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How to answer the following question?

Ramesh is a cashier at Canara Bank. He has notes of denominations of Rs 100. 50 and 10 respectively. The ratio of the number of these notes are is  2: 3: 5 respectively.

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in this test how is the answer for the question: what is the probability of a coin getting tail and the answer came 1 i nstead of  the correct answer 1/2

 

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Good Evening teacher,

In Vedic Maths, adding time, at the last 1.35+3.55=490 and the last 2 digits greater than 60 we added 40 and we got the answer 530. My question how do we get "40" ?

Abhijith C J

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Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

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Reliable latest lines and angles question 6

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Reliable latest lines and angles question 6

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Question No - 20

 

In the below figure ABleft | CDleft | EF: : and : : : GHleft | KL. Find angle HKL ?

In this question from lines and angles exercise.in the solution it was said that:-

-----------------------------X----------------------------X------------------------X------------X-----

Now, alternate angles are equal

angle CHG=angle HGN=60^0

angle HGN=angle KNF=60^0           [ corresponding angles]

Hence, angle KNG=180^{o}-60^{o}=120^{o}

Rightarrow angle GNK=angle AKL=120^0      [ corresponding angles]

angle AKH=angle KHD=25^0           [ alternative angles]

Therefore, angle HKL=angle AKH+angle AKL=25+120=145^0

-------------X-------------------------X--------------------------X------------------------------X-----------------------X----------

In the above said solution where did the 60 degree come from?

 

about 4 months ago 0 Answer 0 views

Question No - 15

 

Draw these figures, Four equal rhombuses

Question No - 16

 

Draw these figures., Five equal rhombuses:

Question No - 17

 

Draw these figures, Four rhombuses around a square:

 

Question No - 18

 

Parallelograms on two sides of a square :

about 2 months ago 1 Answer 12 views

(i)Find the six rational  numbers bbetween 3 and 4 

(ii) Which mathematical concept is used in this problem

(iii) Which value is depticted in this question 

 

(i) We known that between two rational numbers  x and y such that x< y there is a rational number frac{x+y}{2}.

               ie,       3< frac{7}{2}< 4

Now a rational number between 3 and   frac{7}{2}< 4   is :

frac{1}{2}left ( 3+frac{7}{2} right )=frac{1}{2}times left ( frac{6+7}{2} right )=frac{13}{4}

A rational nummber berween  frac{7}{2}: : and : : 4 : : is

frac{1}{2}left ( frac{7}{2}+4 right )=frac{1}{2}times left ( frac{7+8}{2} right )=frac{15}{4}

 3< frac{13}{4}< frac{7}{2}< frac{15}{4}< 4

Further a rational number between 3 and  frac{13}{4}: : is :

frac{1}{2}left ( 3+frac{13}{4} right )=frac{1}{2}left ( frac{12+13}{4} right )=frac{25}{8}

A rational number between  frac{15}{4}: : and: : 4: : is :

frac{1}{2}left ( frac{15}{4}+4 right )=frac{1}{2}times frac{15+16}{4}=frac{31}{8}

A rational number between  frac{31}{8}: : and: : 4: : is

frac{1}{2}left ( frac{31}{8}+4 right ) = frac{1}{2}times left ( frac{31+32}{8} right )=frac{63}{16}

therefore 3< frac{25}{8}< frac{13}{4}< frac{7}{2}< frac{15}{4}< frac{31}{8}< frac{63}{16}< 4

Hence , six rational numbers between 3 and 4 are 

frac{25}{8}, frac{13}{4},frac{7}{2}, frac{15}{4}, frac{31}{8}: : and: : frac{63}{16}

(ii) Number syatem 

(iii) Rationality is always welcomed 

The auto rikshaw fare in a city is charged RS . 10 for first kilometer and @Rs.4 per kilometer for subsequent distance covered. Write the linear equation to express the above statment. Draw the graph of the linear  equation. 

 

Total distance covered =  x km. 

Total fare = y km. 

Fare for the first kilometer 

Subsequent distance  = (x-1) km

Fare for the subsequent distance = Rs. 4(x-1)

According to the question, 

 y = 10+4 (x-1)

Rightarrowy = 10+4x -4 

Rightarrow y =4x +6

Table of solutions

X 0 1 -1
Y 6 10 2

 

A student amit of class IX is unable to write in his examination, due to fracture in his arm. Akhil a student of a class VI writes for him. The sum of their ages is  25 years.

(i) Write a linear equation for the above situation and represent it graphically. 

(ii) Find the age of of Akhil from the graph, when age of Amit is 14 years.

Let  Age of Amit = x years

Age of Akhil =  Y years 

(i) According to the question the linear equation  for above situation is 

Rightarrow x+ y = 25

y = 25-x

X 0 10 15
Y 25 15 10

 

(ii) From the graph when Amit's age = 14  years, then Akhil's age = 11 years. 

 

If the zeroes of the polynomial x^{2}+px+q   are double in value to the zeroes of  2x^{2} -5x -3 , find the value

of  p and q. 

Let, f (x) = 2x^{2}-5x -3

Let the zeroes of polynomial be alpha  and beta then,

Sum of zeroes   =alpha+ beta = frac{5}{2}

product: : of: : zeros=alpha beta = -frac{3}{2}

According to the question, zeroes of  x^{2}+px+q are 2 alpha and 2 beta

Sum of zeroes  = frac{coeff. of x}{coeff.of x ^{2}} = frac{-p}{1}

                  -p=2alpha +2beta = 2 (alpha +beta )

                 -p=2 times frac{5}{2} = 5 Rightarrow p=-5

Product of zeroes  =frac{constant: : term}{coeff.: of : x ^{2}} = frac{q}{1} 

                                

                      Rightarrow q = 2alpha +2beta = 4alpha beta

                      Rightarrow q = 4 left ( -frac{3}{2} right ) = -6

                          p = -5  and q = -6

 

 

4 chairs and  3 tables cost Rs. 2100 and 5 chairs and 2 tables cost Rs. 1750. Find the cost of one chair and one table separately. 

 

Let cost of one chair = Rs  x   and    cost of 1 table = Rs  y 

According to the question, 

4x+3 y = 2100   .....................(1)

5x + 2 y = 1750   .....................(2)

Multiplying eqn. (1)  2 and  eqn (2)  by 3,

 8x +6y = 4200   ....................(3)

15x +6y= 5250   ...................(4)

eqn (4) -eqn (3)

7x = 1050

  x = 150

Substituting the value of x in (1), y  = 500

Cost of chair and table  = Rs 150 , Rs 500 respectively.

Two pipes running together can fill a tank in  11 frac{1}{9}  minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.  

 Let time taken  by pipe A be x minutes. and time taken by pipe  B be x +5 minutes.

In one minute pipe A will fill frac{1}{x} tank 

In one minute pipe B will fill frac{1}{x+5}, tank 

Both pipes A+B  will fill  frac{1}{x}+ frac{1}{x+5}   tank in one minute 

Then according to the question 

                      frac{1}{x}+ frac{1}{x+5} = frac{9}{100}

                       frac{x+5+x}{x(x+5)} = frac{9}{100}

                    100 (2x+5) = 9x (x+5)

                             200x + 500 = 9x^{2}+45x

           9x ^{2}-155x -500 = 0

 9x ^{2}-180 x + 25x -500 = 0

9x (x-20)+25(x-20) = 0

        (x-20)+ (9x+25) = 0

                      x = 20, frac{-25}{9}

 

rejecting negative value, x = 20 minutes

and  x+5 = 25 minutes  

Hence pipe A  will fill the tank in  20  minutes  and pipe B will fill in  25  minutes. 

The digits of a positive number of three digits are in AP and their sum is  15. The number obtained by reserving the digits is  594 less than the original number find the number.  

Let the three digits be  a -d, a , a+d

Sum  = a-d+a+a+d = 3a=15 given 

therefore the  three digits are  5 -d, 5,5+d.

Original number = 100 (5-d) +10 x 5 + 1(5+d)

                          = 555+99d

Revered number = 100 (5+d)+10 x 5+1 (5-d)

                          = 555+99 d

According to question , 

(555-99d) - (555+99d)  = 594

                         -198 d = 594

                          Rightarrow  d = frac{594}{-198} = -3

The three digits are 

5-(-3),5,5+(-3)

8,5 and 2

therefore Original number is  8 x 100 + 5x 10 +2 x 1 = 852

A solid piece of metal, cuboidal in shape, with dimensions 24 cm, 18cm and 4cm is recast into a cube. calculate the lateral surface area of the cube.

  Vol. of cuboid = lbh 

                        = 24times 18times 4

                        = 1728 cu.cm.

Edge of a cube = sqrt[3]{1728}

                       = 12 cm    

          LSA       = 4a^{2}

                       = 4times 12times 12

                       = 576 cm2

   

Alternative method :

Let x be the edge of cube.

According to the question, we have 

Rightarrow Volume of cube =  Volume of the cuboid

                        x^{3}= 24times 18times 4

                        x^{3}=1728

                     Rightarrow x = 12

therefore Edge of the cube =12 cm2

Lateral surface area of the cube 

                            = 4x2 = 4(12)2

                            = 576 cm2

If the coordinates of  a point M are (-2,9) which can also be expressed as (1+x,y2) and y >0, then  find in which quadrant do the following points lie :  P (y,x) Q (2,x) , R (x2,y-1), S (2x,-3y)

According to the question, 

(-2,9) = (1+x,y2)

            -2 = 1+x or x = -3

and        9 =y2 or y

                =3, as y>0

So,      P(y,x) = P (3,-3)

Which will be in IV quadrant  Q (2,x) = Q (2,-3)

Which will be in IV quadrant  R (x2,y-1) = R (9, 2 )

Which will be in  I quadrant  S (2x, -3y) = S (-6,-9)

Which will be in III quadrant  

 

 

In the figure ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively  show that the line segment AF and  EC trisect the diagonal BD.

According to the question E and F are the midpoints of sides AB and CD.

therefore AE = frac{1}{2} AB

CF = frac{1}{2}CD

therefore  In the parallelogram opposite sides are equal so, 

AB = CD

AE= CF

Again AB II CD

So,  AE II FC

Hence AECF  is a parallelogram 

In Delta ABP

E is the mid point of AB.EQ II AP

therefore Q is the mid point of BP 

Similarity P is the mid-point of DQ

DP = PQ =QB

thereforeLine segment AF and EC trisect the diagonal BD. 

Aftab tell his daughter, 7 years ago, I Was seven times as old as you were then. Also, 3 years from now, I shall be three times as old as you will be represented this situation algebraically and graphically .

 

Let the present age of father be x years and the age of daughter be y years.

7 years ago father's age = (x-7) years 

7 years ago daughter's age = (y -7)years 

According to the question 

(x-7) = 7 ( y-7)

x-7y = -42

After 3 years father's age = (x+3) years

After 3 years  daughter's age = (y +3)years 

According to the condition,

x+3 = 3 (y+3)

x- 3y = 6

To fid the equivent geometric representation we find some points on the line represnting each question, these solutions are given below in the table

From eq : 

x-7y = -42

x 0 7 14 12
y =frac{x+42}{7} 6 7 8 12
Points A B C G

 

 

x 6 12 18 42
y =frac{x-6}{3} 0 2 4 12
Points  D E F G

Plot the points A(0,6), B (7,7) and C(14,8) and join them to get a straight line ABC. Similarly, plot the points D (6,0), E (12,2) and F (18,4) and join them to get a straight line DEF. These two lines intersect at point G (42,12). Thus we conclude that the present ages of afthaf and his daughter are 42 years and 12 years respectively.

A lady went to a bank with Rs. 1,00,000. She asked the cashier to give her Rs.500 and Rs. 1,000 Notes in return. She got 175 currency notes in all. Find the number of each kind of currency notes?

                                                         OR
There are 40 passengers on a bus. Some with Rs.3/- ticket and remaining with Rs. 10/- ticket.

The total collection from these passengers is Rs.295/- Find how many passengers have the ticket worth Rs.3/-

 

Let the no of notes of Rs. 500 = x

x+y = 175 ----------(1)

500 x + 1000 y = 100,000 ............ (2)

Multiplying equation  (1) by 500

500x + 500y = 27500 ..............(3)

Solving equation (2) and (3)  we get

500y = 12, 500

y = 25

Putting this value to equation  (1)

x = 175-25 = 150

OR

There are 40 passengers in a  bus. Some with Rs.3/- ticket and remaining with Rs. 10/- ticket. The total .......from these passengers is Rs. 295/-. Find how many passengers have the ticket worth Rs.3/-

                                                  OR

Let the no of passengers with Rs. 3 ticket is x

Then the no: of passengers with Rs.10 will be (40-x)

According to the question 

3x + 10 (40-x) = 295

3x+ 400- 10x = 295

7x  = 105

x = 15

 

 

 

 

 

Read the following understand the mathematical idea expressed in it and answer the questions that follow : 

1,4,9,16,............ are the squares of the counting numbers. The remainders got by dividing the square numbers with natural numbers  have a cyclic property. For example, the remainders on dividing these numbers by 4 are tabulated here.

Number       1      4      9        16      25      -          -        -

Remainder   1      0      1         0        1       -          -        -

On dividing by 4 perfect squares leave only 0 and 1 as remainders. From this we can conclude that an arithmetic sequence whose terms leaves remainders 2 on divideing by 4do not have a perfect square.

a. Which are the possible remainders on dividing any number with 4?

b. Which are the numbers we would not get on dividing a perfect square by 4?

c. What is the remainder that leaves on dividing the terms of the arithmetic sequence 

    2,5,8,11,............ by 4?

d. Does the arithmetic sequence 3,7,11,..............contain perfect squares?

e. Write a sequence with common differnce 4 which contains many perfect squares?

a. 0, 1 and 3.

b. 2 and  3.

c. 2, 1, 0 and 3.

d. No, 3,7,11,...... is divided by 4 we get the remainder 3. When we dividing the perfect square          by 4, we get the remainder 0 and 1.

e. 4,8,12,16.................

A bag contains 18 balls out of which x balls are red.

i. If one ball is drawn at random from the bag, what is the probability that it is not red?

ii. If 2 more red balls are put in the bag, the of drawing a red ball will be frac{9}{8} times the probability of drawing a red ball in the first case. Find the value of x.

P(red ball) = frac{x}{8}

i) P(no red ball) = 1 - frac{x}{8} = frac{18-x}{18}

ii) Total number of balls  = 18 + 2 = 20

    Red balls are = x  + 2 

    P(red balls) = frac{x+2}{20}

    Now, According to the question, 

    frac{x+2}{20} = frac{9}{8}times frac{x}{18}

    180x = 144x + 288

    36 x = 288

    x = frac{288}{6} = 8

A trader was moving along a road selling eggs. An idler who did not have much  work to do, started to get the trader into  a wordly duel. This grew into a fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke. The trader requested the panchayath to as the idler to pay for broken eggs. The panchayath asked the trader, How many eggs were broken?  He gave the following response: if counted in pairs one will remain; If counted in  3 two will remain; If counted in 4, 3 will remain  if counted 5, 4 will remain; If counted 6,5  will renmain; if counted in 7 nothing all remain  my basket cannot accomodate more than 150 eggs So, 

i. How many eggs were there?

ii. Which Mathematical concept is used to solve the above question?

iii. Which values are hidden in the above question?

i. 

Let the number of eggs = a

If counted in 7, nothing will remain 

a  = 7 p + 0, for som enatural number p.

If counted in  6, 5  will remain for some natural number is q.

a = 6q +  5

If counted in  5, 4 will remain for some natural number w

a = 5w + 4

If counted in 4, 3  will remain, for some natural number s

a = 4s + 3

If counted in  3 , 2  will remain, for some natural number t.

a = 3t + 2

If counted in pairs, one will remain, for some natural number 

a = 2u + 1

That is in each case, we have a and positive ineteger  (b takes a value 7,6,5,4,3 and 2 respectively) which divides 'a' and leaves a remainder r  (in case r is 6,5,4,3, 2 and 1 respectively ), that is smaller than 'b'. 

We must look for the multiple of 7 which satisfy all the conditions. By trial and error (using the consept of LCM ) we will get total number of eggs  = 199 

 

ii. 

Euclid's division lemma  (Real numbers)

 

iii. 

The values of of the trader are honesty and faith in the panchayath system.