What is the $$100^{th}$$ term of a sequence ?
I. The sequence is an Arithmetic Progression.
II. The first term of the sequence is 5.
AP ICET 2013 Question Paper
In the following questions a question is followed by data in the form of two statements labelled as I and Il. 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 :
Can we find the quadratic equation $$ax^2 + bx + c = 0$$?
I. The sum of the roots is given.
II. The difference of the roots is given.
Do the two circles with their centres at A and B and having radii $$r_1$$, and $$r_2$$, intersect ?
I. Distance between A and B is ‘a’ units.
II. $$\mid r_1 - r_2 \mid < a < r_1 + r_2$$
Are two triangles similar ?
I. They are equilateral.
II. Their areas are equal.
What is the area of a regular hexagon ?
I. A circle of radius ‘a’ units is inscribed in the hexagon.
II. Each side of the hexagon subtends an angle of $$60^\circ$$ at the centre ofthe circle.
What is the value of $$\log_{10}x$$?
I. x > 3
II. $$\log_{10}x + \log_{10}(x - 3) = 1$$
Is ABCD a cyclic quadrilateral ?
I. The lengths of the diagonals are given.
II. A circle with mid point of a diagonal as centre passes through the vertices.
What is the angle subtended by the chord AB of a circle at a point P onthe circumference ?
I. The radius of the circle is 5 cm.
II. The length of the chord AB is 5 cm.
What is the value of 119x + 247y ?
I. 118x + 246y = 482
II. x is greater than y by 2
What is the percentage of profit on the sale of 50 books ?
I. The cost price of each bookis ₹ 100.
II. The sale price of each bookis ₹ 125.
What is the value of n ?
I. $$1^3 + 2^3 + 3^3 + ..... + n^3 = 225$$
II. n is positive integer
What is the day of $$31^{st}$$ December of an year ?
I. In that year the first of March was Monday.
II. That year was a leap year.
What is the Arithmetic Mean of the numbers x, y, z and t ?
I. x + 2y - z - 3t = K
II. 2x + y + 4z + 6t = L
What is the equation of a straight line ?
I. The line is parallel to 2x + 3y = 5.
II. The line does not pass through the origin.
What is the value of $$8a^3 + b^3 - 27c^3$$ ?
I. abc = k
II. 2a + b = 3c
What is the distance between two ships ?
I. Seen from the top of a mountain their angles of depression are $$60^\circ$$ and $$30^\circ$$.
II. Height of the mountain is 60 metres.
Is (x - a) a factor of the polynomial f(x) ?
I. f(a) ≠ 0
II. The sum of the coefficients of f(x) is constant.
What is the simple interest eamed yearly on a deposit in a bank ?
I. The amount deposited is ₹ 10,000.
II. The rate of interest is 8%.
What is the surface area of a sphere ?
I. The sphere is made of iron.
II. The radius of the sphere is given.
What is the value of $$x^4 + y^4$$, if $$xy = 1$$?
I. x + y = 5
II. x > O, y > 0
In the following Questions a sequence of numbers and letters that follow a definite pattern is given. Each question has a blank space. This blank space is to be filled by the correct answer from one of the four given options to complete the sequence without breaking the pattern.
122 : 145 :: 257 : ...........
1001 : 1332 :: ........ : 2198
ABCD : ZYXW :: ....... : VTUS
Tiger : Goat :: ........ : Herbivore
Quadrilateral : Square :: Triangle : .........
Wood : Chair :: .......... : Ornament
Ice : Water :: ....... : Ghee
$$\frac{n(n + 1)}{2} : \sum n :: ....... : \sum n^2$$
ABCDEFG : STUVXYZ :: GECA : .......
Volume of a cylinder : Area of curved surface of the cylinder :: $$\pi r^2 h$$ : .........
In the following questions pick the odd thing out :
Each of the questions follows definite pattern. Observe the same and fill in the blanks with suitable answers.
$$\frac{1}{5}, \frac{5}{10}, \frac{10}{17}, \frac{17}{26}, ..........., \frac{37}{50}$$
6, 12, 20, 30, ......, 56
Solution
The logic here is
12 - 6 = 6
20-12 = 8
30-20 = 10
The difference between the consecutive numbers is increasing by 2.
The next difference should be 12.
$$\therefore\ $$The missing number is 30+12 = 42
Hence, the correct answer is Option D
$$\frac{1}{2}, \frac{8}{5}, \frac{27}{10}, ...... \frac{125}{26}, \frac{37}{50}$$
Solution
Numerator
1^3 =1
2^3 = 8
3^3 = 27
4^3 = 64
Denominator
5-2 = 3
10-5 = 5
next difference should be 7
Hence 10 + 7 = 17
4, 1, 9, 3, 16, 6, 25, 10, 36, ............
1, 4, 27, 256, .........
Solution
The logic here is
$$1^1=1$$
$$2^2=4$$
$$3^3=27$$
$$4^4=256$$
Similarly, the next number in the series = $$5^5=3125$$
Hence, the correct answer is Option D
$$1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, .........., \frac{1}{720}, \frac{1}{5040}$$
4, 7, 12, 19, 28, ......., 52
Solution
The pattern followed is: the difference between the numbers is increasing by 2 starting from 3.
i.e 4+3 = 7
7+5 = 12
12+ 7 =19
19 + 9 = 28
28 + 11 = 39
1 x 8, 3 v 6, 5 t 4, .........
1, 2, 3, 5, 8, 13, 21, ..........
Solution
The logic here is
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
8 + 13 = 21
From 3rd number, every number is the sum of the previous two numbers.
Hence the next number would be 21 + 13 = 34.
5, 6, 10, 19, 35, 60, ........
The following table gives the details of the 5 commodities A, B, C, D, E required, with their costs, for a family in a month.
Study the table and answer the questions.
The total amount(in ₹) spent on all the 5 commodities by the family in 1990 is
The percentage of increase(per kg) in the rate of the commodity E from 1990 to 1995 is
Solution
From the table,
The percentage of increase(per kg) in the rate of the commodity E from 1990 to 1995 is = $$\frac{\left(62-58\right)}{58}\times100$$ = $$\frac{4}{58}\times100$$ = 6.89%
Hence, the correct answer is Option B
The amount (in ₹) spent extra on commodities B and C in the year 1995 than in the year 1990 is
Solution
Money spent on B in 1990 = 15*40 = 600
Money spent on C in 1990 = 8 * 35 = 280
Money spent on B in 1995 = 15 * 45 = 675
Money spent on C in 1995= 8*40 = 320
The amount (in ₹) spent extra on commodities B and C in the year 1995 than in the year 1990 = (675+320) - (600 + 280) = 115
The amounts spent by a person under various heads in a month is given in the following Pie Chart. Based on this information answer the following questions.
If the monthly incomeof the person is ₹ 54,000, then the difference between the amounts spent on travel and food per month (in ₹) is
Solution
Given, monthly income of the person is ₹ 54,000 = $$360^{\circ\ }$$
difference between the amounts spent on travel and food per month (in ₹) is = $$70^{\circ\ }-30^{\circ\ }=40^{\circ\ }$$ = $$40\ \cdot\ \frac{54000}{360}=\ 6000$$
If the amounts spent on savings and children education put together is ₹ 22,000, then the monthly incomeof the person(in ₹) is
Solution
Savings + children's education = $$50^{\circ\ }+60^{\circ\ }=110^{\circ\ }$$
Given, $$110^{\circ\ }=\ 22000$$
monthly income = $$360^{\circ\ }=360\cdot\frac{22000}{110}=72000$$
The amount spent on children education is equal to the amounts spent on which of the following two heads ?
Solution
amount spent on children education = 60
A: Entertainment + misc = 50
B: Travel + mis = 70
C : savings + travel = 80
D : savings + enter = 60
If the amount spent on house rent in a month is ₹ 5,000 more than the amount spent on travel and savings put together, then the monthly income(in ₹) of the person is
Solution
Let total amount = 360x
amount spent on house rent = 100x
amount spent on travel and savings put together = 80x
Given, 100x-80x = 5000
=> x = 250
Monthly income = 360 x 250 = 90000
Monthly savings percentage is
Solution
From the pie chart,
Monthly savings percentage = $$\frac{50^{\circ}}{360^{\circ}}\times100$$ = $$\frac{125}{9}$$ = $$13\frac{8}{9}\%$$
Hence, the correct answer is Option A
In a group of 75 students 28 students study Mathematics (M), 32 study Physics (P), 30 study Chemistry (C) and 15 students study none of these; 13 study both M and P, 12 study P and C and 10 study C and M. Based on this information answer the questions.
The number of students who study only Chemistry is
Solution
Let a = only M
b = only P
c= only C
d = only M and P
e = only M and C
f = only P and C
g = all three
n = none = 15
a+b+c+d+e+f+g+n = 75 => a+b+c+d+e+f+g = 60
X = a+b+c
Y = d+e+f
Z = g
X+2Y + 3Z = 28 + 32 + 30 = 90 ----------------- (1)
d+g = 13
e+g = 10
f+g = 12
Adding we get Y + 3Z = 35 -------------(2)
Subtracting 1 and 2, we get X + Y = 55
=> Z = g = 5
Hence d= 8, e=5 and f = 7 a = 10, b = 12, c= 13
Only chemistry = c = 13
The number of students who study Physics alone is
Solution
Let a = only M
b = only P
c= only C
d = only M and P
e = only M and C
f = only P and C
g = all three
n = none = 15
a+b+c+d+e+f+g+n = 75 => a+b+c+d+e+f+g = 60
X = a+b+c
Y = d+e+f
Z = g
X+2Y + 3Z = 28 + 32 + 30 = 90 ----------------- (1)
d+g = 13
e+g = 10
f+g = 12
Adding we get Y + 3Z = 35 -------------(2)
Subtracting 1 and 2, we get X + Y = 55
=> Z = g = 5
Hence d= 8, e=5 and f = 7 a = 10, b = 12, c= 13
Only physics = b = 12
The letters of English alphabet are arranged cyclically. Then the letters are coded as follows : (i) a vowel is to be coded as the second vowel after it in the clockwise direction. and (ii) a consonant is to be coded as the second consonant after it in the clockwise direction. For decoding, the inverse process is to be followed. Based on this coding and decoding system answer the questions.
Code word for HAND is
The word that is coded as RIUT is
The word that is coded as RNIVO is
Solution
Based on the given conditions,
R will be the code for P
N will be the code for L
I will be the code for A
V will be the code for S
O will be the code for E
$$\therefore\ $$The word that is coded as RNIVO is PLASE.
Hence, the correct answer is Option C
The code word for LEAVE is
Solution
Based on the given conditions,
L will be coded as N
E will be coded as O
A will be coded as I
V will be coded as X
E will be coded as O
$$\therefore\ $$The code word for LEAVE is NOIXO.
Hence, the correct answer is Option A
The word that is coded as QUEUE is
For the following questions answer them individually
If the word COMPUTER is coded as RFUVQNPC, the code word for MEDICINE is
If the word DWUKPGUU is coded as BUSINESS, the word MANAGEMENT is coded as
If the word CENTRAL is coded as LARTNEC then the code word for SEMINAR is
If the word FLUTE is coded as EJRPZ, then the code word of PLANE is
If the code word for VICTORY is XKEVQTA then the code word for SUCCESS is
If the first day of March in a year happens to be Friday, then the day on which Indian Independenceday falls in that year is
If a clock shows 15 minutes past 3 O’clock then the angle between the hours and minutes hands is
If the minutes hand of a clock is facing South, then the direction of the minutes hand after 210 minutes is
A is the mother of B and C. D is the husband of C. Then A is related to D as
Four buses $$B_1, B_2, B_3, B_4,$$ depart from the station $$S_1$$, and arrive the stations $$S_2$$, and $$S_3$$,. The bus numbers and the order in which they depart or arrive is not the same. The first bus to leave $$S_1$$, is the second to reach $$S_3$$, and the third to reach $$S_2$$. The first bus to reach $$S_3$$, is the second bus to leave $$S_1$$, and the last bus to reach $$S_2$$.
Based on this information answerthe following questions.
The first bus to reach the station $$S_1$$ is
The second bus to reach the station $$S_2$$ is
For the following questions answer them individually
In a queue, Anitha is the $$10^{th}$$ from the front while Meena is $$25^{th}$$ from the last and Mohan is just in the middle of the two. If there are 50 persons in the queue, the position of Mohan from the front is
If $$a * b = a + b + ab$$ for $$a, b \epsilon Q$$ then the value of $$x$$ satisfying $$(1 * 2) * x = -13$$ is
Solution
Given, $$a * b = a + b + ab$$ and
$$(1 * 2) * x = -13$$
$$\Rightarrow$$ $$(1 + 2 + 1.2) * x = -13$$
$$\Rightarrow$$ $$5 * x = -13$$
$$\Rightarrow$$ $$5 + x + 5x= -13$$
$$\Rightarrow$$ $$5 + 6x = -13$$
$$\Rightarrow$$ $$6x = -18$$
$$\Rightarrow$$ $$x = -3$$
Hence, the correct answer is Option B
$$a * b = \frac{1}{ab} + 1 \Rightarrow \sum_{n = 1}^{2013} n * (n + 1) =$$
If $$a * b = a + b - \frac{ab}{2}\forall a, b \epsilon R$$ and if c is a non-zero real number, then the value of x satisfying x * c = x is
Solution
Given, $$a * b = a + b - \frac{ab}{2}$$
If x * c = x
$$\Rightarrow$$ x + c - $$\frac{\text{xc}}{2}$$ = x
$$\Rightarrow$$ c - $$\frac{\text{xc}}{2}$$ = 0
$$\Rightarrow$$ $$\frac{\text{xc}}{2}$$ = c
$$\Rightarrow$$ x = 2
$$\therefore\ $$The value of x satisfying x * c = x is x = 2
Hence, the correct answer is Option B
If $$\frac{8^3. (27)^4. 6^5}{(36)^2.9^4.(18)^2} = 2^{\alpha}.3^{\beta}$$ then $$\alpha + \beta =$$
Solution
Solution
$$\frac{8^3. (27)^4. 6^5}{(36)^2.9^4.(18)^2}$$ =$$\frac{\left(\left(2^3\right)^3\ \left(3^3\right)^4\left(2\cdot3\right)^5\right)}{\left(2^2.3^2\right)\left(3^2\right)^4\left(2\cdot3^2\right)}\ =\frac{\left(2^9.3^{12}.2^5.3^5\right)}{\left(2^4.3^4.3^8.2^2.3^4\right)}=\frac{\left(2^{9+5}.3^{12+5}\right)}{\left(2^6.3^{16}\right)}\ =\ 2^{14-6}.3^{17-16}\ =\ 2^8.3^1$$
On comparing,
$$\alpha\ \ =\ 8,\ \beta\ \ =\ 1.\ And\ \alpha\ \ +\ \beta\ \ =\ 9$$
Hence 9 Answer
The smallest number among
$$\sqrt[3]{4}, \sqrt[5]{7}, \sqrt[4]{5}, \sqrt{3}$$ is
Solution
Solution :
we need to find the LCM of roots' power i.e. LCM of 3,5,4,2 i.e. 60
$$\sqrt[3]{4}$$ =$$4^{\frac{1}{3}}\ =\ 4^{\left(\frac{20}{60}\right)}$$ = $$\left(4^{50}\right)^{\frac{1}{60}}$$
$$\sqrt[5]{7}$$ =$$7^{\frac{1}{5}}\ =\ 7^{\left(\frac{12}{60}\right)}$$ = $$\left(7^{12}\right)^{\frac{1}{60}}$$
$$\sqrt[4]{5}$$ =$$5^{\frac{1}{4}}\ =\ 5^{\left(\frac{15}{60}\right)}$$ = $$\left(5^{15}\right)^{\frac{1}{60}}$$
$$\sqrt[2]{3}$$ =$$3^{\frac{1}{2}}\ =\ 3^{\left(\frac{30}{60}\right)}$$ = $$\left(3^{30}\right)^{\frac{1}{60}}$$
As all the external powers are same, we just need to arrange the internal powers in ascending order to find the lowest one.
On careful observation, and also relying on the fact that 7 has the lowest power. $$\sqrt[5]{7}$$ is the lowest one.Answer.
If (a + b) : (b + c) : (c + a) = 2 : 3 : 4 and a + b + c = 9,then the value of c =
Solution
Solution:
(a + b) : (b + c) : (c + a) = 2 : 3 : 4
Let the common factor of the ratio be x
So,
a+b = 2x
b+c = 3x
c+a = 4x
Adding all the 3 equations
2(a+b+c) = 9x
A/q
a + b + c = 9
So, 2.9 = 9x
=> x=2
So a+b = 2.2 =4
c = (a+b+c) - (a+b) = 9 -4 = 5 Answer
A circle and a square have the same area. Then the ratio of the side of the square to the radius of the circle is
Solution
Let circle's radius be r and square's side be a
Circle's area = $$\pi r^2$$
Square's Area = $$a^2$$
A both are same ,
$$\pi r^2$$ = $$a^2$$
$$\ \frac{a^2\ }{r^2}$$ = $$\pi\ $$
$$\ \frac{a\ }{r}$$ = $$\sqrt{\ \pi\ }$$
= $$\sqrt{\ \pi\ }$$ :1 Answer
$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ =
Solution
(Hint:taking the cue from the options we should get the term as a square of 3 added terms.)
Now
$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ =
$$=\sqrt{\ \ \left(\left(1+2+3\right)+2.1.\sqrt{\ 2}+2.1.\sqrt{\ 3}+2.\sqrt{\ 2}.\sqrt{\ 3}\right)}$$
As $$\sqrt{\ \left(a+b+c\right)^2}=\sqrt{a^2+b^2+c^2+2.a.b+2.b.c+2.c.a\ \ }$$
On comparison,
$$a=1\ ,\ b=\sqrt{\ 2\ },c\ =\sqrt{\ 3}$$
Hence,
$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ = $$1 +\sqrt{2} + \sqrt{3}$$ Answer
$$\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{5} - \sqrt{7}} + \frac{\sqrt{5} - \sqrt{7}}{\sqrt{5} + \sqrt{7}}\right)^2$$ =
Solution
$$\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{5} - \sqrt{7}} + \frac{\sqrt{5} - \sqrt{7}}{\sqrt{5} + \sqrt{7}}\right)^2$$ = {$$\frac{\left(\left(\sqrt{\ 5}+\sqrt{\ 7}\right)^2-\ \left(\sqrt{\ 5}-\sqrt{\ 7}\right)^2\right)}{\left(\sqrt{\ 5}^{ }+\sqrt{\ }7\right)\left(\sqrt{\ 5}^{ }-\sqrt{\ }7\right)}$$}^2
= $$\left(\frac{\left(\left(5\ +7\ +2.\sqrt{\ 5}.\sqrt{\ 7}\right)+\left(5\ +7\ -2.\sqrt{\ 5}.\sqrt{\ 7}\right)\right)}{5-7}\right)^2$$ = $$\left(\frac{\left(12+12\right)}{-2}\right)^2=\ \left(-12\right)^2\ =144\ Answer$$
The number that is exactly divisible by 11 among the following is
Solution
Solution'
Concept used: Divisibility rule of 11:The difference of sum of digits at even and odd places should be 0 or multiple of 11
Lets take each no. one by one
So the required no. is 1873410 Answer
The remainder when 73 x 79 x 81 is divided by 11 is
Solution
Principle used to solve :Rem (A x B) = Rem. (A) x Rem. (B)
A/q
When divided by 11
Rem. (73 x 79x 81) = Rem. (73) x Rem.(79) x Rem.(81)
Rem. (73) = 7
Rem. (79) = 2
Rem. (81) = 4
So Rem. (73 x 79x 81) = Rem (7 x 2 x 4) = Rem (56) = 1
Hence Remainder =1 Answer
The greatest 4-digited number which is exactly divisible by each of the numbers 18, 24 and 36 is
Solution
Solution
18 = 6.3
24 = 6.4
36 = 6.6
LCM = 6.4.3 = 72
So the no. should be divisible by 72. Out of the given options, the Highest such no. before 9999 is 9936.
If the l.c.m. of the positive integers a and b is 60 and $$a^2 . b^2 = 32400$$, then the g.c.d. of a and b is
Solution
Solution
Concept Used: Product of two numbers = (L.C.M of numbers) x (G.C.D of numbers)
Here
$$a^2 . b^2 = 32400$$
Hence a.b = $$\sqrt{\ 32400}$$ = 180 =Product of Numbers
Given L.C.M. = 60
So axb = (l.c.m) x( g.c.d)
180 = 60 x g.c.d
g.c.d = 3 Answer
If $$\mid x - 6 \mid = 5$$ and $$\mid 3y - 12 \mid = 6$$ then the maximum value of $$\frac{x}{y} =$$
Solution
Solution
$$\mid x - 6 \mid = 5$$ |x-6|=5
if x-6 > 0 ; x>6
so the mod function will become, x-6 = 5
x=11......x max.
If x-6 < 0 ; x<6
so the mod function will become, -(x-6) = 5
6 -x = 5
x=6-5 =1 .....x min
Similarly for y
$$\mid 3y - 12 \mid = 6$$ |3y-12|= 6
If 3y-12 < 0, y < 4
so the mod function will become, - (3y-12) = 6
12 -3y = 6 ; y =2 ........y min.
If 3y-12 > 0, y > 4
so the mod function will become, (3y-12) = 6
3y = 12+6 = 18
y = 6 .......y max
For,maximum value of $$\frac{x}{y} =$$ Numerator (x) should be maximum and denominator (y) should be minimum.
$$\frac{x\ \left(\max\right)}{y\left(\min\right)}\ =\ \frac{11}{2\ }$$ Answer
If 3 is added to the denominator of a rational number then that number becomes $$\frac{1}{3}$$ and if 4 is added to numerator of the same rational number, then it becomes $$\frac{3}{4}$$ then that rational number is
Solution
Solution:
Let the rational No. be $$\frac{A}{B}$$
A/Q
$$\frac{A}{B+3}$$ = $$\frac{1}{3}$$ =>3A = B+3
3A = B+3 => A=$$\frac{1}{3}$$( B+3) ....(i)
$$\frac{A +4}{B}$$ = $$\frac{3}{4}$$ => 4A +16 = 3B ....(ii)
4A +16 = 3B
$$\frac{4}{3}$$( B+3) + 16 = 3B
$$\frac{4B}{3} $$ + 4 + 16 = 3B
20 = 3B - $$\frac{4B}{3} $$
20 = $$\frac{5B}{3} $$
B=12
Putting in Equation (i)
A = 5
Hence required rational No. is $$\frac{5}{12}$$ Answer
The difference between the biggest and smallest fractions among $$\frac{1}{2}, \frac{5}{6}, \frac{7}{9}, \frac{4}{5}$$ is
Solution
Solution
$$\frac{1}{2}, \frac{5}{6}, \frac{7}{9}, \frac{4}{5}$$
LCM of denominators = LCM(2,6,9,5) = 90
Converting all the fractions to common base
$$\frac{1}{2}$$ = $$\frac{1 . 45}{2. 45}$$ = $$\frac{45}{90}$$
$$\frac{5}{6}$$ = $$\frac{5 . 15}{6 .15}$$ = $$\frac{75}{90}$$
$$\frac{7}{9}$$ = $$\frac{7 . 10}{9 . 10}$$ = $$\frac{70}{90}$$
$$\frac{4}{5}$$ = $$\frac{4 . 18}{5 . 18}$$ = $$\frac{72}{90}$$
As denominators are same, so fraction with biggest numerator is biggest and with smallest numarator is smallest.
Biggest fraction = $$\frac{75}{90}$$
Smallest fraction = $$\frac{45}{90}$$
Difference = $$\frac{75}{90}$$ - $$\frac{45}{90}$$ = $$\frac{30}{90}$$ = $$\frac{1}{3}$$ Answer
The ascending order of the rational numbers $$\frac{13}{14}, \frac{11}{12}, \frac{15}{16}, \frac{8}{11}$$ is
Solution
Solution:
$$\frac{13}{14}, \frac{11}{12}, \frac{15}{16}, \frac{8}{11}$$
We can proceed with approximation;
$$\frac{13}{14}$$ = 0.92..
$$\frac{11}{12}$$ = 0.91...
$$\frac{15}{16}$$ = 0.93...
$$\frac{8}{11}$$ = 0.72....
So , Ascending Order will be 0.72,0.91,0.92,0.93
i.e. $$\frac{8}{11}$$ , $$\frac{11}{12}$$ , $$\frac{13}{14}$$ , $$\frac{15}{16}$$
If 7% of 5% of 8% of x is 105, then x =
Solution
Solution:
A/q
7% of 5% of 8% of x is 105
$$\frac{7}{100}$$ x $$\frac{5}{100}$$ x $$\frac{8}{100} x$$ = 105
7.5.8.$$x\ $$ = 105000 x 1000
280$$x\ $$ = 105000000
$$x\ $$ = 375000 Answer
If the length and breadth of a rectangle are increased vy 20% each, then the area of the rectangle increases by x%, then x =
Solution
Solution:
Let Original dimensions be l and b.
Area = l x b
New Dimension be L and B
% Increase in l = 20
L = (100+20)% Of l = 1.2l
% Increase in b = 20
B = (100+20)% Of b = 1.2b
New Area = L x B = (1.2l)(1.2b) = 1.44 lb
% change increase in area = $$\frac{Difference In Area}{Old Area}$$ x 100
= $$\frac{1.44lb-lb}{lb}$$ x 100 =44% Answer
An article when sold at a gain of 5% yields rupees 15 more than when it is sold at a loss of 5%. Then the cost price of that article in rupees is
Solution
Solution:
Gain % = 5
Loss % = 5
Difference in profit percentages = 5-(-5) = 10%
This 10%(of CP) corresponds to monetary value of Rs 15
1% of CP corresponds to Rs. 15/10 = Rs.1.5
100% of CP = Rs. 1.5* 100 = Rs. 150 Answer
By selling 15 items, a seller recovers the cost price of 20 items. Then his profit percent is
Solution
Solution
A/q
20 C.P. = 15 S.P.
15 C.P. + 5 C.P. = 15 S.P.
5 C.P. = 15 S.P. - 15 C.P.
C.P = $$\frac{15}{5}$$ (S.P. - C.P.) = 3 (S.P. - C.P.)
C.P = 3 (S.P. - C.P.)
$$\frac{1}{3}$$ = $$\frac{(S.P. - C.P.)}{CP}$$ = Fractional Profit
Profit percentage = Fractional Profit x 100 = $$\frac{100}{3}$$ %
=33$$\frac{1}{3}$$ % Answer
A, B, C entered into a business with a total capital of ₹ 50,000/-. A invested ₹ 4,000/- more than B and B invested ₹ 5,000/- more than C. If there is a total profit of ₹ 35,000/- then A’s share in the profit, in rupees is
Solution
Solution
Let C's investment be C
So
B = C+ 5000
A = B+ 4000 = (C+ 5000)+ 4000 = C+ 9000 =A
So A +B +C = 50000
C+ 9000 +C+ 5000 + C = 50000
14000 + 3C = 50000
3C = 36000
C = 12000
Hence A = C+ 9000 = 21000
B = C+ 5000 = 17000
Ratio of A:B:C = 21:17:12
Profit = 35000
As share = $$\frac{21}{21+17+12}$$ x 35000 = $$\frac{21}{50}$$ x 35000 = 21 x 700
= 14700 Answer
A, B and C are running a business by investing the capitals in the ratio 2 : 4 : 5. If C’s sharein the profit is ₹ 2,200/-, then A’s share in the profit, in rupees, is
Solution
Solution:
A, B and C 's investments are in the ratio 2 : 4 : 5.
So profits will be shared in the same ratio.
For every Rs.(2+4+5=)11 profit, C gets Rs 5 and A get Rs 2
So if C gets Rs 5...A gets Rs 2
if C gets Rs 1...A gets Rs $$\frac{2}{5}$$
if C gets Rs 2200...A gets Rs $$\frac{2}{5}$$ x 2200 = Rs $$\frac{4400}{5}$$
= Rs.880 Answer
Two pipes A and B can fill a tank independently in 20 and 30 minutes respectively. Both the pipes are opened simultaneously and after 6 minutes tap B is closed. Then the total time taken, in minutes,to fill the tank, is (assume that the tank is emptyinitially)
Solution
Solution:
Let total work be 60 units [LCM of 20,30]
A does 60 units in 20 minutes. Hence in 1 minute A does (60/20)units = 3 units.
B does 60 units in 30 minutes. Hence in 1 minute B does (60/30)units =2 units.
Total work in 1 minute = 5 units
Now, for the given scenario,
First,
A and B work for 6 minutes together @ 5 units/min
Work done = (5 x 6) units = 30
Work remaining =Total units of work - work done = 60-30 =30 UNITS .
Then,
A work alone @ 3 units/min.
Time taken = (Work remaining / speed) = 30/3 mins. = 10 minutes
Total time = 6 +10 = 16 minutes
Tap A can fill an empty tank in one hour, while a drain pipe B can empty that full tank in 6 hours. If both A and B are opened at the same instant, the total time taken to fill that empty tank,in hours, is
Solution
Solution:
Work done by A in 1 hour = 1 unit
Work done by B in 1 hour = - 1/6 unit
Total work done by A and B in 1 hr. = 1- $$\frac{1}{6}$$ = $$\frac{5}{6}$$
Time taken to fill the tank = $$\frac{1}{WorkIn1hour}$$ = 1/$$\frac{5}{6}$$ $$\frac{1}\frac{5}{6}$$
= $$\frac{6}{5}$$ Answer
A train travels a distance of 60 km in 40 minutes. If its velocity is decreased by 15 km/hour, then the time required by that train to travel the same distance, in minutes, is
Solution
Solution
Distance = 60km
Time = 40 minutes = $$\frac{40}{60}$$ hour = $$\frac{2}{3}$$ Hour
Speed = $$\frac{Distance}{Time}$$
Speed = $$\frac{60}{\frac{2}{3}}$$ = $$\frac{(60)(3)}{2}$$ = 90 kmph
Reduced speed = 90-15 = 75 kmph
Distance = 60 km
Speed = $$\frac{Distance}{Speed}$$
Time = $$\frac{60}{75}$$ = $$\frac{4}{5}$$ hr.= 48 minutes.
A person travels from A to B at a speed of 75 kin/hour in a car and returns from B to A by reducing his speed by 15 km/hour. If the total time taken is 3 hours, then the distance between A and B, in kilometers, is
Solution
Solution:
Let,The distance between A-B be $$d$$.
While going from A to B Speed = 75 kmph
Time = $$\ \ \frac{\ Dis\tan ce}{Speed}$$ = $$\ \ \frac{\ d}{75}$$ hours
While going from B to A Speed = 75 kmph
Time = $$\ \ \frac{\ Dis\tan ce}{Speed}$$ = $$\ \ \frac{\ d}{75 - 15}$$ = $$\ \ \frac{\ d}{60}$$ hours
Total time = 3 hours
=> $$\ \ \frac{\ d}{75}$$ + $$\ \ \frac{\ d}{60}$$ = 3
$$\ \frac{\ d}{75}+\frac{d}{60\ }\ =\ 3$$
$$\ \frac{\ 4d\ +5d}{300}\ =\ 3$$
$$\ \frac{\ 9d}{300}\ =\ 3$$
$$\ \ d=\ 100\ km$$ Answer
Two persons A, B can together complete a piece of work in 12 days, B and C together complete the same work in 8 days. If A, B, C work together, then that work can be finished in 6 days. Then the number of days B alone can finish the same work is
Solution
Solution
A, B can together complete the work in 12 days
Hence efficiency of (A+B) = $$\left(\frac{100}{12}\right)\%$$ = $$8\frac{1}{3\ }\%$$
B and C together complete the same work in 8 days.
Hence efficiency of (B+C) = $$\left(\frac{100}{8}\right)\%$$ = $$12\frac{1}{2\ }\%$$
A, B, C work together, then that work can be finished in 6 days
Hence efficiency of (A+B+C) = $$\left(\frac{100}{6}\right)\%$$ = $$16\frac{2}{3\ }\%$$
Effeciencies
A+B = $$8\frac{1}{3\ }\%$$ .......(i)
B+C = $$12\frac{1}{2\ }\%$$ ......(ii)
A+B+C =$$16\frac{2}{3\ }\%$$ .....(iii)
Solving for B,
iii - ii
(A+B+C) - (B+C) = A= $$16\frac{2}{3\ }\%$$ - $$12\frac{1}{2\ }\%$$ = $$4\frac{1}{6\ }\%\ =\ \frac{25}{6\ }\%$$
Putting this in (i)
$$\ \frac{25}{6\ }\%$$ + B = $$8\frac{1}{3\ }\%$$
B =$$8\frac{1}{3\ }\%$$ - $$\ \frac{25}{6\ }\%$$ = $$\frac{25}{6\ }\%$$ %
Time taken by B = $$\left(\frac{1}{\ efficicency}\right)$$ = $$\frac{100}{\frac{25}{6\ }}\ =\ 24$$
24 days Answer
A, B, C can individually complete a work in 20 days, 15 days, 12 days respectively. B and C start the work and they worked for 3 days and left. Then the number of days required by A to finish the remaining work is
Solution
Solution
A, B, C can individually complete a work in 20 days, 15 days, 12 days respectively
Efficiency of (A + B + C) = $$\left(\frac{1}{20}+\frac{1}{15}\ +\frac{1}{12}\right)$$ = $$\left(\frac{3+4+5}{60}\right)\ =\ \frac{12}{60\ }$$
Efficiency of only (B + C) =$$\left(\frac{1}{15}+\frac{1}{12}\right)$$ = $$\left(\frac{4+5}{60}\right)\ = \ \frac{9}{60\ }$$
So efficiency of A = $$\frac{12}{60}-\frac{9}{60}\ =\ \frac{3}{60}$$
While working, all 3 work for 3 days.
So work completed in 3 days = $$3\ \times\ \frac{12}{60}\ =\ \frac{36}{60}$$
Work Remaining = $$1\ -\frac{36}{60}\ =\ \frac{24}{60}$$
If A works alone @ $$\frac{3}{60}$$ work per day. = $$\frac{\left(\frac{24}{60}\right)}{\left(\frac{3}{60}\right)}$$ = 8 days.
If the length of the hypotenuse of right angled Iscoceles triangle is 12 cm , then the area of the triangle in sqaure centimeters is
Solution
Solution
Let Perpendicular be p, base be b and Hypotenuse be h
As it is isosceles right angle triangle,
so P=B
According to Pythogaras Theorem
$$h\ =\ \sqrt{\ p^{2\ }+b^2}$$
$$12=\ \sqrt{\ b^{2\ }+b^2}$$
$$12= \ \sqrt{\ 2b^{2\ }}$$
$$12=\ \sqrt{\ 2}\ b$$
$$b\ =\ 6\sqrt{\ 2}$$
Area = $$\ \frac{\ 1}{2}\ \times\ product\ of\ sides\ \times\ \sin\ of\ included\ angle\ $$
= $$\ \frac{\ 1}{2}\ \times\ b\times\ b\ \times\ \sin\ 90^{\circ\ }$$
= $$\ \frac{\ 1}{2}\ \times\ \left(6\sqrt{\ 2}\ \right)^2\times\ 1$$
= $$\ \frac{\ 1}{2}\ \times\ 72\times\ 1$$
=36 Answer
If the length of the diagonals of a rhombus are 18 cm and 24 cm then the area of a rhombus in square centimeters is
Solution
Solution
Area of Diagonal = $$\frac{1}{2\ }\left(product\ of\ diagonals\right)$$ = $$\frac{1}{2\ }\left(18\ \times\ 24\right)$$ = 216 $$cm^2$$
A cylindrical bar of height 1 metre and base radius 0.75 metres is melted and cast as a sphere and cast as a sphere. Then the diameter of a sphere in meters is
Solution
Solution:
Volume of Cylindrical bar = Volume of cast sphere
$$\pi\ r^2\ h\ =\ \frac{4}{3\ }\pi\ R^3$$
$$\pi\ \left(0.75\right)^2\ 1\ =\ \frac{4}{3\ }\pi\ R^3$$
$$\left(\frac{3}{4}\right)^2\ =\ \frac{4}{3}R^3$$
$$\left(\frac{3}{4}\right)^2\ \ .\ \frac{3}{4}=\ R^3$$
$$\left(\frac{3}{4}\right)^3=\ R^3$$
R= 3/4 m
D = 2R = 3/2 = 1.5m
Two boxes have square bases. Thesides of these bases are in the ratio 2 : 1. If the heights of the boxes are in the ratio 1 : 2 respectively, then the ratio of the volumesofthe boxes in that order is
Solution
Solution:
Bases (b) are in the ratio 2:1
Height (h) are in the ratio 1:2
As, volume's formula is defined as $$b^2h$$
So Volume ratio => $$\left(2\right)^21\ :\ \left(1\right)^22\ =\ 4:2\ =\ 2:1$$
Answer = 2:1
The radius of a coneis 3 times the radius of a cylinder and their heights are same. Then the ratio of their volumesin that order is
Solution
Solution:
The radius of the cone is 3 times the radius of the cylinder and their heights are same.
Volume of cone = $$\frac{1}{3}\pi\ R^2h$$
Volume of cylinder = $$\pi r^2h$$
Given, R= 3r
So,Volume of cone = $$\frac{1}{3}\pi\ (3r)^2h$$
Ratio of volumes = $$\frac{1}{3}\pi\ (3r)^2h$$ : $$\pi r^2h$$ = 3:1 Answer
A window is in the shape of a rectangle surmounted by a semicircle. If the length of the rectangle is 6 feet and the radius of the semicircle is 2 feet, then the area of the window,in square feet, is
Solution
Solution:
Area of the window = Area of rectangular part + area of semicircular part
= $$l.b\ +\ \ \frac{\ \pi\ r^2}{2}$$ ......(Radius of semicircle = 0.5b)
= $$6\ \times\ \left(2\cdot2\right)\ +\ \ \frac{\pi\ \left(2\right)^{2\ }\ }{2}\ =\ 24\ +2\pi\ $$
= $$24\ +2\pi\ $$
11 cubic meters of steel is cast in to cylindrical bars of diameters 10 cm and the length 1.4 metres. Then the number of such bars that can be cast with the given metal is (Take $$\pi = \frac{22}{7})$$
Solution
Solution
Volume of given steel = 11$$m^3$$
Volume of the bar = $$\pi\ \left(\frac{0.1}{2}\right)^2\ 1.4\ m^3$$ = $$\frac{22}{7}\frac{0.01}{4}\ 1.4\ m^3\ =\ 11\cdot\frac{0.01}{2}\cdot0.2$$ = $$11\cdot0.01\cdot0.1\ =\ 0.011$$
No. of bars, n = $$\left(\frac{\left(Volume\ of\ steel\right)}{Volume\ of\ a\ bar}\right)\ =\ $$
$$\left(\frac{11}{11\ \times\ 10^{-3}}\right)\ =\ 10^{3\ }\ =\ 1000\ Answer$$
The difference between the largest three digit natural number x satisfying x=(mod 8)and the smallest three digit natural number y satisfying y = 2(mod 5) is
Solution
Solution
y will be in the form of 5x+2
If $$3^{48} = x(mod 10)$$ and if $$0 \leq x \leq 9$$ then $$x =$$
Solution
For,
$$3^{48} = x(mod 10)$$
We need to find remainder when $$3^{48\ }$$ is divided by 10.
3 has a cyclicity if 4. So as 48 is a multiple of 4, so last digit will be 1.
Hence remainder on division by 10 =1
1 Answer
If p and q are two statements, then which oneofthe followingis a tautology ?
Solution
Solution
Hence
Option A Answer
Which of the following is equivalent to the statement $$P \rightarrow Q$$?
Solution
None of the options
Set A has 5 elements and set B has 10 elements. Then the possible maximum number of elements in $$A \cup B $$ is
Solution
Solution
For maximum no. of elements, they have to be disjoint
n(AUB)= n(A) + n(B)
= 5+10 =15 Answer
Let A be non empty set and P(A) be the power set of (A). Let R be a relation defined on P(A) by $$ X R Y \Leftrightarrow X \subseteq Y \forall X, Y \epsilon P(A)$$. Then R is not
If a function $$f : R \rightarrow R$$ is defined by $$f(x) = x^{2} + 1$$, then $$f^{-1}(26) =$$
Solution
Solution
$$f(x) = x^{2} + 1$$, then $$f^{-1}(26) =$$
Let f(x) = y
$$y = x^{2} + 1$$, then $$f^{-1}(26) =$$
$$x\ =\ \sqrt{\ \left(y-1\right)}$$
putting y =26
$$f^{-1}(26) =$$ = $$\sqrt{\ \left(26-1\right)}=\sqrt{\ 25}\ $$
Range = (-5,+5) Answer
If the line passes through the points (3,-5) and (-2, 7), then x is the intercept is
Solution
Solution
$$y\ -y1\ =\ \frac{\left(y2-y1\right)}{\left(x2-x1\right)}\left(x-x1\right)$$
Putting the values,
y2 = 7, x2 = -2
y1=-5, x1= 3
$$y\ -\left(-5\right)\ =\ \frac{\left(7-\left(-5\right)\right)}{\left(-2-\left(3\right)\right)}\left(x-3\right)$$
$$y\ +5\ =\ \frac{\left(12\right)}{\left(-5\right)}\left(x-3\right)$$
$$-5\left(y\ +5\right)\ =\ \left(12\right)\left(x-3\right)=>-5y-25=12x-36$$
on reaarranging,
12x + 5y =11
For ,x intercept
putting y=0
$$x=\frac{11}{12}$$ Answer
The length of the intercept made by the line which passes through the points (2, 5) and (-4, 8) between the coordinate axesis
Solution
Solution
A/q
Here A = (2,5) and B = (-4,8)
intercept lenght = CD = AE.
Also , AEB is a right angle triangle right angled at E.
So ACC. to pythagoras theorem
$$\sqrt{\ BA^{2\ }-\ BE^2}\ =\ AE$$ = CD
=
If $$\theta$$ lies in the first quadrant and $$5 \tan \theta = 4$$, then $$\frac{5 \sin \theta - 3 \cos \theta}{\sin \theta + 2 \cos \theta} =$$
Solution
Solution
Given $$5 \tan \theta = 4$$
So $$ tan \theta $$ = (4/5)
P = 4
B = 5
H= $$\sqrt{\ \left(4^2\right)\ +\left(5^2\right)}$$ = $$\sqrt{41\ }$$
$$ sin \theta $$ = (4/$$\sqrt{41\ }$$)
$$ cos \theta $$ = (5/$$\sqrt{41\ }$$)
So
= {5 (4/$$\sqrt{41\ }$$) - 3 (5/$$\sqrt{41\ }$$) } / {(4/$$\sqrt{41\ }$$) + 2 (5/$$\sqrt{41\ }$$) }
={20 $$\sqrt{41\ }$$ - 15 $$\sqrt{41\ }$$} / {4 $$\sqrt{41\ }$$ + 10 $$\sqrt{41\ }$$}
= $$\frac{5}{14}$$ Answer
If $$\theta +\phi = \frac {\pi}{4}$$, then $$(1 + \tan \theta) (1 + \tan \phi) =$$
Solution
Solution
If $$\theta +\phi = \frac {\pi}{4}$$, then $$(1 + \tan \theta) (1 + \tan \phi) =$$
We can assume, $$\theta\ $$ = 0
So $$\phi$$ = $$\frac{\pi}{4}$$
then putting values in,
$$(1 + \tan \theta) (1 + \tan \phi) =$$
= $$(1 + 0) (1 + \tan \frac{\pi}{4}) =$$ $$\left(1+0\right)\left(1+1\right)=1\times\ 2\ =\ 2$$
= 2
$$(\cos^{2}15^\circ - \cos^{2} 75\circ) (\sin 65^\circ . \cos 35^\circ — \cos 65^\circ . \sin 35^\circ) =$$
Solution
Solution:
$$(\cos^{2}15^\circ - \cos^{2} 75\circ) (\sin 65^\circ . \cos 35^\circ — \cos 65^\circ . \sin 35^\circ) =$$
$$\left(\cos\ 15\ +\cos\ 75\right)\left(\cos\ 15\ -\cos\ 75\right)\left(\sin65\cos35-\cos65\sin35\right)$$
$$\left(2\cos\ \left(\frac{90}{2}\right)\cos\left(\frac{60}{2}\right)\right)\left(2\sin\ \left(\frac{90}{2}\right)\sin\ \left(\frac{60}{2}\right)\right)\left(\sin65\cos35-\cos65\sin35\right)$$
$$\left(2\cos\ \left(45\right)\cos\left(30\right)\right)\left(2\sin\ \left(45\right)\sin\ \left(30\right)\right)\left(\sin65\ -35\right)$$
$$\left(2\ \left(\frac{1}{\sqrt{\ 2}}\right)\left(\ \frac{\sqrt{\ 3}}{2}\right)\right)\left(2\left(\frac{1}{\sqrt{\ 2}}\right)\left(\frac{1}{2}\right)\right)\left(\sin30\right)$$
$$\left(2\ \left(\frac{1}{\sqrt{\ 2}}\right)\left(\ \frac{\sqrt{\ 3}}{2}\right)\right)\left(2\left(\frac{1}{\sqrt{\ 2}}\right)\left(\frac{1}{2}\right)\right)\left(\frac{1}{2}\right)$$
On solving
$$\frac{\sqrt{\ 3}}{4}$$ Answer
Two poles of height 15 metres and 30 metres stand upright on a play ground .If the feet of the poles are 36 meters a part , then the distance between their tops are
Solution
Solution
A/q
AB = DE = 15m
So CD = CE - AB = 15m
Also given AE = 36m. It implies BD = 36.
Hence in right angle triangle BCD.
Base = 36m
Height = 15 m
So according to Pythagoras theorem Hypotenuse (or distance between their tops) will be 39m. Answer
If the polynomial $$ax^{4} + bx^{3} + 3x^{2} - 4x - 4$$ is divisible by $$(x^{2} - 1)$$ then $$(a, b) =$$
Solution
SolutionsThe Polynomial is divisible by ( x² -1 ), so it will have ( x² -1 ) as one of its factors.
Also, ( x² -1 ) = (x+1)(x-1)
So both +1 and - 1 will be the roots of the polynomial.
Hence f(1) and f(-1) will be equal to 0.
f(1) = a(1) + b (1) + 3 (1) - 4 - 4 = a + b + 3 - 4 - 4 = a + b - 5 = 0 ….(i)
f(1) = a(1) + b (-1) + 3 (1) + 4 - 4 = a - b + 3 = 0 ……..(ii)
On solving (i) and (ii)
a = 1 and b = 4 .Hence Answer (1,4)
If $$(x - \alpha)$$ and $$(x - \beta )$$ are the two factors of the polynomial $$f(x) = ax^{2} + bx + c$$,then the quadratic polynomial whose factors are $$(x - \frac{1}{\alpha})$$ and $$(x - \frac{1}{\beta})$$ is
Solution
Solution:
If factors are (x- α) and (x - β) then roots of the expression will be α , β.
Product of roots = α x β = (c/a) ..........Eqn (i)
Sum of roots = α + β = -(b/a) ..........Eqn (ii)
In the new case,
Let polynomial be Mx² + NX + O
when factors are $$(x - \frac{1}{\alpha})$$ and $$(x - \frac{1}{\beta})$$Roots will be $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$
So product of roots = O/M = (1/α) x (1/β) = (1/αβ) = a/c …….from equation (i)
Sum of roots = -N/M = (1/α) + (1/β) = {(α + β) / αβ} = (-b/a) / (c/a) = - b/c…….from equation (ii)
So product of roots = a/c = O/M
Sum of roots = - b/c = - N/M
On comparison, the polynomial Mx2 + NX + O turns out to be
If $$63 \times 65 \times 67 \times 69$$ is divided by 12 then the remainder obtained is
Solution
Solution:
Principle used to solve :Rem (A x B) = Rem. (A) x Rem. (B)
A/q
When divided by 12
Rem. (63 x 65 x 67 x 69) = Rem. (63) x Rem.(65) x Rem.(67) x Rem.(69)
The common multiple of 12 samller than all the above nos. is 60. So taking 60 as reference.
Rem. (63) x Rem.(65) x Rem.(67) x Rem.(69) = Rem. (3 x 5 x7 x9) = Rem.(945)
945 leaves a remainder of 9 when divided by 12, as 936 is the closest multiple of 12 before 945.
Hence Answer 9
A polynomial in x leaves remainders 8, 4 when divided by (x + 2) and (x - 2) respectively. If the same polynomialis divided by $$x^{2} - 4$$ then the remainder is
Solution
A polynomial in x leaves remainders 8, 4 when divided by (x + 2) and (x - 2) respectively. If the same polynomialis divided by $$x^{2} - 4$$ then the remainder is
Let fn. be f(x).
As f(x) leaves remainder, 8 when divided by (x+2)
Therefore, f(-2)= 8
When f(x) is divided by (x−2), remainder = 4
Therefore, f(2)=4
Now, polynomial is divided by $$x^{2} - 4$$ or (x+2)(x-2).
the remainder will be linear. in the form of ax+b.
f(x)=q(x)(x+2)(x−2)+r(x)
When, x=2,
⇒f(2)=0.(x−2)q(2)+r(2)=4⇒f(1)=0.(x−2)q(1)+r(1)=4
r(2) = 2a+b = 4......(i)
When, x=-2
⇒f(-2)=0.(x+2)q(-2)+r(-2)=8⇒f(-2)=0.(x+2)q(-2)+r(2)=8
r(-2) = (-2a)+b = 8......(i)
Solving above two equations, we get remainder r(x) = 6-x Answer
If x + y = 12 y + z = 20 and z + x = 18 then x - y + z =
Solution
given that
x + y = 12 equation 1
y + z = 20 equation 2
z + x = 18 equation 3
on adding equation 1 + 2 + 3 we get
x + y + z = 25 equation 4 now perform operation eq4 - eq1, eq4 - eq2, and eq4 - eq3
we get
x = 5 ,y = 7 and z = 13 then put the value in eq following
x - y + z = 5 - 7 + 13 = 11 Answer
A Person has a certain number of two rupee coins and some five rupee coins.If the total number of coins is 16 and thwe value of coins is ₹50 ,then the number by which the number of two rupee coins exceeds number of five rupee coins is
Solution
let he have x number of 2 Rs coin and y number of 5 Rs coin
then we can say that
2x + 5y = 50 equation 1
x + y = 16 equation 1 on solving equation 1 and 2 we get
x = 10 number of 2 Rs coin
y = 6 number of 2 Rs coin
then by question x - y = 10 - 6 = 4 Answer
If thew first and sixth terms of geometric progessions are $$\frac{2}{3}$$ and 162 respectively then the $$8^{th}$$ term of the progression is
Solution
let the first term is a and common ratio is r then
$$T_{1}$$= {a}$$\times{r}^{1-1}$$ = a = $$\frac{2}{3}$$ equation 1
$$T_{6}$$= {a}$$\times{r}^{6-1}$$ = {a}$$\times{r}^{5}$$ = 162 equation 2
equation 1 \equation 2
$$\frac{2}{3}$$\162 = 1\r^5
243 = r^5
3 = r
r = 3
put the value of a and r in equation following equation
$$T_{8}$$= {a}$$\times{r}^{8-1}$$ = $$\frac{2}{3}$$ $$\times{3}^{7}$$
$$T_{8}$$ = $$\frac{2}{3}$$ $$\times243$$ $$\times3$$ $$\times3$$
$$T_{8}$$ = 1458 Answer
Three numbers are in an arithematic progession. their sum is 3 The third number is greater than first number by 16 ,then the product of those three number is
Solution
Let x-d,x and x+d be the three numbers.
Then in question given that
x - d + x + d + x = 3 or x = 1 and also given that
x + d - x + d = 16 or d = 8
then we get the sequence = -7,1 and 9 so then multiplication on all term
we get -63 Answer
The term independent of x in the expansion of $$\left(3x^{3} - \frac{4}{x}\right)^8$$
Solution
let $$(r+1)^{th}$$ term be the independent of x which is given by bio nominal theorem
$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ $$a^{(8-r)}$$ $$\times$$ $$b^{r}$$
here we have a = 3$$x^{3}$$ and b = -4\x
put the value we get
$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3$$x^{3}$$)^(8-r) $$\times$$ (-4\x)^r
$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{3}$$)^(8-r) $$\times$$ (-4)^r(x)^(-r)
$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{24-3r}$$) $$\times$$ (-4)^r(x)^(-r)
since the term is independent of x, we have
24 - 4r = 0
r = 6
so the (r+1) = 7 term is independent so we can say that
$$T_{7}$$ = 8$$c_{6}$$ $$\times$$ $$3^{3-1}$$ $$\times$$ (-4)^2(6-1)
$$T_{7}$$ = 8 $$\times$$ 7 $$\times$$ $$3^{2}$$ $$\times$$ $$2^{11}$$
$$T_{7}$$ = $$2^{14}$$ $$\times$$ 7 $$\times$$ $$3^{2}$$ Answer
If P is the sum of odd terms and Q is the sum of even terms in the expansion of $$(x + y)^n$$ then $$P^{2} - Q^{2} =$$
Solution
we know that the expansion of sum of even term and sum of odd term are respective given below
$$(a + x)^{n}$$ = n$$C_{0}$$ $$a^{n}$$ + n$$C_{1}$$ $$a^{n-1}$$x + n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ + ............. + n$$C_{n}$$ $$x^{n}$$
$$(a - x)^{n}$$ = n$$C_{0}$$ $$a^{n}$$ - n$$C_{1}$$ $$a^{n-1}$$x + n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ - ............. + $$(-1)^n$$ n$$C_{n}$$ $$x^{n}$$
adding both we get
$$(a + x)^{n}$$ + $$(a - x)^{n}$$ = 2( n$$C_{0}$$ $$a^{n}$$ + n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ + ............. + n$$C_{n}$$ $$x^{n}$$)
hence
P = ($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2 equation 1
Q = ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2 equation 2
there fore PQ we get
PQ = ($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2 $$\times$$ ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2 = ($$(a + x)^{2n}$$ + $$(a - x)^{2n}$$)\4
or 4PQ = ($$(a + x)^{2n}$$ + $$(a - x)^{2n}$$)
$$P^{2}$$ - $$Q^{2}$$ = (($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2)^2 - (($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2)^2
$$P^{2}$$ - $$Q^{2}$$ = $$(a + x)^{n}$$ $$(a - x)^{n}$$
$$P^{2}$$ - $$Q^{2}$$ = ($$(a)^{2}$$ - $$(x)^{2}$$)^n Answer
If A is a 3 x 3, square matrix and if A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$, then det(2A) =
Solution
given that A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$
means that
A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$ = |A|I = 4 $$\begin{bmatrix}1 & 0 & 0\\0 & 1&0 \\ 0 & 0 & 1\end{bmatrix}$$
∣A∣=4
det (2Adj A) = ∣A∣^(n−1) , where n is number of rows in a matrix
= 2$$\times$$4^(3-1)
= 2$$\times$$ 16
= 32 Answer
$$\lim_{x \rightarrow 5} \frac{x^{2} - 3x - 10}{x^{2} - 7x + 10} =$$
Solution
given that
$$\lim_{x \rightarrow 5} \frac{x^{2} - 3x - 10}{x^{2} - 7x + 10} $$
we can write this as
$$\lim_{x \rightarrow 5} \frac{x^{2} - 5x + 2x - 10}{x^{2} - 5x -2x + 10} =$$ $$\lim_{x \rightarrow 5} \frac{x(x-5) + 2(x - 5)}{x(x-5) - 2(x- 5)} =$$
$$\lim_{x \rightarrow 5} \frac{(x - 5)(x +2)}{(x - 5)(x -2)} =$$ $$\lim_{x \rightarrow 5} \frac{(x + 2)}{(x - 2)} =$$
put the value of x we get
7\3 Answer
If $$x = t - \frac{1}{t}, y = t + \frac{1}{t}$$ where t is a parameter then $$\frac{dy}{dx} =$$
Solution
given that
$$x = t - \frac{1}{t}, y = t + \frac{1}{t}$$
then
$$\frac{dy}{dx}$$ = $$\frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}}$$
then $$\frac{\text{d}y}{\text{d}t}$$ = 1\y and $${\text{d}x}{\text{d}t}$$ = 1\x
$$\frac{dy}{dx}$$ = 1\y\1\x = x\y Answer
If A is a square matrix and A^{T} denotes transpose then the matrix $$A. A^{T}$$ is always
Solution
we know that the
$$A. A^{T}$$ = $$A^{T}.A$$
so when we solve this always get symmetric matrix we found
When a piece of wire is bent in the form of equilateral triangle then the area of the triangle is $$121 \sqrt{3} cm^{2}$$. If the same piece of wire is bent in the form circle , then the area of the circle in square centimeters is (Take $$\pi = \frac{22}{7})$$
Solution
we know that the area of the eqitriangle = $$a^{2}$$ $$\times$$ $$\sqrt{3}$$ \4 = $$121 \sqrt{3}$$
a = 22
then the length of the wire = perimeter of the triangle
= 3a = 3*22 = 66
then the radius of the circle = 2$$\pi$$r = 2*$$\frac{22}{7})$$*r = 66
r = 21\2
then area of the circle = $$\pi$$ $$\times$$ $$r^{2}$$ = $$\frac{22}{7})$$ $$\times$$ $$\frac{21}{2})$$ $$\frac{21}{2})$$
then area of the circle = 346.5 cm square Answer
The sum of all the internal angles of a regular hexagon, in radians, is
Solution
we know that the sum of all the internal angles of a regular hexagon, in radians, is = 720 = 4$$\pi$$ Answer
If two circles of radii, 7 cm and 10 cm respectively touch each other internally, then the distance between their centres, in centimeters, is
Solution
If circles are touching internally the distance between their centers will be = radius of bigger circle - radius of smaller circle
= 10 - 3
= 3 cm Answer
note -
If circles are touching externally the distance between their centers will be = = radius of bigger circle + radius of smaller circle
If the mid points of the sides AB and AC of a triangle ABC are (4, -2), (-8, 14) respectively, then the length of the side BC, in units, is
Solution
let the mid points of the slides AB and AC of the triangle is D(4,-2) and F(-8,14)
which is parallel to the BC and BC = 2DF
then DF Length is DF = (-8 +4) + (!4 + 2) = 20
then BC = 2* 20 = 40 Answer
The distance between the lines $$\frac{x}{3} + \frac{y}{4} = 1$$ and 8x + 6y = 5 in units, is
Solution
we know the fourmula of distance between two line
$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$ = $$\pm$$ $$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$
then we have here $$(A_{1}$$ = 1\3 $$(B_{1}$$ = 1\4 and $$(c_{1}$$ = -1 and
$$(A_{2}$$ = 8 $$(B_{2}$$ = 8 and $$(c_{2}$$ = -5
then
|$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$| = $$\pm$$ |$$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$|
|$$\frac{1\3 + 1\4 - 1}\sqrt{{1\3}^{2}+{1\4}^{2}}$$| = $$\pm$$ |$$\frac{8 +6 -5}\sqrt{{8}^{2}+{6}^{2}}$$|
1 + 9\10 = 10\19 Answer
The mean of 6 observations is calculated as 24 and later it was noted that one of the observations is wrongly taken as 24 intstead of -24 then the correct mean is
Solution
when we calculated mean we ascending data
here if we take -24 instead of the 24 so it comes in first place as it is smallest value and affect first 3 value of 6 observation
so (24-(-24))\3 = 16 new mean Answer
The median of 15 observations is 32 .Each of the obeseravation is greater than the median are increased by 8 and each of the observations leaas than the median are decreased by 6 Then the median of new data is
Solution
Since 8 is less than 32 and 32 is more than 6, so the middle value remains unchanged and hence the median of new data observations taken together will be 32.
Answer
The mode of the following distribution: 19, 13, 16, 13, 19, 24, 13, 9, 24, 13, 26 is
Solution
Mode is the observation that occurs most frequently in the data
here 13 is repeated 3 times so 13 is our answer
If the standard deviation of the first n natural numbers is t then $$12t^{2} + 1 =$$
Solution
The standard deviation of n natural numbers = t
Therefore, but we know that
First n natural numbers are 1,2,3,…,n.
Mean=(1+2+3+…+n)÷n
=[n(n+1)/2]/n
(n+1)/2 …(i)
And, 1²+2²+3²+…+n²=n(n+1)(2n+1)/6
Or, (1²+2²+3²+…+n²)/n= {(n+1)(2n+1)/6} …(ii)
Variance of first n natural numbers $$\sqrt{t}$$ = {(1²+2²+3²+…+n²)/n} - (Mean)²
= {(n+1)(2n+1)/6}-{(n+1)²/4}
= (n+1)/12{4n+2-3(n+1)}
= (n+1)(n-1)/12
$$\sqrt{t}$$ = (n²-1)/12
and we have the equation given
$$12t^{2} + 1 =$$
t^2 = -1\12 put the of t we get
n^2 Answer
If the standard deviation of a variable x is 12 then the variance of the variable 2x + 17 is
Solution
Let say
Y = 2X + 17
Taking variance of both sides:
V(Y) = V(2X) + V(17)
V(Y) = 2 V(X) + 0
V(Y) = 2 * 12
V(Y) = 24 Answer
V(aX) = a^2 * V(X) .. where a is a constant
V(a) = 0 .. where a is a constant
If 3 coins are tossed, then the probability of getting at least 2 heads is
Solution
when the 3 cion are tossed the total number of cases are = $$2^3$$ = 8
and at least two heads are there in the following = HHT, HTH, THH, HHH = 4
then the probility of getting at least two heads = number of favourable cases \ total number of cases
then the probility of getting at least two heads = 4\8 = 1\2 Answer
When two unbiased dice are thrown , the probability that they show up different numbers is
Solution
when the two unbiased dice are thrown the number of total cases are = 36
let the both the dice show the same number then cases are = (1,1), (2,2), (3,3),(4,4),(5,5) and (6,6) = 6
then the probility of getting different number = 1 - probility of getting same number on both dice
then the probility of getting different number= 1 - 6\36
then the probility of getting different number = 1-1\6
then the probility of getting different number = 5\6 Answer
In a bivariate distribution $$X_i, y_i, i = 1, 2, ....., 8$$, if $$d_i$$ is the deviation between the the ranks of $$X_i$$ and $$Y_i$$ and and $$\sum_{i = 1}^{8} d_i^{2} = 21$$ then the coeffefficient of rank correlation between $$X_i$$ and $$Y_i$$ is
Solution
If X1, X2, ... , Xn are mutually independent normal random variables with means μ1, μ2, ... , μn and variances σ21,σ22,⋯,σ2nσ12,σ22,⋯,σn2, then the linear combination
of ranks $$X_i$$ and $$Y_i$$ means have equal value 4 and 4 as i = 1,2,3,4......8
Y = $$\sum_{i = 1}^{8} d_i^{2} $$ = 21
Then, finding the probability that X is greater than Y reduces to a normal probability calculation:
P(X>Y )= P(X − Y > 0) = P( Z > 0 − 1\4 ) = P(Z>−0.75 ) = P(Z<0.75) = 0.75 Answer
If 5 boys and 4 girls sit in a row at random then the probability the boys and girls sit alternatively is
Solution
Given that 5 boys and 4 girls sit in a row at random.
Total number of ways in which 9 persons can be arranged in a row = 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1
= 5 boys can be arranged themselves in 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.
B 1 B 2 B 3 B 4 B
Now,
Then there are 4 positions left(1,2,3,4,).
The 4 girls can be seated in (1,2,3,4) = 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.
The girls can be arranged themselves in 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.
Therefore, the number of ways both girls and boys can sit alternatively = 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1
Hence,
Required probability =
(5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\\times 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)\( 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)
= (120 $$\times$$ 24)\(9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)
= 4/504
= 1/126. Answer
When a leap year is selected at random then the probability that it has exactly 52 Sundays is
Solution
The question is asking the probability of getting "exactly" 52 Sundays and not at least 52 Sundays.
There are 52 weeks in a year, so there will be 52 Sundays, no doubts about it. However, 52 $$\times$$ 7 = 364 days
so that still leaves a day for a normal year, and two days for a leap year.
Now, let us look at the combinations of the two days possible.
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
Now, to get exactly 52 Sundays, none of the last two days should be a Sunday. As you can see from the combinations listed above, that is possible in 5 out of 7 cases. So the Probability would be 5/7. Answer
Choose the correct answer :
A situation in which goods or shares are plentiful and buyers can keep prices down is called
A differnce in wages between industries or between categories of Employees in the same industry referred to as
A large corportion formed by the merging of separate firms is known as
Celebraties who endorsed a product reffered to as
Stocks ,Shares ,Bonds are collectively known as
A computer language that is used to create websites is known as
connection of networks that can be joined together is possible through
ISDN stands for
ELDstands for
The connection between a computer and another piece of equipmentis
Choose the Correct Answer.
A: “How is your business ?”
B: “I’m broke!”
B’s answerindicates that
A: "you should not have spoken so harshly to him. you know that the faulty was yours "
B:"I know. I'll have to eat humble pie "
A: " I have two tickets for the movie tommorrow .Will you come with me ? "
B: "could you invite some one else ? movies give me a headache ."
"I dont think his jokes ever come off"
The speakers implies that the jokes were
The passive form of the sentence "They are not going to play the match today " is
A: “I’m sorry I couldn’t come to your party as I was out of town.”
B: “It’s alright. Don’t worry about it”.
In this conversation, the speaker “B”is
“When it comes to grammar, she’s really on the ball.”
The speaker implies that
Fill in the blanks with the appropriate phrase/verb/preposition.
A: “Are you going to the bank today ?”
B: “No. I've already to the bank.”
I hope the enquiry will ....... truth
The workers .......... their resignation when their demands were not met by the managing committee
Here is the watch you asked .....
The loan will be repaid ........ two months
It ...... raining for a while , but now its raining again.
They are ......... the whole family for Christmas.
A subordinate has to abide .......... the decisions ofhis superiors.
Read the following passage and answer questions :
In times of social fragmentation, vulgarity often becomes a wayoflife. To be shocking becomes more important than to be civil or creative or truly original. Given the degree of vulgarity in our society, cynicism seems almost irresistible. But cynicism represents a secession from society, a dissolution of the bonds between people and families and communities, an indifference to the fate of anything, beyond the self Cynicism is deadly. It bites everything it can reach-like a dog with a foot caughtin a trap. And then it devours itself. It drains us of the will to improve; it diminishes our public spirit; it saps our inventiveness; it withers our souls. Cynics often see themselves as merely being world-weary. There is no new thing under the sun, the cynics say. They claim their weariness is wisdom. Butit is usually mere posturing . Their weariness seems to be most effective when they consider the aspirations of those beneath them, who have neither powernor influence nor wealth. For these unfortunates, nothing can be done, the cynics declare. Hope is considered an affrontto rationality : the notion that the individual has a responsibility for the community is considered a dangerousradicalism. Ultimately, however, the life of a cynic is lonely and self - destructive. It is our human nature to make connections with other human beings. The gift of symapthy for another is one of the most powerful sentiments we ever feel.If we do not have it, we are not have human .Indeed it is so powerful that the cynics who denies it goes to war with himself.
What is Considered normal in today's society?
What have divisions in the society led to
Why cynism dangerous
What charecterises the behaviour of a cynic ?
What quality makes us human?
Read the following passage and answer questions
Production may not be a dominant ideological leverage in economic decision making, but it cannot be altogether dismissed from economic calculations at whatever level. The difference that globalization has brought about is not that it has made production concerns immaterial, but that the leading factor is now the consumer and production must do what it can to meet the demands of the market. So every nowand again, one comes up against serious production problems at the national level which are resolved at the global level by importing skills or machinery. Thus while it is true that consumer tastes are getting more homogenized across the world, production facilities have not kept up. Economic difficulties at the national level arise from this mismatch.
What cannot be kept out of economic calculations ?
What is the most prominent consideration today ?
Which word does the writer use to describe buyer’s tastes ?
What has caused the change in approach to economic decision-making ?
At the national level, whatis the effect of the discrepancy between production and demand ?
Read the following passage and answer questions
The land in Egypt is very fertile when it ts irrigated, and water for irrigation has to come mainly from the one big river which Egypt has, namely, the Nile. Therefore, most of the farming is done near this river. As the Nile nears the sea, it splits into several branches to form a delta. The land here is very flat; ploughing and irrigation are easy andit is, therefore, the most highly cultivated arca. Further inland, away from the sea, only the strips of land on either side of the river are cultivated, There are, of course, a few irrigation projects and model farms, but a large proportion of the Egyptian country side remains a dry, sandy desert. In the delta, the main crops are cotton, rice, fruit such as banana, dates, figs and melons, and vegetables. Cattle, buffalo, and goats are also kept for milk and meat. Further inland, however, where the land is not so flat, and water is scarce, the cultivation is less intense; in this part of the country, there is a scarcity of fresh food, such as fruit and vegetables, because there is not enough water to grow it. Sugarcane and wheat are the main crops here. In the huge, barren deserts, there are wandering herdsmen who keep goats and camels. Thusit is clear that vast areas of land are lying unused at present only because ofthe lack of irrigation. The country can double its production if the irrigation projects multiply.
What is the meaning of ‘irrigation’ ?
What does the Nile form whenit reaches the sea ?
What are cultivated towards the interior of the city ?
Who are the ‘wandering herdsmen’ ?
How can agricultural production be increased in Egypt ?
Choose the correct meaning for the word given:
bulwark
cower
plebiscite
bizarre
Platitude
feign
Fill in the blank choosing the correct word:
Conspirators were satirically called ............ men.
The ........ of informationin the organization will be given high priority.
Bitter experiences make us ............ though wiser.
The ......... of the flowers enhanced the beauty of the garden.
