How to Prepare for Quantitative Aptitude for CAT (2014)
SOLVED PAPERS
XAT 2014
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Note: XAT 2014 had 31 questions on the Quantitative Aptitude section. Of these there were 21 questions directly on Quantitative Aptitude and 9 questions were on Data Interpretation. In this book, the paper consists of the 22 questions from QA part. The 9 question on DI for XAT 2014 can be found in my book—How to Prepare for Data Interpretation for the CAT also published by McGraw Hill. |
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1. |
x, 17, 3x – y2 – 2, and 3x + y2 – 30, are four consecutive terms of an increasing arithmetic sequence. The sum of the four numbers is divisible by: |
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(a) 2 |
(b) 3 |
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(c) 5 |
(d) 7 |
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(e) 11 |
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2. |
In quadrilateral PQRS, PQ = 5 units, QR = 17 units, RS = 5 units, and PS = 9 units. The length of the diagonal QS can be: |
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(a) > 10 and < 12 |
(b) > 12 and < 14 |
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(c) > 14 and < 16 |
(d) > 16 and < 18 |
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(e) Cannot be determined |
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3. |
Consider the formula, |
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, where all the parameters are positive integers. If w is increased and a, t and r are kept constant, then S: |
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(a) increases |
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(b) decreases |
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(c) increases and then decreases |
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(d) decreases and then increases |
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(e) Cannot be determined |
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4. |
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. |
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Ramesh is preparing for the last quiz and he realises that he will score a minimum of 97 to get an A grade. After the quiz, he realises that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take? |
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(a) 6 |
(b) 7 |
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(c) 8 |
(d) 9 |
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(e) None of these |
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5. |
A polynomial “ax3 + bx2 + cx + d ” intersects x-axis at 1 and –1, and y-axis at 2. The value of b is: |
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(a) – 2 |
(b) 0 |
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(c) 1 |
(d) 2 |
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(e) Cannot be determined |
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6. |
The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is: |
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(a) 0 |
(b) 1 |
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(c) 2 |
(d) 3 |
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(e) None of the above |
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7. |
There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the ladder is at Point A. After the first window is fixed, the foot of the ladder is pushed backwards to Point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is |
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(a) < 9 m |
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(b) ≥ 9 m and < 9.5 m |
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(c) ≥ 9.5 m and < 10 m |
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(d) ≥ 10 m and < 10.5 m |
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(e) ≥ 10.5 m |
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8. |
Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let ‘x’ be the smallest initial number that results in a win for Amitabh. The sum of the digits of ‘x’ is: |
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(a) 3 |
(b) 5 |
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(c) 7 |
(d) 9 |
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(e) None of these |
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9. |
Consider four natural numbers: x, y, x + y, and x – y. Two statements are provided below: |
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I.All four numbers are prime numbers. II.The arithmetic mean of the numbers is greater than 4. |
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Which of the following statements would be sufficient to determine the sum of the four numbers? |
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(a) Statement I (b) Statement II (c) Statement I and Statement II (d) Neither Statement I nor Statement II (e) Either Statement I or Statement II |
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10. |
Triangle ABC is a right-angled triangle. D and E are mid points of AB and BC respectively. Read the following statements. |
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I.AE = 19 II.CD = 22 III.Angle B is a right angle. |
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Which of the following statements would be sufficient to determine the length of AC? |
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(a) Statement I and Statement II |
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(b) Statement I and Statement III |
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(c) Statement II and III |
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(d) Statement III alone |
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(e) All three statements |
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11. |
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 X. The distance of X from the center of the smaller circle is 5 units. What is the length of the line segment P1P2? |
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(a) ≤ 13 |
(b) > 13 and ≤ 14 |
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(c) > 14 and ≤ 15 |
(d) > 15 and ≤ 16 |
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(e) > 16 |
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12. |
The probability that a randomly chosen positive divisor of 1029 is an integer multiple of 1023 is: a2/b2, then ‘b – a’ would be: |
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(a) 8 |
(b) 15 |
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(c) 21 |
(d) 23 |
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(e) 45 |
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13. |
Circle C1 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: |
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(a) 1/ |
(b) 1/ |
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(c) 1/ |
(d) 1/ |
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(e) None of these |
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14. |
There are two squares S1 and S2 with areas 8 and 9 units, respectively. S1 is inscribed within S2, with one corner of S1 on each side of S2. The corners of the smaller square divides the sides of the bigger square into two segments, one of length ‘a’ and the other of length ‘b’, where, b > a. |
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A possible value of ‘b/a’, is: |
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(a) ≥ 5 and < 8 |
(b) ≥ 8 and < 11 |
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(c) ≥ 11 and < 14 |
(d) ≥ 14 and < 17 |
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(e) > 17 |
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15. |
Diameter of the base of a water-filled inverted right circular cone is 26 cm. A cylindrical pipe, 5 mm in radius, is attached to the surface of the cone at a point. The perpendicular distance between the point and the base (the top) is 15 cm. The distance from the edge of the base to the point is 17 cm, along the surface. If water flows at the rate of 10 meters per minute through the pipe, how much time would elapse before water stops coming out of the pipe? |
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(a) < 4.5 minutes |
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(b) ≥ 4.5 minutes but < 4.8 minutes |
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(c) ≥ 4.8 minutes but < 5 minutes |
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(d) ≥ 5 minutes but < 5.2 minutes |
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(e) ≥ 5.2 minutes |
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16. |
Aditya has a total of 18 red and blue marbles in two bags (each bag has marbles of both colors). A marble is randomly drawn from the first bag followed by another randomly drawn from the second bag, the probability of both being red is 5/16. What is the probability of both marbles being blue? |
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(a) 1/16 |
(b) 2/16 |
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(c) 3/16 |
(d) 4/16 |
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(e) None of these |
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17. |
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 ? |
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(a) > 9.5 and ≤ 10 |
(b) > 10 and ≤ 10.5 |
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(c) > 10.5 and ≤ 11 |
(d) > 11 and ≤ 11.5 |
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(e) > 11.5 |
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18. |
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: |
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(a) 11 |
(b) 12 |
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(c) 15 |
(d) 17 |
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(e) 19 |
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19. |
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 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: |
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(a) 50 |
(b) 51 |
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(c) 53 |
(d) 56 |
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(e) 57 |
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20. |
Read the following instruction carefully and answer the question that follows: |
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Expression can also be written as |
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What would be the remainder if x is divided by 11? |
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(a) 2 |
(b) 4 |
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(c) 7 |
(d) 9 |
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(e) None of the above |
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21. |
A rectangular swimming pool is 48 m long and 20 m wide. The shallow edge of the pool is 1 m deep. For every 2.6 m that one walks up the inclined base of the swimming pool, one gains an elevation of 1 m. What is the volume of water (in cubic meters), in the swimming pool? Assume that the pool is filled up to the brim. |
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(a) 528 |
(b) 960 |
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(c) 6790 |
(d) 10560 |
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(e) 12960 |
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22. |
The value of the expression: |
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is: |
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(a) 0.01 |
(b) 0.1 |
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(c) 1 |
(d) 10 |
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(e) 100 |
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ANSWER KEY |
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1. (a) |
2. (b) |
3. (a) |
4. (d) |
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5. (a) |
6. (b) |
7. (e) |
8. (c) |
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9. (a) |
10. (e) |
11. (c) |
12. (d) |
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13. (a) |
14. (d) |
15. (d) |
16. (c) |
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17. (d) |
18. (e) |
19. (e) |
20. (d) |
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21. (d) |
22. (c) |
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Solutions
1.Given that one of the terms of the AP is 17, we can identify that the AP would either have all four odd numbers or have two odd and two even numbers.
Either ways the sum of the AP for four numbers would be even and hence would be divisible by 2.
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Note: we can safely assume here that since the terms of the AP are based on values of the variables x and y, all the four terms of the AP would be integers and there would be no possibility of some of the terms having decimal values. |
Hence, Option (a) is correct.
2.Since, PQ, QR, RS and PS are consecutive sides in the quadrilateral, we would get a figure for the quadrilateral as:
In this figure we have two relationships for the range of the value of SQ (based on the fact that the sum of two sides of a triangle would always be greater than the third side).
From Triangle SRQ: 5 + QS >17 Æ 5 + x >17 Æ x > 12
From Triangle PQS: x < 5 + 9 Æ x < 14.
Hence, 12 < x < 14. Hence, Option (b) is correct.
3.The expression can be modified to:
It can be clearly seen that when w is increased, would decrease. Hence, S would increase if we increase w.
Hence, Option (a) is correct.
4.The difference between just getting an A grade and just getting a B grade would be equal to (number of quizzes × 3). The difference between the required score to manage an A grade and the score achieved to just manage a B grade as given in the problem’s information is equal to 97 – 70 = 27.
Thus, number of quizzes × 3 = 27. Hence, number of quizzes is 9.
Hence, Option (d) is correct.
5.When a polynomial intersects the x-axis, the value of y = 0. When a polynomial intersects the y-axis, the value of x = 0.
If we take the polynomial as y = ax3 + bx2 + cx + d, we get:
Intersects x-axis at x = 1 implies: a + b + c + d = 0;
Intersects x-axis at x = –1 implies: –a + b – c + d = 0;
Hence, 2(b + d) = 0 or simply put (b + d) = 0.
Intersects y-axis at y = 2 implies: 0 + d = 2 Æ d = 2. Hence, b = –2. Hence, Option (a) is correct.
6.You need to consider the value of the expression for three ranges of values of X.
If X > 8, |X + 7| and |X – 8| are both non-negative.
If –7 < X < 8, |X + 7| is positive and |X – 8| is negative.
& If X < –7, |X + 7| and |X – 8| are both negative.
For the first case:
If X > 8, |X + 7| and |X – 8| are both non-negative.
|X + 7| = X + 7 and |X – 8| = X – 8.
Hence, |X + 7| + |X – 8| = 16;
Implies X + 7 + X – 8 = 16
Æ 2X – 1 = 16
\ x = 8.5.
For the second case:
If –7 < X < 8, |X + 7| is positive and |X – 8| is negative.
|X + 7| = X + 7 and |X – 8| = 8 – X.
Hence, X + 7 + 8 – X = 16. This leads to the ridiculous outcome 15 = 16.
Of course this is not possible and hence we can rule out this range – i.e. there would be no value of X between –7 £ x < 8 that would satisfy this equation.
For the third case:
If X < –7, |X + 7| and |X – 8| are both negative.
|X + 7| = –7–X and |X – 8| = 8 – X.
Thus, we get:
–7 – X + 8 – X = 16 Æ 2X = –15 Æ X = –7.5.
Thus, we have two possible values of X : i.e. 8.5 and –7.5 and their sum = 8.5 – 7.5 = 1.
Hence Option (b) is correct.
7.
From the figure we have:
cos q = y/ 30 and cos 2q = x/30.
Solving using Pyhtagoras theorem we get:
x = ª 15. Thus, the value of the angle 2q is approximately equal to 60o. Hence, q ª 30o.
Solving cos 30 = y/30, we get y = m.
The distance between A and B would be – 15
= 15 × 0.73 ª 10.95.
Hence, Option (e) is correct.
8.Let the number chosen by Amitabh be X. Then the flow of the numbers would go as follows:
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Amitabh |
Sashi |
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Round 1 |
2X |
2X + 50 |
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Round 2 |
4X + 100 |
4X + 150 |
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Round 3 |
8X + 300 |
8X + 350 |
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Round 4 |
16X + 700 |
16X + 750 |
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Round 5 |
32X + 1500 (Obviously the game cannot continue beyond this point as the value would definitely have crossed 1000 by this time.) |
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The latest that Amitabh can win is when Sashi gets a value of 16X +750. For 16X +750 > 1000, the smallest possible value of X = 16. Hence, sum of the digits of X = 1 + 6 = 7. Hence, Option (c) is correct.
9.If we take only Statement I, for all four numbers to be prime one of them must be even and hence equal to 2. Only in such an event do we get x + y and x–y as odd numbers and only if they are odd can all the four numbers be prime. A little bit of trial and error then gives us we get x = 5, y = 2, x + y = 7 and x – y = 3. There is not other case of x–y, x and x+y being prime as if we take y as 2, these numbers become x–2, x and x+2 and hence represent three consecutive odd numbers. (After 3, 5, 7 there is no situation where three consecutive odd numbers are all prime.)
Hence, Statement I is sufficient.
Statement II can be easily rejected as all it is giving us is that the sum of the four numbers is greater than 16. As we can easily imagine there are infinite sets of four such numbers, which have a sum greater than 16.
Hence, Option (a) is correct.
10.First things first while solving this. If we do not include Statement III, we do not know which angle is a right angle and hence cannot uniquely calculate the value of the Side AC.
Also, Statement III alone does not give us anything. Hence, we can reject Options (a) and (d). Options (b) and (c) give us similar set of information – i.e., the value of 1 median and the fact that B is the right angle in the triangle. This is also clearly not sufficient to answer the question.
For Option (e): We have 2 medians of a right-angled triangle, and we know B is the right angle. Hence, we can find AC.
Hence, Option (e) is the correct answer.
11.Let A, B be the centre of the two circles with radius 3 cm and 8 cm respectively.
The use of the Pythagoras triplet 3, 4, 5 is self evident from the figure.
In order to find P1P2 we need to find the value of XP2.
The similarity between triangles AP1X and BP2X implies that .
Since, we know three of the lengths in the above equation we get:
Æ P2 X = 4 × 2.66 = 10.66.
Thus, the length of T = 4 + 10.66 = 14.66
Hence, Option (c) is correct.
12.1029 = 529 × 229. Thus, the number of divisors of 1029 = (29 + 1)(29 + 1) = 30 × 30.
Now, if we look at divisors of 1029, which satisfy the given condition, we need to get divisors of the form: as N × 1023. The number of such divisors can be traced based on the number of divisors of N, such that 1029 = N ×1023.
N = 106 = 56 × 26 which would have 7 × 7 = 49 divisors. Hence, 1029 would have 49 divisors that are also divisible by 1023.
The probability that a randomly chosen divisor of 1029 is divisible by 1023 is 49/900 = a2/b2.
Thus, a = 7 and b = 30 and consequently b – a = 23.
Hence, Option (d) is correct.
13.The crux of the solution to this problem lies in understanding the fact that the Points P and Q would have the same Y-coordinates (since the diameter PQ is parallel to the X-axis) while their X co-ordinates would differ by 6 (since the diameter of the circle is 6).
We thus have two possible cases:
Case 1: If P lies on the curve y = ax and Q lies on the curve y = 2ax
Then, the difference between the x-coordinates of being 6, we get:
y = 2ax – a(x + 6). We get a = 21/6
Case 2: If Q lies on the curve y= ax and P lies on the curve y = 2ax
Then, the difference between the x-coordinates of being 6, we get:
y = ax – 2a(x + 6). We get a = (1/2)1/6
Since, we are given that a < 1, only Case 2 is possible and hence Option (a) is correct.
14.If you visualise a figure for this situation, you would be able to see something as follows:
Solving through Pythagoras theorem, we will get a = 0.18 and b = 3 – 0.18 = 2.82.
Hence, the value of b/a = 2.82/0.18 = 15.666. Hence, Option (d) is correct.
15.The following figure would exemplify the situation, with the pipe attached at a height of h from the apex (bottom) of the cone.
In the above figure the cones with height h and the cone with height h + 15 are similar to each other. Hence using similarity we will get:
Æ 10h + 150 = 26h Æ 16h = 150 Æ = 9.375 cm.
Based on this information we can then calculate the volume of water that would flow out from the pipe as: Total volume of the cone with height (h + 15) – Volume of cone with height h.
=
Calculating this we get the volume of water that overflows = 1295p cm3
Further, the rate at which the water flows out of the hole per minute is given by: p × (0.52 × 1000) cm3
Hence, the required time can be given as,
= 5.18 min
Hence, Option (d) is the correct answer.
16.Let there be ‘r’ red marbles in the first bag and ‘R’ red marbles in the second bag.
Also, let there be ‘b’ blue marbles in the first bag and ‘B’ blue marbles in the second bag.
Then (r + b) + (R + B) = 18
The number of ways of selecting 1 marble from the first bag and one marble from the second bag = (r + b) × (R + B).
Also, the number of ways of selecting a red marble from the first bag and a red marble from the second bag = r × R.
Hence, the probability of selecting two red marbles would be given by:
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This equation shows us that the value of (r + b) × (R + B) must be a multiple of 16 while the value of r × R must be a multiple of 5.
We now know that (r + b) + (R + B) = 18 and (r + b) × (R + B) = a multiple of 16.
If we try to break down 18 into two parts, such that their product is a multiple of 16, we get two cases:
Case I: 16 × 2 = 32 and Case II: 10 × 8 = 80.
Analysing Case I:
If (r + b) × (R + B) = 32, then r × R = 10 (in order to maintain the ratio of 5/16 between the two).
There would be 16 balls in one bag and 2 balls in the second bag.
The second bag must contain 1 red and 1 blue ball. i.e., R = 1 and B = 1
Since r × R = 10 we get r = 10 and b = 6.
The probability of both balls being blue in this case would turn out as:
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If we try to solve the other case in the same way, we get r = 5, R = 5 and b = 3 and B = 5 and the probability of both balls being blue becomes
We get the same answer in both the cases. Hence, Option (c) is correct.
17.The given situation can be visualised based on the following figure:
The area of the quadrilateral PQNM = Area of PQFE – Area Triangle MVS + Area Triangle VFN.
So our focus to find the area of the required quadrilateral should shift to the area of the three individual components on the right hand side of the above equation.
Finding the area of quadrilateral PQFE: Being a paralleogram, the required area would be given by:
× sum of parallel sides × perpendicular distance between the parallel sides.
In the figure, let side AB = 3x and side BC = y. Then, for the quadrilateral PQFE, the perpendicular distance between the parallel sides would be y/2.
Further, the sum of parallel sides would be equal to PQ + EF =
Area of PQFE =
We know that the area of the rectangle is 3x × y = 90.
Hence, area of PQFE = 90 ÷ 8 =11.25.
Finding the area of the triangle MSV= Finding the area of the triangle MEV.
× base × height = × EV × height
Using similarity between APM and VEM, we can see that since AP = x and EV = x/4, the ratios of the lengths of APM and VEM would be 4:1.
Thus, if the height of APM = 4h, the height of VEM = h and also 4h + h = y/2 Æ h = height of VEM with base VE = y/10.
Hence, area triangle MEV = × EV × height =
= 0.375.
Similarly, the area of triangle VFN ª 0.27.
Thus, the required area = 11.25 – 0.375 + 0.27
ª 11.145.
Hence, Option (d) is correct.
18.Solve this question through options. The correct option would give the number 792B as a multiple of the number 297B
Now 792B = B2 × 7 + B × 9 + 2
And 297B = B2 × 2 + B × 9 + 7.
If, we put B = 19 from Option (e) we get:
792B = B2 × 7 + B × 9 + 2 = 27000 and 297B = B2 × 2 + B × 9 + 7 = 900. This option satisfies the condition given in the problem. Hence, Option (e) is the correct answer.
19.Let the number of students scoring 6, 8 and 20 be a, b, and c respectively.
So, 6a + 8b + 20c = 504
Also, since ‘The number of students in the 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’ we get: a + 2c = b.
Thus, 14a + 36c = 504.
By trying out values we get that:
c =7 and a = 18 which gives us b = 32.
Therefore, total number of students = 18 + 32 + 7 = 57
Hence, Option (e) is the correct answer.
20.By equating the summation to we get:
x =
All the terms in x are divisible by 11 except 13!/11
Thus, the remainder of x divided by 11 would be given by the remainder of:
1.2.3.4.5.6.7.8.9.10.12.13 ÷ 11 = Rem (10! × 12 × 13) ÷ 11 = Rem (–1 × 1 × 2) ÷ 11. The required remainder is –2 = 9. Hence, Option (d) is correct.
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Note: Rem |
21.For every 2.6 m that one walks along the slanting part of the pool, there is a height of 1 m that is gained. Also, since the length of the pool is 48 m we get the following dimensions of the pool.
The pool would look as given in the figure below:
The volume of water in the pool = volume of the upper part + volume of the slanted triangular vessel
× 20 + (48 × 20 × 1)
fi 48 × 20 × 11
fi10560 m3
Hence, Option (d) is the correct answer.
22.The given expression would be equal to: log100!2 + log100!3 + log100!4 + …+ log100!100
= log100!100! = 1.
Hence, Option (c) is the correct answer.