Adv
  1. The Doon school does not need any special introduction. Everyone is well aware of the school. It is a well-known school located in Dehradun, India. It is like a dream come true if you get admission in the Doon school. If you also dream to get admission to the Doon school then you need to do little hRead more

    The Doon school does not need any special introduction. Everyone is well aware of the school. It is a well-known school located in Dehradun, India. It is like a dream come true if you get admission in the Doon school. If you also dream to get admission to the Doon school then you need to do little hard work to get through the admission procedure. Talking about the admission procedure, you need to qualify the entrance examination as well as the personal interview stage. However, the entrance exam is not that easy as per the status of the Doon school.

    The school selects the best students to provide admission in the 7th and 8th standards. Start preparation today itself if you wish to qualify the entrance test with good scores. The best option is to practice from the sample question paper to get an idea about the entrance test. Basically, you need to prepare five papers- English, Mathematics, Hindi, and two Abstract Reasoning test (Verbal & Non-Verbal Reasoning). Focus on practicing all these papers equally because all the papers will be considered for the further selection procedure. The question paper will be set in such a manner that a student from any Board will not get any difficulty in attempting the question paper.

    Qualifying the entrance test is not the end of it, you have to put much effort to qualify the next round of selection procedure i.e. personal interview. Well, most of the students find the personal interview stage even more difficult as compared to the entrance test. Even though you are very good in studies, this is not just enough to get admission in the Doon school. You need to be good at some kinds of extra-curricular activities like sports, arts, music, etc. The students having extra-curricular activities will get advantages in selection.

    So, the gist of the whole story is that there is no specific blueprint to get admission in the Doon school. Keep practicing the subjects and also prefer to join any extra-curricular activity to get an edge over other students. You can also join a coaching institute to prepare for the Doon school entrance examination. Truemaths coaching academy is one of the best options for the Doon school entrance exam coaching. It will provide you complete guidance to fulfill your dream to be a part of the Doon school.

     

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  2. Let x be the ones digit and y be the tens digit. Then Two digit number before reversing = 10y + x Two digit number after reversing = 10x + y As per the question (10y + x) + (10x + y) = 121 ⇒11x + 11y = 121 ⇒x + y = 11 …….(i) Since the digits differ by 3, so x – y = 3 ……….(ii) Adding (i) and (ii), weRead more

    Let x be the ones digit and y be the tens digit.

    Then

    Two digit number before reversing = 10y + x

    Two digit number after reversing = 10x + y

    As per the question (10y + x) + (10x + y) = 121

    ⇒11x + 11y = 121

    ⇒x + y = 11 …….(i)

    Since the digits differ by 3, so

    x – y = 3 ……….(ii)

    Adding (i) and (ii), we get

    2x = 14

    ⇒ x = 7

    Putting x = 7 in (i), we get

    7 + y = 11

    ⇒ y = 4

    Changing the role of x and y, x = 4 and y = 7

    Hence, the two-digit number is 74 or 47.

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  3. Let the larger number be x and the smaller number be y. Given that 3 times, the larger number is divided by smaller we get 4 as Quotient and 8 as remainder. = > 3x = y * 4 + 8 = > 3x = 4y + 8 = > 3x - 4y = 8   --------- (1) Given that if five times the smaller is divided by larger we get 3Read more

    Let the larger number be x and the smaller number be y.

    Given that 3 times, the larger number is divided by smaller we get 4 as Quotient and 8 as remainder.

    = > 3x = y * 4 + 8

    = > 3x = 4y + 8

    = > 3x – 4y = 8   ——— (1)

    Given that if five times the smaller is divided by larger we get 3 as Quotient and 5 as remainder.

    = > 5y = 3x + 5

    = > 5y – 3x = 5   

    = > -3x + 5y = 5  ——— (2)

    On solving (1) & (2), we get

    = >  3x – 4y = 8

    = > -3x + 5y = 5

         ——————–

                  y = 13

    Substitute y = 13 in (1), we get

    = > 3x – 4y = 8

    = > 3x – 4(13) = 8

    = > 3x – 52 = 8

    = > 3x = 52 + 8

    = > 3x = 60

    = > x = 60/3

    = > x = 20.

    Therefore the numbers are 20, 13.

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  4. Let x and y be the two numbers. according to the question, If 2 is added to each of two given numbers, their ratio becomes 1:2. => ( (x + 2) / ( y+2) ) = 1/2 => 2x - y = - 2 => y = 2x + 2....................................................(1) if 4 is subtracted from each of the given numberRead more

    Let x and y be the two numbers.

    according to the question,
    If 2 is added to each of two given numbers, their ratio becomes 1:2.
    => ( (x + 2) / ( y+2) ) = 1/2
    => 2x – y = – 2
    => y = 2x + 2…………………………………………….(1)

    if 4 is subtracted from each of the given numbers, the ratio becomes 5:11.
    => ( ( x – 4 ) / ( y-4 ) ) = 5 /11
    => 11x – 5y= 24………………………………………(2)

    substituting the value of y from eq (1) in eq(2) , we get
    => 11 x – 5( 2x+2) = 24
    => 11x – 10x  = 34
    => x= 34

    y = 2x + 2
    => y = 2(34) +2 = 68 + 2
    => y = 70

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  5. Let the tens and the units digits of the required number be x and y, respectively. Required number = (10x + y) 10x + y = 7(x + y) 10x + 7y = 7x + 7y or 3x – 6y = 0 ……….(i) Number obtained on reversing its digits = (10y + x) (10x + y) - 27 = (10y + x) ⇒10x – x + y – 10y = 27 ⇒9x – 9y = 27  ⇒9(x – y)Read more

    Let the tens and the units digits of the required number be x and y, respectively.

    Required number = (10x + y)

    10x + y = 7(x + y)

    10x + 7y = 7x + 7y or 3x – 6y = 0 ……….(i)

    Number obtained on reversing its digits = (10y + x)

    (10x + y) – 27 = (10y + x)

    ⇒10x – x + y – 10y = 27

    ⇒9x – 9y = 27  ⇒9(x – y) = 27

    ⇒x – y = 3 ……..(ii)

    On multiplying (ii) by 6, we get:

    6x – 6y = 18 ………(iii)

    On subtracting (i) from (ii), we get:

    3x = 18

    ⇒ x = 6

    On substituting x = 6 in (i) we get

    3 × 6 – 6y = 0

    ⇒ 18 – 6y = 0

    ⇒ 6y = 18

    ⇒ y = 3

    Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63

    Hence, the required number is 63.

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  6. We can conclude that the first term a = 25 And the common difference(d) d = 20-25 d = - 5 For finding the A. P. :- We can use the formula An = a+(n-1)d As we know:- For the first negative term it should be less than 0. So the required equation will be : => a+(n-1)d<0 => 25+(n-1)*(-5)<0 =Read more

    We can conclude that the first term

    a = 25

    And the common difference(d)

    d = 20-25

    d = – 5

    For finding the A. P. :

    We can use the formula

    An = a+(n-1)d

    As we know:

    For the first negative term it should be less than 0.

    So the required equation will be :

    => a+(n-1)d<0

    => 25+(n-1)*(-5)<0

    => (n-1)*(-5)<-25

    =>(n-1)>5

    => n>6

    Therefore-:

    The first negative term is it’s 7th term.

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  7. 201 , ............ 9 , 7 , 5 a = 201  ,   d = 7 - 9 =  - 2   a6 =  a + (5d) a6 = 201 + (5) (-2) a6 =  201 - 10 a6 =  191

    201 , ………… 9 , 7 , 5

    a = 201  ,   d = 7 – 9 =  – 2

     

    a6 =  a + (5d)

    a6 = 201 + (5) (-2)

    a6 =  201 – 10

    a6 =  191

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  8. Sn​=1+2+3+....+n Sn​=2n​[2a+(n−1)d] where a=1,d=2−1=1 ⇒Sn​=2n​[2×1+(n−1)1] ⇒Sn​=2n​(n+1) ∴ Sum of first n natural numbers=2n(n+1)​

    Sn=1+2+3+....+n

    Sn=2n[2a+(n1)d] where a=1,d=21=1

    Sn=2n[2×1+(n1)1]

    Sn=2n(n+1)

     Sum of first n natural numbers=2n(n+1)

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  9. Even natural numbers = 2,4,6,8,.... d=4-2=2 Sn=n/2[2a+(n-1)d] Sn=n/2[2×2+(n-1)2] Sn=n/2[4+2n-2] Sn=n/2[2n+2] Sn=2n^2+2n/2 Sn=2(n^2+n)/2 Sn=n^2+n   Sum of first n even natural numbers = n square +n

    Even natural numbers = 2,4,6,8,….

    d=4-2=2

    Sn=n/2[2a+(n-1)d]

    Sn=n/2[2×2+(n-1)2]

    Sn=n/2[4+2n-2]

    Sn=n/2[2n+2]

    Sn=2n^2+2n/2

    Sn=2(n^2+n)/2

    Sn=n^2+n

     

    Sum of first n even natural numbers = n square +n

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  10. First n natural numbers 1 , 3 , 5 , 7 , 9 ... ( 2n - 1 ).  

    First n natural numbers
    1 , 3 , 5 , 7 , 9 … ( 2n – 1 ).

     

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