Question 1 1.2 / -0
Select the correct mirror image from the options for the given combination of letters and numbers, if the mirror is placed along MN.
Question 2 1.2 / -0
If '-' denotes '+', '×' denotes '
', '+' denotes '×', '=' denotes '<' and '
' denotes '-', then which of the following options is correct?
Solution
After interchanging the signs, we get
(1) 6 × 4 - 20 + 12
6 < 2
24 - 20 + 2 < 2
6 < 2 (Incorrect)
(2) 3 - 3
3 + 4 × 5 < 20
3 - 1 + 20 < 20
22 < 20 (Incorrect)
(3) 12
6 × 4 - 3 + 6 < 10
2 × 4 - 3 + 6 < 10
8 - 3 + 6 < 10
11 < 10 (Incorrect)
(4) 4 + 20
5 × 4 - 4 < 20
4 + 4 × 4 - 4 < 20
4 + 16 - 4 < 20
16 < 20 (Correct)
Question 3 1.2 / -0
Select a figure from the options which will continue the same series as established by the Problem Figures.
Solution
Black and white arrows are added to the figure alternately. Also, only previous arrow rotates 90° clockwise.
Question 4 1.2 / -0
Pointing to a photograph, Monika said, ''He is the brother of the son of my grandfather's only child.'' How is the man related to Monika?
Solution
Monika's grandfather's only child is Monika's father and the son of her father is her brother.
Question 5 1.2 / -0
Which of the following Venn diagrams best illustrates the relationship among Sun, Moon and Stars?
Question 6 1.2 / -0
Directions:
Solution
Eight friends A, B, C, D, E, F, G and H are sitting around a circle facing the centre. E is third to the left of G, who is to the immediate right of B, who is third to the left of A.
H is second to the right of F, who is not an immediate neighbour of E.
D is not an immediate neighbour of B.
As H is sitting to the immediate left of D, option (4) is correct.
Question 7 1.2 / -0
lf 'a - b' = a × b, 'a + b' = a
b and 'a * b' = b - a, then the value of (10 + 2) × (5 - 2) × (2 * 5) is _____.
Solution
According to the given conditions, the given expression becomes:
Question 8 1.2 / -0
There is a certain relationship between figures (i) and (ii). Establish a similar relationship between figures (iii) and (iv) by selecting a suitable figure from the options which will replace '?' in figure (iii).
Solution
From figure (i) to figure (ii), shaded parts of the outer elements become unshaded and vice versa. Also, the number of sides of the inner element gets increased by 1.
Question 9 1.2 / -0
A man walked 25 m towards east. Then, he turned to his right and walked 20 m, and again he turned to his right and walked 25 m. He then turned to his left and walked 10 m. How far and in which direction is he from the starting point?
Solution
Here, PQ = 20 + 10 = 30 m
So, the man is 30 m far and in the south direction from the starting point.
Question 10 1.2 / -0
How many squares are there in the given figure?
Solution
The squares formed are:
E, H, I, L, Q, N, O, T, U, NOUT, EF, GH, IJ, KL, MN, OP, QR, ST, UV, EFGHNMKL, HIGJMPON, MPNOTUSV, KMNLQRST, ABCD, i.e. 24 in number
Question 11 1.2 / -0
Select a figure from the options which will replace '?' to complete the pattern in the given figure.
Question 12 1.2 / -0
Three positions of a dice are shown below. When 5 is at the top, then which number will be at the bottom?
Solution
Taking the first and second positions, number 5 is common to both of them.
Taking the numbers in clockwise direction,
4 is opposite to 6 and 2 is opposite to 3.
So, 5 is opposite to 1.
Numbers on the opposite faces are: (1, 5), (2, 3) and (4, 6).
Question 13 1.2 / -0
The following question consists of a set of three figures P, Q and R showing a sequence of folding of a piece of paper. Figure R shows the manner in which the folded paper has been cut. Select the figure from the options which would most closely resemble the unfolded form of figure R.
Question 14 1.2 / -0
Select a figure from the options which satisfies the same condition of placement of dots as in Fig. (X).
Solution
One dot is inside rectangle and triangle, one dot is inside circle and triangle only ,and one dot is inside triangle, square and circle only.
Only option 3 satisfies the above condition of placement of dots.
Question 15 1.2 / -0
Select the correct water image of the given figure.
Question 16 1.2 / -0
The square root of 0.64 - 2 × 0.48 + 0.36 is _______.
Solution
We have
0.64 - 2 × 0.48 + 0.36 = (0.8)
2 - 2 × 0.8 × 0.6 + (0.6)
2 = (0.8 - 0.6)
2 = (0.2)
2 So,
= 0.2
Question 17 1.2 / -0
In the given figure (not drawn to scale), ABCD is a rhombus. Find the values of (i) DO + AC, and (ii)
DAB.
Solution
In
BOC,
BC
2 = BO
2 + OC
2 (By Pythagoras theorem)
OC =
=
= 4cm
Since ABCD is a rhombus;
OC = OA and OD = OB
(i) DO + AC = (3 + 8) cm = 11 cm
(ii) Since AB || DC and AD is a transversal;
BAD =
ADE = 60° (Alternate Interior angles)
Question 18 1.2 / -0
Given below are the steps of construction to construct a quadrilateral KLMN, where KL = 6 cm, LM = 8 cm, MN = 9.5 cm,
L = 105° and
LM = 75°. Which of these steps is incorrect?
Step-1: Draw LM = 8 cm and construct
MLX = 105°.
Step-2: With L as centre and radius 6 cm, draw an arc which cuts LX at K.
Step-3: Construct
LMY = 75°. With M as centre and radius 9.5 cm, draw an arc which cuts MY at N.
Step-4: Now join N and X to obtain the required quadrilateral KLMN.
Solution
To construct a quadrilateral KLMN, where KL = 6 cm, LM = 8 cm, MN = 9.5 cm,
L = 105° and
LM = 75°:
Step-1: Draw LM = 8 cm and construct
MLX = 105°.
Step-2: With L as centre and radius 6 cm, draw an arc which cuts LX at K.
Step-3: Construct
LMY = 75°. With M as centre and radius 9.5 cm, draw an arc which cuts MY at N.
Step-4: Now join N and X to obtain the required quadrilateral KLMN. It is incorrect as we need to join N and K to obtain the required quadrilateral KLMN.
Question 19 1.2 / -0
Given that the number 67y19 is divisible by 9, where y is a single digit, what is the least possible value of y?
Solution
67y19 is divisible by 9, so sum of its digits is also divisible by 9.
6 + 7 + y + 1 + 9 = 23 + y is divisible by 9.
So, the least possible value of y is 4.
Question 20 1.2 / -0
Simple interest on a sum of money for 1 year at 12% per annum is Rs. 1200. What will be the compound interest when compounded half-yearly on that sum at the same rate for the same period?
Solution
Let the principal be P.
S.I. = Rs. 1200, T = 1 year, R = 12% p.a.
∴
= Rs. 10,000
Now, amount =
= 10,000
= 10,000
= Rs. 11,236
C.I. = A - P = 11,236 - 10,000 = Rs. 1236
Question 21 1.2 / -0
Find the ratio of the area of the shaded portion to that of the unshaded portion in the given figure, if ABCD is a rectangle (not drawn to scale).
Solution
Area of rectangle ABCD = (20 × 30) cm
2 = 600 cm
2 Area of the shaded portion = 4 ×
× Area of the circle
=
cm
2 Area of the unshaded portion =
cm
2 =
cm
2 Required ratio =
= 176 : 349
Question 22 1.2 / -0
If a and b vary inversely with each other, find the value of
.
Solution
Since a and b varies inversely,
6q = 90 × 2
q =
= 30
Also, 5p = 90 × 2
p =
= 35
∴
= 33
Question 23 1.2 / -0
The cube root of a number when divided by the smallest prime number gives the square of the smallest prime number. Find the number.
Solution
Let x be the required number.
According to the question,
x
1/3 = 8
On cubing both sides, we get
x = 8
3 = 512
Question 24 1.2 / -0
Study the given graph and choose the letters that represent the points (5, 0) and (3, 2), respectively.
Solution
In the graph, E represents the point (5, 0) and D represents the point (3, 2).
Question 25 1.2 / -0
The number of faces in a square pyramid is _________.
Solution
A square pyramid has 5 faces (4 lateral and 1 bottom).
Question 26 1.2 / -0
The order of rotational symmetry of the given figure is _____.
Solution
Thus, order of rotational symmetry =
= 1
Question 27 1.2 / -0
Subtract the sum of
and
from the sum of
and
.
Solution
Sum of
Sum of
So, required value =
Question 28 1.2 / -0
If
× 8
x - 2 = (0.25)
x , find the value of
.
Solution
We have 8
x- 2 ×
= (0.25)
x 2
3x - 6 - 16 + 12x = 2
-2x 2
15x - 22 = 2
-2x 15x - 22 = -2x
17x = 22
x =
= 1 + 1 = 2
Question 29 1.2 / -0
The tens digit of a two-digit number exceeds its units digit by 6. The number itself is 10 times the sum of its digits. Find the number.
Solution
Let the units digit of the two-digit number be x. So, its tens digit = x + 6
According to the question,
10(x + 6) + x = 10 (x + (x + 6))
10x + 60 + x = 10(2x + 6)
11x + 60 = 20x + 60
9x = 0
x = 0
Units digit = 0 and tens digit = 0 + 6 = 6
Hence, the number is 60.
Question 30 1.2 / -0
Directions For Questions
Directions:
...view full instructions
The average number of students who qualified the examination from schools C and D is what percent of the average number of students who appeared for the examination from the same schools? (Rounded off to 2 digits after decimal)
Solution
Number of students who qualified the examination from school C = 2250
Number of students who qualified the examination from school D = 2000
Average =
= 2125
Number of students who appeared for the examination from school C = 3250
Number of students who appeared for the examination from school D = 2500
Average =
= 2875
Required percentage =
= 73.91%
Question 31 1.2 / -0
Directions For Questions
Directions:
...view full instructions
What is the ratio of the number of students who qualified the scholarship examination from school A and the number of students who qualified the examination from school B?
Solution
Number of students who qualified the examination from school A = 1750
Number of students who qualified the examination from school B = 1250
Required ratio =
, i.e. 7 : 5
Question 32 1.2 / -0
If
= 2, then the value of
is
Question 33 1.2 / -0
Select the incorrect match of the given solids with the product of their number of faces and number of vertices.
Solution
i) For rectangular pyramid: 5 (Vertices) x 5 (Faces) = 25
ii) For triangular prism: 6 (Vertices) x 5 (Faces) = 30
iii) For octahedron: 6 (Vertices) x 8 (Faces) = 48
iv) Number of faces of a triangular pyramid = 4
Number of vertices of a triangular pyramid = 4
Required product = 4 × 4 = 16
Question 34 1.2 / -0
The radii of the bases as well as the heights of a cylinder and a cone are equal to r cm and the radius of a hemisphere is also equal to the height of the cone. Find the ratio of the volumes of the cone, cylinder and hemisphere, respectively.
Solution
As radius of the cylinder = radius of the cone = height of the cylinder = height of the cone = r cm, and radius of the hemisphere = r cm;
Volume of the cone : Volume of the cylinder: Volume of the hemisphere
=
= 1 : 3 : 2
Question 35 1.2 / -0
Divide (64x2 + 48xy + 9y2 )(x + 2) by (8x2 + 16x + 3xy + 6y).
Question 36 1.2 / -0
Atul has a field in the form of a rectangle. The area of the field is
m
2 . Find the breadth of the field, if its length is
m.
Solution
Area of the rectangle =
m
2 Length of the rectangle =
m
Now, area of a rectangle = length × breadth
Breadth of the rectangle =
=
Question 37 1.2 / -0
A theatre in a town is built in the form of a kite. Its perimeter is 100 m. If one of its sides is 20 m, then what are the lengths of other sides?
Solution
Let ABCD be the theatre in which AB = AD = 20 m and let BC = CD = x m.
Perimeter of the theatre = 100 m
100 = AB + BC + CD + DA
100 = (20 + x + x + 20) m
40 + 2x = 100
2x = 60
x = 30
So, AB = AD = 20 m and BC = CD = 30 m
Question 38 1.2 / -0
The weight of Moon is (7.346 × 1022 ) kg and the weight of Earth is (5.9724 × 1024 ) kg. What is the total weight of both in standard form?
Solution
Weight of Moon = (7.346 ×10
22 ) kg
Weight of Earth = (5.9724 × 10
24 ) kg
Total weight = 7.346 × 10
22 + 5.9724 × 10
24 = 0.07346 × 10
24 + 5.9724 × 10
24 = (0.07346 + 5.9724) × 10
24 = (6.04586 × 10
24 ) kg
Hence, the standard form of the total weight is (6.04 × 10
24 ) kg.
Question 39 1.2 / -0
A box contains 9 green jellies, 11 blue jellies, 13 pink jellies and 14 orange jellies. Akshit takes out a jelly from the jar at random. What is the probability that the chosen jelly is of pink colour?
Solution
Total number of jellies in the box = 9 + 11 + 13 + 14 = 47
Probability of getting a pink coloured jelly =
Question 40 1.2 / -0
Two pipes A and B can till a tank in 36 hours and 45 hours, respectively. If both the pipes are opened simultaneously, how much time will they take to fill the tank?
Solution
Work done by pipe A =
Work done by pipe B =
Total work done by both the pipes =
Required time = 20 hours
Question 41 1.2 / -0
A man bought a house for Rs. 5 lakhs and rents it. He puts 12
% of each month's rent aside for repairs, pays Rs. 1,660 as annual taxes and gets 10% of the investment made for buying the house annually. What is the monthly rent of the house?
Solution
Let the rent received annually = 12x
Amount for repairs =
Total annual tax = Rs. 1,660
According to question,
12x -
= 10% of Rs. 5,00,000
= 4,920
The monthly rent of the house is Rs. 4,920.
Question 42 1.2 / -0
How many bricks of size 22 cm × 10 cm × 7 cm are required to construct a wall 11 m long, 3.5 m high and 40 cm thick, if the cement and sand used in the construction occupy (1/10)th part of the wall?
Solution
Volume of a brick = (22 × 10 × 7) cm
3 = 1,540 cm
3 Now, length of the wall = 11 m = 1,100 cm
Height of the wall = 3.5 m = 350 cm
Width of the wall = 40 cm
Volume of the wall = (1,100 × 350 × 40) cm
3 = 1,54,00,000 cm
3 It is given that cement and sand occupy
th part of 1,54,00,000 i.e. 15,40,000 cm
3 .
Volume of the wall occupied by the bricks
= 1,54,00,000 - 15,40,000 = 1,38,60,000
Number of bricks required
=
=
= 9,000
Question 43 1.2 / -0
Principal of a college arranges his 8,341 students in the form of a square. He found that he still had 60 students left. Find the number of students in the middle row.
Solution
Let x be the number of students in each row. To form a perfect square, number of students in each row must be equal to the number of rows.
According to the question, number of students who form square
= 8,341 - 60 = 8281
Now, x × x = 8281
x
2 = 8281
x =
= 91
Number of students in the middle row = 91
Question 44 1.2 / -0
Mini buys
kg ghee at the rate of Rs. 166 per kg,
kg butter at the rate of Rs. 90 per kg and
kg peas at the rate of Rs. 75 per kg. If she gives a Rs. 1,000 note to the shopkeeper, then how much change should she get back?
Solution
Cost of
kg ghee = Rs.
= Rs. 323.70
Cost of
kg butter = Rs.
= Rs. 54
Cost of
kg peas = Rs.
= Rs. 135
Total amount = Rs. (323.70 + 54 + 135)
= Rs. 512.70
Amount of money Mini gave to the shopkeeper = Rs. 1,000
Amount of money she received back
= Rs. (1,000 - 512.70)
= Rs. 487.30
Question 45 1.2 / -0
A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk to water becomes 5 : 2. Find the quantity of milk in the original mixture.
Solution
Let the quantity of milk in the mixture be 5x litres and the quantity of water in the mixture be x litres.
Now, according to the question
10x = 5x + 25
5x = 25
x = 5
Hence, the quantity of milk in the original mixture = 5x = 5 × 5
= 25 litres
Question 46 1.2 / -0
Directions:
Solution
(i) True: Sum of interior angles of a hexagon = (6 - 2) × 180° = 720°False: Diagonals of a square are equal and bisect at 90°.False: A parallelogram in which all sides are equal is a rhombus.True: A trapezium is a quadrilateral with exactly one pair of parallel sides and other two are non-parallel sides.
Question 47 1.2 / -0
Which of the following is correct with respect to the given solids?
Solution
Matching with the given options, we get option (2).
Question 48 1.2 / -0
Directions: Statement-1: The amount of extension in an elastic spring varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then the weight that would produce an extension of 17.4 cm is 900 gm.Statement-2: x and y vary directly means that if x increases, then y decreases.
Solution
Statement-1 : Let x gm of weight produce an extension of 17.4 cm.
Weight 150 x Extension 2.9 17.4
It is the case of direct variation.
∴
= 900 gm
So, Statement-1 is true.
Statement-2: x and y vary directly means that if x increases, then y increases.
So, Statement-2 is false.
Question 49 1.2 / -0
Directions: Study the given histogram representing the weights of students of class VIII and answer the following question.
(i) How many students weigh at least 30 kg, but less than 45 kg?
(ii) Which of the following class intervals has the highest frequency?
Solution
(i) Number of students that weigh at least 30 kg, but less than 45 kg
Question 50 1.2 / -0
Directions: P times.-16 is Q .R S
Solution
(i) Let edge of the cube be a cm.
Surface area = 6a
2 Edge of new cube = 2a cm
New surface area = 6(2a)
2 = 24a
2 = 4(6a
2 )
i.e. Four times the initial surface area.
(ii) 2
-16 =
and
× 2
16 = 1
i.e. 2
16 is the multiplicative inverse of 2
-16 .
(iii) Since 7
3 = 343 i.e. units digit of 7
3 is 3.
Now, units digit of 1,787 is 7.
So, units digit of 1,787
3 is also 3.
(iv) Let fourth angle be x.
x + 70° + 70° + 70° = 360°
x = 360° - 210°
= 150°
×
Sign in
Profile
Signout
Get latest Exam Updates 
