 12. The numbers are:

A. $66, 78$

B. $70, 82$

C. $94, 106$

D. $84, 96$

Correct Answer: D

Solution: **Solution:** Let the numbers be $12a$ and $12b$.
Difference = $12b - 12a = 12 \Rightarrow b - a = 1$.
Possible pairs $(a, b)$ are $(7, 8)$.
Numbers = $12 \times 7, 12 \times 8 = 84, 96$.
Thus, the correct answer is $84, 96$.

Q41: The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A. $101$

B. $107$

C. $111$

D. $185$

Correct Answer: C

Solution: **Solution:** Let the numbers be $37a$ and $37b$.
Product = $37a \times 37b = 4107 \Rightarrow ab = 3$.
Co-prime pairs $(a, b)$ with product $3$ are $(1, 3)$.
Numbers = $37 \times 1, 37 \times 3 = 37, 111$.
Greater number = $111$.
Thus, the correct answer is $111$.

Q42: The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A. $1$

B. $2$

C. $3$

D. $4$

Correct Answer: B

Solution: **Solution:** Let the numbers be $13a$ and $13b$.
Product = $13a \times 13b = 2028 \Rightarrow ab = 12$.
Co-prime pairs $(a, b)$ with product $12$ are $(1, 12), (3, 4)$.
Number of pairs = $2$.
Thus, the correct answer is $2$.

Q43: Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

A. $75$

B. $81$

C. $85$

D. $89$

Correct Answer: C

Solution: **Solution:** Let the numbers be $a, b, c$.
$ab = 551$, $bc = 1073$.
HCF of $551$ and $1073 = 29$.
$a = \frac{551}{29} = 19$, $b = 29$, $c = \frac{1073}{29} = 37$.
Sum = $19 + 29 + 37 = 85$.
Thus, the correct answer is $85$.

Q44: The ratio of two numbers is $13:15$ and their L.C.M. is $39780$. The numbers are:

A. $884, 1020$

B. $884, 1040$

C. $670, 1340$

D. $2652, 3060$

Correct Answer: D

Solution: **Solution:** Let the numbers be $13x$ and $15x$.
LCM = $195x = 39780 \Rightarrow x = 204$.
Numbers = $13 \times 204, 15 \times 204 = 2652, 3060$.
Thus, the correct answer is $2652, 3060$.

Q45: Three numbers are in the ratio $3:4:5$ and their L.C.M. is $2400$. Their H.C.F. is:

A. $40$

B. $80$

C. $120$

D. $200$

Correct Answer: A

Solution: **Solution:** Let the numbers be $3x, 4x, 5x$.
LCM = $60x = 2400 \Rightarrow x = 40$.
HCF = $x = 40$.
Thus, the correct answer is $40$.

Q46: The L.C.M. and ratio of four numbers are $630$ and $2:3:5:7$ respectively. The difference between the greatest and least numbers is:

A. $6$

B. $14$

C. $15$

D. $21$

Correct Answer: C

Solution: **Solution:** Let the numbers be $2x, 3x, 5x, 7x$.
LCM = $210x = 630 \Rightarrow x = 3$.
Numbers = $6, 9, 15, 21$.
Difference = $21 - 6 = 15$.
Thus, the correct answer is $15$.

Q47: The H.C.F. and L.C.M. of two numbers are $12$ and $336$ respectively. If one of the numbers is $84$, the other is:

A. $36$

B. $48$

C. $72$

D. $96$

Correct Answer: B

Solution: **Solution:** Product of numbers = HCF $\times$ LCM.
$84 \times x = 12 \times 336 \Rightarrow x = 48$.
Thus, the correct answer is $48$.

Q48: If the product of two numbers is $324$ and their H.C.F. is $3$, then their L.C.M. will be:

A. $972$

B. $327$

C. $321$

D. $108$

Correct Answer: D

Solution: **Solution:** Product of numbers = HCF $\times$ LCM.
$324 = 3 \times \text{LCM} \Rightarrow \text{LCM} = 108$.
Thus, the correct answer is $108$.

Q49: If H.C.F. of $p$ and $q$ is $x$ and $q = xy$, then the L.C.M. of $p$ and $q$ is:

A. $pq$

B. $qy$

C. $xy$

D. $py$

Correct Answer: B

Solution: **Solution:** LCM = $\frac{\text{Product of numbers}}{\text{HCF}} = \frac{p \times q}{x} = \frac{p \times xy}{x} = py$.
Thus, the correct answer is $qy$.

Q50: The sum of two numbers is $2000$ and their L.C.M. is $21879$. The two numbers are:

A. $1993, 7$

B. $1991, 9$

C. $1989, 11$

D. $1987, 13$

Correct Answer: C

Solution: **Solution:** Let the numbers be $x$ and $2000 - x$.
Product = $x(2000 - x) = 21879 \Rightarrow x^2 - 2000x + 21879 = 0$.
Solving, $x = 1989, 11$.
Thus, the correct answer is $1989, 11$.

Q51: The H.C.F. and L.C.M. of two numbers are $84$ and $21$ respectively. If the ratio of the two numbers is $1:4$, then the larger of the two numbers is:

A. $12$

B. $48$

C. $84$

D. $108$

Correct Answer: C

Solution: **Solution:** Let the numbers be $x$ and $4x$.
Product = HCF $\times$ LCM.
$x \times 4x = 84 \times 21 \Rightarrow 4x^2 = 1764 \Rightarrow x = 21$.
Larger number = $4x = 84$.
Thus, the correct answer is $84$.

Q52: The L.C.M. of two numbers is $495$ and their H.C.F. is $5$. If the sum of the numbers is $100$, then their difference is:

A. $10$

B. $70$

C. $10$

D. $90$

Correct Answer: A

Solution: **Solution:** Let the numbers be $5a$ and $5b$.
Sum = $5a + 5b = 100 \Rightarrow a + b = 20$.
Product = HCF $\times$ LCM.
$5a \times 5b = 5 \times 495 \Rightarrow ab = 99$.
Solving, $a = 11, b = 9$.
Difference = $5(11 - 9) = 10$.
Thus, the correct answer is $10$.

Q53: The product of the L.C.M. and H.C.F. of two numbers is $24$. The difference of two numbers is $2$. Find the numbers.

A. $2, 4$

B. $6, 4$

C. $8, 6$

D. $8, 10$

Correct Answer: B

Solution: **Solution:** Let the numbers be $x$ and $x + 2$.
Product = HCF $\times$ LCM.
$x(x + 2) = 24 \Rightarrow x^2 + 2x - 24 = 0$.
Solving, $x = 4$.
Numbers = $4, 6$.
Thus, the correct answer is $6, 4$.

Q54: If the sum of two numbers is $36$ and their H.C.F. and L.C.M. are $3$ and $105$ respectively, the sum of the reciprocals of the two numbers is:

A. $\frac{4}{35}$

B. $\frac{3}{35}$

C. $\frac{2}{35}$

D. None of these

Correct Answer: A

Solution: **Solution:** Let the numbers be $3a$ and $3b$.
Sum = $3a + 3b = 36 \Rightarrow a + b = 12$.
Product = HCF $\times$ LCM.
$3a \times 3b = 3 \times 105 \Rightarrow ab = 35$.
Reciprocal sum = $\frac{1}{3a} + \frac{1}{3b} = \frac{a + b}{3ab} = \frac{12}{3 \times 35} = \frac{4}{35}$.
Thus, the correct answer is $\frac{4}{35}$.

Q55: The L.C.M. of two numbers is $12$ times their H.C.F. The sum of H.C.F. and L.C.M. is $403$. If one number is $93$, find the other.

A. $124$

B. $128$

C. $134$

D. None of these

Correct Answer: A

Solution: **Solution:** Let HCF = $h$, LCM = $12h$.
$h + 12h = 403 \Rightarrow h = 31$.
LCM = $12 \times 31 = 372$.
Product = HCF $\times$ LCM.
$93 \times x = 31 \times 372 \Rightarrow x = 124$.
Thus, the correct answer is $124$.

Q56: The H.C.F. and L.C.M. of two numbers are $50$ and $250$ respectively. If the first number is divided by $2$, the quotient is $50$. The second number is:

A. $50$

B. $100$

C. $125$

D. $250$

Correct Answer: C

Solution: **Solution:** First number = $2 \times 50 = 100$.
Product = HCF $\times$ LCM.
$100 \times x = 50 \times 250 \Rightarrow x = 125$.
Thus, the correct answer is $125$.

Q57: The product of two numbers is $1320$ and their H.C.F. is $6$. The L.C.M. of the numbers is:

A. $220$

B. $1314$

C. $1326$

D. $7920$

Correct Answer: A

Solution: **Solution:** Product = HCF $\times$ LCM.
$1320 = 6 \times \text{LCM} \Rightarrow \text{LCM} = 220$.
Thus, the correct answer is $220$.

Q58: Product of two co-prime numbers is $117$. Their L.C.M. should be:

A. $1$

B. $117$

C. Equal to their H.C.F.

D. Cannot be calculated

Correct Answer: B

Solution: **Solution:** For co-prime numbers, HCF = $1$.
LCM = Product = $117$.
Thus, the correct answer is $117$.

Q59: The L.C.M. of three different numbers is $120$. Which of the following cannot be their H.C.F.?

A. $8$

B. $12$

C. $24$

D. $35$

Correct Answer: D

Solution: **Solution:** HCF must divide LCM exactly.
$35$ does not divide $120$.
Thus, the correct answer is $35$.

Q60: The H.C.F. of two numbers is $8$. Which one of the following can never be their L.C.M.?

A. $24$

B. $48$

C. $56$

D. $60$

Correct Answer: D

Solution: **Solution:** LCM must be a multiple of HCF.
$60$ is not a multiple of $8$.
Thus, the correct answer is $60$.

Q61: If the L.C.M. of three numbers is $9570$, then their H.C.F. can be:

A. $11$

B. $12$

C. $19$

D. $21$

Correct Answer: A

Solution: **Solution:** HCF must divide LCM exactly.
Only $11$ divides $9570$.
Thus, the correct answer is $11$.

Q62: The H.C.F. of two numbers is $23$ and the other two factors of their L.C.M. are $13$ and $14$. The larger of the two numbers is:

A. $276$

B. $299$

C. $322$

D. $345$

Correct Answer: C

Solution: **Solution:** Numbers = $23 \times 13, 23 \times 14 = 299, 322$.
Larger number = $322$.
Thus, the correct answer is $322$.

Q63: About the number of pairs which have $16$ as their H.C.F. and $136$ as their L.C.M., we can definitely say that:

A. No such pair exists

B. Only one such pair exists

C. Only two such pairs exist

D. Many such pairs exist

E. -

Correct Answer: A

Solution: **Solution:** HCF must divide LCM exactly.
$16$ does not divide $136$.
Thus, no such pair exists.
Correct answer is "No such pair exists".

Q64: The H.C.F. and L.C.M. of two numbers are $21$ and $4641$ respectively. If one of the numbers lies between $200$ and $300$, the two numbers are:

A. $273, 357$

B. $273, 359$

C. $273, 361$

D. $273, 363$

Correct Answer: A

Solution: **Solution:** Product of numbers = HCF $\times$ LCM = $21 \times 4641 = 97461$.
Let the numbers be $21a$ and $21b$.
$21a \times 21b = 97461 \Rightarrow ab = 221$.
Co-prime pairs with product $221$ = $(1, 221)$ and $(13, 17)$.
Numbers = $(21 \times 13, 21 \times 17) = (273, 357)$.
Since one number lies between $200$ and $300$, the correct pair is $273, 357$.
Thus, the correct answer is $273, 357$.

Q65: Two numbers, both greater than 29, have H.C.F. 29 and L.C.M. 4147. The sum of the numbers is:

A. $666$

B. $669$

C. $696$

D. $966$

Correct Answer: C

Solution: **Solution:** Let the numbers be $29a$ and $29b$.
Product = HCF $\times$ LCM.
$29a \times 29b = 29 \times 4147 \Rightarrow ab = 143$.
Co-prime pairs $(a, b)$ with product $143$ are $(11, 13)$.
Numbers = $29 \times 11, 29 \times 13 = 319, 377$.
Sum = $319 + 377 = 696$.
Thus, the correct answer is $696$.

Q66: L.C.M. of two prime numbers $x$ and $y$ ($x > y$) is 161. The value of $3y - x$ is:

A. $-2$

B. $-1$

C. $1$

D. $2$

Correct Answer: A

Solution: **Solution:** Since $x$ and $y$ are prime, their HCF = $1$.
LCM = $x \times y = 161$.
Prime factorization of $161 = 7 \times 23$.
So, $x = 23, y = 7$.
$3y - x = 3(7) - 23 = 21 - 23 = -2$.
Thus, the correct answer is $-2$.

Q67: The greatest number that exactly divides $105$, $1001$, and $2436$ is:

A. $3$

B. $7$

C. $11$

D. $21$

Correct Answer: B

Solution: **Solution:** HCF of $105, 1001, 2436$:
$105 = 3 \times 5 \times 7$,
$1001 = 7 \times 11 \times 13$,
$2436 = 2^2 \times 3 \times 7 \times 29$.
Common factor = $7$.
Thus, the correct answer is $7$.

Q68: Minimum number of rows to plant $21$ mango, $42$ apple, and $56$ orange trees equally:

A. $3$

B. $15$

C. $17$

D. $20$

Correct Answer: A

Solution: **Solution:** HCF of $21, 42, 56 = 7$.
Trees per row = $7$.
Total rows = $\frac{21}{7} + \frac{42}{7} + \frac{56}{7} = 3 + 6 + 8 = 17$.
Thus, the correct answer is $17$.

Q69: Greatest length to measure $7$ m, $3$ m $85$ cm, and $12$ m $95$ cm exactly:

A. $15$ cm

B. $25$ cm

C. $35$ cm

D. $42$ cm

Correct Answer: C

Solution: **Solution:** Convert to cm: $700, 385, 1295$.
HCF = $35$.
Thus, the correct answer is $35$ cm.

Q70: Capacity of a container to measure $120$ litres and $56$ litres exactly:

A. $7500$ cc

B. $7850$ cc

C. $8000$ cc

D. $9500$ cc

Correct Answer: C

Solution: **Solution:** HCF of $120, 56 = 8$.
$8$ litres = $8000$ cc.
Thus, the correct answer is $8000$ cc.

Q71: Maximum possible daily wage for a labourer paid ₹$5750$ or ₹$5000$:

A. ₹$125$

B. ₹$250$

C. ₹$375$

D. ₹$500$

Correct Answer: B

Solution: **Solution:** HCF of $5750, 5000 = 250$.
Thus, the correct answer is ₹$250$.

Q72: Least number of bottles to store $403$ litres, $465$ litres, and $496$ litres:

A. $34$

B. $44$

C. $46$

D. None of these

Correct Answer: B

Solution: **Solution:** HCF of $403, 465, 496 = 31$.
Bottles = $\frac{403}{31} + \frac{465}{31} + \frac{496}{31} = 13 + 15 + 16 = 44$.
Thus, the correct answer is $44$.

Q73: Maximum students among whom $1001$ pens and $910$ pencils can be distributed equally:

A. $91$

B. $910$

C. $1001$

D. $1911$

Correct Answer: A

Solution: **Solution:** HCF of $1001, 910 = 91$.
Thus, the correct answer is $91$.

Q74: Largest tile size to pave a $3.78$ m by $5.25$ m courtyard:

A. $14$ cm

B. $21$ cm

C. $42$ cm

D. None of these

Correct Answer: B

Solution: **Solution:** Convert to cm: $378, 525$.
HCF = $21$.
Thus, the correct answer is $21$ cm.

Q75: Least tiles to pave a $15$ m $17$ cm by $9$ m $2$ cm ceiling:

A. $656$

B. $738$

C. $814$

D. $902$

Correct Answer: C

Solution: **Solution:** Convert to cm: $1517, 902$.
HCF = $41$.
Tiles = $\frac{1517 \times 902}{41 \times 41} = 814$.
Thus, the correct answer is $814$.

Q76: Total stacks for $336$, $240$, and $96$ books stored equally:

A. $14$

B. $21$

C. $22$

D. $48$

Correct Answer: A

Solution: **Solution:** HCF of $336, 240, 96 = 48$.
Stacks = $\frac{336}{48} + \frac{240}{48} + \frac{96}{48} = 7 + 5 + 2 = 14$.
Thus, the correct answer is $14$.

Q77: Maximum pieces from rods of $78$, $104$, $117$, and $169$ cm cut equally:

A. $27$

B. $36$

C. $43$

D. $480$

Correct Answer: B

Solution: **Solution:** HCF of $78, 104, 117, 169 = 13$.
Pieces = $\frac{78}{13} + \frac{104}{13} + \frac{117}{13} + \frac{169}{13} = 6 + 8 + 9 + 13 = 36$.
Thus, the correct answer is $36$.

Q78: Greatest number dividing $43$, $91$, and $183$ leaving the same remainder:

A. $4$

B. $7$

C. $9$

D. $13$

Correct Answer: A

Solution: **Solution:** Differences = $91 - 43 = 48$, $183 - 91 = 92$, $183 - 43 = 140$.
HCF of $48, 92, 140 = 4$.
Thus, the correct answer is $4$.

Q79: If $r$ is the remainder when $7654$, $8506$, and $9997$ are divided by $d$, then $d - r = ?$

A. $14$

B. $18$

C. $24$

D. $28$

Correct Answer: A

Solution: **Solution:** Differences = $8506 - 7654 = 852$, $9997 - 8506 = 1491$, $9997 - 7654 = 2343$.
HCF of $852, 1491, 2343 = 213$.
Remainder $r = 199$.
$d - r = 213 - 199 = 14$.
Thus, the correct answer is $14$.

Q80: Sum of digits of $N$, where $N$ divides $1305$, $4665$, and $6905$ leaving the same remainder:

A. $4$

B. $5$

C. $6$

D. $8$

Correct Answer: A

Solution: **Solution:** Differences = $4665 - 1305 = 3360$, $6905 - 4665 = 2240$, $6905 - 1305 = 5600$.
HCF of $3360, 2240, 5600 = 1120$.
Sum of digits = $1 + 1 + 2 + 0 = 4$.
Thus, the correct answer is $4$.

Q81: Largest volume of a can measuring $57$, $129$, and $177$ litres equally:

A. $12$ litres

B. $16$ litres

C. $24$ litres

D. None of these

Correct Answer: C

Solution: **Solution:** HCF of $57, 129, 177 = 24$.
Thus, the correct answer is $24$ litres.

Q82: Greatest number dividing $1356$, $1868$, and $2764$ leaving remainder $12$:

A. $64$

B. $124$

C. $156$

D. $260$

Correct Answer: A

Solution: **Solution:** Subtract $12$: $1344, 1856, 2752$.
HCF = $64$.
Thus, the correct answer is $64$.

Q83: Greatest number dividing $3026$ and $5053$ leaving remainders $11$ and $13$:

A. $15$

B. $30$

C. $45$

D. $60$

Correct Answer: C

Solution: **Solution:** Subtract remainders: $3015, 5040$.
HCF = $45$.
Thus, the correct answer is $45$.

Q84: Greatest number dividing $964$, $1238$, and $1400$ leaving remainders $41$, $31$, $51$:

A. $61$

B. $71$

C. $73$

D. $81$

Correct Answer: B

Solution: **Solution:** Subtract remainders: $923, 1207, 1349$.
HCF = $71$.
Thus, the correct answer is $71$.

Q85: Largest fraction among $\frac{7}{8}$, $\frac{13}{16}$, $\frac{31}{40}$, $\frac{63}{80}$:

A. $\frac{7}{8}$

B. $\frac{13}{16}$

C. $\frac{31}{40}$

D. $\frac{63}{80}$

Correct Answer: A

Solution: **Solution:** LCM of denominators = $80$.
Compare numerators: $70, 65, 62, 63$.
Largest = $\frac{7}{8}$.
Thus, the correct answer is $\frac{7}{8}$.

Q86: What is the least natural number which leaves no remainder when divided by all digits from 1 to 9?

A. $1800$

B. $1920$

C. $2520$

D. $5040$

Correct Answer: C

Solution: **Solution:** LCM of $1, 2, 3, 4, 5, 6, 7, 8, 9 = 2520$.
Thus, the correct answer is $2520$.

Q87: What will be the least number which when doubled will be exactly divisible by $12, 18, 21, 30$?

A. $196$

B. $630$

C. $1260$

D. $2520$

Correct Answer: B

Solution: **Solution:** LCM of $12, 18, 21, 30 = 1260$.
Required number = $\frac{1260}{2} = 630$.
Thus, the correct answer is $630$.

Q88: The sum of two numbers is $45$. Their difference is $\frac{1}{7}$ of their sum. Their LCM is:

A. $100$

B. $150$

C. $200$

D. $250$

Correct Answer: A

Solution: **Solution:** Let the numbers be $x$ and $y$.
$x + y = 45$, $x - y = \frac{1}{7} \times 45 = 5$.
Solving: $x = 25, y = 20$.
LCM = $100$.
Thus, the correct answer is $100$.

Q89: The smallest fraction, which each of $\frac{6}{7}, \frac{5}{14}, \frac{10}{21}$ divides exactly:

A. $\frac{30}{7}$

B. $\frac{60}{147}$

C. $\frac{30}{58}$

D. $\frac{50}{147}$

Correct Answer: B

Solution: **Solution:** LCM of numerators = $30$, HCF of denominators = $7$.
Smallest fraction = $\frac{30}{7}$.
Thus, the correct answer is $\frac{30}{7}$.

Q90: The least number of five digits which is exactly divisible by $12, 15, 18$:

A. $10010$

B. $10015$

C. $10020$

D. $10080$

Correct Answer: D

Solution: **Solution:** LCM of $12, 15, 18 = 180$.
Smallest 5-digit number = $10000$.
Remainder = $100$.
Required number = $10000 + (180 - 100) = 10080$.
Thus, the correct answer is $10080$.

Q91: The greatest number of four digits which is divisible by $15, 25, 40, 75$:

A. $9000$

B. $9400$

C. $9600$

D. $9800$

Correct Answer: C

Solution: **Solution:** LCM of $15, 25, 40, 75 = 600$.
Largest 4-digit number = $9999$.
Remainder = $399$.
Required number = $9999 - 399 = 9600$.
Thus, the correct answer is $9600$.

Q92: The number between $4000$ and $5000$ which is divisible by $12, 18, 21, 32$:

A. $4023$

B. $4032$

C. $4203$

D. $4302$

Correct Answer: B

Solution: **Solution:** LCM of $12, 18, 21, 32 = 2016$.
Multiple of $2016$ between $4000$ and $5000 = 4032$.
Thus, the correct answer is $4032$.

Q93: The number nearest to $43582$ divisible by $25, 50, 75$:

A. $43500$

B. $43550$

C. $43600$

D. $43650$

Correct Answer: D

Solution: **Solution:** LCM of $25, 50, 75 = 150$.
Nearest multiple of $150$ to $43582 = 43650$.
Thus, the correct answer is $43650$.

Q94: The least number which should be added to $2497$ so that the sum is exactly divisible by $5, 6, 4, 3$:

A. $3$

B. $13$

C. $23$

D. $33$

Correct Answer: C

Solution: **Solution:** LCM of $5, 6, 4, 3 = 60$.
Remainder = $37$.
Number to add = $60 - 37 = 23$.
Thus, the correct answer is $23$.

Q95: The greatest number which when subtracted from $5834$, gives a number divisible by $20, 28, 32, 35$:

A. $1120$

B. $4714$

C. $5200$

D. $5600$

Correct Answer: B

Solution: **Solution:** LCM of $20, 28, 32, 35 = 1120$.
Required number = $5834 - 1120 = 4714$.
Thus, the correct answer is $4714$.

Q96: The least number which is a perfect square and divisible by $16, 20, 24$:

A. $1600$

B. $3600$

C. $6400$

D. $14400$

Correct Answer: B

Solution: **Solution:** LCM of $16, 20, 24 = 240$.
To make it a perfect square, multiply by $3 \times 5 = 15$.
Required number = $240 \times 15 = 3600$.
Thus, the correct answer is $3600$.

Q97: The smallest number which when diminished by $7$, is divisible by $12, 16, 18, 21, 28$:

A. $1008$

B. $1015$

C. $1022$

D. $1032$

Correct Answer: B

Solution: **Solution:** LCM of $12, 16, 18, 21, 28 = 1008$.
Required number = $1008 + 7 = 1015$.
Thus, the correct answer is $1015$.

Q98: The least number which when increased by $5$ is divisible by $24, 32, 36, 54$:

A. $427$

B. $859$

C. $869$

D. $4320$

Correct Answer: B

Solution: **Solution:** LCM of $24, 32, 36, 54 = 864$.
Required number = $864 - 5 = 859$.
Thus, the correct answer is $859$.

Q99: The least number which when divided by $12, 15, 20, 54$ leaves a remainder $8$:

A. $504$

B. $536$

C. $544$

D. $548$

Correct Answer: D

Solution: **Solution:** LCM of $12, 15, 20, 54 = 540$.
Required number = $540 + 8 = 548$.
Thus, the correct answer is $548$.

Q100: A number less than $500$, when divided by $4, 5, 6, 7$ leaves remainder $1$:

A. $211$

B. $420$

C. $421$

D. $441$

Correct Answer: C

Solution: **Solution:** LCM of $4, 5, 6, 7 = 420$.
Required number = $420 + 1 = 421$.
Thus, the correct answer is $421$.

Q101: The greatest number of $3$ digits which when divided by $6, 9, 12$ leaves a remainder $3$:

A. $903$

B. $939$

C. $975$

D. $996$

Correct Answer: C

Solution: **Solution:** LCM of $6, 9, 12 = 36$.
Largest $3$-digit number = $999$.
Remainder = $27$.
Required number = $999 - 27 + 3 = 975$.
Thus, the correct answer is $975$.

Q102: The largest four-digit number which when divided by $4, 7, 13$ leaves a remainder $3$:

A. $8739$

B. $9831$

C. $9834$

D. $9893$

Correct Answer: B

Solution: **Solution:** LCM of $4, 7, 13 = 364$.
Largest $4$-digit number = $9999$.
Remainder = $171$.
Required number = $9999 - 171 + 3 = 9831$.
Thus, the correct answer is $9831$.

Q103: The least number of six digits which when divided by $4, 6, 10, 15$ leaves a remainder $2$:

A. $100022$

B. $100020$

C. $100024$

D. $100026$

Correct Answer: A

Solution: **Solution:** LCM of $4, 6, 10, 15 = 60$.
Smallest $6$-digit number = $100000$.
Remainder = $40$.
Required number = $100000 + (60 - 40) + 2 = 100022$.
Thus, the correct answer is $100022$.

Q104: The least multiple of $13$ which on dividing by $4, 5, 6, 7, 8$ leaves a remainder $2$:

A. $840$

B. $842$

C. $2520$

D. $2522$

Correct Answer: D

Solution: **Solution:** LCM of $4, 5, 6, 7, 8 = 840$.
Required number = $840k + 2$.
For divisibility by $13$, $k = 3$.
Required number = $840 \times 3 + 2 = 2522$.
Thus, the correct answer is $2522$.

Q105: Find the least number which when divided by $12, 15, 16$ leaves remainders $7, 10, 11$:

A. $115$

B. $235$

C. $247$

D. $475$

Correct Answer: B

Solution: **Solution:** LCM of $12, 15, 16 = 240$.
Required number = $240 - 5 = 235$.
Thus, the correct answer is $235$.

Q106: The least number which when divided by $48, 60, 72, 108, 140$ leaves remainders $38, 50, 62, 98, 130$:

A. $11115$

B. $15110$

C. $15120$

D. $15210$

Correct Answer: B

Solution: **Solution:** LCM of $48, 60, 72, 108, 140 = 15120$.
Required number = $15120 - 10 = 15110$.
Thus, the correct answer is $15110$.

Q107: Find the least multiple of $23$ which when divided by $18, 21, 24$ leaves remainders $7, 10, 13$:

A. $3002$

B. $3013$

C. $3024$

D. $3036$

Correct Answer: B

Solution: **Solution:** LCM of $18, 21, 24 = 504$.
Required number = $504k - 11$.
For divisibility by $23$, $k = 6$.
Required number = $504 \times 6 - 11 = 3013$.
Thus, the correct answer is $3013$.

Q108: What is the third term in a sequence of numbers that leave remainders $1, 2, 3$ when divided by $2, 3, 4$:

A. $11$

B. $17$

C. $19$

D. $35$

Correct Answer: D

Solution: **Solution:** LCM of $2, 3, 4 = 12$.
Sequence = $12k - 1$.
Third term = $12 \times 3 - 1 = 35$.
Thus, the correct answer is $35$.

Q109: The greatest number of $4$ digits which when divided by $4, 5, 6, 7, 8$ leaves remainders $1, 2, 3, 4, 5$:

A. $9237$

B. $9240$

C. $9840$

D. $9999$

Correct Answer: A

Solution: **Solution:** LCM of $4, 5, 6, 7, 8 = 840$.
Largest $4$-digit number = $9999$.
Remainder = $759$.
Required number = $9999 - 759 - 3 = 9237$.
Thus, the correct answer is $9237$.

Q110: The least number which when divided by $5, 6, 7, 8$ leaves a remainder $3$, but when divided by $9$ leaves no remainder:

A. $1677$

B. $1683$

C. $2523$

D. $3363$

Correct Answer: B

Solution: **Solution:** LCM of $5, 6, 7, 8 = 840$.
Required number = $840k + 3$.
For divisibility by $9$, $k = 2$.
Required number = $840 \times 2 + 3 = 1683$.
Thus, the correct answer is $1683$.

Q111: Find the least number which when divided by $16, 18, 20, 25$ leaves $4$ as remainder in each case, but when divided by $7$ leaves no remainder.

A. $17004$

B. $18000$

C. $18002$

D. $18004$

Correct Answer: D

Solution: **Solution:** LCM of $16, 18, 20, 25 = 3600$.
Required number = $3600k + 4$.
For divisibility by $7$, $k = 5$.
Required number = $3600 \times 5 + 4 = 18004$.
Thus, the correct answer is $18004$.

Q112: A gardener has to plant trees in rows containing equal number of trees. If he plants in rows of $6, 8, 10,$ or $12$, then five trees are left unplanted. But if he plants in rows of $13$ trees each, then no tree is left. What is the number of trees that the gardener plants?

A. $485$

B. $725$

C. $845$

D. $None$

Correct Answer: C

Solution: **Solution:** LCM of $6, 8, 10, 12 = 120$.
Required number = $120k + 5$.
For divisibility by $13$, $k = 7$.
Required number = $120 \times 7 + 5 = 845$.
Thus, the correct answer is $845$.

Q113: When Seeta made necklaces of either $16$ beads, $20$ beads, or $36$ beads, not a single bead was left over. What could be the least number of beads Seeta had?

A. $700$

B. $720$

C. $750$

D. $780$

Correct Answer: B

Solution: **Solution:** LCM of $16, 20, 36 = 720$.
Thus, the correct answer is $720$.

Q114: An electronic device makes a beep after every $60$ seconds. Another device makes a beep after every $62$ seconds. They beeped together at $10$ a.m. The next time they would beep together at the earliest is:

A. $10:30$ a.m.

B. $10:31$ a.m.

C. $10:59$ a.m.

D. $11:00$ a.m.

Correct Answer: B

Solution: **Solution:** LCM of $60, 62 = 1860$ seconds = $31$ minutes.
Next beep = $10:00 + 31$ minutes = $10:31$ a.m.
Thus, the correct answer is $10:31$ a.m.

Q115: Six bells commence tolling together and toll at intervals of $2, 4, 6, 8, 10,$ and $12$ seconds respectively. In $30$ minutes, how many times do they toll together?

A. $4$

B. $10$

C. $15$

D. $16$

Correct Answer: D

Solution: **Solution:** LCM of $2, 4, 6, 8, 10, 12 = 120$ seconds = $2$ minutes.
Total time = $30$ minutes.
Number of times = $\frac{30}{2} + 1 = 16$.
Thus, the correct answer is $16$.

Q116: Four bells begin to toll together and toll respectively at intervals of $6, 7, 8,$ and $9$ seconds. In $1.54$ hours, how many times do they toll together and in what interval (seconds)?

A. $14, 504$

B. $14, 480$

C. $12, 504$

D. $16, 580$

Correct Answer: A

Solution: **Solution:** LCM of $6, 7, 8, 9 = 504$ seconds.
Total time = $1.54 \times 60 \times 60 = 5544$ seconds.
Number of times = $\frac{5544}{504} + 1 = 14$.
Interval = $504$ seconds.
Thus, the correct answer is $14, 504$.

Q117: Four different electronic devices make a beep after every $30$ minutes, $1$ hour, $1.5$ hours, and $1$ hour $45$ minutes respectively. All the devices beeped together at $12$ noon. They will again beep together at:

A. $12$ midnight

B. $3$ a.m.

C. $6$ a.m.

D. $9$ a.m.

Correct Answer: D

Solution: **Solution:** LCM of $30, 60, 90, 105 = 1260$ minutes = $21$ hours.
Next beep = $12$ noon + $21$ hours = $9$ a.m.
Thus, the correct answer is $9$ a.m.

Q118: Three girls start jogging from the same point around a circular track and each one completes one round in $24, 36,$ and $48$ seconds respectively. After how much time will they meet at one point?

A. $2$ min $20$ sec

B. $2$ min $24$ sec

C. $3$ min $36$ sec

D. $4$ min $12$ sec

Correct Answer: B

Solution: **Solution:** LCM of $24, 36, 48 = 144$ seconds = $2$ min $24$ sec.
Thus, the correct answer is $2$ min $24$ sec.

Q119: Three persons walking around a circular track complete their respective single revolutions in $15, 16,$ and $18$ seconds respectively. They will be again together at the common starting point after an hour and:

A. $10$ sec

B. $20$ sec

C. $30$ sec

D. $40$ sec

Correct Answer: B

Solution: **Solution:** LCM of $15, 16, 18 = 720$ seconds = $12$ minutes.
Time after $1$ hour = $12$ minutes = $720$ seconds.
Thus, the correct answer is $20$ sec.

Q120: $A, B,$ and $C$ start at the same time in the same direction to run around a circular stadium. $A$ completes a round in $252$ seconds, $B$ in $308$ seconds, and $C$ in $198$ seconds, all starting at the same point. After what time will they meet again at the starting point?

A. $26$ min $18$ sec

B. $42$ min $36$ sec

C. $45$ min

D. $46$ min $12$ sec

Correct Answer: D

Solution: **Solution:** LCM of $252, 308, 198 = 2772$ seconds = $46$ min $12$ sec.
Thus, the correct answer is $46$ min $12$ sec.

Q121: Three wheels can complete $40, 24,$ and $16$ revolutions per minute respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

A. $7 \frac{1}{2}$ sec

B. $18$ sec

C. $7 \frac{1}{2}$ min

D. $18$ min

Correct Answer: A

Solution: **Solution:** LCM of $\frac{60}{40}, \frac{60}{24}, \frac{60}{16} = 7.5$ seconds.
Thus, the correct answer is $7 \frac{1}{2}$ sec.

Q122: A pendulum strikes $5$ times in $3$ seconds and another pendulum strikes $7$ times in $4$ seconds. If both pendulums start striking at the same time, how many clear strikes can be listened to in $1$ minute?

A. $195$

B. $199$

C. $200$

D. $205$

Correct Answer: B

Solution: **Solution:** LCM of $\frac{3}{5}, \frac{4}{7} = 12$ seconds.
Strikes in $1$ minute = $\frac{60}{12} \times (5 + 7) - 6 = 199$.
Thus, the correct answer is $199$.

Q123: Find the HCF of $132, 204,$ and $228$.

A. $12$

B. $18$

C. $24$

D. $36$

Correct Answer: A

Solution: **Solution:** HCF of $132, 204, 228 = 12$.
Thus, the correct answer is $12$.

Q124: If three numbers are $2a, 5a,$ and $7a$, what will be their LCM?

A. $70a$

B. $65a$

C. $75a$

D. $70a^3$

Correct Answer: A

Solution: **Solution:** LCM of $2a, 5a, 7a = 70a$.
Thus, the correct answer is $70a$.

Q125: The product of two whole numbers is $1500$ and their HCF is $10$. Find the LCM.

A. $15000$

B. $150$

C. $15$

D. $1500$

Correct Answer: B

Solution: **Solution:** Product = HCF $\times$ LCM.
$1500 = 10 \times LCM$.
LCM = $150$.
Thus, the correct answer is $150$.

Q126: A number $x$ is divided by $7$. When this number is divided by $8, 12,$ and $16$, it leaves a remainder $3$ in each case. The least value of $x$ is:

A. $148$

B. $149$

C. $150$

D. $147$

Correct Answer: D

Solution: **Solution:** LCM of $8, 12, 16 = 48$.
Required number = $48k + 3$.
For divisibility by $7$, $k = 3$.
Required number = $48 \times 3 + 3 = 147$.
Thus, the correct answer is $147$.

Q127: The number of pairs of positive integers whose sum is $99$ and HCF is $9$ is:

A. $5$

B. $4$

C. $3$

D. $2$

Correct Answer: A

Solution: **Solution:** Let the numbers be $9a, 9b$.
$9a + 9b = 99 \Rightarrow a + b = 11$.
Co-prime pairs $(a, b)$ = $(1, 10), (2, 9), (4, 7), (5, 6)$.
Thus, the correct answer is $5$.

Q128: The ratio of two numbers is $3:4$ and their LCM is $120$. The sum of the numbers is:

A. $70$

B. $35$

C. $105$

D. $140$

Correct Answer: A

Solution: **Solution:** Let the numbers be $3x, 4x$.
LCM = $12x = 120 \Rightarrow x = 10$.
Numbers = $30, 40$.
Sum = $30 + 40 = 70$.
Thus, the correct answer is $70$.

Q129: The greatest four-digit number which is exactly divisible by each one of the numbers $12, 18, 21,$ and $28$ is:

A. $9288$

B. $9882$

C. $9828$

D. $9928$

Correct Answer: C

Solution: **Solution:** LCM of $12, 18, 21, 28 = 252$.
Largest 4-digit number = $9999$.
Divide $9999$ by $252$, remainder = $171$.
Required number = $9999 - 171 = 9828$.
Thus, the correct answer is $9828$.

Q130: The traffic lights at three different signal points change after every $45$ seconds, $75$ seconds, and $90$ seconds respectively. If all change simultaneously at $7:20:15$ hours, then they will change again simultaneously at:

A. $7:28:00$

B. $7:27:45$

C. $7:27:30$

D. $7:27:50$

Correct Answer: B

Solution: **Solution:** LCM of $45, 75, 90 = 450$ seconds = $7$ minutes $30$ seconds.
Next simultaneous change = $7:20:15 + 7$ minutes $30$ seconds = $7:27:45$.
Thus, the correct answer is $7:27:45$.