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Sample Question Paper Quantitative Solution 4

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
3.
will be released shortly.
4.
Under Construction
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

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1 comment

  1. Biswajit Malakar

    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

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