Question 1 1.2 / -0
What will be the next term in the following series?
Solution
256 × (32 - 22 ) = 1280
1280 × (42 - 32 ) = 8960
8960 × (52 - 42 ) = 80640
80640 × (62 - 52 ) = 887040
887040 × (72 - 62 ) = 11531520
Question 2 1.2 / -0
Rohit moved 8 km towards north before turning to his right and walking 11 km straight. He then took a left turn and walked straight for 1 km before taking another left turn and then walking straight for 5 km. From there, he took a left turn and walked straight for 1 km to reach a shop. Find the shortest distance between the shop and the starting point.
Solution
Required shortest distance is PQ.
It forms a right-angled triangle.
Using Pythagoras Theorem,
(PQ)
2 = (8
2 + 6
2 ) km
= 64 km + 36 km = 100 km
Starting point to shop distance = 10 km
Question 3 1.2 / -0
Find out the wrong term in the given series:
Solution
The given series follows the pattern:0 + 1 = 1 + 1 = 21 + 2 = 6 + 2 = 82 + 3 = 36 + 3 = 393 + 4 = 216 + 4 = 2204 + 5 = 1296 + 5 = 1301
Question 4 1.2 / -0
If '×' stands for 'addition', '÷' stands for 'subtraction', '+' stands for 'multiplication' and '-' stands for 'division', then which of the following equations is correct?
Solution
204 - 17 + 13 x 12 ÷ 4 means 204 ÷ 17 x 13 + 12 - 4
= 12 x 13 + 12 - 4 = 164
Question 5 1.2 / -0
Count the number of cubes in the following figure:
Solution
Columns 1, 2, 3, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 19, 20, 21, 22, 23, 24 have 4 blocks each.
So, total number of blocks in these columns is 76.
Now, columns 4, 5, 11, 14, 17, 25, 26, 27, 28 have 3 blocks each.
So, total number of blocks in these columns is 27.
Now, columns 29, 30, 31 and 32 have 2 blocks each.
So, total number of blocks in these columns is 8.
Hence, total number of blocks in the given figure is 76 + 27 + 8 = 111.
Question 6 1.2 / -0
How many faces are there in the following figure?
Solution
There are 5 + 4 + 1 + 1 + 2 = 13 faces in the above figure.
Question 7 1.2 / -0
Directions: In the following question, a set of three figures X, Y and Z shows a sequence in which a paper is folded and finally cut from a particular section. Below these figures, a set of answer figures marked (a), (b), (c), (d) shows the design which the paper actually acquires when it is unfolded. You have to select the answer figure that most closely resembles the unfolded piece of paper.
Solution
The paper is folded along the edges and then 2 holes are cut out. So, when we unfold the paper, we will get
.
Question 8 1.2 / -0
Directions: In the following question, there are two sets of figures - the Problem figures and the Answer figures. The Problem figures are presented in two units. The first unit has two figures and the second unit has one figure and a question mark. Find out which one of the Answer figures marked (A) to (D) should come in place of the question mark.
Solution
Option (4) is correct. In the first problem unit, two horizontal lines are followed by three vertical lines. So, in the second problem unit, four vertical lines should be followed by five horizontal lines, which is seen in Answer figure (D).
Question 9 1.2 / -0
In a joint family, there is a father, a mother, three married sons and one unmarried daughter. Out of the sons, two have two daughters each and one has a son. How many female members are there in the family?
Solution
Circles denote females.
Squares denote males.
There are total 9 female members in the family.
Question 10 1.2 / -0
How many edges are there in the given figure?
Solution
There are 8 edges on the top face + 8 edges on the bottom face + 8 edges on the face in between the top and the bottom face (sides) = 24 edges.
Question 11 1.2 / -0
Directions:
Solution
In the same way,
Question 12 1.2 / -0
Directions :
Solution
Except option (3), all the others are planets.
Question 13 1.2 / -0
Directions: Study the following arrangement carefully and answer the question given below.
B M % R 3 J @ K © D F 6 9 W 4
N E P 2 $ A Y 5 I Q Z # 7 U G
If all the symbols and all the vowels are dropped from the above arrangement, then which of the following will be twelfth from the right end?
Solution
New arrangement after dropping symbols and vowels:
Question 14 1.2 / -0
Directions: The first figure in the first unit of the problem figures bears a certain relationship to the second figure. Similarly, one of the figures in the answer figures bears the same relationship to the second figure of the second unit of the problem figures. Identify the figure which would fit in place of the question mark.
Solution
Only the central small line is extended and rest of the pattern remains the same.
Question 15 1.2 / -0
Directions : Statements: Some ropes are walls.Conclusions: I. Some tables are walls.
Question 16 1.2 / -0
An A.P. consists of 53 terms. If the last term and the first term be 216 and 8 respectively, then find the value of 48th term.
Solution
Given, n = 80, a1 = 8, a53 = 216
Now, a1 + 52d = 216
8 + 52d = 216
52d = 216 - 8
52d = 208
d = 4
Hence, a48 = a1 + 47d = 8 + 47(4)
a48 = 196
Question 17 1.2 / -0
If
where R is a point on the line segment P(2, 1) and Q(- 2, 3), then find out the ordinate of the point R.
Solution
ATQ
Let co-ordinates of R be (x, y)
As
Therefore,
PR : RQ = 4 : 5
Now,
Therefore:
(x, y) =
(x, y) =
Now, ordinate means the y-value which is
.
Question 18 1.2 / -0
Evaluate tan2 θ - sec2 θ (1 - sin2 θ)
Solution
According to the question:2 θ - sec2 θ (1 - sin2 θ)2 θ - sec2 θ (sin2 θ + cos2 θ - sin2 θ) [By using identity: sin2 θ + cos2 θ = 1]2 θ - 2 θ) = tan2 θ - 1
Question 19 1.2 / -0
A wall around a circular park needs to be build. The cost of building the wall around the park is Rs. 1.76/cm. If the area of the park is 962.5 m2 , then find the total cost to build the wall around the park.
Solution
To find the cost, we have to find the circumference.
Area of park = πr
2 962.5 m
2 =
x r
2 6737.5 m
2 = 22 x r
2 r = 17.5 m
Now, circumference of circle = 2
r = 2 x
x 17.5 m = 110 m
Required cost = Rs. (1.76 x 11000) = Rs. 19,360
Question 20 1.2 / -0
If
, then find the value of
.
Solution
As
So,
or,
--eq. (1)
Now,
=
[By using identity:
]
Now, by using eq. (1), we get
=
Question 21 1.2 / -0
Find the area of the shaded region from the given figure in which AB = 24 cm, AC = 7 cm, and O is the centre of the circle. BC is the diameter of the circle.
Solution
BC
2 = AB
2 + AC
2 BC
2 = 24
2 cm
2 + 7
2 cm
2 BC = 25 cm
So, radius OC = 12.5 cm
Area of a triangle =
x base x height
=
x 24 x 7 sq. cm
=84 sq. cm
Area of a semicircle =
Area = 245.31 sq. cm
So, area of the shaded region = (245.31 – 84) sq. cm = 161.31 sq. cm
Question 22 1.2 / -0
If the length of AO is 5 cm, A is the centre of the bigger circle and O is the centre of the smaller circle, then find the area of the remaining portion if the smaller circle is taken out of the bigger one.
Solution
Area of circle = π r
2 So, area of smaller circle =
= 78.57 cm
2 Area of bigger circle =
= 314.28 cm
2 So, remaining area = 314.28 – 78.57 = 235.71 cm
2
Question 23 1.2 / -0
The Y-axis divides the line joining points (- 3, - 2) and (5, 4) in the ratio 3 : 5. Find the coordinates of Y axis.
Solution
By using internal section formula:
The co-ordinates are given by
As on Y-axis, x-coordinate is zero
So, y =
So, the answer is (0, 0.25)
Question 24 1.2 / -0
What will be the H.C.F of three numbers which are in ratio of 3 : 5 : 7 and their L.C.M is 1575?
Solution
Let the numbers be 3X, 5X and 7X.
Then, the L.C.M is 105X.
So, 105X = 1575
=> X = 15
So, the numbers are 3 x 15, 5 x 15 and 7 x 15.
So, H.C.F is 15.
Question 25 1.2 / -0
Find the centroid of the triangle whose vertices are (2, 1), (5, 2), (3, 4).
Solution
Centroid of a triangle =
=
Question 26 1.2 / -0
Find the area of the shaded portion in the given figure, if the radius of each small circle is 14 cm.
Solution
Area of shaded portion = Area of square – 4 × area of a sector (of each small circle)
= 282 cm2 - 4 × cm2
= 784 cm2 – 616 cm2
= 168 cm2
Question 27 1.2 / -0
In triangle PQR, UV || QR, PU : UQ = 3 : 7. Find the ratio of area of triangle PUV to area of PQR.
Solution
PUV ∽
PQRPU = 3aUQ = 7aPU + UQ = 10a = PQ
Thus answer option (1) is correct.
Question 28 1.2 / -0
The figure given below shows an octagon with each interior angle measuring 135°. What is the area of the octagon?
Question 29 1.2 / -0
The point which divides the line joining (1, 2) and (3, 4) internally in the ratio of 1 : 1 lies in which quadrant?
Solution
Since the coordinates of both points are positive, the line joining them lies in the first quadrant. The division of the line in the ratio of 1 : 1 means the mid-point of this line also, therefore, lies in the first quadrant, with both coordinates positive.
Question 30 1.2 / -0
If in ΔABC, DE
BC, then what can be the possible value(s) of x?
Solution
or
(x – 3)(x – 4) = 4(3x – 19)
Or x
2 – 7x + 12 = 12x – 76
i.e. x
2 – 19x + 88 = 0
Or (x – 8)(x – 11) = 0, giving x = 8 or 11
Question 31 1.2 / -0
In a quadrant of radius 6a, two semicircles with centres D and F are cut as shown in the figure. If a circle with centre E is cut as shown in the figure, then what will be the area of the remaining part of the quadrant?
Solution
Required area = Area of the quadrant – Area of the bigger semicircle – Area of the smaller semicircle – Area of the circle
Given: OA = 6a
Area of the quadrant =
Area of the bigger semicircle =
Let the radius of the smaller semicircle be 'x' and the radius of the circle be 'y'.
From the figure, DE = OF = x + y = 3a, and 3a + y = EF = OD; and in triangle ODF, DF = 3a + x. As the triangle is a right-angled triangle, we have
OD
2 + OF
2 = DF
2 (y + 3a)
2 + 9a
2 = (3a + x)
2 (y + 3a)
2 + 9a
2 = (6a – y)
2 6ay + 18a
2 = 36a
2 – 12ay
18ay = 18a
2 y = a
Hence, x = 2a
Area of the circle =
Area of the second semicircle =
Required area =
Question 32 1.2 / -0
In the figure given below, AB is a diameter of the circle, TD is a tangent to the circle and AB = 2AD. If
AHD = 36°
and
DBA = 30°, then what is the measure of
CDT?
Solution
BDA = 90°,
AHD = 36°, AB = 2AD
DBA = 30°
So,
DCA = 30°
(because angles by the same segment at the circumference are equal)
Now, we know that triangle ABD is a right triangle.
So,
DAB = 60° and
DBA = 30°
TDA =
DBA (Angles in alternate segment)
DCA = 30°
CDH = 6°
(Exterior angle = Sum of interior opposite triangles for triangle DCH)
CDT = 6 + 90 + 30 = 126°
Hence, 126° is the correct answer.
Question 33 1.2 / -0
x = 1712 + 1012 and y = 166 + 96 . If x - y is divided by 13, then the remainder is R1 and if it is divided by 14, then the remainder is R2 . What is (R1 , R2 )?
Solution
x - y = 17
12 + 10
12 - 16
6 - 9
6 x - y = 17
12 + 10
12 - 4
12 - 3
12 x - y = (17
12 - 4
12 ) + (10
12 - 3
12 )
17
12 - 4
12 is divided by 17 - 4 = 13 and 10
12 - 3
12 is divided by 10 + 3 = 13
So, R
1 = 0
x - y can also be written as
x - y = (17
12 - 3
12 ) + (10
12 - 4
12 )
Now, both (17
12 - 3
12 ) and (10
12 - 4
12 ) are divided by 14.
So, R
= 0
Question 34 1.2 / -0
If the LCM of two whole numbers is twice their HCF, then what is the ratio of the larger number to the smaller number?
Solution
Suppose, k is the HCF of the two numbers. Then, the numbers are ak and bk.
The LCM of these two numbers is 2k.
We know: Product of the numbers = HCF
× LCM
ak × bk = HCF × LCM
ak × bk = k × (2k)
ab = 2
This is possible only if either a = 1, b = 2 or a = 2, b = 1.
So, the required ratio of larger to smaller number =
= 2 : 1
Question 35 1.2 / -0
On 2nd July 1985, it was Wednesday. The day of the week on 2nd July, 1984 was
Solution
There are 365 days between 2nd July, 1984 and 2nd July, 1985.nd July, 1985 was a Wednesday, thennd July, 1984 was the day that comes before Wednesday which means Tuesday.
Question 36 1.2 / -0
Rina participated in a quiz which consisted of 30 questions, out of which some were multiple choice questions and the rest were subjective questions. Each multiple choice question carried 2 marks and each subjective question carried 5 marks. The quiz was of 75 marks. Find out the number of multiple choice questions asked in the quiz.
Solution
Let the number of MCQs be m.
Question 37 1.2 / -0
For which of the following values of (p), does the given quadratic equation 2x2 + 3x + p = 0 have real roots?
Solution
General quadratic equation is ax
2 + bx + c = 0.
Given quadratic equation is
2x
2 + 3x + p = 0
So, a = 2 , b = 3 , p = c
Given, Discriminant ≥ 0
So, b
2 – 4ac ≥ 0
(3)
2 - 4 (2) (p) ≥ 0
9 – 8p ≥ 0
9 ≥ 8p
≥ p
Question 38 1.2 / -0
On a busy day, 2000 sweets were sold by a confectionery. Total amount collected on that day was Rs. 9600. If the cost of mint flavoured sweets is Rs. 8/sweet, the cost of chocolate sweets is Rs 4/sweet, then how many mint sweets and chocolates sweets were sold that day?
Solution
Number of sweets sold = 2000
Cost of mint sweet = Rs. 8, and cost of chocolate sweet = Rs. 4.
Let x be number of mint sweets sold, and y be the number of chocolate sweets sold.
x + y = 2000 (i)
and, 8x + 4y = 9600 (ii)
Multiplying the equation (i) by 4 and subtracting from eq. (ii)
4x = 9600 - 8000
4x = 1600 , Hence x = 400
Putting the value in equation (i),
400 + y = 2000
y = 2000 - 400 = 1600
Hence, the number of mint sweets sold = 400, and number of chocolate sweets sold = 1600
Question 39 1.2 / -0
Liquor is being poured into a vessel with two circular ends of radii 14 cm and 7 cm. The height of the bucket being filled is 60 cm. What is the maximum limit of alcohol which the vessel can hold?
Solution
Height = 60 cm
Radii = 14 cm and 7 cm
Capacity of bucket =
[R2 + r2 ]
=
= 22 x 20 x 48
= 21,120 cm3
Question 40 1.2 / -0
The angles of elevation of a plane flying at a constant altitude of 10,000 ft are found to be 60o and 30o at an interval of 1 minute. What is the speed of the plane?
Solution
In the figure,
BE = CD = 10,000 ft
BAE = 60o and
CAD = 30
o In triangle ABE,
AB =
In triangle ACD,
Distance travelled in one minute = ED = BC = AC - AB =
Hence, speed =
ft /sec
Question 41 1.2 / -0
A toy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The base radius of the cone is 3.5 m and its volume is (2/3)rd of the volume of the hemisphere. Calculate its height.
Solution
Radius of the hemisphere = Base radius of the cone = 3.5 m
Volume of the hemisphere =
=
m
3 Volume of the cone =
m
3 Let the height of the cone be h metres.
Volume =
R
2 h =
m
3 h = 4.67 m
Question 42 1.2 / -0
An acrobat climbs up a rope stretched from a point 120 metres above the ground to a point on the ground. The angle made by the rope with the ground is 60°. Find the length of the rope.
Solution
Here, AB = 120 m, ∠ACB = 60° and let AC =
be the length of the rope.
In Δ ABC,
sin 60° =
⇒
=
× 2 m =
×
m = 80
m
Question 43 1.2 / -0
Two cyclists met each other at 10 a.m. on Ferozepur road. After their meeting, one of them proceeded in the north direction and the other proceeded in the east direction. Exactly at noon, they were 60 km apart. If the difference between their speeds was 6 km/hr, then find the speed of the faster cyclist.
Solution
Let the speed of the slower cylclist be x km/hr.
So, the speed of the faster one will be (x + 6) km/hr.
The distance travelled in 2 hr is 2x and 2(x + 6) km.
From Pythagoras theorem:
(2x)
2 + [2(x + 6)]
2 = (60)
2 4x
2 + 4(x
2 + 36 + 12x) = 3600
8x
2 + 48x + 144 = 3600
x
2 + 6x - 432 = 0
(x - 18)(x + 24) = 0
x = 18 or -24 (neglected)
So, the speed of the faster cyclist is 24 km/hr.
Question 44 1.2 / -0
Three plots having areas 132, 204 and 228 square metres are to be sub-divided into equal sized flower beds. If the breadth of a bed is 3 metres, find the maximum length that each bed can have.
Solution
HCF of 132, 204 and 228 = 2 × 2 × 3 = 12
Question 45 1.2 / -0
A pencil is cylindrical in shape with a length of 20 cm and a cross-sectional radius of 1 cm. It is sharpened using a good sharpener on both the sides such that the shape on either end of the cylindrical pencil is that of a right circular cone with a height of 3 cm. If there is no change in the net length of the pencil due to sharpening, what is the volume of the material that was removed from the pencil to sharpen the edges?
Solution
Initial volume of pencil = 12 π 20 = 20π 3 2 × 3 3
Question 46 1.2 / -0
Two farmers Ravi and Ram have 40 goats in total. They sell their goats at different prices, but each of them receives the same amount of money. If Ravi had sold his goats at Ram's price, he would have received Rs. 5,400. If Ram had sold his goats at Ravi's price, he would have received Rs. 2,400. How many goats does Ravi have in total?
Solution
Assume Ravi and Ram have n and 40 – n goats, respectively.
Question 47 1.2 / -0
How many real solutions of the equation
= 1 exist?
Solution
= 1
Or
Squaring both sides, we get
(y + 3) = 1 + y - 2
2
= -2
1
, which is not possible as a square root is always positive.
Hence, there are no solutions which are real numbers.
Thus, answer option 1 is correct.
Question 48 1.2 / -0
In the figure, PQ is parallel to BC. If the ratio between the areas of the triangles APQ and ABC is 4 : 9, and ABCD is a parallelogram, then find the ratio between the areas of parallelograms APRD and PRCB.
Solution
Draw AF
BC intersecting PR at E.
Since
~
~
Now,
~
Therefore,
Or
… (2)
From (1) and (2),
Now,
Question 49 1.2 / -0
A cylinder, having a diameter of 27 units and a height of 30 units, contains two lead spheres of radii 6 units and 9 units, with the larger sphere sitting at the bottom of the cylinder, as shown. Water is poured into the cylinder so that it just covers both spheres. Find the volume of water required (in cubic units).
Solution
We examine a vertical cross-section of the cylinder and the spheres such that it contains the vertical axis of the cylinder and the centres of the spheres. (There is such a cross-section since the spheres will be pulled into this position by gravity and because of rotational symmetry.)
Let the centres of the spheres be O
1 and O
2 , as shown.
Join the centres of the spheres to each other and to the respective points of tangency of the spheres to the walls of the cylinder.
Then, O
1 O
2 = (6 + 9) units = 15 units (O
1 O
2 passes through the point of tangency of the two spheres), O
1 P = 6 units and O
2 Q = 9 units.
Next, draw
O
1 RO
2, so that RO
1 is perpendicular to O
1 P and
O
1 RO
2 = 90
o .
Then, looking at the width of the cylinder, since PO
1 = 6 units and O
2 Q = 9 units, we have
PO
1 + RO
2 + O
2 Q = 27 units, RO
2 = 12 units
By Pythagoras theorem in
O
1 RO
2 , we see that O
1 R = 9 units.
Then, the depth of the water will be:
Radius of lower sphere + RO
1 + Radius of upper sphere = (9 + 9 + 6) units = 24 units
So, Volume of water = (Volume of cylinder to height of 24 units) - (Volume of spheres)
=
units
3 = (4374 π - 288 π - 972 π ) units
3 = 3114π units
3 Therefore, the volume of water required is 3114 π cubic units.
Question 50 1.2 / -0
If t
n = 1 +
, n = 1, 2, 3, ____, then find the value of t
1 t
2 t
3 ____
t
25 .
Solution
t
1 = 1 +
= 2
t
2 = 1 +
=
t
3 = 1 +
=
t
1 x t
2 = 2
= 3
t
1 x t
2 x t
3 = 3
= 4
Similarly, t
1 x t
2 …….. t
n = n + 1
Thus, t
1 t
2 t
3 ……. x t
25 = 25 + 1 = 26.
×
Sign in
Profile
Signout
Get latest Exam Updates 
