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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

\(x\) and \(y\) are 2 different digits. If the sum of the two digit numbers formed by using both the digits is a perfect square, then find \(x+y\).

x and y are 2 different digits. If the sum of the two-digit numbers formed by using both the digits is a perfect square, then find x+y. (a) 10 (b) 11 (c) 12 (d) 13

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 3:24 pm

    Let's analyze the given information to solve the problem: Given that \(x\) and \(y\) are two different digits forming two-digit numbers \(xy\) and \(yx\), the sum of these two numbers can be represented as: \[ 10x + y + 10y + x = 11x + 11y \] This sum is stated to be a perfect square. Given that \(xRead more

    Let’s analyze the given information to solve the problem:

    Given that \(x\) and \(y\) are two different digits forming two-digit numbers \(xy\) and \(yx\), the sum of these two numbers can be represented as:

    \[
    10x + y + 10y + x = 11x + 11y
    \]

    This sum is stated to be a perfect square. Given that \(x\) and \(y\) are digits, the sum \(11(x + y)\) suggests that \(x + y\) itself must be a number that, when multiplied by 11, results in a perfect square.

    Looking at the options and considering that \(x + y\) must be small enough to fit the constraints of single digits (1 through 9), let’s analyze the options directly:

    – (a) 10: \(11 \times 10 = 110\), not a perfect square.
    – (b) 11: \(11 \times 11 = 121\), which is a perfect square (\(11^2\)).
    – (c) 12: \(11 \times 12 = 132\), not a perfect square.
    – (d) 13: \(11 \times 13 = 143\), not a perfect square.

    The only option where \(11(x + y)\) forms a perfect square is when \(x + y = 11\), making \(11 \times 11 = 121\), which is indeed a perfect square (\(11^2\)).

    Therefore, the correct answer is (b) **11**.

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

A number lies between 300 and 400 . If the number is added to the number formed by reversing the digits, the sum is 888 and if the unit’s digit and the ten’s digit change places, the new number exceeds the original number by 9 . Find the number.

A number lies between 300 and 400 . If the number is added to the number formed by reversing the digits, the sum is 888 and if the unit’s digit and the ten’s digit change places, the new number exceeds ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 3:23 pm

    Given the constraint that the sum of the number and its reversed form is 888 and focusing on the clue that the units digit in the sum scenario must add up to 8, we look directly at the options provided: - (a) 339: The sum of 3 and 9 does not lead to an end digit of 8 in the sum. - (b) 341: The sum oRead more

    Given the constraint that the sum of the number and its reversed form is 888 and focusing on the clue that the units digit in the sum scenario must add up to 8, we look directly at the options provided:

    – (a) 339: The sum of 3 and 9 does not lead to an end digit of 8 in the sum.
    – (b) 341: The sum of 1 and 4 does not lead to an end digit of 8 in the sum.
    – (c) 378: The sum of 8 and 7 does not directly address the end digit sum condition.
    – (d) 345: When 345 is added to its reverse 543, the unit digits 5 and 3 indeed add up to 8, meeting the immediate condition.

    The specific insight about the unit’s digits adding up to 8 being satisfied only by option (d) “345” (since \(3+5=8\)) simplifies the approach significantly. However, to validate this option fully in the context of the entire problem:

    – If the original number is 345 and its reverse is 543, their sum is indeed 888 (\(345 + 543 = 888\)), which satisfies one of the problem’s conditions.
    – The second condition mentioned is that swapping the unit’s and ten’s digit of the original number results in a number that is 9 more than the original. Swapping the units and tens digit of 345 gives 354, which is indeed 9 more than 345 (\(354 – 345 = 9\)).

    Therefore, considering both conditions and the insight provided about the sum leading to the last digits adding up to 8, the correct number is indeed option (d) **345**. This choice fulfills both specified conditions of the problem: the sum with its reversed form equals 888, and swapping the tens and units digits results in a number that is 9 more than the original.

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

The least number by which we multiply to the 11760 , so that we can get a perfect square number:

The least number by which we multiply to the 11760 , so that we can get a perfect square number:

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 2:19 pm

    To find the least number by which 11760 must be multiplied to become a perfect square, we first factorize 11760 into its prime factors. This will help us determine which primes and in what quantity are needed to make all exponents even, as a perfect square has even exponents in its prime factorizatiRead more

    To find the least number by which 11760 must be multiplied to become a perfect square, we first factorize 11760 into its prime factors. This will help us determine which primes and in what quantity are needed to make all exponents even, as a perfect square has even exponents in its prime factorization.

    Prime Factorization of 11760

    \[11760 = 2^4 \times 3 \times 5^1 \times 7^2\]

    A perfect square requires all exponents in its prime factorization to be even. Here, the prime factor \(3\) and \(5\) has an exponent of \(1\) (which is odd).

    Finding the Least Number to Multiply

    To make the exponent of \(3\) and \(5\) even, we need to multiply 11760 by another \(3\) and \(5\) = \(15\) .

    Conclusion

    Therefore, the least number by which 11760 must be multiplied to become a perfect square is \(\boldsymbol{15}\).

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

If \(p \times q=p+q+\frac{p}{q}\), then value of \(8 \times 2=\) ?

If p * q = p + q + p/q, then the value of 8 * 2 = ?

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 2:16 pm

    Given the operation defined by \(p \times q = p + q + \frac{p}{q}\), let's calculate the value of \(8 \times 2\) using this operation. Calculating \(8 \times 2\) Using the Given Operation Substitute \(p = 8\) and \(q = 2\) into the given formula: \[ 8 \times 2 = 8 + 2 + \frac{8}{2} \] Simplify the eRead more

    Given the operation defined by \(p \times q = p + q + \frac{p}{q}\), let’s calculate the value of \(8 \times 2\) using this operation.

    Calculating \(8 \times 2\) Using the Given Operation

    Substitute \(p = 8\) and \(q = 2\) into the given formula:

    \[
    8 \times 2 = 8 + 2 + \frac{8}{2}
    \]

    Simplify the expression:

    \[
    8 \times 2 = 10 + 4 = 14
    \]

    Therefore, according to the given operation, \(8 \times 2 = \boldsymbol{14}\).

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

If \(a\) and \(b\) are positive ingegers, such that \(a^{b}=125\), then \((a-b)^{a+b-4}=\) ?

If a and b are positive integers, such that a^b = 125, then (a – b)^(a + b – 4) = ?

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 2:13 pm

    Given \(a^b = 125\) and knowing \(a\) and \(b\) are positive integers, let's first identify \(a\) and \(b\). Finding \(a\) and \(b\) The number 125 is a perfect cube, specifically \(5^3\). Therefore, \(a = 5\) and \(b = 3\). Calculating \((a-b)^{a+b-4}\) Now, substitute \(a = 5\) and \(b = 3\) intoRead more

    Given \(a^b = 125\) and knowing \(a\) and \(b\) are positive integers, let’s first identify \(a\) and \(b\).

    Finding \(a\) and \(b\)

    The number 125 is a perfect cube, specifically \(5^3\). Therefore, \(a = 5\) and \(b = 3\).

    Calculating \((a-b)^{a+b-4}\)

    Now, substitute \(a = 5\) and \(b = 3\) into the expression:

    \[
    (a-b)^{a+b-4} = (5-3)^{5+3-4}
    \]

    This simplifies to:

    \[
    (2)^{4} = 16
    \]

    Therefore, the value of \((a-b)^{a+b-4}\) is \(\boldsymbol{16}\).

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

The no. plate of a bus had peculiarity. The bus number was a perfect square. It was also a perfect square when the plate was turned upside down. The bus company had only five hundred buses numbered from 1 to 500 . What was the number?

The no. plate of a bus had peculiarity. The bus number was a perfect square. It was also a perfect square when the plate was turned upside down. The bus company had only five hundred buses numbered from 1 to ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 1:37 pm

    Identification of the Unique Bus Number The task is to identify a bus number within the range of 1 to 500 that is not only a perfect square but also remains a perfect square when the digits are viewed upside down. Following the clarified criteria: Criteria for the Bus Number The number must be a perRead more

    Identification of the Unique Bus Number

    The task is to identify a bus number within the range of 1 to 500 that is not only a perfect square but also remains a perfect square when the digits are viewed upside down. Following the clarified criteria:

    Criteria for the Bus Number

    • The number must be a perfect square.
    • When turned upside down, the number must still represent a perfect square.
    • The number must include only the digits 0, 1, 6, 8, and 9, as these are the digits that can represent other numbers or themselves when flipped.

    Analysis

    We consider the range of perfect squares from \(1^2\) to \(22^2\) as these are the perfect squares within the 1 to 500 range:

    \[
    1, 4, 9, \ldots, 484\ (22^2)
    \]

    Among these, the number that meets the specific requirement of being a perfect square that, when turned upside down, also becomes a perfect square, is 169:

    – Original Number: \(169 = 13^2\), a perfect square.
    – Upside Down: When 169 is flipped upside down, it becomes 961. Notably, 6 and 9 flip, while 1 remains the same, making the new figure, 961, which is \(31^2\), another perfect square.

    Conclusion

    Therefore, the bus number with the described peculiarity is 169. This number is unique in that it satisfies the condition of being a perfect square and also transforms into another perfect square (961) when viewed upside down.

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

The value of \(\left[\frac{1}{\sqrt{9}-\sqrt{8}}\right]-\left[\frac{1}{\sqrt{8}-\sqrt{7}}\right]+\left[\frac{1}{\sqrt{7}-\sqrt{6}}\right]\) \(-\left[\frac{1}{\sqrt{6}-\sqrt{5}}\right]+\left[\frac{1}{\sqrt{5}-\sqrt{4}}\right]\) is:

The value of [1/(sqrt(9)-sqrt(8))] – [1/(sqrt(8)-sqrt(7))] + [1/(sqrt(7)-sqrt(6))] – [1/(sqrt(6)-sqrt(5))] + [1/(sqrt(5)-sqrt(4))]:

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 12:35 pm

    To simplify the given expression, we can use the conjugate of each denominator to rationalize it. The conjugate of a binomial \(\sqrt{a} - \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\). Multiplying both the numerator and denominator by the conjugate, we get: \[ \left[\frac{1}{\sqrt{9}-\sqrt{8}}\right] - \leRead more

    To simplify the given expression, we can use the conjugate of each denominator to rationalize it. The conjugate of a binomial \(\sqrt{a} – \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\). Multiplying both the numerator and denominator by the conjugate, we get:

    \[
    \left[\frac{1}{\sqrt{9}-\sqrt{8}}\right] – \left[\frac{1}{\sqrt{8}-\sqrt{7}}\right] + \left[\frac{1}{\sqrt{7}-\sqrt{6}}\right] – \left[\frac{1}{\sqrt{6}-\sqrt{5}}\right] + \left[\frac{1}{\sqrt{5}-\sqrt{4}}\right]
    \]

    \[
    = \left[\frac{\sqrt{9} + \sqrt{8}}{(\sqrt{9} – \sqrt{8})(\sqrt{9} + \sqrt{8})}\right] – \left[\frac{\sqrt{8} + \sqrt{7}}{(\sqrt{8} – \sqrt{7})(\sqrt{8} + \sqrt{7})}\right] + \left[\frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7} – \sqrt{6})(\sqrt{7} + \sqrt{6})}\right]
    \]

    \[
    – \left[\frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6} – \sqrt{5})(\sqrt{6} + \sqrt{5})}\right] + \left[\frac{\sqrt{5} + \sqrt{4}}{(\sqrt{5} – \sqrt{4})(\sqrt{5} + \sqrt{4})}\right]
    \]

    \[
    = \left[\frac{\sqrt{9} + \sqrt{8}}{9 – 8}\right] – \left[\frac{\sqrt{8} + \sqrt{7}}{8 – 7}\right] + \left[\frac{\sqrt{7} + \sqrt{6}}{7 – 6}\right] – \left[\frac{\sqrt{6} + \sqrt{5}}{6 – 5}\right] + \left[\frac{\sqrt{5} + \sqrt{4}}{5 – 4}\right]
    \]

    \[
    = (\sqrt{9} + \sqrt{8}) – (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) – (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + \sqrt{4})
    \]

    Now, notice that the terms cancel out in pairs:

    \[
    = \sqrt{9} + \sqrt{4} = 3 + 2 = 5
    \]

    Therefore, the value of the given expression is 5.

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

In a group of buffaloes and ducks, the number of legs are 24 more than twice the number of heads. What is the number of buffaloes in the group?

In a group of buffaloes and ducks, the number of legs are 24 more than twice the number of heads. What is the number of buffaloes in the group?

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 12:23 pm

    Let's denote the number of buffaloes as \(b\) and the number of ducks as \(d\). Since each buffalo has 4 legs and each duck has 2 legs, the total number of legs in the group is \(4b + 2d\). The total number of heads, which is also the total number of animals, is \(b + d\). According to the problem,Read more

    Let’s denote the number of buffaloes as \(b\) and the number of ducks as \(d\).

    Since each buffalo has 4 legs and each duck has 2 legs, the total number of legs in the group is \(4b + 2d\). The total number of heads, which is also the total number of animals, is \(b + d\).

    According to the problem, the number of legs is 24 more than twice the number of heads. Therefore, we can write the equation:

    \[4b + 2d = 2(b + d) + 24\]

    Simplifying this equation:

    \[4b + 2d = 2b + 2d + 24\]
    \[2b = 24\]
    \[b = 12\]

    So, there are 12 buffaloes in the group.

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N.K. Sharma
N.K. Sharma
Asked: April 5, 2024In: SSC Maths

Which of the following represents the numeral for 2949 (a) MMMIXL (b) MMXMIX (c) MMCMIL (d) MMCMXLIX

Which of the following represents the numeral for 2949 (a) MMMIXL (b) MMXMIX (c) MMCMIL (d) MMCMXLIX

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 12:18 pm

    To represent the number 2949 in Roman numerals, we can break it down into its components: - 2000 = MM - 900 = CM (1000 - 100) - 40 = XL (50 - 10) - 9 = IX (10 - 1) Putting these components together, we get MMCMXLIX. Therefore, the numeral for 2949 is MMCMXLIX, which corresponds to option (d) MMCMXLIRead more

    To represent the number 2949 in Roman numerals, we can break it down into its components:

    – 2000 = MM
    – 900 = CM (1000 – 100)
    – 40 = XL (50 – 10)
    – 9 = IX (10 – 1)

    Putting these components together, we get MMCMXLIX.

    Therefore, the numeral for 2949 is MMCMXLIX, which corresponds to option (d) MMCMXLIX.

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N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

\[ \text { if } 7 \sin ^2 \theta+3 \cos ^2 \theta=4,\left(0^{\circ}<\theta<90^{\circ}\right) \text {. then value of } \theta \text { is } \]

If `7 sin^2(theta) + 3 cos^2(theta) = 4, (0 < theta < 90)`, then the value of `theta` is.

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 5:52 pm

    Given the equation: \[7 \sin^2 \theta + 3 \cos^2 \theta = 4\] Since \(\cos^2 \theta = 1 - \sin^2 \theta\), we can substitute this into the equation: \[7 \sin^2 \theta + 3 (1 - \sin^2 \theta) = 4\] Expanding and simplifying: \[7 \sin^2 \theta + 3 - 3 \sin^2 \theta = 4\] \[4 \sin^2 \theta = 1\] \[\sinRead more

    Given the equation:
    \[7 \sin^2 \theta + 3 \cos^2 \theta = 4\]

    Since \(\cos^2 \theta = 1 – \sin^2 \theta\), we can substitute this into the equation:
    \[7 \sin^2 \theta + 3 (1 – \sin^2 \theta) = 4\]

    Expanding and simplifying:
    \[7 \sin^2 \theta + 3 – 3 \sin^2 \theta = 4\]
    \[4 \sin^2 \theta = 1\]
    \[\sin^2 \theta = \frac{1}{4}\]
    \[\sin \theta = \pm \frac{1}{2}\]

    Since \(0^\circ < \theta < 90^\circ\), we know that \(\sin \theta\) is positive in this interval. Therefore, we can disregard the negative solution, leaving us with:
    \[\sin \theta = \frac{1}{2}\]

    The value of \(\theta\) that satisfies this equation in the given interval is:
    \[\theta = 30^\circ \text{ or } \frac{\pi}{6} \text{ radians}\]

    So, the value of \(\theta\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.

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