Is the positive integer n odd?
I. 2n is even.
II. n(n + 1) = 6
AP ICET 2018 Question Paper Shift-2 (2nd May)
In the following questions each question is followed bydata in the form of two statements labeled as I and II. You must decide whether the data given in the statements are sufficient to answer the questions. Using the data make an appropriate choice from (a) to (d) as per the following guidelines:
Is x an integer?
I. 5x is a positive integer.
II. 10x = 1
Is $$3^n > 2^k ?$$
I. k = n + 1
II. n is positive integer
If x is a positive integer, is the HCF of 150 and x a prime number?
I. x is a prime number.
II. x < 4
If n and m are real numbers, is n < m ?
I. $$n^2 < m^2$$
II. $$\mid m \mid < \mid n \mid$$
Is the integer n divisible by 45?
I. n is a miultiple of 36
II. The units digit in n is 0
Is the integer n a prime?
I. 4 > n
II. n > 1
Is the positive integer n even?
I. $$n^4$$ is odd
II. $$2n^5$$ is even
What is the cost of laying an outside road all around the circular field ?
I. The breadth of the road is 5 m.
II. The diameter of the field is 500m and the cost of laying the road is Rs. 40 per sq.m
$$0 \leq \theta \leq 90^\circ \Rightarrow \tan \theta = ?$$
I. $$\theta$$ is an angle in a triangle.
II. $$\sin \theta = \frac{\sqrt{3}}{2}$$
Is the integer p a prime number?
I. 2p has exactly 3 factors
II. p is an even number
Is $$\mid 15 - m \mid + \mid m - 15 \mid > 15 ?$$
I. $$m > 6$$
II. $$m < 7$$
For a positive real number x, is $$x^3 = 125?$$
I. $$x > 4$$
II. $$x < 6$$
What is the area of the rectangle ABCD?
I. The circumference of the rectangle is 24 cm.
II. The sides of the rectangle are in the ratio 1 : 2
What is the cost of each table?
I. 2 tables and 2 chairs cost Rs 10,000/-
II. 3 chairs and 3 tables cost Rs 15,000/-
What is the two digit number xy?
I. xy is a multiple of 97
II. x + y = 6
TA and TB are tangents to a circle from an external point T. Whatis the length of the chord AB?
I. Radius of the circle is 6 cm
II. AT = 8 cm
What is the sum of the two integers a and b?
I. LCM of {a, b} = 576
II. HCF {a, b} = 4
What is the value of $$\theta$$?
I. The matrix $$\begin{bmatrix}\tan \theta & 1 \\1 & 1 \end{bmatrix}$$
II. $$0 < \theta < \frac{\pi}{2}$$
For three non empty sets A, B and C, what is $$n(A \cup B \cup C)$$?
I. $$n(A \cup B) = 64$$
II. $$n(B \cup C) = 124$$
In each of the Following questions a sequence of numbers or letters that follow a definite pattern is given. Each question has a blank space. This has to befilled bythe correct answer from the four given options to complete the sequence without breaking the pattern.
(5, 14) : (25, 196) :: ............. : (81, 49)
625 : ........ :: 1296 : 6
122 : 290 :: 442 : ..........
20736 : 144 :: ....... : 288
WU : FD :: ZX : .........
$$X^2 - 2X + 1 : X^3 - 3X^2 + 3X - 1 :: X^2 - 4X + 4 : ............$$
$$5 + 2\sqrt{6} : \sqrt{3} + \sqrt{2} :: 18 + 2\sqrt{77} : .............. $$
$$(3^5 \times 7^2 \times 11) : (36) :: (5^3 \times 11^2 \times 13) : ..............$$
$$\begin{bmatrix}7 & 2 \\-3 & 3 \end{bmatrix} : 27 :: ........ : 22$$
(11, 12) : 265 :: ......... : 313
In the Following questions pick the odd thing out.
The following questions followa definite pattern. Observe the same and fill in the blanks with suitable answers.
1297, 626, 257, ......... 17, 2.
4, 18, 48, ......, 180, 294.
ZIA, Y4C, X9E, ........., V25I
10, 12, 15, ......., 24, 30
0, 6, 20, ......, 72, 110
$$\frac{1}{2}, \frac{1}{24}, \frac{1}{720}, ..............$$
3, 10, 29, 66, ......., 218
$$20, 16\frac{2}{3}. 14\frac{2}{7}, 12\frac{1}{2}, ............, 10$$
3, 5, 9, 17, .........., 65, ......
MNPT, JKMQ, GHJN, ......, ABDH
The following table shows the production of different types of two wheelers from 1993 to 1998. Study the table carefully and answer the Following questions.
What is the maximum difference in the production of any two types of two wheelers for the years 1997 and 1998?
The approximate percentage increase in the total production of all types of two wheelers in 1997 in comparison to 1995 was
How many types of two wheelers have shown a continuous growth in the production for the given period?
The pie diagram given here shows the distributions of students in six schools A, B, C, D, E and F in a town. Answer the following questions based on it.
If the number of students in all the schools of the town is 5400 in a year then the number of the students in school E is
In a year the total number of students in schools Cand B together is 1800 then the number of students in the school D that year is
If the difference in the number of students in B and C in anyear is 250, then the total number of students in that year all the schools in the town is
If the school A has 350 students in a year then the numberof students in the school E (in thousands) in that year is
In which schools in an year the number of students in B and C together is double the number of students
A Survey is conducted in 400 persons with regard to the readership of three national dailies E, D and T revealed the following information:
170 read E, 170 read D, 190 read T, 50 read E and D, 70 read D and T, 60 read E and T, 30 Read none and20 read all the three papers
Draw a Venndiagram for this data and answer the following questions
How many read either D only or T only?
How many read only E?
In a code, vowels of English alphabets are coded to $$\alpha, \beta, \gamma, \delta, \eta$$ respectively which each of the consonant is coded to the third successive consonant cyclically. following this coding process answer the following questions.
Which letter is coded as H ?
The code for the word ‘CYCLE’ is
The code for the word ‘DOZEN’ is
Which word is coded as $$F \delta \delta O$$?
Which word is coded as $$P \gamma \delta R$$?
For the following questions answer them individually
In a code ‘SROLFH’ is coded for ‘POLICE’, then the code for ‘STATION’
In a code ‘BUS‘ is coded as ‘YFH’. The code for ‘CAR’ in their code is
If CRICKET is coded as BOQHBJDS then the word that is coded as GNBJDX is
If AMKNSRCP is the code for COMPUTER then the word coded as KMSQC is
If FDSLWDO is the code word for CAPITAL then the code for HOTEL is
It was Thursday on Feb $$15^{th}$$ 2018. Whatday of the week will be December $$15^{th}$$ 2018?
A clock strikes once at 1’o clock. twice at 2’o clock and so on. A person started counted from 5’o clock in the morning and heard 69 strikes, the time when he heard the last strike is
The acute angle between the two hands of a clock when the time is fifteen minutes past five is
A is father of B and C; E is the mother of B while D is wife of F. If F is the father of E then what is the relation between D and E
Three trains A, B and C start at a station in different routes to reach their destination 90kms, 70kms and 85kms away respectively. If they start at 10 a.m and moveat respective speeds 45kms, 40kms and 35kms per hour. then the order in which they reach their respective
destinations is
Anita left home for the bus stop 20 minutes late than usual. It takes 15 minutes to reach the bus stop. If she reached the bus stop at 8.30 am. at what time does she usually leave home for the busstop.
Five persons A, B, C, D and E sit around a table such that A is diagonally opposite to E; C and D are oneitherside of A. If there is only one person D to the right A, the anti clockwise order in which they sits is
If $$x \triangle y = (x - 2y) + 2xy$$, then a zero value of x such that $$x \triangle x = 0$$is
Consider the string of letters given below:
BCDEOEOONQRSOVOVOOBKBOBOO
How many letters are immediately preceded by two O’s and immediately followed by no O’s
If $$\alpha$$ means $$( )$$
$$\beta$$ means $$\div$$
$$\gamma$$ means $$+$$
$$\delta$$ means $$-$$
$$\wedge$$ means $$X$$
then find the value of $$2 \wedge 3 \gamma 4 \delta 8 \beta 2 \gamma 4 \wedge 3$$
$$\frac{1}{a^{x - y} + a^{x - z} + 1} + \frac{1}{a^{y - z} + a^{y - x} + 1} + \frac{1}{a^{z - x} + a^{z - y} + 1} =$$
For $$a \neq 0$$, $$\left(\frac{1}{1 + a^2 + a^{-2}} + \frac{1}{1 + a^{-2} + a^{-4}}\right)^{-1} - 1 =$$
$$a : b :: 2 : 3; b : c :: 4 : 5 \Rightarrow a + b : b + c =$$
Solution
Given, a:b = 2:3 and b:c = 4:5
$$\Rightarrow$$ a:b = 8:12 and b:c = 12:15
$$\Rightarrow$$ a:b:c = 8:12:15
Let a = 8k, b = 12k and c=15k
$$\therefore\ $$a+b : b+c = 8k + 12k : 12k + 15k
= 20k : 27k
= 20 : 27
Hence, the correct answer is Option D
In an alloy of 35 kgs the ratio of two metals A and B is 4 : 1. If 1 kg of metal B is added then the ratio of A and B in the new allow is
$$\sqrt[4]{17 + 12\sqrt{2}} =$$
Solution
$$\sqrt[4]{17+12\sqrt{2}}=\sqrt[4]{9+8+12\sqrt{2}}$$
$$=\sqrt[4]{3^2+\left(2\sqrt{2}\right)^2+2\left(3\right)\left(2\sqrt{2}\right)}$$
$$=\sqrt[4]{\left(3+2\sqrt{2}\right)^2}$$
$$=\sqrt{3+2\sqrt{2}}$$
$$=\sqrt{1+2+2\sqrt{2}}$$
$$=\sqrt{1^2+\left(\sqrt{2}\right)^2+2\left(1\right)\left(\sqrt{2}\right)}$$
$$=\sqrt{\left(1+\sqrt{2}\right)^2}$$
$$=1+\sqrt{2}$$
Hence, the correct answer is Option B
If $$x = 2 + \sqrt{3}, xy = 1$$, then $$\frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}} =$$
The number of integers between 100 and 1000 whichare divisible by eachof 4, 5 and6 is
Solution
Numbers divisible by eachof 4, 5 and 6 = Numbers divisible by LCM of (4,5,6) = 60
Number of numbers between 100 and 1000 divisible by 60 = 15
The number of natural numbers k such that $$\frac{3k^2 + 4k + 12}{k}$$ is a prime is
Solution
For the result to be prime, the expression must firstly yield a whole number.
The expression can be broken up as = 3*k + 4 + (12/k)
Hence, k has to be a whole number factor of 12.
Probable values of k = 1,2,3,4,6,12
Of these, the values of k = 1,3,4,12 yield prime numbers for the expression.
All the numbers 411, 752 and 1031 leave same remainder 8 whendivided by n. The value of n is
Solution
The given information means that all the following numbers must be divisible by n
411 - 8 = 403
752 - 8 = 744
1031 - 8 = 103
403 = 13 * 31
744 = 24 * 31
1023 = 33 * 31
From this, we can say that the only common factor of all three is 31 and hence, n = 31
The sum of the greatest common divisor (GCD) andthe least common multiple (LCM) is 403 and their LCMis 12 times their GCD. If one of those numbers is 124, then the other number is
Solution
We are given that LCM = 12*GCD and also LCM + GCD = 403
Combining the two, we get GCD = 31 and hence, LCM = 12*31
Now, we know that 124*Other number = 31*(12*31)
Hence, the other number = 93
$$5 + \frac{1}{6 + \frac{1}{8 + \frac{1}{10}}} =$$
Solution
$$5+\frac{1}{6+\frac{1}{8+\frac{1}{10}}}=5+\frac{1}{6+\frac{1}{\frac{81}{10}}}$$
$$=5+\frac{1}{6+\frac{10}{81}}$$
$$=5+\frac{1}{\frac{496}{81}}$$
$$=5+\frac{81}{496}$$
$$=\frac{2561}{496}$$
Hence, the correct answer is Option D
0.6666 ... ... ... + 0.7777 ... ... ... + 0.8888 ... ... ... =
Solution
Recurring decimals can be expressed as fractions having 9 or multiples of 9 in some form.
0.6666.... can be expressed as 6/9
0.77777... can be expressed as 7/9
0.8888..... can be expressed as 8/9
Hence, 0.6666... + 0.77777.... + 0.8888.... = (6/9)+(7/9)+(8/9) = 21/9 = 7/3
If $$a_1, a_2, a_3$$ and $$a_4$$ is the decreasing order of the elements in $$\left\{\frac{2}{7}, \frac{7}{15}, \frac{5}{8}, \frac{9}{23}\right\}$$ then $$a_1 - a_4 =$$
Solution
$$\frac{2}{7}=0.285$$
$$\frac{7}{15}=0.47$$
$$\frac{5}{8}=0.625$$
$$\frac{9}{23}=0.391$$
Decreasing order of the fractions is $$\frac{5}{8}$$, $$\frac{7}{15}$$, $$\frac{9}{23}$$, $$\frac{2}{7}$$
$$\Rightarrow$$ $$a_1=\frac{5}{8}$$ and $$a_4=\frac{2}{7}$$
$$\therefore\ $$ $$a_1-a_4=\frac{5}{8}-\frac{2}{7}=\frac{35-16}{56}=\frac{19}{56}$$
Hence, the correct answer is Option D
The largest integer in the set $$\left\{x \epsilon R: \mid x - 2 \mid < 5\right\}$$ is
Solution
Given, $$\mid x-2\mid<5$$
$$\Rightarrow$$ $$-5<x-2<5$$
$$\Rightarrow$$ $$-3<x<7$$
The values of $$x$$ would range from -2 to +6 to satisfy the given condition.
$$\therefore\ $$The largest integer of the set = 6
Hence, the correct answer is Option A
If 16% of a property is worth Rs 3.52 lakhs then 25% of the property is worth (in lakhs of rupees)
Solution
This is a simple case of direct variation and we can say :
$$\frac{?}{3.52}=\frac{25}{16}$$
This brings us the ? value to be = 5.5 lakhs
If 25% of a number m is added to another number n then n increases by 10% then m : n =
Solution
According to the problem,
$$\frac{25}{100}$$m + n = $$\frac{110}{100}$$n
$$\Rightarrow$$ $$\frac{25}{100}$$m = $$\frac{110}{100}$$n - n
$$\Rightarrow$$ $$\frac{25}{100}$$m = $$\frac{10}{100}$$n
$$\Rightarrow$$ m : n = 2 : 5
Hence, the correct answer is Option A
By selling a mobile for Rs. 5700, a trader would lose 5%. The price at which he sell it to gain 5% is
Solution
By selling at Rs. 5700, if the loss is 5%, it means that 5700 = 0.95*CP
Hence, CP = 5700/0.95
Now, to make a gain of 5%, the new SP = 1.05*CP = 1.05 * (5700/0.95) = Rs. 6300
The cost price of anarticle is Rs 800/-. Its marked price is Rs 1000/-. The shopkeeper has announced a discount because of which he incurs a loss of 10%. The discount percentage is
Solution
After the discount, the loss incurred is = 10% i.e. 10% of 800 = 80 rupees
Hence, the product is sold for 800-80 = 720 rupees
The marked price of article was Rs. 1000 on which the discount D% was given
The actual selling price Rs. 720 is = (720/1000) = 72% of the marked price
This means, the discount percentage D = 100-72 = 28%
P, Q, R are partners in a business. P invested half as much as Q and Q invested twice as much as R. If R’s investment is Rs. 5630, then the investment (in Rupees) of P and Q together is
Solution
Given, R’s investment is Rs. 5630
Q invested twice as much as R
$$\Rightarrow$$ Q’s investment = 2 x 5630 = 11260
P invested half as much as Q
$$\Rightarrow$$ P's investment = $$\frac{11260}{2}$$ = 5630
$$\therefore\ $$Investment of P and Q together = 5630 + 11260 = 16890
In a property A, B and C have shares in the ratio 5:6:11. If C gets Rs 3 lakhs more than A then 50% of the share of B (in lakhs of rupees) is
Solution
Let the actual shares of A,B,C be 5x, 6x, 11x respectively
We are given that 11x-5x = 3 lakhs ===> x = 0.5 lakhs
Hence, share of B = 3 lakhs and 50% of his share = 1.5 lakhs
Two pipes P and Q can separately fill a cistern in 20 and 30 minutes respectively. There is another exhaust pipe R at the bottom of the cistern. If all the three pipes are opened at the same time, the cistern will be full in 24 minutes. The time (in minutes) for the pipe R to
empty the full cistern is
Solution
Let the capacity of the cistern be 600 liters
Hence, capacity of pipe P = 30 liters/min and capacity of Q = 20 liters/min
Let the draining capacity of R be -r liters per minute.
Now, when all three are opened together, the cistern gets filled in 24 minutes.
Effective capacity of all three together = 600/24 = 25 liters/min
Hence, 30+20-r = 25 ===> r = 25 liters/min
Hence, time required by R to empty complete cistern = 600/25 = 24 minutes
Two taps A and B can fill a tank independently in 10 and 15 minutes respectively while C empties the full tank in 20 minutes. If all the three are opened for 1 minute and C is closed then extra time (in minutes) to get the tank filled is
Solution
Let capacity of the tank = 60 liters.
Hence, capacities of the taps respectively are:
A = 6 liters/min
B = 4 liters/min
C = -3 liters/min (minus because it drains the tank)
When all three are opened for a minute, tank filled = 6+4-3 = 7 liters
Tank remaining to be filled = 53 liters
Time needed now to fill the tank = 53/10 = 5.3 minutes
The speed ( in kmph) of a person who crosses a 1200 meter long bridge in 12 minutes is
Solution
Speed of the person = $$\frac{1200}{12}=100\ \frac{meters}{\min}$$
Multiplying both numerator and denominator by 60 (because we want speed in kmph),
We get Speed = $$6000\ \frac{meters}{hour}=6\ kmph$$
A person travels from P to Q at a speed of 60 kmph and returns at a speed of 45 kmph If total time taken is $$3\frac{1}{2}$$ hours then the distance (in kms) between P and Q is
Solution
Let the distance between P and Q be D km.
Total time = $$3\frac{1}{2}$$ hours = 3.5 hours
According to the problem, $$\frac{D}{60}+\frac{D}{45}=3.5$$
$$\Rightarrow$$ $$\frac{1}{15}\left(\frac{D}{4}+\frac{D}{3}\right)=3.5$$
$$\Rightarrow$$ $$\frac{1}{15}\left(\frac{7D}{12}\right)=3.5$$
$$\Rightarrow$$ D = 90 km
$$\therefore\ $$Distance between P and Q = 90 km
P and Q can do a work individually in 50 and 40 days respectively. They together agreed to complete that work for Rs. 900. Then the share of P ( in rupees) in this amount is
Solution
It is known that the share of any single worker in the complete pay of the workers is inversely proportional to the number of days each worker takes to complete the task alone.
Here, P and Q take 50 and 40 days respectively. Hence, the pay off for P and Q will be in the ratio of 40:50 i.e. 4:5 respectively
Hence, for a total pay of Rs. 900, P's share will be = (4/9)*900 = 400 rupees.
A and B can do a piece of work in 20 days.The same work can be done by B and C in 30 days; C and A in 40 days. If A works alone the numberof days required to complete the work is
Solution
Let the work capacity of A,B,C be a,b,c units/day respectively.
Let the amount of work be 120 units
Hence, we can frame the following equations:
a+b = 6
b+c = 4
a+c = 3
Adding up all these 3, we get: a+b+c = 6.5
From this, we can say that a = 2.5
Hence, the time required by A alone to get work done = 120/2.5 = 48 days
The perimeter of a semicircle is 144 cm. Then its area ( in sq.cms) is ( Here take $$\pi = \frac{22}{7}$$)
Solution
The perimeter of this semicircle is given by
$$2r+\pi\ r\ =\ 144$$
Solving this equation, we get: r = 28 cm
Hence, area of the semicircle is = $$\pi\ \frac{r^2}{2}=1232\ sq\ cm$$
The area of the pentagon ABCDE, given that $$AB = BC = CD = AE = 2, DE = 2\sqrt{2}$$ and $$\angle BAE = \angle BCD = 90^\circ$$ is
Solution
Draw a rough figure, write length of sides alongside and draw diagonals BE and BD.
You can notice that BE is hypotenuse of triangle BAE and hence, BE = 2*(root 2)
Similarly, BD is also the hypotenuse of BCD and BD = 2*(root 2)
Next, we can see that BDE becomes an equilateral triangle of side 2*(root 2)
Area of the entire pentagon = Area of the 2 right angled triangles BAE and BCD + Area of equilateral triangle BED
= 4 + 2*(root 3) sq units
A hallow iron pipe of 63 cms length having outer circumference of 440 cmis of thickness 4 cms. The volume(in cubic cms) of the iron used is ( Here take $$\pi = \frac{22}{7}$$)
Solution
Let the outer radius of pipe be R and inner radius be r.
Also, we are given that 2*(22/7)*R = 440 cm ===> R = 70 cm
Now, Volume of iron used in making the pipe = (22/7)*63*[(70*70)-(66*66)].......(as thickness is given as 4 cm)
Hence, Volume = 1,07,712 cubic cm
The surface area of a hemisphere is 16632 sq.cm. Then its volume (in cubic centimeters) is ( Here take $$\pi = \frac{22}{7}$$)
Solution
Area of a closed hemisphere = $$3\pi\ r^2$$
$$3\pi\ r^2$$ = 16632
r = 42 cm
Now, Volume of the hemisphere = $$\frac{2}{3}\pi\ r^3$$
Volume = $$\frac{2}{3}\cdot\frac{22}{7}\ \cdot42^3$$ $$=\frac{2}{3}\cdot\frac{22}{7}\ \cdot42^3\ =\ 88\cdot42^2$$
A rectangle of length l meters and breadth 50 meters has an area which is equal to the area of a square with 90 meters diagonal. Then l =
Solution
Given, diagonal length of the square = 90 m
Side of the square = $$\frac{90}{\sqrt{2}}$$ m
Area of the square = $$\left(\frac{90}{\sqrt{2}}\right)^2$$ = 90 x 45 m$$^2$$
Area of rectangle = Area of square
$$\Rightarrow$$ l x 50 = 90 x 45
$$\Rightarrow$$ l = 81 m
Hence, the correct answer is Option D
In 10 minutes a person can walk around a rectangular field of 300 m X 200 m. The speed (in kmph) of the person is
Solution
Perimeter of the rectangular = 2(300+200) = 2 x 500 = 1000 m
Distance walked in 10 mins = 1000 m
Hence, Distance walked in 60 mins = 1000 x 6 = 6000 m = 6km
$$\therefore\ $$Speed of the person = 6kmph
If the radius of a circular wheel is 14 cm. then the number of revolutions necessary to cover a distance of 105.6 meters is ( Here take $$\pi = \frac{22}{7}$$)
Solution
Radius of the circular wheel = 14 cm
Circumference of the circular wheel = $$2\times\pi\times14$$ = $$2\times\frac{22}{7}\times14$$ = 88 cm
$$\therefore\ $$The number of revolutions required to cover a distance of 105.6 m = $$\frac{\text{Distance}}{\text{Circumference of the wheel}}$$
$$=\frac{105.6\times100}{88}$$
$$=120$$
Hence, the correct answer is Option A
The largest 3-digit number x satisfying the congruence $$3x + 2 \equiv 4 (mod 13)$$ is
Solution
We need a three digit number which is of the form 3x + 2 and when this number is divided by 13, the remainder must be 4.
The largest 3 digit number of this form can be when x = 332 and the number comes out to be 998. When divided by 13, the remainder is not 4.
The next can be when x = 331 which makes the number as 995. When divided by by 13, remainder is not 4.
The next can be x = 330 which makes the number as 992. When divided by by 13, remainder is 4.
Hence, answer is 992.
Another approach can be directly taking options from largest to smallest and actually dividing by 13 to arrive at the answer.
For real numbers a and b write $$a * b = a^2 + ab - b^2$$. If $$4 * x = 6$$ then $$(x - 2)^2 =$$
Solution
Given, $$a * b = a^2 + ab - b^2$$
$$4 * x = 6$$
$$\Rightarrow$$ $$4^2+4x-x^2=6$$
$$\Rightarrow$$ $$16+4x-x^2=6$$
$$\Rightarrow$$ $$x^2-4x-10=0$$
$$\Rightarrow$$ $$x^2-4x+4-14=0$$
$$\Rightarrow$$ $$\left(x-2\right)^2-14=0$$
$$\Rightarrow$$ $$\left(x-2\right)^2=14$$
Hence, the correct answer is Option C
The inverse of $$(p \wedge \sim q) \rightarrow r$$ is
Solution
Introduce negation to both sides of relation.
p gets changed to "not p"
"And" function gets changed to "Or"
"Not q" gets changed to "q"
"r" gets changed to "not r"
Hence, option C is the answer.
Consider the statement S: “If 10 < 8 then 2 + 4 = 8". Then S is
Solution
S:$$p\longrightarrow\ q$$
where, p: 10<8 is a mathematically incorrect/false statement
and q: 2+4=8 is also a mathematically false statement.
False implying another false statement. This makes S as a true statement.
S would have been false, if either of p or q had been true and the other one false.
For any set S let n(S) denote the number of elements in S. If A, B and C are sets such that
$$n(A \cup B \cup C) = 31, n(A) = 20, n(b) = 16, n(C) = 8, n(A \cap B \cap C') = 3, n(A' \cap B \cap C) = 2$$ and $$n(A \cap B' \cap C) = 4$$ then $$n(A \cap B \cap C) =$$
If $$P(A)$$ is the power set of the non-empty set $$A$$, define a relation $$R$$ by $$XRY$$ if and only if $$X \subseteq Y$$ for $$X,Y \epsilon P(A)$$, the relation R is not
The number of injective functions from a 5-elementset into a 8-element is
Solution
This can be determined by doing some mapping:
The first member of the 5-member set can take 8 values
The second member of the 5-member set can take 7 values
The third member of the 5-member set can take 6 values
The fourth member of the 5-member set can take 5 values
The fifth member of the 5-member set can take 4 values
hence, the total number of injective functions possible = 8*7*6*5*4 = 6720
The intercept made on the X-axis by the line passing through (3, -5) and (-2, 7) is
Solution
Slope of the line = -12/5
Equation of the line:
y - 7 = (-12/5) * (x + 2)
X-intercept is obtained by putting y = 0 in above equation
Hence, 35/12 = x + 2 ==> x = 11/12
The equation of the perpendicular bisector of the line segment joining the points (-2, 3) and (1, -2) is
Solution
Points to remember: The perpendicular bisector will pass through the mid-point of the segment obtained by joining the given points.
The product of slope of the bisector and the slope of the segment = -1
Firstly, mid-point of the segment = [(-2+1)/2, (3-2)/2] = (-1/2, 1/2)
hence, perpendicular bisector passes through this point.
Slope of the segment = (-2-3)/(1+2) = -5/3
Hence, slope of perpendicular bisector = 3/5
Now, equation of perpendicular bisector:
(y - 1/2) = (3/5)*(x+1/2)
Solving the equation gives us option B
$$\sin^2 60^\circ + \cos^2 225^\circ + \tan^2 210^\circ + \cot^2 240^\circ =$$
Solution
Sin 60 = (root 3)/2
Cos 225 = cos (180 + 45) = - cos 45 = -1/(root 2)
Tan 210 = tan (180 + 30) = tan 30 = 1/(root 3)
Cot 240 = cot (180 + 60) = cot 60 = 1/(root 3)
Squaring up all terms and adding, we get answer as 23/12
For $$\alpha$$ and $$\beta$$ with $$0 \leq \alpha, \beta < \frac{\pi}{2}$$ if $$\cos \alpha = \sin \beta = \frac{1}{2}$$ then $$\alpha + 2\beta = $$
Solution
From the given relations, it becomes clear that $$\alpha\ =\frac{\pi}{3},\ \beta\ =\frac{\pi}{6}\ \ $$
Hence, the given relation results in option D
$$\tan \theta = \frac{3}{4} \Rightarrow \tan 2\theta + \sec 2\theta = $$
Solution
$$\tan\ 2\theta\ +\sec\ 2\theta\ $$
$$=\frac{\left(1+\sin2\theta\ \right)}{\cos\ 2\theta\ }$$
$$\frac{\left(1+2\sin\ \theta\ \cos\ \theta\ \right)}{\cos\ ^2\theta\ -\sin\ ^2\theta\ }$$
Dividing both numerator and denominator by $$\cos\ ^2\theta$$ , we get
$$\frac{\left(\frac{1}{\cos\ ^2\theta}+2\tan\ \theta\ \right)}{1-\tan\ \theta\ }=\frac{\left(1+\tan^2\theta\ +2\tan\ \theta\ \right)}{1-\tan^2\theta\ }$$
Substitute the value of tan from the given data, we get option D as answer.
Fromthe top of a building of height 60 feet which is on the bank of a river, the angle of depression of a house onthe other side of the river is observed to be $$60^\circ$$. The widthofthe river (in feet) is
Solution
The diagram can be constructed like this:
As we can see, triangle HBB' is a a 30-60-90 triangle with one of the sides known.
HB is the width of the river.
HB = BB'/(root 3) = 20*(root 3)
If $$x - \alpha, x - \beta$$ and $$x - \gamma$$ are factors of the polynomial $$3x^3 + 4x^2 + 2x + 5$$ then the polynomial having factors $$x - \frac{1}{\alpha}, x - \frac{1}{\beta}$$ and $$x - \frac{1}{\gamma}$$ is
Solution
As $$\left(x-\alpha\ \right)$$ is a factor of $$3x^3 + 4x^2 + 2x + 5$$,
$$3x^3 + 4x^2 + 2x + 5$$ = 0 at $$x=\alpha\ $$
I.e. $$3\alpha\ ^3+4\alpha\ ^2+2\alpha\ +5=0$$
Dividing both sides of equation by $$\alpha^3$$ , we get
$$\frac{5}{\alpha^3}+\frac{2}{\alpha^2}+\frac{4}{\alpha\ }+3=0$$
This result is equivalent to saying that if $$x\ =\ \frac{1}{\alpha\ }$$ , then
$$5x^3+2x^2+4x+3=0$$
If $$x^2 - 4$$ divides $$x^4 + ax^3 + bx^2 - 5x + 4$$ then $$a - b =$$
Solution
It means that both (x-2) and (x+2) are factors of $$x^4 + ax^3 + bx^2 - 5x + 4$$
This implies that $$x^4 + ax^3 + bx^2 - 5x + 4$$ at x = 2 is 0
i.e. 4a + 2b = -5
Also, it implies that $$x^4 + ax^3 + bx^2 - 5x + 4$$ at x = -2 is 0
i.e. 4a - 2b = 15
Adding up the two equations and solving them gives us
a = 5/4, and b = -5
Hence, a - b = 25/4
A factor of $$2x^4 - 3x^2 + 2x - 1$$, among the following is
Solution
Applying remainder theorem or performing actual division/factorization is the theoretical approach in such questions.
However, one can perform a quick addition of the coefficients and see that their sum = 0
This is one of the most basic tests and it means that (x-1) is a factor of the given polynomial.
If (x - 2) is a factor of $$p(x^2 + 4)$$. where p(x) is a polynomial. then a factor of p(x) is
Solution
If (x-2) is a factor of $$p\left(x^2+4\right)$$, it means that at x = 2, $$p\left(x^2+4\right)$$ = 0
That is, $$p\left(2^2+4\right)=0$$ i.e. p(8) = 0
This is equivalent to saying that whatever polynomial function p(x) is, (x-8) is a factor of p(x) as we can see that p(8) = 0
Two numbers are in the ratio 5:6 and when 3 is added to each of them. their ratio becomes 7:8. Then the sum of those two numbers is
Solution
Let the two numbers be 5x and 6x. When 3 is added to both, ration becomes 7:8
$$\frac{5x+3}{6x+3}=\frac{7}{8}$$
Cross multiplying, we get: x = 3/2
Hence, sum of the two numbers = 11x = 33/2
If $$\begin{bmatrix}3 & -2 \\-1 & 2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}11 \\-5 \end{bmatrix}$$ then 3x + 7y =
Solution
Multiplying the matrices on the LHS and equating with the one on RHS, we get:
3x - 2y = 11
-x + 2y = -5
Adding the 2 equations, we get x = 3, y = -1
Hence, 3x - 7y = 9-7 = 2
If $$S_1, S_2$$ and $$S_3$$ denote the sums of the first n, 2n and 3n terms respectively of an arithmetic progression then $$\frac{S_3}{S_2 - S_1} =$$
Solution
$$S_1=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$$
$$S_2=n\left[2a+\left(2n-1\right)d\right]$$
$$S_3=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]$$
The desired operation results in:
$$\frac{3\left[2a+\left(3n-1\right)d\right]}{2\left[2a+\left(2n-1\right)d\right]-2a-\left(n-1\right)d}$$
=$$\frac{\left[6a+9nd-3d\right]}{2a+3nd-d}=3$$
If S is the sum of five consecutive even integers then the least among them is
Solution
Let the five consecutive even integers in ascending order be x, x+2, x+4, x+6, x+8
$$\Rightarrow$$ x + x + 2 + x + 4 + x + 6 + x + 8 = S
$$\Rightarrow$$ 5x + 20 = S
$$\Rightarrow$$ x = $$\frac{S-20}{5}$$
$$\therefore\ $$Least number among them = x = $$\frac{S-20}{5}$$
The coefficient of $$x^{n - 1}$$ in $$(x - 1)(x - 2)(x - 3) ................. (x - n)$$ is
Solution
Consider the most basic case of (x-1)*(x-2)
The co-efficient of $$x^{2-1}=x$$ is -3
We can reject option A and C outright. Option D gives value -1, hence, incorrect.
Only option B gives correct answer.
The term independent of x in the binomial expansion of $$\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9$$ is
Solution
The power of x in the first term of expansion would be 18
The power of x in the second term would be 16(of the first term in original bracket) - 1 (of the second term in original bracket) = 15
The power of x in the third term would be 14 - 2 = 12
The power of x in the fourth term would be 12 - 3 = 9
The power of x in the fifth term would be 10 - 4 = 6
The power of x in the sixth term would be 8 - 5 = 3
The power of x in the seventh term would be 6-6 = 0
The seventh term would be independent of x
The value of seventh term would be (9C6)*$$\left(\frac{3}{2}\right)^3\cdot\left(\frac{1}{3}\right)^6$$ = 7/18
$$A = \begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix} \Rightarrow \mid A^5 \mid =$$
Solution
Upon successive multiplication, obtain the matrix $$A^5$$
Its terms are:
$$A_{11}=6140$$, $$A_{12}=8097$$, $$A_{21}=10796$$, $$A_{22}=14237$$
Cross multiplication and subtraction of terms results in determinant = -32
If $$i^2 = -1$$ and $$A = \begin{bmatrix}i & 0 \\0 & -i \end{bmatrix}$$ then $$A^{-1} =$$
Solution
As can be seen from the options, $$A^{-1}$$ is going to be a power of A itself.
We can continuing multiplying A successively to its own powers till we arrive at an identity matrix.
If the identity matrix is arrived at $$A^n$$ , then we can say that $$A^{n-1\ }=\ A^{-1}$$
Starting with the multiplication, we can see that a 2x2 identity matrix is arrived at at $$A^4$$
Hence, $$A^3$$ is the inverse of matrix A.
$$\lim_{x \rightarrow 0}\frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$$
Solution
Multiplying numerator and denominator by $$\sqrt{\ 1+x}+\sqrt{\ 1-x}$$, we get:
$$\lim\ x\longrightarrow\ 0\ $$ $$\frac{\left(1+x\right)-\left(1-x\right)}{x\cdot\left(\sqrt{\ 1+x}+\sqrt{\ 1-x}\right)}$$
$$=\frac{2}{\left(\sqrt{\ 1+x}+\sqrt{\ 1-x}\right)}$$
Putting x = 0 now, we get
Result = 2/2 = 1
$$y = x \sin x \Rightarrow \frac{dy}{dx} =$$
Solution
Given, $$y = x \sin x$$
$$\Rightarrow$$ $$\frac{dy}{dx}=x\cdot\left(\frac{d}{dx}\sin\ x\right)+\sin\ x\cdot\left(\frac{d}{dx}x\right)=x\cdot\cos\ x+\sin\ x$$
If the hypotenuse of a right angled isosceles triangle is 8 cm. then its area, in square centimeters, is
Solution
Let the length of equal sides of right angled isosceles triangle = a
In the triangle,
a$$^2$$ + a$$^2$$ = 8$$^2$$
$$\Rightarrow$$ 2a$$^2$$ = 64
$$\Rightarrow$$ a$$^2$$ = 32
$$\Rightarrow$$ a = $$4\sqrt{2}$$ cm
$$\therefore\ $$Area of the right angled isosceles triangle = $$\frac{1}{2}\times a\times a$$
$$=\frac{1}{2}\times4\sqrt{2}\times4\sqrt{2}$$
$$=$$ 16 cm$$^2$$
In the quadrilateral ABCD, it is given that AB = 40, BC = 250, CD = 50, 2BAD = ZABC = $$90^\circ$$ and AD is parallel to BC. The perimeter of the quadrilateral ABCD is
Solution
We can draw the figure as:
We can conclude that AD will be 280 units in length and hence, perimeter = 620
A, B and C are three points on the circle with centre O. In $$\triangle$$ABC if $$\angle$$B = $$60^\circ$$, $$\angle$$C = $$70^\circ$$ then $$\angle$$BOC =
Solution
In $$\triangle$$ABC,
$$\angle$$A + $$\angle$$B + $$\angle$$C = $$180^\circ$$
$$\angle$$A + $$60^\circ$$ + $$70^\circ$$ = $$180^\circ$$
$$\angle$$A = $$50^\circ$$
Angle subtended by the arc at the centre is twice the angle subtended by the arc at any point circle.
$$\Rightarrow$$ Angle subtended by the arc BC at centre O is twice the angle subtended by arc BC at point A.
$$\Rightarrow$$ $$\angle$$BOC = 2$$\angle$$A
$$\Rightarrow$$ $$\angle$$BOC = $$100^\circ$$
The length of the median through A of $$\triangle$$ABC where A(4, 2), B(6, 5) and C(1, 4) is
Solution
The midpoint of BC to which the median will connect from A : (3.5, 4.5)
Length of median = Distance between A(4,2) and (3.5, 4.5) = $$\sqrt{\left(3.5-4\right)^2+\left(4.5-2\right)^2}$$ = $$\sqrt{6.5}=\frac{\sqrt{\ 26}}{2}$$
If A(4, 1), B(7, 4), C(13, 2), D(x, y) are the vertices of a rectangle ABCD then(x, y) =
The mean deviation of the scores 3, 5, 7, 9, 11 and 13 from their arithmetic mean is
Solution
The mean of the given numbers = 8
The absolute deviations are : 5,3,1,1,3,5
Sum of absolute deviations = 18
Mean deviation = 18/6 = 3
In 12 consecutive odd numbers the arithmetic mean of the first five numbers is 85. The difference of the medians ofthe first five numbers and the last five numbers is
Solution
Let the first odd number be 2n+1.
Hence, the first five odd numbers would be: 2n+1, 2n+3, 2n+5, 2n+7, 2n+9
Median of these: 2n+5
The last 5 odd numbers would be: 2n+15, 2n+17, 2n+19, 2n+21, 2n+23
The median of these numbers: 2n+19
Hence, the difference between the medians = 2n+19 - 2n - 5 = 14
The mode of the following observations is:
14, 21, 64, 78, 42, 21, 39, 42, 21, 18, 17, 21, 10, 21
Solution
Mode is simply that entity which occurs with the highest frequency in the set of observations
Here, 21 occurs the most: 5 times out of 14 total observations.
Hence, it is the mode here.
If $$\sigma$$ is the standard deviation of $$x_1, x_2, ... ... x_n$$ then the standard deviation of $$9 + 3x_1, 9 + 3x_2, ..... ..... 9 + 3x_n$$ is
Solution
Let the mean of the first series be M
Hence, M = $$\frac{x_1+x_2+......+x_n}{n}$$
Now, for the second series, let its mean be M'
Hence, M' = $$\frac{3\cdot\left(3n+x_1+x_2+......+x_n\right)}{n}=9+3M$$
Now, respective deviation of the individual terms from the mean M' are:
3M - 3$$x_1$$ = 3(M-$$x_1$$)
3M - 3$$x_2$$ = 3(M-$$x_2$$)
.
.
.
3M - 3$$x_n$$ = 3(M-$$x_n$$)
Variation = $$\frac{\left[9\cdot\left(M-x_1^{ }\right)\left(M-x_1\right)+9\cdot\left(M-x_2\right)\left(M-x_2\right)+....+9\left(M-x_n\right)\left(M-x_n\right)\right]}{n}$$
Standard deviation = Square root of this Variance = 3*$$\sigma\ $$
The variance of the first five odd natural numbers is
Solution
The first 5 odd natural numbers are: 1 3 5 7 9
Average of these = 5
Respective deviation from mean: -4 -2 0 2 4
Variance = $$\frac{\left(Sum\ of\ squares\ of\ individual\ deviation\right)}{Number\ of\ terms\ }$$ = (16+4+0+4+16)/5 = 8
The rank correlation coefficient r of a distribution lies in the interval
A box contains nine tickets numbered 1 to 9. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either odd. even. odd or even. odd, even is
Solution
Consider the first case of Odd-Even-Odd:
$$\frac{5}{9}\cdot\frac{4}{8}\cdot\frac{4}{7}=\frac{20}{126}$$
Consider the second case of Even-Odd-Even:
$$\frac{4}{9}\cdot\frac{5}{8}\cdot\frac{3}{7}=\frac{15}{126}$$
Hence, The net probability = $$\frac{20}{126}+\frac{15}{126}=\frac{35}{126}=\frac{5}{18}$$
A number is chosen at random from the first 100 natural numbers. The probability that the chosen number is divisible by 7 is
Solution
There are a total of 14 numbers from 1 to 100 which are multiples of 7.
$$\therefore\ $$The probability that the chosen number is divisible by 7 = $$\frac{14}{100}$$ = $$\frac{7}{50}$$
Hence, the correct answer is Option C
If A and B are independent events such that $$P(A \cap B) = \frac{3}{25}, P(B) = \frac{11}{25}$$ then $$P(\overline{A}) = $$
If three unbiased coins are tossed then the probability that at least two coins give heads is
Solution
The order of the appearance of heads and tails does not matter here.
So we can have only 4 distinct possible combinations: HHH, HTH, HTT, TTT
From this, our desired situations are HHH and HTH.
Hence, the probability = 2/4 = 1/2
Choose the correct meaning of the word given:
importunate
decorum
connoisseur
subsist
enfeeble
myriad
Fill in the blank choosing the correct word:
He took a .............. of beer.
This region is known for its ............... vegetation.
We must .......... from smoking.
It is difficult to ................. these old documents.
Choose the correct answer:
In IP address, IP stands for
...................... is a mobile operating system developed by Google
The smallest unit of a digital image that can be displayed on a digital display device is ................
MMS stands for ................
Compressing one or more items into a smaller archive is called .....................
Comparing one’s products /services to those of competitors in order to improve quality and performance is called ..........
The monetary authority of our country is the ...........
In the sale or supply of goods or services, something that is ready for immediate use is called ................
The written record of what is said at a meetingis called ................
A person who buys shares in companies hoping the price will rise, so that they can be sold later at a profit is ................
“Had he not been a dreamer. he would not have gonethis far”. The statement implies that ...............
I am afraid you will have to put up ................ this inconvenience.
Let us do it ourselves ............ expecting the governmentto doit forus.
We thought Sindhu would lose the match but she scraped through in the final set. The underlined phrase means that she ............
He asked me, “Have you seen that film?” The sentence may be rewritten in indirect speech as: He asked me ............
To look after the transport, .............. special officer has been appointed.
I ............. Ravi since 1993.
Fill in the blanks with the appropriate phrase/verb/preposition:
He felt that he ........... by his own people.
I am thinking .......... a novel.
If you .............. games regularly. you would become strong and healthy.
Many accidents ............... rash driving.
The child was ................ short to reach the top shelf.
We went to the park ................
Can you tell me ................?
His apology took the edge off her anger. The underlined phrase means that his apology .................
Read the passage below and answerthe following questions.
Mascowis the world’s most expensive city for expatriates followed by Tokyo whichalso is Asia’s most expensivecity.
The cost of Living Survey 2008 has shown Mumbai and NewDelhi taking the bottom two spots of Asia’s ten most expensive cities. Mumbai is on 48th spot in the world wide list and NewDelhi is on 55th spot. New Delhi incidentally shares its 55th spot on the list with Los Angeles.
The survey covers 143 cities across six continents and measures the comparative cost of over 200 items in each location including housing, transport, food, clothing, household goods and entertainment.
It is the world’s most comprehensive cost of living survey and is used to help multinational companies and governments determine compensation allowances for their expatriate employees.
Which word means “persons living outside one’s own country”?
Amongthe following, whichcity is the most expensive in the world?
What is Mumbai’s rank in Asia’s ten most expensive cities?
Which item is not mentioned in the passage for measuring the comparative cost of different items?
Whom does the survey help the most?
Read the passage below and answer the following questions
Stories have survived for centuries in the formof folklore because man never tired of hearing themrepeated. Childrenstill delight in stories told to them by their elders, and it may well be that a primary motive for learning to read is the insatiable desire of a child for more stories than his parents or grandparents can supply. Narration, as a mode of discourse, arises from the universal desire of humanbeings to hear a story, whetherit is new or old. The media of mass communication — television, radio, moving pictures — have vastly multiplied the means of satisfying this desire, but have added little to the art of narration itself.
Good narration depends upon sequence, unity and point. The imagined sequence must be chronological, for narrative must present actions in time. The written sequence, however, does not have to be in the same orderas the imagined sequence — often, infact , a story maystart in the middle, “flash back” to earlier related events, and then conclude with later events that give a sense of resolution there is a sense of past, present and future to keep the action going,to give the readerthat feeling of pace which1s so vital in keeping his interest alive.
What is the main reasonfora child to learn reading?
Which word means “too great to be satisfied”?
Which statement is true according to the author?
The media of mass communication has .............. the art of narration.
Which is not stated as essential for good narration?
Read the passage below and answerthe following questions.
Manis the Reasoning Animal. Such is the claim. I think it is open to dispute. Indeed, my experiments have proven to me that he is the Unreasoning Animal. In truth, manis incurably foolish. Simple things whichother animals easily learn, he is incapable of learning. Among my experiments was this. In an hourI thought themto be friends with a rabbit. In the course of two days I wasable to add a fox, a goose, a squirrel and some doves and finally a monkey. They lived together in peace, even affectionately.
Next, in another cage I confined an Irish Catholic from Tipperary, and as soon as he seemed tame I added a Scotch Presbyterian from Aberdeen. Next a Turk fromthe wilds of Arkansas: a Buddhist from China: a Brahman from Benares. Finally, a Salvation Army Colonel from Wapping. Then I stayed away for two days. When I came back to note results, the cage of Higher Animals wasall right, but in the other there was but a chaos of
gory odds and ends of turbans and plaids and bones and flesh-not a specimenleft alive. These Reasoning Animals had disagreed ona theological detail and carried the matter to a Highercourt.
The writer ........... the view that manis a reasoning animal
Which description of manis Not true according to the author?
The higher animals referred to in the passage are ...............
What did the writer find when he went for noting the results?
The “Higher Court” refers to
