### 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. SolutionLet 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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