Question 1 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 given options that would replace the (?) in figure (iv).
Solution
From first figure to second; the outer elements move anticlockwise by 90°. Also, the shaded elements get unshaded and vice-versa.
Question 2 1.2 / -0
Select the correct water image of Fig. (X).
Question 3 1.2 / -0
A is the son of E's brother B. D is the daughter of B. E is the mother of C. How is D related to C?
Solution
D is the daughter B and C is the child of B's sister E. So, D is the cousin of C.
Question 4 1.2 / -0
Which of the following options best illustrates the given Venn diagram?
Question 5 1.2 / -0
Ali walks 10 km towards north. He turns to his right and after walking 2 km, he turns left. Finally after walking 6 km, he turns left again. In which direction is he moving now?
Solution
Now, Ali is moving in the west direction.
Question 6 1.2 / -0
If it is possible to make only one meaningful word with the third, sixth and eighth letters of the word ORDINATE, then the third letter of the word formed is the answer. If no meaningful word is formed, then give 'X' as the answer. If more than one such words can be formed, then give 'Z' as the answer.
Solution
The third, sixth and eighth letters of the given word, respectively, are D, A and E.
Question 7 1.2 / -0
In a certain code language, 'HOLIDAY' is written as 'ACFKNQJ'. How will 'STUDENT' be written in the same code language?
Solution
We have, H O L I D A Y
On reversing the letters,
Similarly, S T U D E N T
On reversing the letters,
Hence, STUDENT will be written as 'VPGFWVU'.
Question 8 1.2 / -0
Consider the following arrangement and answer the question that follows:
If the middle digit of the least number is subtracted from the middle digit of the greatest number after adding ten to each of the numbers of the given arrangement, then which of the following numbers will be obtained?
Solution
Given arrangement:
Adding 10 to each of the numbers in the given arrangement, we get
Greatest number = 882
Middle digit of 882 = 8
Smallest number = 303
Middle digit of 303 = 0
Required number = 8 - 0 = 8
Question 9 1.2 / -0
The given sheet of paper is folded to form a box. Choose the box from the given options that is similar to the box formed.
Solution
Only box shown in option (3) is similar to the box formed when the given sheet of paper is folded to form a box.
Question 10 1.2 / -0
Count the number of squares in the given figure.
Solution
Number of squares formed = 18.
Question 11 1.2 / -0
Select a figure from the given options that will complete the given figure matrix.
Solution
First and second element of each row have been merged and the third figure is the combination of uncommon part of first two figures.
Question 12 1.2 / -0
Which of the following figures does not satisfy the same conditions of placement of the dots as in the given figure?
Solution
Hence, option 4 is the correct answer.
Question 13 1.2 / -0
The given question consists of a set of three figures X, Y and Z showing a sequence of folding of a piece of paper. Fig. (Z) shows the manner in which the folded paper has been cut. Select a figure from the options which would most closely resemble the unfolded form of Fig. (Z).
Question 14 1.2 / -0
In which of the following figures is the given figure exactly embedded as one of its parts?
Solution
Option 1.
Question 15 1.2 / -0
In a queue of 28 girls, Kuhu is 6 ranks ahead of Pihu. If Pihu's rank is 15th from the last, then what is the rank of Kuhu from the starting?
Solution
Rank of Kuhu from starting of the queue = 28 - (6 + 15) + 1
= 28 - 21 + 1 = 8
th
Question 16 1.2 / -0
In a
ABC, if AB + BC = 10 cm, BC + CA = 12 cm and CA + AB = 16 cm, then the perimeter of the triangle is
Solution
We have AB + BC = 10 cm ... (i)
BC + CA = 12 cm ... (ii)
CA + AB = 16 cm ... (iii)
Adding (i), (ii) and (iii), we get
AB + BC + BC + CA + CA + AB = 10 + 12 + 16
2(AB + BC + CA) = 38
AB + BC + CA =
= 19 cm
Perimeter of
ABC = 19 cm
Question 17 1.2 / -0
In the given figure (not drawn to scale), AB || CE and ACD is a straight line. Which of the following options is not correct?
Solution
ACD is a straight line.
30° + z + 100° = 180° (Linear pair)
z = 180° - 130° = 50°
As AB || CE and BC is a transversal
So, z = y = 50° (Alternate interior angles)
Also, x = 100° (Corresponding angle)
(1) y + z = 50° + 50° = 100° = x
(2)
= 50° = y = z
(3) 2x + y = 2 × 100° + 50° = 200° + 50° = 250°
150°
Question 18 1.2 / -0
In triangle XYZ, XP is the perpendicular bisector of the base YZ. Which of the following congruence criteria can be used to conclude that
XYP
XZP?
Solution
In
XYP and
XZP,
XP = XP (Common)
XPY =
XPZ = 90°
YP = ZP (XP bisects YZ)
XYP
XZP (By SAS congruence criterion)
Question 19 1.2 / -0
Which of the following statements is CORRECT?
Solution
Only option (1) is correct. As length of the third side must be greater than the difference of lengths of the given two sides and less than the sum of the lengths of the given two sides. So, length of the third side lies between 3 cm and 17 cm.
Question 20 1.2 / -0
Directions For Questions
Directions:
...view full instructions
In which music category is the difference between DVD buyers and CD buyers the minimum?
Solution
Jazz Folk Classical Pop DVD buyers 80 60 40 60 CD buyers 70 50 50 55 Difference 10 10 10 5
Hence, pop music has the minimum difference between DVD buyers and CD buyers.
Question 21 1.2 / -0
Directions For Questions
Directions:
...view full instructions
What is the ratio of Pop music CD buyers to the Jazz music DVD buyers?
Solution
Number of Pop music CD buyers = 55
Number of Jazz music DVD buyers = 80
Required ratio = 55 : 80 = 11 : 16
Question 22 1.2 / -0
P and Q are two integers such that P is the multiplicative inverse of (-4) × 5 and Q is the additive inverse of -5 + 4. Find the value of P - Q.
Solution
We have, (-4) × 5 = -20
P = multiplicative inverse of (-20) =
.
Also, -5 + 4 = -1
Q = additive inverse of (-1) = 1
P - Q =
Question 23 1.2 / -0
If the product of two fractions is
and their quotient is
, then the sum of two fractions is _______.
Solution
Let the two fractions be x and y.
According to the question,
x × y =
… (i)
x
y =
… (ii)
From (i)
x =
Putting value of x in (ii), we get
12 × 5 = 3 × 5y × y
60 = 15y
2 y
2 = 4
y
2 = 2
2 y = 2
Now,
Required sum = x + y =
Question 24 1.2 / -0
What is the least number of squares that must be added so that the line AB becomes the line of symmetry?
Question 25 1.2 / -0
Mr. Gupta asked two of his students to solve the expression
and write the answer on the blackboard. Aryan wrote the answer as 3 a
-4 b
5 , whereas Ayan wrote the answer as
. Who wrote the correct answer?
Solution
We have,
= (9a
-5 a
-3 b
2 b
8 )
1/2 = (9a
-8 b
10 )
1/2 = 3 a
-4 b
5 i.e.,
Hence, both Aryan and Ayan wrote the correct answer.
Question 26 1.2 / -0
In the given diagram, PQU is an equilateral triangle, QRVU is a rhombus and RSTU is a square. Find the perimeter (in cm) of the whole diagram.
Solution
Since PQU is an equilateral triangle
PQ = QU = UP = 4 cm
Also, QRVU is a rhombus.
QR = RV = VU = UQ = 4 cm
Also, RSTU is a square.
RS = ST = TU = UR = 5 cm
Perimeter of the given figure
= PQ + QR + RS + ST + TU + UP
= (4 + 4 + 5 + 5 + 5 + 4) cm = 27 cm
Question 27 1.2 / -0
If a : b = 2 : 3, then (3a + 2b) : (5a + 3b) is equal to ______.
Solution
We have,
Now,
Question 28 1.2 / -0
If a number is increased by 20% and then the increased number is decreased by 20%, then the overall change in number is
Solution
Let the number be x.
Increased number = x + 20% of x.
=
=
Decreased number =
Overall decrease in number = x -
Overall percentage decrease in number
=
Question 29 1.2 / -0
In the given figure, PQ || RS. Find the value of b.
Solution
In
URT,
URT +
RTU +
TUR = 180° (Angle sum property)
90° + a + 40° = 180°
a = 180° - 90° - 40° = 50°
As PQ || RS and PT is a transversal,
80° + c = 180° (Co-interior angles)
c = 180° - 80° = 100°
Now, RTS is a straight line.
a + b + c = 180°
50° + b + 100° = 180°
b = 180° - 100° - 50° = 30°
Question 30 1.2 / -0
Which of the following options represents the value of P shown on the number line?
Solution
Since, P = -1
=
A.
B.
C.
D.
Hence, option 3 is the correct answer.
Question 31 1.2 / -0
Which of the following figures is the top view of the given figure?
Solution
Top view of the given figure is
Question 32 1.2 / -0
If
= 1, then find the value of
.
Question 33 1.2 / -0
lf twice of A is equal to thrice of B and four times of B is equal to five times of C, then find the ratio of A to C.
Solution
We have,
2A = 3B ... (i)
and 4B = 5C ... (ii)
On multiplying equation (i) by 4 and equation (ii) and 3, we get
8A = 12B ....... (iii)
12B = 15C .......(iv)
From (iii) and (iv), we get
8A = 15C
A : C = 15 : 8
Question 34 1.2 / -0
ABCD is a rectangle (not drawn to scale) having length 30 cm and breadth 25 cm. P, Q, R and S are midpoints of AB, BC, CD and AD, respectively. Find the area of the shaded region of the figure.
Solution
Area of rectangle ABCD = Length × breadth = (30 × 25) cm
2 = 750 cm
2 AB is the length and AD is the breadth.
Now, DS =
= 12.5 cm
DR =
= 15 cm
Area of
DSR = Area of
RCQ = Area of
QBP
= Area of
DSR =
× DS × DR
=
cm
2 = 93.75 cm
2 So, area of shaded region = Area of rectangle - (area of 4 triangles)
= (750 - 4 × 93.75) cm
2 = 375 cm
2
Question 35 1.2 / -0
If A = 9t2 + 14xt - 3x2 , B = 16xt + 4x2 - 5t2 and C = -7x2 - 19xt - 8t2 , then find the value of B - A - C.
Solution
B - A - C2 - 5t2 - (9t2 + 14xt - 3x2 ) - (-7x2 - 19xt - 8t2 )2 - 5t2 - 9t2 -14xt + 3x2 + 7x2 + 19xt + 8t2 2 -9t2 + 8t2 ) + (4x2 + 3x2 + 7x2 ) + (16xt - 14xt + 19xt)2 + 14x2 + 21xt
Question 36 1.2 / -0
Manvi says that she got 16 more marks than 4 times the number of marks scored by Aashi. Form an equation which gives the marks scored by Aashi, if Manvi scored 72 marks.
Solution
Let number of marks scored by Aashi be x.
Number of marks scored by Manvi = 4x + 16
Thus, according to the question, 4x + 16 = 72
Question 37 1.2 / -0
Kirti bought a plot of land in the outer of the city for Rs. 32,50,000. She built a wall around it for which she spent Rs. 1,75,000. Then she wants to sell it at Rs. 40,00,000 by making an advertisement in the television which costs her Rs. 15,000. Find her approximate profit percent.
Solution
Cost of land = Rs. 32,50,000
Expenditure on the plot of land = Rs. (1,75,000 + 15,000) = Rs. 1,90,000
Now, C.P. = Rs. (32,50,000 + 1,90,000) = Rs. 34,40,000
Profit = Rs. 40,00,000 - 34,40,000 = Rs. 5,60,000
Profit % =
× 100 = 16.28%
Question 38 1.2 / -0
In a circular park of diameter 200 m, a flower bed is made in the form of a circle with radius 20 m. Find the approximate area of the park left out. (Take
= 22/7)
Solution
Area of the circular park =
r
2 =
Area of the flower bed =
Area of the park left out =
=
= 30171.43 m
2
Question 39 1.2 / -0
Ram needs to pay his bill. After deducting 6% of the bill, the amount still to be paid is Rs. 282. How much was the original bill?
Solution
Let the original bill be Rs. x.
According to the question,
6% of x + 282 = x
+ 282 = x
282 = x -
282 =
= 300
The original bill was Rs. 300
Question 40 1.2 / -0
The ratio of the present ages of two friends is 1 : 2 and six years back, the ratio was 1 : 4. What will be the ratio of their ages after six years from now?
Solution
Let the present ages of two friends be x years and 2x years, respectively.
According to the question,
4x - 24 = 2x - 6
4x - 2x = -6 + 24
2x = 18
x = 9
So, present ages are 9 years and 18 years, respectively.
Ages after six years will be 15 years and 24 years, respectively.
Required ratio =
, i.e. 5 : 8
Question 41 1.2 / -0
In Shimla the minimum temperature on Tuesday was 25°C. On Wednesday it fell by 3°C and due to rainfall on Thursday it further fell by 4°C. What was the the minimum temperature on Thursday?
Solution
Minimum temperature on Tuesday = 25°C
Question 42 1.2 / -0
Priya invests
of her capital at 7% p.a.
at 8% p.a. and the remainder at 10% p.a. If her annual income is Rs. 561, then find the capital.
Solution
Let the capital be Rs. x.
Capital invested at 7% p.a. =
x
Interest earned at 7% p.a. =
x × 7 ×
Capital invested at 8% p.a. =
x
Interest earned at 8% p.a. =
Capital invested at 10% p.a. =
=
Interest earned at 10% p.a. =
Total interest earned =
Now, Priya's annual income = Rs. 561
x =
Question 43 1.2 / -0
A rectangular piece of land is to be sold off in smaller pieces. The total area of the land is 217 sq. miles. The pieces to be cut out are 162 sq. miles in size. How many smaller pieces of the land can be sold at the given size?
Solution
Total area of the land = 2
17 sq. miles
Area of smaller pieces of the land = 16
2 sq. miles
Number of pieces of the land can be sold
=
= 2
17- 8 = 2
9
Question 44 1.2 / -0
In a class there are 20 girls and 30 boys. The mean weight of 20 girls is 35 kg and mean weight of 30 boys is 60 kg. Find the mean weight of the class.
Solution
Mean weight of girls =
Sum of weights of all girls = 35 × 20 = 700 kg ...(i)
Also, mean weight of boys =
60 =
Sum of weights of all boys = 60 × 30 = 1800 kg ... (ii)
Mean weight of class =
=
[Using (i) and (ii)]
=
= 50 kg
Question 45 1.2 / -0
The average height of 20 students in a class is 105 cm. If ten more students of average height 120 cm join the class, then the average height of the class is _____________
Solution
Average height of 20 students = 105 cm
Sum of height of 20 students = (105 × 20) cm
= 2100 cm
Now, average height of 10 new students = 120 cm
= 120 cm
Sum of height of 10 new students = (120 × 10) cm = 1200 cm
Now, total number of students = 20 + 10 = 30
And, sum of height of all 30 students = (2100 + 1200) cm = 3300 cm
Average height of class =
cm = 110 cm
Question 46 1.2 / -0
Fill in the blanks and select the correct option.(P) ____(Q) ____ as well as ___(R) ___(S) ___
Solution
(i) If two lines intersect each other, then the vertically opposite angles are equal.
(ii) The angles of a linear pair are adjacent as well as supplementary.
∠AOC + ∠COB = 180°
(iii) If a transversal is perpendicular to two parallel lines, then sum of each pair of corresponding angles is 180°.
∠MNB + ∠DON = 180°
Question 47 1.2 / -0
The ratio of the outer and the inner circumferences of a circular path as shown in the figure is 26 : 25. If the path is 5 m wide, then find
(i) the area enclosed by the path
(ii) the diameter of the inner circle
Solution
Let the radius of inner and outer circle be r and R, respectively.
Let the outer circumference and inner circumference be 26x and 25x, respectively.
Since, 2
r = 25 x .... (i)
And 2
R = 26 x
2
(r + 5) = 26x
2
r + 10
= 26x
25x + 10
= 26x (Using (i))
x = 10
Putting x = 10
in (i), we get
2
r = 25 × 10
r =
= 125
So, R = 125 + 5 = 130 m
(i) Area enclosed by the path =
R
2 -
r
2 =
(R
2 - r
2 )
=
(130
2 - 125
2 ) =
(16900 - 15625)
=
× 1275 = 4007.14 m
2 (ii) Diameter of inner circle = 2r = 2 × 125 = 250 m
Question 48 1.2 / -0
Solution
(P) 2700 × x
6 = 27 × 10
8 x
6 =
x
6 = 10
6 x = 10
(R) =
=
(S)
=
Question 49 1.2 / -0
Read the following statements carefully and select the correct option.
Statement-1: If Rs. 880 has to be divided among three daughters of Shyam such that Sunita receives
of what Rekha and Jyoti together receive, then Sunita's share is Rs. 264.
Statement-2: If the mean of 8, 9, x, 10, 5 is 15, then the value of x is 43.
Solution
Statement-1:
Let the amount of money Rekha and Jyoti together receive be Rs. x.
Amount of money Sunita receives =
of x.
= Rs.
Total amount of money = Rs. 880
+ x = 880
= 880
= 616
So, Sunita's share = Rs.
= Rs. 264
Hence, Statement-1 is true.
Statement-2:
Mean =
15 =
15 × 5 = 32 + x
75 = 32 + x
x = 75 - 32 = 43
Hence, Statement-2 is also true.
Question 50 1.2 / -0
In the given figure (not drawn to scale), x || y || z and
|| m || n. Find the values of
(i) 4c + 6a + 3b + 5d
(ii) 2(a + b + c + d)
Solution
As,
|| m || n and z is a transversal
4c = 120° (Corresponding angles)
c =
= 30°
Also, 5d = 120°
d =
= 24°
Now, y || z ad
is a transversal
3b = 4c = 120° (Corresponding angles)
b =
= 40°
Also, x || z and m is a transversal
6a = 120° (Corresponding angles)
a =
= 20°
(i) 4c + 6a + 3b + 5d
= 120° + 120° + 120° + 120°
= 480°
(ii) 2(a + b + c + d)
= 2(20° + 40° + 30° + 24°)
= 2 × 114° = 228°
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