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Home/SSC Maths Practice Questions with Solution/Page 5

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Bhulu Aich
Bhulu AichExclusive Author
Asked: April 5, 2024In: SSC Maths

If \(x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) and \(y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\), then the value of \(x^{3}+y^{3}\) is: (a) 950 (b) 730 (c) 650 (d) 970

If x=(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2)) and y=(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2)), then the value of x^3 + y^3 is: (a) 950 (b) 730 (c) 650 (d) 970

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 7:16 pm

    Evaluation of \(x^3 + y^3\) Given \(x\) and \(y\) For the given values of \(x\) and \(y\), defined as: \[x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\] \[y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\] We aim to find the value of \(x^3 + y^3\). Step 1: Simplify \(x\) and \(y\) First, rationalize \(Read more

    Evaluation of \(x^3 + y^3\) Given \(x\) and \(y\)

    For the given values of \(x\) and \(y\), defined as:

    \[x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\]

    \[y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\]

    We aim to find the value of \(x^3 + y^3\).

    Step 1: Simplify \(x\) and \(y\)

    First, rationalize \(x\) by multiplying the numerator and denominator by the conjugate of the denominator:

    \[
    x=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{3 – 2\sqrt{6} + 2}{1} = 5 – 2\sqrt{6}
    \]

    Using a similar process for \(y\), we rationalize by multiplying by the conjugate, yielding:

    \[
    y=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = 5 + 2\sqrt{6}
    \]

    Step 2: Calculate \(x+y\)

    Adding \(x\) and \(y\) gives us:

    \[
    x+y = (5 – 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10
    \]

    Step 3: Calculate \(xy\)

    Multiplying \(x\) and \(y\) provides:

    \[
    xy = (5 – 2\sqrt{6})(5 + 2\sqrt{6}) = 25 – (2\sqrt{6})^2 = 25 – 24 = 1
    \]

    Step 4: Derive \(x^3 + y^3\)

    Using the identity \(x^3 + y^3 = (x + y)^3 – 3xy(x + y)\) and substituting the calculated values:

    \[
    x^3 + y^3 = 10^3 – 3 \cdot 1 \cdot 10 = 1000 – 30 = 970
    \]

    Conclusion

    The value of \(x^3 + y^3\) is 970, corresponding to:

    (d) 970

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

If \(a^{2}+b^{2}=5 a b\), then the value of \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\) is : (a) 32 (b) 16 (c) 23 (d) -23

If a^2 + b^2 = 5ab, then the value of (a^2/b^2 + b^2/a^2) is : (a) 32 (b) 16 (c) 23 (d) -23

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 7:09 pm

    Calculation of the Given Expression Given the equation \(a^{2}+b^{2}=5ab\), we are tasked with determining the value of the expression \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\). Step 1: Simplify the Given Relation Starting with the given equation, we divide both sides by \(ab\) to simRead more

    Calculation of the Given Expression

    Given the equation \(a^{2}+b^{2}=5ab\), we are tasked with determining the value of the expression \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\).

    Step 1: Simplify the Given Relation

    Starting with the given equation, we divide both sides by \(ab\) to simplify:

    \[
    \frac{a^2 + b^2}{ab} = 5
    \]

    This leads to:

    \[
    \frac{a}{b} + \frac{b}{a} = 5
    \]

    Step 2: Square Both Sides

    To find the value of \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\), we square both sides of the simplified equation:

    \[
    \left(\frac{a}{b} + \frac{b}{a}\right)^2 = 5^2
    \]

    This yields:

    \[
    \frac{a^2}{b^2} + 2\left(\frac{a}{b}\cdot\frac{b}{a}\right) + \frac{b^2}{a^2} = 25
    \]

    Given that \(\frac{a}{b}\cdot\frac{b}{a} = 1\), we simplify further:

    \[
    \frac{a^2}{b^2} + \frac{b^2}{a^2} + 2 = 25
    \]

    Step 3: Isolate the Target Expression

    Subtracting 2 from both sides to isolate the expression gives us:

    \[
    \frac{a^2}{b^2} + \frac{b^2}{a^2} = 25 – 2 = 23
    \]

    Conclusion

    Therefore, the value of the expression \(\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)\) is \(\textbf{23}\), making the correct answer:

    (c) \(\textbf{23}\)

    This solution methodically derives the value of the given expression by leveraging the initial condition and algebraic manipulation, leading to a clear and logical conclusion.

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

If the square of the sum of two numbers is equal to 4 times of their product. then the ratio of these numbers is : (a) \(2: 1\) (b) \(1: 3\) (c) \(1: 1\) (d) \(1: 2\)

If the square of the sum of two numbers is equal to 4 times their product, what is the ratio of these numbers? (a) 2:1 (b) 1:3 (c) 1:1 (d) 1:2

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 7:04 pm

    To find the ratio of two numbers given that the square of their sum is equal to four times their product, let's denote the two numbers as \(a\) and \(b\). According to the given condition, we have: \[ (a + b)^2 = 4ab \] Expanding the left side of the equation gives: \[ a^2 + 2ab + b^2 = 4ab \] RearrRead more

    To find the ratio of two numbers given that the square of their sum is equal to four times their product, let’s denote the two numbers as \(a\) and \(b\). According to the given condition, we have:

    \[
    (a + b)^2 = 4ab
    \]

    Expanding the left side of the equation gives:

    \[
    a^2 + 2ab + b^2 = 4ab
    \]

    Rearranging the terms to bring them all to one side:

    \[
    a^2 + 2ab – 4ab + b^2 = 0
    \]

    Simplifying:

    \[
    a^2 – 2ab + b^2 = 0
    \]

    Notice that the left side of the equation now represents the square of the difference between \(a\) and \(b\):

    \[
    (a – b)^2 = 0
    \]

    For a square to equal zero, the quantity being squared must itself be zero:

    \[
    a – b = 0
    \]

    This implies:

    \[
    a = b
    \]

    Therefore, the ratio of \(a\) to \(b\) is \(1:1\), which means the correct answer is:

    (c) \(1: 1\)

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

Equal amounts of water were poured into two empty jars of different capacities, which made one jar \(\frac{1}{4}\) full and the other jar \(\frac{1}{3}\) full. If the water in the jar with lesser capacity is then poured into the jar with greater capacity, then the part of the larger jar filled with water is:

Equal amounts of water were poured into two empty jars of different capacities, which made one jar \(\frac{1}{4}\) full and the other jar \(\frac{1}{3}\) full. If the water in the jar with lesser capacity is then poured into the jar ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 5:28 pm

    Calculation of Water Levels in Jars with Different Capacities This problem presents a scenario where equal volumes of water are introduced into two jars of distinct capacities. Here's a breakdown of the situation and the outcome when the water from one jar is transferred to the other. Initial SetupRead more

    Calculation of Water Levels in Jars with Different Capacities

    This problem presents a scenario where equal volumes of water are introduced into two jars of distinct capacities. Here’s a breakdown of the situation and the outcome when the water from one jar is transferred to the other.

    Initial Setup

    – Larger Jar: When filled with a certain volume of water, it reaches \(\frac{1}{4}\) of its full capacity.
    – Smaller Jar: The same volume of water fills this jar to \(\frac{1}{3}\) of its capacity, indicating its smaller size compared to the larger jar.

    Transfer Process and Outcome

    Upon transferring the water from the smaller jar (which is \(\frac{1}{3}\) full) into the larger jar (\(\frac{1}{4}\) full), we aim to understand how the water level changes in the larger jar.

    – Observation: Since the initial amounts of water in both jars are equal, transferring the water from the smaller jar to the larger one effectively doubles the amount of water in the larger jar.

    – Mathematical Representation: The act of pouring water from the smaller jar doubles the water volume in the larger jar, leading to the equation \(2 \times \frac{1}{4} = \frac{1}{2}\).

    Conclusion

    Therefore, after the water from the smaller jar is poured into it, the larger jar becomes \(\frac{1}{2}\) full.

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Abstract Classes
Abstract ClassesPower Elite Author
Asked: April 5, 2024In: SSC Maths

Last year my age was a perfect square number. Next year it will be a cubic number. What is my present age?

Last year my age was a perfect square number. Next year it will be a cubic number. What is my present age? (a) 25 years (b) 27 years (c) 26 years (d) 24 years

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 5:22 pm

    Finding the Present Age Based on Mathematical Properties Given the intriguing conditions about the nature of one's age in relation to mathematical figures: Stated Conditions: The age last year was a perfect square number. The age next year will be a cubic number. Evaluation of Options: Considering tRead more

    Finding the Present Age Based on Mathematical Properties

    Given the intriguing conditions about the nature of one’s age in relation to mathematical figures:

    Stated Conditions:

    • The age last year was a perfect square number.
    • The age next year will be a cubic number.

    Evaluation of Options:

    Considering the options provided and applying the given conditions to each, we meticulously analyze to find the correct age:

    – Option (a) 25 years: Not viable, as 24 (last year) is not a perfect square and 26 (next year) is not a cube.
    – Option (b) 27 years: Not viable, as 26 (last year) is not a perfect square and 28 (next year) is not a cube.
    – Option (c) 26 years: This is the correct choice. If the present age is 26, then:
    – Last year’s age was 25 (\(5^2\)), a perfect square.
    – Next year’s age will be 27 (\(3^3\)), a perfect cube.
    – Option (d) 24 years: Not viable, as 23 (last year) is not a perfect square and 25 (next year) is not a cube.

    Correct Answer:

    The logical deduction based on the conditions clearly points to Option (c) 26 years as the present age. At 26 years old:

    • Last year, the individual was 25 years old, aligning with the property of being a perfect square (\(5^2\)).
    • Next year, they will be 27 years old, fulfilling the characteristic of being a cubic number (\(3^3\)).

    Conclusion:

    The individual’s current age, which perfectly transitions from a perfect square to a cubic number, is unequivocally 26 years. This finding not only satisfies the unique mathematical conditions presented but also underscores the harmonious relationship between sequential numerical properties and real-life scenarios.

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Bhulu Aich
Bhulu AichExclusive Author
Asked: April 5, 2024In: SSC Maths

The sum of three consecutive odd numbers is 1383 . What is the largest number? (a) 463 (b) 49 (c) 457 (d) 461 (e) None of these

The sum of three consecutive odd numbers is 1383 . What is the largest number? (a) 463 (b) 49 (c) 457 (d) 461 (e) None of these

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

    To find the three consecutive odd numbers whose sum is 1383, let's denote the smallest of these numbers as \(n\), the next one as \(n + 2\), and the largest as \(n + 4\) (since odd numbers differ by 2). The sum of these numbers is given as: \[n + (n + 2) + (n + 4) = 1383\] Simplifying, we get: \[3nRead more

    To find the three consecutive odd numbers whose sum is 1383, let’s denote the smallest of these numbers as \(n\), the next one as \(n + 2\), and the largest as \(n + 4\) (since odd numbers differ by 2).

    The sum of these numbers is given as:
    \[n + (n + 2) + (n + 4) = 1383\]

    Simplifying, we get:
    \[3n + 6 = 1383\]

    Subtracting 6 from both sides:
    \[3n = 1377\]

    Dividing by 3:
    \[n = 459\]

    So, the three consecutive odd numbers are 459, 461, and 463. The largest number among them is \(463\).

    Therefore, the correct answer is (a) 463.

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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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Bhulu Aich
Bhulu AichExclusive Author
Asked: April 5, 2024In: SSC Maths

If \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\), then (a) \(a\) must equal \(c\) (b) \(a+b+c+d\) must equal zero (c) either \(a=c\) or \(a+b+c+d=0\), or both (d) \(a(b+c+d)=c(a+b+d)\)

If (a) (a+b)/(b+c)=(c+d)/(d+a), then (b) a+b+c+d must equal zero (c) either a=c or a+b+c+d=0, or both (d) a(b+c+d)=c(a+b+d)

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 3:17 pm

    Starting from the given proportional relationship: \[ \frac{a+b}{b+c} = \frac{c+d}{d+a} \] Multiplying across to eliminate the denominators, we have: \[ (a + b)(d + a) = (b + c)(c + d) \] Expanding both sides: \[ ad + a^2 + bd + ab = bc + c^2 + cd + bd \] Rearranging to group like terms: \[ a^2 - c^Read more

    Starting from the given proportional relationship:

    \[
    \frac{a+b}{b+c} = \frac{c+d}{d+a}
    \]

    Multiplying across to eliminate the denominators, we have:

    \[
    (a + b)(d + a) = (b + c)(c + d)
    \]

    Expanding both sides:

    \[
    ad + a^2 + bd + ab = bc + c^2 + cd + bd
    \]

    Rearranging to group like terms:

    \[
    a^2 – c^2 + ad – cd + ab – bc = 0
    \]

    Factoring by grouping, where appropriate, using the difference of squares for \(a^2 – c^2\) and factoring out the common terms in the other parts:

    \[
    (a – c)(a + c) + (a – c)d + (a – c)b = 0
    \]

    Factoring \(a – c\) from each term:

    \[
    (a – c)(a + c + d + b) = 0
    \]

    For this product to equal zero, at least one of the factors must be zero. Therefore:

    \[
    a – c = 0 \quad \text{or} \quad a + b + c + d = 0
    \]

    This means:

    – \(a = c\), or
    – \(a + b + c + d = 0\), or
    – Both conditions could be true in certain scenarios.

    Therefore, the correct interpretation is option (c) either \(a = c\) or \(a+b+c+d = 0\), or both.

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Ramakant Sharma
Ramakant SharmaInk Innovator
Asked: April 5, 2024In: SSC Maths

If sum of two numbers is 42 and their product is 437 , then find their difference.

If sum of two numbers is 42 and their product is 437 , then find their difference.

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 5, 2024 at 2:36 pm

    Let's denote the two numbers as \(x\) and \(y\). We are given that: 1. The sum of the two numbers is 42: \[x + y = 42\] 2. Their product is 437: \[xy = 437\] We are asked to find their difference, which is \(|x - y|\). To find \(x\) and \(y\), we can use the given sum and product. These can be consiRead more

    Let’s denote the two numbers as \(x\) and \(y\). We are given that:

    1. The sum of the two numbers is 42:
    \[x + y = 42\]

    2. Their product is 437:
    \[xy = 437\]

    We are asked to find their difference, which is \(|x – y|\).

    To find \(x\) and \(y\), we can use the given sum and product. These can be considered as the sum and product of the roots of a quadratic equation, which can be written as:

    \[x^2 – (x + y)x + xy = 0\]

    Substituting the given values:

    \[x^2 – 42x + 437 = 0\]

    Now, we solve this quadratic equation for \(x\). The solutions to the equation will give us the values of \(x\) and \(y\). Instead of directly solving the quadratic equation, let’s use another approach to find the difference, as solving it might get a bit complex without a calculator.

    The difference squared can be found using the formula derived from the sum and product of roots:
    \[(x – y)^2 = (x + y)^2 – 4xy\]

    Given \(x + y = 42\) and \(xy = 437\), we substitute these values into the equation:
    \[(x – y)^2 = 42^2 – 4 \cdot 437\]

    Calculating the right side:
    \[(x – y)^2 = 1764 – 1748\]

    \[(x – y)^2 = 16\]

    Taking the square root of both sides:
    \[|x – y| = 4\]

    Therefore, the difference between the two numbers is \(\boldsymbol{4}\).

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