## numerical

 Question 1

Aptitude       Numerical       Gate 2018
Question 1 Explanation:

→ ∠1 = ∠5 + ∠4 ………(i)
According to triangular property:
Angle of exterior = sum of interior angles
→ ∠4 = ∠3 + ∠2 ……….(ii)
By substituting (ii) in (i)
→ ∠1 = ∠5 + ∠3 + ∠2
→ ∠1 = ∠BCD
→ ∠BCD - ∠BAD = ∠1 - ∠2
= ∠5 + ∠3 + ∠2 - ∠2
= ∠ 5 + ∠3
= ∠DEC + ∠BFC
 Question 2
In a party, 60% of the invited guests are male and 40% are female. If 80% of the invited guests attended the party and if all the invited female guests attended, what would be the ratio of males to females among the attendees in the party?
 A 2:3 B 1:1 C 3:2 D 2:1
Aptitude       Numerical       Gate 2018
Question 2 Explanation:
Let us consider total no. of guests = 100
→ 60% invited guests are males i.e., 60 males
→ 40% invited guests are females i.e., 40 females
→ 80% invited guests are attended i.e., total guests attended to a party = 80
→ All the invited females are attended then remaining people are males = 80 – 40 = 40 males
40 males : 40 females
1 : 1
 Question 3
In appreciation of the social improvements completed in a town, a wealthy philanthropist decided to gift Rs 750 to each male senior citizen in the town and Rs 1000 to each female senior citizen. Altogether, there were 300 senior citizens eligible for this gift. However, only 8/9th of the eligible men and 2/3rd of the eligible women claimed the gift. How much money (in Rupees) did the philanthropist give away in total?
 A 1,50,000 B 2,00,000 C 1,75,000 D 1,51,000
Aptitude       Numerical       Gate 2018
Question 3 Explanation:
Male senior citizen’s gift = Rs.750
Female senior citizen’s gift = Rs. 1000
No. of males = a say
No. of females = b say
Altogether no. of persons eligible for gift = 300
i.e., a+b = 300
Total amount be given = (8x/9 × 750) + (2y/3 × 1000)
= (2000x/3) + (2000y/3)
= 2000/3 (x+y)
= 2000/3 (300)
= 2,00,000
 Question 4
A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment?
 A Three green faces and four red faces. B Four green faces and three red faces. C Five green faces and two red faces. D Six green faces and one red face.
Aptitude       Numerical       Gate 2018
Question 4 Explanation:
→ Each side of a unbiased die can have equal probability i.e., = 1/6
→ If we roll a die for six time then we get 4 green faces and 2 red faces.
→ And if we roll for seventh time green face can have more probability to become a outcome.
→ Then most likely outcome is five green faces and two red faces.
 Question 5
The area of a square is d.  What is the area of the circle which has the diagonal of the square as its diameter?
 A πd B πd2 C (1/2)πd2 D (1/2)πd
Aptitude       Numerical       Gate 2018
Question 5 Explanation:
In general,
Area of a square A = a2 (where a is side)
In the question,
Area of a square = d
The side of a square = √d
Diagonal of a square = Diameter of circle

From Pythagoras theorem,
 Question 6
What is the missing number in the following sequence? 2, 12, 60, 240, 720, 1440, _______, 0
 A 2880 B 1440 C 720 D 0
Aptitude       Numerical       Gate 2018
Question 6 Explanation:

2×6=12
12×5=60
60×4=240
240×3=720
720×2=1440

1440×0=0
 Question 7

What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of '7' in each case?

 A 3047 B 6047 C 7987 D 63847
Aptitude       Numerical       Gate 2018
Question 7 Explanation:
Method-I:
Take the LCM of 20,42,76 i.e., 7980.
⟹ Remainder = 7
⟹ Smallest number divisible by 20 (or) 42 (or) 72 which leaves remainder 7 is = 7980+7 = 7987
Method- II:
Option I:
3047 – 7 = 3040
3040 not divisible by 42
Option II:
6040 not divisible by 42
Option III:
7980 divisible by 20, 42, 76
Option IV:
63840 divisible by 20, 42, 76
→ Smallest (7980, 63840) = 7980
→ 7980 + remainder 7 = 7987
 Question 8
If pqr≠0and p -x=1/q, q -y=1/r, r -z=1/p, what is the value of the product xyz?
 A -1 B 1/pqr C 1 D pqr
Aptitude       Numerical       Gate 2018
Question 8 Explanation:
pqr≠0
→ p-x = 1/q ⟹ 1/px = 1/q
⟹ q = px
⟹ log q = log px
⟹ x log p = log q ⟹ x = log q/ log p → q-y = 1/r ⟹ 1/qy = 1/r
⟹ qy = r
⟹ y log q = log r
⟹ y = log r/ log q
→ r-z = 1/p ⟹ 1/rz = 1/p
⟹ p = rz
⟹ log rz = log p
⟹ z log r = log p
⟹ z = log p/ log r
XYZ = log q/ log p * log r/ log q * log p/ log r =1
 Question 9
Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali,Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other?
 A Rahul and Murali B Srinivas and Arul C Srinivas and Murali D Srinivas and Rahul
Aptitude       Numerical       Gate 2017 set-01
Question 9 Explanation:
Assuming, they face the center of the table

Srinivas and Murali and Arul and Rahul are opposite to each other.
They face away from the center of table

Even in this case,
Srinivas and Murali and Arul and Rahul are opposite to each other.
 Question 10
Find the smallest number y such that y×162 is a prefect cube.
 A 24 B 27 C 32 D 36
Aptitude       Numerical       Gate 2017 set-01
Question 10 Explanation:
Prime factorize: 162
⇒162 =2×3×3×3×3 = 33×(2×3)
For (2×3) to be a perfect cube, it should be multiplied by (22×32)
∴ Required number = y = 22×32 = 36
 Question 11
The probability that a k-digit number does NOT contain the digits 0, 5 or 9 is
 A 0.3k B 0.6k C 0.7k D 0.9k
Aptitude       Numerical       Gate 2017 set-01
Question 11 Explanation:
Possibilities of doesn't contain 0,5,9 = (7)k
Total no.of possibilities = (10)k
Required probability = (7)k / (10)k = 0.7k
 Question 12
Six people are seated around a circular table. There are atleast two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is
 A 2 B 3 C 4 D Cannot be determined
Aptitude       Numerical       Gate 2017 set-01
Question 12 Explanation:
There are atleast two men and two women.
Atleast three right- handed persons.
None of the women are right-handed.
⇒ All the right-handed persons are men. Every woman has a left-handed person to her immediate right.
⇒ For this to happen, there must be three left-handed-persons and one of them should be a male.
As three persons are right-handed and those being men.
The number of women at the table = 2
 Question 13
The expression (x+y)-|x-y|/2 is equal to
 A the maximum of x and y B the minimum of x and y C 1 D none of the above
Aptitude       Numerical       Gate 2017 set-01
Question 13 Explanation:
As the given expression has modulus, let us consider:
Case-I: x>y
⇒ |x-y| = x-y
Expression ⇒ (x+y)-(x-y)/ 2 = y
In this case value of expression is minimum of y.
Case-II: x ⇒ |x-y| = -(x-y)
Expression ⇒ (x+y) - (x-y) / 2
⇒ (x+y) - [-(x-y)]/2 ⇒ x
In this case value of expression is minimum of x.
Finally, the value of expression is minimum of x & y.
 Question 14
Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts coloured red, pink, blue and white respectively. Arun dislikes the colour red and Shweta dislikes the colour white. Gulab and Neel like all the colours. In how many different ways can they choose the shirts so that no one has a shirt with a colour he or she likes?
 A 21 B 18 C 16 D 14
Aptitude       Numerical       Gate 2017 set-01
Question 14 Explanation:
Total number of ways that each select one shirt without any condition = 4P4 = 4! = 24
Number of ways Arun chooses red or Shweta chooses white
= Number of ways Arun chooses red + Number of ways Shweta chooses white - Number of ways Arun chooses red and Shweta chooses white
= 3P3 + 3P3 - 22P2
= 3! + 3! - 2!
= 10
∴ Required number of ways = 24 - 10 = 14
 Question 15

 A P,Q B P,Q,T C R,S,T D Q,R,S
Aptitude       Numerical       Gate 2017 set-01
Question 15 Explanation:
The contour lines are shown at 25m interval in this plot.
∴ From the plot,
P=575m
Q=525m
R=475m
S=425m
T=500m
For a village to be submerged, the water level should rise to level greater than the village.
If water level rises to 525m, villages which are at heights below 525m, get submerged.
Here, R, S, T get submerged.
 Question 16
There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to the West of W, Z is to the East of X and the West of V, W is to the West of Y. Which is the building in the middle?
 A V B W C X D Y
Aptitude       Numerical       GATE 2017(set-02)
Question 16 Explanation:
V is to west of W = VW .........(i)
Z is to East of X and west of V = XZV ........(ii)
W is to west of Y = WY .........(iii)
From (i) and (ii) ⇒ VWY .........(iv)
From (ii) and (iv) ⇒ XZVWY
→While building V is in the middle.
 Question 17
A test has twenty questions worth 100 marks in total. There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have?
 A 12 B 15 C 18 D 19
Aptitude       Numerical       GATE 2017(set-02)
Question 17 Explanation:
No. of 3 marks questions = X say
No. of 4 marks questions = Y say
No. of questions X+Y = 20
Total no. of marks = 100
⇒ 3X+11Y = 100
Option I: If X=12; Y=8 ⇒ 3(12)+11(8) ⇒ 36+88 ≠ 100
Option II: If X=15; Y=5 ⇒ 3(15)+11(5) ⇒ 100 = 100
Option III: If X=18; Y=2 ⇒ 3(18)+11(2) ≠ 100
Option IV: If X=19; Y=1 ⇒ 3(19)+11(1) ≠ 100
 Question 18
There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is
 A 1⁄5 B 7⁄30 C 1⁄4 D 4⁄15
Aptitude       Numerical       GATE 2017(set-02)
Question 18 Explanation:
Totally there are 3 red socks, 4 green socks, 3 blue socks. In those we need to select 2 socks
The probability of selecting same colour is
3C2 / 10C2 * 4C2 / 10C2 * 3C2 / 10C2
⇒ 3/45 + 6/45 + 3/45
⇒ 12/45
= 4/15
 Question 19
There are three boxes. One contains apples, another contains oranges and the last one contains both apple and oranges. All three are known to be incorrectly labeled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes?
 A The box labeled ‘Apples’ B The box labeled ‘Apples and Oranges’ C The box labeled ‘Oranges’ D Cannot be determined
Aptitude       Numerical       GATE 2017(set-02)
Question 19 Explanation:
Correct Answer is option B i.e., the box labeled Apples an Oranges.
If we open a box labeled as apples and oranges that contains either apples (or) oranges:
Case I: If it contains apples, then the box labeled oranges can contain apples and oranges and box labeled apples can contain oranges.
Case II: If it contains oranges, then box labeled apples can contains apples and oranges and box labeled oranges can contains apples.
 Question 20
X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X3 will have
 A 90 digits B 91 digits C 92 digits D 93 digits
Aptitude       Numerical       GATE 2017(set-02)
Question 20 Explanation:
X is a 30 digit number starting with digit 4 and followed by the digit 7
X = 4777......... (29 times (7))
X = 4.777........ ×1029
X = 5×1029 (4.777 rounded as 5)
X3 = 125×1087
Total no. of digits = 3+87 [125=3 digit, 1087 = 87 digit]
= 90
 Question 21
The number of roots of ex + 0.5x2 – 2 in the range [-5, 5] is
 A 0 B 1 C 2 D 3
Aptitude       Numerical       GATE 2017(set-02)
Question 21 Explanation:
ex+0.5x2-2=0
ex=-0.5x2+2
If we draw a graph for ex and -0.5x2+2 in between [-5, 5]
 Question 22
 A P B Q C R D S
Aptitude       Numerical       GATE 2017(set-02)
Question 22 Explanation:
Region R can have more air pressure as compared to other regions because it has maximum number of lines crossing the area.
 Question 23
A cube is built using 64 cubic blocks of side one unit. After it is built, one cubic block is removed from every corner of the cube. The resulting surface area of the body (in square units) after the removal is __________.
 A 56 B 64 C 72 D 96
Aptitude       Numerical       2016 set-01
Question 23 Explanation:
64 cubic blocks are used to make the cube
⇒ Side of cube = 4
Surface area of cube = 6s2 = 6 × 42 = 96
Each corner block is associated with three faces of cube. When they are removed, three new faces are exposed, thus there appears no net change in the exposed surface area.
Hence, Surface area = 96
 Question 24

 A Elegance B Executive C Smooth D Soft
Aptitude       Numerical       2016 set-01
Question 24 Explanation:
Revenue from Elegance = (27300+25222+28976+21012)×48 = Rs. 4920480
Revenue from Smooth = (20009+19392+22429+18229)×63 = Rs. 5043717
Revenue from Soft = (17602+18445+19544+16595)×78 = Rs.5630508
Revenue from Executive = (9999+8942+10234+10109)×173 = Rs. 6796132
Clearly, Executive contributes the greatest fraction to the revenue of the company as the revenue from it is the highest.
 Question 25
In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.
 A 40 B 46.02 C 60.01 D 92.02
Aptitude       numerical       2016 set-01
Question 25 Explanation:
In a process, the number of cycles to failure decrease exponentially with an increase in load.
General exponential function = a⋅e-bx
i.e., No. of cycles to failure = a⋅e-bx
At a load of 80 units, it takes 100 cycles to failure.
So, no. of cycles to failure = 100
i.e., 100 = a⋅e-b(80) --------(1)
When the load is halved, it takes 10000 cycles to failure.
No. of cycles to failure = 10,000
i.e., 10,000 = a⋅e-b(40) ---------(2)

No. of cycles to failure = 5,000
i.e., 5000 = a⋅e-bx
Multiply with 2 on both sides,
10,000 = 2⋅a⋅e-bx ------- (4)
 Question 26

 A (i) only B (ii) only C (i) and (ii) D neither (i) nor (ii)
Aptitude       numerical       2016 set-01
Question 26 Explanation:
From statements I, II & IV, we can say that
All three can beat S.
But from statement III,
S loses to P only sometimes.
(ii) Cannot be inferred.
And in poker, the transitive law does not apply. This can be seen from statement III.
As S loses to P only sometimes which states that wins against P most of the time.
So, (i) cannot be logically inferred.
 Question 27
If f(x) = 2x7 + 3x - 5, which of the following is a factor of f(x)?
 A (x 3 +8) B (x-1) C (2x-5) D (x+1)
Aptitude       Numerical       2016 set-01
Question 27 Explanation:
f(x)=2x7+3x - 5
For (x - a) to be a factor of f(x)
f(a) = 0
From options,
Only a=1, satisfies the above equation
∴ (x - 1) is a factor of f(x).
 Question 28
Pick the odd one from the following options.
 A CADBE B JHKIL C XVYWZ D ONPMQ
Aptitude       Numerical       GATE 2016 set-2
Question 28 Explanation:
In all the given options except D, the 1st, 3rd, 5th alphabets are consecutive (increasing order) and 2nd and 4th are consecutive (increasing order).
But in D, 2nd and 4th are in decreasing order.
 Question 29

 A n4 B 4n C 22n-1 D 4n-1
Aptitude       Numerical       GATE 2016 set-2
Question 29 Explanation:
Given,
Product of roots (α, β) = 4
⇒ αβ = 4

(αβ)n = 4n
 Question 30
Among 150 faculty members in an institute, 55 are connected with each other through Facebook® and 85 are connected through WhatsApp®. 30 faculty members do not have Facebook® or WhatsApp® accounts. The number of faculty members connected only through Facebook® accounts is ______________.
 A 35 B 45 C 65 D 90
Aptitude       Numerical       GATE 2016 set-2
Question 30 Explanation:
Total number of faculty = 150
Number of faculty members connected through Facebook = 55
Number of faculty members connected through Whatsapp = 85
Number of faculty members with Facebook (or) Whatsapp accounts = 30
Number of faculty members with either Facebook (or) Whatsapp accounts = 150 - 30 = 120
Number of faculty members with both Facebook and Whatsapp accounts = 85 + 55 - 120 = 20
Number of faculty members with only Facebook accounts = 55 - 20 = 35
 Question 31

 A 21 B 27 C 30 D 36
Aptitude       Numerical       GATE 2016 set-2
Question 31 Explanation:
Number of rectangles in a grid with 'm' horizontal and 'n' vertical lines = 5C2 × 3C2 = 10 × 3 = 30
 Question 32

 A f(x)=1-|x-1| B f(x)=1+|x-1| C f(x)=2-|x-1| D f(x)=2+|x-1|
Aptitude       Numerical       GATE 2016 set-2
Question 32 Explanation:
From the graph,
x = -1 ⇒ f(x) = 0
x = 0 ⇒ f(x) = 1
We have check, which option satisfies both the conditions.
Only option (C) satisfies both of them.
 Question 33
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16?
 A 0.2 B 0.25 C 0.3 D 0.33
Aptitude       Numerical       GATE 2015 (Set-01)
Question 33 Explanation:
Favourable outcomes = {2,14}, {3,13}, {4,12}, {5,11} = 4
Total outcomes = 4C1 × 5C1 = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
 Question 34

 A Statement I alone is sufficient, but statement II alone is not sufficient. B Statement II alone is sufficient, but statement I alone is not sufficient. C Both statements together are sufficient, but neither statement alone is sufficient. D Statement I and II together are not sufficient.
Aptitude       Numerical       GATE 2015 (Set-01)
Question 34 Explanation:
When you climbing, only height matters not the width.
Statement I:
Each step is 3/4 foot high.
No. of steps = 9/(3/4) = 12 steps
Statement I alone is sufficient.
Statement II:
Each step is 1 foot wide.
Statement II alone is not sufficient.
 Question 35

 A 32 B 33 C 34 D 35
Aptitude       Numerical       GATE 2015 (Set-01)
Question 35 Explanation:
Male to female students in each department = 5 : 4
There are 40 males in Electrical Engineering.
⇒ Number of females in Electrical Engineering = 4/5 × 40 = 32
Total number of students in Electrical Engineering = 40+32 = 72
This constitutes 20% of the strength of college.
Number of students in Civil department = 30/100 × 72 = 108
Number of female students in Civil department = 4/(4+5) × 108 = 48
Number of students in Mechanical department = 10/20 × 72 = 36
Number of female students in Mechanical department = 4/(4+5) × 36 = 16
Required Difference = 48 - 16 = 32
 Question 36

 A Only relation I is true B Only relation II is true C Relations II and III are true. D Relations I and III are true.
Aptitude       Numerical       GATE 2015 (Set-01)
Question 36 Explanation:
75% of students have a chance of passing atleast one subject = 1 - (1 - m) (1 - p) (1 - c) = 0.75 ----(i)
50% of students have a chance pass in atleast two
(1 - m)pc + (1 - p)mc + (1 - c) mp + mpc = 0.5 ----(ii)
40% of students have a chance of passing exactly two
(1 - m)pc + (1 - p)mc + (1 - c)mp = 0.4 ----(iii)
From equation (ii) and (iii) we can get
mpc = 0.1
⇒ m*p*c = 1/10
Statement (III) is correct.
→ Simplify eq (i), we get ⇒ p+c+m - (mp+mc+pc) + mpc = 0.75
⇒ p+c+m - (mp+mc+pc) = 0.65 ----(iv) → Simplify equation (iii), we get
⇒ pc + mc + mp - 3mpc = 0.4
From (iv) and (v)
p + c +m - 0.7 = 0.65
⇒ p + c + m = 1.35 = 27/20
Statement (I) is correct.
 Question 37
 A 2.29 B 2.97 C 6.795 D 8.795
Aptitude       Numerical       GATE 2015 (Set-01)
Question 37 Explanation:
Total marks obtained = 21×2+15×3+11×1+23×2+31×5=299
Average marks = 299/44 = 6.795
 Question 38

 A Statement I alone is not sufficient B Statement II alone is not sufficient C Either I or II alone is sufficient D Both statements I and II together are not sufficient.
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 38 Explanation:
Statement-I:
One fourth of the weight of a pole is 5Kg. ⇒ Weight of pole is 4×5=20Kg
Weight of 10 poles each of same weight = 10×20 = 200 Kg
∴Statement I alone is sufficient.
Statement-II:
Let, Weight of each pole = W Kg
Given,
10W = 2W + 160
⇒ 8W = 160
W = 20Kg
∴ Weight of each pole = 20 Kg
∴ Weight of 10 poles = 10×20 Kg = 200 Kg
∴ Statement II alone is sufficient.
Either I or II alone is sufficient.
 Question 39
Consider a function f(x= 1 - |x| on -1 ≤ x ≤ 1. The value of x at which the function attains a maximum and the maximum value of the function are
 A 0, -1 B -1, 0 C 0, 1 D -1, 2
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 39 Explanation:
f(x) = 1 - |x| on -1 ≤ x ≤ 1
In the given function, it is given as
-|x| ⇒ To obtain the maximum of this function we have to minimize the value -|x| and the minimum value is 0.
∴ Maximum value of f(x) is at f(0) and i.e., f(x) = 1
Maximum value is 1 at x=0.
 Question 40

 A B C D
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 40 Explanation:
 Question 41
 A I only B I and II C II and III D I and III
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 41 Explanation:
If any set of numbers are in a arithmetic sequence and if a common number if added (or) subtracted from each of these numbers, the new set will also be in arithmetic sequence.
Hence, II is an arithmetic sequence.
If the set of numbers is multiplied by the common number, even then the new set will also be in arithmetic sequence.
Hence, I is an arithmetic sequence.
 Question 42

 A 8 B 9 C 7 D 6
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 42 Explanation:
h(2,5,7,3) = remainder of (7×3)/(2×5)
= remainder of 21/10
=1
fg(1,4,6,8) = f(1,4,6,8)×g(1,4,6,8)
= max(1,4,6,8) × min(1,4,6,8)
= 8×1
= 8
 Question 43
Four branches of a company are located at M, N, O and P. M is north of N at a distance of 4km; P is south of O at a distance of 2 km; N is southeast of O by 1 km. What is the distance between M and P in km?
 A 5.34 B 6.74 C 28.5 D 45.49
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 43 Explanation:
 Question 44
An unordered list contains n distinct elements. The number of comparisons to find an element in this list that is neither maximum nor minimum is
 A Θ(nlog n) B Θ(n) C Θ(log n) D Θ(1)
Aptitude       Numerical       GATE 2015 -(Set-2)
Question 44 Explanation:
Consider first three element of the list, atleast one of them will be neither minimum nor maximum. ∴ Θ(1)
 Question 45
A function f(x) is linear and has a value of 29 at x = -2 and 39 at x = 3. Find its value at x = 5.
 A 59 B 45 C 43 D 35
Aptitude       Numerical       GATE 2015(Set-03)
Question 45 Explanation:
f(x)=2x+33
 Question 46
If ROAD is written as URDG, then SWAN should be written as:
 A VXDQ B VZDQ C VZDP D UXDQ
Aptitude       Numerical       GATE 2015(Set-03)
Question 46 Explanation:
R+3=U, O+3=R, A+3=D, D+3=G;
S+3=V, W+3=Z, A+3=D, N+3=Q.
 Question 47
 A 2006 B 2007 C 2008 D 2009
Aptitude       Numerical       GATE 2015(Set-03)
Question 47 Explanation:
Increase in exports in 2006 = 100 - 70/70 = 42.8%
Increase in imports in 2006 = 120 - 90/90 = 33.3% which is more than any other year.
 Question 48
 A P-Home, Q-Power, R-Defense, S-Telecom, T-Finance B R-Home, S-Power, P-Defense, Q-Telecom, T-Finance C P-Home, Q-Power, T-Defense, S-Telecom, U-Finance D Q-Home, U-Power, T-Defense, R-Telecom, P-Finance
Aptitude       Numerical       GATE 2015(Set-03)
Question 48 Explanation:
Since U does not want any portfolio, (C) and (D) are ruled out. R wants Home, or Finance or No portfolio, (A) is not valid. Hence option (B) is correct
 Question 49
 A x=y-|y| B x=-(y-|y|) C x=y+|y| D x=-(y+|y|)
Aptitude       Numerical       GATE 2015(Set-03)
Question 49 Explanation:
Verify which of the options satisfies the x=2 when y=-1.
Then the Answer is x = - (y - |y|)
 Question 50
If (z + 1/z)= 98, compute (z2+ 1/z2).
 A 96 B 97 C 98 D 99
Aptitude       Numerical       GATE 2014(Set-01)
Question 50 Explanation:
z2+1/z2+2∙z∙ 1/z=98⇒z2 + 1/z2=96
 Question 51
The roots of ax2+ bx + c = 0 are real and positive. a, b and c are real. Then ax+ b|x|+ c = 0 has
 A no roots B 2 real roots C 3 real roots D 4 real roots
Aptitude       Numerical       GATE 2014(Set-01)
Question 51 Explanation:
ax2+bx+c=0
For roots to be real & positive b2-4ac>0
This will have 2 real positive roots.
ax2+b|x|+c=0
Discriminant =b2-4ac>0
ax2+bx+c
(-b)2-4ac
⇒b2-4ac
Is also > 0. This will have real roots.
⇒ This will have 4 real roots.
 Question 52
Round-trip tickets to a tourist destination are eligible for a discount of 10% on the total fare.  In addition, groups of 4 or more get a discount of 5% on the total fare.  If the one way single person fare is Rs 100, a group of 5 tourists purchasing round-trip tickets will be charged Rs _________.
 A 850 B 851 C 852 D 853
Aptitude       Numerical       GATE 2014(Set-01)
Question 52 Explanation:
One way fare for a single person = Rs.100
Round trip, there is a discount of 10% of total fare.
Group of 4 or more, Discount of 5% of total fare.
∴ 5 tourists ⇒ Total fare = 5 × 200 = 1000
Total discount = 10% of 1000 + 5% of 1000
= 100 + 50
= ₹150
∴ Fare charged = 1000 - 150 = ₹850
 Question 53

 A 48 B 49 C 50 D 51
Aptitude       Numerical       GATE 2014(Set-01)
Question 53 Explanation:
Total respondents = 300
[No. of respondents who do not own a scooter] = [No. of respondents who own's only a car + No. of respondents who do not own any vehicle]
= 40 + 34 + 70
= 144
Percentage = 144/300 × 100 = 48%
 Question 54
When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines? _____________
 A 6 B 7 C 8 D 9
Aptitude       Numerical       GATE 2014(Set-01)
Question 54 Explanation:
Tetrahedron is a pyramid like structure with 4 corner points (vertical).
→ If you take a point inside a tetrahedron, then you have 5 points.
→ An internal plane is formed by joining any of the three points.
→ No. of planes = 5 C 3 = 10
But 4 of them will be faces of tetrahedron.
∴ New planes = 10 - 4 = 6
 Question 55
 A ∀x:glitters(x) ⇒¬gold(x) B ∀x:gold(x) ⇒glitters() C ∃x: gold(x)∧¬glitters(x) D ∃x:glitters(x)∧¬gold(x)
Aptitude       Numerical       GATE 2014(Set-01)
Question 55 Explanation:
Method 1:
Not all that glitters is gold.
Option A:
∀x:glitters(x) ⇒¬gold(x)
which means that every item (x), which glitters is not gold.
Option B:
∀x:gold(x) ⇒glitters()
Every item (x) which is gold is a glitter.
(or)
Every golden item glitters.
Option C:
∃x: gold(x)∧¬glitters(x)
There are some gold items which does not glitters.
Option D:
∃x:glitters(x)∧¬gold(x)
There exists some glitters which are not gold.
(or)
Not all glitters are gold.

Method 2:

⇒ (∼∀x) (∼ (glitters(x) ⇒ gold(x))
⇒ ∃x (∼ (∼glitters(x) ∨ gold(x))
⇒ ∃x (glitters ∧ gold(x))
 Question 56
Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ .
 A 0.25 B 0.26 C 0.27 D 0.28
Aptitude       Numerical       GATE 2014(Set-01)
Question 56 Explanation:
We break a unit length rod into two pieces at a uniformly chosen point.

If we break at point x, length of the one piece x and the other piece is 1 – x.
Length of the shorter stick is between 0 to 0.5. If it is more than 0.5 then it will be longer stick.
The random variable (l) follows a uniform distribution. The probability function of l is
1/(b-a)=1/(0.5-0)=2 (length is between 0 to 0.5)
Expected value of length
(where P(l) is the probability density function)
 Question 57
 A 1 B 2 C 3 D 4
Aptitude       Numerical       GATE 2014(Set-01)
Question 57 Explanation:
Given:
3x + 2y = 1
4x + 7z = 1
x + y + z = 3
x – 2y + 7z = 0
This is a non-homogeneous equation system.
The Augmented matrix for above set of equations is

For matrix (A):
R4→ R4+R1

R4→ R4-R2

Rank = 3
For matrix (AB):

R4→(R4+R1 )-R2

Rank = 3
Rank of (A) = Rank (AB), so Unique solution
rank = 3 = no. of variables
Working Rule for Non-homogeneous equation:
(1) rank (A) < rank (AB), Inconsistent solution
(2) is rank (A) = rank (AB) = r then
if r = n, Unique solution
r < n, Infinite solution
 Question 58
What is the average of all multiples of 10 from 2 to 198?
 A 90 B 100 C 110 D 120
Aptitude       Numerical       Gate 2014 Set -02
Question 58 Explanation:
Average of 10, 20, 30, ..., 180, 190
Average = 10+20+30+...+180+190 / 19 = (19/2 [10+190]) / 19 = 100
 Question 59

 A 3.464 B 3.932 C 4 D 4.444
Aptitude       Numerical       Gate 2014 Set -02
Question 59 Explanation:
Let,

Squaring on both sides,
x2 = 12+x
x2 - x - 12 = 0
(x-4) (x+3) = 0
∴ x = 4 (x ≠ -3)
 Question 60
If x is real and |x2 - 2x + 3| = 11, then possible values of |- x3 + x2 - x| include
 A 2, 4 B 2, 14 C 4, 52 D 14, 52
Aptitude       Numerical       Gate 2014 Set -02
Question 60 Explanation:
Given,
|x2 - 2x + 3| = 11, x is real
x2-2x+3 = 11
x2-2x+8 = 0
(x-4)(x+2) = 0
x = 4, -2
x2-2x+3 = -11
x2-2x+14 = 0
x is not real in this case.
|-x3+x2-x|
when x=-2
⇒ |-(-2)3+(-2)2-(-2)|
= |(-(8) + (4) + 2| = 14
x=4
⇒ |-(4)3+(4)2-(4)|
= |-64 + 16 -4|
= 52
Possible values of |-x3+x2-x| include 14, 52.
 Question 61

 A 140 B 141 C 142 D 143
Aptitude       Numerical       Gate 2014 Set -02
Question 61 Explanation:
Number of male in 2008 = x
Number of female in 2008 = y
Number of female in 2009 = 2y
Given,
x/y = 2.5 ⇒ y = 2x/5
Let, number of male in 2009 = M
Given, M/2y = 3
M/2(2x/5) = 3 ⇒ M = 12x/5
Percentage increase = M-x/x × 100
= 12x/5-x/ x × 100
= 7/5 × 100
= 140%
 Question 62
At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand of a clock make an angle closest to 60°?
 A 6:22 am B 6:27 am C 6:38 am D 6:45 am
Aptitude       Numerical       Gate 2014 Set -02
Question 62 Explanation:
At 6 a.m., the angle between minute hand and hour hand is 180°.
And for every minute, the angle between them decreases by 6° - (1/2)° = 5 (1/2)°
∴ For the angle to be closest to 60°, the angle must be reduced by atmost 120°.
1 min - 5(1/2)°
x min - 120°
x = 2/11 × 120 = 240/11 = 21.81 m ≈ 22 min
i.e. 6.22 a.m. the angle between minute hand and hour hand will be closest to 60°.
 Question 63
 A 17 B 37 C 64 D 26
Aptitude       Numerical       Gate 2014 Set -03
Question 63 Explanation:
2,5, 10, 17, 26, 37, 50, 64
2 = 12+1
5 = 22+1
10 = 32+1
17 = 42+1
26 = 52+1
37 = 62+1
50 = 72+1
64 = 82+0
64 does not belong to the series.
 Question 64
 A 1.34 B 1.74 C 3.02 D 3.91
Aptitude       Numerical       Gate 2014 Set -03
Question 64 Explanation:
Number of students = 21+17+6 (or) 15+27+2 (or) 23+18+3 = 44
Total marks obtained = (21×2)+(15×3)+(23×2) = 133
Average marks = 133/44 = 3.02
 Question 65
A dance programme is scheduled for 10.00 a.m. Some students are participating in the programme and they need to come an hour earlier than the start of the event. These students should be accompanied by a parent. Other students and parents should come in time for the programme. The instruction you think that is appropriate for this is
 A Students should come at 9.00 a.m. and parents should come at 10.00 a.m. B Participating students should come at 9.00 a.m. accompanied by a parent, and other parents and students should come by 10.00 a.m. C Students who are not participating should come by 10.00 a.m. and they should not bring their parents. Participating students should come at 9.00 a.m. D Participating students should come before 9.00 a.m. Parents who accompany them should come at 9.00 a.m. All others should come at 10.00 a.m.
Aptitude       Numerical       Gate 2014 Set -03
Question 65 Explanation:
Students who are particularly in the program and they need to come an hour earlier i.e., 09.00 am because the program is start of 10.00 am.
→ All other students and parents should come in time for the programme i.e. 10.00 am.
→ Option B is correct answer.
→ In option D, they gave, all other should come at 10.00 am that includes student's parents, staff and all ohers. So this is not correct option.
 Question 66
The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For international comparison, the GDP is compared in US Dollars (USD) after conversion based on the market exchange rate.  During the period 2012-2013 the exchange rate for the USD increased from Rs. 50/ USD to Rs. 60/ USD. India’s GDP in USD during the period 2012-2013
 A increased by 5% B decreased by 13% C decreased by 20% D decreased by 11%
Aptitude       Numerical       Gate 2014 Set -03
Question 66 Explanation:
Let,
GDP in rupees = x
GDP in dollars = x/50
Increase in GDP in rupees = 7%
∴ New GDP in rupees = 1.07x
New GDP in dollars = 1.07x/60
Change = ((1.07x/60) - (x/50))/(x/50) = -6.5/60 = -10.83%
As it is negative, the value has decreased.
GDP in VSD has decreased by 11%.
 Question 67

 A 1:1 B 2:1 C 1.5:1 D 2.5:1
Aptitude       Numerical       Gate 2014 Set -03
Question 67 Explanation:
Given,
Male to female students ratio in 2011 = 1 : 1
Male to female students ratio in 2012 = 1.5 : 1 = 3 : 2

⇒ M1/F1 = 1:1
M1 = F1 ------- (1)
⇒ M2/F2 = 1:1
2M2 = 3F2 ------- (2)
Given,
F1 = F2 ------- (3) From (1) & (2)
M1/M2 = F1/(3F2/2) = 2F1/3F2
But from (3)
M1/M2 = 2/3
We need to find
M2 : M1 = 3 : 2 = 1.5 : 1
 Question 68
Consider the equation: (7526)8 - (Y)= (4364)8, where (X)N stands for X to the base N. Find Y.
 A 1634 B 1737 C 3142 D 3162
Aptitude       Numerical       Gate 2014 Set -03
Question 68 Explanation:
(7526)8 - (Y)8 = (4364)8
⇒ 1/8 = (7526)8 - (4364)8
Base 8 ⇒ 0 to 7 digits

When you are borrowing you will add the value of the base, hence 2 becomes (2+8) = 10
Y = 3142
 Question 69
What will be the maximum sum of 44, 42, 40, ...... ?
 A 502 B 504 C 506 D 500
Aptitude       Numerical       Gate 2013
Question 69 Explanation:
44, 42, 40, ......
44, 42, 40, ...... 0
2(22 + 21 + 20 + …. 1)
Sum(n)=(n(n+1))/2=(22×23)/2=253
⇒2(253)
=506
 Question 70
 A 7 B 8 C 9 D 10
Aptitude       Numerical       Gate 2013
Question 70 Explanation:

Multiply on numerator & denominator

=-1+9
=8
 Question 71
Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?
 A 13/90 B 12/90 C 78/90 D 77/90
Aptitude       Numerical       Gate 2013
Question 71 Explanation:
⇨ Total number of 2 digit numbers from 1 to 100 = 90
⇨ Find which can be divisible by 7 upto 100
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98
Total = 13
Probability of divisible by 7 is = 13/90
Which can’t be divisible by 7 is
⇒ 1 – (probability of divisible by 7)
⇒ 1 – (13/90)
⇒ 77/90
 Question 72
A tourist covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and the rest by cycle at 10 km/h. The average speed of the tourist in km/h during his entire journey is
 A 36 B 30 C 24 D 18
Aptitude       Numerical       Gate 2013
Question 72 Explanation:

Let ‘x’ be the total distance,
Speed = distance/ time
Total speed = (total distance)/(total time); Time =distance/speed
Time taken for train journey = (x⁄2)/60kmph=x/120
Time taken for bus journey = (x⁄4)/30kmph=x/120
Time taken for cycle journey = (x⁄4)/10kmph=x/40
Total time = x/120+x/120+x/40=5x/120
Total speed = (total distance)/(total time)=x/(5x⁄120)=(120×x)/5x=24kmph
 Question 73
The current erection cost of a structure is Rs. 13,200. If the labour wages per day increase by 1/5 of the current wages and the working hours decrease by 1/24 of the current period, then the new cost of erection in Rs. is
 A 16,500 B 15,180 C 11,000 D 10,120
Aptitude       Numerical       Gate 2013
Question 73 Explanation:
Current wages is increased by (1/5) ⇒ 1+1/5 ⇒ 6/5
Working hours decreased by (1/24) ⇒ 1-1/24 ⇒ 23/24
The new cost of erection = 13,200 × 6/5 × 23/24 =15,180
 Question 74
The cost function for a product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of ₹50 per unit. The number of units to be produced by the firm such that the profit is maximized is
 A 5 B 10 C 15 D 25
Aptitude       Numerical       Gate 2012
Question 74 Explanation:
Total selling price =50q
Profit = 50q - 5q2
This is maximum at q = 5.
 Question 75
A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x - 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is
 A 8 meters B 10 meters C 12 meters D 14 meters
Aptitude       Numerical       Gate 2012
Question 75 Explanation:
Given,
y = 2x - 0.1x2
Differentiating both sides,
dy/dx = 2 - 0.2x
For maximum, dy/dx = 0
2 - 0.2x = 0
0.2x = 2
x = 10
 Question 76

 A 0.288 B 0.334 C 0.667 D 0.72
Aptitude       Numerical       Gate 2012
Question 76 Explanation:
Probability that a shock absorber from X is reliable = 0.6×0.96 = 0.576
Probability that a shock absorber from Y is reliable = 0.4×0.72 = 0.288
Probability that a randomly selected reliable absorber is from Y is = 0.288/(0.576+0.288) = 0.334
 Question 77

 A P, Q B Q, R C P, R D R, S
Aptitude       Numerical       Gate 2012
Question 77 Explanation:
If a value of 'x' is addered (or) multiplied to everagery element in the list, the average also changes by same value.
⇒ P, R are correct
 Question 78
Given the sequence of terms: AD  CG  FK  JP, the next term is
 A OV B OW C PV D PW
Aptitude       Numerical       Gate 2012
Question 78 Explanation:

OV is the next term.
 Question 79
If Log (P) = (1/2) Log (Q) = (1/3) Log (R), then which of the following options is TRUE?
 A P2 = Q3R2 B Q2 = PR C Q2 = R3P D R = P2Q2
Aptitude       Numerical       Gate 2011
Question 79 Explanation:
logP = 1/2 logQ = 1/3 log (R) = k
∴ P = bk, Q = b2k, R = b3k
Now, Q2 = b4k = b3k bk = PR
 Question 80

 A P B Q C R D S
Aptitude       Numerical       Gate 2011
Question 80 Explanation:
By observation of the table, we can say S
 Question 81
The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quantity produced. The fixed cost (F) of production of same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V+F)?
 A 5 B 4 C 7 D 6
Aptitude       Numerical       Gate 2011
Question 81 Explanation:
Checking with all options in formula: (4q+100/q) i.e. (V+F). Option A gives the minimum cost.
 Question 82
A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day?
 A 4 B 5 C 6 D 7
Aptitude       Numerical       Gate 2011
Question 82 Explanation:
Let each truck carry 100 units.
2800 = 4n + e; n = normal
300 = 10n + 2; e = excess/pending
n = 100/3, e = 8000/3
5 days: 500 ×(5⋅100/3)+ 8000/3
Minimum possible = 6
 Question 83
A container originally contains 10 litres of pure spirit. Form this container 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container?
 A 7.58 litres B 7.84 litres C 7 litres D 7.29 litres
Aptitude       Numerical       Gate 2011
Question 83 Explanation:
10(729/1000×1=7.29 litres)
 Question 84
25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is:
 A 2 B 17 C 13 D 3
Aptitude       Numerical       2010
Question 84 Explanation:

Total number of persons = a+b+c = 25 → (1)
Number of persons who play hockey = a+b = 15 → (2)
Number of persons who play football = b+c = 17 → (3)
Number of persons who play hockey and football = b = 10 → (4)
From (2) ⇒ a=5
From (3) ⇒ c=7
From (1) ⇒ d = 3
Number of persons who play neither hockey nor football = d = 3
 Question 85
If 137+276 = 435 how much is 731+672?
 A 534 B 1403 C 1623 D 1513
Aptitude       Numerical       2010
Question 85 Explanation:
Let, base = x
(137)x + (276)x = (435)x
x2 + 3x + 7 + 2x2 + 7x + 6 = 4x2 + 3x + 5
x2 - 7x - 8 = 0
x = 8 (or) -1
∴ x = 8 (-1 cannot be a base)
(731)x + (672)x = (731)8 +( 672)8
= 7 × 82 + 3× 8 + 1×1 + 6 × 82 + 7 × 8 + 2 × 1
= 915
From the options, 915 can be written as 1623 in base 8.
 Question 86

 A HSIG B SGHI C IGSH D IHSG
Aptitude       Numerical       2010
Question 86 Explanation:
Let us solve using option elimination approach .
Option A: HSIG
from (ii) , we can say that Gita and Saira are successive siblings.
Hence, option A is not true.
Option C: IGSH

In any of the above 4 cases, (i) is not true.
Hence, option C is not true.
Option D: IHSG

In any of the above 4 cases, (i) is not true.
Hence, option D is not true.
Option B: SGHI

In last two cases, all the facts are true.
∴ The order is SGHI.
 Question 87
5 skilled workers can build a wall in 20 days: 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?
 A 20 B 10 C 16 D 15
Aptitude       Numerical       2010
Question 87 Explanation:
5 skilled workers can build the wall in 20 days ⇒ 1 skilled worker can build the wall in 100 days
Capacity = 1/100
8 semi-skilled workers can build the wall in 25 days
⇒ 1 semi-skilled worker can build the wall in 200 days
Capacity = 1/200
10 un-skilled workers can build the wall in 30 days
⇒ 1 un-skilled worker can build the wall in 300 days
Capacity = 1/300
1 day work (2 skilled + 6 semi-skilled + 5 unskilled) = 2(1/100) + 6(1/200) + 5(1/300) = 1/15
∴ They can complete the work in 15 days.
 Question 88
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed?
 A 50 B 51 C 52 D 54
Aptitude       Numerical       2010
Question 88 Explanation:
Greater than 3000
⇒ First digit: 3 or 4
(i) First digit - 3:
We have to choose 3 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222, 333) is not possible as we have only two 2's and two 3's.
Total = 3 × 3 × 3 - 2 = 25
(ii) First digit - 4:
We have to choose 4 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222) is not possible we have only two 2's.
Total = 3 × 3 × 3 - 1 = 26
∴ Total number possible = 25 + 26 = 51
 Question 89

In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?

 A B C D
Aptitude       Numerical       Gate 2004-IT
Question 89 Explanation:
Let us consider total no. of families = 100
In that 50% of families having 3 childrens
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 childrens = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23
 Question 90

In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses.How many students have not taken any of the three courses?

 A 15 B 20 C 25 D 35
Aptitude       Numerical       Gate 2004-IT
Question 90 Explanation:
n = 200
PL = 125
DS = 85
CO = 65
PL & DS = 50
DS & CO = 35
CO & PL = 30
PL & DS & CO = 15

⇒ Not taken any course = 200 - (60 +15+15+35+15+15+20)
= 200 - 175
= 25
There are 90 questions to complete.