1. Given, P={2, 3, 4,….. 100} Q={101, 102, 103, ………200} To calculate the number of elements of Q which do not have any element of P as factor i.e. we have to find all the prime numbers from 101 to 200 Total number of prime numbers between 101 and 200 is 21. ( D) |

2. Firstly we have to find the pattern of numbers for calculating the sum. If we analyse the question for getting a pattern, it is clearly mentioned that number will give a remainder of 6 when divided by 8 i.e. 8k+6 (k is a constant) First number = 14 Second number = 22 Last term (two digit)= 94 It will form an arithmetic progression with first term= 14 and common difference=8. Sum of first 11 terms = 594. ( B)Reference : https://en.wikipedia.org/wiki/Arithmetic_progression |

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5. Under construction |

7. As we have to form the number between 3000 and 9000. Thousand’s place can be filled in 3 ways. (3, 5, 7) Hundred’s place can be filled in 4 ways (repetition is not allowed) Tens place can be filled in 3 ways.(repetition is not allowed) Unit Place can be filled in 2 ways. Total number of possible ways= 3 x 4 x 3 x 2= 72 ( D) |

8. Let the distance between the point P and Q= a When B overtakes A for the first time, both of them cover 3a/10. When B meets A after that, it (B) covers (7a/10+7a/30) or 28a/30, while A covers (23a/30 – 9a/30) or 14a/30. Therefore, B is twice as fast as A. A starts 1 H after B, it catches up with in 1 H. Therefore, covers 0.3a in 1 H. or x in 10/3H. ( D) |

9. Given equation is ||x|-3|<2 and ||x|-2|<3 Let |x|=p (p>=0 i.e. value of p cannot be negative) —- (1) Now we have to put the values in the given equation Then, |p-3|<2 and |p-2|<3 For |p-3|<2 Possible value of p (1<p<5) For |p-2|<3 Possible value of p(-1<p<5) Considering them as equation (2) and (3) i.e. 1<p<5 — (2) and -1 <p<5— (3) The conditions (1), (2) and (3) are satisfied by 1 < p < 5. i.e. -5 < x < -1 or 1 < x < 5. Therefore the x belongs to (-5, -1)U(1, 5) ( B)Reference : https://en.wikipedia.org/wiki/Absolute_value |

10. Let the taxi number will be x The smallest number divisible by 2, 3, 4, 5 and 6 is their LCM ( 2, 3, 4, 5, 6)=60 As, After dividing the x by ( 2, 3, 4, 5, 6) remainder will be 1. Hence taxi number will follow the pattern as 60k+1 Now we will use option to find the required taxi number. Only option (a) and (d) satisfies the above criteria. Further, we have to choose the smallest value. Hence ( A)Reference : https://en.wikipedia.org/wiki/Least_common_multiple |

Sample Question Paper Quantitative 04

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● Number System |

● Permutations and Combinations |

● Data Sufficiency Problems |

● Cryptarithmetic Multiplication Problems |

Solution of question number 6.

We have to find out Rem[2^1040/131]

As per Fermat theorem we know that a^p-1/p=1

where p is prime number and HCF of(a,p)=1;

Rem[2^130/131]=1

=> Rem [(2^130)^8/1131]=1

=>Rem [2^1040/131]=1

Solution to question number 4:

https://www.quora.com/P-is-a-prime-and-m-is-a-positive-integer-How-many-solutions-exist-for-the-equation-p-6-p-m-2%2Bm%2B6-p-1/answer/Prashant-Pokhriyal-4?srid=JfXg

please review the answer of question no.8

It clearly states that “car B reverses and meets car A, after covering 70/3 of the distance QP”.Then how could B travel a distance of 7a/30.It should be 70a/3