XAT 2015 QA | Previous Year XAT Paper
What is the sum of the following series?
–64, –66, –68,……..….., –100
- A.
-1458
- B.
-1558
- C.
-1568
- D.
-1664
- E.
None of the above
Answer: Option B
Explanation :
Number of terms = (100 – 64)/2 + 1 = 19
Sum = n/2 (a + a_{n})
= 19/2 × (–64 –100) = –1558
Hence, option (b).
Workspace:
The Maximum Retail Price (MRP) of a product is 55% above its manufacturing cost. The product is sold through a retailer, who earns 23% profit on his purchase price. What is the profit percentage (expressed in nearest integer) for the manufacturer who sells his product to the retailer? The retailer gives 10% discount on MRP.
- A.
31%
- B.
22%
- C.
15%
- D.
13%
- E.
11%
Answer: Option D
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).
Workspace:
Ramesh plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively.
Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?
- A.
0.004
- B.
0.006
- C.
0.216
- D.
0.994
- E.
0.996
Answer: Option E
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).
Workspace:
The figure below has been obtained by folding a rectangle. The total area of the figure (as visible) is 144 square meters. Had the rectangle not been folded, the current overlapping part would have been a square. What would have been the total area of the original unfolded rectangle?
- A.
128 square meters
- B.
154 square meters
- C.
162 square meters
- D.
172 square meters
- E.
None of the above
Answer: Option C
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).
Workspace:
Find the equation of the graph shown below.
- A.
= 3x – 4
- B.
y = 2x^{2} – 40
- C.
x = 2y^{2} – 40
- D.
y = 2x^{2} + 3x – 19
- E.
x = 2y^{2} + 3y – 19
Answer: Option E
Explanation :
From the graph, observe that at y = 0, x
= –19
Only x = 2y^{2} + 3y – 19 satisfies the above criteria.
Hence, option (e).
Workspace:
A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?
- A.
9%
- B.
16%
- C.
25%
- D.
50%
- E.
None of the above
Answer: Option D
Explanation :
Flat surface area of the cylinder = 2 × π × 7^{2} = 98 π cm^{2}
Volume of the cylinder = π × 7^{2} ×10
= 490 π cm^{3}
Volume of cone A = (3/7) × 490 π = 210 π cm^{3} = (1/3) × flat surface area of cone
A × 10
Flat surface area of cone A = 63 π cm^{2}
Volume of cone B = (4/7) × 490 π = 280 π cm^{3} = (1/3) × flat surface area of cone
B × 10
Flat surface area of cone B = 84 π cm^{2}
Total flat surface area of cones
= (63 + 84) π = 147 π cm^{2}
Percentage change in the flat surface area
= (147 – 98)/98 × 100 = 50%
Hence, option (d).
Workspace:
A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs. 100 per sq. ft. Find the lowest possible cost to construct 50% of the total road.
- A.
Rs. 70,400
- B.
Rs. 125,400
- C.
Rs. 140,800
- D.
Rs. 235,400
- E.
None of the above
Answer: Option B
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 cm^{2}
Cost = 1254 ×100 = Rs. 125,400
Hence, option (b).
Workspace:
Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?
- A.
328 units
- B.
368 units
- C.
392 units
- D.
392 units
- E.
616 units
Answer: Option B
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).
Workspace:
Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals.
- A.
15 and 30
- B.
15 and 40
- C.
17 and 30
- D.
17 and 40
- E.
Multiple solutions are possible
Answer: Option D
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).
Workspace:
In the diagram below, CD = BF = 10 units and ∠CED = ∠BAF = 30°. What would be the area of triangle AED? (Note: Diagram below may not be proportional to scale.)
- A.
100 × (√2 + 3)
- B.
$\frac{100}{(\sqrt{3}+4)}$
- C.
$\frac{50}{(\sqrt{3}+4)}$
- D.
$50\times (\sqrt{3}+4)$
Answer: Option D
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
$\Rightarrow \mathrm{FC}=\frac{20}{\sqrt{3}}$ units and BC = $\frac{10}{\sqrt{3}}$
In ∆AFC,
$\mathrm{FC}=\frac{20}{\sqrt{3}}\mathrm{units}\Rightarrow \mathrm{AC}=\frac{40}{\sqrt{3}}\mathrm{units}$
$\therefore \mathrm{AD}=\left(10+\frac{40}{\sqrt{3}}\right)\mathrm{units}$
$\mathrm{A}(\u2206\mathrm{ADE})=\frac{1}{2}\times 10\sqrt{3}\times \left(10+\frac{40}{\sqrt{3}}\right)$
= 50(√3 + 4) sq.units
Hence, option (d).
Workspace:
If f(x^{2} – 1) = x^{4} – 7x^{2} + k_{1} and f(x^{3} – 2) = x^{6} – 9x^{3} + k_{2} then the value of (k_{2} – k_{1}) is
- A.
6
- B.
7
- C.
8
- D.
9
- E.
None of the above
Answer: Option C
Explanation :
Substitute x = 0 in f(x^{2} – 1) = x^{4} – 7x^{2} + k_{1}
⇒ f(–1) = k_{1} … (i)
Substitute x = 1 in f(x_{3} – 2) = x^{6} – 9x^{3} + k_{2}
⇒ f(–1) = 1 – 9 + k_{2} … (ii)
From (i) and (ii), k_{1} = –8 + k_{2}
∴ k_{2} – k_{1} = 8
Hence, option (c).
Workspace:
In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs. 10,000 and in the beginning of 2010, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is:
- A.
Rs. 16,000
- B.
Rs. 18,000
- C.
Rs. 20,000
- D.
Rs. 25,000
- E.
None of the above
Answer: Option C
Explanation :
Let P be the principal.
By the given conditions,
$P\times {\left(1+\frac{r}{100}\right)}^{3}-P=10000$
$\mathrm{}\phantom{\rule{0ex}{0ex}}P\times {\left(1+\frac{r}{100}\right)}^{6}-P=25000$
Let ${\left(1+\frac{r}{100}\right)}^{3}=X$
Thus,
P(X - 1) = 10000 ...(i)
P(X^{2 }- 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).
Workspace:
The tax rates for various income slabs are given below.
There are 15 persons working in an organization. Out of them, 3 to 5 persons are falling in each of the income slabs mentioned above. Which of the following is the correct tax range of the 15 persons? (E.g. If one is earning Rs. 2000, the tax would be: 500 × 0 + 1500 × 0.05)
- A.
1350 to 7350, both excluded
- B.
1350 to 9800, both included
- C.
2175 to 7350, both excluded
- D.
2175 to 9800, both included
- E.
None of the above
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).
Workspace:
If a, b, c and d are four different positive integers selected from 1 to 25, then the highest possible value of ((a + b) + (c + d))/((a + b) + (c – d)) would be:
- A.
47
- B.
49
- C.
51
- D.
96
- E.
None of the above
Answer: Option C
Explanation :
Let a + b + c = X
We need to find the maximum value of:
$\frac{X+d}{X-d}=\frac{(X-d)+2d}{X-d}=1+\frac{2d}{X-d}$
$\frac{}{}\phantom{\rule{0ex}{0ex}}\frac{2d}{X-d}$
$\frac{\mathrm{is\; maximum\; when\; d\; is\; maximum}}{}$ 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).
Workspace:
Devanand’s house is 50 km West of Pradeep’s house. On Sunday morning, at 10 a.m., they leave their respective houses.
Under which of the following scenarios, the minimum distance between the two would be 40 km?
Scenario I: Devanand walks East at a constant speed of 3 km per hour and Pradeep walks South at a constant speed of 4 km per hour.
Scenario II: Devanand walks South at a constant speed of 3 km per hour and Pradeep walks East at a constant speed of 4 km per hour.
Scenario III: Devanand walks West at a constant speed of 4 km per hour and Pradeep walks East at a constant speed of 3 km per hour.
- A.
Scenario I only
- B.
Scenario II only
- C.
Scenario III only
- D.
Scenario I and II
- E.
None of the above
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).
Workspace:
The median of 11 different positive integers is 15 and seven of those 11 integers are 8, 12, 20, 6, 14, 22, and 13.
Statement I: The difference between the averages of four largest integers and four smallest integers is 13.25.
Statement II: The average of all the 11 integers is 16.
Which of the following statements would be sufficient to find the largest possible integer of these numbers?
- A.
Statement I only.
- B.
Statement II only.
- C.
Both Statement I and Statement II are required.
- D.
Neither Statement I nor Statement II is sufficient.
- E.
Either Statement I or Statement II is sufficient.
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).
Workspace:
For a positive integer x, define f(x) such that f(x + a) = f(a × x), where a is an integer and f(1) = 4. If the value of f(1003) = k, then the value of ‘k’ will be:
- A.
1003
- B.
1004
- C.
1005
- D.
1006
- E.
None of the above
Answer: Option E
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).
Workspace:
An ascending series of numbers satisfies the following conditions:
- When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
- When divided by 11, the numbers leave no remainder.
The 6th number in this series will be:
- A.
242
- B.
2882
- C.
3542
- D.
4202
- E.
None of the above
Answer: Option C
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).
Workspace:
The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find Sin ∠ABC, given 2∠DAB = ∠BCD.
- A.
4/5
- B.
16/25
- C.
5/6
- D.
24/25
- E.
A single solution is not possible
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,
$\frac{100}{\mathrm{sin}\mathrm{\theta}}=\frac{120}{\mathrm{sin}(180-2\mathrm{\theta})}$
$\frac{}{}\phantom{\rule{0ex}{0ex}}\frac{100}{\mathrm{sin}\mathrm{\theta}}=\frac{120}{\mathrm{sin}2\mathrm{\theta}}.....\left(\mathrm{i}\right)$
sin 2θ = 2 × sin θ × cos θ
∴ (i) becomes
$\frac{100}{\mathrm{sin}\theta}=\frac{120}{2\times \mathrm{sin}\theta \times \mathrm{cos}\theta}$
$\frac{}{}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cos}\theta =\frac{3}{5}\Rightarrow \mathrm{sin}\theta =\frac{4}{5}$
$\frac{\mathrm{}}{}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{sin}\angle ABC=\mathrm{sin}(180-\theta )=\mathrm{sin}\theta =\frac{4}{5}$
Hence, option (a).
Workspace:
In an examination, two types of questions are asked: one mark questions and two marks questions. For each wrong answer, of one mark question, the deduction is 1/4 of a mark and for each wrong answer, of two marks question, the deduction is 1/3 of a mark. Moreover, 1/2 of a mark is deducted for any unanswered question. The question paper has 10 one mark questions and 10 two marks questions. In the examination, students got all possible marks between 25 and 30 and every student had different marks. What would be the rank of a student, who scores a total of 27.5 marks?
- A.
5
- B.
6
- C.
7
- D.
8
- E.
None of theabove
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).
Workspace:
The centre of a circle inside a triangle is at a distance of 625 cm. from each of the vertices of the triangle. If the diameter of the circle is 350 cm. and the circle is touching only two sides of the triangle, find the area of the triangle.
- A.
240000
- B.
387072
- C.
480000
- D.
506447
- E.
None of the above
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).
Workspace:
If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) × (M – N)?
Both (M) and (N) are positive integers and M > N. (M)! is factorial M.
- A.
150
- B.
180
- C.
200
- D.
225
- E.
234
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:
- N = 0 to 4 (no zeroes in N!)
- N = 5 to 9 (1 zero in N!)
- N = 10 to 14 (2 zeroes in N!)
- 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) = M^{2} – NM can take four of the five given values.
Hence, there are five possible equations – one for each option.
Consider option 1: M^{2} – 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
M^{2} – 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
M^{2} – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.
Option 4:
For N = 16, the equation becomes
M^{2} – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.
Option 5:
For N = 17, the equation becomes
M^{2} – 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 M^{2} – 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.
Workspace:
A person is standing at a distance of 1800 meters facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at 30 degree elevation from his eye-level. Immediately, the person starts walking towards the tower. At 5.10 pm., the person noticed that the tip of the minute hand made an angle of 60 degrees with respect to his eye-level. Using three-dimensional vision, find the speed at which the person is walking. The length of the minutes hand is 200√3 meters (√3 = 1.732).
- A.
7.2 km/hour
- B.
7.5 km/hour
- C.
7.8 km/hour
- D.
8.4 km/hour
- E.
None of the above
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).
Workspace:
Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?
- A.
Pipe A was open for 19 hours.
- B.
Pipe A was open for 19 hours 30 minutes.
- C.
Pipe B was open for 19 hours 30 minutes.
- D.
Pipe C was open for 19 hours 50 minutes.
- E.
The situation is not possible.
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.
$\frac{1}{3}\pi {r}^{2}h=24x$
∴ πr^{2}h = 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).
Workspace:
A three-digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three-digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?
- A.
5
- B.
6
- C.
7
- D.
8
- E.
None of the above
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).
Workspace:
Answer the following questions on the basis of information given below.
Twitter allows its users to post/share and read short messages known as tweets. Tweets can be of three types – Positive Tweets (in support), Negative Tweets (against) and Neutral Tweets. The following table presents the Number of Votes and Tweets received by certain political parties.
* Any party which has secured less than 2% of the total votes falls under ‘Other Parties’ category. For example, Party E secured less than 2% of total votes, in the year 2000.
Note: If the vote share (%age of total votes) of a party changes from 15% to 40%, gain in vote share would be 25% (= 40%, –15%).
Which of the following options correctly arranges the political parties in descending order of gain in vote share from the year 2005 to the year 2010?
- A.
EBDCA
- B.
EBCDA
- C.
EBCAD
- D.
BCEDA
- E.
BCEAD
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.
${A}_{G}=\left(\frac{364450}{985000}-\frac{343200}{880000}\right)\times 100=-2$
$\mathrm{}\phantom{\rule{0ex}{0ex}}{D}_{G}=\left(\frac{54175}{985000}-\frac{48400}{880000}\right)\times 100=0$
Since A_{G} < D_{G}
Hence, option (e) is eliminated.
Hence, option (d).
Workspace:
Which of the following parties received maximum number of “neutral tweets” in the year 2010?
- A.
Party B
- B.
Party C
- C.
Party D
- D.
Party E
- E.
One of the parties categorised under ‘Other Parties’
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).
Workspace:
Between 2000 and 2010, in terms of gain in vote share which of the following cannot be a possible value (approximated to one decimal place) for any party?
- A.
2.0%
- B.
2.5%
- C.
3.5%
- D.
4.5%
- E.
7.5%
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).
Workspace:
In 2010, which of the following options has maximum difference between the vote share and tweet share?
- A.
Party B
- B.
Party C
- C.
Party D
- D.
Party E
- E.
Other Parties
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).
Workspace:
Answer the questions on the basis of information given below.
As a part of employee improvement programs, every year an organization conducts a survey on three factors: 1. Number of days (in integers) of training undergone, 2. Amount of bonus (in lacs) received by an employee and 3. Employee effectiveness score (on the scale of 1 to 10). Survey results for last two years are given below for the same seven employees.
In Survey 1, what was the average bonus earned by employees who underwent training for more than 17 days?
- A.
Between 16 and 17 lacs
- B.
Between 17 and 18 lacs
- C.
Between 18 and 19 lacs
- D.
Between 19 and 20 lacs
- E.
None of the above
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.
Workspace:
Identify the number of employees whose employee effectiveness score was higher than 7 in Survey 1, but whose bonus was lower than 20 lacs in Survey 2.
- A.
2
- B.
3
- C.
4
- D.
5
- E.
None of the above
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).
Workspace:
From Survey 1 to Survey 2, how many employees underwent more days of training but their annual bonus decreased?
- A.
1
- B.
2
- C.
3
- D.
4
- E.
None of the above
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).
Workspace:
From Survey 1 to Survey 2: for how many employees training days increased along with an increase of employee effective score by at least 1.0 rating?
- A.
2
- B.
3
- C.
4
- D.
7
- E.
None of the above
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).
Workspace:
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