For a > 1, b > 1, what is the LCM of a and b?
I. Product of a and b is 1763
II. a and b are coprimes
TS ICET 2019 Question Paper Shift-1 (24th May)
In the following questions a question is followed by data in the form of two statements labeled as I and II. You must decide whether the data given in the statements are sufficient to answer the questions. Using the data make an appropriate choice from (a) to (d) as per the following guidelines:
If a, b, c are positive integers, is their product abc even?
I. a + b + c is odd
II. a + c is odd
Is a > b?
I. $$x - 3$$ is a factor of $$x^3 - 27^a$$
II. $$x - 2$$ is a factor of $$x^4 - 16^b$$
Is the porallelogram ABCD a rectangle?
I. AC = BD
II. $$\angle A + \angle B = 180^\circ$$
What is the area of the circle?
I. The circle is inscribed in a square
II. The length of the side of the square is 8 cm
What is the three digit number P, where 100 < P < 210?
I. P gives remainder 9 when divided by 10
II. P is divisible by 11
What is the cost of one chair?
I. 1 Chair and 1 Table cost Rs. 12.000
II. The cost of Table is twice the cost of the Chair
What is selling price of each article?
I. The trader made a profit of Rs. 100
II. Two successive discounts of 10% each were given
What are the two natural numbers a and b?
I. LCM of a and b = 240
II. HCF of a and b = 20
How many people in the city speak Hindi?
I. 10% of people in the city can speak two languages
II. 5% of people in the city who can speak two languages can speak Hindi
Does the point(x, y) lie in the first quadrant?
I. xy < -10
II. $$x + y \leq 10$$
Is 6 a factor of n + 3 ?
I. n is even and divisible by 3
II. n is divisible by a prime number
What is the height of the tree?
I. The height of mountain is 300mts
II. The angles of depression of the top of the tree and the bottom of the tree are $$45^\circ$$ and $$60^\circ$$ respectively as seen from the top of the mountain.
How many men are required to complete the work in 40 days?
I. 8 men and 10 women can complete the work in 20 days.
II. 12 women can complete the work in 30 days.
What is the value of x in the matrix $$A = \begin{bmatrix}x & 1 & 0 \\1 & 2 & 3 \\3 & 4 & x \end{bmatrix}$$
I. A is a singular matrix
II. A is a non-singular matrix
C is a point on the line joining B and E. D and A are point above BC such that $$\lfloor DCE = \frac{1}{2} \lfloor ACE$$. Is CD parallel to AB ?
I. ABC is a scalene triangle
II. BC = CE
What are the co-ordinates of the point Q ?
I. Q lies on the straight line l whose slopeis 3
II. l makes an intercept of 4 on y-axis
What is the number of divisors of n?
I. $$1 \leq n \leq 10,000$$
II. $$n = 6000$$
How many positive integers are less than n and relatively prime to n?
I. $$n = 2^3. 5^2. 7^3$$
II. n is composite
What is the first term of the Geometric Progression $$t_1, t_2, t_3 ..... ?$$
I. $$t_4 : t_3 = 2$$
II. $$t_5 = 64$$
In each of the questions a sequence of numbers or letters that follow a definite pattern is given. Each question has a blank space. This has to be filled by the correct answer from the four given options to complete the sequence without breaking the pattern.
27, 343, ...... 6859.
7, 3, 8, ......, 9, 5.
I, L, S, ........, Y
Z, W, Q, ....., V
170, 1332, 50, ........, 10.
6FL : 9IR :: ..... : 5EJ
Book : Publisher :: Film : .........
RAM : UGV :: CAT : ........
$$\frac{Z}{M} : 2 :: \frac{T}{E} : .........$$
OP : 30 :: CX : ........
Pick the odd thing out:
Each of the questions follow a definite pattern. Observe the same andfill in the blanks withsuitable answers.
9, ....., 126, 48, 1332, 168
DOG, HDN, PHB, .........
Z, U, Q, ........, L
1, 4, 8, 11, 22, ......
A2A, D9E, G16I, ........
EDU, IGS, ........, QMO
BAT, EVA, ......, QFU, ........
6, 3, 1, 0.25, .........
{X, Y}, {X + Y, X - Y}, {2X, 2Y}, {2(X + Y), 2(X - Y)}, {........, ...........}
$$2 + \sqrt{6}, 4 + \sqrt{20}, 6 + \sqrt{42}, ...........$$
Study the following Bar diagram carefully and answer questions
The average food production (in Tonnes) from 2001 to 2005 is
The ratio between the increase in production between 2000 and 2003 and 2000 and 2005 is:
The approximate percentage increase in production of food grains from 2001 to 2003 is.
The marks obtained by a student in the examination in (A) Mathematics (B) English (C) Telugu. (D) Social Science (E) Science are depicted in the following Pie Diagram. Based on this data answer the questions
If the student secured 150 marks in A. how manytotal marks did he secure in the complete examination?
The ratio of the marks obtained by the student in Mathematics and Telugu.
If a student has scored 150 marks in Mathematics, how many marks did he score in social science?
What is the percentage of the total marks scored by the student in C, D and E put together?
If a student got 150 marks in A. how many marks did he get in D & E put together?
The following passage to be analyzed for answering the questions:
In a class of 60 students. 23 play Hockey, 15 play Basketball and 20 plays Cricket. 7 play Hockey and Basketball. 5 play Cricket and Basket Ball. 4 play Hockey and Cricket. and 15 students do not play any of these games. Using this data. answer the questions
The number of students who play Hockey. Basketball and Cricket is.
The number of students who play Cricket and Hockey but not Basketball is
The letters A, B, C, ....., Z of the English Alphabet are numbered 1, 2 ......... 26 respectively. A Code is designed by the following rule:
Rule: The i th the letter is coded as the jth letter where $$j \equiv 3i + 2 (mod 26)$$. The reverse process is followed for decoding.
Example: The $$9^{th}$$ letter I is coded as C because, 3 X 9 + 2 (mod 26) is 29 (mod 26) which is 3 and the letter C is numbered as 3. Based on this coding
procedure answer the questions
The code for COMPUTER is?
What is the code for STATE?
Which word is coded as F U L L Y?
The number of letters of the alphabet that remain unchanged in this coding is
What is the code for M A N A G E R?
A word or a group of letters to be coded or decoded based on the certain codes. Using the information, answer the questions
If “CHENNAI” is coded as “XSVMMZR” than “MUMBAI” is coded as:
If “PUNE’ is coded as “QXSL”, then the code for “CUTE” is
If “MAHATMA" is coded as “PDKAQJX” then “KRISHNA” is coded as
If “ RING” is coded as “SJMF” then which of the following is coded as “GSNF”?
If “2018” is coded as “7981” then “1947” is the code for
For the following questions answer them individually
If today is Friday, then the day after 185 days will be
At what time between 7'O clock and 8’O clock will the two hands in the clock be at right angles to each other?
If a clock takes 20 seconds to strike 11, how many seconds will it take to strike 5?
P says “my father is the only son of Q’s mother and A.Q aresisters”. If R is grandfather of A: B is Q’s son then how is P related to B?
Ram left office for the bus stop 15 minutes early. It takes 25 minutes for him to reach bus stop. Due to traffic he reached 10 minutes late and missed the bus. How much time did he take to come to bus stop?
A train leaves Hyderabad at 6:10 a.m. and reaches Warangal at 9:25 a.m. The average speed of the train is 44 Kmph. Then the distance from Hyderabad to Warangal in km is:
A, F are in different rows and columns. B is opposite to F and H is opposite to D, who is in the same rowas that of A, E and G are neighbors
of F. If the personsitting in the rows faces each other and C is immediate right to D, then who1s diagonally nearest to D?
If $$+$$ means $$\div$$, $$-$$ means $$\times$$, $$\div$$ means $$+$$ and $$\times$$ means $$-$$. Then the value of $$16 + 4 \div 5 - 3 \times 1$$ is
$$a * b = \frac{a^2 + b^2}{ab}, a > 0, b > 0 \Rightarrow \frac{(1 * 2) * (1 * 1)}{1 * 3} =$$
If the symbols *, #, @, £ represent addition, subtraction, multiplication and division respectively then:
$$\frac{(15 * x) @ (12 \# 8) £ (15 \# 2)}{(15 \# 3) @ (12 * 8) £ (5 * 5)} =$$
$$\frac{\left(x + \frac{1}{y}\right)^a . \left(x - \frac{1}{y}\right)^b}{\left(y + \frac{1}{x}\right)^a . \left(y - \frac{1}{x}\right)^b} =$$
Solution
$$\left(\frac{x + \frac{1}{y}}{y + \frac{1}{x}}\right)^a\left(\frac{x - \frac{1}{y}}{y - \frac{1}{x}}\right)^b$$
$$= \left(\frac{xy + 1}{xy + 1} \times \frac{x}{y}\right)^a \left(\frac{xy - 1}{xy - 1} \times \frac{x}{y}\right)^b$$
$$= \left(\frac{x}{y}\right)^a \left(\frac{x}{y}\right)^b$$
$$= \left(\frac{x}{y}\right)^{a+b}$$
Hence the answer is option A
$$3^a = 5^b = 75^c \Rightarrow c(2a + b) =$$
Solution
$$3^a = 5^b = 75^c = k$$
$$3 = k^{\frac{1}{a}}, 5 = k^{\frac{1}{b}}, 75 = k^{\frac{1}{c}}$$
$$75 = 3 \times 5^2$$
$$k^{\frac{1}{c}} = k^{\frac{1}{a}} \times k^{\frac{2}{b}}$$
$$\frac{1}{c} = \frac{1}{a} + \frac{2}{b}$$
$$c = \frac{ab}{2a + b}$$
$$c(2a + b) = ab$$
Hence the answer is option B
$$ab : bc : ca = 3 : 1 : 2 \Rightarrow \frac{a}{bc} : \frac{b}{ca} =$$
Solution
Shortcut
$$ab : bc : ca = 3 : 1 : 2$$
c = 2 a = 6 b = 3
$$\frac{a}{bc} : \frac{b}{ca}$$
$$\Rightarrow \frac{6}{6} : \frac{3}{12}$$
$$1 : \frac{1}{4}$$
$$\Rightarrow 4 : 1$$
Explanation:
$$ab : bc :ca = 3 : 1 : 2$$
$$\Rightarrow \frac{1}{c} : \frac{1}{a} : \frac{1}{b} = 3 : 1 : 2$$
$$c = \frac{k}{3}, a = k, b = \frac{k}{2}$$
Ratio $$c = 2k, a = 6k, b = 3k$$
$$\frac{a}{bc} : \frac{b}{ca}$$
$$\frac{6}{6} : \frac{3}{12}$$
4 : 1
Hence the answer is option B
$$a : b = 3 : 5 :: ab = 2535 \Rightarrow a + b =$$
Solution
a = 3k
b = 5k
$$15k^2 = 2535$$
$$k^2 = 169$$
$$k = 13$$
sum $$\Rightarrow 8k = 104$$
Hence the answer is option A
$$x = 5 + \sqrt[3]{5} + \sqrt[3]{25} \Rightarrow x^3 - 15x^2 + 60x + 10 =$$
Solution
$$(x - 5)^3 = 5 + 25 + 3 \times 5 (x - 5)$$
$$(x - 5)^3 = 30 + 15 (x - 5)$$
$$(x - 5)[x^2 + 25 - 10x - 15] = 30$$
$$(x - 5)[x^2 - 10x + 10] = 30$$
$$x^3 - 15x^2 + 60x - 50 = 30$$
$$x^3 - 15x^2 + 60x = 30 + 15$$
$$x^3 - 15x^2 + 60x + 10 = 90$$
Hence the answer is option C
$$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + ........}}}} =$$
Solution
short cut $$2 \times 3 = 6$$
Longer Method
$$\sqrt{6 + \sqrt{6}.....} = x$$
$$6 + x = x^2$$
$$x^2 - x - 6 = 0$$
$$(x - 3)(x + 2) = 0$$
$$x = 3, x = -2$$
x = -2 not possible
So answer is x = 3
Hence the answer is option C
What is the least value of x such that 43x4567 is divisible by 11?
Solution
43x4567 is divisible by 11
So, 4+x+5+7 - (3+4+6) is divisible by 11
So, x+16 - 13 is divisible by 11
So, x+3 is divisible by 11
So, x = 8
Hence the answer is option D
The number of positive integers between 1 and 2087 that are divisible by 31 is:
Solution
First no 31
Last no
31) 2087 (67
186
227
217
10
$$1 \rightarrow 67$$
Total no = 67
Hence the answer is option C
What is the largest four digit number that leaves a remainder 4 when it is divided by 36, 48 and 64?
Solution
No = LCM (36, 48, 64)k + 4
= 576k + 4
= 576) 9999 ( 17
576
4239
4032
207
9999 - 207 + 4
9999 - 203
= 9796
Hence the answer is option C
What is the minimum number of square tiles required to pave a floor of dimensions 24.78 meters x 22.26 meters?
Solution
HCF 2478 2226
252
$$2 \times 126$$
$$2 \times 18 \times 7$$
HCF = $$2 \times 3 \times 7$$
= 42
No.of tiles $$2478 \times 2226$$
42 $$\times$$ 42
$$\Rightarrow 59 \times 53$$
= 3127
Hence the answer is option B
$$\frac{3.61}{0.361} = \frac{361}{x} \Rightarrow x =$$
If the recurring decimal number $$1.\overline{27}$$ is equal to the rational number $$\frac{p}{q}$$ then p - q =
The difference between the largest and the smallest of the following rational numbers is:
$$\frac{3}{5}, \frac{3}{8}, \frac{2}{3}, \frac{4}{9}, \frac{1}{2}, \frac{6}{11}$$
What is the least fraction to be added to the sum of the following fractions to make it an integer?
$$3\frac{1}{3}, 6\frac{11}{12}, 4\frac{7}{36}, 5\frac{1}{12}, 7\frac{1}{4}$$
If the length of a rectangle is increased by 15%and the breadth decreased by 10% then the percentage change in its area is:
A spends 70% of his income and savesthe rest. If his income increases by 15% and expenses increase by 20% the percentage change in his saving will be:
A person buys anarticle for Rs.18000 and after one yearsells it for less than 25%of its cost price. Whatis the sale price (in Rs.) ?
A buys a piece of land for Rs. 75 lakhs andsells it to B at a profit of 10%. B in turn sells it to C at a profit of 20%. The amount of money paid by C (in Rs. lakhs) is:
A and B started a business with capitals of Rs. 12 lakhs and Rs. 24 lakhs respectively. A receives a monthly salary for running the business. At the end of the year they receive Rs. 1.80.000 each. What was the monthly salary of A (in Rs.)?
Solution
A’s capital : B’s capital = Rs. 12 lakhs : Rs. 24 lakhs = 1 : 2.
Hence after paying A’s fixed salary, the remaining profit must be shared in the ratio 1 : 2.
Let the monthly salary paid to A be $$x$$ rupees.
Salary for one year = $$12x$$ rupees.
Let the total profit of the firm before paying salary be $$P$$ rupees.
Amount left for distribution of profit after paying salary = $$P - 12x$$.
Share of A in this residual profit = $$\frac{1}{1+2}\,(P - 12x) = \frac{1}{3}\,(P - 12x)$$.
Share of B in this residual profit = $$\frac{2}{3}\,(P - 12x)$$.
According to the question, each partner finally receives Rs. 1,80,000.
For B (who gets no salary):
$$\frac{2}{3}\,(P - 12x) = 1,80,000$$
$$\Rightarrow P - 12x = 1,80,000 \times \frac{3}{2} = 2,70,000$$ $$-(1)$$
For A (who gets salary plus profit share):
$$12x + \frac{1}{3}\,(P - 12x) = 1,80,000$$ $$-(2)$$
Substitute the value of $$P - 12x$$ from $$(1)$$ into $$(2)$$:
$$12x + \frac{1}{3}\,(2,70,000) = 1,80,000$$
$$12x + 90,000 = 1,80,000$$
$$12x = 1,80,000 - 90,000 = 90,000$$
$$x = \frac{90,000}{12} = 7,500$$.
Therefore, A’s monthly salary is Rs. 7,500.
Option B which is: 7,500
A and B enter into a partnership business with capitals in the ratio 4:5 respectively. At the end of 4 months 4 withdraws and B continued. At the end of the business they share profits in the ratio 1:5. The number of months B’s Capital was used is:
2 pipes canfill a tank in 30 minutes and 36 minutes respectively. A third pipe empties the tank at the rate of 11 liters per minute. If all the three pipes are opened simultaneously it take $$22\frac{1}{2}$$ minutes for the tank to be full. The capacity of the tank in litres is:
Solution
Two pipes 4 and 8B canfill a tank in 2 hours and 3 hours respectively. If both the pipes are opened together the time required in munutes to fill the tank is:
Solution
A man is walking at a speed of 10 kmph. After every kilometer he takes rest for 5 minutes. How muchtime he takes in minutes to cover a distance of 7 kms?
Solution
OPTION E= 67
What is the time taken in hours by a cart to cover a distance of 0.9 km at 0.25 (meters per second) ?
Solution
R and S can finish a job in 10 days, S and M can finish it in 12 days and M and R can finish it in 15 days. In how many days will R. S and M together finish the job?
Solution
P can do a work in 12 days and O can do the same work in 16 days. Both start the work together and QO left after 4 days. How many more days are required by P to finish the work?
Solution
The length of a plot is 5 times is its breadth. A playground measuring 500 sq.m in occupied by one fourth of the total area of the plot. The length of the plot in meters is:
Solution
Area of the given figure.
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Solution
What is the volume of a cube (in cu.cms) whose total surface area in 726 sq cms?
Solution
The length, breadth and the height of a cuboid are in the ratio 6 : 5 : 4 and its total surface area is 33,300 sq.m. What is the length of the cuboid (in meters)?
Solution
If the length and breadth of a rectangular plot area 120m and 76m respectively, then the cost of fencing it at Rs.8 per metre (in Rs) is:
The area (in sq.cm) of a regular hexagon of side 4 cm is:
Two equal cubes of side of length 18 cm are joined together to form a cuboid. Then the total surface area of this cuboid (in sq.cm) is:
Solution
L=36, B=18 H=18
TSA=2(LB +BH+HL)
=2(36*18+18*18+18*36)
=2*1620=3240
$$\left\{x : \mid 8 - 4x \mid > 12 \right\} =$$
If $$5x + 7 \equiv 12(mod 18)$$, then the smallest integral value of such an x is:
If p, q are any two statements, the negation of the statement $$((\sim p \vee q) \wedge (p \vee \sim q)) \vee (p \wedge q)$$ is
For any two statements p, q a tautology amongthe following is:
Let A, B be subsets of a set X and let n(A) denote the number of elements in A. If $$n(A) = 15, n(A \cap B) = 4, n(A \cup B) = 21$$ then n(B - A) =
If a set X has four elements. then the number of relations on X that are symmetric is:
Let $$R = \left\{(m, n) \epsilon N^2 : m + n$$ is even}. Then the relation R on N is:
A line l which makes equal intercepts with the coordinate axes, passes through the point $$(3, - 4)$$. Then it also passes through the point:
If A(2, -3), B(3, 4) and C(-1, 5) are vertices of a triangle, then the length of median through A is:
If $$\sin \theta = \frac{5}{13}$$ and $$\theta$$ is not in the first quadrant, then $$\sec \theta + \tan \theta =$$
$$\sec \theta - \tan \theta = x \Rightarrow \frac{1 + x^2}{1 - x^2} =$$
$$A + B = \frac{\pi}{4} \Rightarrow (1 + \tan B)(1 + \tan B) =$$
The angles of elevation of a tower fromthe top and bottomofa building of height 30 meters are $$30^\circ$$ and $$60^\circ$$ respectively. Then the height of the tower(in meters) is:
$$x + \frac{1}{x} = 5 \Rightarrow x^3 + \frac{1}{x^3} =$$
$$3x - 4y = 7; xy = 5 \Rightarrow 9x^2 + 16y^2 =$$
The remainder when the polynomial $$7x^3 + 13x^2 - 19x - 20$$ is divided by x + 2 is:
If a polynomial in x leaves remainders -19, 17 respectively when divided by x + 4 and x - 5, then the remainder it leaves when divided by $$x^2 - x - 20$$ is
The value of k for which the system ky + 3y = k - 3 and 12x + ky = k of equations has infinitely many solutions is:
Solution
KX + 3Y = K -3
KX + 3Y - ( K -3) = 0 ------------(1)
And,
12X + KY = K
12X + KY - K = 0
These equations are of the form of A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0
Where,
A1 = K , B1 = 3 and C1 = -K +3
And,
A2 = 12 , B2 = K and C2 = -K
For no solution we must have,
A1/A2 = B1/B2 =C1/C2 [ Where # stand for not equal]
K / 12 = 3/K
$$K^2 = 12 × 3$$
$$K^2 = 36$$
$$K = \sqrt{36} = 6$$
A fraction becomes $$\frac{4}{5}$$ if 1 is added to both its numerator and denominator while it becomes $$\frac{1}{2}$$ if 5 is subtracted from its numerator and denominator. The fraction is
Solution
$$\frac{x+1}{y+1}$$=$$\frac{4}{5}$---------equation 1
$$\frac{x-5}{y-5}$$=$$\frac{1}{2}$$---------equation 2
from equation 1 and 2,
5x-4y=-1------ multiply with 2
2x-y=5----------multiply with 5
therefore
10x-8y=-2-----(1)
10-5y=25------(2)
by subtracting we get Y=9
subsitute y value in any equation we get X=7
If the sum of first 21 terms of an arithmetic progression is 609, then the $$11^{th}$$ term of progression is
Solution
Sn=n/2[2a+(n-1)d]
609=21/2[2a+(21-1)d]
by taking 2 common and eliminate 2
609/21=a+10d i.e a11
a11=29
The harmonic mean of $$\frac{1}{x}$$ and $$\frac{1}{y}$$ is
Solution
If X1,X2,X3 are in harmonic mean
harmonic mean=$$\frac{n}{x1}±\frac{n}{x2}±\frac{n}{x3}$$
Therefore harmonic mean of $$\frac{1}{x}$$ and $$\frac{1}{y}$$
is $$\frac{x+y}2$
The coefficient of $$x^{12}$$ in the expansion of (x - 1)(x - 2)(x - 3)......(x - 13) is
Solution
The sum of all the coefficients in the expansion of $$(1 - 2x + 3x^2)^9$$ is
Solution
$$(1 - 2x + 3x^2)^9$$
substitute x=1
$$(1 - 2\times1 + 3\times1^2)^9$$
$$(1 - 2 + 3)^9$$
$$(2)^9$$
$$A = \begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \Rightarrow A^{2019} =$$
Solution
$$A = \begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \
by help of $$$$$$A = \begin{bmatrix}a1 & a2 & a3 \\b1 & b2 & b3\\c1 & c2 & c3 \end{bmatrix} \
$$$$$$A = \begin{bmatrix}a2*b3+b2*a3 & a1*c3+c1*a3 & a1*b2+a2*b1 \\\end{bmatrix} \
$$$$$$A = \begin{bmatrix}1*1+1*1 & 1*1+1*1 & 1*1+1*1 \\\end{bmatrix} \
$$$$$$A = \begin{bmatrix}2 & 2 & 2 \\\end{bmatrix} \
$$$$$$A = 2\begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \
A^{2019} = 3^{2018}\begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \
A^{2019} = 3^{2018}\begin{bmatrix}A \end{bmatrix} \
$$A = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}, B = \begin{bmatrix}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0 \end{bmatrix} \Rightarrow A^{99} B^{100} + B^{99} A^{100} =$$
$$\lim_{x \rightarrow 0} \frac{\sin x - x}{x^3} =$$
Solution
If $$y = x \sin x$$, then at $$x = \frac{\pi}{2}, \frac{dy}{dx} =$$
Solution
$$y = x \sin x$$, then at $$x = \frac{\pi}{2
$$y = x \sin x$$, then at $$x = \frac{180}{2}
\frac{dsin90}{dx}
1
In a triangle $$ABC, \angle A = \angle B = \angle C$$. The bisectors of the angles $$\angle B$$ and $$\angle C$$ interect at D. Then $$\angle BDC =$$
Solution
In a $$\ \triangle\ $$ ABC,
$$\ \angle\ $$A+$$\ \angle\ $$B+$$\ \angle\ $$C=180$$\ ^{\circ\ }$$
$$\angle\ $$A=$$\angle\ $$B=$$\angle\ $$C=60$$^{\circ\ }$$
since, bisector of $$\angle\ $$B and$$\angle\ $$C meet at D
$$\therefore\ $$ $$\angle\ $$DBC=$$\angle\ $$DCB=30$$^{\circ\ }$$
In$$\triangle\ $$DBC,
$$\angle\ $$DBC+$$\angle\ $$DCB+$$\angle\ $$BDC=180$$^{\circ\ }$$
30$$^{\circ\ }$$+30$$^{\circ\ }$$+$$^{ }\angle\ $$BDC=180$$^{\circ\ }$$
$$\angle\ $$BDC=120$$^{\circ\ }$$
If the diagonals of a rhombus are 9 cm and 12 cm, then the perimeter of the rhombus(in cm) is:
Solution
Let ABCD be rhombus with AC and BD it's diagonals bisecting each other at O ,where
AC=9cm,AO=OC=4.5cm
BD=12cm, BO=OD=6 cm
In $$\triangle\ $$AOB,
$$\angle\ $$AOB=90$$^{\circ\ }$$(we know diagonals of rhombus bisect each other at 90$$^{\circ\ }$$)
Apply Pythagoras theorem,
$$AO^{ 2}$$+$$OB^{2 }$$=$$AB^{ 2}$$
$$4.5^{ 2}$$+$$6^{ 2}$$=$$AB^{ 2}$$
20.25$$\ +\ $$36=$$AB^{2 }$$
$$AB^{ 2}$$=56.25
AB=7.5
Perimeter of rhombus=4$$\times\ $$7.5=30
Let P be an external point to a circle and PT be a tangent drawn from P meeting the circle at 7. If AB is a chord ofthe circle such that A, B, P are collinear, PT = 5 cm and PB = 4 cm then PA =
Solution
Using tangent secant theorem,
$$PT^{2 }$$=PA$$\times\ $$PB
5$$\times\ $$5=PA$$\times\ $$4
PA=$$\ \frac{25}{ 4}$$=6.25
If the area of the triangle with vertices (1, 2), (2, 3) and (x, 4) is 40 sq. units then the value of $$\mid x - 3 \mid$$ is:
Solution
Consider a traingle ABC,Let A(1,2)=$$\ (x_1,y_1)$$
B(2,3)=$$\ (x_2,y_2)$$
C(x,4)=$$\ (x_3,y_3)$$
Area of traingle=
$$\ \frac{1 }{ 2}$$ [$$\ x_1*(y_2$$-$$y_3)$$+$$\ x_2*(y_3$$-$$y_1) $$+$$\ x_3*(y_1$$-$$y_2) $$]
40=$$\ \frac{1 }{2 }$$[1*(3$$\ -\ $$4)+2*(4$$\ -\ $$2)+x*(2$$\ -\ $$3)]
80=$$\ -\ $$1$$\ +\ $$4$$\ +\ $$x($$\ -\ $$1)
x=$$\ -\ $$77
$$\mid x-3 \mid$$ = $$\mid -77-3\mid$$=80
Area (in sq. units) of the quadrilateral ABCD with vertices A(2, -1), B(4, 3), (-1, 2), D(-3, -2) is:
Solution
Area of quadrilateral ABCD= Area of $$\triangle\ $$DAB+ Area of $$\triangle\ $$BCD
Area of $$\triangle\ $$DAB=$$\ \frac{1 }{ 2}$$*[$$\ -\ $$3($$\ -\ $$1$$\ -\ $$3)+2(3$$\ -\ $$($$\ -\ $$2))+4($$\ -\ $$2$$\ -\ $$($$\ -\ $$1))]
=$$\ \frac{1}{ 2}$$[12$$\ \ +\ $$10$$\ -\ $$4]
=9sq.units
Area of $$\ \triangle\ $$BCD= $$\ \frac{1 }{ 2}$$*[4(2$$\ -\ $$($$\ -\ $$2))+($$\ -\ $$1)($$\ -\ $$2$$\ -\ $$3)+($$\ -\ $$3)(3$$\ -\ $$2)]
=$$\ \frac{1 }{2 }$$*[16$$\ +\ $$5$$\ -\ $$3]=9sq.units
Area of quadrilateralABCD=9$$\ +\ $$9=18sq.units
Arithmetic mean of 56, 57, 58, ..., 100 is:
Solution
Arithmatic mean=$$\ \frac{sumof observations }{ total no. Of observations}$$
sum of first n natural no.=$$\ \frac{n(n+1)}{ 2}$$
sum of first 100 natural no.=$$\ \frac{100*101 }{2 }$$=5050
sum of first 55 natural no.=$$\ \frac{55*56}{2 }$$=1540
sum of 56,57,...…,100=5050$$\ \ \ -\ $$1540
=3510
Arithmatic mean=$$\ \frac{3510 }{45 }$$=78
The median of some observations is 12. If two additional observations 10, 15 are added to these observations. then the median of the new series
of observations is:
If the arithmetic mean of some observations is 26.5 and their median is 25.4. then their mode is:
Solution
.
Arithmetic mean=26.5
Median=25.4
Using empirical formula,
mode=3$$\times\ $$median$$\ -\ $$2$$\times\ $$mean
=3$$\times\ $$25.4$$\ -\ $$2$$\times\ $$26.5
=76.2$$\ -\ $$53
=23.2
Standard deviation of first 2n natural numbers is:
Forthe ten observations X1, X2, ......, X10 it is given that $$\sum_{i = 1}^{10} x_i = 12$$ and $$\sum_{i = 1}^{10} x_i^2 = 18$$ Then the standard deviation of the data is:
Solution
Mean($$\overline{x}$$)=$$\frac{1}{n}$$ $$\Sigma_{i=1}^{10}x_i\ $$=$$\ \frac{12 }{10 }$$=1.2
Standard deviation= $$\sqrt{\frac{\Sigma_{i=1}^{10}x_i^{2}{n}}}\$$
If the sum of the squares of the deviations of ranks of two students in seven subjects is 21. then the coefficient of rank correlation is:
Solution
here we have
number of subject n = 7
square of sum of deviation are $$ \sum_di^2$$= 21
now we know the fourmula of the coefficient of rank correlation is rR = 1- $$\times {n-1}\sum_di^2\frac\times{n}{n^2-1}$$
the coefficient of rank correlation is rR = 1- $$\times {7-1}\sum_21 \frac\times{7}{7^2-1}$$
rR = 1- $$\frac{3}{8}$$
rR= $$\frac {5}{8}$$ answer
A problem in mathematics is given to three students P, Q and R whose chances of solving it are $$\frac{1}{3}, \frac{1}{4}$$ and $$\frac{1}{2}$$ respectively. Then the probability that the problem will be solved is:
Solution
probility of solving problems by P, is= $$\frac{1}{3}$$
probility of not solving the problem by P , is= $$\frac{2}{3}$$
probility of solving problems by q, is= $$\frac{1}{4}$$
probility of not solving the problem by Q , is= $$\frac{3}{4}$$
probility of solving problems by r, is= $$\frac{1}{2}$$
probility of not solving the problem by R , is= $$\frac{1}{2}$$
so the probility of solving the problem are =problem is solved by p * problem is not solved by q and r + problem is solved by q * problem is not solved by p and r + problem is solved by r * problem is not solved by p and q + problem is solved by p and q and not solved by r + problem is solved by q and r and not solved by p + problem is solved by p and r and not solved by q + problem is solved by all three of them
P{solved}= $$\frac{1}{3}$$* $$\frac{3}{4}$$*$$\frac{1}{2}$$ + $$\frac{1}{4}$$* $$\frac{2}{3}$$*$$\frac{1}{2}$$ + $$\frac{1}{2}$$* $$\frac{3}{4}$$*$$\frac{2}{3}$$ + $$\frac{1}{3}$$* $$\frac{1}{4}$$*$$\frac{1}{2}$$ + $$\frac{1}{4}$$* $$\frac{1}{2}$$*$$\frac{2}{3}$$ + $$\frac{1}{3}$$* $$\frac{1}{2}$$*$$\frac{3}{4}$$ + $$\frac{1}{3}$$* $$\frac{1}{4}$$*$$\frac{1}{2}$$
p{soleed}= $$\frac{18}{24}$$
p{solved}=$$\frac{3}{4}$$ answer
or
probility of not solving problems by P,Q and R are = $$\frac{2}{3}, \frac{3}{4}$$ and $$\frac{1}{2}$$ respectively
so the probility of problem is solved is = 1-probility of not solving problems by P,Q and R are = $$\frac{2}{3}, \frac{3}{4}$$ and $$\frac{1}{2}$$
=1- $$\frac{2}{3}* \frac{3}{4}$$* $$\frac{1}{2}$$
=1-$$\frac{6}{24}$$
so the probility of problem is solved is = $$\frac{3}{4}$$
Four dice are thrown simultaneously. The probability that all of them show the same numbers is:
Solution
when 4 dices are thrown the number of total event occurs = 6*6*6*6= 1296
showing same number event are ={1,1,1,1] {2,2,2,2],{3,3,3,3],{4.4.4.4},{5,5,5,5},{6,6,6,6}
total number of favorable events are =6
Probility of getting same number on four dices are = $$\frac{number of favorable cases}{total number of events occurs}$$
P{e4}= $$\frac{6}{6^4}$$
P{e4}=$$\frac{1}{6^3}$$ answer
If two cards are drawn at randomfrom a well shuffled pack of 52 cards, then the probability of getting both cards from same suit is:
Solution
total number of event occurs when 2 card are drawn from pack of 52 cards = 52*51= 2652
number of suits are there in packs of cards = 4
in a suit to total number of cards are there = 13
total number of favorable events when two cards are drawn from suits = 13*12*4 =156*4
now the probility of getting both cards are from same suits are =$$ \frac{total number of favorable events}{total number of event occurs when 2 card are drawn from pack of 52 cards}$$
$$p{e2}=\frac{156*4}{2652}$$
$$p{e2}=\frac{4}{17}$$ answer
When two dice are rolled at random, then the probability of getting same prime number on both dice is
Solution
now the total number of event occurs when two unbiased dice are thrown =36
we have to choose prime number which are 2,3 and 5
so the favorable outcomes pair are $${{2, 2}}$$, pair $${{3, 3}}$$,pair $${{5, 5}}$$
then the probility of getting same prime number are
$$P{ {same- prime}}$$=$$\frac{number of favorable outcomes}{total number of possible outcomes}$$
$$p{ {same- prime}}$$=$$\frac{3}{36}$$
$$p {{same- prime}}$$=$$\frac{1}{12}$$ answer
Choose the correct meaning for the word given:
Slipshod
Recalcitrant
Placid
Frugal
Conciliate
Wholesome
Fill in the blank choosing the correct word:
A printed list of questions to be answered by a number of people is called ..............
Many would not prefer to live a ............. life like that of a hermit.
The old people in the village still ......... the old traditions.
Although he is shy it certainly has not ............ his career in any way
Choose the correct answer
A storage area in memory that stores information temporarily while in use is known as
Pictorial representation of programming procedure is called
An inter connection of two or more computing devices is generally known as
Which of the following is not a guided transmission media?
The storing and accessing of data and programs over the internet instead of on another type of hardware drive is known as
When do you conduct an induction of a new employee in an organisation?
A report of accounts that represents ending balance of each account in the list at the end of the accounting period is known as
The economic concept that each consecutive unit of good consumed or resource used. is less useful or productive than the preceding one is known as
One of the assumptions in economics is that human wants are
A legal document by which a bank lends money at an interest in exchange for the title of a debtor’s property is known as
A: "Look at what you have done! You have knocked me down with your cycle.”
B: “Well you should be more careful while crossing the road.” ‘B’ is
A: “Myfriend went through many ups and downsbefore he could reach the present stage."
B: “Patience has its rewards, afterall”. ‘B’ is
Change the following sentence into active voice:
My pocket has been picked.
“You had better hold your tongue.” The underlined expression means
“Membership in the sports club can cost you an arm and a leg.” The underlined expression means, it is
There is so much red tape involved in getting official permits. The underlined phrase means
“Nobody appreciated the manager throwing his weight around”. underlined expression means
Fill in the blanks with the appropriate phase/verbs/ preposition:
I must protest ............ this course of action
He delights ......... being helpful.
............ being a gentleman, he was also a reputed singer.
I ............... waiting here for the last three hours.
I wish she .......... dead.
You are walking too fast. I can’t ......... with you.
I love to ......... the stars in the sky at night.
A crisis can ............. the best in someone.
Read the following passage and answer questions
To avoid the various foolish opinions to which mankind is prone, no superhuman genius is required. A few simple rules will keep you, not from error, but from silly error!
If the matter is one that can be settled by observation, make the observation yourself. Aristotle could have avoided the mistake of thinking that women have fewer teeth than men, by the simple device of asking Mrs. Aristotle to keep her mouth open while he counted. He did not do so because he thought he knew.
Thinking that you know when in fact you don’t is a fatal mistake to which we are all prone. I believe myself that hedgehogs eat black beetles, because I have been told that they do: but if I were writing a book on the habits of hedgehogs I should not commit myself until I had seen one enjoying this unappetizing diet. Aristotle. however, was less cautious. Ancient and medieval authors knewall about unicorns: not one of them thought it necessary to avoid dogmatic statements about them because he had never seen one of them.
What do humans require to avoid being dogmatic?
What was a common personality trait in Aristotle?
What does the author believe in?
What would the author do if he had to write a book onthe eating habits of hedgehogs?
According to the author, what did the ancient and medieval authors feel about their observations?
Read the following passage and answer questions
Most managers don’t do things in the order of priority that they have rationally selected. They do things according to feelings. That’s how their day is run. Professional managers fall into two categories. There are doers and there are feelers. Doers do what needs to be done to reach a goal that they themselves have set. They come to work having planned out what needs to be done. Feelers. on the other hand. do what they feel like doing. Feelers take their emotional temperature throughout the day. checking in on themselves, figuring out what they feel like doing right now. Their lives, their outcomes, their financial security are all dictated by the fluctuation of their feelings. Their feelings will change constantly. of course, so it’s hard for a feeler to follow anything through to a successful conclusion. By contrast. a doer has high self-esteem. A doer enjoys many satisfactions throughout the day. even though some of them were preceded by discomfort. A feeler is almost always comfortable, but never really satisfied. A doer experiences more and more power every year of his life. A feeler feels less and less powerful as the years go on.
What is the major shortcoming in most managers?
Who is a ‘feeler’?
What is the meaning of “emotional temperature” ?
What does a ‘doer’ have?
What does a ‘feeler’ experience as the years go by?
Read the following passage and answer questions:
It requires a sense of superiority assurance, and self-confidence, to write about bores at all, except as one of them. But since a true bore is always unconscious of his ‘borishness’, and indeed usually thinks of himself as the most companionable of men. to write as one of them is to acquit oneself of the stigma.
Bores are happy largely because they have so much to tell and they can find people to tell it to. The tragedy is they can always find their listeners. me almost first And why can they? Why can even notorious bores always be sure of an audience? The answer is the ineradicable kindness of human nature. Few men are strong enough to say “For Heaven’s sake go away. you weary me’
One of the bore’s greatest assets is his simplicity which disarms. Astute crafty men are seldombores: very busy men are seldom bores.
Which quality is NOT required to write about bores?
Which of the following is NOT true of a true bore?
Why are bores happy?
Which quality stands out in a bore?
What does ‘ineradicable’ mean?
