# 1
| |||

| |||

Quantitative Aptitude For Elitmus |

Similar Threads | ||||

Thread | Thread Starter | Forum | Replies | Last Post |

IBPS Specialist Officer Quantitative Aptitude Paper | Unregistered | General Discussion | 1 | May 30th, 2016 02:33 PM |

Quantitative aptitude Sample question paper | Unregistered | General Discussion | 1 | February 10th, 2016 06:42 PM |

Quantitative Aptitude for IIFT | Unregistered | General Discussion | 1 | January 19th, 2016 01:34 PM |

Quantitative aptitude for SBI clerk exam | Unregistered | General Discussion | 0 | June 19th, 2015 02:51 PM |

CA CPT Quantitative Aptitude Video Lectures | Unregistered | General Discussion | 0 | June 18th, 2015 12:48 PM |

Tips to solve quantitative aptitude questions | Unregistered | General Discussion | 2 | August 14th, 2014 04:55 PM |

Bank PO Books for Quantitative Aptitude | Unregistered | General Discussion | 1 | July 18th, 2014 09:36 AM |

CAT Quantitative Aptitude Books | Unregistered | General Discussion | 1 | July 18th, 2014 09:29 AM |

List of quantitative aptitude shortcut methods | Unregistered | General Discussion | 1 | July 10th, 2014 05:37 PM |

EIL Management Trainee Quantitative Aptitude Question Papers | Unregistered | General Discussion | 1 | July 10th, 2014 01:15 PM |

Quantitative Aptitude solved Questions | Unregistered | General Discussion | 1 | July 10th, 2014 09:02 AM |

SBI Clerks General Quantitative Aptitude Exam questions | Unregistered | General Discussion | 1 | July 1st, 2014 01:17 PM |

How to Solve Quantitative Aptitude Section | Unregistered | General Discussion | 1 | June 24th, 2014 05:50 PM |

Books on Quantitative Aptitude for Bank Exam | Unregistered | General Discussion | 1 | June 22nd, 2014 02:26 PM |

R.S Agarwals Quantitative Aptitude book | Unregistered | General Discussion | 1 | June 21st, 2014 12:14 PM |

# 2
| |||

| |||

Re: Quantitative Aptitude For Elitmus
eLitmus Evaluation Private Limited was founded in the year 2005 by former employees of Infosys About the eLitmus PH test pH Test or Hiring Potential Test is an assessment exam conducted by eLitmus. Companies use pH Test as a criterion for hiring employees. The test consists of three parts- the Quantitative section, Verbal section Reasoning section The Sample questions of the Quantitative Section eLitmus PH test is given below . If v,w,x,y,z are non negative integers each less than 11, then how many distinct combinations are possible of(v,w,x,y,z) which satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z = 151001 ? Ans : Changing 151001 to base 11 number we get, it will be A34A4 i.e. 10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4 12. In a certain examination paper there are n questions. For j=1,2,3,.....n, there are 2^(n-1) students whoanswered j or more question wrongly. If the total number of wrong answers is 4096 then the value of n is a)13 b)11 c)10 d)9 Ans Given that, 2^(n-1) = 4096 = 2^12 i.e)n-1=12 ==> n=13 How many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at it's unit's place? Ans As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3. The numbers ending with digit 1 will be divisible 1. The numbers ending with digit 2 will be divisible 2. The numbers ending with digit 3 are divisible 3. The numbers ending with digit 5 will be divisible 5. The numbers ending with digit 6 will be divisible 6. Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4. The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4) so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648 . A natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have? Ans Consider the no is 512, the divisors of this no. are 1,2,4,8,16,32,64,128,256,512; so only one prime factors are there i.e. 2; if the no is 48 the divisors are 1,2,3,4,6,8,12,16,24,48; so the prime factors are 2,3; 15. If m and n are two positive integers, then what is the value of mn? Given: (1)7m + 5n= 29 (2) m + n= 5 Ans 7m + 5n = 29 m + n = 5(multiply both sides by 7) subtract equation second from first,we get 7m + 5n =29 7m + 7n = 35 n = 3 substitute value of n in any equation. m = 2 mn = 6. 16. A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can this natural number have? A. Either 1 or 2 B. Either 1 or 3 C. Either 2 or 3 D. Either 1,2 or 3 Solution: Ans) Either 1 or 2 Check on 29 i.e 512 , 39, 59 which have only 1 prime factor and 80 , 48 which are having 2 prime factors and total of 10 divisors. Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have minimum 8 such factors with xyz as the nos or the factor of the nos. When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then we will get at least 12 divisors. So we donâ€™t get 3 prime factors with 10 divisors. 17. What is the remainder when 128^1000 is divided by 153? 128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153 = [(2*3*17+1)^125]mod153 At this point,observe that 153=17*(3^2); Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153. Actually,the above line can be written since only except the last two terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e, [153*68] as one of its factors. Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1; [125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153); which means [125*2*3*17 + 1] = 83*153 + 52; which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ; So, remainder is 52. 18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6? log(abc)=6 abc=10^6 because of b^2=ac=100, b^3=10^6, b=10^2=100 and the combinations are (2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500). 19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48times^2 is divided by 9? First, let us consider a general case : (222222222222.....{2 is repeated n times})^2 =[2(111111111111........{1 is repeated n times})]^2 =4(111111111111........{1 is repeated n times})^2 =4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)] Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times); Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer]. Now, looking at the [(eqn1)] , we can find that here k =(n-1), (n-2), (n-3),....,1. Now,for 10^(n-1) = 9*(a1) + 1; 10^(n-2) = 9*(a2) + 1; 10^(n-3) = 9*(a3) + 1; 10^(n-4) = 9*(a4) + 1; . . . . 10^1 =9 +1; 1 = 1 So,now in [(eqn1)] , we can write (222222222222.....{2 is repeated n times})^2 =4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2 =4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer. =4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)] Now this a general expression for (222222222222.....{2 is repeated n times})^2. For the given problem,we can find that n=1,2,3,4,5,6,....48. In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have to consider the last term,i.e, 4*(n^2); Summing up [4*(n^2)] for n=1,2,...48. we get 4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)} =4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6] =4*8*49*97 =4(9- 1)(9*5 +4)(9*11 -2) =9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important =9*B + 32 =9*B + 9*3 + 5; So, clearly the remainder will be 5. 20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 givesremainder 2? for 5 7 12 17.........197....39 terms sum=3978 for 7 9 16 23 30....... 198...28 terms sum =2848 but some terms are counted more than 1 ..ie which are common in 5 and 7 eg 37 72 107 142 177..sum=535 so 6876-535=6341 |