Numerical
Question 1 
∠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 2 
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 3 
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 4 
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 5 
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 6 
2880  
1440  
720  
0 
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?
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 8 
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 9 
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 10 
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 11 
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 12 
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 13 
the maximum of x and y  
the minimum of x and y  
1  
none of the above 
CaseI: x>y
⇒ xy = xy
Expression ⇒ (x+y)(xy)/ 2 = y
In this case value of expression is minimum of y.
CaseII: x
Expression ⇒ (x+y)  (xy) / 2
⇒ (x+y)  [(xy)]/2 ⇒ x
In this case value of expression is minimum of x.
Finally, the value of expression is minimum of x & y.
Question 14 
21  
18  
16  
14 
Question 15 
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 16 
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 17 
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 18 
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 19 
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 20 
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 21 
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 22 
P  
Q  
R  
S 
Question 23 
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 24 
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 25 
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 26 
(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 27 
(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 28 
CADBE  
JHKIL  
XVYWZ  
ONPMQ 
But in D, 2nd and 4th are in decreasing order.
Question 29 
n^{4}  
4^{n}  
2^{2n1}  
4^{n1} 
Product of roots (α, β) = 4
⇒ αβ = 4
(αβ)^{n} = 4^{n}
Question 30 
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 31 
21  
27  
30  
36 
Question 32 
f(x)=1x1  
f(x)=1+x1  
f(x)=2x1  
f(x)=2+x1 
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 
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 34 
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 35 
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 36 
Only relation I is true  
Only relation II is true  
Relations II and III are true.  
Relations I and III are true. 
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 
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 38 
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 39 
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 41 
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 42 
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 43 
5.34  
6.74  
28.5  
45.49 
Question 44 
Θ(nlog n)  
Θ(n)  
Θ(log n)  
Θ(1) 
Question 45 
59  
45  
43  
35 
Question 46 
VXDQ  
VZDQ  
VZDP  
UXDQ 
S+3=V, W+3=Z, A+3=D, N+3=Q.
Question 47 
2006  
2007  
2008  
2009 
Increase in imports in 2006 = 120  90/90 = 33.3% which is more than any other year.
Question 48 
PHome, QPower, RDefense, STelecom, TFinance  
RHome, SPower, PDefense, QTelecom, TFinance  
PHome, QPower, TDefense, STelecom, UFinance  
QHome, UPower, TDefense, RTelecom, PFinance 
Question 49 
x=yy
 
x=(yy)  
x=y+y
 
x=(y+y) 
Then the Answer is x =  (y  y)
Question 50 
96  
97  
98  
99 
Question 51 
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 52 
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 53 
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 54 
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 55 
∀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 56 
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 57 
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 58 
90  
100  
110  
120 
Average = 10+20+30+...+180+190 / 19 = (19/2 [10+190]) / 19 = 100
Question 59 
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 60 
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 61 
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 62 
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 63 
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 64 
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 65 
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 66 
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 67 
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 68 
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 69 
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 70 
7  
8  
9  
10 
Multiply on numerator & denominator
=1+9
=8
Question 71 
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 72 
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 73 
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 74 
5  
10  
15  
25 
Profit = 50q  5q^{2}
This is maximum at q = 5.
Question 75 
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 76 
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 77 
P, Q  
Q, R  
P, R  
R, S 
⇒ P, R are correct
Question 78 
OV  
OW  
PV  
PW 
OV is the next term.
Question 79 
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 80 
P  
Q  
R  
S 
Question 81 
5  
4  
7  
6 
Question 82 
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 83 
7.58 litres  
7.84 litres  
7 litres  
7.29 litres 
Question 84 
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 85 
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 86 
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 87 
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 88 
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 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?
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?
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