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.

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 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.

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

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)

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:–

Forthe first negative term it should be less than 0.

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

## How to get admission into The Doon School

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

admissionin theDoon 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 theentrance examinationas 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.

## The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.

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.

See less## If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. find the numbers.

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 = 13Substitute 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.

See lessTherefore the numbers are 20, 13.## If 2 is added to each of two given numbers, their ratio becomes 1:2 However, if 4 is substracted from each of the given numbers, the ratio becomes 5:11 Find the numbers.

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

See less=> y = 2(34) +2 = 68 + 2

=> y = 70

## A number consisting of two digits is seven times the sum of its digits. When 27 is substracted from the number, the digits are reversed. Find the number.

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.

See less## Which term of the AP 25,20,15,…is the first negative term ?

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:–Forthe 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.

See less## Find the 6th term from the end of the AP 5,7,9,…..,201.

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

See less## Find the sum of first n natural numbers.

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+(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)

See less## Find the sum of first n even natural numbers.

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

See less## Find the sum of first n odd natural numbers.

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

First n natural numbers

1 , 3 , 5 , 7 , 9 … ( 2n – 1 ).