# XAT 2014 QA | Previous Year XAT Paper

XAT 2014 QA

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**1. XAT 2014 QA | Algebra - Simple Equations**

*x*, 17, 3*x* – *y*^{2} – 2, and 3*x* + *y*^{2} – 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

Answer: Option A

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**Explanation** :

Since it is an A.P.,

Therefore, 17 – *x* = 3*x* +* y*^{2} – 30 – 3*x* + *y*^{2} + 2

Or, *x* + 2*y*^{2} = 45 … (i)

Again, 17 −* x* = 3*x* − *y*^{2} − 2 −17

Or, 4*x* – *y*^{2} = 36 … (ii)

Solving equations (i) and (ii),

we get *x* = 13.

Sum of the A.P. = *x* + 17 + 3*x* – *y*^{2} – 2 + 3*x* +* y*^{2} – 30

= 7*x* – 15 = 7(13) – 15 = 76

Out of the given options, 76 is only divisible by 2.

Hence, option (a).

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**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

Answer: Option B

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**Explanation** :

Let QS = x, we get the following figure

In any triangle, the sum of any two sides must be greater than the third side. Similarly, the difference between any two sides must be smaller than the third side. Hence,

In ∆QRS,

*x* + 5 > 17

⇒ *x* > 12 …(i)

In ∆PQS,

*x* < 9 + 5

⇒ *x* < 14 …(ii)

Combining (i) and (ii), we get

12 < *x* < 14

Hence, option (b).

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**3. XAT 2014 QA | Algebra - Simple Equations**

Consider the formula,

$S=\frac{\alpha \times \omega}{\tau +\rho \times \omega},\; where\; all\; the\; parameters\; arepositive\; integers.\; If\; \u2375\; is\; increased\; and\; \u237a,\; \tau \; and\; \rho \; are\; kept\; constant,\; then\; S:$

- 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** :

$\frac{1}{S}=\frac{\tau +\rho \times \omega}{\alpha \omega}=\frac{\tau}{\alpha \omega}+\frac{\rho}{\alpha}=\frac{{K}_{1}}{\omega}+{K}_{2}$

Where K_{1 }and K_{2} are constants

⇒ 1/S decreases when ω increases.

⇒ S increases when ω increases.

Hence, option (a).

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**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

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

$\mathrm{}\phantom{\rule{0ex}{0ex}}\frac{y+70}{x}=87$

$\mathrm{}\phantom{\rule{0ex}{0ex}}\frac{y}{x}+\frac{97}{x}=90$

$\mathrm{}\phantom{\rule{0ex}{0ex}}\frac{y}{x}+\frac{70}{x}=87$

$\mathrm{}\phantom{\rule{0ex}{0ex}}\frac{y}{x}=90-\frac{97}{x}$

$\frac{}{}\phantom{\rule{0ex}{0ex}}\frac{y}{x}=87-\frac{70}{x}$

$\frac{}{}\phantom{\rule{0ex}{0ex}}\therefore 90-\frac{97}{x}=87-\frac{70}{x}$

$\frac{\mathrm{}}{}\phantom{\rule{0ex}{0ex}}3=\frac{-70}{x}+\frac{97}{x}=\frac{27}{x}$

$\frac{\mathrm{}}{}\phantom{\rule{0ex}{0ex}}x=9$

Hence, option (d).

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**5. XAT 2014 QA | Algebra - Simple Equations**

A polynomial “*ax*^{3} + *bx*^{2} + *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

Answer: Option A

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**Explanation** :

*ax*^{3 }+ *bx*^{2 }+ *cx *+ *d* intersects *x* axis at 1 & –1

∴ *a *+ *b *+ *c *+ *d *= 0

–*a *+ *b *– *c *+ *d *= 0

∴ 2(*b *+ *d*) = 0

∴ *b *+ *d *= 0

*ax*^{3 }+ *bx*^{2 }+ *cx *+ *d* intersects *y* axis at 2

0 + *d *= 2

∴ *d *= 2

From (1) & (2), *b* = –2

Hence, option (a).

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**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

$\therefore X=\frac{17}{2}$

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).

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**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,

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

$\therefore \mathrm{cos}2\theta =\frac{15}{30}=\frac{1}{2}$

$\Rightarrow 2\theta =60\xb0\Rightarrow \theta =30\xb0$

$\mathrm{cos}30\xb0=\frac{y}{30}$

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

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

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

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

Hence, option (e)

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**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

Answer: Option C

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**Explanation** :

Let the smallest number be ‘X’

Amitabh has to win and X is the least possible number in the range 1 -999 ∴ step 4 has to be the last step.

⇒ 16X + 750 > 1000

The least possible value of X = 16

Sum of the digit = 1 + 6 = 7

Hence, option (c).

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**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.

Answer: Option A

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**Explanation** :

Considering Statement I:

Since this is the only possible solution, statement I is sufficient.

Considering Statement II:

Since no unique solution is possible, statement II is not sufficient.

Hence, option (a).

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**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.

Answer: Option E

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**Explanation** :

Option 1: If we take statements I and II, we have 2 medians of a right angled triangle, without knowing which angle is a right angle.

Option 2: If we take statements I and III, we have 1 median of a right angled triangle, and B is the right angle.

Option 3: If we take only statements II and III, we have 1 medians of a right angled triangle, and B is the right angle.

Option 4: If we only know that B is a right angle, we cannot determine the length of AC.

Option 5: If we have all 3 statements, then 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).

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**11. XAT 2014 QA | Algebra - Simple Equations**

There are two circles C_{1} and C_{2} of radii 3 and 8 units respectively. The common internal tangent, T, touches the circles at points P_{1} and P_{2} 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 P_{1}P_{2}?

- A.
≤ 13

- B.
> 13 and ≤ 14

- C.
> 14 and ≤ 15

- D.
> 15 and ≤ 16

- E.
> 16

Answer: Option C

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**Explanation** :

Let A, B be the centre of the two circles with radius 3cm, 8cm respectively.

AX = 5cm, AP_{1} = 3cm

Using Pythagoras theorem, P_{1}X = 4cm

Now ∆ A P_{1}X ≈ ∆ B P_{2}X

⇒ A P_{1}/ B P_{2} = P_{1}X/ P_{2}X

⇒ 3/8 = 4/ P_{2}X

P_{2}X = 10.66

P_{1} P_{2} = P_{1}X + P_{2}X = 14.66

Hence, option (c).

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**12. XAT 2014 QA | Algebra - Simple Equations**

The probability that a randomly chosen positive divisor of 10^{29} is an integer multiple of 10^{23} is: *a*^{2}/*b*^{2}, 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

10^{29} = 5^{29} × 2^{29}

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

We need to find all the divisors of *K* such that 10^{29}

= *K* × 10^{23}

K = 10^{6} = 5^{6} × 2^{6}

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

The probability that a randomly chosen positive divisor of 10^{29 }is an integer multiple of 10^{23}

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

∴ a = 7 and b = 30

∴ b – a = 30 – 7 = 23

Hence, option (d).

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**13. XAT 2014 QA | Algebra - Simple Equations**

Circle C_{1} 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*= 2

*a*respectively, where

^{x}*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:**

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

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

**Case 2:**

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

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

But it is given that a < 1

∴ Case 2 can be discarded

Hence, option (a).

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**14. XAT 2014 QA | Algebra - Simple Equations**

There are two squares S_{1} and S_{2} with areas 8 and 9 units, respectively. S_{1} is inscribed within S_{2}, with one corner of S_{1} on each side of S_{2}. 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** :

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

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

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

⇒ a = 0.18, b = (3 – a) = 2.82

∴ b/a = 15.66

Hence, option (d).

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**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,

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

$\Rightarrow h=9.375\mathrm{cm}$

Volume of water that overflows can be given as,

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

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

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

$\Rightarrow 1295\pi {\mathrm{cm}}^{3}$

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

The volume of water flowing in 1 minute,

(0.5^{2} × 1000)π cm^{3}

Hence, the required time can be given as,

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

Hence, option (d).

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**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

Answer: Option C

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**Explanation** :

(*X*_{1} + *Y*_{1}) + (*X*_{2} + *Y*_{2}) = 18 … (i)

Selecting a marble from the first bag and then from the second bag can be done in (*X*_{1} + *Y*_{1}) × (*X*_{2} + *Y*_{2}) ways.

Selecting a red marble from the first bag and then a red marble from the second bag can be done in (*X*_{1}) × (*X*_{2}) ways.

∴ Probability of selecting red marbles from both the bags = (X_{1}) × (X_{2})/ (X_{1} + Y_{1}) × (X_{2} + Y_{2}) = 5/16

Let (*X*_{1} + *Y*_{1}) × (*X*_{2} + *Y*_{2}) = 16*a *… (ii)

∴ (*X*_{1}) × (*X*_{2}) = 5*a *… (iii)

Considering (i) and (ii), *a* = 2 or 5

**Case I:** *a* = 2

(*X*_{1} + *Y*_{1}) × (*X*_{2} + *Y*_{2}) = 32 … (iv)

∴ From (i) and (iv),

(*X*_{1} + *Y*_{1}) = 2 and (*X*_{2} + *Y*_{2}) = 16

∴ *X*_{1} = *Y*_{1} = 1 ⇒ *X*_{2} = 10 (∵ (*X*_{1}) × (*X*_{2}) = 10)

∴ *Y*_{2} = 6

Probability of both marbles being blue = (1 × 6)/32

= 3/16

**Case II:** *a* = 5

(*X*_{1} + *Y*_{1}) × (*X*_{2} + *Y*_{2}) = 80 … (v)

∴ From (i) and (v),

(*X*_{1} + *Y*_{1}) = 8 and (*X*_{2} + *Y*_{2}) = 10

∴ *X*_{1} = *X*_{2} = 5 (∵ (*X*_{1}) × (*X*_{2}) = 25)

*Y*_{1} = 3 and *Y*_{2} = 5

∴ *Y*_{2} = 6

Probability of both marbles being blue = (5 × 3)/80

= 3/16

Hence, option (c).

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**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 (*h*_{1})of ∆MEV with respect to EV = AP : EV = *x* : *x*/4 = 4 : 1

Let height of ∆MPA = 4*k* and height (*h*_{1}) of ∆MEV = *k*

∴ 4*k* + *k* = 5*k* = *y*/2

∴ *k* = *y*/10

∴ Height (*h*_{1}) of ∆MEV = *y*/10

$A(\u25b3MEV)=\frac{1}{2}\times EV\times {h}_{1}=\frac{1}{2}\times \frac{x}{4}\times \frac{y}{10}=\frac{30}{60}=0.375$

Similarly, ∆VFN ∼ ∆CRN

Height (*h*_{2}) of ∆VFN with respect to VF : Height of ∆CRN with respect to CR = VF : CR = *x*/4 : 3*x*/2 = 1 : 6

Let height (*h*_{2}) of ∆VFN = *m* and height of ∆CRN = 6*m*

∴ *m* + 6*m* = 7*m* = *y*/2

∴ *m* = *y*/14

∴ Height (*h*_{2}) of ∆VFN = *y*/14

$A(\u25b3VFN)=\frac{1}{2}\times FV\times {h}_{2}=\frac{1}{2}\times \frac{x}{4}\times \frac{y}{14}=\frac{30}{112}\approx 0.27$

$A(\square PQE)=\frac{1}{2}\times (PQ+EF)\times \frac{y}{2}$

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

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

= 11.25 – 0.375 + 0.27 ≈ 11.145

Hence, option (d).

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**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

Answer: Option E

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**Explanation** :

Solving by options,

By option(1), we get,

297_{B} = 11^{2} × 2 + 11 × 9 + 7 = 348

792_{B }= 11^{2} × 7 + 11 × 9 + 2 = 948

Since 348, is not a factor of 948, option(1) is eliminated.

By option(2), we get,

297_{B} = 12^{2} × 2 + 12 × 9 + 7 = 403

792_{B} = 12^{2} × 7 + 12 × 9 + 2 = 1118

Since 403, is not a factor of 1118, option(2) is eliminated.

By option(3), we get,

297_{B} = 15^{2 }× 2 + 15 × 9 + 7 = 592

792_{B} = 15^{2} × 7 + 15 × 9 + 2 = 1712

Since 592, is not a factor of 1712, option(3) is eliminated.

By option(4), we get,

297_{B} = 17^{2} × 2 + 17 × 9 + 7 = 738

792_{B} = 17^{2} × 7 + 17 × 9 + 2 = 2178

Since 738, is not a factor of 2178, option(4) is eliminated.

By option(5), we get,

297_{B} = 19^{2} × 2 + 19 × 9 + 7 = 900

792_{B} = 19^{2} × 7 + 19 × 9 + 2 = 2700

Since 900, is a factor of 2700,

Hence, option (e).

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**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

Answer: Option E

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**Explanation** :

Let the number of students scoring 6, 8 and 20 be *x*, *y* and *z* respectively.

So, 6*x* + 8*y* + 20*z* = 504

*x* + 2*z* = *y*

or, 14*y* + 8*z* = 504

or, 7*y* + 4*z* = 252

By hit and trial we get *y* = 32 and *z* = 7

Therefore,* x* = 18

Therefore, total number of students = 32 + 7 + 18 = 57

Hence, option (e).

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**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** :

$\sum _{n=1}^{13}\frac{1}{n}=\frac{x}{13!}\phantom{\rule{0ex}{0ex}}\therefore x=\frac{13!}{1}+\frac{13!}{2}+\frac{13!}{3}+...+\frac{13!}{11}+\frac{13!}{12}+\frac{13!}{13}$

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,

$rem\left(\frac{(p-1)!}{p}\right)=-1$

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

$\therefore rem\left(\frac{{\displaystyle \frac{13!}{11}}}{11}\right)=\frac{10!\left(12\right)\left(13\right)}{11}=(-1).1.2=-2=9$

Hence, option (d).

Workspace:

**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,

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

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,

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

⇒ 48× 20 × 11

⇒ 10560 m³

Hence, option (d).

Workspace:

**22. XAT 2014 QA | Algebra - Simple Equations**

The value of the expression:

$\sum _{i=2}^{100}\frac{1}{{\mathrm{log}}_{i}100!}\mathrm{is}:$

- A.
0.01

- B.
0.1

- C.
1

- D.
10

- E.
100

Answer: Option C

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**Explanation** :

$\sum _{i=2}^{100}\frac{1}{{\mathrm{log}}_{i}100!}=\sum _{i=2}^{100}{\mathrm{log}}_{100!}2+{\mathrm{log}}_{100!}3+{\mathrm{log}}_{100!}4+...+{\mathrm{log}}_{100!}100$

$={\mathrm{log}}_{100!}100!=1$

Hence, option (c).

Workspace:

**Answer the questions based on the trends lines from the following graphs.**

Note: Left side of X axis represents countries that are “poor” and right side of X axis represents countries that are “rich”, for each region. GDP is based on purchasing power parity (PPP).

These are World Bank (WB) estimates.

**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

Answer: Option D

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**Explanation** :

Observe that Scandinavia and Western Europe have all their democratic score points as 20. SO, they should be the last two regions in ascending order. However, Scandinavia has no difference between rich and poor. So, it should be the last region and Western Europe should be second last.

Only option 4 satisfies this condition.

Hence, option (d).

Workspace:

**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

Answer: Option B

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**Explanation** :

The greatest disparity of democratic participation between rich and poor is obtained from the graph with the steepest slope and having some point as close as possible to full democracy and another point as close as possible to full authoritarianism. This condition is satisfied for C and E Europe where the slope of the line is the steepest and the difference between the maximum and minimum points is around 18.

Hence, option (b).

Workspace:

**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

Answer: Option E

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**Explanation** :

The maximum GDP of Africa and South America has the same value.

Hence, option (e).

Workspace:

**Answer the questions based on the given data on the tourism sector in India.**

**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

Answer: Option C

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**Explanation** :

Observe that the required percentage is to be found for all years from 2004 to 2008.

The percentage increase in the number of Indians going abroad and the percentage increase in the number of domestic tourists for each year from 2004 to 2008 is as shown below:

Thus, the percentage increase in the number of Indians going abroad is greater than the percentage increase in the number of domestic tourists in 2005, 2007 and 2008.

Hence, option (c).

Workspace:

**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

Answer: Option B

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**Explanation** :

The rupee is cheapest with respect to the dollar when the numerical value of rupees corresponding to 1 dollar is the highest

i.e. If 1 US $ = X INR, then the rupee is cheapest when X takes its largest numerical value.

The value of the rupee with respect to the dollar for the given years is:

Thus, the rupee is cheapest in 2002.

Hence, option (b).

Workspace:

**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

Answer: Option C

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**Explanation** :

The value of R for each year is as shown below:

When a pie-chart is made using the values of R, the value for 2011 forms the largest sector. Also, the total value of R is 26415.49.

Observe that the value of R in 2011 is approximately 10% of the total. So, the angle that it subtends at the centre is 10% of 360 i.e. 36 degrees.

Hence, option (c).

Workspace:

**Answer the questions based on the following information:**

The exhibit given below compares the countries (first column) on different economic indicators (first row), from 2000-2010. A bar represents data for one year and a missing bar indicates missing data. Within an indicator, all countries have same scale.

**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

Answer: Option C

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**Explanation** :

It is observed from the bar-graphs that India has the second-highest spending on military as percentage of GDP in the period 2000-2010.

Hence, option (c).

Workspace:

**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

Answer: Option D

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**Explanation** :

Among the options given, it can be seen that Malaysia in 2008-09 has witnessed the maximum year-on-year decline in “industry as percentage of GDP”.

Hence, option (d).

Workspace:

**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

Answer: Option B

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**Explanation** :

It can be seen that India has shown the maximum increase in the “services, value added as percentage of GDP” from 2000 to 2010”.

Hence, option (b).

Workspace:

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