# XAT 2019 QADI

Paper year paper questions for XAT 2019 QA

**1. XAT 2019 QADI | Arithmetic - Ratio, Proportion & Variation**

Two numbers a and b are inversely proportional to each other. If a increases by 100%, then b decreases by

- A.
200%

- B.
50%

- C.
100%

- D.
80%

- E.
150%

Answer: Option B

**Explanation** :

Given that ‘a’ and ‘b’ are inversely proportional

∴ a ∝ 1/b

⇒ a × b = constant

Hence, if a is doubled (i.e., increased by 100%) b needs to become half.

∴ % change in b = (1/2 - 1) × 100% = 50%.

Hence, option (b).

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**2. XAT 2019 QADI | Arithmetic - Percentage**

A, B, C, D and E are five employees working in a company. In two successive years, each of them got hikes in his salary as follows:

A : p% and (p + 1)%,

B : (p + 2)% and (p - 1)%,

C : (p + 3)% and (p - 2)%,

D : (p + 4)% and (p - 3)%,

E : (p + 5)% and (p - 4)%.

If all of them have the same salary at the end of two years, who got the least hike in his salary?

- A.
E

- B.
B

- C.
D

- D.
A

- E.
C

Answer: Option A

**Explanation** :

Since final salary for all five employees is same, least hike will be for that employee who has the highest initial salary.

Again, since final salary for all five employees is same, initial salary will be highest for that employee who has least overall % change.

Let us calculate % change for all the employees.

We know when a number is successively increased b a% and then b%, overall % change is a + b + ab/100.

Here let us assume the value of p as 10%

Therefore, % change for

A: 10 + 11 + 10 × 11/100 = 21 + 1.1 = 22.1%

B: 12 + 9 + 12 × 9/100 = 21 + 1.08 = 22.08%

C: 13 + 8 + 13 × 8/100 = 21 + 1.04 = 22.04%

D: 14 + 7 + 14 × 7/100 = 21 + 0.98 = 21.98%

E: 15 + 6 + 15 × 6/100 = 21 + 0.90 = 21.90%

∴ Least % change is for E, hence E has the highest initial salary hence the least hike.

Hence, option (a).

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**3. XAT 2019 QADI | Geometry - Mensuration**

A gold ingot in the shape of a cylinder is melted and the resulting molten metal molded into a few identical conical ingots. If the height of each cone is half the height of the original cylinder and the area of the circular base of each cone is one ﬁfth that of the circular base of the cylinder, then how many conical ingots can be made?

- A.
40

- B.
30

- C.
60

- D.
20

- E.
10

Answer: Option B

**Explanation** :

Volume of a cylinder = π × (base area) × height

Volume of a cylinder = 1/3 × π × (base area) × height

Volume of a cone with same base and height as that of a cylinder is 1/3rd the volume of the cylinder.

If height of the cone is half that of the cylinder, volume will also become half. Also, if the base area is 1/5th the volume will also be 1/5th.

∴ Volume of the given cone = 1/3 × 1/2 × 1/5 = 1/30^{th} the volume of the original cylinder.

∴ 30 such cones can be made from the original cylinder.

**Alternately,**

Let A and h be the base area and height of the cylinder.

Base area of the cone = A/5 and height = h/2

Volume of a cylinder (V) = πAh

Volume of a cylinder = 1/3 × π × A/5 × h/2 = πAh/30 = V/30

∴ Volume of the given cone is 1/30^{th} the volume of the cylinder.

⇒ 30 such cones can be made from the original cylinder.

Hence, option (b).

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**4. XAT 2019 QADI | Arithmetic - Average**

A ﬁrm pays its ﬁve clerks Rs. 15,000 each, three assistants Rs. 40,000 each and its accountant Rs. 66,000. Then the mean salary in the ﬁrm comprising of these nine employees exceeds its median salary by rupees

- A.
14600

- B.
14000

- C.
15480

- D.
15200

- E.
14720

Answer: Option B

**Explanation** :

To get the median salary, we arrange the salaries of the 9 employees in numerical ascending order.

The salaries when arranged in ascending numerical order will be as follows

15000, 15000, 15000, 15000, 15000, 40000, 40000, 40000, 66000

Now the median of these 9 values will be the middle value i.e., 5^{th} value which is 15,000.

∴ Median = 15000

Mean Salary of these 9 employees = (15000×5 + 40000×3 + 66000)/(5 + 3 + 1) = 29,000

∴ Mean salary exceeds Median Salary by 29000 -15000 = 14000.

Hence, option (b).

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**5. XAT 2019 QADI | Algebra - Surds & Indices**

If $\sqrt[3]{{7}^{a}\times (35{)}^{(b+1)}\times (20{)}^{(c+2)}}$ is a whole number then which one of the statements below is consistent with it?

- A.
a = 3, b = 2, c = 1

- B.
a = 3, b = 1, c = 1

- C.
a = 2, b = 1, c = 2

- D.
a = 2, b = 1, c = 1

- E.
a = 1, b = 2, c = 2

Answer: Option A

**Explanation** :

Let us prime factorize all the terms under the cube root sign.

7^{a }× (35)^{(b+1) }× (20)^{(c+2)} = 7^{a} × 5^{b+1} × 7^{b+1} × 2^{2(c+2)} × 5^{c+2}

= 2^{2(c+2)} × 5^{b+c+3} × 7^{a+b+1}

We have to calculate cube-root of this expression, hence powers of all prime numbers should be divisible by 3 for cube-root to be a whole number.

∴ 2(c + 2), (b + c + 3) and (a + b + 1) all should be multiples of 3.

Only option (a) satisfies this given condition.

Hence, option (a).

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**6. XAT 2019 QADI | Geometry - Circles**

Let C be a circle of radius √20 cm. Let l_{1}, l_{2} be the lines given by 2x − y −1 = 0 and x + 2y −18 = 0, respectively. Suppose that l_{1} passes through the center of C and that l_{2} is tangent to C at the point of intersection of l_{1} and l_{2}.

If (a, b) is the center of C, which of the following is a possible value of a + b?

- A.
11

- B.
14

- C.
17

- D.
8

- E.
20

Answer: Option C

**Explanation** :

l_{1}: 2x – y = 1

l_{2}: x + 2y = 18

Slope for l_{1} i.e., m_{1} = -2/-1 = 2 and slope for l_{2} i.e., m_{2} = -1/2

Here, m_{1} × m_{2} = -1, hence the two given lines are perpendicular

Let us first represent the figure and the 2 lines l_{1} and l_{2}.

Let O (a, b) be the centre of the circle and A be the point of intersection of the 2 lines l_{1} and l_{2}.

The coordinates of A can be found out by solving the simultaneous equations of the lines l_{1}(2x – y = 1) and l_{2} (x + 2y = 18).

Solving both these equations we get the value of x and y as 4 and 7 respectively.

∴ Coordinates of A are (4, 7).

Also, distance OA is the radius of the circle i.e., √20 units.

∴ (a - 4)^{2} + (b - 7)^{2} = 20 …(1)

Also (a, b) lies on l_{1} hence, 2a − b = 1

⇒ b = 2a – 1 ...(2)

From (1) and (2) we get

(a - 4)^{2} + (2a - 1 - 7)^{2} = 20

⇒ 5a^{2} – 40a + 60 = 0

⇒ a^{2} – 8a + 12 = 0

⇒ a = 2 or 6.

If a = 2, b = 3, hence a + b = 5

If a = 6, b = 11, hence a + b = 17

∴ (a + b) is either 5 or 17.

From 5 and 17 only 17 is listed in option (c).

Hence, option (c).

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**7. XAT 2019 QADI | Arithmetic - Profit & Loss**

An article is marked x% above the cost price. A discount of 2/3x% if given on the marked price. If the profit is 4% of the cost price and the value of x lies between 25 and 50, then the value of 50% of x is:

- A.
15

- B.
13

- C.
20

- D.
16

- E.
12

Answer: Option A

**Explanation** :

Let the CP of the article be Rs 100.

∴ M.P = 100 + x

As a discount of 2x/3 % is given on the Marked Price.

SP = (100 + x)$\left(1-\frac{2x/3}{100}\right)$ …(1)

Also, since profit is 4%, it means selling price = 100 + 4 = 104 …(2)

Equating (1) and (2)

⇒ (100 + x)$\left(1-\frac{2x/3}{100}\right)$ = 104

⇒ 100 + x – $\frac{2x}{3}$ – $\frac{2{x}^{2}}{300}$ = 104

⇒ $\frac{2{x}^{2}}{300}$ - $\frac{x}{3}$ + 4 = 0

⇒ x^{2} – 50x + 600 = 0

⇒ x = 20 or 30.

Since it is given that x is between 25 and 50 hence x = 30.

∴ 50% of 30 = 15.

Hence, option (a).

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**8. XAT 2019 QADI | Geometry - Coordinate Geometry**

Let P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018 and Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1. If the line through P and Q has slope 2, the value of a is:

- A.
1/2

- B.
1

- C.
4035

- D.
1009

- E.
3026

Answer: Option E

**Explanation** :

P be the point of intersection of the lines 3x + 4y = 2a and 7x + 2y = 2018.

Solving these two equations we get the coordinates of P in terms of a.

∴ P ≡ $\left(\frac{4036-2a}{11},\frac{7a-3027}{11}\right)$

Q the point of intersection of the lines 3x + 4y = 2018 and 5x + 3y = 1

Solving these two equations we get the coordinates of Q.

∴ Q ≡ (-550, 917)

Now slope of line connecting P and Q is 2

∴ $\frac{{\displaystyle \frac{7a-3027}{11}}-917}{{\displaystyle \frac{4036-2a}{11}}+550}$ = $\frac{2}{1}$

⇒ $\frac{7a-13114}{10086-2a}$ = $\frac{2}{1}$

⇒ 11a = 33286

⇒ a = 3026

Hence, option (e).

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**9. XAT 2019 QADI | Algebra - Number Theory**

When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.

During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted. When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)

- A.
14

- B.
13

- C.
15

- D.
11

- E.
12

Answer: Option E

**Explanation** :

In the given triangular array, 1st row will have 1 apple, 2nd row will have 2 apples and so on.

∴ n^{th} row will have n apples.

⇒ Total number of apples initially = 1 + 2 + 3 + … + n = $\frac{n(n+1)}{2}$

Initially total number of apples should be greater than 126 but less than or equal to 150

∴ 126 < $\frac{n(n+1)}{2}$ ≤ 150

Only integral value of n satisfying this is 16.

⇒ $\frac{16\times 17}{2}$ = 136

∴ Initially there are 16 rows and 136 apples arranged in a triangular array in these 16 rows.

As the customer observes 126 apples, it would mean that 10 apples were removed.

Now, apples are removed starting from Row 1, which has 1 apple. So apples must have been removed in such a way with 1 apple from Row 1, 2 apples from Row 2, 3 apples from Row 3 and 4 apples from Row 4.

∴ Total 10 apples were removed the first 4 rows.

As out of 16 rows, apples are removed till Row 4, it would mean that only 16 - 4 or 12 rows of apples are visible to the customer.

Hence, option (e).

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**10. XAT 2019 QADI | Algebra - Logarithms**

$\frac{\mathrm{log}\left(97-56\sqrt{3}\right)}{\mathrm{log}\left(\sqrt{7+4\sqrt{3}}\right)}$ equals which of the following?

- A.
-5

- B.
-3

- C.
None of the others

- D.
-4

- E.
-2

Answer: Option D

**Explanation** :

Given $\frac{\mathrm{log}\left(97-56\sqrt{3}\right)}{\mathrm{log}\left(\sqrt{7+4\sqrt{3}}\right)}$

= $\frac{\mathrm{log}{\left(7-4\sqrt{3}\right)}^{2}}{\mathrm{log}{\left(7+4\sqrt{3}\right)}^{1/2}}$

Now, $7+4\sqrt{3}$ = $\frac{1}{7-4\sqrt{3}}$

⇒ $\frac{\mathrm{log}{\left(7-4\sqrt{3}\right)}^{2}}{\mathrm{log}{\left(7+4\sqrt{3}\right)}^{1/2}}$ = $\frac{\mathrm{log}{\left(7-4\sqrt{3}\right)}^{2}}{\mathrm{log}{\left(7-4\sqrt{3}\right)}^{-1/2}}$ = $\frac{2}{-1/2}$ = -4

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**11. XAT 2019 QADI | Geometry - Quadrilaterals & Polygons**

In the trapezium ABCD the sides AB and CD are parallel. The value of $\frac{\mathrm{sin}\angle BAC}{\mathrm{sin}\angle BAD}$ is

- A.
AD/AC

- B.
AB/CD

- C.
BC/AD

- D.
AC/AD

- E.
AC/CD

Answer: Option A

**Explanation** :

Refer the diagram below:

In ∆ACL,

Sin∠BAC = Sin∠LAC = perpendicular/hypotenuse = CL/AC …(1)

Similarly, in ∆ADM,

Sin∠BAD = Sin∠MAD = DM/AD …(2)

(1) ÷ (2)

⇒ Sin∠BAC : Sin∠BAD = CL/AC : DM/AD

Now, CL = DM

Hence, Sin∠BAC : Sin∠BAD = 1/AC : 1/AD = AD : AC

Hence, option (a).

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**12. XAT 2019 QADI | Arithmetic - Simple & Compound Interest**

A computer is sold either for Rs.19200 cash or for Rs.4800 cash down payment together with five equal monthly installments. If the rate of interest charged is 12% per annum, then the amount of each installment (nearest to a rupee) is

- A.
Rs. 2965

- B.
Rs. 2990

- C.
Rs.3016

- D.
Rs. 2896

- E.
Rs. 2880

Answer: Option A

**Explanation** :

Price of computer = Rs 19,200

Down payment = Rs. 4,800

∴ Amount borrowed (i.e., loan taken) = 19200 – 4800 = Rs. 14,400.

This amount has to be repaid equally over 5 months at 12% p.a.

Let Rs. x is paid at the end of every month.

Interest charged for 1 month = 12%/12 = 1%

∴ 14400$\left(1+\frac{1\times 5}{100}\right)$ = x$\left(1+\frac{1\times 4}{100}\right)$ + x$\left(1+\frac{1\times 3}{100}\right)$ + x$\left(1+\frac{1\times 2}{100}\right)$ + x$\left(1+\frac{1\times 1}{100}\right)$ + x

⇒ 15120 = 1.04x + 1.03x + 1.02x + 1.01x + x

⇒ 15120 = 5.1x

⇒ x = 15120/5.1 ≈ 2965.

Hence, option (a).

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**13. XAT 2019 QADI | Geometry - Quadrilaterals & Polygons**

In the picture below, EFGH, ABCD are squares, and ABE, BCF, CDG, DAH are equilateral triangles. What is the ratio of the area of the square EFGH to that of ABCD?

- A.
√2 + 2

- B.
3 + √2

- C.
√2 + √3

- D.
√3 + √2

- E.
1 + √3

Answer: Option D

**Explanation** :

In □EFGH, EG is the diagonal.

Also, EI and GJ are the perpendicular bisectors of the equilateral triangles AEB and GCD.

Let us suppose AB = ’a’ units.

BC will also be ‘a’ units since AB and BC are sides of the same square.

In ∆ABE, EI = a√3/2 (perpendicular of an equilateral triangle is √3/2 times the side)

Similarly, GJ = a√3/2

Also, IJ = BC = a

∴ EG = EI + IJ + GJ = a√3/2 + a + a√3/2 = a(√3 + 1)

∴ Side of square EFGH = a(√3 + 1)/√2

⇒ Area of square EFGH = [a(√3 + 1)/√2]^{2} = (2 + √3)a^{2}

Also, area of square ABCD = a^{2}

Area of square ABCD = a^{2}

Ratio of area of EFGH to area of ABCD = (2 + √3)a^{2} : a^{2} = (2 + √3) : 1

Hence, option (d).

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**14. XAT 2019 QADI | Geometry - Triangles**

The figure below shows two right angled triangles ∆OAB and ∆OQP with right angles at vertex A and P, respectively, having the common vertex O, The lengths of some of the sides are indicated in the figure. (Note that the figure is not drawn to scale.) AB and OP are parallel. What is ∠QOB?

- A.
tan

^{–1}(2/3) - B.
45°

- C.
60°

- D.
30°

- E.
tan

^{–1}(3/2)

Answer: Option B

**Explanation** :

Consider the figure given below.

In ∆POQ, OQ = $\sqrt{{1}^{2}+{2}^{2}}$ = √5

In ∆AOB, OB = $\sqrt{{1}^{2}+{3}^{2}}$ = √10

In ∆DQB, QB = $\sqrt{{2}^{2}+{1}^{2}}$ = √5

Now, in ∆QOB, OB^{2} = OQ^{2} + QB^{2}

⇒ ∆QOB is a right isosceles triangle, right angles at Q.

∴ ∠QOB = 45°

Hence, option (b).

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**15. XAT 2019 QADI | Algebra - Functions & Graphs**

Consider the function f(x) = (x + 4)(x + 6)(x + 8) ⋯ (x + 98). The number of integers x for which f(x) < 0 is:

- A.
48

- B.
26

- C.
49

- D.
24

- E.
23

Answer: Option D

**Explanation** :

(x + 4)(x + 6)(x + 8) ⋯ (x + 98) < 0

Critical points are -98, -96, -94, … -8. -6 and -4

For x > -4, the given expression will be positive. Hence, we will reject all values of x > -4.

For -6 < x < -4, the given expression will be negative. The only integral value for x in this range is -5.

For -8 < x < -6, the given expression will be positive.

For -10 < x < -8, the given expression will be negative. The only integral value for x in this range is -9.

The given expression will be negative for x = -5, -9, -13, and so on

Now the only integral value of x between -98 and -96 is -97

∴ f(-97) = -93 × - 91 × -89 × … × -1 × 1

Here we have product of 47 negative terms and 1 positive term, hence the product will be negative.

∴ For -98 < x < -96, the given expression will be negative. The only integral value for x in this range is -97.

Finally, for x < -98, f(x) is product of 48 negative terms, which will give us a positive number. Hence, we will reject all value of x < -98.

The given expression will be negative for x = -5, -9, -13, and so on till -97

Number of possible values of x for which f(x) < 0 = (-5 + 97)/4 + 1 = 24

Hence, option (d).

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**16. XAT 2019 QADI | Geometry - Triangles**

Let ABC be an isosceles triangle. Suppose that the sides AB and AC are equal and let the length of AB be x cm. Let b denote the angle ∠ABC and sin b = 3/5. If the area of the triangle ABC is M sq. cm, then which of the following is true about M?

- A.
x

^{2}/4 ≤ M < x^{2}/2 - B.
3x

^{2}/4 ≤ M < x^{2} - C.
M ≥ x

^{2} - D.
x

^{2}/2 ≤ M < 3x^{2}/4 - E.
M < x

^{2}/4

Answer: Option A

**Explanation** :

Now ∆ABC is an isosceles triangle where AB = AC. Let a perpendicular from A meet BC at D. As ∆ABC is isosceles, AD is a perpendicular bisector and BD = CD.

Given, Sin b = 3/5

⇒ Cos b = 4/5

In ∆ABD,

Sin b = $\frac{3}{5}$ = $\frac{AD}{AB}$

⇒ AD = $\frac{3}{5}AB$ = $\frac{3x}{5}$

Also, Cos b = $\frac{4}{5}$ = $\frac{BD}{AB}$

BD = $\frac{4}{5}AB$ = $\frac{4x}{5}$

∆ABC, BD = CD = 4x/5

∴ BC = 8x/5

⇒ Area of ∆ABC = ½ × BC × AD = ½ × 8x/5 × 3x/5 = 12x/25 = 0.48x.

Looking at the options we can see that M lies between x^{2}/4 and x^{2}/2.

Hence, option (a).

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**17. XAT 2019 QADI | Algebra - Number Theory**

If x^{2} + x + 1 = 0, then x^{2018 }+ x^{2019} equals which of the following:

- A.
-x

- B.
x

- C.
None of the others

- D.
x - 1

- E.
x + 1

Answer: Option A

**Explanation** :

x^{2} + x + 1 = 0 …(1)

⇒ x^{2} = - x – 1 …(2)

Multiplying x in (2)

⇒ x^{3} = - x^{2} – x

⇒ x^{3} = 1 …(3) [From (1)]

Now,

x^{2018} + x^{2019} = x^{2018}(1 + x)

= x^{2018} × -x^{2}

= -x^{2020}

= - (x^{3})^{673} × x

= - 1 × x

= - x

∴ x^{2018} + x^{2019 }= - x

Hence, option (a).

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**18. XAT 2019 QADI | Geometry - Circles**

What is the maximum number of points that can be placed on a circular disk of radius 1 metre (some of the points could be placed on the bounding circle of the disk) such that no two points are at a distance of less than 1 metre from each other?

- A.
5

- B.
8

- C.
6

- D.
9

- E.
7

Answer: Option E

**Explanation** :

If we take one point at the centre of the circle, the remaining points can only be at the circumference of the circle as the minimum distance between any 2 points is at least 1 m (which is the radius of the circle).

Also, the remaining points on the circumference of the circle have to be such that 2 points are at a distance of less than one.

Now circumference of the circle = 2 × 22/7 × 1 = 44/7 ≈ 6.28 m

As the circumference or the length of the boundary is 6.28 m, we can have a maximum of 6 points on the circumference such that the distance between any 2 points is at least 1 m.

So, in all we can have a maximum of 6 points on the circumference and 1 point at the centre of the circle, making a total of 7 points.

Hence, option (e).

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**19. XAT 2019 QADI | Modern Math - Permutation & Combination | Data Sufficiency**

A bag contains marbles of three colours-red, blue and green. There are 8 blue marbles in the bag. There are two additional statement of facts available:

- If we pull out marbles from the bag at random, to guarantee that we have at least 3 green marbles, we need to extract 17 marbles.
- If we pull out marbles from the bag at random, to guarantee that we have at least 2 red marbles, we need to extract 19 marbles.

Which of the two statements above, alone or in combination shall be sufficient to answer the question "how many green marbles are there in the bag"?

- A.
Both statements taken together are suﬃcient to answer the question, but neither statement alone is suﬃcient.

- B.
Each statement alone is suﬃcient to answer the question.

- C.
Statements 1 and 2 together are not suﬃcient, and additional data is needed to answer the question.

- D.
Statement 2 alone is suﬃcient, but statement 1 alone is not suﬃcient to answer the question.

- E.
Statement 1 alone is suﬃcient, but statement 2 alone is not suﬃcient to answer the question.

Answer: Option D

**Explanation** :

We know that in all there are 8 blue marbles.

Let us first look at statement I.

As per statement I if we are to pull out 17 marbles from the bag, we will ensure that there are at least 3 greeen marbles. Now out of 17 marbles removed, 8 are blue .So from the 9 marbles that are removed, if at least 3 are green, it would mean maximum possible no of red marbles removed are 9-3 or 6.Which means that the red marbles in the bag are 6.However, this statement alone gives us no information about the no of green marbles. Hence statement I alone is not sufficient to answer the question.

Let us next look at statement II.

As per this statement, if we are to pull out 19 marbles from the bag, we would have at least 2 red marbles. Out of the 19 marbles removed, suppose 8 are blue. Now out of the remaining 11 marbles removed, if we have at least 2 red marbles, it would mean that the maximum possible no of green marbles removed is 9.This means that in all there are 9 green marbles in the bag .

Hence statement 2 alone is sufficient to answer the question.

Hence, option (d).

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**20. XAT 2019 QADI | Algebra - Number Theory | Data Sufficiency**

We have two unknown positive integers m and n, whose product is less than 100.

There are two additional statement of facts available:

- mn is divisible by six consecutive integers { j, j + 1,...,j + 5 }
- m + n is a perfect square.

Which of the two statements above, alone or in combination shall be sufficient to determine the numbers m and n?

- A.
Statements 1 and 2 together are not suﬃcient, and additional data is needed to answer the question.

- B.
Both statements taken together are suﬃcient to answer the question, but neither statement alone is suﬃcient.

- C.
Each statement alone is suﬃcient to answer the question.

- D.
Statement 2 alone is suﬃcient, but statement 1 alone is not suﬃcient to answer the question.

- E.
Statement 1 alone is suﬃcient, but statement 2 alone is not suﬃcient to answer the question.

Answer: Option B

**Explanation** :

Looking at the initial statement , we know that mn < 100.

Looking at statement I ,we get to know that the product mn has to be a multiple of 10 since it is divisible by 6 consecutive integers .So the product can be either 10, 20, 30 ….90

Now of all these numbers only 60 is divisible by 6 consecutive numbers i.e. numbers 1 to 6.60 can be expressed as a product of 2 nos. in the following ways : 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, 6 ×10

So from statement I alone we cannot determine values of m and n. Looking at statement II alone determine values of m and n as the only information provided to us is that “m + n” is a perfect square. So we can have numerous possibilities for m and n [e.g (7, 9), (2, 7), (1, 3) etc ]

Combining both statements out of (1, 60), (2, 30), (3, 20), (4,15), (5, 12), (6, 10), the only pair of values such that “m+n” is a perfect square is (6, 10). Hence both statements are required to answer the question.

Hence, option (b).

Workspace:

**Answer the next 3 questions based on the information given below**.

Given below is the time table for a trans-continental train that cuts across several time zones. All timings are in local time in the respective cities. The average speed of the train between any two cities is the same in both directions.

**21. XAT 2019 QADI | LR - Mathematical Reasoning**

Which of the following pairs of cities are in the same time zone?

- A.
Vaq and Sab

- B.
No pair of cities are in the same time zone.

- C.
Sab and Raz

- D.
Sab and Raz

- E.
Zut and Yag

Answer: Option E

**Explanation** :

When travelling between 2 cities in different time zones, the actual travelling time is the average of two time taken both ways between the cities when calculated on local basis.

Also, the time difference between the two cities is half of difference of time taken both ways between the cities when calculated on local basis.

Let’s take the example of journey between Vaq and Yag.

Time taken to go from Vaq to Yag, on local basis (11:55 pm to 10:50 am) = 10 hours 55 mins.

Time taken to go from Yag to Vaq, on local basis (7:50 am to 2:45 pm) = 6 hours 55 mins.

∴ The actual travelling time between Yag and Vaq is average of 10 hours 55 mins and 6 hours 55 mins i.e., 8 hours 55 mins.

Also, the time difference between the two cities is half of (10 hours 55 mins - 6 hours 55 mins) i.e., 2 hours.

Similarly, we can calculate the travelling time and time difference between all the cities.

From the table we can see that Zut and Yag are in the same time zone.

Hence, option (e).

Workspace:

**22. XAT 2019 QADI | LR - Mathematical Reasoning**

What is the total time taken in minutes by the train to go from Zut to Raz?

- A.
28 hours, 40 minutes

- B.
22 hours, 40 minutes

- C.
24 hours, 40 minutes

- D.
20 hours, 40 minutes

- E.
16 hours, 40 minutes

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Journey time of train from Zut to Raz = 1 hour 45 mins + 8 hours 55 mins + 7 hours 50 mins + 3 hours 55 mins = 22 hours 25 mins.

Waiting time at stations Yag, Vaq and Sab = 5 + 5 + 5 = 15 mins.

Total travel time between Zut to Raz = 22 hours 25 mins + 15 mins = 22 hours 40 mins.

Hence, option (b).

Workspace:

**23. XAT 2019 QADI | LR - Mathematical Reasoning**

What time is it at Yag when it is 12:00 noon at Sab?

- A.
5:00 pm

- B.
7:00 pm

- C.
9:30 pm

- D.
2:30 pm

- E.
12:00 pm

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Time difference between Yag and Sab = 2 + 3 = 5 hours..

∴ Yag is 5 hours ahead of Sab.

∴ If it is 12 noon at Sab, it will be 5 pm at Yag.

Hence, option (a).

Workspace:

**Answer the next 2 questions based on the information given below.**

The break-up of the students in a university by subject major is given in the polar pie-chart. The bar chart shows the number of students who major in physics by geographic location.

**24. XAT 2019 QADI | DI - Tables & Graphs**

How many students major in chemistry?

- A.
200

- B.
180

- C.
190

- D.
170

- E.
175

Answer: Option E

**Explanation** :

Total no of majors in Physics across all geographical locations = 18 + 36 + 80 + 40 + 23 + 28 = 225

Now students majoring in Physics account for 18% of all students

∴ 18% of total number of students = 225

∴ Total number of students = 225/0.18 = 1250

∴ No of students majoring in Physics = 14% of 1250 = 125 + 50 = 175

Hence, option (e).

Workspace:

**25. XAT 2019 QADI | DI - Tables & Graphs**

12% of all students are from Chennai. What is the largest possible percentage of economics students that can be from Chennai, rounded oﬀ to the nearest integer?

- A.
69%

- B.
71%

- C.
75%

- D.
73%

- E.
77%

Answer: Option D

**Explanation** :

Consider the solution to previous question.

Total number of students = 1250

12% of these are from Chennai = 12% of 1250 = 125 + 25 = 150

Out of these 150 students 40 are definitely majoring in Physics.

∴ Maximum students from Chennai who can major in Economics = 150 – 40 = 110.

Total number of students in Economics = 12% of 1250 = 150.

∴ Highest possible percentage of Economics students that can be from Chennai = 110/150 × 100% = 73.33% ≈ 73%.

Hence, option (d).

Workspace:

**26. XAT 2019 QADI | DI - Tables & Graphs**

If the proportion of physics majors who are from Delhi is the same as the proportion of engineering majors who are from Delhi, how many engineering majors are from Delhi?

- A.
22

- B.
26

- C.
18

- D.
20

- E.
24

Answer: Option D

**Explanation** :

Total number of majors = 1250.

Number of majors in Engineering ≈ 20% 0f 1250 = 250.

Proportion of Physics major from Delhi = 18/225

Let number of Engineering majors from Delhi = d

∴ d/250 = 18/225

⇒ d = 18 × 250/225 = 20

Hence, option (d).

Workspace:

**27. XAT 2019 QADI N/A**

Given that a and b are integers and that 5x + 2√7 is a root of the polynomial x^{2} - ax + b + 2√7 in x, what is the value of b?

- A.
14

- B.
9

- C.
7

- D.
2

- E.
5

Answer: Option A

**Explanation** :

No solution is being provided as question is incorrect.

Full Marks were awarded to all candidates.

Workspace:

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