Qu@ntose

Mr. Trump has a number of coins with him. Each coin has two sides, named ‘H’ and ‘T’. He plays a game, which begins with a number of coins all facing either ‘H’ or ‘T’. If the coins face ‘H’ initially, the target of the game is to make all the coins face ‘T’ at the end. Similarly, if the coins face ‘T’ initially, the target of the game is to make all the coins face ‘H’ at the end. The game consists of a number of moves. In each move, Mr. Trump has to flip exactly three coins (from ‘H’ to ‘T’ or from ‘T’ to ‘H’, as the case might be). 1) Mr. Trump begins the game with four coins all facing "T". What is the minimum number of moves he need to meet the target of the game? (a) 2    (b) 3    (c) 4      (d) More than 4 2)Mr. Trump begins the game with five coins,all facing "H".What is the minimum number of moves he need to meet the target of the game? (Write "0" if the target cannot be achieved). 3) ADDITIONAL INFORMATION(for question 3 and 4):-  Mr Trump now f

Find the sum of the series- 7*11 + 11*15 + 15*19 + 19*23 +........+ 95*99 a)80707 b)80705 c)80703 d)80701 SOLUTION- 7*11 + 11*15 + 15*19 + 19*23 +........+ 95*99 General term = (4n+3)(4n+7) Sum=Σ(4n+3)(4n+7)        => Σ(16n^2 + 28n +12n + 21)        => Σ(16n^2 + 40n + 21)        =>16*(n)(n+1)(2n+1)/6 +40(n)(n+1)/2 +21n ------------(i) Number of terms = (99-11)/4 + 1 = 23  Sum=(16*23*24*47)/6 + (20*23*24) +21*23          = 69184 + 11040 + 483 =80707 ANS Approach-2 Last digit of first term = 7 Last digit of the second term =5 Last digit of third term = 5 Last digit of fourth term = 7 Last digit of fifth term = 1 Now from the sixth term, the cycle repeats, Sum of last digits of five terms =7+5+5+7+1=25 Now, w e know that there are total 23 terms in the series 5*4+3,i.e, the sum of last digits of first 20 terms 5*4=0 Now the sum of the last three terms,7+5+5=7 Hence, the last digit of require sum is 7, only option (A) satisfies.

Find the roots of the equation  x^4 - 10x^3 + 5x^2 + 100x + 100 = 0 ? SOLUTION- x^4 -10x^3 +5x^2 +100x +100 = 0 Dividing both sides by x^2, we get, => x^2 -(10*x) + 5 + (100/x) +(100/x^2) = 0 => x^2 + (10/x)^2 + 5 + (100/x) - (10*x) = 0 => (x- 10/x)^2 + 20 + 5 - 10( x - 10/x) = 0 Now, let (x - 10/x) = K => K^2 + 25 -10K = 0 => (K-5)^2 = 0 => K=5 Now substituting the value of K, we get,       x - 10/x = 5 => x^2 - 5x - 10 = 0 => x = [5 +√(25+40)]/2 and  x = [5-√(25+40)]/2 => x = [5 + √65]/2 and  x = [5-√(65)]/2    ANS Solution credit - Atyant Yadav One can also solve it by breaking this eqaution into two quadratic but this approach is more feasible.

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A number N, when divided by a divisor D, gives a remainder of 52. The number 5N, when divided by D, gives a remainder of 4.How many values of D are possible? SOLUTION- We know that, Dividend = Divisor*Quotient + Remainder So, a number N will be of a type, N=D*k + 52 Now, from the above equation, we can say that D>52 Now, when 5N is divided by D, i.e, 5N = 5(D*k+52) = Dk 1  + 4 =>5Dk + 260 = Dk 1  + 4 Now, 5Dk is divisible by D and when 260 is divided by D remainder is 4, =>5Dk+256=Dk 1 =>5k + 256/D=k 1 So, D must be 256 or its factors as k 1  is always a naturak number, Factors of 256 = 2^8 Now, 52<D<=256 (where D is a natural number) Values of D satisfying are 64,128,256 Hence, 3 values of D are possible ANS

In  a  certain  series , the  nth  term  T n   equals   4T n+1   + n-1.  If  T 1  = 4 ,then   find  the  value  of  T 200. Solution- T 1 =4 T 2  = 4T 1  + (2-1) = 4^2 +1 T 3  = 4T 2  + (3-1) = 4^3 +4+2 T 4 = 4T 3  + (4-1) = 4^4 +4^2 +2*4 +3 T 5  = 4T 4  + (5-1) = 4^5 +4^3 +2*4^2 +3*4 +4 .... ...... ......... ........... T 200  = 4^200 + 4^198 +2*4^197 +3*4^196+......+198*4 +199        =4^200 + S n Now, S n =4^198 +2*4^197 +3*4^196 +.......+198*4 + 199------(i) S n  /4 = 4^197 +2*4^196 +3*4^195+..........+198 + 199/4--------(ii)  Eqn (i)-(ii),  3S n /4  = 4^198 +4^197 +4^196+.........+4+1-199/4   3S n /4  = (4^199 - 1)/(4-1) - 199/4  3S n /4  = (4^199 - 1)/3 - 199/4       S n  =(4^200-4)/9 -3*199/9       S n   =(4^200 - 4-597)/9       S n   = (4^200-601)/9 Substituting this value,we get, T 200  = 4^200 +(4^200 - 601)/9 T 200  = (10*4^200 - 601)/9 ANS

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How many integral values can the expression (15x^2+2x+1)/(x^2-2x-1) not take? Solution- As we have to find the values that the equation cannot take,we can find range of possible values that equation will attain and then remove this range from R(R->Real numbers) =>(15x^2+2x+1)/(x^2-2x-1) = y =>(15x^2+2x+1)=y*(x^2-2x-1) =>(15x^2+2x+1)=y*x^2-2yx-y =>(15-y)x^2 +2*(y+1)*x + (1+y) = 0 Now for x to be real discriminant must be >=0 So, =>(2^2)*(y+1)^2 - 4(15-y)(y+1) >=0 =>(y+1)[y+1-15+y]  >=0 =>(y+1)(y-7) >=0 So, y = (-inf,-1] u [7,inf) So the given equation will not take integral values between (-1,7) i,e; 0,1,2,3,4,5,6 Hence,the given equation cannot take 7 integral values. 

A party was organized in a college in which three types of cold drinks were served. The served cold drinks were Pepsi, Coke and Sprite. There were 500 students in the college but 10% students did not drink any of the three cold drinks. A total of 750 bottles of cold drinks were served at the party. It is known that cold drinks were served in bottles only and no student drank more than one bottle of the same type of cold drink. The additional information is given below: (i) The number of students who drank Pepsi and Coke both was equal to the number of students who drank Pepsi and Sprite both. (ii) The number of students who drank Coke and Sprite but not Pepsi was twice the number of students who drank all the three cold drinks. (iii) The number of students who drank Pepsi and Sprite but not Coke was more than 50. (iv) The number of students who drank all the three cold drinks was at least 1. #Q1 If the difference between the numbers of students who drank Pepsi and Sprite but n

abc  is a three-digit number having 5 factors ,what would be the number of factors for abcabc and find the value of abc ? Solution- abcabc=abc*(1001) abcabc = abc*(7*11*13) As abc has odd factors abc must be  (prime)^4, where prime number must be less than 7, only such prime number is 5. abc = (5^4)=625 ANS Numbers of factors of abcabc = 5*2*2*2= 40 ANS 

 N filling pipes, P 1 , P 2 , P 3 ,.....P n (where n>10)  attached to a tank.The rate of filling of P i , where  i >1, is equal to thrice the combined rate of filling of all the lower numbered pipes. If the time taken by P n-3 to fill the tank is 128 minutes, find the time (in minutes) taken by P n   to fill the tank. Solution- It is given that, R i  = 3*(R 1  +R 2  +R 3  +...+R i-1 ) Where R i   is the rate of ith pipe. Now, T n-3  = 128 min R n-3 = 3*(R 1  +R 2  +R 3  +....R n-4 ) Let R 1  +R 2  +R 3  +...+R n-4 = x R n-3 = 3x R n-2  = 3*(x+R n-3 ) = 12x R n-2 = 3*(x + R n-3 + R n-2 ) = 48 R n = 3*(x + R n-3   + R n-2 +R n-1 ) = 192x Now we know that, rate*time=work R n-3  *time = work (constant) 3x*128 = work Now, for pipe P n , R n  *time = work 192x*time = 3x*128 time = (3x*128)/192x time = 2 minutes ANS

The currency of the country Banama is Rupas (Rp). There are exactly five denominations in which the currency is available – Rp 2, Rp 3, Rp 7, Rp 17 and Rp 19. Each of five persons, A through E, had notes of exactly one denomination with him and no two persons among the five had notes of the same denomination. On a particular day, the five persons visited a shop and each of them purchased a different product. The amounts that A through E paid the shopkeeper were Rp 102, Rp 357, Rp 399, Rp 238 and Rp 147, in that order. Further, the number of notes that each person gave the shopkeeper was distinct. However, the amount that each person paid the shopkeeper was more than the price of the product that he purchased. Therefore, the shopkeeper returned the excess amounts to each of A through E in the form of 2 notes, 4 notes, 1 note, 2 notes and 1 note respectively, but in denomination(s) different from that in which the person had paid the shopkeeper. Q1)What is the total numb

INTRODUCTION-                                  Let's go through a small story to understand all (most😛😛) the terminologies and concepts related to PROFIT-LOSS. Honey( a fashion freak) goes to a local shop to buy a T-shirt. The price of the T-shirt is Rs.500. After an altercation with the shopkeeper, Honey was able to buy that T-shirt for Rs.400. Although shopkeeper reduced the price he was able to save Rs.100 by selling that T-shirt to Honey. TERMINOLOGIES AND CONCEPTS- 1)Selling Price= Rs.400    The price at which an article is sold. 2)Cost Price = Rs.300    The actual price of the article. 3)Marked Price(Tag Price) = Rs.500    The price increased by shopkeepers to earn an extra profit is known as MARK UP(Rs.100 in this case). 4)Profit = Rs.100    Any extra money that the seller gets while selling i.e;   From Cost price to Selling Price;   Selling Price - Cost Price = Profit 5)Loss = Any extra money that seller loses while selling like selling an article that costs him Rs100

INTRODUCTION-                                   A percentage  is a number or ratio  expressed as a fraction  of 100. The sign for percent is  " %". Let's take an example. Ram scores 60 marks in a test whose maximum marks is 100 whereas Shyam scores 70 marks in a test of 140 marks.Who scored better between the two? Total maximum marks in both the cases are different so we cannot give a verdict about their performances. That's where percentage comes into the picture. Ram scored 60 out of 100 i.e; (60*100)/100 %=60% Shyam scored 70 out of 140 i.e; (70*100)/100 %=50% Now we can tell who scored better among the two i.e; RAM. From the above example, we deduce that the one use of percentage is the comparison. Let's take an example. B’s salary is 25% more than A’s salary. By what percent is A’s salary less than B’s salary? This question can be solved by 3 methods.Choose wisely😛  METHOD 1- Let A's salary be x B's salary= x+ 25%x = x+ 25x/100 =1.25 x D

INTRODUCTION-               Suppose Champu is given a task of constructing a building of 30 floors. He can construct 2 floors in a day. In how many days he will complete his task? Seems Simple I guess. Yes, 15 Days. Most of you have got it right. As he is building 2 floors in a day, in 15 days he will complete his task. Now, does it conclude anything????????? 30(Task) = 2(Work completed per day)*15(Number of days it took to complete the task) This takes us towards a formula; WORK DONE=WORK RATE * TIME BUT DO WE REQUIRE THIS FORMULA?????????????? Let's try a question. Question- Ram can do a piece of work in 10 days and his friend Shyam can do the same piece of work in 20 days.How long will it take for both of them to complete the work? General Method-                   Ram's 1-day work(work rate)=1/10 Shyam's 1-day work= 1/20 Now the question is why 1/10 or 1/20? Why we have taken Work Done to be 1 unit? So, it's just for simplification.