Numerical
Question 1 
Two cars start at the same time from the same location and go in the same direction. The speed of the first car is 50 Km/h and the speed of the second car is 60 Km/h. The number of hours it takes for the distance between the two cars to be 20 Km is _____.
2  
3  
1  
6 
Speed of the second car = 60 km/h
Let no. of hours = x say
⇒ From the question we can write
(60)x  (50)x = 20
10x = 20
∴ x = 2 hrs
Question 2 
In a college, there are three student clubs. Sixty students are only in the Drama club, 80 students are only in the Dance club, 30 students are only in the Maths club, 40 students are in both Drama and Dance clubs, 12 students are in both Dance and Maths clubs, 7 students are in both Drama and Maths club, and 2 students are in all the clubs. If 75% of the students in the college are not in any of these clubs, then the total number of students in the college is _____.
900  
975  
225  
1000 
No. of students present in three student clubs
= 60 + 80 + 30 + 38 + 2 + 10 + 5
= 225 [i.e., 25% of the students in the college]
Total no. of students in the college = 225 × 4 = 900
Question 3 
Three of the five students allocated to a hostel put in special requests to the warden. Given the floor plan of the vacant rooms, select the allocation plan that will accommodate all their requests.
Request by X: Due to pollen allergy, I want to avoid a wing next to the garden. Request by Y: I want to live as far from the washrooms as possible, since I am very sensitive to smell. Request by Z: I believe in Vaastu and so want to stay in the Southwest wing.The shaded rooms are already occupied. WR is washroom.
Question 4 
In the given diagram, teachers are represented in the triangle, researchers in the circle and administrators in the rectngle. Out of the total number of the people, the percentage of administrators shall be in the range of _____.
16 to 30  
0 to 15  
46 to 60  
31 to 45 
From the given diagram:
Total no. of people = 70 + 10 + 20 + 20 + 40 = 160
No. of Administrators = 50
% of Administrators = 50/160 = 31.25
Question 5 
∠BCD  ∠BAD  
∠BAD + ∠BCF
 
∠BAD + ∠BCD  
∠CBA + ADC 
→ ∠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
∠2 = ∠BAD
→ ∠BCD  ∠BAD = ∠1  ∠2
= ∠5 + ∠3 + ∠2  ∠2
= ∠ 5 + ∠3
= ∠DEC + ∠BFC
Question 6 
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?
2:3
 
1:1
 
3:2
 
2:1

→ 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 7 
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?
1,50,000
 
2,00,000  
1,75,000  
1,51,000 
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 8 
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?
Three green faces and four red faces.  
Four green faces and three red faces.  
Five green faces and two red faces.  
Six green faces and one red face. 
→ 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 9 
πd  
πd^{2}  
(1/2)πd^{2}  
(1/2)πd 
Area of a square A = a^{2} (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 10 
What is the missing number in the following sequence?
2880  
1440  
720  
0 
2 × 6 = 12
12 × 5 = 60
60 × 4 = 240
240 × 3 = 720
720 × 2 = 1440
1440 × 0 = 0
Question 11 
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?
3047  
6047  
7987  
63847 
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 12 
1  
1/pqr  
1  
pqr 
→ p^{x} = 1/q ⟹ 1/p^{x} = 1/q
⟹ q = p^{x}
⟹ log q = log p^{x}
⟹ x log p = log q
⟹ x = log q/ log p
→ q^{y} = 1/r ⟹ 1/q^{y} = 1/r
⟹ q^{y} = r
⟹ y log q = log r
⟹ y = log r/ log q
→ r^{z} = 1/p ⟹ 1/r^{z} = 1/p
⟹ p = r^{z}
⟹ log r^{z} = 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 13 
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?
Rahul and Murali  
Srinivas and Arul  
Srinivas and Murali  
Srinivas and Rahul 
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 14 
Find the smallest number y such that y × 162 is a prefect cube.
24  
27  
32  
36 
⇒ 162 = 2×3×3×3×3 = 3^{3}×(2×3)
For (2×3) to be a perfect cube, it should be multiplied by (2^{2}×3^{2})
∴ Required number = y = 2^{2}×3^{2} = 36
Question 15 
0.3^{k}  
0.6^{k}  
0.7^{k}  
0.9^{k} 
Total no.of possibilities = (10)^{k}
Required probability = (7)^{k} / (10)^{k} = 0.7^{k}
Question 16 
Six people are seated around a circular table. There are atleast two men and two women. There are at least three righthanded persons. Every woman has a lefthanded person to her immediate right. None of the women are righthanded. The number of women at the table is
2  
3  
4  
Cannot be determined 
Atleast three right handed persons.
None of the women are righthanded.
⇒ All the righthanded persons are men. Every woman has a lefthanded person to her immediate right.
⇒ For this to happen, there must be three lefthandedpersons and one of them should be a male.
As three persons are righthanded and those being men.
The number of women at the table = 2
Question 17 
the maximum of x and y  
the minimum of x and y  
1  
none of the above 
CaseI: x > y
⇒ x  y = x  y
Expression ⇒ (x + y)  (x  y) / 2 = y
In this case value of expression is minimum of y.
CaseII: x < y
⇒ 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 18 
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?
21  
18  
16  
14 
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
= ^{3}P_{3} + ^{3}P_{3}  ^{2}P_{2}
= 3! + 3!  2!
= 10
∴ Required number of ways = 24  10 = 14
Question 19 
A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot. If in a flood, the water level rises to 525 m, which of the villages P, Q, R, S, T get submerged?
P, Q  
P, Q, T  
R, S, T  
Q, R, S 
∴ 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 20 
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?
V  
W  
X  
Y 
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 21 
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?
12  
15  
18  
19 
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 22 
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
1⁄5  
7⁄30  
1⁄4  
4⁄15 
The probability of selecting same colour is
⇒ ^{3}C_{2} / ^{10}C_{2} * ^{4}C_{2} / ^{10}C_{2} * ^{3}C_{2} / ^{10}C_{2}
⇒ 3/45 + 6/45 + 3/45
⇒ 12/45
= 4/15
Question 23 
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?
The box labeled ‘Apples’  
The box labeled ‘Apples and Oranges’  
The box labeled ‘Oranges’  
Cannot be determined 
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 24 
X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X^{3} will have
90 digits  
91 digits  
92 digits  
93 digits 
X = 4777......... (29 times (7))
X = 4.777........ ×10^{29}
X = 5×10^{29} (4.777 rounded as 5)
X^{3} = 125×10^{87}
Total no. of digits = 3+87 [125=3 digit, 10^{87} = 87 digit]
= 90
Question 25 
The number of roots of e^{x} + 0.5x^{2} – 2 in the range [5, 5] is
0  
1  
2  
3 
e^{x} = 0.5x^{2} + 2
If we draw a graph for e^{x} and 0.5x^{2} + 2 in between [5, 5]
Question 26 
An air pressure contour line joins locations in a region having the same atmospheric pressure. The following is an air pressure contour plot of a geographical region. Contour lines are shown at 0.05 bar intervals in this plot.
If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region, which of the following regions is most likely to have a thunderstorm?
P  
Q  
R  
S 
Question 27 
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 __________.
56  
64  
72  
96 
⇒ Side of cube = 4
Surface area of cube = 6s^{2} = 6 × 4^{2} = 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 28 
A shaving set company sells 4 different types of razors, Elegance, Smooth, Soft and Executive. Elegance sells at Rs. 48, Smooth at Rs. 63, Soft at Rs. 78 and Executive at Rs. 173 per piece. The table below shows the numbers of each razor sold in each quarter of a year.
Which product contributes the greatest fraction to the revenue of the company in that year?
Elegance  
Executive  
Smooth  
Soft 
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 29 
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 ________.
40.00  
46.02  
60.01  
92.02 
General exponential function = a⋅e^{bx}
i.e., No. of cycles to failure = a⋅e^{bx}
load = x
At a load of 80 units, it takes 100 cycles to failure.
So, no. of cycles to failure = 100
load = 80
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
load = 40
i.e., 10,000 = a⋅e^{  b(40)} (2)
No. of cycles to failure = 5,000
load = x?
i.e., 5000 = a⋅e^{  bx}
Multiply with 2 on both sides,
10,000 = 2⋅a⋅e^{  bx}  (4)
Question 30 
Consider the following statements relating to the level of poker play of four players P, Q, R and S.

I. P always beats Q
II. R always beats S
III. S loses to P only sometimes
IV. R always loses to Q
Which of the following can be logically inferred from the above statements?

(i) P is likely to beat all the three other players
(ii) S is the absolute worst player in the set
(i) only  
(ii) only  
(i) and (ii)  
neither (i) nor (ii) 
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 31 
If f(x) = 2x^{7} + 3x  5, which of the following is a factor of f(x)?
(x ^{3} +8)  
(x1)  
(2x5)  
(x+1) 
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 32 
Pick the odd one from the following options.
CADBE  
JHKIL  
XVYWZ  
ONPMQ 
But in D, 2nd and 4th are in decreasing order.
Question 33 
n^{4}  
4^{n}  
2^{2n1}  
4^{n1} 
Product of roots (α, β) = 4
⇒ αβ = 4
(αβ)^{n} = 4^{n}
Question 34 
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 ______________.
35  
45  
65  
90 
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 35 
In a 2 × 4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?
21  
27  
30  
36 
Question 36 
f(x) = 1  x  1  
f(x) = 1 + x  1  
f(x) = 2  x  1  
f(x) = 2 + x  1 
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 37 
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?
0.20  
0.25  
0.30  
0.33 
Total outcomes = ^{4}C_{1} × ^{5}C_{1} = 20
Probability = Favourable outcomes/ Total outcomes = 4/20 = 0.20
Question 38 
Based on the given statements, select the most appropriate option to solve the given question.
If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building?
Statements:
(I) Each step is 3/4 foot high.
(II) Each step is 1 foot wide.
Statement I alone is sufficient, but statement II alone is not sufficient.  
Statement II alone is sufficient, but statement I alone is not sufficient.
 
Both statements together are sufficient, but neither statement alone is sufficient.  
Statement I and II together are not sufficient.

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 39 
The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?
32  
33  
34  
35 
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 40 
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c:

(I) p + m + c = 27/20
(II) p + m + c = 13/20
(III) (p) × (m) × (c) = 1/10
Only relation I is true  
Only relation II is true  
Relations II and III are true.  
Relations I and III are true. 
= 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 41 
The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.
What is the average of the marks obtained by the class in the examination?
2.290  
2.970  
6.795  
8.795 
Total marks obtained = 21 × 2 + 15 × 3 + 11 × 1 + 23 × 2 + 31 × 5 = 299
Average marks = 299/44 = 6.795
Question 42 
Statement I alone is not sufficient  
Statement II alone is not sufficient  
Either I or II alone is sufficient  
Both statements I and II together are not sufficient. 
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.
StatementII:
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.
Option (C) is the answer.
Either I or II alone is sufficient.
Question 43 
0, 1
 
1, 0
 
0, 1
 
1, 2

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 45 
I only  
I and II  
II and III  
I and III 
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 46 
8  
9  
7  
6 
= 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 47 
5.34  
6.74  
28.5  
45.49 
Question 48 
Θ(nlog n)  
Θ(n)  
Θ(log n)  
Θ(1) 
Question 49 
59  
45  
43  
35 
Question 50 
VXDQ  
VZDQ  
VZDP  
UXDQ 
S+3=V, W+3=Z, A+3=D, N+3=Q.
Question 51 
2006  
2007  
2008  
2009 
Increase in imports in 2006 = 120  90/90 = 33.3% which is more than any other year.
Question 52 
PHome, QPower, RDefense, STelecom, TFinance  
RHome, SPower, PDefense, QTelecom, TFinance  
PHome, QPower, TDefense, STelecom, UFinance  
QHome, UPower, TDefense, RTelecom, PFinance 
Question 53 
x=yy
 
x=(yy)  
x=y+y
 
x=(y+y) 
Then the Answer is x =  (y  y)
Question 54 
96  
97  
98  
99 
Question 55 
no roots  
2 real roots  
3 real roots  
4 real roots 
For roots to be real & positive b^{2}4ac>0
This will have 2 real positive roots.
ax^{2}+bx+c=0
Discriminant =b^{2}4ac>0
ax^{2}+bx+c
(b)^{2}4ac
⇒b^{2}4ac
Is also > 0. This will have real roots.
⇒ This will have 4 real roots.
Question 56 
850  
851  
852  
853 
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 57 
48  
49  
50  
51 
[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 58 
6  
7  
8  
9 
→ 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 59 
∀x:glitters(x) ⇒¬gold(x)  
∀x:gold(x) ⇒glitters()  
∃x: gold(x)∧¬glitters(x)  
∃x:glitters(x)∧¬gold(x) 
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.
The answer is Option (D).
Method 2:
⇒ (∼∀x) (∼ (glitters(x) ⇒ gold(x))
⇒ ∃x (∼ (∼glitters(x) ∨ gold(x))
⇒ ∃x (glitters ∧ gold(x))
Question 60 
0.25  
0.26  
0.27  
0.28 
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/(ba)=1/(0.50)=2 (length is between 0 to 0.5)
Expected value of length
(where P(l) is the probability density function)
Question 61 
1  
2  
3  
4 
3x + 2y = 1
4x + 7z = 1
x + y + z = 3
x – 2y + 7z = 0
This is a nonhomogeneous equation system.
The Augmented matrix for above set of equations is
For matrix (A):
R_{4}→ R_{4}+R_{1}
R_{4}→ R_{4}R_{2}
Rank = 3
For matrix (AB):
R_{4}→(R_{4}+R_{1} )R_{2}
Rank = 3
Rank of (A) = Rank (AB), so Unique solution
rank = 3 = no. of variables
Working Rule for Nonhomogeneous 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 62 
90  
100  
110  
120 
Average = 10+20+30+...+180+190 / 19 = (19/2 [10+190]) / 19 = 100
Question 63 
3.464  
3.932  
4.000  
4.444 
Squaring on both sides,
x^{2} = 12+x
x^{2}  x  12 = 0
(x4) (x+3) = 0
∴ x = 4 (x ≠ 3)
Question 64 
2, 4  
2, 14  
4, 52  
14, 52 
x^{2}  2x + 3 = 11, x is real
x^{2}2x+3 = 11
x^{2}2x+8 = 0
(x4)(x+2) = 0
x = 4, 2
x^{2}2x+3 = 11
x^{2}2x+14 = 0
x is not real in this case.
x^{3}+x^{2}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 x^{3}+x^{2}x include 14, 52.
Question 65 
140  
141  
142  
143 
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 = Mx/x × 100
= 12x/5x/ x × 100
= 7/5 × 100
= 140%
Question 66 
6:22 am  
6:27 am  
6:38 am  
6:45 am 
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 67 
17  
37  
64  
26 
2 = 1^{2}+1
5 = 2^{2}+1
10 = 3^{2}+1
17 = 4^{2}+1
26 = 5^{2}+1
37 = 6^{2}+1
50 = 7^{2}+1
64 = 8^{2}+0
64 does not belong to the series.
Question 68 
1.34  
1.74  
3.02  
3.91 
Total marks obtained = (21×2)+(15×3)+(23×2) = 133
Average marks = 133/44 = 3.02
Question 69 
Students should come at 9.00 a.m. and parents should come at 10.00 a.m.  
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.  
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.  
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. 
→ 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 70 
increased by 5%  
decreased by 13%  
decreased by 20%  
decreased by 11% 
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 71 
1:1  
2:1  
1.5:1  
2.5:1 
Male to female students ratio in 2011 = 1 : 1
Male to female students ratio in 2012 = 1.5 : 1 = 3 : 2
⇒ M_{1}/F_{1} = 1:1
M_{1} = F_{1}  (1)
⇒ M_{2}/F_{2} = 1:1
2M_{2} = 3F_{2}  (2)
Given,
F_{1} = F_{2}  (3) From (1) & (2)
M_{1}/M_{2} = F_{1}/(3F_{2}/2) = 2F_{1}/3F_{2}
But from (3)
M_{1}/M_{2} = 2/3
We need to find
M_{2} : M_{1} = 3 : 2 = 1.5 : 1
Question 72 
1634  
1737  
3142  
3162 
⇒ 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 73 
502  
504  
506  
500 
44, 42, 40, ...... 0
2(22 + 21 + 20 + …. 1)
Sum(n)=(n(n+1))/2=(22×23)/2=253
⇒2(253)
=506
Question 74 
7  
8  
9  
10 
Multiply on numerator & denominator
=1+9
=8
Question 75 
13/90  
12/90  
78/90  
77/90 
⇨ 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 76 
36  
30  
24  
18 
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 77 
16,500  
15,180  
11,000  
10,120 
Working hours decreased by (1/24) ⇒ 11/24 ⇒ 23/24
The new cost of erection = 13,200 × 6/5 × 23/24 =15,180
Question 78 
5  
10  
15  
25 
Profit = 50q  5q^{2}
For Profit to be maximum,
d(Profit)/dq = 0
d(50q  5q^{2}) /dq = 0
50  10q = 0
50 = 10q
q = 5
Question 79 
8 meters  
10 meters  
12 meters  
14 meters 
y = 2x  0.1x^{2}
Differentiating both sides,
dy/dx = 2  0.2x
For maximum, dy/dx = 0
2  0.2x = 0
0.2x = 2
x = 10
Question 80 
0.288  
0.334  
0.667  
0.720 
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 81 
P, Q  
Q, R  
P, R  
R, S 
⇒ P, R are correct
Question 82 
OV  
OW  
PV  
PW 
OV is the next term.
Question 83 
P^{2} = Q^{3}R^{2}  
Q^{2} = PR  
Q^{2} = R^{3}P  
R = P^{2}Q^{2} 
∴ P = b^{k}, Q = b^{2k}, R = b^{3k}
Now, Q^{2} = b^{4k} = b^{3k} b^{k} = PR
Question 84 
P  
Q  
R  
S 
Question 85 
5  
4  
7  
6 
Question 86 
4  
5  
6  
7 
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 87 
7.58 litres  
7.84 litres  
7 litres  
7.29 litres 
Question 88 
2  
17  
13  
3 
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 89 
534  
1403
 
1623  
1513 
(137)_{x} + (276)_{x} = (435)_{x}
x^{2} + 3x + 7 + 2x^{2} + 7x + 6 = 4x^{2} + 3x + 5
x^{2}  7x  8 = 0
x = 8 (or) 1
∴ x = 8 (1 cannot be a base)
(731)_{x} + (672)_{x} = (731)_{8} +( 672)_{8}
= 7 × 8^{2} + 3× 8 + 1×1 + 6 × 8^{2} + 7 × 8 + 2 × 1
= 915
From the options, 915 can be written as 1623 in base 8.
Question 90 
HSIG  
SGHI  
IGSH  
IHSG 
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 91 
20  
10  
16  
15 
Capacity = 1/100
8 semiskilled workers can build the wall in 25 days
⇒ 1 semiskilled worker can build the wall in 200 days
Capacity = 1/200
10 unskilled workers can build the wall in 30 days
⇒ 1 unskilled worker can build the wall in 300 days
Capacity = 1/300
1 day work (2 skilled + 6 semiskilled + 5 unskilled) = 2(1/100) + 6(1/200) + 5(1/300) = 1/15
∴ They can complete the work in 15 days.
Question 92 
50  
51  
52  
54 
⇒ 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 93 
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?
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 94 
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?
15  
20  
25  
35 
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