Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Math Assignment Class X Ch-4 | Quadratic Equations

 Math Assignment / Class X / Chapter 4 /  Quadratic Equations

Extra questions of chapter 4 class 10 Quadratic Equations with answer and  hints to the difficult questions. Important and useful math assignment for the students of class 10

For better results

  • Students should learn all the basic points of  Quadratic Equations 
  • Student should revise N C E R T book thoroughly with examples.
  • Now revise this assignment. This assignment integrate the knowledge of the students.

ASSIGNMENT FOR 10 STANDARD QUADRATIC EQUATIONS

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Question 28. 

Out of a number of Saras birds, (1/4)th of the number are moving about in Lotus plants, (1/9)th coupled with (1/4)th,  as well as 7 times the square root of the number moving on a hill, 56 birds remain on trees.   What is the total number of words.

Solution

Let total birds = x

No. of birds moving in lotus plant = (1/4)x

No. of birds coupled with = (1/9 +1/4)x = (13/36)x

No of birds moving on the hill = 7x

No. of birds on the tree = 56

Acoording to the question

\[ \frac{1}{4}x+\frac{13}{36}x+7\sqrt{x}+56=x\]

\[ \frac{22}{36}x+7\sqrt{x}+56=x\]

\[ x-\frac{22}{36}x-7\sqrt{x}-56=0\]

\[ \frac{14}{36}x-7\sqrt{x}-56=0\]

\[\frac{2}{36}x-\sqrt{x}-8=0\]

\[ \frac{1}{18}x-\sqrt{x}-8=0\]

\[ x-18\sqrt{x}-144=0\]

\[ Putting \: \: \: \: \sqrt{x}=y \: \: \: \Rightarrow\: \: x=y^{2}\]

\[ y^{2}-18y-144=0\]

\[ y^{2}-24y+6y-144=0\]

\[ y(y-24)+6(y-24)=0\]

\[ (y-24)(y+6)=0\]

\[ \Rightarrow y=24, -6\]

\[ y=-6 \: \: is \: \: rejected\]

\[ y=\sqrt{x}=24\Rightarrow x=(24)^{2}=576\]

Hence Total Number of birds = 576

Question : 38

7 years ago, age of Varun was five times the square of the age of Swati. After 3 years, age of Swati will be 2/5 of the age of Varun. Find their present ages.

Solution :

Let 7 years ago the age of Swati = x years

Let 7 years ago the age of Varun = y years

ATQ       y = 5x2 …………………(1)

Present age of Swati = x + 7

Present age of Varun = y + 5

After 3 years age of Swati = x + 10

After 3 years age of Varun= y + 10

Again ATQ :  Age of Swati = (2/5) age of Varun

x + 10 = (2/5)[y + 10]

5(x + 10) = 2y + 20

5x + 50 = 2y + 20

Putting  y = 5x2  we get

5x + 50 = 2(5x2 ) + 20

10x2  - 5x + 20 - 50 = 0

10x2 - 5x - 30 = 0

2x2 - x - 6 = 0    (Dividing by 5)

(2x + 3)(x - 2) = 0

x = - 3/2 (Rejected)     and   x = 2

Putting x = 2 in equation (1) we get

y = 5 (2)2    ⇒     y = 20

Present age of Swati = x + 7  = 2 + 7 = 9

Present age of Varun = y + 7 =  20 + 7 = 27


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