Friday, August 30, 2019

Quantitative Aptitude: Simplification Questions Set 79

Directions(1-10): What approximate value will come in place of question mark '?' in the following questions.
  1. [(๐Ÿ–๐Ÿ’) ^๐Ÿ ÷ ๐Ÿ๐Ÿ– × ๐Ÿ๐Ÿ] ÷ ๐Ÿ๐Ÿ’ = ๐Ÿ• ×?
    20
    22
    18
    24
    10
    Option C
    [(๐Ÿ–๐Ÿ’) ^๐Ÿ ÷ ๐Ÿ๐Ÿ– × ๐Ÿ๐Ÿ] ÷ ๐Ÿ๐Ÿ’ = ๐Ÿ• ×?
    =>7 ×? = 84×84/28 × 12 × 1/24
    => ? = 18

     


  2. (7.9% of 134) – (3.4% of 79) =?
    12
    8
    10
    6
    13
    Option B
    (7.9% of 134) – (3.4% of 79) =?
    ? = 7.9/100 × 134 − 3.4/100 × 79
    = 7.9 == 8

     


  3. [(๐Ÿ๐Ÿ—๐Ÿ)^ ๐Ÿ ÷ ๐Ÿ”๐Ÿ’ × ๐Ÿ๐Ÿ’] ÷ ๐Ÿ’๐Ÿ– = √?
    84490
    87389
    81020
    89300
    82944
    Option E
    [(๐Ÿ๐Ÿ—๐Ÿ)^ ๐Ÿ ÷ ๐Ÿ”๐Ÿ’ × ๐Ÿ๐Ÿ’] ÷ ๐Ÿ’๐Ÿ– = √?
    =>√? = 36864/64 × 24 × 1/48 = 288
    => ? = 82944

     


  4. 30% of ๐Ÿ/๐Ÿ• of ๐Ÿ/๐Ÿ— of ๐Ÿ/๐Ÿ“ of ๐Ÿ/๐Ÿ‘ of 9450 =?
    46
    42
    44
    40
    48
    Option E
    30% of ๐Ÿ/๐Ÿ• of ๐Ÿ/๐Ÿ— of ๐Ÿ/๐Ÿ“ of ๐Ÿ/๐Ÿ‘ of 9450 =?
    =>? = 30/100 × 2/7 × 2/9 × 2/5 × 2/3 × 9450
    = 48

     


  5. (๐Ÿ/๐Ÿ๐ŸŽ๐Ÿ๐Ÿ’)^( −๐Ÿ/๐Ÿ“) + ( ๐Ÿ/๐Ÿ‘๐Ÿ’๐Ÿ‘)^( −๐Ÿ/๐Ÿ‘) = ? × 5
    18
    13
    10
    15
    17
    Option B
    (๐Ÿ/๐Ÿ๐ŸŽ๐Ÿ๐Ÿ’)^( −๐Ÿ/๐Ÿ“) + ( ๐Ÿ/๐Ÿ‘๐Ÿ’๐Ÿ‘)^( −๐Ÿ/๐Ÿ‘) = ? × 5
    =>?× 5 = (4)^{( −5)×( −2/5)} + (7) ^{(−3)×( −2/ 3)}
    => ? = (16+49)/5 = 13

     


  6. 3.5% of 40 + 3.5% of 80 =? % of 10
    40
    45
    48
    42
    46
    Option D
    3.5% of 40 + 3.5% of 80 =? % of 10
    =>? = (1.4 + 2.8)/10 × 100 = 42

     


  7. ๐Ÿ( ๐Ÿ•/๐Ÿ—) + ๐Ÿ(๐Ÿ“/๐Ÿ‘) + ๐Ÿ‘(๐Ÿ/๐Ÿ—) – ๐Ÿ’(๐Ÿ/๐Ÿ“) = ?
    1(15/41)
    3(11/46)
    4(16/45)
    3(10/45)
    4(16/41)
    Option C
    ๐Ÿ( ๐Ÿ•/๐Ÿ—) + ๐Ÿ(๐Ÿ“/๐Ÿ‘) + ๐Ÿ‘(๐Ÿ/๐Ÿ—) – ๐Ÿ’(๐Ÿ/๐Ÿ“) = ?
    ? = (1 + 2 + 3 − 4) + ( 7/9 + 5/3 + 1/9 – 1/5 )
    = 2 + 106/45 = 4(16/45)

     


  8. (๐Ÿ × ๐Ÿ‘)^๐Ÿ‘ ÷ (๐Ÿ’ × ๐Ÿ—)^๐Ÿ × (๐Ÿ๐Ÿ• × ๐Ÿ–)^๐Ÿ = (๐Ÿ”)^?
    3
    5
    2
    6
    7
    Option B
    (๐Ÿ × ๐Ÿ‘)^๐Ÿ‘ ÷ (๐Ÿ’ × ๐Ÿ—)^๐Ÿ × (๐Ÿ๐Ÿ• × ๐Ÿ–)^๐Ÿ = (๐Ÿ”)^?
    =>(6)^? = (6) ^3 ÷ 6^4 × 6^6
    ⇒ (6)^? = 6^(3−4+6)
    ⇒ ? = 5

     


  9. ๐Ÿ’๐Ÿ“๐Ÿ’. ๐Ÿ“๐Ÿ– − ๐Ÿ‘๐Ÿ•๐Ÿ”. ๐Ÿ–๐Ÿ— + ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ’๐Ÿ“ − ๐Ÿ—๐Ÿ“. ๐Ÿ’๐Ÿ = ?
    104
    110
    100
    107
    117
    Option A
    ๐Ÿ’๐Ÿ“๐Ÿ’. ๐Ÿ“๐Ÿ– − ๐Ÿ‘๐Ÿ•๐Ÿ”. ๐Ÿ–๐Ÿ— + ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ’๐Ÿ“ − ๐Ÿ—๐Ÿ“. ๐Ÿ’๐Ÿ = ?
    ? = 576.03 – 472.31 = 103.72
    == 104

     


  10. √๐Ÿ“๐Ÿ•๐Ÿ” ÷ (๐Ÿ’)^๐Ÿ × ๐Ÿ•. ๐Ÿ’ + (๐Ÿ•)^๐Ÿ‘ − ๐Ÿ๐Ÿ‘๐Ÿ = ?
    110
    111
    120
    127
    123
    Option E
    √๐Ÿ“๐Ÿ•๐Ÿ” ÷ (๐Ÿ’)^๐Ÿ × ๐Ÿ•. ๐Ÿ’ + (๐Ÿ•)^๐Ÿ‘ − ๐Ÿ๐Ÿ‘๐Ÿ = ?
    = ? = 24 ÷ 16 × 7.4 + 343 − 231
    = 11.1 + 112 = 123.1 == 123

     




No comments:

Post a Comment