Question 1 1.2 / -0
In a certain code language, '256' means 'you are beautiful', '637' means 'they are bad' and '358' means 'beautiful and bad'. Which of the following numbers represents 'they' in that code language?
Solution
In the first and second statements, the common word is 'are' and common code is '6'. In the second and third statements, the common word is 'bad' and the common code is '3'. So, from the second statement, '7' represents 'they'.
Question 2 1.2 / -0
Seven persons are standing in a row. Ritul is standing left to Jasmine but right to Priyanka, Om is standing right to Neha and left to Priyanka. Similarly, Sneha is standing right to Jasmine and left to Tanmay. Which of the following persons is standing in the middle?
Solution
According to the given information, we have:
Question 3 1.2 / -0
How many semicircles are there in the given figure?
Solution
There is no division in the middle circle. So, Number of semicircles in the given figure = 4 + 4 + 4 + 4 + 0 = 16
Question 4 1.2 / -0
Which of the following is the correct water-image of the given combination?
Question 5 1.2 / -0
Gunjan is facing west. She walks 20 m towards the right and then 80 m towards the left. At last she turns to the left and walks 20 m. In which direction is she facing now?
Solution
We have,
So, Gunjan is facing towards south now.
Question 6 1.2 / -0
Select a figure from the options which will continue the same series as established by the given Problem Figures.
Question 7 1.2 / -0
There is a definite relationship between figures (i) and (ii). Establish a similar relationship between figures (iii) and (iv) by selecting a figure from the options that will replace (?) in figure (iv).
Solution
From figure (i) to (ii), every corner element moves 1 step in clockwise direction. Also, the upper right corner and the bottom left corner elements rotate 90° clockwise.
Question 8 1.2 / -0
Select the figure from the options in which Fig. (X) is exactly embedded as one of its parts.
Question 9 1.2 / -0
Select a figure from the options which completes the figure matrix.
Solution
In each row, the first figure has the lower part shaded, the second figure has the left part shaded and the third figure has its diagonally lower left part shaded.
Question 10 1.2 / -0
If it is possible to make only one meaningful word with the second, the seventh, the tenth and the eleventh letters of the word TRADITIONAL, what will be the second letter of the word? If no such word can be formed, give 'X' as the answer. If only two such words can be formed, give 'Y' as the answer and if more than two such words can be formed give 'Z' as the answer.
Solution
The second, seventh, tenth and eleventh letters of the word TRADITIONAL are R, I, A and L, respectively. So, the meaningful words formed are RAIL, LIAR, LAIR, ..., etc.
Question 11 1.2 / -0
Sonam is 9 ranks ahead of Riya, who has the seventeenth rank in the class in the annual examinations. What is Sonam's rank from the bottom, if there are 40 students in the class?
Solution
Rank of Riya from beginning = 17th So, rank of Sonam from beginning = 17 - 9 = 8th Total number of students = 40 Hence, rank of Sonam from last = 40 - 8 + 1 = 33rd
Question 12 1.2 / -0
Two rows of numbers are given. Resultant number in each row is to be worked out separately based on the following rules and the question below the row of numbers is to be answered. The operations of numbers progress from left to right.
Rules: (i) If an even number is followed by a prime number, then they are to be multiplied.
(ii) If an even number is followed by a composite odd number, then the odd number is to be subtracted from the even number.
(iii) If a composite odd number is followed by a prime number, then the first number is to be divided by the second number.
(iv) If an odd number is followed by an even number, which is a perfect square, then they are to be added.
(v) If an odd number is followed by another odd number, then they are to be added.
If m is the resultant of the first row, then what is the resultant of the second row?
Solution
First row: 86 45 13 21 86 - 45 13 21 = 41 13 21 = 41 + 13 21 = 54 21 = 54 - 21 = 33 So, m = 33 Second row: 33 11 36 21 = 33 ÷ 11 36 21 = 3 36 21 = 3 + 36 21 = 39 21 = 39 + 21 = 60
Question 13 1.2 / -0
Which of the following numbers will be obtained if the second digit of the smallest number is added to the third digit of the greatest number after adding 5 to each number of the given arrangement? 152 453 279 536 785
Solution
After adding 5 to each number, the new arrangement becomes: 157 458 284 541 790 Now, the smallest number is 157 and the second digit of 157 is 5. Also, the greatest number is 790 and the third digit of 790 is 0. So, required value = 5 + 0 = 5
Question 14 1.2 / -0
Which of the following Venn diagrams best illustrates the relationship among ''Rings", "Ornaments", and "Diamond rings''?
Question 15 1.2 / -0
How many consonants are there in the given sequence which are immediately preceded by a digit and immediately followed by a symbol?
E G 4 B H 7 5 @ K 8 D N
Q Z $ W 3 C 1 9 * L B 2 S 6
Solution
There is no consonant in the given sequence which is immediately preceded by a digit and immediately followed by a symbol.
Question 16 1.2 / -0
Solve for x :
Question 17 1.2 / -0
8 pumps can empty a water tank in 42 hours. How many hours would 56 pumps take to do the same work?
Solution
It is a case of inverse proportion.
∴ 8 × 42 = x × 56
So, 56 pumps take 6 hours to empty the water tank.
Question 18 1.2 / -0
What is the smallest natural number by which 7056 must be divided so that the quotient becomes a perfect cube?
Solution
We have,
7056 =
× 2 × 3 × 3 × 7 × 7
To make it a perfect cube, we divide the number by 2 × 3 × 3 × 7 × 7, i.e. 882.
Question 19 1.2 / -0
Find the values of the letters A, B, C and D.
Solution
We have,
∴ A = 4, B = 1, C = 0, D = 2
Question 20 1.2 / -0
The vertices and faces of a polyhedron are 32 and 15, respectively. Find the number of edges of this polyhedron.
Solution
By Euler's formula, we have: Edges = Vertices + Faces - 2 ∴ Edges = 32 + 15 - 2 = 45
Question 21 1.2 / -0
To construct a unique rectangle, the minimum number of measurements required is
Solution
For a unique rectangle to be constructed, we require (length and width) a minimum of 2 measurements.
Question 22 1.2 / -0
A bag contains 3 red, 1 pink and 9 white balls. One ball is drawn at random. Find the probability that the ball drawn is not white.
Solution
Total number of balls = 3 + 1 + 9 = 13
Number of balls which are not white = 3 + 1 = 4
∴ Required probability =
Question 23 1.2 / -0
Which of the following expressions shows that rational numbers are associative under multiplication?
Solution
The expression
shows that rational numbers are associative under multiplication.
Question 24 1.2 / -0
A hat contains 5 red balls and 7 green balls. If a ball is drawn at random, then what is the probability that it is a red ball?
Solution
Total number of balls = 5 + 7 = 12
Number of red balls = 5
∴ Probability (drawing a red ball) =
Question 25 1.2 / -0
The two adjacent sides of a rectangle are 5x2 - 3y2 and x2 + 2xy. Find the perimeter.
Solution
Adjacent sides of the rectangle are 5x2 - 3y2 and x2 + 2xy. So, perimeter of rectangle = 2(length + breadth) = 2(5x2 - 3y2 + x2 + 2xy) = 2(6x2 - 3y2 + 2xy) = 12x2 - 6y2 + 4xy
Question 26 1.2 / -0
Factorise: 4a2 + 12ab + 9b2 - 8a - 12b
Solution
We have, 4a2 + 12ab + 9b2 - 8a - 12b = (2a)2 + (3b)2 + 2 × 2a × 3b - 8a - 12b = (2a + 3b)2 - 4(2a + 3b) = (2a + 3b) (2a + 3b - 4)
Question 27 1.2 / -0
In the given figure, DE is the internal bisector of
D. If
D = 60°, then find the value of x.
Solution
ACD =
ABC +
BAC
[Exterior angle property]
ACD = 75° + 25° = 100°
Now, in ΔDCF,
DCF +
FDC +
CFD = 180°
(Angle sum property)
100° + 60° +
CFD = 180°
CFD = x = 180° - 160° = 20°
Question 28 1.2 / -0
If x =
, then find the value of x
-2 y.
Solution
We have,
x =
=
And,
y =
=
Now,
=
Question 29 1.2 / -0
The difference between the compound interest (compounded annually) and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.
Solution
Let:
Principal = Rs. P
Rate (R) = 10% p.a.
Time (T) = 2 years
Since, C.I. - S.I. = P
P = 631 × 100 = Rs. 63,100
Question 30 1.2 / -0
lf (t + 2) units and (t2 + 4) units are length and breadth of a rectangle, respectively, then find its area (in sq. units).
Solution
Length of the rectangle = (t + 2) units Breadth of the rectangle = (t2 + 4) units ∴ Area of the rectangle = Length × Breadth = (t + 2) (t2 + 4) = t3 + 2t2 + 4t + 8
Question 31 1.2 / -0
Using Euler's formula, find the respective values of P, Q and R.
Faces : P 5 20 Vertices : 6 Q 12 Edges : 12 9 R
Solution
By Euler's formula,
F + V = E + 2
So, P + 6 = 12 + 2
P = 14 - 6 = 8
Also, 5 + Q = 9 + 2
Q = 11 - 5 = 6
Similarly, 20 + 12 = R + 2
R = 32 - 2 = 30
Question 32 1.2 / -0
If 31y5 is divisible by 15, where y is a single digit, then what would be the values of y?
Solution
31y5 is divisible by 15. So, it must be divisible by 3 and 5 both. Since, its unit digit is 5, it is divisible by 5. Now, to be divisible by 3, sum of its digits must be divisible by 3. Sum of digits = 3 + 1 + y + 5 = 9 + y Hence, for 9 + y to be divisible by 3, the values of y are 0, 3, 6, 9.
Question 33 1.2 / -0
In the given parallelogram XYZW, find the values of x, y and z respectively.
Solution
In parallelogram XYZW,
WXY =
WZY
[Opposite angles of a parallelogram are equal]
y = 102°
In ΔXWY,
y + x + 50° = 180° [Angle sum property]
102° + x + 50° = 180°
x = 180° – 152° = 28°
Now, x = z [Alternate angles]
z = 28°
Question 34 1.2 / -0
Directions For Questions
Directions: The given double column graph compares the average number of hours for which an electrical appliance is used on weekdays and weekends. Study the graph carefully and answer the question that follows.
...view full instructions
Which appliance is used half as many hours on the weekdays as it is used on the weekends?
Solution
Number of hours for which lights are used in weekdays = 12 Number of hours for which lights are used in weekends = 18 Number of hours for which refrigerators are used in weekdays = 24 Number of hours for which refrigerators are used in weekends = 24 Number of hours for which A.C. is used in weekdays = 10 Number of hours for which A.C. is used in weekends = 16 Number of hours for which computer is used in weekdays = 4 Number of hours for which computer is used in weekends = 8 So, computer is used half as many hours on the weekdays as it is used on weekends.
Question 35 1.2 / -0
Directions For Questions
Directions: The given double column graph compares the average number of hours for which an electrical appliance is used on weekdays and weekends. Study the graph carefully and answer the question that follows.
...view full instructions
What is the total number of appliances used in weekdays?
Solution
Total number of appliances used in weekdays = 12 + 24 + 4 + 10 + 2 = 52
Question 36 1.2 / -0
The monthly incomes of Komal and Asha are in the ratio of 4 : 3. Their monthly expenses are in the ratio of 3 : 2. However, both save Rs. 600 per month. What is their total monthly income?
Solution
Let the monthly incomes of Komal and Asha be 4x and 3x, respectively.
Both save Rs. 600 per month.
∴ According to question, we have:
2(4x - 600) = 3(3x - 600)
8x - 1200 = 9x - 1800
x = 600
So, their total monthly income = 4x + 3x = 7x = Rs. 7 × 600 = Rs. 4,200
Question 37 1.2 / -0
Sam, Trishu and Sanchi begin to jog around a circular stadium. They complete their revolutions in 42 seconds, 56 seconds and 63 seconds, respectively. After how many seconds will they be together at the starting point?
Solution
Time after which they will be together at the starting point = L.C.M. (42,56,63) = 504 seconds
Question 38 1.2 / -0
The outer dimensions of a closed box are 10 cm by 8 cm by 7 cm. Thickness of the wood is 1 cm. Find the total cost of wood required to make box, if 1 cm3 of wood costs Rs. 2.00.
Solution
Outer volume of box = 10 × 8 × 7 = 560 cm3 Inner dimensions of the box are (10 - 2) by (8 - 2) by (7 - 2), i.e. 8 cm by 6 cm by 5 cm. ∴ Inner volume of box = 8 × 6 × 5 = 240 cm3 Now, volume of wood used = Outer volume - Inner volume = (560 - 240) cm3 = 320 cm3 Cost of 1 cm3 wood = Rs. 2 So, cost of 320 cm3 wood = Rs. (2 × 320) = Rs. 640
Question 39 1.2 / -0
Gopal wants to plant 99,225 shrubs and arranges them in such a way that there are as many rows as there are shrubs in a row. Find the number of shrubs in a row.
Solution
Let the number of shrubs in each row be x.
Total number of shrubs = 99,225
According to the question,
x
2 = 99,225
x
2 =
= 315
∴ Number of shrubs in each row = 315
Question 40 1.2 / -0
37
% of the candidates in an examination were girls. 75% of the boys and 62
% of the girls passed and 342 girls failed. Find the number of boys who failed.
Solution
Let the total number of candidates = x
∴ Number of girls = 37
Number of girls who passed =
=
∴ Number of girls who failed =
∴ Total number of candidates = 2432
Number of girls =
= 912
∴ Number of boys = 2432 - 912 = 1520
Now, number of boys who passed = 75% of 1520
=
= 1140
∴ Number of boys who failed = 1520 - 1140 = 380
Question 41 1.2 / -0
The cost of papering the four walls of a room is Rs. 480. Each one of the length, breadth and height of another room is double that of this room. What is the cost of papering the walls of the new room?
Solution
Let the length, breadth and height of the first room be l, b and h, respectively. Then, the length, breadth and height of the new room will be 2l, 2b and 2h, respectively. ∴ Area of four walls of first room = 2(l + b)h Cost of papering the four walls of first room = Rs. 480 ... (i) Also, area of four walls of new room = 2(2l + 2b) × 2h = 4(2(l + b)h) ∴ Cost of papering the walls of new room = 4 × 480 = Rs. 1,920 (Using (i))
Question 42 1.2 / -0
Harshad bought 15 pieces of DVD players at Rs. 4,500 each and sold all of them at the total price of Rs. 81,000. What is the percent profit earned in the deal?
Solution
Since, C.P. of 1 DVD player = Rs. 4500
C.P. of 15 DVD players = Rs. (4500 × 15) = Rs. 67,500
S.P. of 15 DVD players = Rs. 81,000
So, profit % =
=
=
Question 43 1.2 / -0
One custard recipe requires
cup of sugar. Another recipe for the same custard requires 2 table-spoons of sugar. If 1 table-spoon is equivalent to
cup, then how much more sugar does the first recipe require?
Solution
1 table-spoon =
cup
Quantity of sugar required for first recipe =
cup
Quantity of sugar required for second recipe = 2 table-spoons
=
So, the excess quantity of sugar the first recipe requires than the second recipe =
Question 44 1.2 / -0
Renuka borrowed Rs. 15,000 from Vipul at 5% per annum at simple interest for 3 years. If she had borrowed this sum at the same rate and time at compound interest, then what extra amount would she have had to pay?
Solution
P = Rs. 15,000
R = 5% p.a.
T = 3 years
Also,
=
So, required difference = Rs. (2364.38 – 2250) = Rs. 114.38
Question 45 1.2 / -0
Mohit is 20 years younger than his father. Three years ago, Mohit's age was one-sixth the age of his father. Find the present age of Mohit.
Solution
Let the present age of Mohit be x years.
Then, the present age of Mohit's father = (x + 20) years
3 years ago,
Mohits age = (x - 3) years
And,
Mohit's fathers age = (x + 20 - 3) = (x + 17) years
According to the question,
x - 3 =
(x + 17)
6x - 18 = x + 17
5x = 35
x = 7
So, present age of Mohit = 7 years
Question 46 1.2 / -0
Read the statements carefully and select the CORRECT option.Statement-1: While solving a mathematical problem, Samidha squared a number and then subtracted 25 from it rather than the required, i.e. first subtracting 25 from the number and then squaring it. But she got the right answer. The given number is 13.Statement-2 : The smallest natural number which is a perfect square and which ends in 3 identical digits lies between 2000 and 3000.
Solution
Statement-1: Let x be the number.
x
2 - 25 = (x - 25)
2 x
2 - 25 = x
2 + 625 - 50x
50x = 650
x = 13
∴ Statement-1 is true.
Statement-2: The smallest natural number which is a perfect square and which ends in 3 identical digits is 10,000 which doesn't lie between 2000 and 3000.
∴ Statement-2 is false.
Question 47 1.2 / -0
Find P + Q - R. (i) If a rectangle of length 44 cm is rolled along its length to form a cylinder, the radius of cylinder is P cm. (ii) The cost of plastering the walls of a cuboidal room of dimensions 12 m × 10 m × 4 m at the rate of Rs. 25 per sq. m is Rs. Q . (iii) The volume of a cuboid of dimensions 14 m × 7 m × 12 m is R m3 .
Solution
(i) Length of rectangle = 44 cm
Since, the rectangle is rolled along its length.
∴ Circumference of base of cylinder = Length of rectangle
P = 7
(ii) Length, breadth and height of the cuboidal room are 12 m, 10 m and 4 m, respectively.
Surface area of walls of room = 2(l + b)h
= 2(12 + 10)4
= 8 × 22 = 176 m
2 Cost of plastering the walls per m
2 = Rs. 25
∴ Cost of plastering the walls = Rs. 25 × 176 = Rs. 4,400
Q = 4400
(iii) Length, breadth and height of the cuboid are 14 m, 7 m and 12 m respectively.
Volume of cuboid = I × b × h
= 14 × 7 × 12 = 1,176 m
3 R = 1,176
So, P + Q - R = 7 + 4,400 - 1,176 = 3,231
Question 48 1.2 / -0
Look at the given data.
22, 10, 29, 22, 32, 5, 37, 14, 12, 23, 42, 15, 38, 26, 32, 18, 26, 28, 19, 27, 35, 31, 24, 37, 19, 20
The frequency distribution is given below:
Which of the following tables is the frequency table of the given data?
Solution
The frequency distribution for the given data is as follows.
So, neither frequency table P nor Q represents the given data.
Question 49 1.2 / -0
Solution
(P) C.S.A. of cylinder = 2
rh
(Q) Area of trapezium =
(Sum of parallel sides × Distance between them)
(R) Volume of cylinder =
r
2 h
(S) Volume of cube = (Side)
3
Question 50 1.2 / -0
Ten separate slips, bearing numbers from 1 to 10 (one number on one slip), are kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of getting: (a) a number greater than 6? (b) a number less than 6? (c) a 1-digit number?
Solution
Total possible outcomes = 10
(a) Numbers greater than 6 are 7, 8, 9 and 10.
∴ Required probability =
(b) Numbers less than 6 are 1, 2, 3, 4 and 5.
∴ Required probability =
(c) 1-digit numbers are 1, 2, 3, ...., 9, i.e. 9 in number.
∴ Required probability =