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Quantitative Aptitude Test - 6

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Quantitative Aptitude Test - 6
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  • Question 1
    1 / -0.25

    In an examination 70% of the candidate passed in English, 80% passed in Mathematics, 10% failed in both subjects. If 144 candidates passed in both, the total number of candidates was

    Solution

    Failed candidates in English =(10070)=30%

    Failed candidates in Mathematics =(10080)=20%

    Candidates who fail in both subject =10%

    Candidates who only fail in English =3010= 20%

    Candidates who only fail in Mathematics =2010=10%

    Percentage of passed students in both subject =100 (Candidates who only fail in English + Candidates who only fail in Mathematics + Candidates who fail in both subject)

    =100(20+10+10)

    =60%

    According to the question,

    60% of students =144

    Total students =14460×100

    =240

  • Question 2
    1 / -0.25

    X and Y enter into a business partnership with capital in the ratio 3:5. After 5 months, X adds 50% of his capital, while Y withdraws 60% of his capital. What is the share (in ₹ lakhs) of X in the annual profit of ₹ 10.26 lakhs?

    Solution

    Given:

    X and Y initial capital=3:5

    X capital after 5 months =150% of initial investment of X

    Y capital after 5 months =40% of initial investment of Y

    Annual profit =10.26 lakhs

    Formula used:

    (X's profit) : (Y's profit) = (X's investment × time period) × (Y's investment × time period) ----- (1)

    Let X's and Y's initial investment be 30 and 50 same as the given ratio.

    X's capital for the last 7 months =150100×30=45

    Y's capital for the last 7 months =40100×50=20

    Using equation (1)

    ⇒ X's profit : Y's profit =(30×5+45×7):(50×5+20×7)

    ⇒ X's profit : Y's profit =(465):(390)

    =31:26

    Total profit =31+26 units

    =57 units

    X's profitTotal profit=3157

    ⇒ X's profit=3157×10.26 lakhs

    =0.54×10.26 lakhs

    =5.58 lakhs

    ∴ The share (in ₹ lakhs) of X is 5.58.

  • Question 3
    1 / -0.25

    If the ratio of A ∶ B is 3 ∶ 6 and ratio of B ∶ C is 7 ∶ 28, then find the ratio of A ∶ B ∶ C.

    Solution

    Given:

    A : B = 3 : 6 = 1 : 2 ----(i)

    B : C = 7 : 28 = 1 : 4 ----(ii)

    We know that:

    If P ∶ Q and Q ∶ R are given, then P ∶ Q ∶ R is given by:

    To equate the value of B in both equation we will multiply equation (ii) by 2 so we get,

    A : B = 1 : 2

    B : C = 2 : 8

    A : B : C = 1 × 2 : 2 × 2 : 2 × 8

    ⇒ A : B : C = 2 : 4 : 16 = 1 : 2 : 8

    ∴ A : B : C = 1 : 2 : 8

     
  • Question 4
    1 / -0.25

    The average marks obtained by 40 students of a class is 86. If the 5 highest digits are removed, the average is reduced by one point. State the average marks of the top 5 students.

    Solution

    As given, the average marks obtained by 40 students of a class is 86.

    Average = Sum of all digits  Number of students 

    First of all, we will just add the sum of the digits

    =86×40=3440

    Now the sum which will be formed after taking out those five

    digits is =35×85=2975

    Difference of both =34402975=465

    Now the average of five digits =4655=93

  • Question 5
    1 / -0.25

    Directions For Questions

    Direction: The table shows the number of national parks and wild life sanctuaries in different states. Study the table carefully and answer the following questions.

    State

     National Parks 

     Wild life Sanctuaries 

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

     Andaman and Nicobar 

    10

    96

    Uttarakhand

    7

    7

    ...view full instructions

    What is the ratio of National parks to Wild life sanctuaries?

    Solution

    State

    National Parks

    Wild life Sanctuaries

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

    Andaman and Nicobar

    10

    96

    Uttarakhand

    7

    7

    Total = 42

    Total = 221

    ∴ Required Ratio = 42 ∶ 221

  • Question 6
    1 / -0.25

    Directions For Questions

    Direction: The table shows the number of national parks and wild life sanctuaries in different states. Study the table carefully and answer the following questions.

    State

     National Parks 

     Wild life Sanctuaries 

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

     Andaman and Nicobar 

    10

    96

    Uttarakhand

    7

    7

    ...view full instructions

    The number of wild life sanctuaries in Maharashtra is how much percent more than the number of wild life sanctuaries in Himachal Pradesh?

    Solution

    Wild life sanctuaries in Maharashtra = 42

    Wild life sanctuaries in Himachal Pradesh = 28

    ∴ Required percentage =1428 × 100 = 50%

  • Question 7
    1 / -0.25

    Directions For Questions

    Direction: The table shows the number of national parks and wild life sanctuaries in different states. Study the table carefully and answer the following questions.

    State

     National Parks 

     Wild life Sanctuaries 

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

     Andaman and Nicobar 

    10

    96

    Uttarakhand

    7

    7

    ...view full instructions

    What is the average of all the national parks together?

    Solution

    State

    National Parks

    Madhya Pradesh

    10

    Himachal Pradesh

    5

    Gujarat

    4

    Maharashtra

    6

    Andaman and Nicobar

    10

    Uttarakhand

    7

    Total = 42

    ∴ Required average =426 = 7

  • Question 8
    1 / -0.25

    Directions For Questions

    Direction: The table shows the number of national parks and wild life sanctuaries in different states. Study the table carefully and answer the following questions.

    State

     National Parks 

     Wild life Sanctuaries 

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

     Andaman and Nicobar 

    10

    96

    Uttarakhand

    7

    7

    ...view full instructions

    What is the total number of wild life sanctuaries in all the states together?

    Solution

    State

    Wild life Sanctuaries

    Madhya Pradesh

    25

    Himachal Pradesh

    28

    Gujarat

    23

    Maharashtra

    42

    Andaman and Nicobar

    96

    Uttarakhand

    7

    Total = 221

    ∴ All states together has total 221 wild life sanctuaries.

  • Question 9
    1 / -0.25

    Directions For Questions

    Direction: The table shows the number of national parks and wild life sanctuaries in different states. Study the table carefully and answer the following questions.

    State

     National Parks 

     Wild life Sanctuaries 

    Madhya Pradesh

    10

    25

    Himachal Pradesh

    5

    28

    Gujarat

    4

    23

    Maharashtra

    6

    42

     Andaman and Nicobar 

    10

    96

    Uttarakhand

    7

    7

    ...view full instructions

    Which state has least number of national parks?

    Solution

    State

    National Parks

    Madhya Pradesh

    10

    Himachal Pradesh

    5

    Gujarat

    4

    Maharashtra

    6

    Andaman and Nicobar

    10

    Uttarakhand

    7

    ∴ Gujarat has least number of National parks.

  • Question 10
    1 / -0.25

    Among a group of students, 50 played cricket, 50 played hockey and 40 played volley ball. 15 played both cricket and hockey, 20 played both hockey and volley ball, 15 played cricket and volley ball and 10 played all three. If every student played at least one game, find the number of students and how many played only cricket, only hockey and only volley ball respectively?

    Solution

    According to the question,

    n(C) = 50, n(H) = 50, n(V) = 40

    n(C∩H) = 15

    n(H∩V) = 20

    n(C∩V) = 15

    n(C∩H∩V) = 10

    No. of students who played at least one game

    n(CUHUV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

    = 50 + 50 + 40 – 15 – 20 – 15 + 10

    Total number of students = 100.

    Let a denote the number of people who played cricket and volleyball only.

    Let b denote the number of people who played cricket and hockey only.

    Let c denote the number of people who played hockey and volleyball only.

    Let d denote the number of people who played all three games.

    Accordingly, d = n (C∩H∩V) = 10

    Now, n(C∩V) = a + d = 15

    n(C∩H) = b + d = 15

    n(H∩V) = c + d = 20

    Therefore, a = 15 – 10 = 5 [cricket and volleyball only]

    b = 15 – 10 = 5 [cricket and hockey only]

    c = 20 – 10 = 10 [hockey and volleyball only]

    No. of students who played only cricket = n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

    No. of students who played only hockey = n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

    No. of students who played only volley ball = n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

    Alternatively, we can solve it faster with the help of a Venn diagram.

    The Venn diagram for the given information looks like this.

    Subtracting the values in the intersections from the individual values gives us the number of students who played only one game.

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