Question 1 1.2 / -0
Directions: Two rows of numbers are given. The resultant number in each row is to be worked out separately based on the following rules and the question below the rows is to be answered. The operations on numbers progress from left to right.
Rules: (i) If an even number is followed by another even number, they are to be added.
(ii) If an even number is followed by a prime number, they are to be multiplied.
(iii) If an odd number is followed by an even number, the even number is to be subtracted from the odd number.
(iv) If an odd number is followed by another odd number, the first number is to be added to the square of the second number.
(v) If an even number is followed by a composite odd number, the even number is to be divided by the odd number.
What is the sum of the resultants of the two rows?
Solution
65 + 112 = 186 (Row 1 : Rule iv)2 = 24 (Row 2 : Rule iv)
Question 2 1.2 / -0
Kalyani is mother-in-law of Veena who is sister-in-law of Ashok. Dheeraj is father of Sudeep, the only brother of Ashok. How is Kalyani related to Ashok?
Solution
The diagrammatic representation of the given relation is shown below.
Kalyani is mother of Ashok.
Question 3 1.2 / -0
Inspector Jatin travelled 400 metres from his police station. He then turned left and travelled 500 metres straight, after which he turned left again and travelled 400 metres straight. Then, he turned right and walked another 600 metres straight. How far is he from the police station?
Solution
BC = Initial displacement (400 m)
CD = Displacement after left turn (500 m)
DA = Displacement after another left turn (400 m)
AE = Displacement after right turn (600 m)
Therefore,
BE = Final displacement from the police station
= 600 + 500 m
= 1100 m = 1.1 km
Question 4 1.2 / -0
Select the correct water image of given combination of letters and numbers.S5L3T8
Solution
The water image of S5L3T8 is shown below.
Question 5 1.2 / -0
Select a figure from the options which satisfies the condition of placement of dots as in Fig. (X).
Solution
In the given figure, one of the dots is placed in the area shared by square and circle only and the other dot is placed in the area shared by circle, square and the triangle.
Out of the given options, only the option three has area shared by square and circle only and area shared by all the three shapes, according to which the dots can be placed as shown below.
Question 6 1.2 / -0
Select a figure from among the options which will continue the same series as established by the five Problem Figures P, Q, R, S and T.
Solution
The Alphabet C remains at same position for two consecutive steps. So, position of C will be same as T box. Therefore, we are left with option 3 and 4. Now looking at triangle it would also remain at the same position for two consecutive steps. So position of triangle will be same as T. Hence, we are left with only option (3).
Question 7 1.2 / -0
Study the following information carefully and answer the question given below.
Solution
Q, a male engineer travels with only W, a teacher in vehicle I.
Vehicle I W (female teacher) Q (male engineer) Vehicle II Vehicle III
R is lady doctor and she does not travel with the pair of sisters P and V. S is a male doctor.
P is not an engineer and travels in vehicle II.
Vehicle I W (female teacher) Q (male teacher) Vehicle II R (female doctor) P (male teacher) Vehicle III P and V( female) S (male doctor)
Therefore, W, P, V and R are four females travelling in the vehicles.
Question 8 1.2 / -0
Rita stands third in a row of twenty students, arranged in ascending order of heights. Five new students join the group, all taller than Rita. What will be Rita's position if the students are now arranged in descending order of their heights?
Solution
Arranged in ascending order of heights, Rita stands third in a row of twenty students.
Question 9 1.2 / -0
Select a figure from the options which will complete the pattern in Fig. (X).
Solution
The complete figure is shown below.
Therefore, the missing part is
.
Question 10 1.2 / -0
If RUST = 9-6-8-7 and BOARD = 25-12-26-9-23, then how will you code "BEAT"?
Solution
A B C D E F G H I J K L M 26 25 24 23 22 21 20 19 18 17 16 15 14 N O P Q R S T U V W X Y Z 13 12 11 10 9 8 7 6 5 4 3 2 1
Therefore,
Question 11 1.2 / -0
Which of the following Venn diagrams best represents the relationship amongst, ''Honesty, Intelligence, and Aptitude''?
Solution
Honesty, intelligence and aptitude are all different qualities or skills of a person. The following diagram represents the best relation between them.
Question 12 1.2 / -0
The letters in the word ULTRAVIOLET are arranged in the alphabetical order and each letter is assigned a numerical value equal to its serial number as in the English alphabet. What will be the difference between the sum of odd-positioned numbers and that of even-positioned numbers?
Solution
The given word ULTRAVIOLET will be written alphabetically as AEILLORTTUV and its serial numbers will be: 1, 5, 9, 12, 12, 15, 18, 20, 20, 21, 22.
Question 13 1.2 / -0
A cube of side 10 cm is colored red with a 2 cm wide green strip along all the sides on all the faces. Now, the cube is cut into 125 smaller cubes of equal size. How many cubes have three green faces each?
Solution
Clearly, upon colouring the cube as stated and then cutting it into 125 smaller cubes of equal size we get a stack of cubes as shown in the following figure.
The figure can be analysed by assuming the stack to be composed of 5 horizontal layers. All the corner cubes are painted green. So there are 8 cubes with 3 sides painted green.
Question 14 1.2 / -0
Code the group of digits as per the scheme and conditions given below.
(i) If the first as well as the last digit is odd their codes are to be interchanged.
(ii) If the first digit is even and the last digit is odd both are to be coded by the code for odd digit.
(iii) If the last digit is '0' it is to be coded by 'X'.
(iv) If the first as well as the last digit is even both are to be coded by '-'.
586403
Solution
5 8 6 4 0 3
3 8 6 4 0 5 (Using (i)
RJQHK
Hence, option (2) is the correct answer.
Question 15 1.2 / -0
A set of three figures X, Y and Z shows a sequence of folding of a piece of paper. Figure (Z) shows the manner in which the folded paper has been cut. Select the figure from the options which would resemble the unfolded form of paper.
Solution
Two squares at the bottom of the unfolded figures are same for all. Observing carefully the last image of the folded one, we find that at the left top corner there is a triangular-shaped cut; so after unfolding, it will give a square shape as shown in option (3).
Question 16 1.2 / -0
The points, whose abscissa and ordinate have different signs, lie in quadrants _______.
Solution
Abscissa is the horizontal 'x' value and ordinate is the vertical 'y' value in a pair of coordinates (x, y) and their signs in all the four quadrants are (x, y), (-x, y), (-x, -y) and (x, -y), respectively. So, if abscissa and ordinate have different signs, then they will lie in quadrant II and IV.
Question 17 1.2 / -0
Which of the following is a true statement?
Solution
Statement (1) is false because there are infinite number of lines that can be drawn to pass through a given point.
Question 18 1.2 / -0
The value of 3
is
Solution
We know (a
2 - b
2 ) + (b
2 - c
2 ) + (c
2 - a
2 ) = 0
So, (a
2 - b
2 )
3 + (b
2 - c
2 )
3 + (c
2 - a
2 )
3 = 3(a
2 - b
2 )(b
2 - c
2 )(c
2 - a
2 )
Also, (a - b) + (b - c) + (c - a) = 0
So, (a - b)
3 + (b - c)
3 + (c - a)
3 = 3(a - b)(b - c)(c - a)
Now,
3
= =
Question 19 1.2 / -0
The expression 2x3 + ax2 + bx + 3, where a and b are constants, has a factor of x - 1 and leaves a remainder of 15 when divided by x + 2. Find the values of a and b, respectively.
Solution
Given: P(x) = 2x3 + ax2 + bx + 3
Question 20 1.2 / -0
In the given figure, AOB is a straight line and
AOX
3 = 57°,
X
1 OX
4 = 97°,
X
3 OB = 123° and
X
4 OB = 68°. Find
AOX
1.
Solution
Given,
AOX
3 = 57°,
X
1 OX
4 = 97°,
X
3 OB = 123° and
X
4 OB = 68°
Now,
AOX
4 = 180° -
X
4 OB = 180° - 68° = 112°
So,
AOX
1 =
AOX
4 -
X
1 OX
4 = 112° - 97° = 15°
Hence, option (4) is correct.
Question 21 1.2 / -0
Find the value of
.
Solution
Consider the numerator:
Consider the denominator:
= 65/16
So, value of the required expression = 15/(65/16) = 48/13
Question 22 1.2 / -0
Which of the following statements is INCORRECT?
Solution
Statement (1) is incorrect because a real number is either rational or irrational, but not both.
A rational number can be expressed as the ratio of two integers.
An irrational number is one that cannot be expressed as the ratio of two integers.
Statement (2) is correct. Consider two irrational numbers
and
. Then, their sum is 6, which is not irrational.
Statement (3) is correct. Consider two positive integers 2 and 3 such that 2 < 3. This implies that 2
2 < 3
2 .
Statement (4) is correct, i.e. every integer is a rational number.
Hence, option (1) is the correct answer.
Question 23 1.2 / -0
The degree of the polynomial 3x
2 + 12 -
+ 12x + 4 is
Solution
3x
2 + 12 -
+ 12x + 4
= 3x
2 + 12 - 3x
2 - 12 - 12x + 12x + 4
= 4
Since this is a constant polynomial and we know that the degree of a constant polynomial is 0, option (4) is correct.
Question 24 1.2 / -0
'L', 'B' and 'H' of a cuboid are increased, decreased and increased by 1%, 3% and 2%, respectively. The volume of the cuboid will ______.
Solution
Volume of a cuboid = LBH
New measurements:
Hence, new volume = l
1 b
1 h
1 =
LBH < LBH
Therefore, the volume will decrease.
Question 25 1.2 / -0
The area of the triangle formed by 2x + 3y = 6 and the coordinate axes is ________.
Solution
The required area is the above shaded region.
So, area of triangle OAB = (1/2) x base x height = (1/2) x 3 x 2 = 3 square units
Question 26 1.2 / -0
The perimeter of a triangle is 6p2 - 4p + 9 and two of its sides are p2 - 2p + 1 and 3p2 - 5p + 3. Find the third side of the triangle.
Solution
A, B and C are the sides of the triangle.2 - 4p + 9 = ( p2 - 2p + 1 ) + ( 3p2 - 5p + 3 ) + C2 - 4p + 9 - p2 + 2p - 1 - 3p2 + 5p - 32 + 3p + 5
Question 27 1.2 / -0
In the given figure, ABCD is a rectangle. BD = BE,
BED = 40° and
EDA = 260°. Find
CDB.
Solution
We know that reflex angle + interior angle = 360°
That is, reflex
EDA + interior
EDA = 360°
260° + interior
EDA = 360°
Interior
EDA = 100°
Now, in triangle BDE, BD = BE, so
BED =
BDE = 40°
From the figure,
Interior
EDA =
ADB +
BDE
ADB = interior
EDA -
BDE = 100° - 40° = 60°
Since
ADC is a right angle;
CDB = 90° -
ADB = 90° - 60° = 30°
Hence, option (2) is correct.
Question 28 1.2 / -0
In the given figure, line l is II line BC and D is the mid-point of BC.
If area (
ABC) = x × area (
EDC), then find the value of x.
Solution
We know that, two triangles on the same base and between the same parallels are equal in area.
So, area of triangle EDC = area of triangle ADC = (1/2) × a × h
Area of triangle ABC = (1/2) × 2a × h = ah
Given that, area of triangle ABC = x × area of triangle EDC
So,
ah = x × (1/2)ah
x = 2
Hence, option (2) is correct.
Question 29 1.2 / -0
Fill in the blanks:
Solution
If a straight line falling on two straight lines make the interior angles on the same side of it, taken together are less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. This is Euclid 5th postulate.
Question 30 1.2 / -0
From the given information, find the probability of people with ages 60, 61 and 64 who can drive.
Age (in years) Number of persons of different ages who can drive the car 60 16,090 61 11,490 62 8,012 63 5,448 64 3,607 65 2,320
Solution
Total persons of different ages = 16,090 + 11,490 + 8,012 + 5,448 + 3,607 + 2,320 = 46,967 = A
Question 31 1.2 / -0
Factors: x4 + 5x3 + 5x2 - 5x - 6
Solution
x4 + 5x3 + 5x2 - 5x - 64 - x3 + 6x3 - 6x2 + 11x2 - 11x + 6x - 63 (x - 1) + 6x2 (x - 1) + 11x(x - 1) + 6(x - 1)3 + 6x2 + 11x + 6)3 + 2x2 + 4x2 + 8x +3x + 6)2 (x + 2) + 4x(x + 2) + 3(x + 2))2 + 4x + 3)2 - 1)(x + 2)(x + 3)
Question 32 1.2 / -0
A certain distance is covered at a certain speed (s1 ). If half of this distance is covered in double the time at speed (s2 ). Find the ratio of the two speeds. Also, if s1 is 60 km/hr, then find s2 .
Solution
Let the distance be d and time be t.1 = d/t and 2 = (d/2)/(2t) = d/(4t) = s1 /41 /s2 = 4/1, this implies that s1 : s2 = 4 : 11 = 60 km/hr, then s2 = 15 km/hr
Question 33 1.2 / -0
'Lines are parallel if they do not intersect' is stated in the form of
Solution
An axiom or postulate is a statement which is taken to be true without proof.
Question 34 1.2 / -0
The two circles shown below have radii x and 3x. A point is chosen, at random, inside the larger circle. Find, in its simplest fractional form, the probability that this point is in the shaded area.
Solution
Probability = Shaded area/Total area = (π(3x)2 - πx2 )/π(3x)2 = 8/9
Question 35 1.2 / -0
Number of players participating in three different games in five different schools.
Number of players participating in Kho-Kho from School-4 is what percent of number of players participating in hockey from School-2?
Solution
Number of players participating in Kho-Kho from School-4 = 32
Question 36 1.2 / -0
The population of a town was 1,60,000 three years ago. If it had increased by 3%, 2.5% and 5% in the last three years, then find its present population.
Solution
The population of the town was 1,60,000.
It had increased by 3%, 2.5% and 5% in the last three years.
Hence, the present population is
=
= 1,60,000 x 1.03 x 1.025 x 1.05 = 1,77,366
Question 37 1.2 / -0
The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution
△BAD is a right angle triangle,
BD
2 = 7
2 + 24
2 = 625
BD = 25
Area of triangular part ABD = 1/2 × base × height = (1/2) × 24 × 7 = 84 sq.m
In triangle BDE ,
BD = 25 = a
BE = 26 = b
DE = 27 = c
s = ( a + b + c)/2 = ( 25 + 26 + 27 )/2 = 39
Area of triangle BDE =
= 291.85
Total area = 291.85 + 84 = 375.85 sq.m
Question 38 1.2 / -0
A alone can complete a work in 16 days and B alone in 12 days. Starting with A, they work on alternate days. The total work will be completed in ______________
Solution
One day work of A + one day work by B = (1/16) + (1/12) = 7/48
As they are doing work on alternate days, so work is be completed in 48/7 or 6 + (6/7) days.
This shows that for 12 days they will work on alternate days.
Then A will work for 1 day and then B will do for 3/4 days to finish (1/16)
th part of the work which is left after 13 days.
So, total number of days = 12 + 1 + (3/4) = 13
Question 39 1.2 / -0
A triangular park in a city has dimensions 100 m × 90 m × 110 m. A contract is given to a company for planting grass in the park at the rate of Rs. 4,000 per hectare. Find the amount to be paid to the company.
(Take
= 1.414)
Solution
Area of a triangle =
where s is the semi-perimeter and a, b and c are the sides of the triangle.
s =
Here, for the triangular park,
a = 100 m, b = 90 m, c = 110 m ands =
= 150 m
Area of the triangular park =
Rate of planting grass = Rs. 4000/ha
So, the amount to be paid = 4000 x 0.4242 = Rs. 1696.80
Hence, option (3) is correct.
Question 40 1.2 / -0
Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire rental of the car, then the share of each of the remaining persons is increased by ___________ of the original share.
Solution
Let the cost of the rental car be x.
Question 41 1.2 / -0
The taxi charges in a city comprise of a fixed charge, together with the charge of the distance covered. For a journey of 16 km, the charges paid are Rs. 156 and for a journey of 24 km, the charges paid are Rs. 204.
Solution
Let the fixed charge be Rs. x and the distance per km is Rs. y.
Question 42 1.2 / -0
If 6 years are subtracted from the present age of Gagan and is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then what is Gagan's present age?
Solution
Let the present age of Gagan is x years.
Question 43 1.2 / -0
A trader purchases 70 kg of tea at Rs. 15 per kg and 30 kg of tea at Rs. 18.50 per kg. If the packing charges are 2%, then at what price he must sell the mixture of two to gain 15%?
Solution
Total price of procurement = 70 × 15 + 30 × 18.50 = 1050 + 555 = 1605
Including packing charge of 2%, CP = 1605 ×
= 1637.10
Total weight of Tea = 70 + 30 = 100 kg
Expected profit = 15%
Therefore, SP = 1637.10 ×
= 18.83 per kg
Question 44 1.2 / -0
The population of a city had increased successively at the rate of 6%, 4% and 2% per annum during last three years. If its present population is 11,24,448, then what was its population three years ago?
Solution
Let the population three years ago was x.
Question 45 1.2 / -0
The cost price of an article A is Rs. 160 and selling price of another article B is Rs. 240. If the selling price of A be equal to the cost price of B, then the profit after selling A is 20%. What is the profit on B?
Solution
Cost price of article A = Rs. 160
Selling price of article A = cost price of article B = Rs. x
Profit of A = 20%
20 =
x = Rs. 192
Selling price of article B = Rs. 240
Profit on B =
= 25%
Question 46 1.2 / -0
The figure below is made up of one big circle, two identical medium circles and two identical small circles.
The ratio of the radius of the small circle to the radius of the medium circle is 2 : 3.
(a) What is the total area of the unshaded part in the figure?
(b) What fraction of the big circle is shaded?
Solution
Let r,
, R be the radii of small, medium and big circles respectively.Radius of small circle = r = 4 cm.Radius of medium circle =
=
= 6 cmRadius of big circle = R = 2 x 6 = 12 cm.(a)Area of unshaded part = 2 x πr
2 + 2 x πρ
2 = 2π (4
2 + 6
2 ) = 104π cm
2 (b)Area of big circle = πR
2 + 122 π = 144 πArea of shaded part = πR
2 - 104 π = 144π - 104 π = 40 π cm
2 Fraction of area shaded =
Question 47 1.2 / -0
Study the statements carefully.
Statement I: If P(x) is a polynomial of degree
1 and ax + b is a factor of P(x), then we have P
= 0.
Statement II: If P(x) is a polynomial of degree
1, then polynomial (x - a) (x - b) is a factor of P(x), if P (a) = 0 and P (b) = 0.
Solution
Statement I: If P(x) is a polynomial of degree
1 and ax + b is a factor of P(x), then we have P
= 0.
Since ax + b is a factor of P(x),
ax + b = 0
x = (-b/a)
This value of x will satisfy P(x), i.e. P(-b/a) = 0
Statement (1) is true.
Statement II: If P(x) is a polynomial of degree
1, then polynomial (x - a) (x - b) is a factor of P(x), if P(a) = 0 and p (b) = 0.
Going by the same factor theorem, (x - a) (x - b) is a factor of P(x), then x -a = 0 and x - b = 0 will satisfy P(x).
i.e. x = a and x = b both will satisfy P(x).
Or, P(a) = 0 and P(b) = 0
Statement (2) is also true.
Question 48 1.2 / -0
ABCDE.... is part of a regular polygon which has interior angles of 160°. CDLM is a square.
Find the value of x and y respectively.
Solution
BCD = 160°
BCM = x° = 160° - 90°
= 70°
∵ BC = CD = CM
In triangle BCM,
BMC =
CBM
BMC +
CBM + 70° = 180°
BMC =
CBM = 55°
y° = 160° - 55°= 105°
Question 49 1.2 / -0
State True (T) or False (F).
(P) In a
ABC, if E is the midpoint of median AD, then ar(
BED) =
ar(
ABC).
(Q) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(R) If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is 1 : 2.
(S) If in a trapezium ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O, then, ar (
AOB) = ar (
COD).
Solution
In a ABC, if E is the midpoint of median AD, then ar(BED) =
ar(ABC).
A parallelogram and a rectangle on the same base and between the same parallels are equal in area
If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is 1 : 2.
If in a trapezium ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O, then,
Since, △ AOB and △COD are not on the same base
∴ ar (AOB) ≠ ar (COD).
Question 50 1.2 / -0
A, B, C are three sets of values of x:
Solution
A: 1, 2, 2, 3, 3 , 3, 7
×
Sign in
Profile
Signout
Get latest Exam Updates 
