Question 1 1.2 / -0
In the given question, 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 of 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 rows?
Solution
Row 1;2 = 186; apply rule (iv)2 = 24; apply rule (iv)
Question 2 1.2 / -0
A child went 90 m in the East to look for his father and then he turned right and went 20 m. Then he turned right and after going 30 m, he reached his uncle's house. His father was not there. From there, he went 100 m to his North and met his father in the shop. How far did he meet his father from the starting point?
Solution
As per the given condition, the following is the sequence of the child's movements;
AC = 90 m, CD = 20 m, DE = 30 m and EF = 100 m
In triangle ABF, AF
2 = AB
2 + BF
2 Therefore, AF = 100 m
Question 3 1.2 / -0
Priya starts her trip from a fixed point. She moves towards East and walks 25 metres. She then turns to her left and walks 30 metres. Next she moves 25 metres to her right. She then turns to her right again and walks 55 metres. Finally she turns to her right and moves 75 meters. In which direction is she now from her starting point?
Solution
She is in the South-west direction from her starting point.
Question 4 1.2 / -0
Choose from the options, the correct water image of the given word.CARING
Solution
Hence, option 4 is the correct answer.
Question 5 1.2 / -0
Choose from the given options which satisfies the same conditions of placement of the dots as in Fig. (X).
Solution
Observe carefully the common points among the option figures. Only option 3 suits all the given placement of dots and hence, the correct option.
Question 6 1.2 / -0
Select a figure from the options which will continue the series as established by the Problem Figures.
Solution
The pattern followed is:
The first step is to rotate the image 90° anticlockwise and then rotating the image vertically and to get the next image again 90° and rotate the image horizontally. The next image will appear by rotating 90° and then rotating the image vertically again. For the final image, we have to rotate the image 90° anticlockwise and then rotate horizontally.
Question 7 1.2 / -0
A, B, C, D, E, F, G, H and K are sitting around a circle facing the centre. F is fourth to the right of A, who is third to the right of B, K is fourth to the left of B and third to the right of D, C is third to the right of H. E is second to the left of G. What is E's position with respect to B?
Solution
E is third to the left of B.
Question 8 1.2 / -0
Select a figure from the options which when placed in the blank space of Fig. (X) would complete the pattern.
Solution
There is a reflection of the box 2 of first row in box 1 of second row.
Hence there would be a reflection of box 2 of second row in box 1 of first row.
Question 9 1.2 / -0
If the code of SENSITIVE is QHLVGWGYC, then what will be the code of MICROSOFT?
Question 10 1.2 / -0
Find the missing number, if a certain rule is followed row-wise or column-wise.
7 4 5 8 7 6 3 3 ? 29 19 31
Solution
7 × 3 + 8 = 29
Question 11 1.2 / -0
Which of the following Venn diagrams best represents the relationship among State, Country and Village?
Solution
Village is the part of state and state is the part of country.
Question 12 1.2 / -0
If it is possible to make a meaningful word with the second, fourth, eight and tenth letters of the word CONFIDENCE, which of the following will be the second letter of that word? If more than one such word can be made, give 'S' as the answer. If no such word can be made, give 'P' as the answer.
Solution
The word is 'CONFIDENCE'. If second, fourth, eighth and tenth letter is taken out, we will be left with O, F, N, E. No meaningful word can be formed by using these letters. So, no such word can be made and the answer is 'P'.
Question 13 1.2 / -0
Three different positions of a dice are shown below. Which of the following colours will be opposite to the face having red colour?
Solution
Red is taken as base and violet is taken as top. From the given figure pink, blue and brown are adjacent to red colour therefore, violet is opposite to red colour.
Question 14 1.2 / -0
A goldsmith has five gold rings, each having a different weight;
Solution
Statement 1: Ring D is weighing twice as much as ring E.
D = 2E
Statement 2: Ring E is weighing four times as much as ring F.
E = 4F
Statement 3: Ring F is weighing half as much as ring G.
F =
G
Statement 4: Ring G is weighing half as much as ring H.
G =
H
Statement 5: Ring H is weighing less than ring D but more than ring F.
H < D
H > F
From all the statements, D > H = E > G > F.
Thus the data is insufficient to arrange D, E, F, G, H in descending order.
Question 15 1.2 / -0
A sheet of paper has been folded (either once or twice) and then it has been cut. You have to select a figure from the options that would most closely resemble the unfolded form of Fig. (X).
Solution
Hence, this is the correct option.
Question 16 1.2 / -0
The term containing the highest power of x in the polynomial f(x) is 2x4 . Two of the roots of the equation f(x) = 0 are -1 and 2. Given that x2 - 3x + 1 is a quadratic factor of f(x), find the remainder when f(x) is divided by 2x - 1.
Solution
Given:2 - x - 2.2 - 3x + 1 is a quadratic factor and 2x4 is a highest power of f(x).2 - x - 2) (x2 - 3x + 1) = 2x4 - 8x3 + 4x2 + 10x - 4.4 - 8(1/2)3 + 4(1/2)2 + 10(1/2) - 4 = 9/8
Question 17 1.2 / -0
PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. Find the ratio ar(||gm XQRY) : ar(△QSR)
Solution
Area (||gm XQRY) = (1/2) × area of parallelogram PQRS
Question 18 1.2 / -0
Select the correct option.
Solution
In first case: Value of xy = 10 × 5 = 50 and 5 × 2.5 = 12.5
Question 19 1.2 / -0
The numbers 7.478478... and 1.101001000100001... are:
Solution
7.478478... = 7.478, as the decimal expansion is non terminating and recurring, so it is a rational number.
Question 20 1.2 / -0
If x =
, find the value of x
3 - 2x
2 - 7x + 5.
Solution
x =
x
3 - 2x
2 - 7x + 5 = x
2 (x - 2) - 7x + 5
=
=
=
= 3
Question 21 1.2 / -0
The base of an isosceles triangle is 4 cm and its area is 16 cm2 . If one of the two equal sides of the triangle is k cm, then the approximate value of k - 1.24 is _________.
Solution
Considering here, k = a
Area = 1/2 (b) (h) where h
2 = (a
2 - b
2 /4) = (a
2 - 16/4)
So, 16 = 1/2 × 4 × h Or a = 8.24
So, a - 1.24 = 7 cm
Question 22 1.2 / -0
The given figure is not drawn to scale. Find the values of
QPS and
TRQ respectively.
Solution
PQ and TR are parallel lines and PR is a transversal.
Thus,
P =
T (Corresponding angles are equal)
RTS =
QPR = 65°
QPS = 65° + 23° = 88°
TSR = 180° - 135° = 45°
TRS = 180°
- (45°
+ 65°) = 180° - 110° = 70°
Thus,
TRQ = 180° - 70° = 110° (Linear pair)
Question 23 1.2 / -0
Study the given graph and answer the following question.
Calculate the area enclosed by the lines I, x = -3, y = -2 and y = -x + 2.
Solution
The area of the rectangle formed by I, x = -3, y = -2 and y-axis = 3 × 4 = 12 sq. units
The area of right triangle formed by y-axis, y = -2 and y = -x + 2 = 1/2 × 4 × 4 = 8 sq. units
So, the total area = 12 + 8 = 20 sq. units
Question 24 1.2 / -0
Find the ratio of the shaded area to the area of the quadrilateral ABCD.
Solution
In triangle ABC,
S = (24 + 28 + 20)/2 = 36
Area of triangle ABC =
=
Area of triangle ACG =
D is the mid point of AG, so CD is the median of triangle ACG.
∴ Area of triangle ACD = (1/2) of area of ACG = 96 × 2 m
2 Required ratio =
Question 25 1.2 / -0
Three statements are given below:
Solution
In a parallelogram, the angle bisectors of two adjacent angles intersect at right angles. The triangle formed by joining the midpoints of an isosceles triangle is also an isosceles triangle. A rectangle is a parallelogram whose opposite sides are equal and form a right angle. Hence, the option (3) is well defined.
Question 26 1.2 / -0
Solve for x:
Solution
2x-3 /8-x = 32/4(1/2)x 2x × 8x = 28 or 22x × 23x = 28
Question 27 1.2 / -0
Simplify:
Question 28 1.2 / -0
The weight, in kg, of 50 students is given below:
40 45 55 62 50 51 56 69 61 36 60 56 69 38 35 63 57 50 57 48 40 63 53 64 47 42 56 51 42 60 55 39 64 57 64 44 66 35 59 59 73 62 49 63 37 63 54 72 44 60
Find the mean, median and mode respectively for the given data.
Solution
Sum of numbers = 2695
Question 29 1.2 / -0
The figure below is made up of a square ABCD and two rhombuses ATCP and DRBV.
Given that
BVD = 135° and AT = BR, find
PCT and
ABD respectively.
Solution
Since ABCD is a square, each angle will be 90°.
And BD is the diagonal, so ∠ABD = 90°/2 = 45°
Given that: ATCP and DRBV are two rhombuses and AT = BR, so sides of both rhombuses are equal.
Given that: ∠BVD = 135°.
We know that, in rhombus all sides have equal length.
Opposite sides are parallel, and opposite angles are equal, so
∠ATC = ∠APC = 135°
In rhombus APCT,
2∠PCT + 2 × 135° = 360°
2∠PCT = 90°
∠PCT = 45°
Hence, option (4) is correct.
Question 30 1.2 / -0
Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm at the speed of 15 km per hour. At what time, will the water be 3 metres deep?
Solution
Volume of tank = 150 × 100 × 3 = 45000 cubic metre
Question 31 1.2 / -0
Directions: Study the graph carefully to answer the question that follows.
Number of Players Participating in three different games in five different schools
25% of the numbers of the players participating in hockey from School-5 are females. What is the number of the hockey players who are males in School-5?
Solution
The numbers of the players participating in hockey from School-5 = 36
Question 32 1.2 / -0
Sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find its area.
Solution
Sides of triangle = 13x, 14x, 15x
Perimeter = 84 cm
13x + 14x + 15x = 84
x = 2
Side a = 13x = 26 cm
Side b = 14x = 28 cm
Side c = 15x = 30 cm
2s = a + b + c = 84
s = 42
Area of triangle =
Question 33 1.2 / -0
Study the figure shown here (not drawn to scale), If ABG is a straight line, then find
ABH and reflex
ABC respectively.
Solution
IA II HB
Question 34 1.2 / -0
The value of
up to 35 decimal places is given below:
3.14159265358979323846264338327950288
Find the probability of occurring 8 in it.
Solution
Probability of occurring 8 = (Number of 8)/(Total number of digits) = 5/36
Question 35 1.2 / -0
In the figure shown, square 2 is formed by joining the mid-points of square 1; square 3 is formed by joining the mid-points of square 2 and so on. In this way, total five squares are drawn. The side of the square 1 is 'a' cm. What is the sum of perimeters of all the five squares?
Solution
Side of the square 1 = 'a' cm
Perimeter of square 1 = 4a cm
Side of square 2 =
Perimeter of square 2 = 4 ×
Side of square 3 =
Perimeter of square 3 = 4 ×
= 2a
Side of square 4 =
Perimeter of square 4 =
Side of square 5 =
Perimeter of square 5 = a
Sum of perimeters of all the five squares =
Question 36 1.2 / -0
Two men and three women finish 25% of a work in 4 days, while six men and fourteen women can finish the whole work in 5 days. In how many days will 20 women finish it?
Solution
2 men and 3 women finish 100% of the work in 16 days.
(2M + 3W)16 = (6M + 14W)5
32M + 48W = 30M + 70W
2M = 22W
M = 11W
20W × D = (6 × 11W + 14W)5
D = 400/20 = 20 days
Question 37 1.2 / -0
The length of the longest pole that can be placed on the floor of a room is 12 m and the length of the longest pole that can be placed in the room is 15 m. The height of the room is
Solution
Let the height of the room be x.2 = (15)2 – (12)2
Question 38 1.2 / -0
The price of rice is reduced by 2% per kg. How many kilograms of rice can now be bought for the money which was sufficient to buy 49 kg of rice earlier?
Solution
Let the initial price be Rs. 100.
Question 39 1.2 / -0
The fluid contained in a bucket can fill four large bottles or seven small bottles. A full large bottle is used to fill an empty small bottle. What fraction of the fluid is left over in the large bottle when the small one is full?
Solution
Let the capacity of the bucket be x litres.
Then,
Capacity of 1 large bottle =
; Capacity of 1 small bottle =
Fluid left in large bottle =
Required fraction =
Question 40 1.2 / -0
A well with 10 m inside-diameter is dug 14 m deep. The earth taken out of it is spread all around to a width of 5 m to form an embankment. The height of the embankment is ________.
Solution
Radius of the well = 5 m
Height of the well = 14 m
Width of the embankment = 5 m
Radius of the embankment = 5 + 5 = 10 m
Let h be the height of the embankment.
So, volume of the embankment = volume of the well
(10
2 - 5
2 ) × h =
× 5
2 × 14
× 75 × h =
× 25 × 14
h =
h = 4.66 m
Question 41 1.2 / -0
In a mixture of 60 litres, the ratio milk to water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is __________.
Solution
Quantity of milk = 60 × (2/3) = 40 litres
Quantity of water = 60 - 40 = 20 litres
As per the question, we need to add (x litres) water to get quantity 1 : 2.
40/(20 + x) = 1/2
20 + x = 80
x = 60 litres
Question 42 1.2 / -0
Suresh travelled 1200 km by air which was (2/5) of his trip. He travelled one-third of the whole trip by car and the rest of the journey by train. Find the distance travelled by train. Also, find the speed of the train if the time taken for the train to travel the whole distance was 8 hr.
Solution
2/5 of total travel = 1200 km
Question 43 1.2 / -0
There are four bells. They ring after every one minute, two and a half minutes, 50 seconds and 5 seconds, respectively. If all the four bells rang together last time at 8:20 p.m., then at what time will they all next ring simultaneously?
Solution
The LCM of 60 seconds, 150 seconds, 50 seconds, 5 seconds is 300 seconds.
Question 44 1.2 / -0
Roma took a loan of Rs. 16,000 against her insurance policy at the rate of 12
% per annum. Calculate the total compound interest that will be paid by Roma after 3 years.
Solution
P = Rs. 16,000, R = 12.5%, T = 3 years
Question 45 1.2 / -0
A fruit seller has 24 kg of apples. He sells a part of these at a gain of 20% and the balance at a loss of 5%. If on the whole he earns a profit of 10%, the amount of apples sold at a loss is ______.
Solution
Let the amount of apples to sell at gain be y kg.
Question 46 1.2 / -0
Select the correct option.Statement 1: If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is not isosceles.Statement 2: If the bisector of the vertical angle of a triangle bisects the base, then the triangle is not isosceles.
Solution
Altitude from the vertex is also the perpendicular bisector of the opposite side. The triangle must be isosceles and may be an equilateral triangle. Both the statements are false.
Question 47 1.2 / -0
Which of the following statements is correct?
Solution
In triangle ABC,
∠A + ∠B + ∠C = 180°
∠ACD = ∠A + ∠B
An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Question 48 1.2 / -0
Fill in the blanks.
Solution
The number of quadrants is 4. The point of intersection is called origin, and in quadrants I and III, the signs of coordinates are same as (+, +) and (-, -), respectively.
Question 49 1.2 / -0
Solution
(i) Let 'a' be the side of equilateral triangle ABC.
That is,
AB = BC = CA = a
Area of triangle ABC =
Given: BD = CD = a/2
Area of equilateral triangle BDE =
th of the area of equilateral triangle ABC =
of the area of triangle ABD
That is, area(△BDE) = (1/4) area(△ABC) = (1/2) area(△ABD) ... (1)
Hence, (i) - (q)
(ii) Let h be the altitude of triangle ABD.
Then, altitude of triangle BDE = (h/2) (from (1))
Now,
Area(△AFC) = (1/2) × FC × h = (1/2) × (FD + DC) × h = (1/2) × (FD + BD) × h = (1/2) × (FD + BF + FD) × h ... (2)
Since altitude of △BEF = (1/2) altitude of (△AFD) and area(△BEF) = area(△AFC); BF = 2FD ... (3)
Using BF = 2FD in equation (2), we get
Area(△AFC) = (1/2) × (2FD + 2FD) × h = 2 × FD × h ... (4)
Now, area(△FED) = (1/2) × FD × (h/2) ... (5)
From equations (4) and (5), we get
8 × area(△AFC) = area(△FED)
Hence, (ii) - (r)
(iii) Area(△BEF) = (1/2) × BF × (h/2) --- From equation (3)
Area(△BEF) = (1/2) × 2FD × (h/2) = 2[(1/2) × FD × (h/2)] = 2 × area(△FED)
Hence, (iii) - (p)
Therefore, option (4) is correct.
Question 50 1.2 / -0
Which of the following statements is INCORRECT for a parallelogram?
Solution
It is not necessary that opposite angles are always bisected by the diagonals.
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