Qu@ntose

  While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is _________? SOLUTION  -                        Let the other two numbers be x and y.                                              Correct Product = x*y*37                       Incorrect Product = x*y*73                      Difference = 720                 => 73xy - 37xy = 720                 => 36xy = 720                 => xy = 20 -----------------(i)              Now, we know that,                             (x^2 + y^2)/2  ≥  (x^2 * y^2)^(1/2)    [ AM ≥ GM ]                        => (x^2 + y^2)  ≥ 2*(xy)            => (x^2 + y^2) ≥ 2*20                     => (x^2 + y^2) ≥ 40 ANS

F(x) is a fourth order polynomial with integer coefficients and with no common factor. The roots of F(x) are -2,-1,1,2.If P is a prime number greater than 97,then the largest integer that divides F(p) for all the values of p is (a)72 (b)120 (c)240 (d)360 (e)None of these Solution  - F(x) = (x-2)(x-1)(x+1)(x+2) P must be of the form - 6k+1/6k-1 Substituting the value of p(6k-1) in the eqn, (6k-3)(6k-2)(6k)(6k+1) 3*2*6=36  Substituting the value of p(6k+1) in the eqn, (6k-1)(6k)(6k+2)(6k+3) 6*2*3=36 so 72 must be a factor  now options (a),(d) and (e) are left, now they are 5 consecutive numbers with middle term missing, starting with an odd number (6k-1)(6k+2)(6k)(6k+3)or (6k-3)(6k-2)(6k)(6k+1) the number will always contain a multiple of 5 36*2*5=360 ANS

The LCM of two numbers "a" and "b" has eight factors. If the sum of these two numbers is 45. Find the maximum product of  "a" and "b". Solution- a + b = 45  for product to be maximum, a and b must be as close as possible, => a = 22, b = 23 LCM(a,b) = 22*23 = 506 Factors of 506 = 2*11*23 =>(1+1)*(1+1)*(1+1) = 8 Hence maximum value of the product = 22*23 = 506 ANS

Solution:- Profit effect = product of all effects                     ={(10/9)*(10/9)}*{(11/10)*(9/10)} Decrease in 10% while sale * decrease in 10% due to winter * increase in 10% to cheat while purchase* decrease in 10% due to winter  while purchase                   =99/81 Profit%= 18/81=22.22%

When they work alone, B needs 25% more time to finish  a job than A does. They two finish the job in 13 days in  the following manner: A works alone till half the job is  done, then A and B work together for four days, and  finally, B works alone to complete the remaining 5% of  the job. In how many days can B alone finish the entire  job? 1) 22 2) 18 3) 20 4) 16 Solution- Let A does 5x units/day B does 4x units/day Let total work = 100% A does 50% of work alone and B does 5% work alone Remaining Work = 100 - (50+5) = 45% This 45% = 4*(4x+5x) => 4x = 5% =>80x = 100% B will take, 80x/4x = 20 days Ans

A laptop was sold at a profit of 15%. If it was sold at a price that was 10% lower, the profit would have been `1050. What is the cost price of the laptop? (A) `21000 (B) `35000 (C) `30000 (D) `42000 Solution:-                                              

An arithmetic progression consists of an even number of terms. The sum of its odd terms is 50 whereas the sum of its even terms is 56. Find the number of terms in  the series if the last term of the series exceeds the first term by 11.25. Solution- Let the total number of terms = 2n Sum of odd terms = 50 a1 + a3 + a5  + a7 + ... + a(2n-1) = 50 - ------------(1) Sum of even terms = 56 a2 + a4 + a6 + a8 + ....  +a(2n) = 56 -------------(2) Subtract eqn (1) from (2), (a2 - a1) + (a4 - a3) + ...... + (a(2n) - a(2n-1) = 6 =>d + d + .....................+ d = 6 =>n*d = 6----------(3) Now, a1 + (2n-1)d - a1 = 11.25 => (2n-1)d = 11.25--------(4) eqn (3)/(4), n/(2n-1) = 6/(11.25) => n = 8 Number of terms = 2n = 2*8 = 16 ANS

A father distributes 76000 unequally among his 3 sons Ram,Sam and Bam-such that Ram gets 1/3rd of Sam,Sam gets 1/5th of Bam. What amount does Bam get? a)50000 b)60000 c)45000 d)40000 Solution:- Ram- 1/3 of Sam        Sam-1/5 of Bam Lcm of (3 and 5)=15 Now, Bam   -----------Sam-----------------Ram 15----------------1/5*15=3-----------1/3*3=1 Total=15+3+1=19 So Bam gets (15/19*76000)=60000

a)50% b)75% c)37.5% d)25% Solution:- let shirts be s and pants be p If I buy 5 shirts I get 3 pants free 5s + 3p ....(i) If I buy 9 pants I get 3 shirts free 9p + 3s.....(ii) Rajesh buys 15 shirts so he gets 9 pants......from equation (i) Balram buys 27 pants so he gets 9 shirts......from equation(ii) let the cost of 1 pant be Re 1 let the cost of 1 shirt be Re 1 So total cost of Rajesh is 15+9=24 So total cost of Balram is 27+9=36 Rajesh/balram=0.25/discount 24/36=0.25/discount discount=37.5

In the presidency college two candidate contested election. 15% of the voters did not vote and 41 votes were invalid. The elected contestant got 341 votes more than the other candidate. If the elected candidate got 45% of the total eligible votes, which is equal to the no. of all the students of the college. The individual votes of each candidate is: a)2250 and 1936 b)3568 and 3254 c)2442 and 2128 d)2457 and 2143 Solutions:- This kind of sums can be time taking ...but we can solve easily by going through the options.... The winner candidate got 45% of the eligible votes.... So the winner got votes which is a multiple of 9 and 5 Only option (a) satisfies here.... and the difference between winner and looser is 314

Example- Example- https://quantose.blogspot.com/2018/07/dilr-2.html Example- https://quantose.blogspot.com/2018/11/dilr-friends-doubt.html 4 or more than 4 set venn-

Ashwin bought an article at 200 and marked it at 300. He offered a discount and then sold it his  profit/loss percentage and discount percentage are in the ratio 3 : 2. Find his profit/loss percentage. (A) 29% profit  (B) 25% profit (C) 20% loss  (D) 25% loss Solution:- CP=200 MP=300   P%=3X  D%=2X We know that MarkUP% and Discount percetage are successive discounts MArk up%=(100/200)*100=50% Profit %= markup% -Discount % - (markup%*-Discount%/100)     3x          =50             -2x                -(50*-2x/100)      x=25/3% profit  3x=25%

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct? (1) 201 ≤ N ≤ 242. (2) N ≥ 253. (3) N ≤ 200. (4) 243 ≤ N ≤ 252. Solution:- Falls sort by 68%..so pass percent is 32% 36=32 % of Pass Mrks 36=32% of (45% of N) N=250

2/5 of voters promise to vote for P and rest promised to vote for Q.Of these on the last day 15% of the voters went back on their promise to vote for P and 25% of voters went back on their promise to vote for Q and P lost by 2 votes.Then the total number of voter is:- a)100 b)110  c)95  d)90 Solution:- Lets do this in a smart way Always  try to use the options to your own advantage so,we can see that           25 % of voters must be an integer and only 100 satisfy in that

Priyanka, Akshay and Salman started out on a journey to watch the newly released movie "Mujhse Shaadi Karogi" which was being shown at wave cine-multiplex. The multiplex was 120 km away from their starting point of the journey. Priyanka and went by car at the speed of 50 km/he, while Akshay travelled by Tonga at 10 km/hr. After a certain distance, Salman got off and travelled rest of the distance by another Tonga at 10 km/hr, while Priyanka went back for Akshay and reached the destination at the same time that Salman arrived. The number of hours required for the trip was: (a)4 hr (b)5hr (c)4.8 hr (d)cannot be determined Solution-

In La-Liga, a football tournament,  there are 15 teams competing in a round robin format i.e, each team plays all the other teams exactly once. The top two teams during the league will play the finals against each other. After every match, points will be awarded to both the teams based on the following criteria:- 1 win = 3 points 1 draw = 1 point 1 loss = 0 points. When 11 of the fifteen teams were done playing all their matches, the following observations were made about the top five scoring teams at that stage: Teams                     Matches Played         Points Barcelona                        13                        32 Athletico Madrid             12                        29 Sevilla                              12                        23 Real Madrid                    11                        26 Valencia                          14                        28 Further, when all teams have played all their league matches, it is noted that:- (i) Real Madrid scored the highest

What is the remainder when ( 1^6 + 2^6 + 3^6 + 7^6 + ........ + 100^6) is divided by 7? Solution:- We can solve it by Euler Method    

In a division sum, the remainder was 71. With the same divisor but twice the dividend, the remainder is 43. Which one of the following is the divisor? (i)86 (ii)93 (iii)99 (iv)104 Solution- We know that, Dividend = Quotient*Divisor + Remainder   Let the Dividend = D Quotient = Q Divisor = d => D = d*Q + 71----------(i) Also, 2D = d*q + 43-------(ii) Subtracting eqn 2*(i) from (ii), 0 = d*(q - 2Q) - 99 => d*(q - 2Q) = 99 Now, 99 = 1*99 or 3*33 or 9*11 d*(q - 2Q) = 1*99 or 3*33 or 9*11 Checking options we can see only one value of d satisfies, Hence, d = 99 ANS

A shopkeeper uses false weights thereby cheats customer by giving 750 gm instead of 1 kg. He also claims that he sells rice at the cost price. In order to increase his sales he offered 125 gm of rice free with every 1 kg of rice. What is his overall percentage(approximate) profit if a customer buys 1 kg of rice and if the shopkeeper uses false weight to measure 125 gm also? Solution- Let the cost price of 1 gm of rice = Re 1 Cost Price of 750 gm = Rs. 750 Selling Price = Rs. 1000 We know that, SP = CP*(1 + Profit%) He is offering 125 gm for free but actually he is just giving 125*3/4 gms for free => SP = 1000*1 + 125*(3/4)*0 =>CP = 750 + 125*(3/4) => 1000 = (750 + 125*0.75)*(1 + Profit%) => Profit% = 18.51% Ans

A, B, C and D started a business with investments in the ratio 3 : 4 : 5 : 6. As B and C were working partners they were paid equal salaries. The ratio of B’s and C’s total annual income is 9 : 10. If the total annual profit is 84000, find B’s salary . Solution:- Let the investments of A, B, C and D be 3x, 4x, 5x and 6x respectively. Let the salary of B be S. (4x +S)/(5x+S)=9/10                     => S=5x Total amount of profit = 18x + 2S = 28x = 84000        x = 3000 B’s salary is 5x = 5*3000=15000

SOLUTION-

A survey was done in a city by an independent body to check the popularity of the bikes - Hero, Honda, Bajaj, Suzuki, TVS and KTM. Total 6300 people participated in the survey. Each person likes at least one of the six bikes. Further, we know the following:- (i)The number of persons who like exactly one bike to the number of persons who like exactly two bikes to the number of persons who like exactly three bikes to the number of persons who likes exactly four bikes is in the ratio 4:3:2:1, respectively. (ii)The persons who like TVS, like neither Hero nor Suzuki. (iii)The persons who like Honda, like neither KTM nor TVS. (iv)The persons who like KTM, like neither Hero nor TVS. (v)The number of persons who like only one bike is equal for each bike. (vi)The number of persons who like exactly two bikes is equal for each possible combination of only two liked bikes. Similarly, the number of persons who like exactly three bikes is equal for each for each possible combination of only t

Abhishek and Rani started a partnership business by investing Rs.1 cr and Rs.2 cr respectively in the beginning of the year. Each of them withdrew one-fourth of their investment after each quarter till 2nd quarter. At this point Susmita and Randeep joined them with Rs.50 lakh and Rs.1 cr respectively. Randeep withdrew his entire investment after 3 months. If the profit at the end of the year is Rs. 41 lakhs, the combined share of Abhishek and Rani is: (1) Rs.22 lakh (2) Rs.33 lakh (3) Rs.8 lakh (4) Rs.30 lakh Solution:-

In a certain school, there are 300 boys and some girls. Each of them is either short or tall, either overweight or underweight. The number of boys who are tall as well as underweight is maximum possible and is 1/3rd of the students who are tall and underweight. Also, the number of girls who are tall and overweight exceeds the number of boys who are short and underweight by 15. The number of boys who are short and overweight exceeds the number of girls who are short and underweight by 25. Further, the number of girls who are short and overweight exceeds the number of boys who are tall and overweight by 50. Also, each category of students has at least 5 students. 1)What is the minimum possible strength of this school? 2)What is the maximum possible difference between the students who are tall and the students who are overweight? 3)What is the maximum possible difference between the tall underweight girls and short overweight boys? 4)What is the approximate percentage of students wh

If  x^2 - x - 1 = 0 , then find the value of (x^8-1)/(x^3+x^5) a)2 b)3 c)4 d)none Solution:- x^2 = x + 1 x^8 - 1 = (x^2)^4 - 1 = (x + 1)^4 - 1 = x^4 + 4x^3 + 6x^2 + 4x + 1 - 1 = x(x^2 + 2x + 2)(x + 2) = x(3x^2)(x^2 + 1) x^3 + x^5 = x^3(x^2 + 1) => (x^8 - 1)/(x^3 + x^5) = x(3x^2)(x^2 + 1)/x^3(x^2 + 1) = 3

Solution-

Ravi's grandfather, as a part of his will, bequeathed to Ravi a safe, which had exactly four combination locks- L1, L2, L3 and L4. In order to open the safe, Ravi must enter exactly four keys K1, K2, K3 and K4, in each of L1, L2, L3 and L4 respectively. While K1  can be entered first, K2 can be entered only after L1 is unlocked; K3 can be entered only after L2 is unlocked and so on. Each key is a four-digit number (with the first digit at the extreme left and the last digit at the extreme right). while Ravi was not aware of the keys of any locks, also given to him along with the box, was parchment containing the following clues; 1)The first digit of K1 is the same as the last digit of K3. 2)K2 and K4 have the same sets of digits but not necessarily in the same order. 3)The number represented by the last two digits of K3 and the number represented by the first two digits of K2 are both squares of odd numbers other than 1. 4)The number represented by the last two digits of K4 appe

There is a number 49A81B.How many pairs of (a,b) are possible when divided by 66? a)0  b)1 c)2 d)None of these Solution:- B=0,2,4,6,8 ---(1) 22+a+b = 3k ---(2) B-a+12=11k ---(3) For (3) to be true, a,b are consecutive numbers where b<a. So possibilities (a,b)= (2,0)(3,2)(5,4)(7,6)(9,8) Now putting these values in eq (2) you can check. Two pairs satisfy (3,2) and (9,8)

A tank is fitted with two filling pipes and one drain pipe. One filling pipe takes four hours more than the other filling pipe to fill the empty tank. The drain pipe can empty a full tank in a time which is double the time taken by the faster filling pipe to fill the tank. How much time is taken by the drain pipe to empty the full tank, if it takes 48/7 hours to fill the empty tank when all the three pipes are functioning simultaneously? Solution- Let the time taken by pipe 1 = x hours Time taken by pipe 2 = x + 4 hours Time taken by drain pipe = 2x hours Approach-1:- Let the capacity of the tank = 2*x*(x+4) Rate at which pipe 1 will fill tank = 2x(x+4)/x = 2(x+4)---------(i) Rate at which pipe 2 will fill tank = 2x(x+4)/(x+4) = 2x------(ii) Rate at which drain pipe will empty tank = 2x(x+4)/(x+4) = (x+4)-------(iii) Now, 48/7*[2(x+4) + 2x - (x+4)] = 2x(x+4)  =>24[3x + 4] = 7x^2 + 28x  =>72x + 96 = 7x^2 + 28x  =>7x^2 - 44x - 96 = 0  =>(x - 8)(7x + 12) = 0  =>x = 8 Ti

If log (20) = A ; log (45) = B ; log (24) = C, what is the value of  log (120)? (a) (3A + 2C + B)/4 (b) (3B + 2A + C)/4 (c) (3C + 2B + A)/4 (d) (3B - 2A + C)/4 Solution- log20 = log 4*5 = log 4 + log 5 = 2log 2 + log 5 = A ---------(i) Similarly log 45 = log 9 + log 5 = 2log 3 + log 5 = B--------(ii) log 24 = log 8 + log 3 = 3log 2 + log 3 = C ---------(iii) Now, log 120 = log (3*5*8) = 3log 2 + log 3 + log 5 Now, all the options are divided by 4,so let's multiply (log 120) by 4 we get, 4* log 120 = 12*log 2 + 4*log 3 + 4*log 5-------(iv) log 3 is in eqn (ii) and (iii), so if we multiply eqn (iii) by 2 and add eqn (ii),we get, 2log 3 + log 5  + 6log 2 + 2log 3 = B + 2C-------(v) =>6log 2+ 4log 3 + log 5 = B + 2C Now if we add 6log 2 + 3log 5, we will get the required result Multiply eqn (i) by 3, 6log 2 + 3log 5 = 3A----------(vi)  Adding eqn(v) and (vi), we get 4* log 120 = 12*log 2 + 4*log 3 + 4*log 5 = 3A + 2C + B Hence, option (a) is the answer.

1)What is the difference between the maximum and minimum possible scores of B? (a)10 (b)15 (c)20 (d)25 2)How many points did E scored in the third round? (a)10 (b)15 (c)20 (d)25 3)Who scored the third highest score in the fourth round? (a)A (b)B (c)C (d)D 4)The scores of how many players in all the rounds could be uniquely determined? (a)2 (b)3 (c)4 (d)5 Solution- As we have completed the table, let's answer the questions 1) Maximum score of B = 80     Minimum score of B = 65     Difference = 80 - 65 = 15 ANS 2)10 points ANS 3) C scored the third highest points in the fourth round 4) A,C,D and E i.e, 4 ANS

All the students of XYZ college are given an option to choose any combination of the subjects from among History, Polity, Geography and Public Administration(PA). It is also known that: 1) A student who wants to choose History can also choose polity or Geography. 2) Out of the possible six combinations of exactly two subjects, the combination having History and PA, PA and Geography cannot be chosen by any student. Each of the remaining combinations of two subjects is chosen by the same number of students. 3) No student can pick a combination having History, Polity and PA. 4) A student who wants to choose PA and Geography must have to choose History. 5) It is known that each of the subjects is chosen by exactly 200 students out of which 12.5% choose only that subject. 1) What is the number of students who choose all four subjects? 2) If 'x' students choose only Geography and 'y' students choose exactly 2 other subjects with geography, then find the value of  "x +

If eggs are removed from a basket two, three, four, five, six at a time, there remain respectively one, two, three, four and five eggs respectively. But if eggs are removed seven at a time, no eggs remain. What is the least number of eggs that could have been in the basket? Solution- Number of eggs in the basket = 2a+1 = 3b+2 =4c+3 = 5d+4 = 6e+5=7k Now,common difference among the remainders is 1, So, number will be of the form = LCM(2,3,4,5,6) - Common Difference                                                      = 60p - 1 Now, we have to find the least number, (60p -1) must be divisible by 7 for p = 2 60p-1 = 119 is divisible by 7 Hence,119 is the least number of eggs that could have been in the basket.

What is the remainder when 1!+2!+3!+4!.....+15! is divided by 6? Solution- N = 1! + 2! +3! + 4! +........+ 15! Factorising 6, we get 1*2*3 Now from 3! onwards, every term has at least a "2" and a "3", hence divisible by 6. We can write N as 6k + 1! + 2! = 6k + 3 So the remainder when N is divided by 6 is 3 ANS

DIRECTIONS for questions 1 to 4: Answer these questions on the basis of the information given below. In a particular year, a group of 300 people visited some of the countries among China, Sri Lanka, Bhutan, Bangladesh and Pakistan. Further, each person visited at least one country. Any person who visited China also visited Bhutan, while any person who visited Pakistan also visited Bangladesh. Any person who visited Sri Lanka also visited China, while no person who visited Bangladesh visited China. It is also known that 1. the number of persons who visited Sri Lanka was twice the number of persons who visited only Bangladesh. 2. the number of persons who visited only Bhutan was twenty less than the number of persons who visited exactly one country. 3. the number of persons who visited Pakistan was three more than the number of persons who visited China. 4. the number of persons who visited at least three countries was 145. 5. the number of persons who visited China was ninety more th

Aman reaches school every day at 5pm to pick up his children. On a Saturday, the school got over at 4pm and the children started walking towards home. Aman met them on their way and returned home 30 min earlier. How long did the children walk? Solution- Let the meeting point be A. H----------A-------------S As Aman is reaching home 30 min earlier, it means these 30 min are saved by not travelling the distance 'AS' twice. Time taken by Aman to travel distance AS = 30/2 = 15 min As Aman reaches school at 5pm and from A it takes 15 min to reach the point S,that means at 4:45 Aman is at point A. Hence, the children must have walked for 45 min ANS

If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7^m + 7^n is divisible by 5 is: (a)1/5 (b)1/4 (c)1/7 (d)1/49 Solution- (7^m + 7^n) mod 5 = 2^m +2^n Now, the number to be divisible by 5, it must end in 5 or 0. 2^m + 2^n can never end in 5 as it is an even number (minimum value of m,n is 1), so the number must end in 0 if it has to be divisible by 5. Cyclicity of 2 is 4 (2,4,8,6), 2+8 or 6+4 are the only cases when 2^m +2^n is divisible by 5. last digit to be 2 and 8, 2^x must be of the form 2^(k+1) and 2^(k+3) respectively Similarly, last digit to be 4 and 6 it must be of the form 2^(k+2) and 2^(k+4) respectively Now in both of these cases we will have 25 pairs each, Required cases = (25*25 + 25*25)*2  [25*25 +25*25 for both cases and multiplied by 2 as m = 1, n=3 or n=1 or m =3 are both valid cases)  Probability = 25*25*2*2/(100*100) = 1/4 ANS

How much time will an express train of length 150 meters and running at a speed of 75 kmph take to cross a man walking at 5 m/s inside a passenger train of length 200 meters and running at 15 kmph in a direction opposite to that of the express train. The man is walking inside the train and in opposite direction to that of the train. (Assume the man does not reach the end of his train meanwhile.) Solution- Speed of man with respect to passenger train = 5m/sec  or 18km/hr Speed of man with respect to ground = 18 - 15 = 3km/hr Speed of man with respect to express train = 75 - 3 = 72 km/hr or 20m/sec Time taken by express train to cross man = 150/20 = 7.5 sec ANS

                                           Solutions:-

Tony and Harry begin to run in opposite directions on a circular path of radius 35 m at 20 m/sec and 11 m/sec respectively from the same point. What is the time taken by them to meet for the third time at the starting point? (A) 11 min (B) 220 sec (C) 8 min 40sec (D) 325 sec Solution:- Radius = 35 m. Circumference = 2 x 22/7 x 35 = 220 m Time taken for them to meet for the first time at the starting point is LCM [220/20, 220/11] = LCM [11, 20] = 220sec To meet for the third time at the staring point from the start, they need 220 x 3 = 660 sec = 11 minutes

Q1)what is the number of children in the age group 12-15 who like basketball? Q2)If there are 12 boys in the age group 16-18,then how many boys are there in the age group 12-15? Q3)How many boys in England like Football? Q4)What is the number of English boys who like Tennis? Solutions:-

Three children, Rekha, Shankar and Mukesh, were playing the game of Tic-Tac-Toe, on a grid of 3 × 3 squares. Each game was played by all the three children, with the rules slightly modified from the usual, as described below. Each person has his/her own symbol, referred to as his/her 'mark' and in every turn, he/she places the same mark on the grid. The columns are labelled A to C, from left to right, and the rows are numbered 1 to 3, from top to bottom. The position of a mark on the grid is given by the combination of the column and row labels of the square in which it is placed. For example, a mark in position C2 means that the mark is in column C and row 2. The first person to play a game can place his mark on any square in the 3 × 3 grid except B2. In every subsequent turn, each person has to place his/her mark in any empty square, which is not in the same row or column as that of the mark that was placed immediately before his turn not in any diag