XAT 2015 QA | Previous Year XAT Paper

Answer: Option B

Video Explanation

Explanation :

Number of terms = (100 – 64)/2 + 1 = 19
Sum = n/2 (a + a_n)
= 19/2 × (–64 –100) = –1558
Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

Let Manufacturing Cost = Rs. 100

M.R.P = Rs. 155

Discount = 10%

S.P (of retailer) = 155 ×.9 = Rs. 139.5

Profit% (of retailer) = 23%

S.P. (of manufacturer) = 139.5 × (100/123) = Rs. 113

Profit % (of manufacturer) = 13%

Hence, option (d).

Answer: Option E

Video Explanation

Explanation :

Required Probability = 1 – P(receiving no gift)

P(receiving no gift) = 0.4 × 0.2 × 0.1 × 0.5

= 0.0040

1 – P(receiving no gift) = 0.996

Hence, option (e).

Answer: Option C

Video Explanation

Explanation :

Had the rectangle not been folded, the overlapping part would have been a square of side 6.

While unfolding, the increase in area

= Area of the triangle =  $\frac{1}{2}\times 6\times 6=18$

Area of the given figure = 144 square meters
∴ Area of unfolded rectangle = 144 + 18 = 162 square meters
Hence, option (c).

Answer: Option E

Explanation :

From the graph, observe that at y = 0, x
= –19
Only x = 2y² + 3y – 19 satisfies the above criteria.
Hence, option (e).

Answer: Option D

Video Explanation

Explanation :

Flat surface area of the cylinder = 2 × π × 7² = 98 π cm²
Volume of the cylinder = π × 7² ×10
= 490 π cm³
Volume of cone A = (3/7) × 490 π = 210 π cm³ = (1/3) × flat surface area of cone
A × 10
Flat surface area of cone A = 63 π cm²
Volume of cone B = (4/7) × 490 π = 280 π cm³ = (1/3) × flat surface area of cone
B × 10
Flat surface area of cone B = 84 π cm²
Total flat surface area of cones
= (63 + 84) π = 147 π cm²
Percentage change in the flat surface area
= (147 – 98)/98 × 100 = 50%
Hence, option (d).​​​​​​​

Answer: Option B

Video Explanation

Explanation :

​​​​​​​

Radius of inner circle = ½ × diagonal of the square = 25√2 cm
Radius of outer circle = Radius of inner circle + width of the road = 32 cm
50% of the area of the road

$=\frac{1}{2}\times \pi \times \left[{\left(32\sqrt{2}\right)}^2-{\left(25\sqrt{2}\right)}^2\right]$

= 1254 cm²

Cost = 1254 ×100 = Rs. 125,400

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

In 864 units of M,
X = 5/9 × 864 = 480 units
Y = 864 – 480 = 384 units
B in 480 units of X = 3/4 × 480 = 360 units
B in 384 units of Y = 2/3 × 384 = 256 units
Total units of B = 360 + 256 = 616
Concentration of B in the final mixture is 50%
Thus, water in the final mixture = (2 × 616) – 864 = 368 units.
Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

(13.5, 16) and (5.5, 7.5) are adjacent points of the parallelogram and (17.5, 23.5) is the point of intersection of two diagonals of the parallelogram.

Property: Diagonals of a parallelogram bisect each other.

Distance between (17.5, 23.5) and (5.5, 7.5):

$\sqrt{{(17.5-5.5)}^2+{(23.5-7.5)}^2}=20$

Length of diagonal that passes through (17.5, 23.5) and (5.5, 7.5) = 20 × 2 = 40 cm
Distance between (17.5, 23.5) and (13.5, 16):

$\sqrt{{(17.5-13.5)}^2+{(23.5-16)}^2}=8.5$

Length of diagonal that passes through (17.5, 23.5) and (13.5, 16) = 8.5 × 2 = 17 cm

Hence, option (d).

Answer: Option D

Video Explanation

Explanation :

​​​​​​​

m∠ECD = m∠BCF = 60°
Also, m∠AFB = 60°, m∠BFC = 30°
∴ m∠AFC = 90°
In a 30° - 60° - 90° triangle, sides are in the ratio $1:\sqrt{3}:2.$

So, in ∆EDC, ED = 10√3 units

Also, in ∆FBC, BF = 10 units

$\Rightarrow \mathrm{FC}=\frac{20}{\sqrt{\mathrm{3 }}}$​​​​​​​ units and BC =  ​​​​​​​$\frac{10}{\sqrt{3}}$

In ∆AFC,

$\mathrm{FC}=\frac{20}{\sqrt{3}} \mathrm{units}\Rightarrow \mathrm{AC}=\frac{40}{\sqrt{3}} \mathrm{units}$

$\therefore \mathrm{AD}=\left(10+\frac{40}{\sqrt{3}}\right) \mathrm{units}$

$\mathrm{A}(∆\mathrm{ADE})=\frac{1}{2}\times 10\sqrt{3}\times \left(10+\frac{40}{\sqrt{3}}\right)$

= 50(√3 + 4) sq.units

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

Substitute x = 0 in f(x² – 1) = x⁴ – 7x² + k₁
⇒ f(–1) = k₁ … (i)
Substitute x = 1 in f(x₃ – 2) = x⁶ – 9x³ + k₂
⇒ f(–1) = 1 – 9 + k₂ … (ii)
From (i) and (ii), k₁ = –8 + k₂
∴ k₂ – k₁ = 8
Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Let P be the principal.

By the given conditions,

$P\times {\left(1+\frac{r}{100}\right)}^3-P=10000$

$$\begin{gathered} \mathrm{} \\ P\times {\left(1+\frac{r}{100}\right)}^6-P=25000 \end{gathered}$$​​​​​​​​​​​​​​

Let  ​​​​​​​${\left(1+\frac{r}{100}\right)}^3=X$

Thus,

P(X - 1) = 10000             ...(i)

P(X² - 1) = 25000           ...(ii)

Dividing (ii) by (i),

X + 1 = 5/2

∴ X = 3/2

Substituting value of X in (i), we get

P = Rs. 20,000

Hence, option (c).

Answer: Option A

Explanation :

​​​​​​​

Total tax will be minimum if income of 5 persons is not more than 500 and income of 4 persons is minimum and also in the range (500, 2000]. Total tax will be maximum if income of 5 persons is maximum possible and income of 4 persons is maximum in the range (2000, 5000].

Minimum tax > (5 × 0) + (4 × 0) + (3 × 75) + (3 × 375) = Rs. 1350

Maximum tax range < (3 × 0) + (3 × 75) + (4 × 375) + (5 × 1125) = Rs. 7350

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

Let a + b + c = X

We need to find the maximum value of:

$\frac{X+d}{X-d}=\frac{(X-d)+2d}{X-d}=1+\frac{2d}{X-d}$

$$\begin{gathered} \frac{}{} \\ \frac{2d}{X-d} \end{gathered}$$

$\frac{\mathrm{is\; maximum\; when\; d\; is\; maximum}}{}$ and X is minimum possible.

∴ d = 25 and X = 26
The highest possible value of the given expression = 1 + (2 × 25) = 51
Hence, option (c).

Answer: Option A

Explanation :

Scenario 1: Devanand walks East at a constant speed of 3 km per hour and Pradeep towards South at a constant speed of 4 km per hour,

Let the two walk for x hours.

Distance travelled by Devanand and Pradeep is 3x km and 4x km respectively.

Thus, we have

​​​​​​​

∴ (AP)² + (CP)² = (50 – 3x)² + (4x)² = 402
Solving this, we get x = 6
Thus, after 6 hours, the minimum distance between the two would be 40 km.
The distance between the two will increase in other two scenarios.
Hence, option (a).

Answer: Option E

Explanation :

Three integers are not known.
Using Statement I:
Average of four smallest integers
= (6 + 8 + 12 + 13)/4 = 39/4
∴ Average of four largest integers

$=\frac{39}{4}+13\frac{1}{4}=\frac{92}{4}$

In order to get the largest possible integer, two of the three unknown integers must be lowest possible i.e., 16 and 17.
So, the largest possible integer
= 92 – 22 – 20 – 17 = 33
Statement I can answer the question independently.

Using Statement II:
Sum of 11 integers = 11 × 16 = 176
Sum of the given integers = 110
∴ Sum of three unknown integers = 66
In order to get the largest possible integer, two of the three unknown integers must be lowest possible i.e., 16 and 17.
So, the largest possible integer
= 66 – 16 – 17 = 33
Statement II also answers the question independently.
Hence, option (e).

Answer: Option E

Video Explanation

Explanation :

f(1) = 4
f(2) = f(1+1) = f(1 × 1) = f(1) = 4
…
Assume that f(n – 1) = 4
f(n) = f(1 + (n – 1)) = f(1 × (n – 1)) = f(n – 1) = 4
Thus, by induction principle, f(n) = 4, for all n.
∴ f(1003) = k = 4
Hence, option (e).

Answer: Option C

Video Explanation

Explanation :

LCM (3, 4, 5, 6) = 60
From (i), numbers are of the form 60k + 2.
From (ii), numbers are of the form 11m.
11m = 60k + 2 = 55k + (5k + 2)
11 divides (5k + 2)
Units digit of (5k + 2) is 2 or 7
So, values of (5k + 2) are 11 × 2, 11 × 7, 11 × 12, 11 × 17, 11 × 22, 11 × 27, … and so on.
We need to find the 6th number of the ascending series.
∴ 5k + 2 = 11 × 27 ⇒ k = 59
The required number = 60 × 59 + 2 = 3542
Hence, option (c).

Answer: Option A

Explanation :

Let m∠DAB = θ ⇒ m∠BCD = 2θ
□PBCD is a parallelogram.
∴ m∠DPB = 2θ
m∠PBC = m∠PDC = (180 – 2θ)
∠DPB is an exterior angle of ∆PAB.
∴ By exterior angle theorem, m∠PBA = θ
as, m∠DAB = θ
Thus, in ∆PAB, PA = PB

​​​​​​​

∴ m∠ABC = θ + (180 – 2θ) = 180 – θ
According to the given conditions, 10x + y = 1120 and 10x = 1000
Solving the two equations, we get x = 100 and y = 120
sin (180 – θ) = sin θ
Applying sine rule to ∆PAB,

$\frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin (180-2\mathrm{\theta })}$

$$\begin{gathered} \frac{}{} \\ \frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin 2\mathrm{\theta }} .....\left(\mathrm{i}\right) \end{gathered}$$

sin 2θ = 2 × sin θ × cos θ

∴ (i) becomes

$\frac{100}{\sin \theta }=\frac{120}{2\times \sin \theta \times \cos \theta }$

$$\begin{gathered} \frac{}{} \\ \therefore \cos \theta =\frac{3}{5}\Rightarrow \sin \theta =\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{}}{} \\ \therefore \sin \angle ABC=\sin (180-\theta )=\sin \theta =\frac{4}{5} \end{gathered}$$

Hence, option (a).

Answer: Option A

Explanation :

Maximum marks = 30

Marks scored if exactly 29 questions were answered correctly.

Case (1): One question is incorrectly answered.

Case (1a): A question of 1 mark is answered incorrectly.

Total marks = 30 – 1 – 0.25 = 28.75

Case (1b): A question of 2 marks is answered incorrectly.

Total marks = 30 – 2 – 0.33 = 27.67

Case (2): One question is not answered.

Case (2a): A question of 1 mark is not answered.

Total marks = 30 – 1 – 0.5 = 28.5

Case (2b): A question of 2 marks is not answered.

Total marks = 30 – 2 – 0.5 = 27.5

If these marks are arranged in descending order, (27.5) comes in the fifth place.

i.e., the rank of a student, who scores a total of 27.5 marks, would be 5.

Hence, option (a).

Answer: Option B

Explanation :

OA ⊥ PQ, OB ⊥ PR

OP = OQ = OR = 625 cm

In ∆OAQ, OA = 175 cm and OQ = 625 cm ⇒ AQ = 600 cm

Similarly, PA = PB = RB = 600 cm

∆PQR is an isosceles triangle and PQ = PR = 1200 cm

So, PC ⊥ QR

In ∆PBO and ∆PCR,

∠OPB ≅ ∠RPC   … (Common angle)

∠PBO ≅ ∠PCR   … (Right angle)

∆PBO ~ ∆PCR       … (AA test of similarity)

$\therefore \frac{\mathrm{PB}}{\mathrm{PC}}=\frac{\mathrm{BO}}{\mathrm{CR}}=\frac{\mathrm{PO}}{\mathrm{PR}}$

$\therefore \frac{600}{\mathrm{PC}}=\frac{175}{\mathrm{CR}}=\frac{625}{1200}$

∴ PC = 1152 cm and CR = 336 cm

∴ QR = 672 cm

A(△PQR) = $\frac{1}{2}\times 672\times 1152$ = 387072

Hence, option (b).

Answer: Option B

Explanation :

M and N are positive integers such that M > N

∴ M! – N! = abc…999000

∴ [M(M − 1)(M − 2)……N!] – N! = abc…999000

∴ N!{[M(M − 1) (M − 2)……] − 1} = abc…999000

Let the term in the square bracket be x.

Since M is a positive integer, the term in the square brackets i.e., x is also a positive integer, and hence, (x – 1) is also a positive integer.

∴ N!(x – 1) = abc…999000

$\therefore (x-1)=\frac{abc...999000}{N!}$

Hence, the maximum number of zeroes in N! is 3.

∴ N! ≤ 19 (because from 20! onwards, each factorial has atleast 4 zeroes)

Now, there are 4 possible ranges for N:

1. ​​​​​​​N = 0 to 4 (no zeroes in N!) 
2. N = 5 to 9 (1 zero in N!) 
3. N = 10 to 14 (2 zeroes in N!) 
4. N = 15 to 19 (3 zeroes in N!) 

Consider case 3, where there are 2 zeroes in N!

Since N!(x – 1) = abc…999000, the third zero on the LHS should come from (x – 1).

For this, x has to be of the form pqrs…1 i.e., x has to be an odd number.

Now, there are two possibilities:

1. M = N + 1

    Here, M! – N! = (M × N!) – N! = N!(M – 1)

    In this case, M = x

    Since N = 10 to 14 and M – 1 should end in 0, M = 11 and N = 10

    Hence, M(M – N) = 11(11 – 10) = 11

    This does not tally with any of the options.

     Hence, M ≠ N + 1

2. There is at least one integer between M and N

    Hence, x comprises a series of consecutive integers (at least two as explained above) multiplied with each other. So, there has to be at least one even number in this series.

    Hence, x can never be odd.

    Hence, the third zero on the LHS can never come from (x – 1).

Hence, there cannot be 2 zeroes in N!. Similarly, it can be proved that there cannot be 1 zero or no zeroes in N!.

Hence, N! has three zeroes i.e. 15 ≤ N ≤ 19

Now, M(M – N) = M² – NM can take four of the five given values.

Hence, there are five possible equations – one for each option.

Consider option 1: M² – NM – 150 = 0

Here, for N = 15 to 19, see if there exists a value of M (> N) that gives positive integral roots in this equation.

For N = 19, the equation becomes

M² – 19M – 150 = 0 i.e. M = 25 or M = −6. Here, M = 25 is valid.

Similarly, consider each option.

Option 3:

For N = 17, the equation becomes

M² – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.

Option 4:

For N = 16, the equation becomes

M² – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.

Option 5:

For N = 17, the equation becomes

M² – 17M – 234 = 0 i.e. M = 26 or M = −9. Here, M = 26 is valid.

However, no value of N from 15 to 19 gives an integral solution for M in M² – NM – 180 = 0

Hence, M(M – N) can never be 180.

Hence, option (b).

Note: There is a logical flaw in this question as the difference of two factorials can never end in 999000. The actual question should have had the last six digits as abc000.

Answer: Option D

Explanation :

​​​​​​​

At 5.00 p.m., position of the person be P

PB = 1800 (Given)

m∠DPB = 30°

⇒ DB = 1800 tan 30° = 600√3

At 5.10 p.m. the minute hand of the clock moves by 60°

​​​​​​​

DC = CF = 200√3 m (Given)

∆FEC is 30° - 60° - 90° triangle.

So, EC = 100√3 m and EF = 300 m

DE = DC – EC = 200√3 – 100√3 = 100√3 m

FG = DB – DE = 600√3 – 100√3 = 500√3 m

In ∆AFG, m∠FAG = 60° and FG = 500√3 m

By theorem of 30°-60°-90° triangle,

AG = 500 m

BG = EF = 300 m

In ∆ABG, AG = 500 m and BG = 300 m ⇒ AB = 400 m

PA = PB – AB = 1800 – 400 = 1400 m = 1.4 km

Time taken = 10 minutes = (1/6) hours

Speed = 1.4 × 6 = 8.4 km/hr

Hence, option (d).

Answer: Option C

Explanation :

Let the capacity of the tank be 24x litres.

Pipes A and B fill 3x and 2x litres per hour while pipe C empties 6x litres in an hour.

Let radius of the cone be r and height be h.

$\frac{1}{3}\pi r^{2}h=24x$

∴ πr²h = 72x

For first 19 hours, water inside the cone = 24x + 57x + 38x – 114x = 5x litres

​​​​​​​

∆ABE ∼ ∆ACD

If AC = 2AB, CD = 2BE

∴ BE = r/2 and AB = h/2

After 50% reduction in the height of the water, volume

$=\frac{1}{3}\pi {(r/2)}^2(h/2)=\frac{\pi r^{2}h}{24}=\frac{72x}{24}=3x$

Option 1: Pipe A was open for 19 hours.

i.e., B and C were open for 1 more hour.

∴ 2x – 6x = –4x

The cone will have 5x – 4x = x litres of water.

∴ Option 1 is eliminated.

Option 2: Pipe A was open for 19 hours 30 minutes.

i.e., B and C were open for 1 more hour and A for 30 more minutes.

∴ 2x – 6x + 1.5x = –2.5x

The cone will have 5x – 2.5x = 2.5x litres of water

∴ Option 2 is eliminated.

Option 3: Pipe B was open for 19 hours 30 minutes.

i.e., A and C were open for 1 more hour and B for 30 more minutes.

∴ 3x – 6x + 1x = –2x

The cone will have 5x – 2x = 3x litres of water.

∴ Option 3 would be the possible option.

Hence, option (c).

Answer: Option B

Explanation :

Units digit of the number must be 0.

Let 100x + 10y is the number such that x > y.

New number obtained by changing the digits is also divisible by 10.

So, only x and y are to be interchanged

∴ New number is of the form = 100y + 10x

Difference = 90x – 90y = 90(x – y)

For the difference to be divisible by 4,

(x – y) has to be divisible by 4.

(x – y) = 4 or 8

So, y = 1 to 5

For y = 1 to 5, x = (1+4) to (5+4) i.e., 5 to 9

One more possibility for y = 1 is x = 9.

Thus, in all 6 numbers satisfy the given conditions.

Hence, option (b).

Answer: Option D

Explanation :

Observe the options.

Either E or B can have the highest gain in vote share.

Let E_G and B_G represent the gain in vote shares of E and B from 2005 and 2010.

$A_G=\left(\frac{364450}{985000}-\frac{343200}{880000}\right)\times 100=-2$

$$\begin{gathered} \mathrm{} \\ {D}_G=\left(\frac{54175}{985000}-\frac{48400}{880000}\right)\times 100=0 \end{gathered}$$

Since A_G < D_G​​

​​​​​​​Hence, option (e) is eliminated.

Hence, option (d).

Answer: Option A

Explanation :

Number of “natural tweets” for B, C, D and E are:

Party B: (100 – 30.4 – 29.7)% of 108128 = 0.399 × 108128

Party C: (100 – 32.5 – 26.6)% of 96620 = 0.409 × 96620 = (0.399 + 0.01) × 96620

Party D: (100 – 30.6 – 36.1)% of 41524  = 0.333 × 41524

Party E: (100 – 21.6 – 41.0)% of 32724 = 0.374 × 32724

Both the total number of tweets and the percentage of neutral tweets of B and C are greater than those of D and E. Hence, options 3 and 4 are eliminated.

Among B and C, number for party B is more than C.

Number of neutral tweets of B ≈ 0.4 × 108128 = 43251.2

Parties classified under ‘Other Parties’ have a total of 15000 tweets, which is significantly lower than the total number of tweets of the rest of the parties. Thus, ‘Other Parties’ cannot have the maximum number of neutral tweets even if all 15000 tweets are neutral. Hence, option (e) is eliminated.

Hence, option (a).

Answer: Option B

Explanation :

E secured less than 2% i.e. (0% to 2%) of the total votes in the year 2000.

The following table represents the individual vote shares and the gain in vote shares of parties A to E,

​​​​​​​

The values 2% and 7.5% already exist in the table.

E’s gain in vote share can have any value from 3% to 5%. The values 3.5% and 4.5% may exist within this range.

However, it is not possible to obtain 2.5% as the gain in vote share for any party.

Hence, option (b).

Answer: Option E

Explanation :

The following table represents the vote shares and tweet shares of parties B to E and ‘Other Parties’ in 2010,

​​​​​​​

Hence, option (e).

Answer: Option D

Explanation :

Observe the radar chart and consider any employee randomly, say employee 2 (E2). The effectiveness score of E2 in Survey 1 is 5 and Survey 2 is between 9 and 10. The scores for all employees for both the surveys can be similarly summarized as follows,

​​​​​​​

Observe the graph of “Days of Training Undergone” (X axis) vs “Employee Effectiveness Scores” (Y axis), and try to identify each employee. E2 can be identified by finding a triangular mark (corresponding to survey 1) against a score of 5 and a cross (corresponding to survey 2) betweenscores of 9 and 10. Similarly, each employee can be identified in this graph.

The same logic also applies to the graph on “Bonus Received” (X axis) versus “Employee Effectiveness Scores” (Y axis).

Again, consider the graphs on training and bonus. Observe that E2 has spent 10 and 21 days in training; and received approximately 27.5 lacs and 22 lacs bonus, based on surveys 1 and 2 respectively.

Similarly, the training and bonus figures for each employee can be found as shown below:

​​​​​​​

Only E4 and E5 underwent more than 17 days of training in Survey 1, and their respective bonuses (as per Survey 1) are 20.5 and 18 lacs.
Hence, their average bonus for Survey 1 is 19.25 lacs.

Answer: Option A

Explanation :

E1, E4, E5 and E7 have effectiveness scores more than 7 in Survey 1. Out of these, E4 and E7 had bonuses below 20 lacs in Survey 2.

Hence, option (a).

Answer: Option B

Explanation :

The number of days of training of E2, E3 and E6 increased from Survey 1 to Survey 2. Out of these, the bonuses of E2 and E3 decreased.

Hence, option (b).

Answer: Option A

Explanation :

The number of days of training of E2, E3 and E6 increased from Survey 1 to Survey 2. Out of these, the employee effective scores of E2 and E3 increased by at least 1.0 rating.

Hence, option (a).