XAT 2014 QA questions and solutions

XAT 2014 QA

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1. XAT 2014 QA | Algebra - Simple Equations

x, 17, 3x – y² – 2, and 3x + y² – 30, are four consecutive terms of an increasing arithmetic sequence. The sum of the four number is divisible by:

A.
2
B.
3
C.
5
D.
7
E.
11

2. XAT 2014 QA | Algebra - Simple Equations

In quadrilateral PQRS, PQ = 5 units, QR = 17 units, RS = 5 units, and PS = 9 units. The length of the diagonal QS can be:

A.
> 10 and < 12
B.
> 12 and < 14
C.
> 14 and < 16
D.
> 16 and < 18
E.
cannot be determined

3. XAT 2014 QA | Algebra - Simple Equations

Consider the formula,

$S = \frac{α \times ω}{τ + ρ \times ω}, where all the parameters are positive integers. If ⍵ is increased and ⍺, τ and ρ are kept constant, then S:$

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

Answer: Option A

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Explanation :

$\frac{1}{S} = \frac{τ + ρ \times ω}{α ω} = \frac{τ}{α ω} + \frac{ρ}{α} = \frac{K_{1}}{ω} + K_{2}$

Where K₁and K₂ are constants
⇒ 1/S decreases when ω increases.
⇒ S increases when ω increases.
Hence, option (a).

4. XAT 2014 QA | Algebra - Simple Equations

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 will 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 these

Answer: Option D

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Explanation :

Total exams given = x

Score of previous exams = y

$\frac{y + 97}{x} = 90$

$\frac{y + 70}{x} = 87$

$\frac{y}{x} + \frac{97}{x} =90$

$\frac{y}{x} + \frac{70}{x} = 87$

$\frac{y}{x} = 90 - \frac{97}{x}$

$\frac{y}{x} = 87 - \frac{70}{x}$

$∴ 90 - \frac{97}{x} = 87 - \frac{70}{x}$

$ 3 = \frac{- 70}{x} + \frac{97}{x} = \frac{27}{x}$

$ x = 9$

Hence, option (d).

5. XAT 2014 QA | Algebra - Simple Equations

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

6. XAT 2014 QA | Algebra - Simple Equations

The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:

A.
0
B.
1
C.
2
D.
3
E.
None of the above

Answer: Option B

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Explanation :

|X + 7| + |X – 8| = 16
For X ≥ 8, |X + 7| = X + 7 and |X – 8| = X – 8
∴ |X – 7| + |X – 8| = X + 7 + X – 8 = 2X – 1
2X – 1 = 16

$∴ X = \frac{17}{2}$

For -7 ≤ x < 8,
∴ |X + 7| = X + 7 & |X – 8| = 8 – X
∴ |X + 7| = |X – 8| = X + 7 + 8 – X = 15 ≠ RHS
0 ≤ x < 8 is not possible.
-7 ≤ x < 0
∴ |X + 7| = 7 + X & |X – 8| = -X + 8
16 = |X + 7| + |X – 8| = 7 + X – X + 8 = 15
∴ -7 ≤ x < 0 is not possible.
Now, x < -8 |X + 7| = - 7 – x |X – 8| = -X + 8
∴ |X + 7| + |X – 8| = - 7 – X – X + 8 = 1 – 2X = 16
⇒ 2X = -15 ⇒ X = -7.5
Hence, option (b).

7. XAT 2014 QA | Algebra - Simple Equations

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

A.
< 9 m
B.
≥ 9 m and < 9.5 m
C.
≥ 9.5 m and < 10 m
D.
≥ 10 m and < 10.5 m
E.
≥ 10.5 m

Answer: Option E

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Explanation :

The following figure represents the length and the position of the ladder,

By Pythagoras Theorem,

$x = \sqrt{30^{2} - 26^{2}} = \sqrt{224} \approx 15 m$

$∴ \cos 2 θ = \frac{15}{30} = \frac{1}{2}$

$\Rightarrow 2 θ = 60 ° \Rightarrow θ = 30 °$

$\cos 30 ° = \frac{y}{30}$

$\Rightarrow \frac{\sqrt{3}}{2} = \frac{y}{30}$

$\Rightarrow y = 15 \sqrt{3} m$

The approximate distance between A and B can be given as,

$y - x = 15 \sqrt{3} - 15 = 10.98 m$

Hence, option (e)

8. XAT 2014 QA | Algebra - Simple Equations

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:

A.
3
B.
5
C.
7
D.
9
E.
None of these

9. XAT 2014 QA | Algebra - Simple Equations

Consider four natural numbers: x, y, x + y, and x – y. Two statements are provided below:

All four numbers are prime numbers.
The arithmetic mean of the numbers is greater than 4.

Which of the following statements would be sufficient to determine the sum of the four numbers?

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.

10. XAT 2014 QA | Algebra - Simple Equations

Triangle ABC is a right angled triangle. D and E are mid points of AB and BC respectively. Read the following statements.

AE = 19
CD = 22
Angle B is a right angle.

Which of the following statements would be sufficient to determine the length of AC?

A.
Statement I and Statement II.
B.
Statement I and Statement III.
C.
Statement II and III.
D.
Statement III alone.
E.
All three statements.

11. XAT 2014 QA | Algebra - Simple Equations

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?

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

12. XAT 2014 QA | Algebra - Simple Equations

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

Answer: Option D

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Explanation :

10 = 5 × 2

10²⁹ = 5²⁹ × 2²⁹

∴ Number of divisors = (29 + 1)(29 + 1) = 30 × 30

We need to find all the divisors of K such that 10²⁹

= K × 10²³

K = 10⁶ = 5⁶ × 2⁶

∴ Number of divisors = (6 + 1)(6 + 1) = 7 × 7

The probability that a randomly chosen positive divisor of 10²⁹is an integer multiple of 10²³

$= \frac{7 \times 7}{30 \times 30} = \frac{a^{2}}{b^{2}}$

∴ a = 7 and b = 30

∴ b – a = 30 – 7 = 23

Hence, option (d).

13. XAT 2014 QA | Algebra - Simple Equations

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 = a^x and y = 2a^x respectively, where a < 1. The value of a is:

A.
$\frac{1}{\sqrt[6]{2}}$
B.
$\frac{1}{\sqrt[6]{3}}$
C.
$\frac{1}{\sqrt[3]{6}}$
D.
1 6
E.
None of the above

Answer: Option A

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Explanation :

Points P and Q have the same Y coordinate but their X coordinates differ by 6 units (Since the diameter = 6 units)

We have two possible cases:

Case 1:

$y = a^{x} = 2 a^{(x + 6)}$

$a = {(1 / 2)}^{(\frac{1}{6})}$

Case 2:

$y = 2 a^{x} = a^{(x + 6)}$

$a = {(2)}^{(\frac{1}{6})}$

But it is given that a < 1

∴ Case 2 can be discarded

Hence, option (a).

14. XAT 2014 QA | Algebra - Simple Equations

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. A possible value of ‘b/a’, is:

A.
≥ 5 and < 8
B.
≥ 8 and < 11
C.
≥ 11 and < 14
D.
≥ 14 and < 17
E.
> 17

Answer: Option D

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Explanation :

$a^{2} + {(3 - a)}^{2} = 8$

$2 a^{2} - 6 a + 1 = 0$

$a = \frac{3}{2} \pm \frac{\sqrt{7}}{2}$

⇒ a = 0.18, b = (3 – a) = 2.82

∴ b/a = 15.66

Hence, option (d).

15. XAT 2014 QA | Algebra - Simple Equations

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?

A.
< 4.5 minutes
B.
≥ 4.5 minutes but < 4.8 minutes
C.
≥ 4.8 minutes but < 5 minutes
D.
≥ 5 minutes but < 5.2 minutes
E.
≥ 5.2 minutes

Answer: Option D

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Explanation :

Applying the concept of similarity,

$\frac{10}{26} = \frac{h}{h + 15}$

$\Rightarrow h = 9.375 cm$

Volume of water that overflows can be given as,

$\frac{1}{3} π [(13^{2} \times (15 + 9.375)) - (5^{2} \times 9.375)]$

$\Rightarrow \frac{1}{3} π [(13^{2} \times 24.375) - (5^{2} \times 9.375)]$

$\Rightarrow \frac{1}{3} π [4119.375 - 234.375]$

$\Rightarrow 1295 π {cm}^{3}$

Since the radius of the hole is 5 mm i.e. 0.5cm,

The volume of water flowing in 1 minute,

(0.5² × 1000)π cm³

Hence, the required time can be given as,

$\frac{1295 π}{0 . 5^{2} \times 1000 π} = 5.18$

Hence, option (d).

16. XAT 2014 QA | Algebra - Simple Equations

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?

A.
$\frac{1}{16}$
B.
$\frac{2}{16}$
C.
$\frac{3}{16}$
D.
4 16
E.
None of the above

17. XAT 2014 QA | Algebra - Simple Equations

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

Answer: Option D

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Explanation :

Let the breadth be 3x and the breadth be y.

3xy = 90 ⇒ xy = 30

V is midpoint of WR. PW || EV ⇒ EV = PW/2

Similarly, FV = WQ/2

∴ EF = PQ/2 = x/2

∆MPA ∼ ∆MEV

Height of ∆MPA with respect to AP: Height (h₁)of ∆MEV with respect to EV = AP : EV = x : x/4 = 4 : 1

Let height of ∆MPA = 4k and height (h₁) of ∆MEV = k

∴ 4k + k = 5k = y/2

∴ k = y/10

∴ Height (h₁) of ∆MEV = y/10

$A (△ M E V) = \frac{1}{2} \times E V \times h_{1} = \frac{1}{2} \times \frac{x}{4} \times \frac{y}{10} = \frac{30}{60} = 0.375$

Similarly, ∆VFN ∼ ∆CRN

Height (h₂) of ∆VFN with respect to VF : Height of ∆CRN with respect to CR = VF : CR = x/4 : 3x/2 = 1 : 6

Let height (h₂) of ∆VFN = m and height of ∆CRN = 6m

∴ m + 6m = 7m = y/2

∴ m = y/14

∴ Height (h₂) of ∆VFN = y/14

$A (△ V F N) = \frac{1}{2} \times F V \times h_{2} = \frac{1}{2} \times \frac{x}{4} \times \frac{y}{14} = \frac{30}{112} \approx 0.27$

$A (□ P Q E) = \frac{1}{2} \times (P Q + E F) \times \frac{y}{2}$

$= \frac{1}{2} \times (x + \frac{x}{2}) \times \frac{y}{2} = \frac{3 x y}{8} = \frac{90}{8} = 11.25 sq . units$

A(□PQMN) = A(□PQEF) – A(∆MEV) + A(∆VFN)

= 11.25 – 0.375 + 0.27 ≈ 11.145

Hence, option (d).

18. XAT 2014 QA | Algebra - Simple Equations

Two numbers, 297_B and 792_B, 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

19. XAT 2014 QA | Algebra - Simple Equations

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

20. XAT 2014 QA | Algebra - Simple Equations

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

Expression

$\sum_{n = 1}^{13} \frac{1}{n}$ can also be written as $\frac{x}{13!}$

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

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

Answer: Option D

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Explanation :

$\sum_{n = 1}^{13} \frac{1}{n} = \frac{x}{13!} ∴ x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + . . . + \frac{13!}{11} + \frac{13!}{12} + \frac{13!}{13}$

All the terms in x are divisible by 11 except 13!/11

$\frac{13!}{11} = 1.2.3.4 . . . 10.12.13$

According to Wilson theorem,

$r e m (\frac{(p - 1)!}{p}) = - 1$

$\frac{13!}{11} = 1.2.3.4 . . . 10.12.13 . = 10! (12) (13)$

$∴ r e m (\frac{\frac{13!}{11}}{11}) = \frac{10! (12) (13)}{11} = (- 1) . 1.2 = - 2 = 9$

Hence, option (d).

21. XAT 2014 QA | Algebra - Simple Equations

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.

A.
528
B.
960
C.
6790
D.
10560
E.
12960

Answer: Option D

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Explanation :

Since the elevation increases by 1 m for every 2.6 m,

$x = \sqrt{2 . 6^{2} - 1} = \sqrt{6.76 - 1} = \sqrt{5.76} = 2.4 m$

Since $\frac{48}{2.4} = 20,$

The height of the deeper end of the pool is 20 m.

This can be represented as follows,

The total volume of water in the pool,

$(\frac{1}{2} \times 48 \times 20) \times 20 + (48 \times 20 \times 1)$

⇒ 48× 20 × 11

⇒ 10560 m³

Hence, option (d).

22. XAT 2014 QA | Algebra - Simple Equations

The value of the expression:

$\sum_{i = 2}^{100} \frac{1}{\log_{i} 100!} is :$

A.
0.01
B.
0.1
C.
1
D.
10
E.
100

Answer: Option C

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Explanation :

$\sum_{i = 2}^{100} \frac{1}{\log_{i} 100!} = \sum_{i = 2}^{100} \log_{100!} 2 + \log_{100!} 3 + \log_{100!} 4 + . . . + \log_{100!} 100$

$= \log_{100!} 100! = 1$

Hence, option (c).

23. XAT 2014 QA | Algebra - Simple Equations

Which of the following could be the correct ascending order of democratic regions for poor?

A.
North America, C and E Europe, South America, Middle East, Asia Pacific
B.
Scandinavia, Western Europe, Africa, Asia Pacific, Middle East
C.
Scandinavia, Western Europe, North America, C and E Europe, Middle East
D.
C and E Europe, Africa, South America, Western Europe, Scandinavia
E.
Africa, South America, Western Europe, North America, Scandinavia

24. XAT 2014 QA | Algebra - Simple Equations

Which region has the highest disparity, of democratic participation, between rich and poor?

A.
North America
B.
C and E Europe
C.
Africa
D.
South America
E.
Western Europe

25. XAT 2014 QA | Algebra - Simple Equations

The maximum GDP of African region is higher than the maximum GDP of South American region by factor of:

A.
10
B.
100
C.
2
D.
4
E.
None of these

26. XAT 2014 QA | Algebra - Simple Equations

In which of the following years the percentage increase in the number of Indians going abroad was greater than the percentage increase in the number of domestic tourists?

A.
2004 and 2005
B.
2005 and 2006
C.
2005 and 2007
D.
2006 and 2008
E.
2004, 2005 and 2006

27. XAT 2014 QA | Algebra - Simple Equations

In which of the following years was the rupee cheapest with respect to the dollar?

A.
2001
B.
2002
C.
2007
D.
2010
E.
2011

28. XAT 2014 QA | Algebra - Simple Equations

Let ‘R’ be the ratio of Foreign Exchange Earnings from Tourism in India (in US $ million) to Foreign Tourist Arrivals in India (in million). Assume that R increases linearly over the years. If we draw a pie chart of R for all the years, the angle subtended by the biggest sector in the pie chart would be approximately:

A.
24
B.
30
C.
36
D.
42
E.
48

29. XAT 2014 QA | Algebra - Simple Equations

Which of the following countries, after United States, has the highest spending on military as % of GDP, in the period 2000-2010?

A.
Vietnam
B.
China
C.
India
D.
Brazil
E.
Thailand

30. XAT 2014 QA | Algebra - Simple Equations

Which country (and which year) has witnessed maximum year-to-year decline in “industry as percentage of GDP”? Given that the maximum value of industry as percentage of GDP is 49.7% and the minimum value of industry as percentage of GDP is 20.02%, in the chart above.

A.
United States in 2002-3
B.
Brazil in 2006-7
C.
India in 2009-10
D.
Malaysia in 2008-9
E.
China in 2008-9

31. XAT 2014 QA | Algebra - Simple Equations

Which of the following countries has shown maximum increase in the “services, value added as % of GDP” from year 2000 to year 2010?

A.
Brazil
B.
India
C.
United States
D.
Philippines
E.
None of the above

XAT 2014 QA | Previous Year XAT Paper

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