Q1: The last digit of the number obtained by multiplying the numbers $81\times82\times83\times84\times85\times86\times87\times88\times89$ will be
A. 0
B. 9
C. 7
D. 2
Correct Answer: A
Solution: We need to find the unit digit of the product $81\times82\times83\times84\times85\times86\times87\times88\times89$. The unit digit of the product is determined by the unit digits of each number in the product. The unit digits are $1, 2, 3, 4, 5, 6, 7, 8, 9$. The product of these unit digits is $1\times2\times3\times4\times5\times6\times7\times8\times9 = 362880$. The unit digit of this product is 0. Therefore, the unit digit of the number obtained by multiplying the given numbers will be 0.
Q2: The sum of the digits of a two-digit number is 10, while when the digits are reversed, the number decreases by 54. Find the changed number.
A. 28
B. 19
C. 37
D. 46
Correct Answer: A
Solution: Let the two-digit number be $10a + b$, where $a$ and $b$ are digits from 0 to 9, and $a \ne 0$. The sum of the digits is 10, so $a + b = 10$. When the digits are reversed, the new number is $10b + a$. The number decreases by 54 when the digits are reversed, so $10a + b - (10b + a) = 54$. This simplifies to $9a - 9b = 54$, which further simplifies to $a - b = 6$. Now we have a system of two linear equations:
$a + b = 10$
$a - b = 6$
Adding the two equations, we get $2a = 16$, so $a = 8$. Substituting $a = 8$ into $a + b = 10$, we get $8 + b = 10$, so $b = 2$. The original number is $10a + b = 10(8) + 2 = 82$. The reversed number is $10b + a = 10(2) + 8 = 28$. The changed number (the reversed number) is 28.
Q3: When we multiply a certain two-digit number by the sum of its digits, 405 is achieved. If you multiply the number written in reverse order of the same digits by the sum of the digits, we get 486. Find the number.1
A. 81
B. 45
C. 36
D. 54
Correct Answer: B
Solution: Let the two-digit number be $10a + b$, where $a$ and $b$ are digits from 1 to 9. The problem states that $(10a + b)(a + b) = 405$. Also, $(10b + a)(a + b) = 486$. Dividing the second equation by the first, we get:
$\frac{(10b + a)(a + b)}{(10a + b)(a + b)} = \frac{486}{405}$
$\frac{10b + a}{10a + b} = \frac{486}{405} = \frac{54}{45} = \frac{6}{5}$
$5(10b + a) = 6(10a + b)$
$50b + 5a = 60a + 6b$
$44b = 55a$
$4b = 5a$
Since $a$ and $b$ are integers, $a$ must be a multiple of 4 and $b$ a multiple of 5. The only single-digit possibilities are $a=4$ and $b=5$. Therefore, the number is $10a + b = 10(4) + 5 = 45$. Let's check:
$(4+5)(45) = 9 \times 45 = 405$. $(5+4)(54) = 9 \times 54 = 486$. Both conditions are satisfied.
Q4: The difference between two numbers is 48 and the difference between the arithmetic mean and the geometric mean is two more than half of $1/3$ of 96. Find the numbers.
A. 49, 1
B. 12,60
C. 50,2
D. 36,84
Correct Answer: A
Solution: Let the two numbers be $x$ and $y$. We are given that the difference between the two numbers is 48, so we can write:
$x - y = 48$ (1)
The arithmetic mean is $\frac{x+y}{2}$ and the geometric mean is $\sqrt{xy}$. The difference between the arithmetic mean and the geometric mean is given as two more than half of $1/3$ of 96:
$\frac{x+y}{2} - \sqrt{xy} = 2 + \frac{1}{2} \times \frac{1}{3} \times 96 = 2 + 16 = 18$
$\frac{x+y}{2} - \sqrt{xy} = 18$ (2)
From (1), $x = y + 48$. Substituting this into (2):
$\frac{2y + 48}{2} - \sqrt{(y+48)y} = 18$
$y + 24 - \sqrt{y^2 + 48y} = 18$
$\sqrt{y^2 + 48y} = y + 6$
Squaring both sides:
$y^2 + 48y = y^2 + 12y + 36$
$36y = 36$
$y = 1$
Then $x = y + 48 = 1 + 48 = 49$
Therefore, the two numbers are 49 and 1.
Q5: If 3814 is divisible by 9, find the value of smallest natural number A.
A. 5
B. 5
C. 7
D. 6
Correct Answer: D
Solution: A number is divisible by 9 if the sum of its digits is divisible by 9. The given number is 381A. The sum of its digits is $3 + 8 + 1 + A = 12 + A$. For this sum to be divisible by 9, $12 + A$ must be a multiple of 9. The smallest multiple of 9 greater than or equal to 12 is 18. Therefore, $12 + A = 18$, which means $A = 18 - 12 = 6$.
Q6: What will be the remainder obtained when $(9^{6}+1)$ will be divided by 8?
A. 0
B. 3
C. 7
D. 2
Correct Answer: D
Solution: We need to find the remainder when $(9^6 + 1)$ is divided by 8. We can rewrite 9 as $8+1$. Therefore, $9^6 = (8+1)^6$. Using the binomial theorem, we have:
$(8+1)^6 = \binom{6}{0}8^6 + \binom{6}{1}8^5 + \binom{6}{2}8^4 + \binom{6}{3}8^3 + \binom{6}{4}8^2 + \binom{6}{5}8^1 + \binom{6}{6}8^0$
$(8+1)^6 = 8^6 + 6(8^5) + 15(8^4) + 20(8^3) + 15(8^2) + 6(8) + 1$
Every term except the last one is divisible by 8. Therefore, $(8+1)^6 \equiv 1 \pmod{8}$
So, $9^6 \equiv 1 \pmod{8}$
Adding 1 to both sides, we get:
$9^6 + 1 \equiv 1 + 1 \pmod{8}$
$9^6 + 1 \equiv 2 \pmod{8}$
Thus, the remainder when $(9^6 + 1)$ is divided by 8 is 2.
Q7: Find the LCM of $5/2$, $8/9$, 11/14.
A. 280
B. 360
C. 420
D. None of these
Correct Answer: D
Solution: To find the LCM of fractions, we first find the LCM of the numerators and the HCF of the denominators. Numerators: 5, 8, 11. The LCM of these numbers is $5 \times 8 \times 11 = 440$, since they are coprime. Denominators: 2, 9, 14. Prime factorization:
$2 = 2^1$
$9 = 3^2$
$14 = 2^1 \times 7^1$
HCF(2, 9, 14) = $2^1 = 2$
LCM($\frac{5}{2}$, $\frac{8}{9}$, $\frac{11}{14}$) = $\frac{LCM(5, 8, 11)}{HCF(2, 9, 14)} = \frac{440}{2} = 220$
Q8: If the number A is even, which of the following will be true?
A. $3A$ will always be divisible by 6
B. $3A+5$ will always be divisible by 11
C. $(A^{2}+3)/4$ will be divisible by 7
D. All of these
Correct Answer: A
Solution: An even number is an integer that is divisible by 2. If $A$ is an even number, it can be represented as $A = 2k$, where $k$ is an integer. The question lacks the "following" options, making a definitive answer impossible. However, several statements would be true about an even number $A$:
\begin{itemize}
\item $A$ is divisible by 2. \item $A/2$ is an integer. \item $A$ can be expressed as $2k$ where $k$ is an integer. \item The last digit of $A$ is 0, 2, 4, 6, or 8. \end{itemize}
Without the options, a specific correct answer cannot be provided. The question needs to be completed.
Q9: A five-digit number is taken. Sum of the first four digits (excluding the number at the units digit) equals sum of all the five digits. Which of the following will not divide this number necessarily?
A. 10
B. 2
C. 4
D. 5
Correct Answer: C
Solution: Let the five-digit number be represented as $abcde$, where $a, b, c, d, e$ are digits from 0 to 9, and $a \ne 0$. The given condition is that the sum of the first four digits equals the sum of all five digits. This can be written as:
$a + b + c + d = a + b + c + d + e$
Subtracting $a + b + c + d$ from both sides, we get:
$0 = e$
This means the units digit ($e$) must be 0. Therefore, the number is divisible by 10, 2, and 5. Since the number is divisible by 2 and 5, it's also divisible by their least common multiple, which is 10. Any number ending in 0 is divisible by 2, 5, and 10. There is no other information given to determine divisibility by any other number. Thus, the number is necessarily divisible by 2, 5, and 10. The question asks which of the following will *not* divide this number necessarily. Without knowing the specific options, it's impossible to answer which option will not necessarily divide the number. However, based on the provided information, a number that will *not* necessarily divide this number is one that is not a factor of 10 (or any number not necessarily a factor of the number based on the given information).
Q10: A number 15B is divisible by 6. Which of these will be true about the positive integer B?
A. B will be even
B. B will be odd
C. B will be divisible by 6
D. Both (a) and (c)
Correct Answer: D
Solution: A number is divisible by 6 if it is divisible by both 2 and 3. For a number to be divisible by 2, its unit digit must be an even number (0, 2, 4, 6, or 8). Therefore, B can be 0, 2, 4, 6, or 8. For a number to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits of 15B is $1 + 5 + B = 6 + B$. For $6 + B$ to be divisible by 3, B can be 0, 3, 6, or 9. To satisfy both conditions (divisible by 2 and 3), B must be in the intersection of {0, 2, 4, 6, 8} and {0, 3, 6, 9}. The common value is 6. Therefore, the only possible value for B is 6.
Q11: Find the units digit of the expression $25^{6251}+36^{528}+73^{54}$
A. 4
B. 0
C. 6
D. 5
Correct Answer: B
Solution: We need to find the units digit of the expression $25^{6251}+36^{528}+73^{54}$. Let's analyze each term separately. For $25^{6251}$: The units digit of $25$ raised to any positive integer power is always $5$. Therefore, the units digit of $25^{6251}$ is $5$. For $36^{528}$: The units digit of $36$ is $6$. Any positive integer power of a number ending in $6$ will also end in $6$. Therefore, the units digit of $36^{528}$ is $6$. For $73^{54}$: We look at the cycle of the units digit of powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
The cycle of units digits is $(3, 9, 7, 1)$, which has a length of 4. We find the remainder when $54$ is divided by $4$: $54 \div 4 = 13$ with a remainder of $2$. Therefore, the units digit of $73^{54}$ is the second digit in the cycle, which is $9$. Now, we add the units digits: $5 + 6 + 9 = 20$. The units digit of $20$ is $0$.
Q12: Find the units digit of the expression $55^{725}+73^{5810}+22^{853}$
A. 4
B. 0
C. 6
D. 5
Correct Answer: C
Solution: We need to find the units digit of $55^{725}+73^{5810}+22^{853}$. Let's analyze the units digit of each term separately. For $55^{725}$, the units digit is always 5, regardless of the exponent, because $5^1=5$, $5^2=25$, $5^3=125$, and so on. Therefore, the units digit of $55^{725}$ is 5. For $73^{5810}$, we look at the cyclicity of the units digit of powers of 3:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
The pattern of the units digit is (3, 9, 7, 1), which repeats every 4 terms. To find the units digit of $73^{5810}$, we find the remainder when 5810 is divided by 4:
$5810 \div 4 = 1452$ with a remainder of 2. Therefore, the units digit of $73^{5810}$ is the same as the units digit of $3^2$, which is 9. For $22^{853}$, we look at the cyclicity of the units digit of powers of 2:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$
The pattern of the units digit is (2, 4, 8, 6), which repeats every 4 terms. To find the units digit of $22^{853}$, we find the remainder when 853 is divided by 4:
$853 \div 4 = 213$ with a remainder of 1. Therefore, the units digit of $22^{853}$ is the same as the units digit of $2^1$, which is 2. Now, we add the units digits: $5 + 9 + 2 = 16$. The units digit of the sum is 6.
Q13: Find the units digit of the expression $11^{1}+12^{2}+13^{3}+14^{4}+15^{5}+16^{6}$
A. 1
B. 9
C. 7
D. 0
Correct Answer: B
Solution: We need to find the units digit of the expression $11^{1}+12^{2}+13^{3}+14^{4}+15^{5}+16^{6}$. We only need to consider the units digit of each term. For $11^1$, the units digit is 1. For $12^2$, the units digit is $2^2 = 4$. For $13^3$, the units digit is $3^3 = 27$, so the units digit is 7. For $14^4$, the units digit is the units digit of $4^4$. The pattern of units digits for powers of 4 is 4, 6, 4, 6,... Since the exponent is 4 (an even number), the units digit is 6. For $15^5$, the units digit is 5 (any positive integer power of a number ending in 5 will end in 5). For $16^6$, the units digit is 6 (any positive integer power of a number ending in 6 will end in 6). Therefore, the units digit of the sum is the units digit of $1 + 4 + 7 + 6 + 5 + 6 = 29$. The units digit of 29 is 9.
Q14: Find the number of zeroes at the end of 1090!
A. 270
B. 268
C. 269
D. 271
Correct Answer: A
Solution: The number of zeroes at the end of $1090!$ is determined by the number of times 10 is a factor in $1090!$. Since $10 = 2 \times 5$, and there will always be more factors of 2 than 5 in a factorial, we need to count the number of factors of 5 in $1090!$. We can use Legendre's formula for this:
The number of factors of 5 in $1090!$ is given by:
$$ \left\lfloor \frac{1090}{5} \right\rfloor + \left\lfloor \frac{1090}{5^2} \right\rfloor + \left\lfloor \frac{1090}{5^3} \right\rfloor + \left\lfloor \frac{1090}{5^4} \right\rfloor $$
$$ = \left\lfloor \frac{1090}{5} \right\rfloor + \left\lfloor \frac{1090}{25} \right\rfloor + \left\lfloor \frac{1090}{125} \right\rfloor + \left\lfloor \frac{1090}{625} \right\rfloor $$
$$ = 218 + 43 + 8 + 1 $$
$$ = 270 $$
Therefore, there are 270 factors of 5 in $1090!$. Since there are more factors of 2 than 5, there are 270 factors of 10, resulting in 270 zeroes at the end of $1090!$.
Q15: If 146! is divisible by $5^{n}$, then find the maximum value of n.
A. 34
B. 35
C. 36
D. 37
Correct Answer: B
Solution: We need to find the largest power of 5 that divides 146!. This is given by Legendre's formula:
$v_p(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor$
where $v_p(n!)$ is the exponent of the largest power of prime $p$ that divides $n!$. In our case, $n = 146$ and $p = 5$. $v_5(146!) = \left\lfloor \frac{146}{5} \right\rfloor + \left\lfloor \frac{146}{25} \right\rfloor + \left\lfloor \frac{146}{125} \right\rfloor + \left\lfloor \frac{146}{625} \right\rfloor + ...$
$v_5(146!) = \left\lfloor 29.2 \right\rfloor + \left\lfloor 5.84 \right\rfloor + \left\lfloor 1.168 \right\rfloor + \left\lfloor 0.2336 \right\rfloor + ...$
$v_5(146!) = 29 + 5 + 1 + 0 + ...$
$v_5(146!) = 35$
Therefore, the maximum value of $n$ is 35.
Q16: Find the number of divisors of 1420.
A. 14
B. 15
C. 13
D. 12
Correct Answer: D
Solution: To find the number of divisors of 1420, we first need to find the prime factorization of 1420. $1420 = 10 \times 142 = 2 \times 5 \times 2 \times 71 = 2^2 \times 5^1 \times 71^1$
The number of divisors is found by adding 1 to each exponent in the prime factorization and then multiplying the results. Number of divisors = $(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12$
Therefore, 1420 has 12 divisors.
Q17: Find the HCF and LCM of the polynomials $(x^{2}-5x+6)$ and $(x^{2}-7x+10)$
A. $(x-2), (x-2)(x-3)(x-5)$
B. $(x-2), (x-2)(x-3)$
C. $(x-3), (x-2)(x-3)(x-5)$
D. $(x-2), (x-2)(x-3)(x-5)^{2}$
Correct Answer: A
Solution: First, we factorize the given polynomials:
$x^{2} - 5x + 6 = (x-2)(x-3)$
$x^{2} - 7x + 10 = (x-2)(x-5)$
To find the HCF, we identify the common factors:
HCF = $(x-2)$
To find the LCM, we take the highest power of each factor present in the factorizations:
LCM = $(x-2)(x-3)(x-5)$
Therefore, the HCF is $(x-2)$ and the LCM is $(x-2)(x-3)(x-5)$.
Q18: Given two different prime numbers P and Q, find the number of divisors of the following: P.Q
A. 2
B. 4
C. 6
D. 8
Correct Answer: B
Solution: The number of divisors of an integer $n$ with prime factorization $n = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}$ is given by $(a_1 + 1)(a_2 + 1)...(a_k + 1)$. Since $P$ and $Q$ are different prime numbers, the prime factorization of $P \cdot Q$ is simply $P^1 \cdot Q^1$. Therefore, the number of divisors is $(1+1)(1+1) = 2 \cdot 2 = 4$. These divisors are 1, P, Q, and P*Q.
Q19: Given two different prime numbers P and Q, find the number of divisors of the following: $P^{2}Q$
A. 2
B. 4
C. 6
D. 8
Correct Answer: C
Solution: The number of divisors of an integer can be found by considering its prime factorization. Let $n$ be an integer with prime factorization $n = p_1^{a_1} p_2^{a_2} ... p_k^{a_k}$, where $p_i$ are distinct prime numbers and $a_i$ are positive integers. The number of divisors of $n$ is given by $(a_1 + 1)(a_2 + 1)...(a_k + 1)$. In this case, we are given the number $P^2Q$, where $P$ and $Q$ are distinct prime numbers. The prime factorization of $P^2Q$ is simply $P^2 \times Q^1$. Using the formula for the number of divisors, we have:
Number of divisors = $(2 + 1)(1 + 1) = 3 \times 2 = 6$
Therefore, the number $P^2Q$ has 6 divisors. These divisors are 1, P, $P^2$, Q, PQ, and $P^2Q$.
Q20: Given two different prime numbers P and Q, find the number of divisors of the following: $P^{3}Q^{2}$
A. 2
B. 4
C. 6
D. 12
Correct Answer: D
Solution: The number of divisors of a number $N = p_1^{a_1} p_2^{a_2} ... p_n^{a_n}$ (where $p_i$ are distinct prime numbers and $a_i$ are positive integers) is given by $(a_1 + 1)(a_2 + 1)...(a_n + 1)$. In this case, we have $N = P^3 Q^2$. Here, $P$ and $Q$ are distinct prime numbers. Thus, we can apply the formula for the number of divisors. The exponents are 3 and 2. Therefore, the number of divisors is $(3+1)(2+1) = 4 \times 3 = 12$.
Q21: There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the minimum total number of sections thus formed.
A. 24
B. 32
C. 16
D. 20
Correct Answer: C
Solution: This problem involves finding the minimum number of sections, which is related to finding the greatest common divisor (HCF) of the number of boys and girls. We need to find the HCF of 576 and 448. We can use the prime factorization method or the Euclidean algorithm. Let's use the prime factorization method:
$576 = 2^6 \times 3^2$
$448 = 2^6 \times 7$
The HCF is found by taking the lowest power of common prime factors:
HCF$(576, 448) = 2^6 = 64$
This means that each section can have a maximum of 64 students (either boys or girls). To find the total number of sections, we add the number of boy sections and the number of girl sections:
Number of boy sections $= \frac{576}{64} = 9$
Number of girl sections $= \frac{448}{64} = 7$
Total number of sections $= 9 + 7 = 16$
Therefore, the minimum total number of sections is 16.
Q22: A milkman has three different qualities of milk. 403 gallons of 1st quality, 465 gallons of 2nd quality and 496 gallons of 3rd quality. Find the least possible number of bottles of equal size in which different milk of different qualities can be filled without mixing.
A. 34
B. 46
C. 26
D. 44
Correct Answer: D
Solution: This problem involves finding the least number of bottles, implying we need to find the highest common factor (HCF) of the quantities of milk. We need to find the HCF of 403, 465, and 496. Let's use the prime factorization method. $403 = 13 \times 31$
$465 = 3 \times 5 \times 31$
$496 = 2^4 \times 31$
The common factor among these three numbers is 31. Therefore, the HCF(403, 465, 496) = 31. This means the largest possible size of a bottle is 31 gallons. To find the least number of bottles, we sum the number of bottles needed for each quality. Number of bottles for 1st quality: $\frac{403}{31} = 13$
Number of bottles for 2nd quality: $\frac{465}{31} = 15$
Number of bottles for 3rd quality: $\frac{496}{31} = 16$
Total number of bottles = $13 + 15 + 16 = 44$
Therefore, the least possible number of bottles is 44.
Q23: What is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1?
A. 9997
B. 9793
C. 9895
D. 9487
Correct Answer: B
Solution: Let the 4-digit number be $N$. The problem states that when $N$ is divided by 6, 9, 12, and 17, the remainder is 1 in each case. This means that $N-1$ is divisible by 6, 9, 12, and 17. Therefore, $N-1$ is a common multiple of 6, 9, 12, and 17. We need to find the least common multiple (LCM) of 6, 9, 12, and 17. First, find the prime factorization of each number:
$6 = 2 \times 3$
$9 = 3^2$
$12 = 2^2 \times 3$
$17 = 17$
The LCM is found by taking the highest power of each prime factor:
LCM$(6, 9, 12, 17) = 2^2 \times 3^2 \times 17 = 4 \times 9 \times 17 = 612$
So, $N-1$ is a multiple of 612. We can write $N-1 = 612k$ for some integer $k$. Then $N = 612k + 1$. Since $N$ is a 4-digit number, we have $1000 \le N \le 9999$. Substituting $N = 612k + 1$, we get $1000 \le 612k + 1 \le 9999$. Subtracting 1 from all parts of the inequality gives $999 \le 612k \le 9998$. Dividing by 612, we get $\frac{999}{612} \le k \le \frac{9998}{612}$. This simplifies to $1.63 \le k \le 16.33$. Since $k$ must be an integer, the possible values of $k$ are 2, 3, ..., 16. The largest value of $k$ is 16. Therefore, the largest 4-digit number is $N = 612(16) + 1 = 9792 + 1 = 9793$.
Q24: Find the least number that when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.
A. 364
B. 2254
C. 2964
D. 2884
Correct Answer: D
Solution: Let the number be $x$. The given condition is that when $x$ is divided by 16, 18, and 20, the remainder is 4 in each case. This means that $x - 4$ is divisible by 16, 18, and 20. Therefore, $x - 4$ is a common multiple of 16, 18, and 20. We need to find the least common multiple (LCM) of 16, 18, and 20. First, find the prime factorization of each number:
$16 = 2^4$
$18 = 2 \times 3^2$
$20 = 2^2 \times 5$
The LCM is found by taking the highest power of each prime factor present in the numbers:
$LCM(16, 18, 20) = 2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720$
So, $x - 4$ must be a multiple of 720. Thus, $x - 4 = 720k$ for some integer $k$. Then $x = 720k + 4$. We are also given that $x$ is completely divisible by 7. Therefore, $x \equiv 0 \pmod{7}$. Substituting $x = 720k + 4$, we get:
$720k + 4 \equiv 0 \pmod{7}$
$720k \equiv -4 \pmod{7}$
Since $720 = 7 \times 102 + 6$, we have $720 \equiv 6 \pmod{7}$. $6k \equiv -4 \pmod{7}$
$6k \equiv 3 \pmod{7}$
Multiplying both sides by 6 (the multiplicative inverse of 6 modulo 7 is 6, since $6 \times 6 = 36 \equiv 1 \pmod{7}$):
$36k \equiv 18 \pmod{7}$
$k \equiv 4 \pmod{7}$
So $k$ can be written as $k = 7n + 4$ for some integer $n$. Substituting this back into $x = 720k + 4$:
$x = 720(7n + 4) + 4 = 5040n + 2880 + 4 = 5040n + 2884$
For the least number, we take $n = 0$, so $x = 2884$.
Q25: Four bells ring at the intervals of 6, 8, 12 and 18 seconds. They start ringing together at $12^{\circ}$ clock. After how many seconds will they ring together again?
A. 72
B. 84
C. 60
D. 48
Correct Answer: A
Solution: This problem involves finding the LCM (Least Common Multiple) of the intervals at which the bells ring. The intervals are 6, 8, 12, and 18 seconds. We need to find the LCM of these numbers. First, find the prime factorization of each number:
$6 = 2 \times 3$
$8 = 2^3$
$12 = 2^2 \times 3$
$18 = 2 \times 3^2$
To find the LCM, we take the highest power of each prime factor present in the numbers:
LCM$(6, 8, 12, 18) = 2^3 \times 3^2 = 8 \times 9 = 72$
Therefore, the bells will ring together again after 72 seconds.
Q26: For Question 31, find how many times will they ring together during the next 12 minutes. (including the 12 minute mark)
A. 9
B. 10
C. 11
D. 12
Correct Answer: B
Solution: This question is incomplete. It requires information about the individual ringing intervals of at least two entities (e.g., bells, alarms). To solve, we need the time intervals at which each entity rings. Let's assume two entities ring at intervals $t_1$ and $t_2$ minutes, respectively. To find when they ring together, we need to find the least common multiple (LCM) of $t_1$ and $t_2$. Let's denote the LCM as $LCM(t_1, t_2)$. Then the number of times they ring together in 12 minutes is given by $\left\lfloor \frac{12}{LCM(t_1, t_2)} \right\rfloor + 1$, where $\lfloor x \rfloor$ represents the floor function (rounding down to the nearest integer). The '+1' accounts for the ringing at the 12-minute mark if it's a multiple of the LCM. Without specific values for $t_1$ and $t_2$, we cannot provide a numerical answer.
Q27: The units digit of the expression $125^{813}\times553^{3703}\times4532828$ is
A. 4
B. 2
C. 0
D. 5
Correct Answer: C
Solution: We need to find the unit digit of $125^{813}\times553^{3703}\times4532828$. The unit digit of $125^{813}$ is 5 because any positive integer power of a number ending in 5 will always end in 5. To find the unit digit of $553^{3703}$, we look at the cycle of unit digits of powers of 3:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
The cycle of unit digits is (3, 9, 7, 1), which has a length of 4. We find the remainder when 3703 is divided by 4:
$3703 = 4 \times 925 + 3$
So the unit digit of $553^{3703}$ is the same as the unit digit of $3^3$, which is 7. The unit digit of 4532828 is 8. Therefore, the unit digit of the expression is the unit digit of $5 \times 7 \times 8 = 280$. The unit digit is 0.
Q28: Which of the following is not a perfect square?
A. $1,00,856$
B. $3,25,137$
C. $x,000$ where x is a natural number
D. All of these
Correct Answer: D
Solution: A perfect square is a number that can be obtained by squaring an integer. To determine if a number is a perfect square, we can find its prime factorization. If all exponents in the prime factorization are even, the number is a perfect square. Let's analyze some examples (assuming the question provides a list of numbers to choose from, which is missing from the prompt). For example, consider the number 36. Its prime factorization is $2^2 \times 3^2$. Since all exponents are even, 36 is a perfect square ($6^2 = 36$). Conversely, consider the number 12. Its prime factorization is $2^2 \times 3^1$. Since the exponent of 3 is odd, 12 is not a perfect square. The question asks to identify the number that is *not* a perfect square; therefore, we need to examine the provided options and identify the one with at least one odd exponent in its prime factorization. Without the options, a specific answer can't be provided.
Q29: The LCM of 5, 8, 12, 20 will not be a multiple of
A. 3
B. 9
C. 8
D. 5
Correct Answer: B
Solution: First, find the prime factorization of each number:
$5 = 5^1$
$8 = 2^3$
$12 = 2^2 \times 3^1$
$20 = 2^2 \times 5^1$
To find the LCM, we take the highest power of each prime factor present in the numbers:
LCM($5, 8, 12, 20$) = $2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$
Now, let's check the options (which are not provided in the question, but we can deduce them). The LCM, 120, will be a multiple of its factors. Any number that is NOT a factor of 120 will not be a multiple of 120. For example, 7, 11, 13, etc., will not be multiples of 120. The question is incomplete without the options to choose from, but the process to find the correct answer is shown above.
Q30: Find the number of divisors of 720 (including 1 and 720).
A. 25
B. 28
C. 29
D. 30
Correct Answer: D
Solution: To find the number of divisors of 720, we first find the prime factorization of 720. $720 = 72 \times 10 = 8 \times 9 \times 2 \times 5 = 2^4 \times 3^2 \times 5^1$
The number of divisors is found by adding 1 to each exponent in the prime factorization and multiplying the results. Number of divisors = $(4+1)(2+1)(1+1) = 5 \times 3 \times 2 = 30$
Therefore, 720 has 30 divisors.
Q31: The LCM of $(16-x^{2})$ and $(x^{2}+x-6)$ is
A. $(x-3)(x+3)(4-x^{2})$
B. $4(4-x^{2})(x+3)$
C. $(4-x^{2})(x-3)$
D. None of these
Correct Answer: D
Solution: First, factorize the given expressions:
$16 - x^2 = (4-x)(4+x)$
$x^2 + x - 6 = (x+3)(x-2)$
Let $A = 16 - x^2 = (4-x)(4+x)$ and $B = x^2 + x - 6 = (x+3)(x-2)$. There is no common factor between A and B. Therefore, their HCF is 1. We know that for two numbers A and B, $HCF(A, B) \times LCM(A, B) = A \times B$. Since $HCF(A, B) = 1$, we have:
$LCM(A, B) = A \times B = (16 - x^2)(x^2 + x - 6) = (4-x)(4+x)(x+3)(x-2)$
The LCM is $(4-x)(4+x)(x+3)(x-2)$ or $-(x-4)(x+4)(x+3)(x-2)$. The question doesn't specify a numerical solution because the LCM depends on the value of $x$. If the question expected a numerical answer, more context would be needed (a specific value of $x$ would be required).
Q32: GCD of $x^{2}-4$ and $x^{2}+x-6$ is
A. $x+2$
B. $x-2$
C. $x^{2}-2$
D. $x^{2}+2$
Correct Answer: B
Solution: First, we factorize each polynomial:
$x^2 - 4 = (x-2)(x+2)$
$x^2 + x - 6 = (x-2)(x+3)$
The greatest common divisor (GCD) is the product of the common factors raised to the lowest power. The only common factor between $(x-2)(x+2)$ and $(x-2)(x+3)$ is $(x-2)$. Therefore, the GCD of $x^2 - 4$ and $x^2 + x - 6$ is $(x-2)$.
Q33: The number A is not divisible by 3. Which of the following will not be divisible by 3?
A. $9\times4$
B. $2\times A$
C. $18\times A$
D. $24\times4$
Correct Answer: B
Solution: The question tests the understanding of divisibility rules, specifically for the number 3. A number is divisible by 3 if the sum of its digits is divisible by 3. If A is not divisible by 3, then the sum of its digits is not divisible by 3. Let's consider the options (which are not provided in the prompt, but the logic remains the same). If any option involves a simple arithmetic operation on A (like adding a multiple of 3, multiplying by a number not divisible by 3 etc.), it will maintain the property that it's not divisible by 3. For example, if the options were $A+6$, $3A$, $A+1$, and $2A$, then only $3A$ would be certainly divisible by 3. $A+6$ may or may not be divisible by 3, depending on the value of A. $A+1$ and $2A$ will likely not be divisible by 3 if A itself is not divisible by 3. The question lacks options, hence we can only give a generic answer. The answer will depend on what is given as an option.
Q34: Find the remainder when the number $9^{100}$ is divided by 8.
A. 1
B. 2
C. 0
D. 4
Correct Answer: A
Solution: We want to find the remainder when $9^{100}$ is divided by 8. We can use the property of modular arithmetic. We have $9 \equiv 1 \pmod{8}$. Therefore, $9^{100} \equiv 1^{100} \pmod{8}$. Since $1^{100} = 1$, we have $9^{100} \equiv 1 \pmod{8}$. Thus, the remainder when $9^{100}$ is divided by 8 is 1.
Q35: Find the remainder of $2^{1000}$ when divided by 3.
A. 1
B. 2
C. 4
D. 6
Correct Answer: A
Solution: We want to find the remainder when $2^{1000}$ is divided by 3. We can use the property of modular arithmetic. We observe the pattern of powers of 2 modulo 3:
$2^1 \equiv 2 \pmod{3}$
$2^2 \equiv 4 \equiv 1 \pmod{3}$
$2^3 \equiv 8 \equiv 2 \pmod{3}$
$2^4 \equiv 16 \equiv 1 \pmod{3}$
The pattern of remainders is (2, 1, 2, 1, ...). The remainders repeat every 2 terms. Since $1000 = 2 \times 500$, the remainder when 1000 is divided by 2 is 0. This means the remainder will be the same as $2^2$, which is 1. Alternatively, we can write $1000 = 2(500)$, so
$$2^{1000} = (2^2)^{500} = 4^{500} $$
Since $4 \equiv 1 \pmod 3$, we have
$$ 4^{500} \equiv 1^{500} \equiv 1 \pmod 3 $$
Thus, the remainder when $2^{1000}$ is divided by 3 is 1.
Q36: Find the number of zeroes at the end of 50!
A. 13
B. 11
C. 5
D. 12
Correct Answer: D
Solution: The number of zeroes at the end of $50!$ is determined by the number of times 10 is a factor in $50!$. Since $10 = 2 \times 5$, we need to count the number of factors of 5 in $50!$ because there will always be more factors of 2 than 5. The number of factors of 5 in $50!$ is given by:
$$ \left\lfloor \frac{50}{5} \right\rfloor + \left\lfloor \frac{50}{5^2} \right\rfloor = 10 + 2 = 12 $$
where $\lfloor x \rfloor$ denotes the floor function (the greatest integer less than or equal to $x$). Therefore, there are 12 factors of 5 in $50!$. Since there are more factors of 2 than 5, there will be 12 factors of 10, resulting in 12 zeroes at the end of $50!$.
Q37: Which of the following can be a number divisible by 24?
A. $4,32,15,604$
B. $25,61,284$
C. $13,62,480$
D. All of these
Correct Answer: C
Solution: A number is divisible by 24 if it is divisible by both 8 and 3 (since $24 = 8 \times 3$). Divisibility rule for 8 states that the last three digits must be divisible by 8. Divisibility rule for 3 states that the sum of the digits must be divisible by 3. The question doesn't provide a list of numbers to choose from, making it impossible to provide a definitive answer without further information. To solve this type of question given options, one would check each option for divisibility by 8 and 3.
Q38: For a number to be divisible by 88, it should be
A. Divisible by 22 and 8
B. Divisible by 11 and 8
C. Divisible by 11 and thrice by 2
D. All of these
Correct Answer: D
Solution: A number is divisible by 88 if it is divisible by both 8 and 11, since $88 = 8 \times 11$. Divisibility rule for 8 states that the last three digits of a number must be divisible by 8. Divisibility rule for 11 states that the alternating sum of digits must be divisible by 11. Therefore, for a number to be divisible by 88, it must satisfy both conditions.
Q39: The product of three consecutive natural numbers, the first of which is an even number, is always divisible by
A. 12
B. 24
C. 6
D. All of these
Correct Answer: D
Solution: Let the three consecutive natural numbers be $2n$, $2n+1$, and $2n+2$, where $n$ is a natural number. Their product is $P = 2n(2n+1)(2n+2)$. We can rewrite this as $P = 2n(2n+1)2(n+1) = 4n(n+1)(2n+1)$. Notice that $n(n+1)$ is always the product of two consecutive integers, which means it is always even. Therefore, $n(n+1)$ can be written as $2k$ for some integer $k$. Substituting this back into the expression for $P$, we have $P = 4(2k)(2n+1) = 8k(2n+1)$. Since $P$ is always a multiple of 8, it is always divisible by 8. Let's test this with an example: If we consider 2, 3, and 4, the product is $2 \times 3 \times 4 = 24$, which is divisible by 8. If we consider 4, 5, and 6, the product is $4 \times 5 \times 6 = 120$, which is also divisible by 8. Therefore, the product of three consecutive natural numbers, where the first is even, is always divisible by 8.
Q40: Question
A. Option A
B. Option B
C. Option C
D. Option D
E. Option E
Correct Answer: A
Solution: Data sufficiency questions test your ability to determine if given information is sufficient to answer a question, without necessarily finding the final answer. This question's subtopic is "Application across all Quant & Reasoning topics" because data sufficiency problems can be framed using various mathematical concepts (algebra, geometry, etc.) or logical reasoning scenarios. The core skill tested remains the same: evaluating the sufficiency of provided data. Solving the problem requires analyzing whether the given statements individually or together provide enough information to definitively answer the question, regardless of the specific mathematical or logical principles involved. The process involves assessing the information's completeness and avoiding unnecessary calculations.
Q41: Some birds settled on the branches of a tree. First, they sat one to a branch and there was one bird too many. Next they sat two to a branch and there was one branch too many. How many branches were there?
A. 3
B. 4
C. 5
D. 6
Correct Answer: A
Solution: Let $b$ be the number of branches and $n$ be the number of birds. When the birds sat one to a branch, there was one bird too many. This can be written as:
$n = b + 1$ (Equation 1)
When they sat two to a branch, there was one branch too many. This means there are $b-1$ pairs of birds. Therefore:
$n = 2(b - 1)$ (Equation 2)
We can now set the two equations equal to each other:
$b + 1 = 2(b - 1)$
$b + 1 = 2b - 2$
$2b - b = 1 + 2$
$b = 3$
Therefore, there are 3 branches. To verify, let's find the number of birds:
$n = b + 1 = 3 + 1 = 4$
If there are 4 birds and 3 branches, with 2 birds per branch, there would be one branch left over. This confirms the solution.
Q42: If $A=(\frac{-3}{4})^{3}$, $B=(\frac{-2}{5})^{2}$, $C=(0.3)^{2}$, $D=(-1.2)^{2}$ then
A. $A>B>C>D$
B. $D>A>B>C$
C. $D>B>C>A$
D. $D>C>A>B$
Correct Answer: D
Solution: We need to calculate the values of A, B, C, and D. $A = (\frac{-3}{4})^3 = \frac{(-3)^3}{4^3} = \frac{-27}{64}$
$B = (\frac{-2}{5})^2 = \frac{(-2)^2}{5^2} = \frac{4}{25}$
$C = (0.3)^2 = (3/10)^2 = \frac{9}{100}$
$D = (-1.2)^2 = (-6/5)^2 = \frac{36}{25}$
The question is incomplete. It requires a comparison or relationship between A, B, C, and D to be established, which is missing from the provided text. Therefore, a definitive "Correct Answer" cannot be provided without the complete question.
Q43: Which of these is greater? (a) $54^{4}$ or $21^{12}$ (b) $(0.4)^{2}$ or $(0.8)^{2}
A. The question does not provide distinct options for (a) and (b).
B. The question does not provide distinct options for (a) and (b).
C. The question does not provide distinct options for (a) and (b).
D. The question does not provide distinct options for (a) and (b).
Correct Answer: B
Solution: (a) To compare $54^4$ and $21^{12}$, we can try to express them with a common base or exponent. It's difficult to find a common base. Let's try to take the 12th root of both numbers to make the exponents equal. We have $(54^4)^{1/12}$ and $(21^{12})^{1/12}$. This simplifies to $54^{1/3}$ and $21$. Now we need to compare $54^{1/3}$ and $21$. $54^{1/3} = \sqrt[3]{54} \approx 3.78$. Since $3.78 < 21$, $54^4 < 21^{12}$. (b) To compare $(0.4)^2$ and $(0.8)^2$, we can calculate the values directly:
$(0.4)^2 = 0.16$
$(0.8)^2 = 0.64$
Since $0.64 > 0.16$, $(0.8)^2 > (0.4)^2$.
Q44: Is it possible for a common fraction whose numerator is less than the denominator to be equal to a fraction whose numerator is greater than the denominator?
A. Yes
B. No
Correct Answer: 4,
Solution: The question explores the relationship between proper and improper fractions. A common fraction with a numerator less than the denominator is a proper fraction ($ \frac{a}{b} $, where $a < b$). An improper fraction has a numerator greater than or equal to the denominator ($ \frac{c}{d} $, where $c \ge d$). The question asks if a proper fraction can be equal to an improper fraction. Consider the example: $ \frac{1}{2} $ is a proper fraction. We can express this fraction as an improper fraction by multiplying the numerator and denominator by the same number (this doesn't change the value of the fraction). For example, multiplying by 2, we get $ \frac{2}{4} $. While this is still a proper fraction, if we multiply by 3, we get $ \frac{3}{6} $ which is still proper, but multiplying by 4 gives $ \frac{4}{8} $ which is an improper fraction where we can write it as a mixed fraction of $1$. If we take any proper fraction $\frac{a}{b}$, we can always find a number $n > \frac{b}{a}$ such that $an \geq bn$, hence making it an improper fraction. The answer will always be yes.
Q45: What digits should be put in place of c in 38c to make it divisible by (1) 2, (2) 3, (3) 4, (4) 5, (5) 6, (6) 9, (7) 10
A. The question asks for a list of digits, not a single option.
B. The question asks for a list of digits, not a single option.
C. The question asks for a list of digits, not a single option.
D. The question asks for a list of digits, not a single option.
Correct Answer: LCM
Solution: The number is 38c. We need to find the values of c that satisfy divisibility rules for different numbers. (1) Divisible by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). Therefore, $c \in \{0, 2, 4, 6, 8\}$. (2) Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. $3 + 8 + c \equiv 0 \pmod{3}$, which simplifies to $11 + c \equiv 0 \pmod{3}$. This means $c \equiv 1 \pmod{3}$. Thus, $c \in \{1, 4, 7\}$. (3) Divisible by 4: A number is divisible by 4 if the last two digits are divisible by 4. Therefore, 8c must be divisible by 4. This implies $c \in \{0, 4, 8\}$. (4) Divisible by 5: A number is divisible by 5 if its last digit is 0 or 5. Therefore, $c \in \{0, 5\}$. (5) Divisible by 6: A number is divisible by 6 if it is divisible by both 2 and 3. From (1) and (2), we need $c \in \{4\}$. (6) Divisible by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. $3 + 8 + c \equiv 0 \pmod{9}$, which simplifies to $11 + c \equiv 0 \pmod{9}$. This means $c \equiv 7 \pmod{9}$. Thus $c = 7$. (7) Divisible by 10: A number is divisible by 10 if its last digit is 0. Therefore, $c = 0$. We need to find a value of $c$ that satisfies all the conditions. Analyzing the possible values of c from each condition, the only value that satisfies conditions (1), (2), (3), (4), (5), (6), and (7) is when $c=0$. For example, if we check for $c = 4$, it satisfies conditions (1), (2), (3), (5), but not (4), (6), and (7). There is no single value of c that satisfies all conditions simultaneously. However, if we consider each condition independently, we have the following solutions: (1) 0,2,4,6,8; (2) 1,4,7; (3) 0,4,8; (4) 0,5; (5) 4; (6) 7; (7) 0.
Q46: Find the LCM and HCF of the following numbers: (54, 81, 135 and 189), (156, 195) and (1950, 5670 and 3900)
A. The question asks for multiple LCM and HCF pairs, not a single option.
B. The question asks for multiple LCM and HCF pairs, not a single option.
C. The question asks for multiple LCM and HCF pairs, not a single option.
D. The question asks for multiple LCM and HCF pairs, not a single option.
Q47: Amitesh buys a pen, a pencil and an eraser for 41. If the least cost of any of the three items is 12 and it is known that a pen costs less than a pencil and an eraser costs more than a pencil, answer the following questions: What is the cost of the pen?
A. 12
B. 13
C. 14
D. 15
Correct Answer: C
Solution: This question tests data sufficiency within a word problem. It requires understanding the constraints provided (minimum cost, relative costs of items) and determining if sufficient information exists to solve for the pen's cost. Let's represent the costs as follows:
Pen = $p$
Pencil = $c$
Eraser = $e$
We are given:
$p + c + e = 41$
$p \ge 12$
$c \ge 12$
$e \ge 12$
$p < c$
$c < e$
The question asks for the value of $p$. We have a system of inequalities and one equation. Let's analyze the constraints:
Since $p < c < e$, and the total cost is 41, we can deduce that the minimum possible cost of each item is greater than 12. If we take the lowest possible value for $p$, which is slightly greater than 12, this restricts the values of $c$ and $e$, due to the inequality constraints. We don't have enough information to solve exactly for $p$, only to find a range of possible values. For example, if $p = 13$, then $c$ can be 14 and $e$ can be 14, giving us 41. However, different values of $p$ will result in different solutions for $c$ and $e$. We don't have sufficient information.
Q48: If it is known that the eraser's cost is not divisible by 4, the cost of the pencil could be:
A. 12
B. 13
C. 14
D. 15
Correct Answer: C
Solution: The question assesses whether the given statement ("the eraser's cost is not divisible by 4") is sufficient to determine the cost of the pencil. The question is incomplete; it lacks the crucial information regarding the relationship between the eraser's cost and the pencil's cost. Without knowing this relationship (e.g., the total cost of both, the difference in cost, a ratio, etc.), it's impossible to determine the pencil's cost even with the given information about the eraser. Therefore, the statement is insufficient. The question needs additional information to solve it.
Q49: A naughty boy Amrit watches an innings of Sachin Tendulkar and acts according to the number of runs he sees Sachin scoring... At the end of the above innings, how many more oranges were there compared to mangoes inside the basket? (The Basket was empty initially).
A. 4
B. 5
C. 6
D. 7
Correct Answer: C
Solution: The question stem is insufficient to determine a solution. It sets up a scenario where Amrit's actions depend on Sachin's score, but provides no information linking Sachin's score to the number of oranges or mangoes. The question is unanswerable even with additional information because it introduces an arbitrary, unquantifiable action ("acts according to the number of runs"). There's no mathematical relationship established, making it impossible to determine a numerical solution. The problem is fundamentally flawed for data sufficiency, as no amount of supplementary information could fix the inherent lack of connection between the variables.
Q50: For Question 72, if it is known that he has left 10 questions unanswered, the number of correct answers are:
A. 45
B. 48
C. 54
D. Cannot be determined
Correct Answer: B
Solution: The question is incomplete. It lacks the crucial information needed to determine the number of correct answers. We only know that 10 questions were left unanswered. To find the number of correct answers, we need additional information such as the total number of questions or the number of incorrect answers. Therefore, the given statement is insufficient to answer the question. The question cannot be solved with only the provided statement.
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