Howmany teachers teach Physics?
I. There are 11 teachers who teach Mathematics or Physics
II. Of these 11 teachers 7 teach Mathematics and 3 teach both Mathematics and Physics
TS ICET 2019 Question Paper Shift-1 (23rd May)
In questions numbered 1 to 20, a question is followed by data in the form of two statements labeled as I and II. You must decide whetherthe data given in the statements are sufficient to answer the questions. Using the data make an appropriate choice from(1) to (4) as perthe following guidelines:
Is the line PQ parallel to X-axis?
I. Coordinates of P are (4, 5)
II. Coordinates of Q are (2, 3)
Whatis sumofinterior angles of the polygon?
I. Polygonis a closed figure
II. The polygonhas7 sides
For integers a and b is $$(a^3 + b^3)^{\frac{1}{3}}$$ an integer?
I. $$a^3 + b^3$$ is an even integer
II. $$a^3 + b^3$$ is equal to the volume of a box with dimensions 12cm, 18cm and 125cm
ABC Dissquare inscribed in a circle. What is the area of the square?
I. Radius of that circle is 1 unit
II. The centerofthat circle is (0. 0)
Amongthe five rivers $$R_1, R_2, R_3, R_4 and R_5$$ whichoneis the short river?
I. $$R_3$$ is the longest river and $$R_1$$ shorter than $$R_2$$ but longer than $$R_3$$.
II. $$R_4$$ is a little shorter than $$R_2$$ and a little longer than $$R_1$$.
Is the positive integer 7 a multiple of 175?
I. 125 divides n
II. 35 divides n
Whatis the integral value of x ?
I. 10 < x < 20
II. 10 < 9 x + 1 < 20
Is N divisible by 48?
I. $$N = (a - 4)((a -3)(a - 2)(a - 1), a \geq 5$$
II. N is divisible by 8 and 6
Are x and y rational numbers?
I. x + yis a rational number
II. x y is a rational number
Is x = 93y6 divisible by 9
I. y is divisible by 9
II. y is divisible by 3
At how manypoints do the twocircles intersect?
I. The distance betweentheir centres is 15 cm
II. The radii of the two circles are equal
What is the digit in the units place of n?
I. $$n = 6^k, k$$ is an integer $$\geq 1$$
II. $$n = 8^k, k$$ is an integer $$\geq 1$$
If a and b are real numbers, is a > b?
I. b is the real root of $$ x^3 - 1 = 0$$
II. a is the smaller of the roots of $$x^2 - 1 = 0$$
If $$A$$ ≠ $$\phi$$ and $$R$$ is relation on set $$A$$, then is $$R$$ Transitive?
I. $$R = A \times A$$
II. $$ R \subseteq A \times A$$
Whatis the selling price of each article?
I. The cost price of eacharticle is Rs.500
II. The loss on one article is equal to the gain on other article
Whatis the shortest distance between villages X and Y?
I. Village X is to the north of Village Z at a distance of 35 km
II. Village Y is at a distance of 20km from Z
Howis Y related to X?
I. Y is the spouse of a parent of X
II. X is the brother of Y’s daughter
X and Yare in partnership business for one year. At the end of the year, a profit of Rs.90.000 was earned. What is Y’s share in the profit?
I. Y invested Rs.1.40,000
II. X withdraw his capital after 9 months
Whatis the relative speed of the boy swimming against the current?
I. Speed of the boy instill waters is 5 kmph
II. Speed of the current in 2 kmph
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.
ACE : FHJ :: LNP:______________
Telangana : Hyderabad: : Jhakhand:_____________
__________:CXDW:: EVFU: GTHS
A9Z: B7Y::_________: D3W
$$\frac{P}{H} : .......... :: \frac{U}{C} : 7$$
___________:L3M4: : P5Q6: R7S8
GAK : 7A11: :PAT:______
LAKE:______: : MILK : HDGF
AROUND:____________: : POLITY : SROLWB
20736: 144::_________ : 288
In the following questions pick the odd thing out:
Each of the questions follow a definite pattern. Observe the same andfill in the blanks withsuitable answers.
$$1, 2, \frac{27}{8}, 4, \frac{125}{32}$$
(6, 8, 7), (10, 12, 17), (14, 16,.15), ............., (22, 24, 23)
$$22\frac{2}{9}, 25, 28\frac{4}{7}, 33\frac{1}{3}, ............$$
256, 4, 196, 16, 144, 36, 100, 64, ...........
$$-\frac{1}{6}, \frac{1}{3}, 1\frac{1}{3}, 2\frac{5}{6}, 4\frac{5}{6}, 7\frac{1}{3}, .............$$
3, 45, 35, 63, 99, 143, ..........
2, 10, 30, 68, 130, 222, .........
24, 35, 48, 63, 80, 99, ..........
(11, Z, 13) : (15, X, 17) : (19, V, 21) : (23, T, 25) : (27, R, 29) : .............
IJLO, GHJM, EFHK. CDFI, ABDG, ............
The results of a study conducted on the Academic background of students (in thousands) who took admission into MBA/MCA in the Six Years 2013 to 2018 in a University is given in the following table. Based on this data. answer the questions
What is the percentage of the science students who took admission into these programmes inall the years put together?
What is the approximate percentage of students from Arts who took admission into the programme in the year 2016 when comparedto all the Arts students admitted inall the years put together?
What is the average number of commerce students per year who took admission into this programme?
The following Pie Charts show the revenue and expenditure of a public sector undertaking under 5 major heads of account. Based on this data, answer questions
In the Pie chart for the Revenue, what is the difference between the sectorial angle of odd and even heads of accounts?
If the total expenditure is Rs.5000 crores. whatis the difference between the expenditure (in Rs. crores) under heads D and E puttogetherand that of the rest the heads together?
If the Revenue under head II is Rs.25 crores. then the total revenue under the heads of I and IV together? (in crores of Rs)
In a particular year the expenditure under the head E is Rs.1240 Crores. Then in the same year, the difference in the expenditure under the heads B and D (in Rs. crores) is
What is the percentage of the expenditure under the head A. when compared to the expenditure on the rest of the heads?
In the following venn diagram square represents Females, triangle represents corporate managers, circle represents IITians and rectangle represents MBAs
What is the numberof female corporate managers who did MBA?
How many male IITians are there?
The English Alphabet, a code is designed bythe following rules:
1. The vowels are rotated cyclically by moving each vowelto its right by one place.
2. Each consonant from B to Mis moved to the mght by one place cyclically.
3. Each consonant from N to Z is movedto left by one place cyclically.
4. The reverse process is followed for decoding.
Whatis the code for ‘YADADRI’?
Whatis the code for ‘EMAIL’?
Whatis the code for the word ‘BUCKET’?
Which word is coded as ‘FUBEOZ’?
Which one of the following is coded as 'NUMODX’?
A wordor a group ofletters to be coded or decoded based on the certain codes. Using the information, answer the questions.
In a certain code. if the word CRIMEis coded as ‘APGKC’, then the code for ‘QUICK’ is.
In a certain code, if ‘BEAT’ is coded as ‘ADZS’ then the code for ‘SONAR’is
If ‘CARTOON’is coded as ‘NOOTRAC’ then whichof the following is coded as ‘DETIMIL’?
If ‘CUPBOARD’ is coded as ‘UCBPAODR’ then the code for ‘ELEVATER’ is
If ‘WATER’is coded as ‘XAUES’ thenthe code for ‘MOBILE’ is
For the following questions answer them individually
If today is Sunday then what dayit was in the weak 124 days ago?
At 5:15 p.m the angle between the hours hand and the minute’s hand of wall clock (in degrees) is
A Clock gains 10 minutes in 24 hours. If the clock is set at the right time at 8:00 a.m on Sunday, the actual tume when the clock shows 1:00 p.m on Monday?
In a family B is the father of A; C is the wife of B; D is the mother of C and E is the husband of D. Then A is related to E as:
A family has a man. his wife. their four sons and their wives. The first son has three sons and one daughter. the second son has two sons and three daughters. the third and fourth sons have three daughters each. The total number of female members in the whole family is.
A superfast express travelling at 100 kmph takes 2 hours more than another train travelling at 120 kmph to cover the distance betweenthe same two stations (in kms). Then the distance between the stations (in kms)is.
A, B, C, D, E, F and G are sitting around a circular table and facing the centre: G is second to the left of C, C is to the immediate left of F, A is third to the left of E, B is between D and E. Which of the following is true?
If * is defined by $$a * b = a^3 + b^3 - 3ab$$ for $$a, b \epsilon R$$, then $$\frac{(1 * 2) * (1 * 2)}{(1 * 1) * (1 * 1)} =$$
If $$x \downarrow y = (x + y) + (x - y)^2$$ and $$x \uparrow y = (x + y)^3 - (x - y)^3$$ then the value $$\frac{(0 \downarrow 1) \uparrow (0 \downarrow 1)}{(1 \uparrow 2) \downarrow (1 \uparrow 2)} =$$
For ab ≠ 0. if * is defined by $$a * b = \frac{a^2 + b^2}{ab}$$ then $$\frac{(1 * 2) * (2 * 1)}{(1 * 3) + (3 * 1)} =$$
If $$a = 2^x, b = 4^y, c = 8^z$$ and $$a^{yz} . b^{zx} . c^{xy} = 4^6$$, then $$xyz = ...........$$
Solution
given that
$$a = 2^x, b = 4^y, c = 8^z$$ and
$$a^{yz} . b^{zx} . c^{xy} = 4^6$$ put the value of a,b and c
$$ 2^{xyz} . 4^{yzx} . 8^{zxy} = 4^6$$
$$ 2^{xyz} . 4^{yzx} . {2$$\times$$4}^{zxy} = 4^6$$
$$ 4^{xyz} . 4^{yzx} . 4^{zxy} = 4^6$$
$$ 4^{3zxy} = 4^6$$ when base are same then power will be equal
3xyz = 6
xyz = 2 Answer
$$x^{x\sqrt{x}} = (x\sqrt{x})^{6x} \Rightarrow x =$$
Solution
given that
$$x^{x\sqrt{x}} = (x\sqrt{x})^{6x}$$
when base are same power will be equal
X^3x\2 = (x^3\2)^6x
x^3\2 = (3\2)6x
x = 81 Answer
$$A + B + C = 427, 3A = 4B = 7C \Rightarrow A =$$
Solution
Share
given that
$$A + B + C = 427, 3A = 4B = 7C \Rightarrow$$
B = A3\4 and C = A3\7 put the value
A + A3\4 + A3\7 = 427
(28A + 21A + 12A)\28=427
A = 196 Answer
There are 50 paise and 25 paise coins in a bag whose value is Rs.45. If the total numberof coins is 100, then the number of 50 paise coins is
Solution
let we have x number of 50 paise cion and y number of 25 paise coin
then given
$$x\cdot\frac{1}{2}+y\cdot\frac{1}{4}=45$$ and x + y = 100
on solving both equation we get
x = 80 and y = 20
$$\frac{2}{\sqrt{10 + 2\sqrt{21}}} - \frac{1}{\sqrt{12 + 2\sqrt{35}}} - \frac{1}{\sqrt{8 + 2\sqrt{15}}} =$$
Solution
given that
= $$\frac{2}{\sqrt{10 + 2\sqrt{21}}} - \frac{1}{\sqrt{12 + 2\sqrt{35}}} - \frac{1}{\sqrt{8 + 2\sqrt{15}}} $$
we know that $$\sqrt{21}$$ = 4.58 , $$\sqrt{35}$$ = 5.91 , $$\sqrt{15}$$ = 3.87
put the value in the equation we get
= $$\frac{2}{\sqrt{19}} - \frac{1}{\sqrt{19}} - \frac{1}{\sqrt{19}} $$
= 0 Answer
$$x = \frac{2\sqrt{2} + \sqrt{7}}{2\sqrt{2} - \sqrt{7}} \Rightarrow x - \frac{1}{x} =$$
Solution
given that
$$x = \frac{2\sqrt{2} + \sqrt{7}}{2\sqrt{2} - \sqrt{7}}$$
then 1\x = $$\frac{2\sqrt{2} - \sqrt{7}}{2\sqrt{2} - \sqrt{7}}$$
then x - 1\x = 8 + 7 - 15 + 2 $$\times$$ 4 $$\sqrt{14}$$
then $$8\sqrt{14}$$ Answer
What is the remainder when $$2^{45}$$ is divided by 5 ?
Solution
given that
$$2^{45}$$\5 = 2 $$\times$$ $$2^{44}$$\5 = 2$$\times$$ $$2^{4}$$ $$\times$$ $$2^{11}$$ \5
$$2^{4}$$ $$\times$$ $$2^{11}$$ \5 = 1
then
2 is the remainder Answer
What is the total number of positive divisors of 12600 ?
Solution
Setup the equation for determining the number of factors or divisors.
The equation is= d(n) = (a + 1)(b + 1)(c + 1)(d + 1)
Where d(n) is equal to the number of divisors of the number and a, b, etc. are equal to the exponents of the prime factorization.
Now substitute the letters in the equation with the the exponents of your prime factorization and then solve to calculate the total number of divisors.
12,600 = 23 x 32 x 52 x 71
d(n) = (a + 1)(b + 1)(c + 1)(d + 1)
d(12600) = (3 + 1)(2 + 1)(2 + 1)(1 + 1)
d(12600) = (4)(3)(3)(2)
d(12600) = 72 Answer
The G.C.D of two numbers is 42 and their L.C.Mis 1260. If one of the numberis 210, the other number is
Solution
let the numbers are $$N_{1}$$and $$N_{2} = 210
then we know that
$$N_{1}$$ $$\times$$ $$N_{2}$$ = G.C.D $$\times$$ LCM
put the value
$$N_{1}$$ $$\times$$ = ( 42 $$\times$$ 1260)\210
$$N_{1}$$ $$\times$$ = 252 Answer
If the G.C.D of 28 and 49 is expressed as 28x + 49y, then a pair (x , y) =
Solution
we know the G.C.D. of 28 and 49 is = 7
then we have given
28x + 49y = 7
but trial and error we have
28.2-49 = 7
7 = 7
X = 2 and y= -1 Answer
What is the value of the rational number which is the sum of the recurring decimals: $$0.\overline{6} + 0.\overline{7} + 0.\overline{8}$$
Solution
given that
= $$0.\overline{6} + 0.\overline{7} + 0.\overline{8}$$
= (6/9) + (7/9) + (8/9)
= 21/9
7/3 Answer
$$8\left[24 - \left\{25 - (17 - \overline{4 + 7}) + 5 \right\} + 2\right] =$$
Solution
given that
= $$8\left[24 - \left\{25 - (17 - \overline{4 + 7}) + 5 \right\} + 2\right] $$
= $$8\left[24 - \left\{25 - (17 - 9) + 5 \right\} + 2\right] $$
= 8[24 - {25 - 1} + 2 ]
= 8[24 - 24 + 2]
= 8$$\times$$2
= 16 Answer
The ascending order of the following numbers is
$$a = 4\sqrt{2}; b = 2\sqrt{5}; c = 3\sqrt{2}; d = 2\sqrt{3}$$
Solution
we knoe the valeu of $$\sqrt{2}$$ = 1.414 , $$\sqrt{3}$$ = 1.7304 , $$\sqrt{5}$$ = 2.23606 put the value we get the sequence d,c,b,a Answer
Which of the following statements is true?
Solution
we know that the value of $$\sqrt{5}$$ = 2.23606 , $$\sqrt{6}$$ = 2.4494 , $$\sqrt{11}$$ = 3.3166 then we get
$$6\sqrt{5}$$ > 4$$\sqrt{11}$$> 5 $$\sqrt{6}$$ Answer
Due to reduction of $$6\frac{1}{4}\%$$ in the price of onion a person was able to buy 1 kg more for Rs. 120. The original price of onion per kg in Rs. is
Solution
let the original price of the onion is X Rs
then reduced price is = X - 25X\400 = 15X\16
it means now he able to buy 16 kg instead of 15 kg due to reduction
so the original price is = 120\15 = 8 Rs\kg Answer
5% of income of A is equal to 15% income of B and 10% income o fB is equal to 20% income of C. If the income of C is ₹ 20,000, then the total income of A, B and C (in thousands if Rs.) is
Solution
let the income of A = x Rs , B = y RS and C = 20000 Rs
given that
x $$\times$$5%} = y $$\times$$ 15% equation 1
and 10% $$\times$$ = 20% $$\times$$ 20000 equation 2
solve eq 2 we get y = 40000 Rs
put value of y in eq 1 we get x= 12000 Rs
so the total income is = 120000 + 40000 + 20000 = 180000 Rs Answer
The cost price of 20 books is equal to the sale price of 16 books. What is the outcome of the transaction?
Solution
Let us assume that cost price of one Book is Rs1
So, The Cost price of 20 Books is = Rs 20
Then,
Selling price of 16 Books = cost price of 20 Books => Rs 20
Now,
SP>CP
Thus there will be profit
Profit = 20-16 => 4
Profit % = 4/16 × 100 = 25 % Answer
Anarticle is sold for a profit of 20%. Had it been sold fora profit of 25% it would have fetched ₹.50 more. The cost price of the article (in ₹ ) is
Solution
let the cost price of the article is x Rs
then 120% of x = 1.2x
and 125% of x = 1.25 x
then in given in question 1.25x - 1.20x = 50
0.05x = 50
x = $$\frac{50}{5}$$ $$\times$$ 100
x = 1000 Rs Answer
A and B entered into a business with capital ₹80.000 and ₹1,20,000 respectively. After 4 months each invested ₹20,000 more. The difference in the share of their profits in the year end profit is ₹68,000 is (in Rs) is
Solution
A and B starting capital respectively 80000 Rs and 120000 Rs
and total capital of A = 80000$$\times$$ 4 + 80000 $$\times$$ 8 = 320000 + 800000 = 11200000
total capital of B = 1200000$$\times$$ 4 + 1200000 $$\times$$ 8 = 480000 + 1120000 = 16000000
then A : B = 11200000 : 1600000
A : B = 112 : 160 = 7 : 10
given that X = 68000\17 = 4000
then the difference in the profit = ( 10 - 7 ) $$\times$$ 4000 = 12000 Rs Answer
X and Y invested in a business with their capitals respectively in the ratio of 3:2. After deducting 10% of the total profit for taxes if Ys share of profit is Rs. 6.480, then the total profit (in thousandsof Rs.) is
Solution
let the total profit is = x Rs
then after the 10% deducting the taxes the profits is = 0.9x
then the ratio of given profit in between A and B = 3 : 2
then in according to the question
0.9x $$\times$$ $$\frac{2}{5}$$ = 6480
x = 72 $$\times$$250
x = 18000 Rs Answer
Two pipes A and B can fill a huge tank in 6 hours and 8 hours respectively. If both the pipes are opened simultaneously and the B is tumed off after $$1\frac{1}{2}$$ hours then the time required for the tank to be full (in hours) is
Solution
given that
pipes A and B can fill a tank in 6 hours and 8 hours respectively
so the tank capacity is LCM of 6 and 8 = 24
efficiency of A is = 24\6 = 4
efficiency of B is = 24\8 = 3
together they can fill in 1 hour = 7
then in 1.5 hour is = 10.5
then the remaining part is = 13.5
so time require to A to fill that = 13.5\4 = 27\8
so the total time is = 27\8 + 3\2 = $$4\frac{7}{8}$$ Answer
If a pump takes 6 hoursto fill $$\frac{3}{7}th$$ of a huge tank then total time required to completely fill the tank in hours is
Solution
in question given that in 6 hours the tank is fill = $$\frac{3}{7}th$$
so the time required to fill the full tank = 6\$$\frac{3}{7}$$ hour
= 14 hour Answer
A person P is travelling at 72 kmph while is travelling at 25 meters per second. What is the difference in their speeds in meters per second?
Solution
person speed in first case in meter \second is = 72$$\times$$5\18 = 20 m\sec
and given that in second case the person speed is = 25 m\sec
so the difference is = 25 - 20 = 5 m\sec Answer
A person takes 6 hours to reach a place. If he reduces his speed by $$\frac{1}{3}$$ then he travels 10 kmless in that time. His speed in kmph is
Solution
let the distance to travel of the person is = x km
then in given in the question $$\Rightarrow$$ x\3 = 10
x = 30 km
then the speed = ( distance \time ) = 30 \ 6 = 5 kmph Answer
Of the two workers 4 and B. the worker 4 works twice as fast as B. If B works alone the work could be completed in 12 days. If both A and B work together the numberof days in which the work can be completed is
Solution
given that B to complete the work = 12 days
so A to finish the work = 12\2 = 6 days ( A is twice faster than B so he require half the time as compared to B)
SO together A and B can finish the work in = (6 $$\times$$ 12 )\(6+12) = 4 days Answer
Three workers can individually complete a work in 7. 14 and 28 days respectively. If all the three work together the number of days needed to complete the same work is
Solution
given that the number of days for individual require are = 7,14 and 28 days respectively
then the total work is LCM of all of them = 56
each have individual efficiency = 56\7 , 56\14, and 56\28 = 8 , 4 and 2
so they together finish the work = 56\(8+4+2) = 4 days
The dimensions of a rectangular filed are 60 meters and 80 meters. Four cowsare tied at the four corners of the field with ropes of lengths 10, 12, 14 and 16 meters respectively. The area of the grass that cows could eat in sq meters is
Solution
let the rectangular length l = 80 m. and width w = 60 m.
then the area of the rectangular length = 4800 sq. m.
and radius of the quadrant $$r_{1}$$,$$r_{2}$$,$$r_{3}$$ and $$r_{4}$$ will be respectively 10,12,14 and 16 m
then area of all quadrant = $$\pi$$ $$\times$$ $$r_{1}^{2}$$ \4 + $$\pi$$ $$\times$$ $$r_{2}^{2}$$\4 + $$\pi$$ $$\times$$ $$r_{3}^{2}$$\4 +$$\pi$$ $$\times$$ $$r_{4}^{2}$$\4
= $$\pi$$ $$\times$$ $$10^{2}$$ \4 + $$\pi$$ $$\times$$ $$12^{2}$$\4 + $$\pi$$ $$\times$$ $$14^{2}$$\4 +$$\pi$$ $$\times$$ $$16^{2}$$\4
= $$\pi$$ \4 $$\times$$ (100 + 144 + 196 + 256)
= 174 $$\pi$$ Answer
The length of a roomis 5.5 meters and its width is 3.75 meters. Find the cost (in ₹) of paving the floorat the rate of ₹ 800 per square meters.
Solution
let the lenth l = 5.5 m. and w = 3.75 m.
then the area of the room = l$$\times$$w = 5.5$$\times$$3.75
= 20.625 sq. m.
then the total cost of paving the floor = 20.625$$\times$$800 = 16500 Rs. Answer
What is the volume of a cube (in cu.cm) whosetotal surface area is 384 sq. cms?
Solution
let the side of cube is = a
then wholesale total area = 6 $$r^{2}$$
given that wholesale total area = 384 sq. cms
then
6 $$a^{2}$$ = 384
a = 8
then we know that the volume of the cube = $$a^{3}$$ = 512 sq. cms Aanswer
A cylinder and a cone have the same height and the same radius of the base. The ratio of the volumes of the cone and cylinder is
Solution
we know that
the volume of the cone is = 1\3 $$\times$$ $$\pi$$ $$\times$$ r^2 $$\times$$ h and
the volume of the cylinder is = $$\pi$$ $$\times$$ r^2 $$\times$$ h then
volume of the cone \ volume of the cylinder = 1\3 $$\times$$ $$\pi$$ $$\times$$ r^2 $$\times$$ h \ $$\pi$$ $$\times$$ r^2 $$\times$$ h
volume of the cone \ volume of the cylinder = 1\3 Answer
The area of square field is $$8450 m^2$$. The time taken by a person to cross the field diagonally with a speed of 3 kmph.(in minutes) is
The perimeter of a semi-circle is 396 cm. Thenits diameter (in cm) 1s:
Solution
given that the perimeter of the semicircle = 396 cm
we know that the perimeter of the semiicircle = (1 + $$\pi$$\2) $$\times{d}$$
equate both the value
(1 + $$\pi$$\2) $$\times{d}$$ = 396 cm put the value $$\pi$$ = 22\7
(1 + 22\14) $$\times{d}$$ = 396
d = 154 cm Answer
If the radius of a circle is increased by 10%, then the percent increase in its area is
Solution
let the radius of the circle is r then area = $$\pi$$ $$r^{2}$$
then in given in the question the new radius is = 1.1r
then the new area = $$\pi$$ $$(1.1r)^{2}$$
= 1.21$$\pi$$ $$r^{2}$$
then the % change in area = (1.21$$\pi$$ $$r^{2}$$ - $$\pi$$ $$r^{2}$$)\$$\pi$$ $$r^{2}$$ $$\times$$100
then the % change in area = 21% Answer
$$\left\{x \epsilon R \mid \mid x - 3 \mid < 2 \wedge \mid x - 1 \mid < 5 \right\} =$$
Solution
We know that if |a| = b then a = $$\pm$$b
|x - 3| = 2
x = 2 $$\pm$$3
x = 5 and 11 here x = 5 , 1
and
|x - 1| = 5 in that equation x = 6 and 4 both are lie 1< X < 5 so x = 1, 5 Answer
If $$993 = r(mod 23)$$ and $$0 \leq r \leq 22$$. then $$r =$$
If p and q are statements, then "$$\sim p \Rightarrow p \wedge q$$" is true whenever
Solution
If p and q are statements, then "$$\sim p \Rightarrow p \wedge q$$"
let solve that through the navigation of "$$\sim p \Rightarrow p \wedge q$$"
p→(p∨∼q)
−(p→(p∨∼q))
∵ − (p → q) = p∧ ∼ q
= p∧ ∼ ( p∨ ∼ q)= p∧(∼ q∧ ∼ q) = (p∧∼p)∧(p∧q)=t∧ ( p ∧ q) =t Answer
For any two statements p, q the statement $$\left(p \Rightarrow q \right) \vee \left(\sim \left(\left(\sim p\right) \Leftrightarrow q \right)\right)$$ is false only when
Solution
| p | $$\rightsquigarrow$$q | $$\rightsquigarrow$$p$$\Rightarrow$$q | p⇒∼p | (p$$\rightsquigarrow$$q |
| T | F | T | F | F |
| F | T | F | T | F |
From the above table, the truth value of the expression (∼p⇒p)∧(p⇒∼p) is always F. Hence it's a contradiction. so the p is always true and q is false Answer
If A, B, C are subsets of a set X and $$A^1 = X - A$$, then $$\left(A \cup \left(B^1 \cap C^1\right)\right) =$$
Solution
$$A^1 = X - A$$, then $$\left(A \cup \left(B^1 \cap C^1\right)\right) = $$
- Let S{x|x∈A−(B∩C)}S{x|x∈A−(B∩C)} Let Q{y|y∈(A−B)∪(A−C)}Q{y|y∈(A−B)∪(A−C)}
- All x∈Ax∈A and x∉B,Cx∉B,C.
- All y∈Ay∈A and y∉B,Cy∉B,C.
- Because all xx fit the definition of QQ then we say S⊆QS⊆Q
- Because all yy fit the definition of SS then we say Q⊆SQ⊆S
- Since S⊆QS⊆Q and Q⊆SQ⊆S then S=Q⟹A−(B∩C)=(A−B)∪(A−C)S=Q⟹A−(B∩C)=(A−B)∪(A−C)
But then quickly realized it was wrong because the xx from set SS must meet the following criteria:
$$\left(A \cup \left(B^1 \cap C^1\right)\right) $$ Answer
A relation R defined on a non-empty set A is called anti-symmetric if
Solution
NOTE-
Whether the empty relation is reflexive or not depends on the set on which you are defining this relation -- you can define the empty relation on any set XX.
1- The statement "RR is reflexive" says: for each x∈Xx∈X, we have (x,x)∈R(x,x)∈R. This is vacuously true if X=∅X=∅, and it is false if XX is nonempty.
2- The statement "RR is symmetric" says: if (x,y)∈R(x,y)∈R then (y,x)∈R(y,x)∈R. This is vacuously true, since (x,y)∉R(x,y)∉R for all x,y∈Xx,y∈X.
3- The statement "RR is transitive" says: if (x,y)∈R(x,y)∈R and (y,z)∈R(y,z)∈R then (x,z)∈R(x,z)∈R. Similarly to the above, this is vacuously true.
4- To summarize, RR is an equivalence relation if and only if it is defined on the empty set. It fails to be reflexive if it is defined on a nonempty set.
If A and B are two sets such that n(A) = 4 and n(B) = 5, then the number of non-constant functions from A into B is
Solution
If X has m elements and Y has n elements, the number if onto functions are,
$$n^m-\left(\begin{array}{c}n\\ 1\end{array}\right)(n-1)^{m} + \left(\begin{array}{c}n\\ 2\end{array}\right)(n-2^{m}) + ................+(-1)^{n-1}\left(\begin{array}{c}n\\ n-1\end{array}\right)1^{m}$$
here m= 5 and n= 4 put the value and we get
620 Answer
Two lines $$L_1$$ and $$L_2$$ make intercepts a. —b and b, —a respectively on the x and y axes. Then angle between $$L_1$$ and $$L_2$$ is
Solution
we know the line fourmula y = mx + c
we have two line equation
ax - yb = 0 and bx - ay = 0
then we get
$$y_{1}$$ = $$m_{1}$$x + $$ c_{1}$$ and $$y_{2}$$ = $$m_{2}$$x + $$ c_{2}$$
put the value of x and y
we get
-b = a$$m_{1}$$ + $$ c_{1}$$ and -a = b $$m_{2}$$ + $$ c_{2}$$
where $$ c_{1}$$ and $$ c_{2}$$ are constant = 0 then
$$m_{1}$$ = -b\a and $$m_{2}$$ = -a\b
then we know that the angle is
$$\tan\theta$$ = |$$\frac {(m_{1}-m_{2})}{(1 +m_{1}m_{2})}$$ |
put the value of $$m_{1}$$ and $$m_{2}$$ we get
$$\tan\theta$$ = $$\frac {(a^{2}-b^{2})}{2ab}$$ Answer
Equation of perpendicular bisector of the line segment joining (13, —2) and (-5, 10) is
Solution
The general line through two points (a,b)(a,b) and (c,d)(c,d) is
(c − a) (y−b) = (d −b) (x − a)
In our case, that is
(-5 - 13)(y + 3) = 12 (x − 13) = -18(y +2 ) = 12 (x−13)
which we can write
12x + 18y = -120
2x + 18y = -20 Answer
$$\tan 480^\circ =$$
Solution
Tan(480) =Tan(360+120) = Tan 120 $$\because$$ tan(360 + $$\theta$$) = tan$$\theta$$
Tan120 = Tan(90+30) = -Cot 30
Tan30 = 1/√3, so, Cot 30 = √3
So, Tan (480) = - cot 30 = -√3 Answer
$$\cos \frac{2 \pi}{6} + \sin \frac{5 \pi}{6} =$$
Solution
$$\cos \frac{2 \pi}{6} + \sin \frac{5 \pi}{6} =$$
we can write this
cos60 + sin150 = equation 1
and sin150 = sin( 90 + 60)
= sin90co60 + cos90sin60
put sin90 = 1 and cos60 = 1\2
we get
sin150 = 1\2 and cos 60 = 1\2 put value in equation 1
cos60 + sin150 = 1\2 + 1\2 = 1 Answer
$$6 \cos \theta - 7 \sin \theta = 0 \Rightarrow (7 \cos 2\theta + 6 \sin 2\theta)^2 =$$
Solution
we have given that
$$6 \cos \theta - 7 \sin \theta = 0 $$ equation 1 and
$$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$ equation 2
then solving equation 1 we get
$$6 \cos \theta - 7 \sin \theta = 0 $$
$$ \tan \theta$$ = 6\7
now solving equation 2
$$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$ here we know that
$$ \cos 2\theta$$ = $$\cos\theta^{2}$$ - $$\sin\theta^{2}$$ and $$ \sin 2\theta$$ = 2$$\sin\theta$$ $$\cos\theta$$
put the value in equation 2 we get
= (7($$\cos\theta^{2}$$ - $$\sin\theta^{2}$$) + 6 ( 2$$ \sin \theta$$ $$\cos\theta$$))^2
on solving ge get
= 7\6$$\times$$ $$tan\theta$$ $$7^{2}$$
= 49 answer
Two boys are on opposite sides of a tower of 80 meters height. They measures the angles of elevation of the top of tower as $$45^\circ$$ and $$60^\circ$$ respectively. The distance between two boys(in meters) is
Solution
let the boys are standing in the triangle $$\triangleABC$$ here D is the point in between AB
so the height is CD = 80 m and $$\angleCAD$$ = 45 and $$\angleCBD$$ = 60
so we can say that
tan45 = CD\AD = 80\AD $$\Rightarrow$$ AD = 80
tan60 = CD\BD = 80\BD $$\Rightarrow$$ BD = 80\$$\sqrt{3}$$
so that the total ditance is AD + BD = 80 + 80\$$\sqrt{3}$$
$$\frac{80}{3}(3+\sqrt{3})$$ answer
If $$3x^2 - 5x - 8 < 0$$, then x lies in the interval
Solution
we have given that
$$3x^2 - 5x - 8 < 0$$,
now factorize the equation we get
$$3x^2 - 8x + 3x - 8 < 0$$
$$3x^2 + 3x - 8x - 8 < 0$$
so we get
(x + 1) (3x - 8) < 0
x = (-1 , 8\3) answer
The polynomial in x of least degree with the roots $$\frac{3}{2}, \frac{2}{3}, \pm \sqrt{3}$$ is
Solution
we have given that the root of the polynomonal equation
$$\frac{3}{2}, \frac{2}{3}, \pm$$ \sqrt{3}$$ is mean that
(x - 3\2 = 0 ) ( x - 2\3 = 0 ) ( x$$ \pm $$ $$\sqrt{3}$$ = 0 )
so we have (3x - 2) (2x - 3) ($$ x^{2}$$ - 3) = 0 is the equation so the
(3x - 2) (2x - 3) ($$ x^{2}$$ - 3) = 0 answer
$$\left(x^2 - 3x + 2\right) \mid \left(x^3 - 6x^2 + Ax + B\right) \Rightarrow A^2 + B^2 =$$
Solution
we have given
$$\left(x^2 - 3x + 2) $$ =0 equation 1
and $$ left(x^3 - 6x^2 + Ax + B) $$ = 0 equation 2
now we have to factorize equation 1 we get the equation
(x - 1) (x - 2) = 0
x = 1, 2
put the value of x in equation 2 we get
2a + b = 16 equation 3
a + b = -5 equation 4
on solving eq 3 and eq 4 we get
a = 11
b = 6
then a^2 + b^2 = 157 answer
If $$x - 3$$ and $$x^4 - 2x^3 + 3x^2 - mx + 3$$, then $$m =$$
Solution
we have given that
$$x - 3$$ = 0 so we get x = 3 put the value of x in the following equation
$$x^4 - 2x^3 + 3x^2 - mx + 3$$ = 0
3^4 - 2$$\times{3^3}$$ + 3$$\times{3^2}$$ - m$$\times{3}$$ + 3 = 0
m$$\times{3}$$ = 57
m = 19 answer
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5, \frac{5}{x - 3} - \frac{3}{y - 4} = 1 \Rightarrow x + y =$$
Solution
we have given that
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5$$ equation 1
$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$ equation 2
now 2$$\times{equation}$$ - equation 1
2{$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$} -
$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5$$
then we get x= 5 put the value of x in equation 1
$$\frac{4}{5 - 3} + \frac{6}{y - 4} = 5$$ we get y = 6
then x + y = 11 answer
If x, y(x < y) are primes satisfying x + y = 30, then the number of such pairs (x, y) is
Solution
we have given
x + y = 30, and x, y(x < y)
then we have
y - x > 0
y > x
then y should be positive integer and the x have the following condition
x > 0
x = 0
x < 0
so we have three pair of satisfying the condition of (x,y) in equation x + y = 30 answer
If seventh and eleventh terms of an arithmetic progression are 31 and 47 respectively. then fifteenth termis
Solution
Let a and d are first term and common difference of an A.P.
$$n^{th}$$ term will be $$a_{n}$$ = a + (n-1)d then we have
$$a_{7}$$ = a + (7-1)d
$$a_{7}$$ = a + 6d equation 1
$$a_{11}$$ = a + (11-1)d
$$a_{11}$$ = a + 10d equation 2
now eq2 - eq1
we get 4d = 16 then d = 4 put the value in eq 1
we get a = 31
so the 15^(th) term will be
$$a_{15}$$ = 31 + (15-1)4
$$a_{15}$$ = 63 answer
If $$6^{th}$$ term and $$13^{th}$$ term of a geometric progression are 24 and $$\frac{3}{16}$$ respectively, then the $$25^{th}$$ term is
Solution
let the first term is a and common ratio is r then
$$T_{6}$$= {a}$$\times{r}^{5}$$ =24 equation 1
$$T_{13}$$= {a}$$\times{r}^{12}$$ =3\16 equation 2
equation 1 \equation 2
24\{3\16} = 1\r^7
128 = 1\r^7
2 = 1\r
r = 1\2
put the value of r in equation 1
a$$\times\frac{1}{2}^{5}$$ =24
so we get a= 768 so the series is
768, 384,192,96,48,24,12,6,3,3\2,3\2^2,.............. so the 25^{th} term is
$$\frac{3}{2^{16}}$$ answer
The term independent of x in the expression of $$\left(2x^2 + \frac{1}{x^2}\right)$$ is
Solution
we know that general term of expansion (a+b)^n is
$$ T_{r+1}$$ = $$nC_{r}$${a}^{n-r}b^n x$$\geq0$$
here in expression $$\left(2x^2 + \frac{1}{x^2}\right)$$ we have given n=1 a=2x^2 b=1\x^2
$$ T_{r+1}$$ = $$1C_{r}$${2x^2}^{1-r}{1\x^2}^1
= $$1C_{r}$${2}^{1-r}{x^2}^{1-r}{1\x^2}^1
= $$1C_{r}$${2}^{1-r}{x}^{2-2r}{x}^-2r
= $$1C_{r}$${2}^{1-r}{x}^-4r
so from that n-r =0 and r=4 so n=$$5^{th}$$ term answer
$$\left(1 + x + x^2\right)^n = a_0 + a_1x + a_1x^2 + ......... + a_{2n}x^{2n} \Rightarrow a_1 + a_3 + ....... + a_{2n -1} =$$
Solution
$$\left(1 + x + x^2\right)^n = a_0 + a_1x + a_2x^2 + ......... + a_{2n}x^{2n}$$
Put x = 1
$$\left(1 + 1 + 1\right)^n = a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n}$$
$$a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n} = 3^{n}$$ ....... (1)
Put x = -1
$$\left(1 - 1 + 1\right)^n = a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n}$$
$$a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n} = 1$$ ....... (2)
From (1) and (2):
$$(a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n})-(a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n}) = 3^{n} - 1$$
$$2a_1 + 2a_3 + 2a_5 + .........+ 2a_{2n-1} = 3^{n}-1$$
$$a_1 + a_3 + a_5 + .........+ a_{2n-1} = \frac{3^{n}-1}{2}$$
$$A = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} \Rightarrow A^{2019} =$$
Solution
$$A = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}$$
$$A^{2} = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} * \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix}$$
$$A^{2} = 2\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = 2A$$
$$A = 2$$
$$A^{4} = \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix} * \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix} = \begin{bmatrix}8 & 8 \\8 & 8 \end{bmatrix}$$
$$A^{4} = (2A)^{2} = 8\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = 8A$$[$$Since, A^{2} = 2A$$]
$$A^{4} = 4A^{2} = 8A = 2^{2}A^{2}$$
$$Now, (A^{4})^{504} = (2^{2})^{504}(A^{2})^{504}$$
$$A^{2016} = 2^{1008}(2^{2})^{504}$$[$$Since; A = 2$$]
$$A^{2016} = 2^{2016}$$
$$A^{2016}A^{3} = 2^{2016}A^{3}$$
$$A^{2019} = 2^{2016}A^{2}A$$
$$A^{2019} = 2^{2016}2^{2}A$$[$$Since; A = 2$$]
$$A^{2019} = 2^{2018}A$$
If A and B are 3 x 3 matrices. such that the sum of elements along principal diagonal of A is 3 and the sum of elements along principal diagonal of B is 4. then the sum of principal diagonal elements of 3A + 5B =
Solution
Sum of elements along principal diagonal of A = 3
Sum of elements along principal diagonal of B = 4
Sum of principal diagonal elements of 3A + 5B = (3 * 3) + (5 * 4)
= 9 + 20
= 29
$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$
Solution
Since, $$cos2x = 1 - 2sin^{2}x$$
$$2sin^{2}x = 1 - cos2x$$
Similarly, we can write;
$$2sin^{2}\frac{x}{2} = 1 - cosx$$
Now,
$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$
= $$\lim_{x \rightarrow 0} \frac{2sin^{2}\frac{x}{2}}{x^2} =$$
= $$\frac{1}{2}\lim_{x \rightarrow 0} (\frac{sin\frac{x}{2}}{\frac{x}{2}})^{2}$$
= $$\frac{1}{2}$$ * 1 [$$\lim_{x \rightarrow 0} \frac{sin x}{x} = 1$$]
= $$\frac{1}{2}$$
$$\lim_{x \rightarrow 1+} \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} =$$
Solution
$$=\frac{x^2 - \sqrt{x}}{\sqrt{x} - 1}$$
After differentiation:
$$\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$
Now,
= $$\lim_{x \rightarrow 1}\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$
= $$\frac{2 - \frac{1}{2}}{\frac{1}{2}}$$
= $$\frac{\frac{3}{2}}{\frac{1}{2}}$$
= $$3$$
The angles of a triangle are $$30^\circ, 60^\circ$$ and $$90^\circ$$. Then the sides of the triangle are in the ratio
Solution
Let $$\angle$$A = $$30^\circ$$, $$\angle$$B = $$60^\circ$$ and $$\angle$$C = $$90^\circ$$
sin$$30^\circ$$ = $$\frac{BC}{AB}$$
$$\frac{1}{2}$$ = $$\frac{BC}{AB}$$
Let BC = x and AB = 2x
AC = $$\sqrt{AB^{2} - BC^{2}}$$
AC = $$\sqrt{3}x$$
Now,
Required ratio = BC: AC: AB = x: $$\sqrt{3}x$$: 2x = 1: $$\sqrt{3}$$: 2
The area of a rhombus is 24 sq. cm and one of its diagonals is 8 cm. Then the length of the other diagonal (in cm)is
Solution
Area of rhombus = $$\frac{1}{2}$$ * d1 * d2
Where;
d1 = First diagonal and d2 = Second diagonal
Now,
24 = $$\frac{1}{2}$$ * 8 * d2
d2 = 6 cm
If two circles have radii 16 cm and 7 cm and the distance between their centers is 4 cm. then the number of commontangents. that can be drawn to the two circles is
Solution
The difference between radii of both the circles = 16 - 7 = 9 cm
Since, the difference between radii of both the circles is more than the difference between their centers. This means that the smaller circle will be completely inside the bigger circle there will be no contact point.
Since, there is no contact point of both the circle which means there will be no common tangent of both the circle.
If G is the centroid of the triangle $$\triangle ABC$$ with vertices A(-2, 3), B(-7. 5) and C(3, -5) then the area of $$\triangle GAB$$ (in sq. units) is
Solution
A = (-2, 3), B = (-7, 5) and C = (3, -5)
G = $$\frac{(-2) + (-7) + 3}{3}$$,$$\frac{3 + 5 + (-5)}{3}$$ = (-2, 1)
Area of $$\triangle GAB$$ = $$\frac{1}{2}$$[(-2)(5 - 1) + (-7)(1 - 3) + (-2)(3 - 5)]
= $$\frac{1}{2}$$[-8 + 14 + 4]
= $$\frac{1}{2}$$ * 10
= 5 sq. units
If (2, -3), (3, 7) and (-4, 6) are the mid points of the sides BC, CA and AB respectively of the $$\triangle ABC$$, then A =
Solution
Let co-ordinate of A = (x1,y1), B = (x2, y2) and C = (x3, y3)
Mid-point of AB = $$\frac{x1 + x2}{2}$$,$$\frac{y1 + y2}{2}$$ = (-4, 6)
x1 + x2 = -8 and y1 + y2 = 12 ............. (1)
Similarly,
x2 + x3 = 4 and y2 + y3 = -6 ............. (2)
x1 + x3 = 6 and y1 + y3 = 14 ............. (3)
From (1), (2) and (3):
x1 + x2 + x3 = 1 and y1 + y2 + y3 = 10 .......... (4)
From (2) and (4):
x1 = -3 and y1 = 16
Hence, A = (-3, 16)
The mean of the squares of the first n natural numbers is
Solution
The mean of first n natural numbers = [n(n+1)]/2
The mean of square of first n natural nos.=[(n)(n+1)(2n+1)]/6
= [2n^2 + 3n + 1]/6
The mean of cube first n natural numbers = {[n(n+1)]/2}^2
The median of the following frequency distribution is
Solution
The median is a simple measure of central tendency. To find the Median, we arrange the observations in order from smallest to largest value.
If there is an odd number of observations, the median is the middle value.
If there is an even number of observations, the median is the average of two middle values.
We also need to calculate the cumulative frequency which is the sum of two frequencies.
x fx CF
3 5 5
5 11 5+11=16
**7 15 16+15=31**
9 18 31+18=49
11 8 49+8=57
12 3 57+3=60
Now, (60/2) + 1 = 31
31 occurs in the 3rd column corresponding to which the number is 7.
Thus, Median = 7 Ans.
Mode of the following distribution is
Solution
The Mode of a set of data values is the value that appears most often.
In this set of data x=2 appears most often i.e., 28 times .
Therefore, Mode of the following distribution is 2. Ans.
If the standard deviation of 5, 7, 9, .... 151 is $$\sigma$$, then the standard deviation of 32, 44, 56. .... 908 is
Solution
Here,
Series 1 is an Arithmetic Progression with common difference = 2
Series 2 is an Arithmetic Progression with common difference = 12
Thus, Series 2 has common difference 6 times of the first.
Also, both the series have same no. of terms
Series 1 = (151-5)/D = (146)/2 = 73
Series 2 = (908 - 32)/D = (876/12)= 73
Therefore , the Standard deviation of the second series will be 6 times of the first.
Hence, 6$$\sigma\ $$ Ans.
Standard deviation of any five consecutive integers is
Solution
As, any five consecutive integers are required, for the sake of easy calculation take the integers as 1,2,3,4,5
Standard deviation , Sigma = $$\sqrt{\ }$$[(x - x bar ) ^ 2]/N
x Deviation
1 (1-3)^2=4
2 (2-3)^2=1
3 (3-3)^2=0
4 (4-3)^2=1
5 (5-3)^2=4
where xbar = 3 (assumed mean)
Sum of X bar = 4+1+0+1+4 = 10
Therefore, $$\sigma\ $$ = $$\sqrt{\ }$$(10/5)
= $$\sqrt{\ }$$2 Ans.
If the sum of the squares of deviations of ranks of two students in five subjects is 16, then the coefficient of rank correlation is
Solution
Given, Squares of deviation, D^2= 16
No. of subjects, N=5
Now, according to Spearmen's rank correlation
Coeffiecient of rank correlation ,R = 1 - (6 *Summation D^2)/[N(N^2 -1)]
= 1 - (6*16)/[5(25-1)]
= 1 - (6*16/(5*24)
= 1 - (4/5)
=1/5 Ans.
A fair coin is tossed repeatedly. If tail appears on first four tosses, then the probability of head appearing onfifth toss is
Solution
Ocurrence of Heads/Tails appearing on a coin tossed is independent of the previous results.
Therefore , the probability of head appearing on fifth toss is same as any other time = (1/2) Ans.
If a number is selected at random out of the first 120 natural numbers, then the probability that it is divisible by 5 or 8 is
Solution
Numbers out of the first 120 that are divisible by 5 = (120/5)=24
Numbers out of the first 120 that are divisible by 8 = (120/8)=15
So, the numbers that are divisible by 5 or 8 = 24+15= 39
But, these 39 numbers also includes the numbers which are divisible by both 5 and 8.
Therefore we need to subtract these numbers.
Numbers out of the first 120 that are divisible by both 5 and 8
= 120/(5*8)= 120/40=3
Thus , Numbers out of the first 120 that are divisible by 5 or 8 = 39-3=36
Probability of the event required =36/120
=3/10 Ans.
If a coin is tossed five times the probability that at least one head appears on the top is
Solution
Probability of an event, P(E)=(No. of favourable outcomes)/(No. of total outcomes)
Now,
Total no. of outcomes of an event with two possible results = 2^n (2 raise to power n)
where n is no. of times the event occurs
Here, as the coin is tossed 5 times
Therefore, the no. of total outcomes = 2^5 = 2*2*2*2*2 = 32
As,Total cases are 32 and we need the cases with atleast one head
So, the only case possible with no head is the case where all the coins have Tails.
Thus, favourable outcomes are 32-1 = 31
where 1 was the event with 5 tails.
Hence, P(E)=31/32 Ans.
If A and B are two events in a random experiment with $$P(A) = \frac{3}{8}, P(B) = \frac{5}{8}$$ and $$P(A \cap B) = \frac{1}{4}$$, then $$P(A \cup B) =$$
Solution
Since,
P(A union B)=P(A) + P(B) - P(A intersection B)
=(3/8) + (5/8) - (1/4)
=1-(1/4)
= (3/4) Ans.
Choose the correct meaning for the word given:
Throes
Choose the correct meaning for the word given:
Altercation
Choose the correct meaning for the word given:
Victual
Choose the correct meaning for the word given:
Insipid
Choose the correct meaning for the word given:
Waylay
Choose the correct meaning for the word given:
Candid
Fill in the blank choosing the correct word:
The principal ___________the good character of the pupil.
Ramya ________ her friend for being lazy
Police investigators ________________ the case with photographs and recorded interviews.
Many African countries are ____________ by infighting and civil war
Choose the correct answer:
Whichof the following is the fastest way to connectto internet?
Whichof the following protocol converts the URL into the corresponding IP address of a webserver?
A Wireless communication technology intended to replace cables is known as
A Software program that has been developed to do harm to other computers in known as
Which one of the following is not a characteristic of computers?
Whichof the following describes the nature of management?
What helps im easyidentification and accounting of transactions at retail stores?
Whois the present Governor of RBI?
Increase in value of an asset over time is known as
A short period of temporary economic decline during which trade and industrial activity is reduced is known as
A: “He’s hoping to set up a travel business on his own.”
B: “Yes. but the likely cost of it all will put himoff’.
‘B’ is
A: “Why don’t you do the cooking today?”
B: “As far as cooking is concerned. I am a square peg in a round hole.”
‘B’ implies that he is
“He told us the story in a nutshell.” The meaning of the underlined phrase is
The passive form of the sentence. ‘Someone might have already dispatched the letters’ is
“Let’s run over the plans again.” The underlined phrase means
“You really should put your foot down or there will be trouble later”. The underlined phrase means
“I wouldn’t like to go through that again”. The underlined phrase means
Fill in the blanks with the appropriate phase/verbs/ preposition:
The intravenous drug will ________ through the patient’s body in about 20 minutes.
Whenhe was teaching, all the students _________ silent.
When I was eating, the phone _____________
They got ______________ their car quickly
We shall see whether they are amenable _________ reason.
There was an accident in the morning. A bus collided _______ a car
Sales have really ___________ now
The summer ________ very early in the South
Readthe following passage and answer questions:
Unconsciously, managers without leadership qualities will often seek, above all else. to be liked. Rather than holding people accountable. they let them off the hook. They give non-performers the uneasy feeling that everything’s fine. They are managers who seek approval rather than respect. But this habit has a severe consequence.It leads to a lack of trust in the work place. the most common“issue” on employee surveys. A true leader does not focus first on trying to be liked. A true leader focuses on the practices and communications that lead to being respected. A true leader does not try to become everybody’s big buddy. although he or she values being upbeat and cheerful in communication. A true leaderis not overly concerned with always being liked and is even willing to engage in very uncomfortable conversations in the name of being straight and thorough. A true leader does not try to downplayresponsibility for leadership.
Whatdoes a bad leader seek?
What does “let them off the hook” mean?
Whatis the most common‘issue’ on employee surveys?
“A true leader does not try to become everybody’s big buddy....”” Explain
Whatis a true leader prepared to do?
Readt he following passage and answer questions.
You are poised at a crucial juncture of your life. when many paths are open to you. You are like a princess who has but to command and your desires are fulfilled. Still it would be prudent to remember that the saying “All that glitters is not gold”, was never truer than it is today. In this age of virtual reality all that your senses perceive will more often than not be anillusion. There are many rainbows that the world will display to entice you: make sure that what you chase will bear closer scrutiny: remember also that you win some and lose some. and above all never forget that hard work will always pay you in the long run-even if Lady Luck does not smile at you at first.
The glamour displayed in the media coupled with the success of some of your friends and acquaintances in beauty contests, modeling, acting and so on must be awfully tempting to emulate. You would do well, however, to peer behind the scenes and decide whether the world of cut-throat competition that exists behind the facade is something you can take in yours tride. Spare a thought for the thousands who were possibly more beautiful, more gifted. more graceful but lost to anonymity having staked andlost all in the viciousrat race.
What does the discussion represent?
What must be remembered?
What does happen even if Lady Luck does not smile on yourface first?
What does the success of some of your friends in beauty contests mean to you?
What must one keep in mind before deciding to enter into beauty contests, modeling, acting and so on?
Read the following passage and answer questions
The Japan Society’s crash course on howto bridge the chasm between Japanese and American managers forces participants to examine their own cultural assumptions, as well as to learn about the other side. Behaviors which Americans consider trustworthy are often precisely that which Japanese associate with shifty character and vice versa.
To Americans, people who pause before replying to a question are probably dissembling. They expect a trustworthy person to respond directly. The Japanese distrust such fluency. They are impressed by somebody who gives careful thought to a question before making a reply. Most Japanese are comfortable with periods of silence. Americans find silence awkward and like to plug only conversational gaps.
The cherished American characteristics of frankness and openness are also misunderstood. The Japanese think it is sensible as well as polite for a person to be discreet until he is sure that a business acquamtance will keep sensitive information confidential. An American who boasts, “I am my own man” can expect to find his Japanese’s hosts anxiously counting chopsticks after business lunch. As the Japanese see it. individuals are anti-social. Team players are sound.
The Japan Society’s crash course does not include one of the following
If someone pauses before replying to a question, Americans think she/he is probably
The Japanese are impressed by
The Americans are embarrassed by
“The Japanese think team players are sound”, ‘sound’ here means
