Is 6 a factor of n + 3?
I. n is even and divisible by 3.
II. n is divisible by a prime number.
AP ICET 2015 Question Paper
In questions a question is followed by data in the form of two statements labelled as I and II. You must decide whether the data given in the statements are sufficient to answer the questions. Using the data make an appropriate choice from (a) to (d) as per the following guidelines :
The product of two integers a and b is 270. What is the LCM of a and b?
I. The GCD of a and b is 9.
II. a and b are divisible by 3.
What is the value of the positive integer n ?
I. $$1^2 + 2^2 + ...... + n^2 = 285$$
II. $$3 < n < 20$$
If a and b are positive integers such that the product ab is 30, then what is the value of a + b ?
I. a > b
II. $$1 < \frac{a}{b} < 2$$
If a, b, c are integers, is a + b + c divisible by 3?
I. a = 2b - 10 and c = 3b + 25
II. a + b + c is an odd integer.
If x is a non-negative integer, is x even ?
I. $$8^x$$ is an even integer.
II. $$9^x$$ is an odd integer.
Is $$\triangle ABC$$ equilateral ?
I. AB = BC
II. $$\angle{ABC} = 60^\circ$$
Is a + b ≠ 0, if a and b are real numbers ?
I. ab = 0
II. $$a^2 + b^2 = 0$$
If x and y are real numbers, is x > y ?
I. x is the smaller of the roots of $$x^2 - 1 = 0$$
II. y is the real root of $$y^3 - 1 = 0$$
If a and b are real numbers, is ab > 0?
I. $$2 \mid a \mid + \mid b \mid = 0$$
II. $$2 \mid a \mid = \mid b \mid$$
At how many points do the two circles intersect ?
I. The radii of the two circles are equal.
II. The distance between their centres is 12 cms.
What is the share of A in a profit of ₹ 1,500 in that year ?
I. A’s capital is ₹ 5,000 more than that of B.
II. A and B are the only partners in the business.
For two sets A and B, what is $$n(A \cap B)$$ ?
I. n(A) = 6 and n(B - A) = 8
II. n(A - B) = 8 = n(B - A) and $$n(A \cup B) = 24$$
If ‘a’ is a positive integer, is it a prime number ?
I. It is odd number,
II. It is an even integer greater than 2.
If x and y are real numbers, what is the value of xy?
I. x + y = 16
II. $$x^2 + y^2 = 160$$
What is the value of $$\sin (2 \theta)$$ ?
I. $$tan \theta + \cot \theta = 3$$
II. $$(\cosec \theta + \cot \theta)(\cosec \theta - \cot \theta) = 1$$
What is the sum of the first 20 terms of the Anthmetic Progression ?
I. The sum of the first term and $$20^{th}$$ term of the AP is 120.
II. The common difference of the AP is 5.
What is the surface area of the cuboid ?
I. Ratio of its length to its breadth is 2 : 5.
II. Volume of the cuboid is 100 cu. units.
If ‘a’ is an integer, what is the value of a ?
I. $$a^2 = 9$$
II. $$a^2 + a = 6$$
If a and b are real numbers, is $$a^2 > b^2$$
I. a > b
II. a > 0
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.
6 : 210 :: ....... : 120
$$\frac{3}{4} : \frac{5}{6} : \frac{7}{8} : .......$$
5 : 3125 :: 6 : .......
BFJ : OSW :: DJR : ......
136 : 358 :: ........ : 489
1331 : 11 :: 1728 : ...........
6FJ : 16PT :: 8HL : ......
64 : 100 :: 72 : ......
9 : 72 :: ...... : 64
Train : Driver :: Class : .........
In following questions pick the odd thing out :
Each of the questions follow a definite pattern. Observe the same and fill in the blanks with suitable answers.
C, F, I, L, ......, R
AZB, EXD, IVF, OTH, ........
$$\frac{2}{3}, \frac{5}{7}, \frac{9}{11}, .........., \frac{19}{23}$$
$$\frac{5}{3}, \frac{29}{6}, .............. \frac{131}{12}, \frac{179}{15}$$
2, 6, 12, 20, ......., 42
2, 9, 28, 65, ......, 217
1, 5, 9, 15, ........
1, 3, 6, 11, 18, .........
0, 7, 26, ......, 124, 215
$$\sqrt{12}, \sqrt{80}, \sqrt{252}, ....., \sqrt{1100}$$
In the following table, production of various crops (in tons) is given from 2000 to 2004. Study this table and answer the questions.
The highest growth rate of sugarcane production over its previous year is recorded in the year
The average production of wheat (rounded off to the nearest integer) in the period given in the table.
The difference between the average production of Barley and average production of Rice is
The percentage growth of sugarcane in the year 2004 over the year 2000 is approximately
In 2001, the crop that recorded the highest growth percent over its previous year, is
In the following Pie diagram, people of difference age groups of a town are represented by the angle made by that sector at the centre. The Total population of that town is 72,000. Based on this information, answer the questions.
In the above diagram :
A stands for less than 10 years of age.
B stands for 10 to 20 years of age.
C stands for 20 to 30 years of age.
D stands for 30 to 40 years of age.
E stands for 40 to 50 years of age.
F stands for above 50 years of age.
The people in the age group 40 to 50 are less than thosein the age group 20 to 30 by
The ratio of people that are more than 50 years to those that are less than 10 yearsis
The People in the age group 10 to 20 and 30 to 40 years put together exceed those of age group 40 to 50 by
The percentage of people in 30 to 40 years age group in the total population is
Number of people in the age group 20 to 30 years is
The letters of the English alphabet are arranged around a circle and are coded as follows:
(1) A vowel is coded as third vowel from it in the clock-wise direction.
(2) A consonant is coded as fifth consonant from it in the clock-wise dorection.
Reverse process is used for decoding. Answer questions using this coding and decoding processes.
The letter that is coded as U is
The code word for MASTER is
The string of letters that is coded as NORMAL is
The code word for TIGER is
The code word for APICET is
The string of letters which is coded as MUSIC is
The Code word for MATCH is
The letters P, T, O are coded respectively as the letters
The code word for PRINCE is
The letter X is coded as
For the following questions answer them individually
The angle between the two hands of a clock at 8.20 AM is
The number of seconds in $$\frac{11}{36}$$ of an hour is
A is father of C, D is son of B and E is brother of A. If C is sister of D, then how is B related to E ?
Sunil reached the venue of a meeting at 8.50 am. He found that he was 30 minutes earlier than the chairperson who came 40 minutes late. The meeting was scheduled at
Given that $$x * y = \frac{x^2 + 4y^2}{xy}$$ and $$x \triangle y = \frac{x^2}{y}$$ for any real numbers x and y. If $$a * b = 4 \triangle 4$$ then b =
For any $$a, b \epsilon R$$, define
a * b = max {a, b} and aob = min {a, b}. Then a * (boc) =
In 20 seconds the hours hand in a clock moves an angle
$$1 + \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + ...... $$ to infinity =
{$$x^2 : x$$ is an integer, $$\mid x - 2 \mid < 4} =$$
If the year 2003 starts with Monday, then it ends with
The greatest number which divides 121, 134 and 147 leaving 4 as the remainder in each case is
The reciprocal of the sum of the reciprocals of $$\frac{5}{8}$$ and $$\frac{3}{4}$$ is
Solution
Sum of the reciprocals of $$\frac{5}{8}$$ and $$\frac{3}{4}=\frac{8}{5}+\frac{4}{3}=\frac{24+20}{15}=\frac{44}{15\ }$$
$$\therefore\ $$Reciprocal of the Sum of the reciprocals of $$\frac{5}{8}$$ and $$\frac{3}{4}=\frac{15}{44}$$
Hence, the correct answer is Option C
The correct order of $$a = \frac{3}{4}, b = \frac{4}{7}, c = \frac{11}{13}$$ and $$d = \frac{13}{15}$$ is
$$2\frac{1}{3} + 3\frac{3}{4} \div \frac{3}{4} - 2\frac{1}{3}\left(6\frac{1}{2} - 2\frac{1}{4}\right) \div 2\frac{1}{4} =$$
Solution
$$\frac{7}{3}+\frac{\frac{15}{4}}{\frac{3}{4}}-\frac{\frac{7}{3}\left(\frac{17}{4}\right)}{\frac{9}{4}}$$
$$\frac{7}{3}+5-\frac{119}{27}$$
$$\frac{63}{27}+\frac{105}{27}-\frac{119}{27}=\frac{79}{27}$$
Two numbers x and y are respectively 20% and 25% more than the third number z. If x is a% of y, then a =
Solution
x is 20% more than z
$$\Rightarrow$$ x = $$\frac{120}{100}$$z
y is 25% more than z
$$\Rightarrow$$ y = $$\frac{125}{100}$$z
Also, x is a% of y
$$\Rightarrow$$ x = $$\frac{a}{100}$$y
$$\Rightarrow$$ $$\frac{120}{100}$$z = $$\frac{a}{100}\times\frac{125}{100}$$z
$$\Rightarrow$$ a = 96
By selling an item for ₹ 754, a trader gets 16% profit. Then the cost price of the item in rupees is
Solution
S.P=754
trader get 16 % profit
so 116 unit =754
1 unit=6.5
100 unit =650
So C.P is 650.
A pipe can fill an empty tank in 15 hours. Due to a leak at the bottom, the tank is filled in 20 hours. If the tank is full, then in how much time (in hours) will the leak take to empty the tank ?
Solution
Pipe can fill in 15 hr
Pipe along with Leak hole take 20 hr
So lcm of both 60
hence efficiency of pipe is 60/15= 4
Efficiency of both is 60/20= 3
So efficiency of leak is 4-3=1 so the leak hole take 60/1 = 60 hr to leak out all water.
Two pipes A and B can fill an empty tank independently in 3 hours and 4 hours, respectively. If the pipes A and B are opened in alternate hours starting with B, then the total time taken, in hours, to fill the tank is
Solution
A and B can fill the tank in 3 hours and 4 hours respectively.
Let the capacity of the tank be 12 units (LCM of 3 and 4)
Efficiency of A = $$\dfrac{12}{3} = 4$$ units/hour.
Efficiency of B = $$\dfrac{12}{4} = 3$$ units/hour.
If A and B are open every alternate hour starting with B, then in 2 hours, 7 units of the tank is filled.
Remaining capacity = 12-7 = 5 units.
3 units will be filled by B in 1 hour.
Remaining 2 units will be filled by A in $$\dfrac{1}{2}$$ hour.
Therefore, Total tank is filled in $$2+1+\dfrac{1}{2} = 3\dfrac{1}{2}$$ hours.
Two cars A and B start at the same time ad move in opposite directions between the places x and y. The time taken by Car A and Car B to reach their destinations after they meet each other on their way is 4 hours and 9 hours respectively. Then the ratio of the speeds of cars A and B is
Solution
A car travels at $$\frac{8}{7}$$ times its usual speed and reaches the destination 15 minutes early. Then the time taken by the car in hours to reach its destination with its usual speed is
Solution
New Speed of car is 8/7 of old speed
So new time is 7/8 of old time
8-7=1 unit =15 min early
So original time is 8×15=120 min or 2 hr
A can turnout twice the amount of work that B can do. If B alone can complete that work in 12 days, then in how many days can A and B complete that work together ?
Solution
A is twice efficient than B
So A =2
B=1
Total efficiency of A & B=3
B can complete the work in 12 days
So total work is 12×1=12
A & B complete the work in together is =12/3=4 days
2 men and 3 women can do a particular work in 10 days, whereas 4 men can do that work in 10 days. Then in how many days can 3 men and 3 women complete the same work ?
Solution
2 men+3 women complete the work in 10 days.
2 men +3 women =10 days
4 men = 10 days
So equate both equation
2 men + 3 women =4 men
3 women = 2 men
men/women=3/2
Total work is= (2 m+3 w) ×10 days =(2×3+3×2)×10=12×10=120
So 3 men + 3 women=3×3+3×2=15
So 120/15=8 days.
The sides of a rectangle of perimeter 52 meter are in the ratio 4 : 9. Then the area in square meters of that rectangle is
Solution
Breadth to Length ratio is 4:9
So let Breadth is 4x & Length is 9x
We know perimeter of rectangle is 2(L+B)
2(4x+9x)=52
26x=52
X=2
so Length is=9x=9×2=18
Breadth=4x=4×2=8
So area of rectangle is L×B
18×8=144 meter square.
The radius of the base of a right circular cylinder is r and the radius of a sphere is $$\frac{r}{2}$$. If the volumes of that cylinder and the sphere are equal, then the height of the circular cylinder in proper units is
Solution
Volume of Cylinder =V=$$\pi r^2 h$$
Volume of sphere V=$$\frac{4}{3}\times \pi (r/2)^3$$
Cylinder Volume = Sphere Volume
So $$\pi \times r^2 \times h = \frac{4}{3} \times \pi \times (r/2)^3$$
$$h=r/6$$
The volume of a cube (in cubic feet) whose total surface area is 384 square feet is
Solution
Total surface area of cube 6a^2=384
So a=8 feet
Where a = side of cube
So volume of cube is a^3=(8)^3=512 cube feet
A copper sphere of 6 cm diameter is melted to prepare a long wire of diameter 0.2 cm. Then the length of the wire (in metres) is
Solution
If $$\mid \frrac{7 - 2x}{4} \mid = 2$$, then x =
Solution
For integers a, b and m > 0, write $$a \equiv b$$ (mod m)if m divides a - b. Thea Which one of the following is true ?
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of lenghth 18 meters of uniform diameter. Then the diameter of the wire in centimeters is
Solution
In a class, the number of boys and girls are in the ratio 2 : 1. If five boys leave the class and five girls join class. Then the ratio becomes 7 : 5. The number of boys in that class at the beginning is
Solution
If $$A : B : C :: 2 : 3 : 4$$, then $$\frac{A}{B} : \frac{B}{C} : \frac{C}{A}$$ is
Solution
$$\frac{1}{\sqrt{3} + \sqrt{2} = x\sqrt{2} + y\sqrt{3} \Rightarrow x - y =$$
The curved surface area of a cone is 4070 sq. cm. If the diameter of the base of that cone is 70 cm, then its slant height in centimeters is
Solution
A solution of $$8x \equiv 3(mod 21)$$ is
If 20 and 24 are factors of n, then which one of the following is a factor is a factor of n ?
The value of $$\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{6}\right)\left(1 - \frac{1}{7}\right) ........ \left(1 - \frac{1}{20}\right) =$$
If 12% of x = 8% of y, then the ratio x : y is
Solution
The rate of simple interest per annum that makes some amount of money double in 12 years is
Solution
$$\frac{3^{\frac{3}{5}}. \sqrt[5]{3^{12}}}{27^{\frac{5}{3}}} =$$
Solution
If a, b, c are real and a + b + c = 0, then $$x^{a^2b^{-1}c^{-1}}.x^{a^{-1}b^{2}c^{-1}}.x^{a^{-1}b^{-1}c^{2}} =$$
Solution
1f $$7^{3x + 1} = 49^{x + 2}$$, then x =
Solution
If $$9x^2 + 16y^2 - 24 xy = 0$$, then the ratio $$y : x$$ is
Solution
If $$\sqrt{19 - 4\sqrt{x}} = \sqrt{12} - \sqrt{7}$$, then $$x - 1 =$$
$$\sqrt{15 + 2\sqrt{15} + 2\sqrt{21} + 2\sqrt{35}} + \sqrt{15 - 2\sqrt{15} - 2\sqrt{21} + 2\sqrt{35}} =$$
The greatest three digit number which whendivided by 6, 9 and 12 leaves the remainder 4 in each case is
Solution
If (-4, 5) is a vertex and $$7x - y + 8 = 0$$ is the equation of a diagonal of a square, then the equation of other diagonal of the square is
Solution
Equation of the straight line passing through the point (1, -5) and making intercepts in the ratio 2 : 3 on the axes is
Solution
If $$\sec \theta + \tan \theta = \frac{2}{3}$$ then the quadrant in which $$\theta$$ lies is
$$2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) =$$
Solution
If $$\cos \theta + \sec \theta = 2$$, then $$\cos^4 \theta + \sec^4 \theta =$$
If the angles of elevation of the top of a tower of height 60 m and the top of a flag staff on the tower from a point on the level ground are $$30^\circ$$ and $$60^\circ$$ respectively, then the length of the flag staff in metres is
$$\cos^2 45^\circ + \cos^2 135^\circ + \cos^2 225^\circ + \cos^2 315^\circ =$$
Solution
The remainder obtained when $$7x^2 - 13x + 5$$ is divided by $$x - 3$$ is
Solution
If the polynomial $$7x^4 + 4x^3 - 6x^2 + ax + b$$ is divisible by $$x - 1$$ and $$x + 1$$, then $$a + b =$$
The remainder obtained when $$\sum_{n = 1}^{100}(n!)$$ is divided by 6 is
The sum of the first 10 terms of the progression 2, 5, 8, 11, .... is
Solution
Solution
Sum of A.P. = $$\frac{n}{2}$$ (2a + (n-1)d)
Here, a = 2
d = 8 - 5 = 5 - 2 = 3
and A/q n = 10.
Putting the values in the Formula
S = $$\frac{10}{2}$$ (2 x 2 + (10-1)3)
=5 (4 + 27)
= 5 x 31
= 165 Answer
If the $$n^{th}$$ term of the G.P. $$2, 2\sqrt{2}, 4, ......$$ is 128, then n =
Solution
Solution
Here a = 2
and r = $$\sqrt{2}$$
nth term = 128
$$T\ \left(n\right)\ =\ a.r^{\left(n-1\right)}$$
Putting the values,
128 =$$\ 2.\sqrt{2}^{\left(n-1\right)}$$
$$64=2^{\frac{\left(n-1\right)}{2}}$$
$$2^6=2^{\frac{\left(n-1\right)}{2}}$$
Comparing the exponents
$$6={\frac{\left(n-1\right)}{2}}$$
On solving,n = 13 Answer
The constant term in the bionomial expansion of $$\left(3x^2 - \frac{1}{4x}\right)^6$$ is
Solution
Solution:
The constant term in the bionomial expansion of $$\left(3x^2 - \frac{1}{4x}\right)^6$$ is
$$^{n}C_r$$ $$\left(3x^2\right)^{\left(6-r\right)}\left(-\frac{1}{4x}\right)^r$$
$$3^{\left(6-r\right)}x^{\left(12-2r\right)}\frac{\left(-1\right)^r}{4^rx^r}$$
= Constant $$x^{\left(12-2r\right)}x^(-r)$$
= Constant$$x^{\left(12-2r\ -r\right)}\ =Constant\ x^{\left(12-3r\right)}$$
For constant term , power of x should be 0
12-3r = 0
r =4
Constant = $$^6Cr\ 3^{\left(6-r\right)}\ \frac{\left(-1\right)^r}{\left(4\right)^r}\ =\ \frac{6!}{\left(6-2\right)!2!}.3^{\left(6-4\right)}\left(-\frac{1}{4}\right)^4$$
=$$=\ \frac{6!}{\left(4\right)!2!}.3^{\left(2\right)}\frac{1}{4^4}^{ }=\frac{5.6}{2}.9.\left(\frac{1}{256}\right)$$
= 135/256
If $$C_r = ^{10}C_r$$, for $$0 \leq r \leq 10$$, then $$2C_0 + 3C_1 + 4C_2 + ......... + 11C_{10} =$$
Solution
Solution
$$2.^{10}C_0$$ +$$3.^{10}C_1$$ + $$4.^{10}C_2$$.....
=$$2.1\ +\ \frac{3.10!}{1!9!}\ +\frac{4.10!}{2!8!}\ +\frac{5.10!}{3!7!}$$
= 2 + 30 + 180 + 600...
If A is a $$4 \times 4$$ matrix and det 2A = 192, then det A =
Solution
Let A be an n × n matrix and c be a scalar then:
det (cA) = $$c^n\ \det\ \left(A\right)$$
det (2A) = $$2^4\ \det\ \left(A\right)$$ = 192
$$\det\ \left(A\right)=\ \frac{192}{2^4}\ =\ \frac{192}{16}\ =\ 12$$
12 Answer
$$\left(\begin{array}{c}2 & 6\\ 1 & 3\end{array}\right)\left(\begin{array}{c}3 & -6\\ -1 & 2\end{array}\right) =$$
Solution
Solution:
$$\left(\begin{array}{c}2 & 6\\ 1 & 3\end{array}\right)\left(\begin{array}{c}3 & -6\\ -1 & 2\end{array}\right) =$$
let the resultant matrix be, $$\left(\begin{array}{c}A11 & A12\\ A21 & A22\end{array}\right)$$
A11 = 2.3 + 6.(-1) = 0
A12 = 2.(-6) + 6.(2) = 0
A21 = 1.3 +3.(-1) = 0
A22 = 1.(-6) + 3.2 = 0
= $$\left(\begin{array}{c}0 & 0\\ 0 & 0\end{array}\right)$$
$$\lim_{x \rightarrow 3} \frac{x^3 - 3x^2}{3x^2 - 4x - 15} =$$
Solution
Solution:
As on putting limit ,$$\lim_{x \rightarrow 3}$$ f(x) -> $$\frac{0}{0}\ format$$
So using L hospitals rule
Differentiating Nr and Dr. $$\lim_{x \rightarrow 3}$$ $$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}$$
Putting the limits,
$$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}\ =\ \frac{\left(3.3^2\ -\ 6.3\right)}{6.3\ -\ 4}=\ \frac{\left(27-18\right)}{14}=\ \frac{9}{14}$$
$$\ \frac{9}{14}$$ Answer
$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$
Solution
Solution
$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$
As the fraction is in the form of $$of\ the\ \frac{0}{0}\ form$$
We can use L' Hospitals rule
Differentiating Nr and Dr.
$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$ $$\frac{d}{dx}\left(4^{3x}\ -\ 1\right)\ =\ \left(\ln4.\ 4^{3x}.3\right)\ -\ 0$$
Now again putting limits $$\left(\ln4.\ 4^{3x}.3\right)\ \ =\ \frac{\left(\ln\ 4.4^0.3\right)}{1}\ =\ 3\ln4$$ Answer
$$\frac{\text{d}}{\text{d}x}{(\cos(3\log_e x))} =$$
Solution
we will use chain rule, moving inwards
there are 3 functions in the form: cos K, 3M, and log Z
For cos K differentiation, -sin K = -sin (3log x)........where K = (3 logx)
For 3M , differentiation, =3 ..................
For log z, differentiation = 1/z = 1/x
Combining all the three
$$\frac{\text{d}}{\text{d}x}{(\cos(3\log_e x))} =$$ = {-sin (3log x)}{-3}{1/x} =
Answer
The internal angle between two adjacent sides of a regular pentagon in radians is
Solution
Solution
Sum of interior angles = S = (n - 2)*180
S = (5-2) X 180 = 540
Each interior angle = 540/5 = 108
in radians $$\frac{108}{180}\ \pi\ \ radian\ =\ \frac{3}{5}\pi\ radian$$ Answer
If the lengths ofthe side and one diagonal of a rhombus are respectively 13 cm and 10 cm, then the area of the rhombus in square centimeters is
Solution
Soution:
The diagonals of rhombus form right angle at the centre.
AO = 10/2 = 5 cm
AD = 13 cm
So DO = $$\sqrt{\ \left(AD^2\right)\ -\ \left(AO^2\right)}$$
= $$\sqrt{\ \left(13^2\right)\ -\ \left(5^2\right)}\ =12\ cm$$
Length of diagonal = 24
Area of diagonals = $$\frac{1}{2\ }\Pr oduct\ of\ diagonals\ =\ \frac{1}{2\ }\times\ \ 10\ \times24\ =120\ cm^2\ $$
The point on x-axis which is equidistant from the points (7, 6), (-3, 4) is
Solution
Solution:
Any point on X-axis is represented as (x,0)
Distance of this point from (7,6) = $$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$=
Distance of this point from (-3,4) = $$\sqrt{\ \left(-3-x\right)^2\ +\left(4-0\right)^2}$$
Both the distances should be same.
$$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$ = $$\sqrt{\left(\left(-3-x\right)^2\ \ +\left(4\right)^2\right)\ }$$
$$\sqrt{\ \left(49+x^2\ -14x\right)+36}\ =\ \sqrt{\ 9+x^2+6x+16}$$
$$\left(49+x^2\ -14x\right)+36\ =\ 9+x^2+6x+16$$
$$60=20x$$
$$x\ =\ 3$$
So, point = $$\left(3,0\right)$$ Answer
The ratio in which x-axis divides the line segment joining the points (3, -4) and (-2, 5) is
Solution
Solution
Let the required point be (x,0). Any point on x axis will have y ordinate 0
The two given points are
(3,-4) and (-2,5)
Using the formula
As y =0, so equating y part to 0.
0 = $$\frac{\left(m1\left(5\right)\ +\ m2\ \left(-4\right)\right)}{m1+m2}\ =\ \frac{\left(5m1+-4m2\right)}{m1+m2}=0$$
$$\ \left(5m1+(-4m)2\right)=0\ ;Hence\ \frac{m1}{m2}\ =\frac{4}{5}$$ External division. Answer
The inverse of the statement $$p \rightarrow (q \wedge r)$$ is
Solution
Solution:
inverse of $$p \rightarrow (q \wedge r)$$
~(p -> (q^r))
~ p -> ~(q^r)
~ p ->(~q) v (~r) Answer
If p, q are two statements, then which one is a tautology among the following ?
Solution
Solution
A/q
So correct option C
The number of reflexive relations that can be defined on a set A with 5 elements is
Solution
Solution
No. of reflexive relation = $$2^{n\left(n-1\right)}$$
here n = 5
A/q
No. of reflexive relation = $$2^{5\left(5-1\right)}$$ = $$2^{20}$$
Let $$A = \left\{x \epsilon R/x^2 < 9\right\}$$ and $$B = \left\{x \epsilon R/4 < x^2 < 16\right\}$$ then $$A \cap B =$$
Solution
A/q
A = (-3,3)
B = (-4,-2) U (2,4)
So $$A \cap B = $$ (-3,-2) U (2,3) Answer
If A, B are two sets such that $$n(A - B) = 40, n(B - A) = 57$$ and $$n(A \cap B) = 16$$ then n(A) =
Solution
Solution
n(A)=n(A-B)+n(A∩B)
=40+16
=56
Therefore, n(A) is 56.
Now n(B)=n(A∩B)+n(B-A)
Therefore, n(B) is 16+57 = 73.
n(B)=73.
Hence n(B)=73 Answer
Let $$f: R \rightarrow R$$ be a function defined by $$f(x) = 2x^3 + 8$$ for all $$x \epsilon R$$, then the function f is
Solution
Solution:
The graph will be similar to that of $$x^3$$, just displaced 8 units upwards.
As, a function is a one-to-one function when it will pass the vertical line test (to make it a function) and also a horizontal line test (to make it one-to-one).
So it is one-one.
Also,
Also, as you progress along the line, every possible y-value is used.So it is also Onto.
Hence it is bijection.
Let Q be the set of all rational numbers and $$f : Q \rightarrow Q$$ be a function defined $$f(x) = \frac{3x - 1}{10}$$ for all $$x \epsilon Q$$. Then $$f^{-1} (5) =$$
Solution
Solution
Given $$y\ =\ \frac{\left(3x-1\right)}{10}$$
$$10\ y\ =\ 3x-1$$
Hence$$f^{-1} (x) =$$ = $$\frac{\left(10x+1\right)}{3}$$
Putting x = 5
$$\frac{\left(10x+1\right)}{3}\ =\ \frac{\left(10\cdot5+1\right)}{3}=\frac{51}{3}=17$$
The standard deviation of first 13 natural numbers is
Solution
Solution:
Formula used
Here, n = 13
$$\sqrt{\frac{\left(\left(\ 13\right)^2\ -\ 1\right)}{12}}\ =\ \sqrt{\ \frac{\left(13+1\right)\left(13-1\right)}{12}}\ =\ \sqrt{\ \frac{14.12}{12}}\ =\ \sqrt{\ 14}$$
$$\sigma\ =\ \sqrt{\ 14}\ $$ Answer
If the mean and median of a data are respectively 56 and 48, then the mode of the data is
Solution
Solution:
Relation used
$$3median=mode+2mean$$
3 median - 2 mean = mode
3 x 48 - 2 x 56 = 32 Answer
The mean deviation of the observations 15, 17, 20, 23 and 25 about the arithmetic mean is
Solution
Solution:
The mean of given set is =$$\frac{\left(15+17+20+23+25\right)}{5}$$ = 20
mean deviation of the observations about the arithmetic mean = $$=\frac{\left(5+3+0+3+5\right)}{5}\ =\ \frac{16}{5\ }=3.2$$
3.2 Answer
If 8 persons sit in a row at random, then the probbility that 3 particular persons among them sit together in a given specified order is
When Two unbiased dice are thrown at random, then the probability that the sum on them is an odd prime number is
If a number is chosen at random from the first 150 natural numbers, then the probability that the number chosen is divisible by 11 is
Solution
Solution:
Nearest multiple of 11 before 150 = 143
So total 13 nos. below 150 are multiples of 11.
Sample space = 150
Probability = $$\frac{13}{150}$$ Answer
If a 4 digit number is formed at random without repetition using the digits 0, 1, 2, 3, 4, then the probability that the number thus formed is even is
Solution
If a 4 digit number is formed at random without repetition using the digits 0, 1, 2, 3, 4, then the probability that the number thus formed is even is
There are 4 places: _ _ _ _
As repetition is not allowed,
Thousands place cannot have 0, so there are only 4 choices for that position.
Hundreds place can have all other remaining 4 choices for that position.
Tens place can have all other remaining 3 choices for that position.
Units places can have only remaining 2 choices
So, through fundamental process of counting
No. of numbers possible = 4 x 4 x3 x 2 =96
For even nos. units place should be either 2 or 4 or 0
In a similar way of counting lets first fix 2 or 4
So for thousands place 3 options are remaining (except 0).
Hundreds place can have all other remaining 3 choices for that position.
Tens place can have all other remaining 2 choices for that position.
Units places can have only remaining 2 choices. (2 or 4)
No. of numbers possible = 3 x3 x 2 x 2 = 36
In a similar way of counting lets first fix 0
So for thousands place 4 options are remaining (except 0).
Hundreds place can have all other remaining 3 choices for that position.
Tens place can have all other remaining 2 choices for that position.
Units places can have only remaining 1 choices. (only 0)
No. of numbers possible = 4 x 3 x 2 x 1= 24
Total Even nos. 36+24 = 60
Probability = 60/96 = 5/8 Answer
The mode of the following frequency distribution is
Solution
Solution:For mode of tabulated data, the observation with highest frequency is the mode.
Here maximum frequency is 24, the correpsonding observation is 8.
Answer 8
The median of the following data is
28, 4, 21, 42, 10, 9, 24, 30, 16, 5, 13
Solution
Data Set: 28, 4, 21, 42, 10, 9, 24, 30, 16, 5, 13
To find median, we first need to sort them in ascending order,
4,5,9,10,13,16,21,24,28,30,42
No, of terms,n= 11,
So median = $$\left(\ \frac{n+1}{2}\right)\ term\ if\ n\ is\ odd$$
= $$\left(\ \frac{11+1}{2}\right)\ =\ \frac{12}{2\ }\ =\ 6th\ term$$
= 16 Answer
Option (C)
If 6, 5, 4 and 3 occur in a data with frequencies 2, 5, 3 and 4 respectively, then the arithmetic mean of the data is
Solution
Arithmtic Mean = $$\frac{\Sigma\ \left(\ distinct\ values\ \times\ frequency\right)}{\Sigma\ \left(frequency\right)}\ $$
= $$\frac{6\ \times\ 2\ \ +\ 5\times\ 5\ +\ 4\times\ 3\ +\ 3\times\ 4}{2+5+3+4}\ $$
= $$\frac{61}{14}$$ Answer
Choose the correct meaning of the word given :
Genesis
Frivolous
Relish
Rescind
Blatant
Malcontent
Fill in the blank Choosing the correct word:
The writer is admired for his ............. style of writing.
A sneering person who always finds fault is a/an .............
A person who knows a variety of subjects is a ...........
Waiter, please ............. me tea.
Choose the correct answer:
The system for manufacturing products in large quantities through effective combinations of employees with specialized skills, mechanization and standardization is often referred to as
MBO stands for
The process of selecting the right mode of transport, optimizing the cost, taking into account the speed of delivery required is the subject of
The individual dots that make up the actual picture on thie MEnitor screen is
Which circuit is used as a Memory Device in Computers ?
Interconnection of computer devices, peripherals, network nodes in series, one after the other, is
A peripheral device used to connect a computer to another over a phone-line is
A step-by-step procedure used to solve a problem is called
The total amount of money received from customers for their purchase of products or services during the specified time period is referred to as
EXIM policy of the country is formulated through the
A: Are you coming for coffee ?
B: Can I take a rain check ? I must get this work finished this evening.
'B' is
A: I'll do my best. And I'll make sure all the presentation slides are ready on the computer.
B: Good. And don't forget to give me the rundown on the rehearsal.
B expects A to
A: What sort of person is Bharat ?
B: He would give the shirt off his back.
Bharat is
These figures do not support his argument. Moreover I noticed that the whole situation does not add up.
The underlined portion implies
A: You look quite cheerful.
B: That's because I passed my road test with flying colours.
The underlined idiom means
Anita: Rama, will you join us for going out?
Rama: I need to take permission of my mother.
What do you understand by the above conversation ?
The passive form of the sentence,
"Jaya is reading a comic book right now" is
Fill in the blanks with the appropriate phrase/verb/preposition :
I remembered him because only recently I ran ............. him.
He has copied word ........... word from his textbook.
.......... the whole, the students seem to like their teacher.
If it rains, I ............... to your house this evening.
She .......... completed the task before I asked for an explanation.
By next month, we ........... in Vijayawada for twelve years.
After a long delay the plane finally ..............
The driver ............ to a flying start.
Read the following passage and answer questions:
The greatest enemy of mankind, as people have discovered, is not science, but war. Science merely reflects the prevailing social forces. It is found that, when there is peace, science is constructive; when there is war, science js perverted to destructive ends.
The weapons which science gives us do not necessarily cause war, they make war increasinglyterrible. Till now, it has brought us to the doorstep of doom. Our main problem, therefore, is not to curb science, but to stop war — to substitute law for force, and international government for anarchy in the relations of one nation with another. That is a job in which everybody must participate, including the scientists.
But the bombing of Hiroshima suddenly woke us up to the fact that we have very little time. The hour is late and our work has scarcely begun. Now we are face-to-face with an urgent question - “Can education and tolerance, understanding and creative intelligence run fast enough to keep us abreast with our Own mounting capacity to destoroy "? That is the question which we shall have to answer one way or the other in this generation. Science must help us in arriving at the answer, but the main decision lies within ourselves.
According to the writer, the main problem we are faced with is to
The expression ‘bring to the doorstep of doom' means
Which of the following is opposite in meaning to the word ‘anarchy’ in the middle of the passage ?
The antidote to the destructive impulse is
According to the writer, the real enemy of mankind is not science but war because
Read the passage below and choose the Correct answer:
He came out of a stormy February night. Two large eyes glared at me through the darkness of my rain-drenched cabin Window and in the gleam oflighting, I saW a large brown body and hugejaws. I feared it was a lion. I had gone to British Columbia, on the Pacific Coast of Canada, to write a novel. For seven months I had lived all alone in my wooden cabin. Scared, I slowly backed into the kitchen for my torch and an axe. I shone the torch through the window to find myself facing a large black and brown dog, his tail wagging wildly. I opened the door slightly and he rushed into the room, bringing pools of water, ‘oing half-mad with delight. In spite of his big head he was very thin. The bones showeu through his coat. But the lookin his eyes said more clearly than any words that he was hungry. I gave him all the meat I could find in the kitchen. Then he spent the next two hours finding different ways to thank me; whining, burying his wet head in my lap, pawing at mylegs, reacting to my every look or movement. Everytime I looked up from my work, his deep brown eyes were on me, and his tail thumped the floor. So that night in 1967, Booto, the wild dog from forests, came into my lonely life.
Why did the author go to Canada ?
Why did the author fetch an axe from the kitchen ?
The dog beat its tail on the floor to show its
From that stormy night in 1967, Booto became the author's .............
Who came to the authors’s house in storm ?
Read the following passage and answer questions:
During Mao Zedong’s rule in the 1960s the Chinese fought a fierce but futile battle against the ‘Sihan’ — the four parts of rats, bed bugs, flies and mosquitoes which have plagued the country for centuries. Now Beijing has declared a nation wide war against the ‘Liuhai’, the six evils of prostitution, pornography, abduction and trading of women and children, drug trafficking, gambling, and profiteering from superstition. Public Security Minister, Wang Fang, explained that those social ills “have seriously polluted society, disturbed public order and undermined the physical and mental health of a vast number of people, especially the youth”.
China's guardians of public order have not exactly been lax. Supreme, Court Vice President, Lin Zhum, revealed last month thatfrom 1983 to 1988 Chinese courtarts punished 89.500 criminals for committing one of the six evils. Of these, 3142 were given sentences of death or life imprisonment. An additional 11,000 were arrested inthe first nine months of this year. But of course, the evi] persists. In fact, officials in the freewheeling province of Guangdong, last week discovered a seventh evil secret —€riminal gangs. Said a Chinese social scientist “The more prosperous and free we become the more evils we face”.
What does ‘freewheeling’ in the passage mean ?
What made a Chinese comiments “The more prosperous and free we become, the more evils we face” ?
Prosperity and evil are
Who was instrumental in the war against 'Sihan' ?
What was the average number of Chinese punished for ‘Liuhai’ from 1982 to 1989 ?
