Quantitative Aptitude - Numbers and Algebra

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1. If x - y = 1, then x3 - y3 - 3xy equals

  • Option : B
  • Explanation : x- y3- 3xy = x3 - y3 - 3xy(x y)....as (x - y) = 1 = (x - y)3= 1.
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2. If 6 ≥ x ≥ -2 and 4 ≥ y ≥ - 4, then limits for

If 6 ≥  x ≥ -2  and 4 ≥ y ≥ - 4, then limits for

where x and y are non zero integers, is

  • Option : D
  • Explanation :

    y 4 3 2 1 -1 -3 -4

    x 6 5 4 3 1 -1 -2

    Hence, minimum value of

    Number & Algebra

    And maximum value of

    Number & Algebra

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3. If (a, n)! is defined as product of n consecutive numbers starting from a, where a and n are both natural numbers, and if H is the HCF of (a, n)! and n!, then what can be said about H?

  • Option : D
  • Explanation : (a. n)! = product of n consecutive natural numbers starting from 'a' which is atleast divisible by n!. (n)! = product of n consecutive natural numbers. For n = 2 : (a. n)! = a(a + 1) and n! = 2 a(a + 1) is divisible by 2!. For n = 3 : (a n)! = a(a + 1)(a + 2) and n! = 6. One of the factors of a(a + 1)(a + 2) is divisible by 3 and other by 2. Thus, proceeding in this manner, (a. n)! and n! have HCF = n! ∴ H = n!.
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4. If a and b are prime numbers, which of the following is true?

I. a2 has three positive integer factors.

II. ab has four positive integer factors.

III. a3 has four positive integer factors.

Codes

  • Option : C
  • Explanation :

    Factors of a2 are 1. a and a2.
    Factors of ab are 1, a, b and ab.
    Factors of a3 are 1. a. a2 and a3.

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5. If x = b + c, y = c a, z = a b, then
x2 + y2 + z2 - 2xy - 2xz + 2yz is equal to

  • Option : B
  • Explanation :

    x+ y2 + z2-2xy - 2xz + 2yz 
    = (x - y - z)2
    =(b+c-c+a-a+b)2= 4b2

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