What is the central idea of the passage ?
MAT 2001 Question Paper
Read the following passage carefully and then answer these questions based on what is stated or implied in the passage.
Passage I :
We call a man irrational when he acts in a passion, when he cuts off his nose to spite his face. He is irrational because he forgets that, by indulging the desire which he happensto feel moststrongly at the moment, he will thwart other desires which in the long run are more important to him. If men were rational, they would take a more correct view of their own interest than they do at present; and if all men acted from enlightened self-interest the world would be a paradise in comparison with whatit is. | do not maintain that there is nothing better than self-interest as a motive to action; but I do maintain that self-interest, like altruism, is better when it is enlightened than when it is
unenlightened. In an ordered communityit is very rarely to a man's interest to do anything which is very harmful to others. The less rational a man is, the oftener he will fail to perceive how what injures others also injures him, because hatred or envy will blind him. Therefore, although I do not pretend that enlightened self-interest is the highest morality, I do maintain that, if it became common, it would make the world an immeasurably better place than it is.
Rationality in practice may be defined as the habit of rememberingall our relevant desires, and not only the one which happens at the momentto be strongest. Like rationality in opinion, it is a matter of degree. Complete rationality is no doubt an unattainable ideal, but so long as we continue to classify some men as lunatics it is clear that we think some men morerational than others. I believe that all solid progress in the world consists of an increase in rationality, both practical and theoretical. To preach an altruistic morality appears to me somewhat useless, because it will appeal only to those who already have altruistic desires. But to preach rationality is some what different, since rationality helps us to realise our own desires on the whole, whatever they may be. A man is rational in proportion as his intelligence informs and controls his desires. I believe that the control of our acts by our intelligence is ultimately what is of most importance, and that alone will make social life remain possible as science increases the means at our disposal for injuring each other. Education, the press, politics, religion—in a word, all the great forces in the world—areat present on the sideofirrationality, they are in the hands of men who flatter King Demosin orderto
lead him astray. The remedy doesnot lie in anything heroically cataclysmic, but in the efforts of individuals towards a more sane and balanced view of our relations to our neighbours and to the world. It is to intelligence, increasingly side-spread, that we must look for the solution of the ills from which our world is suffering.
Solution
The main idea of the passage is that the world's problems can be solved by people acting rationally and with enlightened self-interest. The author believes that if people made decisions by considering both their own long-term needs and the impact on others, the world would be a much better place. Rational thinking and understanding the consequences of our actions are seen as key to improving society.
What is the author's attitude to modern development in science and communications ?
Rationality, according to the passage means mainly
The author feels thatit is impractical to appeal to altruism because
The King Demos'refers to
Read the following passage carefully and then answer these questions based on it.
Passage II :
Over four hundred years after his death, scholars are still travelling the mysteries of Michelangelo's art. Recently one mystery that was revealed was that his famous drawing of a pensive Cleopatra included a hidden drawing of a different Cleopatra on the reverse side. This hidden Cleopatra shows a tormented woman, whose eyes stare out at the viewer and whose mouth is open, screaming in horror. The two images, drawn on two sides of the same paper, can be viewed simultaneously. A second mystery concerns Michelangelo's architectural plan for the dome of St Peter's Basilica in Rome. Did he intend for the dome to look like the model he built between 1558 and 15617? Or did he change his mind after building the model and decide to elevate the dome in the way it is today? Scholars do not agree on the answer. A third mystery about one of the greatest artists who ever .lived was why he destroyed hundreds or thousands of his drawings before he died. Did he feel they were unimportant? Did he want posterity to see only his finished products?
It can be inferred from the passage that the most unusual aspect of the Cleopatra drawingis that
The word ‘pensive’ (underlined) can best be substituted with the word
The dome of St Peter's Basilica
According to the passage, Michelangelo is
Why did Michelangelo destroy so many drawings before he died?
In these questions, choose the word from the alternatives (a), (b), (c) and (d) that is similar in meaning to the word given in capital letters.
GIST
Solution
Gist means the main or central idea of something.
Substance refers to the core or essential part of something, which is why it is the closest in meaning to "gist."
The other options (contribution, prestige, accessory) are not related to the central idea, making them incorrect.
SOLICITOUS
Solution
The word SOLICITOUS means showing concern, care, or anxious attention about something.
Worried → Means anxious or concerned, which is similar in meaning to solicitous.
HOMELY
LOQUACIOUS
PACIFY
In these qeustions, choose the word from the alternatives (a), (b), (c) and (d) that is opposite in meaning to the word given in capital letters.
PERTINENT
SLOPPY
WANTON
Solution
Option D is the correct answer,
'Wanton' means reckless, unrestrained, or senselessly excessive, often used to describe cruelty, destruction, or behaviour lacking discipline.
Option D) 'Discreet' means careful, cautious, or showing good judgment, which is the opposite of wanton behaviour.
Option A: Sportive means playful or lighthearted, which is not exactly the opposite of 'wanton.'
Option B: Ardent means passionate or enthusiastic, which is unrelated.
Option C: Fragile means delicate or easily broken, which is not the opposite of 'wanton'
FINESSE
IGNITE
In each of these questions, a related pair of words is followed by four pairs of words (a), (b), (c) and (d). Choose the pair that best expressesa relationship similar to that expressed in the given pair.
DRILL : BORE ::
Solution
The given pair DRILL: BORE describes a tool (drill) and its function (boreholes).
Now, let's analyze the options:
(A) Painter: Brush → A painter uses a brush, but "painter" is a person, not a tool. (Incorrect)
(B) Sieve: Sift → A sieve is used to sift materials, just like a drill is used to bore holes. (Correct)
(C) Helmet: Head → A helmet protects the head, but this does not represent a tool-function relationship. (Incorrect)
(D) Mason: Wall → A mason builds a wall, but a mason is a person, not a tool. (Incorrect)
DOE : STAG ::
PHILATELIST : STAMPS ::
UMPIRE : GAME ::
NUCLEUS : CELL ::
Each of the sentences in these questions has a blank indicating that something has been omitted. Beneath each sentence, four choices of words or phrases (a), (b), (c) and (d) are given. Choose the one that bestfits the meaning of the sentence as a whole.
Criticism that tears down without suggesting areas of improvementis not ........... and should be avoided if possible.
Many educators believe that bilingual education has proved to have definite .......... education in any one tongue.
Ballet dancers, ........... actors, must spend many hoursa daypractising before a performance.
The weather in the far north is not .......... it is down south.
Language, culture and personality may be considered independently of each other in thought, but they are .......... in fact.
In these questions, each sentence has four underlined portions marked A, B, C and D. Choose that portion which must be changed so that the sentence becomes correct.
River water pollution (A) / is often indicate (B) / by (C) / algae distribution. (D)
The ways of communication (A) / has (B) /changed dramatically (C) / since (D) / the last century.
Which (A) / determines a (B) /good meal varies (C) / from country to (D) / country.
Gandhiji lived a noble life of fasting (A) / and poverty (B) / in order to work for peaceful (C) / and independence (D)
The first year of child's life is (A) / characterised (B) / in (C) / rapid physical (D) / growth.
In each of these questions, an idiomatic expression is given. Choose the correct choice which expresses the meaning to the given expression from the four options (a), (b), (c) and (d).
To smell a rat
To be above board
Solution
The idiom "to be above board" means to be honest, open, and legitimate, free from any deceit or hidden agendas.
Hence, the answer is Option B.
To have the gift of the gab
Solution
"To have the gift of the gab" means to have a natural ability to speak in an eloquent and persuasive way.
To fall flat
Right-hand man
For the following questions answer them individually
Walking at $$\frac{3}{4}$$ of his usual pace, a man reaches his office 20 minutes late. Find his usual time.
Solution
Let usual speed = $$4$$ m/min and usual time taken be $$t$$ min
=> New speed = $$\frac{3}{4}\times4=3$$ m/min and new time = $$(t+20)$$ min
Speed is inversely proportional to time
=> $$\frac{4}{3}=\frac{t+20}{t}$$
=> $$4t=3t+60$$
=> $$t=60$$
$$\therefore$$ Usual time = 60 min = 1 hour
=> Ans - (B)
If the difference between the simple and the compound interests on some principal amount at 20% for 3 years is Rs 48, then the principal amount must be
Solution
Difference between SI and CI for 3 years = $$\frac{Pr^2}{100^2}(3+\frac{r}{100})$$ and for 2 years = $$\frac{Pr^2}{100^2}$$
Now, rate of interest = $$r=20\%$$ and time = $$t=3$$ years
Let principal amount be Rs. $$P$$
=> Difference = $$P\times\frac{(20)^2}{(100)^2}(3+\frac{20}{100})=48$$
=> $$P\times\frac{1}{25}\times\frac{16}{5}=48$$
=> $$P=\frac{48\times125}{16}=3\times125$$
=> $$P=Rs.$$ $$375$$
=> Ans - (C)
In a zoo, there are rabbits and pigeons. If their heads are counted, these are 90 while their legs are 224. Find the number of pigeons in the zoo.
Solution
Let us assume all 90 are rabbits, thus there must be 360 legs (Rabbits have 4 legs and pigeons have 2)
But there are $$360-224=136$$ less legs, so rabbits must be less.
=> Number of pigeons = $$\frac{136}{2}=68$$
Alternate Method : Let number of pigeons be $$x$$ and number of rabbits be $$y$$
=> $$x+y=90$$ -----------(i)
and $$2x+4y=224$$ ------(ii)
Solving the two equations, we get : $$x=68$$ and $$y=22$$
=> Ans - (B)
The sides of a rectangular field are in the ratio 3 : 4 with its area as 7500 sq m. The cost of fencing the field @ 25-paise per metre is
Solution
Let sides of rectangular field be $$3x$$ and $$4x$$ m respectively.
=> Area = $$lb=7500$$
=> $$3x\times4x=7500$$
=> $$x^2=\frac{7500}{12}=625$$
=> $$x=\sqrt{625}=25$$
=> Perimeter of field = $$2(3x+4x)=14x$$
= $$14\times25=350$$ m
$$\therefore$$ Cost of fencing = $$350\times0.25=Rs.$$ $$87.50$$
=> Ans - (A)
A and B together can do a piece of work in 6 days. A alone can doit in 10 days. What time will B require to do it alone ?
Solution
Time required by B alone = $$\frac{1}{6}-\frac{1}{10}$$
= $$\frac{5-3}{30}=\frac{2}{30}=\frac{1}{15}$$
$$\therefore$$ B alone finishes the work in 15 days.
=> Ans - (B)
A square and an equilateral triangle have the same perimeter. If the diagonal of the square is $$12 \surd2$$ cm, then the area of the triangle is
Solution
Let the perimeter of square and equilateral triangle be $$12x$$ cm
=> Each side of square = $$\frac{12x}{4}=3x$$ cm and each side of equilateral triangle = $$4x$$ cm
Diagonal of a square = $$d=\sqrt2\times$$ side
=> $$3x\times\sqrt2=12\sqrt2$$
=> $$x=\frac{12}{3}=4$$
=> Side of equilateral triangle = $$4\times4=16$$ cm
$$\therefore$$ Area of equilateral triangle = $$\frac{\sqrt3}{4} s^2$$
= $$\frac{\sqrt3}{4}\times(16)\times(16)$$
= $$64\sqrt3$$ $$cm^2$$
=> Ans - (C)
Acistern is filled in 5 hours and it takes 6 hours when thereis a leak in its bottom. If the cistern is full, in what time shall the leak empty it?
Solution
Let capacity of cistern = L.C.M. (5,6) = 30 units
Efficiency to fill it = $$\frac{30}{5}=6$$ units/hr
Let efficiency of leak = $$-x$$ units/hr
According to ques, => $$(6-x)\times6=30$$
=> $$6-x=5$$
=> $$x=1$$
$$\therefore$$ Time taken by leak to empty it = $$\frac{30}{1}=30$$ hours
=> Ans - (C)
The radius of the circumcircle of an equilateral triangle of side 12 cm is
Solution
Side of the equilateral triangle = $$s=12$$ cm
=> Area of triangle = $$\triangle=\frac{\sqrt3}{4}s^2$$
= $$\frac{\sqrt3}{4}\times12\times12=36\sqrt3$$ $$cm^2$$
$$\therefore$$ Circumradius = $$R=\frac{abc}{4\triangle}$$
= $$\frac{12\times12\times12}{4\times36\sqrt3}$$
= $$\frac{12}{\sqrt3}=4\sqrt3$$ cm
=> Ans - (B)
The sides of a triangle are 6 cm, 11 cm and 15cm. The radiusofits incircle is
Solution
Sides of the triangle are $$a=6,b=11,c=15$$ cm
Area of a triangle = $$\triangle=rs$$, where $$r$$=in radius and $$s$$ is semi perimeter
=> Semi perimeter = $$s=\frac{a+b+c}{2}=\frac{6+11+15}{2}=16$$ cm
Area of triangle using Heron's Formula = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{16\times10\times5\times1}$$
= $$\sqrt{2^5\times5^2}=20\sqrt2$$ $$cm^2$$
$$\therefore$$ $$r=\frac{\triangle}{s}$$
= $$\frac{20\sqrt2}{16}=\frac{5\sqrt2}{4}$$ cm
=> Ans - (A)
The sum of the interior angles of a polygon is 1620. The number of sides of the polygon must be
Solution
Let the number of sides of the polygon be $$n$$
Sum of interior angles of any polygon = $$(n-2)\times180=1620$$
=> $$n-2=\frac{1620}{180}=9$$
=> $$n=9+2=11$$
=> Ans - (B)
The distance between the tops of two trees 20 m and 28 m high is 17 m. The horizontal distance between the two trees is
Solution
AD and CE are two trees of height 28 m and 20 m respectively. AC = 17 m
In right $$\triangle$$ ABC,
=> $$(BC)^2=(AC)^2-(AB)^2$$
=> $$(BC)^2=(17)^2-(8)^2$$
=> $$(BC)^2=289-64=225$$
=> $$BC=\sqrt{225}=15$$ m
=> Ans - (C)
Rs 770 have been divided among A, B and C such that A receives 2/9th of what B and C together receive. Then A's share is
Solution
Let B and C together have = Rs. $$9x$$
=> A's share = $$\frac{2}{9}\times9x=Rs.$$ $$2x$$ ------------(i)
Also, amount that A receive = $$(770-9x)$$ -----------(ii)
Comparing equations (i) and (ii), => $$2x=770-9x$$
=> $$11x=770$$
=> $$x=\frac{770}{11}=70$$
$$\therefore$$ A's share = $$2\times70=Rs.$$ $$140$$
=> Ans - (A)
What least number must be subtracted from each of the numbers 14, 17, 34 and 42 so that the remainders are proportional ?
Solution
Let $$x$$ must be subtracted.
=> $$\frac{14-x}{17-x}=\frac{34-x}{42-x}$$
=> $$(14-x)(42-x)=(34-x)(17-x)$$
=> $$588-14x-42x+x^2=578-34x-17x+x^2$$
=> $$56x-51x=588-578$$
=> $$5x=10$$
=> $$x=2$$
=> Ans - (C)
$$\frac{3\pi}{5}$$ radians is equal to
Solution
Expression : $$\frac{3\pi}{5}$$ radians
= $$\frac{3}{5}\times180$$
= $$3\times36=108^\circ$$
=> Ans - (A)
If $$\sin A : \cos A = 4 : 7$$, then the value of $$\frac{(7 \sin A - 3 \cos A)}{(7 \sin A+ 2 \cos A)}$$ is
Solution
Let $$sin A=4$$ and $$cos A=7$$
Expression : $$\frac{(7 \sin A - 3 \cos A)}{(7 \sin A+ 2 \cos A)}$$
= $$\frac{7(4)-3(7)}{7(4)+2(7)}=\frac{28-21}{28+14}$$
= $$\frac{7}{42}=\frac{1}{6}$$
=> Ans - (D)
A tree breaks due to storm and the broken part bends so that the top of the tree first touches the ground, making an angle of 30 with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is 10 m. The height of the tree is
Solution
Given : BC = 10 m and $$\angle ACB=30^\circ$$
To find : AC + AB = ?
Solution : In right $$\triangle$$ ABC
=> $$tan(30^\circ)=\frac{AB}{BC}$$
=> $$\frac{1}{\sqrt3}=\frac{AB}{10}$$
=> $$AB=\frac{10}{\sqrt3}$$ ---------------(i)
Similarly, $$cos(30^\circ)=\frac{BC}{AC}$$
=> $$\frac{\sqrt3}{2}=\frac{10}{AC}$$
=> $$AC=\frac{20}{\sqrt3}$$ ------------------(ii)
$$\therefore$$ Height of tree = $$AB+AC$$
= $$\frac{10}{\sqrt3}+\frac{20}{\sqrt3}$$
= $$\frac{30}{\sqrt3}=10\sqrt3$$ m
=> Ans - (B)
If $$pqr = 1$$ then
$$\left(\left(\frac{1}{(1 + p + q^{-1})}\right) + \left(\frac{1}{(1 + q + r^{-1})}\right) + \left(\frac{1}{(1 + r + p^{-1})}\right)\right)$$ is equal to
Solution
Given : $$pqr=1$$
Expression : $$\left(\left(\frac{1}{(1 + p + q^{-1})}\right) + \left(\frac{1}{(1 + q + r^{-1})}\right) + \left(\frac{1}{(1 + r + p^{-1})}\right)\right)$$
= $$\frac{1}{(1+p+\frac{1}{q})}+\frac{1}{(1+q+\frac{1}{r})}+\frac{1}{(1+r+\frac{1}{p})}$$
= $$\frac{q}{(1+q+pq)}+\frac{1}{(1+q+pq)}+\frac{1}{(1+\frac{1}{pq}+\frac{1}{p})}$$
= $$\frac{q}{(1+q+pq)}+\frac{1}{(1+q+pq)}+\frac{pq}{(1+q+pq)}$$
= $$\frac{1+q+pq}{1+q+pq}=1$$
=> Ans - (A)
An edge of a cube measures 10 cm. If the largest possible cone is cut out of this cube, then the volume of the cone is
Solution
Edge of cube = 10 cm
Radius of the largest possible cone = 5 cm and height of cone = 10 cm
Volume of cone = $$\frac{1}{3}\pi r^2h$$
= $$\frac{1}{3}\times\frac{22}{7}\times(5)^2\times10$$
= $$\frac{25\times220}{21}=261.9$$ $$cm^3$$
=> Ans - (C)
If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball is
Solution
Let radius of each small spherical ball = $$r$$ cm
Volume of a sphere = $$\frac{4}{3}\pi r^3$$
=> $$8\times\frac{4}{3}\pi (r)^3=\frac{4}{3}\pi (10)^3$$
=> $$r^3=\frac{(10)^3}{8}$$
=> $$r=\frac{10}{2}=5$$
$$\therefore$$ Surface area of each ball = $$4\pi r^2$$
= $$4\times\pi \times(5)^2=100\pi$$ $$cm^2$$
=> Ans - (D)
If a and b are the roots of the equation $$x^2 - 6x + 6 = 0$$, then the value of $$a^2 + b^2$$ is
Solution
Equation : $$x^2 - 6x + 6 = 0$$
Sum of roots = $$a+b=6$$ and Product of roots = $$ab=6$$ --------------(i)
Now, squaring both sides, => $$(a+b)^2=(6)^2$$
=> $$a^2+b^2+2ab=36$$
=> $$(a^2+b^2)+2(6)=36$$
=> $$(a^2+b^2)=36-12=24$$
=> Ans - (B)
If $$a : b = 2 : 5$$ then $$(3a + 4b) : (5a + 6b)$$ is equal to
Solution
Let $$a=2$$ and $$b=5$$
To find = $$(3a + 4b) : (5a + 6b)$$
= $$\frac{3(2)+4(5)}{5(2)+6(5)}=\frac{(6+20)}{(10+30)}$$
= $$\frac{26}{40}=\frac{13}{20}$$
=> Ans - (A)
The value of
$$(1^3 + 2^3 + 3^3 + ........ + 15^3) - (1 + 2 + 3 + ......... + 15)$$ is
Solution
Sum of $$n$$ consecutive natural number cubes = $$[\frac{n(n+1)}{2}]^2$$
and sum of $$n$$ consecutive natural numbers = $$\frac{n(n+1)}{2}$$
Expression : $$(1^3 + 2^3 + 3^3 + ........ + 15^3) - (1 + 2 + 3 + ......... + 15)$$
= $$(\frac{15\times16}{2})^2-(\frac{15\times16}{2})$$
= $$(120)^2-120$$
= $$14400-120=14280$$
In the figure given below, O is the centre of the circle. If $$\angle OBC = 37^\circ$$, the $$\angle BAC$$ is equal to
Solution
In the circle, OB=OC (radii), => $$\angle OBC=\angle OCB=37^\circ$$
In $$\triangle$$ OBC, using angle sum property
=> $$\angle BOC+\angle OBC+\angle OCB=180^\circ$$
=> $$\angle BOC+37^\circ+37^\circ=180^\circ$$
=> $$\angle BOC=180^\circ-74^\circ=106^\circ$$
Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.
=> $$\angle BOC=2\times\angle BAC$$
=> $$\angle BAC=\frac{106^\circ}{2}=53^\circ$$
=> Ans - (C)
A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze ?
Solution
Area of the field that the horse can gaze = area of sector having radius $$r=14$$ m
= $$\frac{\theta}{360^\circ} \pi r^2$$
= $$\frac{90^\circ}{360^\circ} \times\frac{22}{7} \times14\times14$$
= $$\frac{1}{4}\times22\times28$$
= $$22\times7=154$$ $$m^2$$
=> Ans - (A)
A two-digit number is such that the product of the digits is 14. When 45 is added to the number, then the digits interchange their places. Find the number.
Solution
Factors of 14 = $$1,2,7,14$$
Thus, only two digit numbers possible are 27 and 72
Now, when 45 is added, number will be reversed, => Number = 27
=> Ans - (B)
Vishal goes to a shop to buy a radio costing Rs 2568. The rate of sales tax is 7%. He tells the shopkeeperto reduce the price of the radio to such an extent that he has to pay Rs 2568, inclusive of sales tax. Find the reduction needed in the price of the radio.
Solution
Let reduced price of the radio = Rs. $$x$$
=> Reduced Price + Sale tax = Rs. 2568
=> $$x+\frac{7x}{100}=2568$$
=> $$\frac{107x}{100}=2568$$
=> $$x=\frac{2568}{107}\times100$$
=> $$x=24\times100=2400$$
$$\therefore$$ Reduction needed = $$2568-2400=Rs.$$ $$168$$
=> Ans - (C)
Dinesh travels 760 km to his home, partly by train and partly by car. He takes 8 hoursif he travels 160 km bytrain and the rest by car. He takes 12 minutes moreif he travels 240 km bytrain and the rest by car. The speedsof the train and the car respectively are
Solution
Let speeds of train and car be $$x$$ km/hr and $$y$$ km/hr respectively.
According to ques,
=> $$\frac{160}{x}+\frac{600}{y}=8$$ ----------(i)
and $$\frac{240}{x}+\frac{520}{y}=8+\frac{12}{60}=8.2$$ -----------(ii)
Applying : 3 x (i)-2 x (ii)
=> $$\frac{1800}{y}-\frac{1040}{y}=24-16.4$$
=> $$\frac{760}{y}=7.6$$
=> $$y=\frac{760}{7.6}=100$$
Similarly, $$x=\frac{160}{2}=80$$
$$\therefore$$ Speeds of train and car are 80 km/hr and 100 km/hr respectively.
=> Ans - (A)
A person on tour has Rs 360 for his daily expenses. If he extends his tour for 4 days, he hasto cut his daily expensesby Rs3. Find the original duration of the tour.
Solution
Let original duration of tour be $$x$$ days and expense per day = Rs. $$\frac{360}{x}$$
According to ques,
=> $$(x+4)(\frac{360}{x}-3)=360$$
=> $$360-3x+\frac{1440}{x}-12=360$$
=> $$-3x^2+1440-12x=0$$
=> $$x^2+4x-480=0$$
=> $$(x+24)(x-20)=0$$
=> $$x=20,-24$$
$$\because x$$ cannot be negative, thus number of days = 20
=> Ans - (C)
Rs 6500 were divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs 30 less. Find the original number of persons.
Solution
Let original number of persons be $$x$$ and amount received by each person = Rs. $$\frac{6500}{x}$$
According to ques,
=> $$(x+15)(\frac{6500}{x}-30)=6500$$
=> $$6500-30x+\frac{15\times6500}{x}-450=6500$$
=> $$-30x^2+(15\times6500)-450x=0$$
=> $$x^2+15x-3250=0$$
=> $$(x+65)(x-50)=0$$
=> $$x=50,-65$$
$$\because x$$ cannot be negative, thus number of persons are 50
=> Ans - (A)
Pipes A and B running together can fill a cistern in 6 minutes. If B takes 5 minutes more than A to fill the cistern, then the times in which A and will fill the cistern separately will be respectively
Solution
Let time taken by A alone to fill the cistern = $$t$$ minutes
=> Time taken by B = $$(t+5)$$ minutes
According to ques, => $$\frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}$$
=> $$\frac{t+t+5}{t(t+5)}=\frac{1}{6}$$
=> $$12t+30=t^2+5t$$
=> $$t^2-7t-30=0$$
=> $$(t-10)(t+3)=0$$
=> $$t=10,-3$$
$$\because$$ Time cannot be negative, => $$t=10$$
Thus, time taken by A and B separately to fill the cistern is 10 min and 15 min respectively.
=> Ans - (C)
In a flight of 3000 km, an aircraft was slowed down by bad weather. Its average speed for the trip was reduced by 100 km/hour and the time increased by one hour. Find the original duration of the flight.
Solution
Let initial speed be $$x$$ km/hr, => New speed = $$(x-100)$$ km/hr
Let original duration of flight = $$t$$ hours and actual time taken = $$(t+1)$$ hours
Total distance = $$xt=3000$$ ------------(i)
Speed is inversely proportional to time,
=> $$\frac{x}{x-100}=\frac{t+1}{t}$$
=> $$xt=xt+x-100t-100$$
=> $$x-100t=100$$
=> $$\frac{3000}{t}-100t=100$$
=> $$-100t^2-100t+3000=0$$
=> $$t^2+t-30=0$$
=> $$(t+6)(t-5)=0$$
=> $$t=5,-6$$
$$\because$$ Time cannot be negative, => Original duration = 5 hours
=> Ans - (A)
Students of a class are made to stand in rows. If 4 students are extra in a row, there would be 2 rows less. If 4 students are less in a row, there would be 4 more rows. Find the numberof students in the class.
Solution
Let number of students in each row be $$x$$ and number of rows = $$y$$
=> Total number of students = $$xy$$
Case 1 : $$(x+4)\times(y-2)=xy$$
=> $$xy-2x+4y-8=xy$$
=> $$2y-x=4$$ ----------(i)
Case 2 : $$(x-4)\times(y+4)=xy$$
=> $$xy+4x-4y-16=xy$$
=> $$x-y=4$$ ------------(ii)
Adding equations (i) and (ii), => $$y=8$$
and similarly, $$x=12$$
$$\therefore$$ Total number of students = $$12\times8=96$$
=> Ans - (C)
Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number ?
Solution
A derangement is a permutation of objects that leave no object in its original position.Number of derangement's of set with n elements is
$$D_n=n![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-.................+(-1)^n\frac{1}{n!}]$$
The probability of derangement = $$\frac{D_n}{n!}$$
Now, in the given case $$n=4$$
$$\therefore$$ Probability that none of the objects occupy the place corresponding to their number
= $$1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}$$
= $$1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}$$
= $$\frac{12-4+1}{24}=\frac{9}{24}=\frac{3}{8}$$
=> Ans - (B)
A boat goes 24 km upstream and 28 km dowstream in 6 hours. If it goes 30 km upstream and 21 km downstream in 6 hours and 30 minutes, find the speed of the stream.
Solution
Let speed of boat = $$x$$ km/hr and speed of stream = $$y$$ km/hr
=> Downstream speed = $$(x+y)$$ km/hr and downstream speed = $$(x-y)$$ km/hr
According to ques,
=> $$\frac{24}{x-y}+\frac{28}{x+y}=6$$ ----------(i)
and $$\frac{30}{x-y}+\frac{21}{x+y}=6.5$$ -------(ii)
Using the operation : 4 x (ii)-3 x (i)
=> $$\frac{48}{x-y}=26-18$$
=> $$x-y=\frac{48}{8}=6$$ --------(iii)
Similarly, $$x+y=14$$ -------------(iv)
Now, subtracting equation (iii) from (iv), => $$2y=8$$
=> $$y=\frac{8}{2}=4$$
$$\therefore$$ Speed of stream = 4 km/hr
=> Ans - (C)
The diameter of a cycle wheel is 70 cm. A cyclist takes 30 hours to reach a destination at the speed of 22 km/hr. How many revolutions will the wheel make during this journey?
Solution
Distance covered at 22 km/hr in 30 hours = $$22\times30=660$$ km = $$66\times10^6$$ cm
Radius of cycle = 35 cm
In one revolution, distance covered by the wheel = Circumference of cycle = $$2\pi r$$
= $$2\times\frac{22}{7}\times35=220$$ cm
$$\therefore$$ Number of revolutions in the total journey = $$\frac{66\times10^6}{220}=3\times10^5$$
= 3 lakh revolutions
=> Ans - (B)
Shyam had 85 currency notes in all, some of which were of Rs 100 denomination and the remaining of Rs 50 denomination. The total amountof all these currency notes was Rs 5000. How much amountin rupeesdid he have in the denomination of Rs 50?
Solution
Let number of notes of Rs. 50 denomination = $$x$$ and number of notes of Rs. 100 denomination = $$(85-x)$$
According to ques,
=> $$50x+100(85-x)=5000$$
=> $$50x+8500-100x=5000$$
=> $$50x=3500$$
=> $$x=70$$
$$\therefore$$ Amount in only Rs. 50 denomination = $$70\times50=Rs.$$ $$3500$$
=> Ans - (A)
Acar owner buyspetrol at Rs 7.50, Rs 8.00 and Rs 8.50 per litre for three successive years. What approximately is the average costper litre of petrol if he spends Rs 4000 each year?
Solution
Since amount spend each year is constant, thus average cost is the harmonic mean of each price
= $$3\div(\frac{1}{7.5}+\frac{1}{8}+\frac{1}{8.5})$$
= $$3\div(\frac{2}{15}+\frac{1}{8}+\frac{2}{17})$$
= $$3\div(\frac{272+255+240}{2040})$$
= $$3\times\frac{2040}{767}=7.98$$
=> Ans - (C)
lf 10 years are subtracted from the present age of Ram and the remainderdivided by 14, then you would get the preset age of his grandson Shyam. If Shyam is 9 years younger to Sunder whose age is 14, then whatis the present age of Ram?
Solution
Sunder's age = 14 years and Shyam's age = $$14-9=5$$ years
Let present age of Ram = $$x$$ years
According to ques, => $$\frac{(x-10)}{14}=5$$
=> $$x-10=70$$
=> $$x=70+10=80$$
$$\therefore$$ Ram's age = 80 years
=> Ans - (A)
Sunder purchased an office bag with a price tag of Rs 600 in a sale where 25%discount was being offered on the tag price. He was given a further discount of 10% on the amount arrived at after giving usual 25%discount. What wasthe final amount paid by Sunder?
Solution
Marked price = Rs. 600
Relative discount after successive discounts of 25 and 10% = $$25+10-(\frac{25\times10}{100})=32.5\%$$
$$\therefore$$ Sale price = $$600-(\frac{32.5}{100}\times600)$$
= $$600-195=Rs.$$ $$405$$
=> Ans - (C)
The mean of 30 values was 150. It was detected on rechecking that one value 165 was wrongly copied as 135 for the computation of the mean. Find the correct mean.
Solution
Mean of 30 values = 150
After replacing 135 by 165, correct mean = $$\frac{(30\times150)+(-135+165)}{30}$$
= $$150+\frac{30}{30}=151$$
=> Ans - (A)
The following table gives the enrolment in Higher Secondary Schools in 1978. Study the table carefully and answer these questions.
What is the approximate percentage of schools, where the enrolment was below 120?
Solution
Percentage of schools where enrolment is below 120 are given by
= $$\frac{(526+620+674+717+681)}{5439}\times100$$
= $$\frac{3218}{5439}\times100\approx59.16\%$$
=> Ans - (A)
What is the approximate percentage of schools, where the enrolment was above 79 but below 180?
Solution
Percentage of schools, with enrolment above 79 & below 180 are given by
= $$\frac{(717+681+612+540+517)}{5439}\times100$$
= $$\frac{3067}{5439}\times100\approx56.39\%$$
=> Ans - (B)
Under which class, the maximum numberof schools fall?
Solution
It is clear after viewing the all the enrolments that under the class 80-99, maximum no. of schools (717) fall.
=> Ans - (B)
What is the approximate percentage of the least number of schools for the classes of enrolment ?
Solution
The least no. of schools (517) are in class 160-179
=> Required % = $$\frac{517}{5439}\times100\approx9.5\%$$
=> Ans - (B)
What is the number of schools where the enrolment is above 99 but below 1607 ?
Solution
No. of schools with enrolment above 99 & below 160 are
= $$681+612+540=1833$$
=> Ans - (C)
What is the average enrolment per HS School?
Solution
Average enrolment is the average of mean enrolment
= $$\frac{(29.5+49.5+69.5+89.5+109.5+129.5+149.5+169.5+189.5)}{9}$$
= $$\frac{985.5}{9}=109.5$$
=> Ans - (C)
The following table gives Population and Activities of Indian Children (1993-94). Study the table carefully and answer these questions.
What is the average of Rural Males Population in millions ?
Solution
Population (in millions) of Rural males (5-9) = 39.7
Population (in millions) of Rural males (10-14) = 36.1
=> Required average = $$\frac{39.7+36.1}{2}=37.9$$
=> Ans - (C)
In which category of populatoin, there is the lowest percentage of children in the school ?
Solution
Percentage of children in school in category :
Urban males (5-9) = $$84.1\%$$
Rural males (5-9) = $$67.2\%$$
Urban females (5-9) = $$80.1\%$$
Rural females (10-14) = $$55.7\%$$ [MIN]
=> Ans - (D)
What is the approximate percentage of children of all categories not in school ?
Solution
Total number of children (in millions) not in school = 57.79
=> Required ratio = $$\frac{57.79}{185.5}\times100\approx31.13\%$$
=> Ans - (D)
What per cent is the ratio between urban males and rural males not in school ?
Solution
Urban males (in millions) not in school = $$1.79+1.5=3.29$$
Rural males (in millions) not in school = $$13.02+8.44=21.46$$
=> Required % ratio = $$\frac{3.29}{21.46}\times100=15.33\%$$
=> Ans - (C)
What is the approximate number of children in millions who are working ?
Solution
The number of children who are working is to be determined from the 4th column from the left.
Rural males (5-9) working in millions = (1.3/100)*39.7 = 0.5161 million
Rural females (5-9) working in millions = (3/100)*35.7 = 1.071 million
And so on, doing approx calculations and adding up all results, the sum comes close to approx 18 million
What is the approximate percentageofall categories of children not in school and not working ?
Solution
Number of children not in school and not working in all categories (in millions) :
= $$(\frac{31.5}{100}\times13.02)+(\frac{40.8}{100}\times15.63)+(\frac{15.2}{100}\times1.79)+(\frac{18.6}{100}\times2.02)+(\frac{10.6}{100}\times8.44)+(\frac{14}{100}\times13.42)+$$
$$(\frac{5.8}{100}\times1.5)+(\frac{5.3}{100}\times1.93)$$
= $$14.08$$
=> Required % = $$\frac{14.08}{57.79}\times100\approx24.36\%$$
=> Ans - (D)
In which category of children, there is maximum number not in school and not working?
Solution
Number of children (in millions) not in school and not working in the category :
Rural females (10-14) = $$\frac{14}{100}\times13.42=1.88$$
Rural males (5-9) = $$\frac{31.5}{100}\times13.02=4.10$$
Rural females (5-9) = $$\frac{40.8}{100}\times15.63=6.37$$ [MAX]
Urban males (10-14) = $$\frac{5.8}{100}\times1.5=0.08$$
=> Ans - (C)
In which category of children, there is maximum number not in school but working?
Solution
Number of children (in millions) not in school but working in the category :
Rural males (10-14) = $$\frac{12.8}{100}\times8.44=1.08$$
Rural females (10-14) = $$\frac{30.3}{100}\times13.42=4.06$$ [MAX]
Urban females (10-14) = $$\frac{13.1}{100}\times1.93=0.25$$
Urban males (10-14) = $$\frac{7}{100}\times1.5=0.10$$
=> Ans - (B)
What percentage of the total population of the children of all categories is in the school?
Solution
Total child population not in school (in millions) = $$57.79$$
Total child population in school (in millions) = $$185.5-57.79=127.71$$
=> Required % ratio = $$\frac{127.71}{185.5}\times100$$
= $$68.87\%$$
=> Ans - (A)
What approximately is the percentage ratio between the total number of children not in school and in school?
Solution
Total child population not in school (in millions) = $$57.79$$
Total child population in school (in millions) = $$185.5-57.79=127.71$$
=> Required % ratio = $$\frac{57.79}{127.71}\times100$$
= $$45.20\%$$
=> Ans - (C)
Study the following graph and answer these questions given below it.
Which year shows the maximum percentage of export with respect to production?
Solution
Percentage of export with respect to production in the year
1992 = $$\frac{180}{540}\times100=33.33\%$$
1993 = $$\frac{288}{720}\times100=40\%$$
1996 = $$\frac{450}{660}\times100=68.19\%$$ [MAX]
1995 = $$\frac{400}{600}\times100=66.67\%$$
=> Ans - (C)
The population of India in 1993 was
Solution
Tea available in India in 1993 = $$720-288=432$$ million kg
Per capita availability in 1993 = 0.4 kg
=> Population = $$\frac{432}{0.4}=1080$$ million
=> Ans - (B)
If the area under tea production was less by 10% in 1994 than in 1993, then the approximate rate of increase in productivity of tea in 1994 was
Solution
Since there is no data given about area, hence it cannot be determined.
=> Ans - (D)
The average proportion of tea exported to the tea produced overthe period is
Solution
Total tea exported (in millions kg) = 96+180+288+340+400+450 = 1754
Total tea produced (in millions kg) = 480+540+720+700+600+660 = 3700
=> Required ratio = $$\frac{1754}{3700}=0.47$$
=> Ans - (B)
What is the first half decade's average per capita availability of tea ?
Solution
First half decade's (1991-1995) average per capita availability of tea
= $$\frac{(390+410+400+450+500)}{5}$$
= $$\frac{2150}{5}=430$$ gm
=> Ans - (D)
In which year was the per capita availability of tea minimum ?
Solution
Clearly, per capita availability of tea is minimum in the year 1991 which was 390.
=> Ans - (C)
In which year was there minimum percentage of export with respect to production?
Solution
Percentage of export with respect to production in the year :
1991 = $$\frac{96}{480}\times100=20\%$$ [MIN]
1992 = $$\frac{180}{540}\times100=33.33\%$$
1993 = $$\frac{288}{720}\times100=40\%$$
1994 = $$\frac{340}{700}\times100\approx50\%$$
=> Ans - (A)
In which year we had maximum quantity of tea for domestic consumption?
Solution
Maximum quantity of tea for domestic consumption was available in the year which has the highest production, i.e. 1993 producing 720 million kg.
=> Ans - (C)
What approximately was the average quantity of tea available for domestic consumption during the period?
Solution
The total production of tea in the 6 year period = 3700 million kg
The total export of tea in the 6 year period = 1754 million kg
Total quantity available for domestic consumption = 3700 - 1754 = 1946 million kg
Average qty available per year = 1946/6 = 324.3 million kg
What was approximately the average population during the period?
Study the table given below and answer these questions.
In which period, did we have the most adverse trade balancefor India ?
Solution
The period with the most adverse trade balance was 1998-99 having lowest trade balance equal to -8199.
=> Ans - (C)
What was the average % growth rate of exports during the entire period ?
Solution
% Growth rate of exports during the entire period.
= $$\frac{33659-16613}{16613}\times100$$
$$\approx102.60\%$$
Thus, average growth for the 10 year interval = $$\frac{102.60}{10}=10.26\%$$
=> Ans - (D)
In which period was the trade balance the best ?
Solution
Clearly, trade balance is best in the period 1993-94 which was -1068 (maximum).
=> Ans - (D)
In which period the growth rate of exports was the highest?
Solution
Clearly, growth rate of exports is highest in the period 1995-96 which was 8.7
=> Ans - (A)
During which period the Export/Import ratio was the highest ?
Solution
Export/Import ratio in the period :
1994-95 = $$\frac{8}{8.7}=0.91$$
1993-94 = $$\frac{8}{8.3}=0.96$$ [MAX]
1991-92 = $$\frac{7.1}{7.8}=0.91$$
1995-96 = $$\frac{8.7}{10.1}=0.86$$
=> Ans - (B)
What is the average growth rate of exports for the period 1992-93 to 1995-96?
Solution
% Growth rate of exports during the period 1992-93 to 1995-96
= $$\frac{31795-18537}{18537}\times100$$
$$\approx71.52\%$$
Thus, average growth for the 4 year interval = $$\frac{71.52}{4}=17.88\%$$
=> Ans - (D)
What is the average Export/ Import ratio for the period 1992-93 to 1998-99?
Using Exports GDP ratio for the period 1989-90, find the GDP for the same period.
Study the table given below carefully and answer these questions.
What is the average growth rate per year of population to be covered from 2001 to 2011?
Solution
Growth rate per year of population to be covered from 2001 to 2011
= $$\frac{8.19-6.23}{6.23}\times100$$
$$\approx31.5\%$$
Now, average growth rate of 10 years = $$\frac{31.5}{10}=3.15\%$$
=> Ans - (A)
In which year was there maximum difference between water requirements and actual or projected supplies?
Solution
Difference in per capita water requirements and the actual or projected supplies in the year :
1994 = $$913-680=233$$
2011 = $$1862-1906=-44$$
2021 = $$2224-1906=318$$ [MAX]
1991 = $$722-545=177$$
=> Ans - (C)
In which year the actual or projected supplies exceeded/exceeds the water requirements?
Solution
Clearly in the year 2011, the actual or projected supplies exceeded/exceeds the water requirements by :
= 1906 - 1862 = 44 mid
=> Ans - (A)
What approximately is the maximum per capita of actual or projected supplies as litres/day ?
Solution
Per capita of water in actual or projected supplies in the year :
1991 = $$\frac{545}{3.3}\approx165$$
1994 = $$\frac{680}{4.35}\approx156$$
2001 = $$\frac{1090}{6.23}\approx175$$
2011 = $$\frac{1906}{8.19}\approx232.7$$ [MAX]
2021 = $$\frac{1906}{10.15}\approx188$$
=> Ans - (C)
In which year, there was the least difference in per capita water requirements and the actual or projected supplies?
Solution
Difference in per capita water requirements and the actual or projected supplies in the year :
2001 = $$1105-1090=15$$
2011 = $$1862-1906=-44$$ [MIN]
1991 = $$722-545=177$$
1994 = $$913-680=233$$
=> Ans - (B)
In which year, there was the lowest per capita of water in actual or projected supplies?
Solution
Per capita of water in actual or projected supplies in the year :
2021 = $$\frac{1906}{10.15}\approx188$$
2001 = $$\frac{1090}{6.23}\approx175$$
1994 = $$\frac{680}{4.35}\approx156$$ [MIN]
1991 = $$\frac{545}{3.3}\approx165$$
=> Ans - (C)
Read the following information carefully to answer these questions.
A sample poll of 200 votes revealed the following information concerning three candidates A, B and C of a certain party who were running for three different offices.
28 in favour of both A and B.
98 in favour of Aor B but not C.
42 in favour of B but not Aor C.
122 in favour of B or C but not A.
64 in favour of C but not Aor B.
14 in favour of A and C but not B.
How many voters were in favour of all the three candidates?
How many voters were in favour of all the three candidates?
How many voters werein favour of B irrespective of A or C?
How many voters were in favour of C irrespective of A or B?
How many voters werein favour of A and B but not C?
How many voters werein favour of only one of the candidates ?
How many voters werein favour of A and C but not B?
How many voters were in favour of C alone?
How many voters were in favour of B and C but not A?
How many voters were in favour of A and C but not B?
For the following questions answer them individually
What is the next letter in the following series ? a, c, b, d, e?
In these qeustions, find the missing number.
Solution
In first diagram, 24+30 = 54
In second diagram, 70+42 = 112
The logic here is the sum of bottom left number and top right number is equal to top left number.
Similarly,
In third diagram, required number = 38+28 = 66
Hence, the correct answer is Option B
Solution
We can see that 6561 is the square of 81.
We will try substituting the blank with the square of 3 i.e. 9 and see whether the relationship holds.
And it does. 3 --> 9 --> 81 --> 6561
Solution
The logic here is
58 = (3 x 4 x 5) - 2
58 = (5 x 6 x 2) - 2
62 = (8 x 4 x 2) - 2
Similarly, the required number = (7 x 6 x 3) - 2 = 126 - 2 = 124
Hence, the correct answer is Option D
6, 15, 36, 75, ?
5 : 7 :: ? : 28
Solution
The logic here is
7 x 4 = 28 $$\longrightarrow\ $$ $$\text{second number}\times4=\text{fourth number}$$
Similarly, $$\text{first number}\times4=\text{third number}$$
$$\therefore\ $$Third number = 5 x 4 = 20
Hence, the correct answer is Option A
15, 45, ?, 405
Solution
We can see that 15*3 = 45
So we can try putting 45*3 = 135 in missing blank and then cross check whether 135*3 equals 405.
We can see that the relationship holds and thus, the blank should be filled with 135.
In these questions, two statements are given followed by two conclusions I and II. You have to consider both the statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions is/are definitely drawn from the given statements.
Statements :
When it rains, usually X does not go out.
X has gone out.
Conclusions:
I. It is not raining.
II. X has some urgent business to transact.
Solution
X may or may not go out if it is raining. So if X went out, we cannot say whether it is raining or not for sure. Also, the reaosn for X going out has nowhere been given to us. Hence, neither of the 2 follow.
Statements :
In a Golf Club, all the members are not active players of the game but all of them are rich.
Mrs X is a member.
Conclusions:
I. She is a golfer.
II. She is rich.
Solution
All members of the club are said to be rich and Mrs. X being a member, has to be rich too. Hence, II follows.
However, it is given that one need not be a golfer compulsorily to be member of the club. Hence, I cannot be said with complete certainty. Hence, only II follows.
Statements :
All employees of company A have identity cards.
Ram is an employee of company A.
Conclusions :
I. Ram has an identity card.
II. Ram is the General Manager of the Company.
Solution
Ram is an employee of the company A and all the employees of company A have identity cards. So Ram has an identity card. Hence conclusion I follows.
There is no information about General Manager of company A from the statements. so conclusion II is unrelated and pointless. Hence conclusion II do not follow.
$$\therefore\ $$Only conclusion I follows the given statements.
Hence, the correct answer is Option A
Statements:
If there is shortage in the production of onions, the price of onions will go up.
Price of onions has gone up.
Conclusions :
I. There is shortage in the production of onions.
II. Onions were exported.
Solution
Prices of onions can go up for reasons beyond and other than shortage in the production of onions. hence, conclusion 1 cannot be made with complete confidence. Conclusion 2 is irrelevant. Hence, C.
Statements:
If all players play to their full potential, we will win the match.
We have won the match.
Conclusions:
I. All players played to their full potential.
II. Some players did not play to their full potential.
Solution
The first statement "If all players play to their full potential, we will win the match." does not mean that the match will be won ONLY if all players play to full potential. The guarantee is only one directional.
Hence, even if the match was won, it is not assured that all players must have played to their full potential.
Also, the second conclusion cannot be said to have occurred definitely. Either conclusion 1 happened or conclusion 2 happened. None can be said to have occurred with complete certainty. Hence, neither of the 2 follow.
Statements :
Some businessmen are rich.
Soman is rich.
Conclusions :
I. Soman is a businessman.
II. Soman has a big farm.
Solution
Even though some businessmen are rich and Sonam is also rich, it doesn't necessarily mean that Sonam is also a businessman.
Also, there is no relevance of owning a farm. Hence, none of the 2 conclusions follow.
Statements :
All persons who own a house or a car should file Income Tax return.
Shiela files her income tax return.
Conclusions:
I. Shiela owns a house or a Car.
II. Shiela neither owns a house nor a Car.
Solution
It is not necessary that only people who own house/car/both file income tax return. So it is perfectly valid that Shiela doesn't own either of these and yet files income tax return.
However, from the given statements of conclusion, we know clearly that either Shiela owns house/car or she doesn't own any at all. There be no common ground nor a situation other than these 2. Hence, either I follows or II.
Statements :
In order to be selected for the local cricket team, a person should be either a good batsman or bowler.
Shyam is selected for the team.
Conclusions:
I. Shyam is a good batsman.
II. Shyam is not a good bowler.
Solution
Shyam got selected in the team means that:
1. He is a good batsman
OR 2. He is a good bowler
OR 3. He is both a good bowler and a good batsman
Any of these 3 situations is possible. If Shyam is a good batsman doesn't mean that he is a bad bowler. We only know that he is a good batsman and thus, got selected to the team. Same logic applies if Shyam is a good bowler.
Hence, neither I or II follows with certainty.
Statements:
Without rains the crops will not be good.
The crops were good.
Conclusions :
I. There were rains.
II. Crops were good due to good fertilizers.
Solution
We are told that crops cannot be good if rains do not happen. And then we are given that crops turned out to be good.
Hence, it follows rationally that rains happened. Hence, I follows.
There is no mention of fertilizers or their effect on crops. Hence, nothing can be said about II. Hence, only I follows.
Statements:
According to the evolution theory, man evolved from a monkey.
X is a monkey.
Conclusions:
I. X can become a man.
II. Man can become a monkey.
Solution
We are given that man evolved from a monkey. However, it is not given how much time the evolution actually took. It can be anything.
So even if X is a particular monkey, it may or may not live long enough to evolve into a man.
Hence, we cannot say that X can become a man.
Also, the question of man becoming a monkey again does not arise because we are not given whether the evolution is unidirectional or reversible. Hence, none follow.
In these questions, select the odd one out.
Solution
In the given figures, Only the figure 1 is a regular polygon with all sides and interior angles are equal. In other figures the sides are irregular and not equal.
$$\therefore\ $$Figure 1 is the odd one among the given figures.
Solution
The arrowed line is perpendicular to the plain segment in all of 1, 2, 4. Only in 3 there is a deviation from this pattern. Hence, odd one out.
Solution
In 1, 3, 4 all the three figures are aligned along the horizontal segment of the figure. Only in 2 are they aligned with the vertical line. Hence, 2 is the odd one out.
For the following questions answer them individually
Four positions of a cube are shown below. Which symbol is opposite to the face having $$'\triangle'$$
Read the following information carefully and answer these questions.
In a certain coding system,
'816321' means "The brown dogfrightened the cat”,
‘64851' means "The frightened cat ran away”,
‘7621' means "The cat was brown",
'341' means "The dog ran"
What is the codefor 'the dog was frightened’?
Solution
'816321' means "The brown dog frightened the cat”,
‘64851' means "The frightened cat ran away”,
‘7621' means "The cat was brown",
'341' means "The dog ran"
All statements have the word "the" and all codes have "1"
Hence, "the" = 1
Sentences 1,2,3 have "cat" in common and the only common number in all 3 codes is 6
Hence, "cat"= 6
Statements 1 and 2 have frightened in common and only "8" left unassigned.
Hence, "frightened" = 8
From statements 1 and 3 we can say that only brown is left common and "2" is unassigned common number.
Hence, "brown" = 2
Consequently, "was" = 7
From statements 1 and 4, dog will stand for "3" as "the" has already been assigned 1.
Hence, the statement 'the dog was frightened’ will be coded as: 1-3-7-8 in any order.
This is reflected in option B
What is the code for ‘frightened’ ?
Solution
'816321' means "The brown dog frightened the cat”,
‘64851' means "The frightened cat ran away”,
‘7621' means "The cat was brown",
'341' means "The dog ran"
All statements have the word "the" and all codes have "1"
Hence, "the" = 1
Sentences 1,2,3 have "cat" in common and the only common number in all 3 codes is 6
Hence, "cat"= 6
Statements 1 and 2 have frightened in common and only "8" left unassigned.
Hence, "frightened" = 8
What is the code for 'away' ?
Solution
The word "away" appears only in the second sentence.
And the only unique number in the code for the second sentence is 5.
hence, "away" = 5
What is the code for 'brown' ?
Solution
'816321' means "The brown dog frightened the cat”,
‘64851' means "The frightened cat ran away”,
‘7621' means "The cat was brown",
'341' means "The dog ran"
All statements have the word "the" and all codes have "1"
Hence, "the" = 1
Sentences 1,2,3 have "cat" in common and the only common number in all 3 codes is 6
Hence, "cat"= 6
Statements 1 and 2 have frightened in common and only "8" left unassigned.
Hence, "frightened" = 8
Sentences 1 and 3 have "brown" in common now and only "2" in common
Hence, "brown" = 2
For the following questions answer them individually
In how many ways can 7 people be seated at a roundtable if 2 particular people must notsit next to each other?
In a certain coding system ETTPI stands for APPLE. Whatis the code for ‘DELHI’ ?
Solution
Let the alphabets be numbered as A = 1, B = 2 up to Z = 26
Thus, if A gets coded to E, P gets coded to T, it means each alphabet moves 4 steps ahead in the coding system.
Hence, D moves forward to H, E moves forward to I and so on.
Hence, DELHI gets coded to H-E-P-L-M
William M Daley who was part of the US President's delegation to India is the
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Salman Khan and Sunil Shetty have been recruited as part of which campaign in the year 2000?
What is the present level of Indo-US trade?
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Who is the Chief Election Commissioner of India at present?
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.......... is not a member of ASEAN.
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The Upper House of Parliament is known as
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Miss World 2000 title has been won by
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Which of the following did India not wantto be part of the last WTO talks ?
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Which artist's autobiography is titled "Autobiography of a Genius"?
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Which of the following cities was not visited by President Mr Bill Clinton during his trip to India?
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Who said ; "I do not fear computers, I fear the lack of them."
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When was the last time that India won the Gold medal in Hockey in the Olympics ?
Recently Bill Gates became the first man to cross the 100 billion dollars mark. According to econo mists this means that only .............. nations have an economic output higher than that of Bill Gates.
