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After successfully completing the CAT 2016 sprint series and the SNAP 2016 sprint series, we are back with the XAT 2017 sprint preparation series – Quant 10 to boost your prep. This series will consist of 10 sets of questions from past year XAT papers, leading to XAT 2017 and covered almost all the question types that you needed to know come the 8th of January.

# XAT 2017 sprint preparation series – Quant 10

1. There are two circles C1, and C2 of radii 3 and 8 units respectively. The common internal tangent, T, touches the circles at points P1 and P2 respectively. The line joining the centers of the circles intersects T at The distance of X from the center of the smaller circle is 5 units. What is the length of the line segment P1P2?

A. ≤13
B. >13 and ≤14
C. >14 and ≤15
D. >15 and ≤16
E. >16

2. Consider the formula, S = (a * v) / (t + r + v) , where all the parameters are positive integers. If v is increased and a, t and r are kept constant, then S:

A. increases
B. decreases
C. increases and then decreases
D. decreases and then increases
E. cannot be determined

3. Prof Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90. Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he must score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?

A. 6
B. 7
C. 8
D. 9
E. None of the above

4. A polynomial “ax3 + bx2 + cx + d” intersects x-axis at 1 and -1, and y-axis at 2. The value of b is:

A. -2
B.  0
C. 1
D. 2
E. Cannot be determined

5. The probability that a randomly chosen positive divisor of 1029 is an integer multiple of 1023 is: a2/b2, then ‘b – a’ would be:

A. 8
B. 15
C. 21
D. 23
E. 45

6. Circle C, has a radius of 3 units. The line segment PQ is the only diameter of the circle which is parallel to the X axis. P and Q are points on curves given by the equations y+ax and y=2ax respectively, where a < 1. The value of a is:

A. 1/6√2
B. 1/6√3
C. 1/3√6
D. 1/√6
E. None of these

7.Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. The diagonal AC intersects the line segments PR and QR at M and N respectively. What is the area of the quadrilateral PQMN?

A. > 9.5 and ≤ 10
B. > 10 and ≤ 10.5
C. > 10.5  and  ≤ 11
D. > 11 and ≤ 11.5
E. > 11.5

8.Two numbers, 297B and 792B, belong to base B number system. If the first number is a factor of the second number then the value of B is:

A.11
B.12
C.15
D.17
E.19

9.A teacher noticed a strange distribution of marks in the exam. There were only three distinct scores: 6, 8 and 20. The mode of the distribution was 8.The sum of the scores of all the students was 504. The number of students in the in most populated category was equal to the sum of the number of students with lowest score and twice the number of students with the highest score. The total number of students in the class was:

A.50
B.51
C.53
D.56
E.57

10. Read the following instruction carefully and answer the question that follows:

Expression  can also be written as x/13!

What would be the remainder if x is divided by 11?

A.2
B.4
C.7
D.9
E.None of the above

1.A. You will get two right angled triangles in this one. Also they will be similar to each other. Easy question.
2.A. Put values and try this one.
3.D. 90n = 97 + (n – 1)k

87n = 70 + (n – 1)k

n = 9.
4.A. a + b + c + d = 0 and – a + b – c + d = 0

2(b + d) = 0. Also, d = 2 and so, b = -2.
5.D. 10^29 has 900 divisors. To be a multiple of 10^23, the number can have powers of 2 and 5 ranging from 23 till 29. So, 49 cases. So, 49/900 = 16/225, a/b = 4/15.
6.A.
7.D. This one is better left alone simply because of the complexity in the figure involved.
8.E.
9.E. 6a + 8b + 20c = 304, b = a + 2c. a = 18, c = 7 and b = 32.
10.D. will be 13!/1 + 13!/2 + 13!/3…. 13!/13 and so, except the case when the term is 13!/11 the others will be divisible by 11. So, 13*12*10! mod 11 is what we have to find out. 10! mod 11 is 10 and so, we get 10*1*2 = 20 mod 11 i.e. 9.

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You can follow the entire sprint series here: XAT 2017 Sprint Preparation Series by Learningroots