Maths By Amiya 246-350.pdf

Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya ©AMIYA KUMAR

Maths By Amiya, QUESTIONS &

Solutions 246-350 except DI 261-280

246.246.246.246. 36C3+35C3+34C3+……………..+3C3= ?

a. 2^36 b. 3^36 c.2^35 d. 37C4 e. None of These Sol: [d] Formula nC(r) + (n+1)C(r) = (n+1)C(r+1)

DirectionDirectionDirectionDirection 247247247247----250250250250:::: Ram has a cuboid of dimension 10*9*8 cm^3. He firstly dipped cuboid in black paint and put under sun . After 1hr he painted red colour to three mutually adjacent sides of the cuboid and then cut in to smaller cubes of dimension 1*1*1 cm^3. Then

247.247.247.247. How many smaller cubes have only Red as their colored surface(s)

a. 138 b. 148 c. 168 d. None of These Ans .[c]Ans .[c]Ans .[c]Ans .[c]

248.248.248.248. How many cubes have Red & Black both color on their surface(s) a. 40 b. 42 c. 48 d. None of These Ans .[c]Ans .[c]Ans .[c]Ans .[c]

249.249.249.249. How many smaller cubes have Red color on their surface(s) a. 210 b. 216 c. 200 d. None of These Ans .[Ans .[Ans .[Ans .[bbbb]]]]

250.250.250.250. How many cubes have Red or Black both color on their surface(s) a. 380 b. 384 c. 240 d. None of These Ans .Ans .Ans .Ans .[b[b[b[b]]]]

251.251.251.251. A(x), B(x) and C(x) are three polynomials in x, such that F(G)

H(G) = IGJKGLM NOP H(G)

Q(G) = M(GRJI)GJM

Which of the following statements are always true? I. A(x) has 1 real root. II. One of the roots of C(x) is 2.


III. If C(x) has 4 real roots, then B(x) has 3 real roots. IV. If A(x) has only 1 real root, then C(x) + B(x) also has only 1 real root. a. I and II b. II and III c. I, II and III d. II, III and IV Sol.Sol.Sol.Sol. dddd F(G)

H(G) = IGJKGLM does not imply that A(x) = 4x + 1 and B(x) = x – 2

A(x) and B(x) could have had other common factors which got cancelled out. Looking at the two relations simultaneously, B(x) = 2(WM + 4) × (W − 2) × any other factor, say Z(W) i.e. [(W) = 2(WM + 4) × (W − 2) × Z(W) From the first relation, \(W) = (4W + 1) × 2(WM + 4) × Z(W) and from the second relation ](W) = (W + 2) × (W − 2) × Z(W). Statement I : The factor (4x+1) gives one real root and the factor (WM + 4) results in two imaginary roots but no real root. However A(x) could also have other real roots because of Z(W). So (I) is not necessarily true. Statement II :: From above we see that one of the roots of C(x) is 2. Statement III: We see that C(x) has 2 real roots 2 and -2. If it has 2 more real roots, then these will be because of the factorZ(W). Because this factor is present in B(x) as well, and B(x) will also have these two roots. B(x) already has one real root i.e. 2 and if it has 2 more, it has 3 real roots. So option (III) is correct. Statement IV: If A(x) has only one real root, it means Z(W) does not have any real root. ](W) + [(W) = (W − 2) × Z(W) × [(W + 2) + 2(WM + 4)] = (W − 2) × Z(W) ×(2WM + W + 10). The last factor has imaginary roots since its determinant is negative. Thus, C(x) + B(x) would also have only 1 real root that obtained from (x – 2). so IV is also true. 252.252.252.252. If ^_

` will terminate after 3 decimal place, where a = Nb ∗ de and a,b,c & d can take only be single digit prime number (repetition is allowed). Then How many different value of N be possible.

a. 20 b. 4^4 c. 19 d. None of These Sol:[d]


Since ^_` is terminating after 3 decimal it means the deniminator should have

either 3rd power of 2 or 5 or both, but if any one have 3rd power than other should have lesser or equal power and no other primes in denomimator otherwise u will get reoccuring decimals. CASE I : a = 2_ ∗ de If c = 5 than d can take only one value which is 2, so numbers = 2_ ∗ 5M = 200 CASE II : a = 5_ ∗ de If c = 2 than d can take only one value which is 2, so numbers = 2M ∗ 5_ = 500 CASE II : a = 2_ ∗ 5_ = 1000, only 1 number in this case. So required Number = 1+1+1 = 3 253.253.253.253. A certain race is made up of three stretches A, B and C, each 2 km long. Charan

completes the race at an average speed of 20 kmph. His average speed over the first two stretches is 4 times that of the last stretch. And his average speed over the first and third stretch is 1.5 times that of the last stretch. Find his speed over stretch B.

a. 30 kmph b. 10 kmph c. 60 kmph d. 40 kmph SolSolSolSol.c.c.c.c Let Charan take time equal to gK, gM, g_ over the three stretches respectively. Thus,

hijJiRJik

= 20l. m. gK + gM + g_ = 3/10 hrs i.e. 18 minutes ……..(i) Also, I

ijJiR= 4 × M

ikl. m. g_ = 2(gK + gM) ……..(ii)

And, IijJik

= _M × M

ikl. m. g_ = 3gK ……..(iii)

Plugging value of g_ from (iii) in (ii), we getgM = 0.5gK. Plugging value of gM and g_ in (i), we get this in 4.5gK = 18 minute’s i.e. gK = 4 minutes. Thus, gM = 2 minutes i.e. 1/30 hour. Hence speed over second stretch = M

K/_p = 60 kmph. 254.254.254.254. If |x| + 2|y| = 1, then line of this curve represent ?

a. sides of a triangle b. sides of a rhombus c. sides of a square d. None of the above Sol: [b]

Maths By Amiya, 3E Learning,

255.255.255.255. rZ Z(W) = mG NOP

Which one of the following statements is not correct ? a. f(x) is always positive b. f(x) > g(x) for all values of x c. g(x) is always positive d. f(x) and g(x) curves never intersect

Sol: (c); log vales are not always positive

256.256.256.256. An isosceles triangle

ordinate axes, then a. |a| = |b| Sol: (a)Sol: (a)Sol: (a)Sol: (a)

257.257.257.257. There are how many such Natural number(s) N have

1,000,063 mod N = 63? Remainder when 13 divided by 5]

Maths By Amiya, 3E Learning, www.facebook.com/MathsByAmiya©AMIYA KUMAR

u(W) = vwuxW Which one of the following statements is not correct ?

f(x) is always positive f(x) > g(x) for all values of x g(x) is always positive f(x) and g(x) curves never intersect

log vales are not always positive

is formed by the straight line ax + by + c = 0 and the

b. |a| = |c| c. |b| = |c| d. None of the

There are how many such Natural number(s) N have the property that 63? [Where mod is a remainder function, 13 mod 5 = 3, =

when 13 divided by 5]

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he straight line ax + by + c = 0 and the co-

None of these

the property that [Where mod is a remainder function, 13 mod 5 = 3, =


a. 29 b. 37 c. 39 d. None of These Sol: (b) 37 Factors of 1,000,000 greater than 63.

258.258.258.258. If yR L _zy J K is an integer then Max(n) =?

a. 36 b. 9 c. 71 d. None of these Sol: (a) 36 We will get remainder -37 when (n^2 - 38) divided by (n + 1). But we want remainder should be zero, so -37 sould be divisible by (n+1) for Max(n) ; n+1 = 37 => Max(n) = 36 ; Hence (a)

259.259.259.259. For which value of “c” , WM – 99 ∗ W + d = 0 has both roots as primes. a. 97 b. 194 c. 332 d. 664 Sol: (b) 194 Let roots be a and b , so a + b = 99, since summation of two numbers is odd, so one should be even and according to condition prime so roots are 2 and 83. So, c = product of roots = 2*97 = 194 ; Hence (b)

260.260.260.260. How many real solutions does the below equation have ?

~��L��

��L��= �

a. 0 b. 1 c. 2 d. 3

Sol: (d) 3 If Nb = 1 then either N = ±1 (lO dN�m wZ − 1 � = m�mO), w� � = 0 (NOP N ≠ 0)

So, given equation ~��L��

��L��= �

Then CASE I: MGRL�

_ = 1 ⇒ W = ±2


CASE II: If MGRL�_ = −1 ⇒ W = ±1 but in this case WM − 2W is odd, so LHS of given

expression is -1 which is not RHS, so no solution from this case. & CASE III: If WM − 2W = 0 ⇒ W = 0 w� 2 From case I and III , W = ±2 w� 0, Three solutions.

Maths By Amiya DI+LR Question 261 -280 : Check Link :-

http://goo.gl/Tp7Mzc & http://goo.gl/mhoJXR

281.281.281.281. What is the value of x+y = ?

Statement I : sin (x+2y) = 1 Statement II : cos (x-y) = 1 a. Question can be solved by using either statement alone b. Question can be solved by using only one statement but not by other c. Question can be solved by using both the statements d. Question cannot be solved

Sol: [d] Question canot be answered, since from any statement we will get infinite many solutions in since range of "x" and "y" is not given. so we can not get unique solution. 282.282.282.282. For which value of k the family of equation is inconsistent or has no solution

K x + 2 y = K+2 8 x + K y = 2*K + 4 a. 4 b. 8 c. ±4 d. None of These Sol: [d] For no solution NKNM

= �K�M

≠ dKdM

From : �j�R

= bjbR

⇒�z = M

� ⇒ �M = 16 ⇒ � = ±4 For � = 4 , �j

�R= bj

bR= K

M & �j�R

= �JMM∗�JI = K

M which contradict the no solution case, so ±4 is not the value of K for no solution case.


For � = − 4 , �j�R

= bjbR

= − KM & �j

�R= �JM

M∗�JI = KM which supports the no solution

case, so -4 is the value of K for no solution case.

Directions for questions Directions for questions Directions for questions Directions for questions 283283283283 to to to to 286286286286:::: Read the passage and answer the questions that follow. Passage is taken from : The Drunkard's Walk by Leonard Mlodinow Looking to the sky on a clear, moonless night, the human eye can detect thousands of twinkling sources of light. Nestled among those haphazardly scattered stars are patterns. A lion here, a dipper there. The ability to detect patterns can be both strength and a weakness. Isaac Newton pondered the patterns of falling objects and created a law of universal gravitation. Others have noted a spike in their athletic performance when they are wearing dirty socks and thenceforth have refused to wear clean ones. Among all the patterns of nature, how do we distinguish the meaningful ones? Drawing that distinction is an inherently practical enterprise. And so it might not astonish you to learn that, unlike geometry, which arose as a set of axioms, proofs, and theorems created by a culture of ponderous philosophers, the theory of randomness sprang from minds focused on spells and gambling, figures we might sooner imagine with dice or a potion in hand than a book or a scroll. The theory of randomness is fundamentally a codification of common sense. But it is also a field of subtlety, a field in which great experts have been famously wrong and expert gamblers infamously correct. What it takes to understand randomness and overcome our misconceptions is both experience and a lot of careful thinking. And so we begin our tour with some of the basic laws of probability and the challenges involved in uncovering, understanding, and applying them. One of the classic explorations of people’s intuition about those laws was an experiment conducted by the pair who did so much to elucidate our misconceptions, Daniel Kahneman and Amos Tversky. Feel free to take part-and learn something about your own probabilistic intuition. Imagine a woman named Linda, thirty-one years old, single, outspoken, and very bright. In college she majored in philosophy. While a student she was deeply concerned with discrimination and social justice and participated in antinuclear demonstrations. Tversky and Kahneman presented this description to a group of eighty-eight subjects and asked them to rank the following statements on a scale of 1 to 8 according to their probability, with 1 representing the most probable and 8 the least. Here are the results, in order from most to least probable: Statement Average Probability Rank Linda is active in the feminist movement. ` 2.1 Linda is a psychiatric social worker. 3.1


Linda works in a bookstore and takes yoga classes. 3.3 Linda is a bank teller and is active in the feminist movement. 4.1 Linda is a teacher in an elementary school. 5.2 Linda is a member of the League of Women Voters. 5.4 Linda is a bank teller. 6.2 Linda is an insurance salesperson. 6.4 At first glance there may appear to be nothing unusual in these results: the description was in fact designed to be representative of an active feminist and unrepresentative of a bank teller or an insurance salesperson. But now let’s focus on just three of the possibilities and their average ranks, listed below in order from most to least probable. This is the order in which 85 percent of the respondents ranked the three possibilities: Statement Average Probability Rank Linda is active in the feminist movement. 2.1 Linda is a bank teller and is active in the feminist movement 4.1 Linda is a bank teller 6.2 If nothing about this looks strange, then Kahneman and Tversky have fooled you, for if the chance that Linda is a bank teller and is active in the feminist movement were greater than the chance that Linda is a bank teller, there would be a violation of our first law of probability, which is one of the most basic of all: The probability that two events will both occur can never be greater than the probability that each will occur individually. Why not? Simple arithmetic: the chances that event A will occur = the chances that events A and B will occur + the chance that event A will occur and event B will not occur. 283.283.283.283. It can be inferred from the passage that a. Great experts have been famously wrong in the area of randomness. b. Gamblers are far more likely to gamble than to read. c. Gamblers are far more likely to read than to gamble d. Gamblers have been mostly correct in the area of randomness. 284.284.284.284. With which of the following ranks is the author most likely to agree? a. Statement Average Probability Rank Linda is active in the feminist movement. 2.1 Linda is a bank teller and is active in the feminist movement. 4.1 Linda is a bank teller. 6.2 b. Statement Average Probability Rank


Linda is active in the feminist movement. 2.1 Linda is a bank teller and is active in the feminist movement 6.2 Linda is a bank teller. 4.1 c. Statement Average Probability Rank Linda is active in the feminist movement. 6.2 Linda is a bank teller and is active in the feminist movement. 4.1 Linda is a bank teller 2.1 d. Both b and c 285.285.285.285. Which of the following words can most appropriately replace the word

‘confiscation’ as used in the first sentence of the 2nd paragraph? a. systematization b. confusion c. organization d. arrangement 286.286.286.286. As per the passage we can most reasonably infer that Daniel Kahneman and

Amos Tversky are: a. Mathematicians b. Economists c. Psychologists d. Neurologists

283.b283.b283.b283.b Question Type: Inference By now you would know that the answer to a n inferential question can’t be directly stated in the passage. It has to be deduced from the stated information. The end of the 1st para says, “And so it might…. The theory of randomness sprang from minds focused on spells and gambling, figures we might sooner imagine with dice or a potion in hand than a book or a scroll.” If we don’t imagine a gambler with a book that implies that he/she is less likely to read than to gamble. So (2) is correct. (a) is directly stated at the beginning of 2nd para): “a field in which great experts have been famously wrong and expert gamblers infamously correct”. The same sentence makes (d) incorrect. Some instances do not imply most instances. (c) is the opposite of (b) and is therefore wrong. 284.b284.b284.b284.b Question Type: Further Application.


Though this question might seem tough, it is actually quite simple. You just have to concentrate on the last para which says, “The probability that two events will both occur can never be greater than the probability that each will occur individually”. Essentially it means that the middle values of the Average Probability Rank should be more than both the 1st and the 3rd value each (thus making it less probable than each one of those individually). Remember from the end of the 3rd para that 1 represents most probable and 8 the least. So (b) is the answer. (a) is obviously wrong because it’s the original about which the author starts the last paragraph in these words: “If nothing about this looks strange, then Kahneman and Tversky have fooled you….” . (c) is wrong because it’s the same as the 1st ; the only difference is that instead of the 2nd value being more than the 3rd one, in this case it’s more than the 1st one – thereby still violating the aforementioned law. 285285285285.a.a.a.a Question Type: Vocab-in-context (b) is the opposite of the intended meaning. Also ‘codification’ cannot mean ‘confusion’ in any context. (a), (c) and (d) can all fit, but remember we have to choose the most appropriate. ‘Systematization’ is the most specific and fitting word in this context. It means ‘arrange according to an organized system’ – so it includes the meaning of (c) and (d) and makes them more specific. What you need to attend to here is that it is the presence of ‘systematization’ that has made the other two wrong, each of which alone would be fine (sort of). 286286286286.c.c.c.c Question Type: Inference/Application Daniel Kahneman and Amos Tversky have been first mentioned in the 2nd paragraph. The biggest clue here is the word ‘intuition’, reading which you need to ask – people in what area would research on ‘intuition’? That would immediately eliminate (a) and (b). Even though the passage deals with probability, making you think of ‘mathematicians’, think that then the passage would be about ‘probability’ – not how/why people get it wrong/right.


Directions for questions 287Directions for questions 287Directions for questions 287Directions for questions 287 to to to to 290290290290:::: The following table gives the percentage break-up of the total marks scored by 8 CAT aspirants across the three sections, QA, DI and EU. The students are identified with their initials as given in the first column.

Present Distribution of Total marksPresent Distribution of Total marksPresent Distribution of Total marksPresent Distribution of Total marks

InitialsInitialsInitialsInitials QAQAQAQA DIDIDIDI EUEUEUEU

VPVPVPVP 16 46 38

RBRBRBRB 28 50 22

PMPMPMPM 42 14 44

SJSJSJSJ 30 40 30

NSNSNSNS 36 24 40

HAHAHAHA 30 55 15

No two test takers have scored the same total marks and nor have they scored equal marks in any of the sections. Each test taker is ranked in each of the sections with the person getting the highest marks in that section being ranked 1, person scoring the second highest marks in the section being ranked 2 and so on. Rank 1 is the best possible rank and rank 6 is the worst possible rank. 287.287.287.287. If SJ has scored the highest marks in QA, then the worst possible rank of SJ in DI

could be: a. 2 b. 3 c. 4 d. 5 e. None of These 288.288.288.288. If marks scored by NS in DI is same as the marks scored by VP in EU and also

same as the marks scored by RB in QA, arrange them in the decreasing order of their rank in DI (Best rank being placed first).

a. NS, VP, RB b. VP, RB, NS c. RB, NS, VP d. RB, VP, NS e. None of These 289.289.289.289. With which of the following data can you accurately determine who has scored

the highest total marks? a. RB has scored the maximum marks in DI


b. HA has scored the highest in EU c. VP has scored the lowest in QA d. PM has scored the lowest in DI e. None of These 290.290.290.290. If the ranking based on marks scored in QA, in decreasing order (best rank is

placed first) is, PM, VP, RB, HA, NS, SJ, which of the following definitely CANNOT be a decreasing order of ranking based on total marks scored?

a. VP, RB, PM, NS, SJ, HA b. VP, PM, RB, HA, SJ, NS c. VP, RB, HA, PM, SJ, NS d. VP, RB, PM, HA, SJ, NS e. None of These Sol 287 cSol 287 cSol 287 cSol 287 c Short-cut: SJ has scored the highest marks in QA among all the six students. Also SJ has scored more marks in DI than that he scored in QA. Thus, he would have scored more marks in DI as compared to all the students who scored lesser marks in DI as compared to their respective scores in QA. There are two such students: PM and NS. Thus, the worst possible rank of SJ in DI is 4. Mathematically: 30% of SJ > 16% of VP, 28% of RB, 42% of PM, 36% of NS, 30% of HA. Multiplying both sides by 4/3, we have 40% of SJ > 21.33% of VP, 37.33% of RB, 56% of PM, 48% of NS, 40% of HA. Thus, in DI, we are absolutely certain that SJ has scored more than PM and NS. As compared to the others, we cannot be certain who has scored higher marks in DI. Thus, the worst possible score of SJ in DI would be 4th. Sol 288. dSol 288. dSol 288. dSol 288. d If k is the marks scored by NS in DI, VP in EU and RB in QA, then finding their marks in DI in terms of k … NS in DI = k marks

VP in DI = Ih_z � = M_

K� �

RB in DI =�pMz � = M�

KI �

Obviously their ranking in DI will be RB, VP, NS. Sol 289.bSol 289.bSol 289.bSol 289.b


If a student has scored in m marks and his percentage – break up in that section is p%, then his total marks will be �� × 100. We can come to a firm conclusion about the 1st rank if a student has scored the highest marks in a section and if his percentage-break-up in that section is the lowest as compared to other students. This is the case only with option (2). Sol 290.aSol 290.aSol 290.aSol 290.a Shortcut: If total marks scored by a student are represented by his initials, 42% of PM > 16% of VP > 28% of RB > 30% of HA > 36% of NS > 30% of SJ If a student has a higher percentage break-up in QA and yet appears lower in ranking in QA compared to any other student, then he would have a lower total score as compared to the other student. Thus, looking at students pair-wise and also keeping an eye on the options….. VP > RB, which is true for all options RB > HA, which is also true for all options HA > NS, which is not true of option (5) We need not compare for all pair of students, the above is some randomly taken pairs to identify which option is not possible. Mathematically: Assuming that all of them scored ‘the same’ marks QA (it can’t be possible but will make our working and understanding very easy), say k marks, then their totals marks can be found in terms of k. The order below is in the ranking order given for QA i.e. to say that the value of k in an expression that appears before another has to be atleast marginally higher (it can also be very very high but HAS to be atleast marginally higher) than the value of k in an expression that appears below …

PM = �IM × 100 = �p

MK � = 2.3�, �� = �Kh × 100 = M�

I � = 6.25�

[ = �Mz × 100 = M�

^ � = 3.57�, ¡\ = �_p × 100 = Kp

_ � = 3.33�

a¢ = �_h × 100 = M�

� � = 2.77�, ¢£ = �_p × 100 = 3.33�

Thus, the order of ranking based on total marks NECESARRILY has to have VP > RB > HA > NS


291.291.291.291.

Ans : 1000 Directions for questions 292 and 293:Directions for questions 292 and 293:Directions for questions 292 and 293:Directions for questions 292 and 293: Step 1: Input a natural number N. Step 2: Find P, the product of the digits of N. Step 3: Is P < 10? Yes: Output P and Stop. No: N = P. Go to Step 2 292.292.292.292. N, the number that is inputted in step 1 is a four digit natural number such that

all its digit are same. For how many different values of N do we get the final output as 0?

a. 9 293.293.293.293. For how many two digit numbers that are inputted as N in Step 1, will the

algorithm run only once i.e. step 2 is performed only once? a. 29 292.d292.d292.d292.d One just need to check for 1111, 2222, 3333… 9999.1111: 1 × 1 × 1 × 1 = 1 2222: 2 × 2 × 2 × 2 = 16: 1


Directions for questions 292 and 293:Directions for questions 292 and 293:Directions for questions 292 and 293:Directions for questions 292 and 293: Given below is an algorithm. Step 1: Input a natural number N. Step 2: Find P, the product of the digits of N.

Yes: Output P and Stop. to Step 2

N, the number that is inputted in step 1 is a four digit natural number such that all its digit are same. For how many different values of N do we get the final

b. 8 c. 7

For how many two digit numbers that are inputted as N in Step 1, will the algorithm run only once i.e. step 2 is performed only once?

b. 30 c. 31

One just need to check for 1111, 2222, 3333… 9999.

× 6 = 6


N, the number that is inputted in step 1 is a four digit natural number such that all its digit are same. For how many different values of N do we get the final

d. 6

For how many two digit numbers that are inputted as N in Step 1, will the

d. 32


3333: 3I = 81 ; 1 × 8 = 8 4444: 4I = 256; 2 × 5 × 6 = 60; 6 × 0 = 0 5555: 5I = 625; 6 × 2 × 5 = 60; 6 × 0 = 0 6666: 6I = 1296; 1 × 2 × 9 × 6 = 108; 1 × 0 × 8 = 0 7777: 7I = 2401; 2 × 4 × 0 × 1 = 0 8888: 8I = 2KM = 4096; 4 × 0 × 9 × 6 = 0 9999: 9I = 6561; 6 × 5 × 6 × 1 = 180; 1 × 8 × 0 = 0 Thus for 4444, 5555, 6666, 7777, 8888, 9999 the final output is 0. 293.d293.d293.d293.d Whenever the product of digits of a two digit number is a single digit number, the algorithm will run only once. Thus is possible when…. If unit digit is zero, any of 1, 2, 3… 9 can be the ten’s digit i.e.9 such numbers (the numbers are 10, 20, 30… 90) If unit digit is 1, any of 1, 2, 3… 9 can be the ten’s digit i.e. 9 such numbers ( the numbers are 11, 21, 32, …, 91) If unit digit is 2, any of 1, 2, 3, or 4 can be the ten’s digit i.e. 4 such numbers (the numbers are 12, 22, 32 and 42) If unit digit is 3, any of 1 or 2 can be the ten’s digit i.e. 3 such numbers (the numbers are 13, 23, and 33) If unit digit is 4, any of 1 or 2 can be the ten’s digit i.e. 2 such numbers (the numbers are 14 and 24) If unit digit is any of 5, 6, 7, 8 or 9 , the ten’s digit can only be 1 i.e. 5 such numbers (the numbers are 15, 16, 17, 18, 19) Total possible numbers are 9 + 9 + 4 + 3 + 2 + 5 = 32 294. I is a point inside the triangle ABC, the bisectors of angle BIC, angle AIC and angle

AIB meet the sides BC, CA and AB in points E,F, and D respectively. The lengths

(in cm) of AD, DB AF, FC and BE are 3, 5, 4, 4, 6 respectively. What is the length of

CE?

a. 2 cm b. 3.6 cm c. 4.4 cm d. None of these

SolSolSolSol.... bbbb


The angle bisector divides the opposite side in ratio of adjacent sides.Thus, F¥

¥H = _� NOP F¥

¥Q = KK

Assuming AI = 3k, we get IB = 5k and IC = 3k.Further, H¥

¥Q = hG ⟹ ��

_� = hG ⟹

295.295.295.295. W_ + 4WM − 3W + � is the product of 3 factors, 2 of which are identical. How

many values can k have? a. 0 Sol. cSol. cSol. cSol. c If the factors are (W − N)M and (x (WM − 2NW + NM) × (W − �) =Equating the constant term we values that this can assume. Equating coefficient of WM on the two sides, we get Equating coefficient of x on the two sides, we get 2ab +Substituting the value of b from (i) in (ii), we get, simplifying this, we get−3NM

need to find the actual values) and according will get two respective values of b. Thus, k could assume 2 distinct values. 296.296.296.296. A and B are playing a game in which they have a empty box and a single chip of

each number from 1 to 1000. Each player, taking turns alternately, has to put a composite numbered chip in the box such that it is coalready in the box. When a player is unable to put in such a number, he loses. What is the maximum number of chips that can be in the box at the end of the game?

a. 10 Sol. bSol. bSol. bSol. b Since the numbers in the box are to be cocommon factor e.g. if 6 is put in the box, no number that is a multiple of 2 or 3 can be


des the opposite side in ratio of adjacent sides.

Assuming AI = 3k, we get IB = 5k and IC = 3k. ⟹ W = Kz

� = 3.6

is the product of 3 factors, 2 of which are identical. How many values can k have?

b. Only 1 c. Only 2

and (x – b), then we have ) = W_ + 4WM − 3W + �

Equating the constant term we find that k = −NM�. And we have to find the number of

on the two sides, we get –b – 2a = 4 i.e. 2a + b = Equating coefficient of x on the two sides, we get 2ab +NM= - 3 …….(ii)

uting the value of b from (i) in (ii), we get, 2N(−4 − 2N) + NM =M − 8N + 3 = 0. Solving, we get two distinct values of a (no

need to find the actual values) and according will get two respective values of b. Thus, k sume 2 distinct values.

A and B are playing a game in which they have a empty box and a single chip of each number from 1 to 1000. Each player, taking turns alternately, has to put a composite numbered chip in the box such that it is co-prime to every nualready in the box. When a player is unable to put in such a number, he loses. What is the maximum number of chips that can be in the box at the end of the game?

b. 11 c. 31

Since the numbers in the box are to be co-prime to each other, they should have no common factor e.g. if 6 is put in the box, no number that is a multiple of 2 or 3 can be


is the product of 3 factors, 2 of which are identical. How

d. Only 4

And we have to find the number of

2a = 4 i.e. 2a + b = - 4 …..(i)

= −3. On . Solving, we get two distinct values of a (no

need to find the actual values) and according will get two respective values of b. Thus, k

A and B are playing a game in which they have a empty box and a single chip of each number from 1 to 1000. Each player, taking turns alternately, has to put a

prime to every number already in the box. When a player is unable to put in such a number, he loses. What is the maximum number of chips that can be in the box at the end of the game?

d. 500

prime to each other, they should have no common factor e.g. if 6 is put in the box, no number that is a multiple of 2 or 3 can be


put. Hence to maximize the number of chips that can be put, numbers having a single prime number as its factor should be used. But since the number itself cannot be prime, it should be a power of a prime number, say 2², 3², 5², 7², 11², etc. We are just taking the squares because cubes would be higher values and fewer of them would be the range given. The largest square of a prime number less than 1000 is 31²=961.The prime numbers less than or equal to 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 i.e. 11 of them. Thus, maximum 11 chips can be put in the box.While we have considered only squares, please understand that instead of putting 2², if 2³, 2I, 2�, …. is put in, the scenario is identical to 2² being put in. And so with others powers of prime. Directions for 2Directions for 2Directions for 2Directions for 297979797 and 29and 29and 29and 298 8 8 8 :::: equal smaller circles have been drawn inside such that the pair of the two circles placed next to each other horizontally are tangential to each other and so are the two circles placed next to each other vertically. The four circles are also tangential to the outer circle internally.

297.297.297.297. The area of intersecting regions of four small circles (shaded by straight lines) is:

a. NM ~©M − 1�

298.298.298.298. The area of the dotted a. a²(ª − 1) Sol 297.dSol 297.dSol 297.dSol 297.d Each of the smaller circle will have a radius of a.Each of the four ‘petals’ will be inside a square of side a. The area of this square is made up of two quarter circles which o


put. Hence to maximize the number of chips that can be put, numbers having a prime number as its factor should be used. But since the number itself cannot be

prime, it should be a power of a prime number, say 2², 3², 5², 7², 11², etc. We are just taking the squares because cubes would be higher values and fewer of them would be the range given. The largest square of a prime number less than 1000 is 31²=961.The prime numbers less than or equal to 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 i.e. 11 of them. Thus, maximum 11 chips can be put in the box.

idered only squares, please understand that instead of putting 2², if is put in, the scenario is identical to 2² being put in. And so with others

In the diagram, the outer circle has radius ‘2a’ cm. Four equal smaller circles have been drawn inside such that the pair of the two circles placed next to each other horizontally are tangential to each other and so are the two circles

h other vertically. The four circles are also tangential to the outer

The area of intersecting regions of four small circles (shaded by straight lines) is:

b. a²(ª − 1) c. a²(2ª − 1)

The area of the dotted regions is: b. 2a²(ª − 1) c. a²(ª − 2)

Each of the smaller circle will have a radius of a. Each of the four ‘petals’ will be inside a square of side a. The area of this square is made up of two quarter circles which overlap to form the petal.


put. Hence to maximize the number of chips that can be put, numbers having a prime number as its factor should be used. But since the number itself cannot be

prime, it should be a power of a prime number, say 2², 3², 5², 7², 11², etc. We are just taking the squares because cubes would be higher values and fewer of them would be in the range given. The largest square of a prime number less than 1000 is 31²=961. The prime numbers less than or equal to 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31

idered only squares, please understand that instead of putting 2², if is put in, the scenario is identical to 2² being put in. And so with others

In the diagram, the outer circle has radius ‘2a’ cm. Four equal smaller circles have been drawn inside such that the pair of the two circles placed next to each other horizontally are tangential to each other and so are the two circles

h other vertically. The four circles are also tangential to the outer

The area of intersecting regions of four small circles (shaded by straight lines) is:

d. a²(2ª − 4)

d. 2a²(ª − 2)

Each of the four ‘petals’ will be inside a square of side a. The area of this square is made


Thus, if area of which petal is x, then©�R

I + ©�R

I − W = NM ⟹ W = NM ~©M − 1�.

Thus, the total shaded area will be 4 × NM ~©M − 1� = 2NM(ª − 2)

Sol 298.dSol 298.dSol 298.dSol 298.d When all the four smaller circles are added. EACH petal will be added twice (because each petal is part of two circles). Thus the area of the four smaller circles without adding the overlapped part twice will be 4 × ªNM − 2NM(ª − 2) = 2ªNM + 4NM. The area of the four petals is found in the earlier question) The dotted region can be found by subtracting the above from the larger circle as ª(2N)M − (2ªNM + 4NM) = 2ªNM − 4NM

299.299.299.299. If Z(W) = √GLMJ√GL√M√GRLI

then f(2.000000000001) approx equal to ? a. 0 b. 1/2 c. 1/4 d. None of These

Sol: [b] 1/2 Just simplify numerator and denominator. 300.300.300.300. The LCM and HCF of two numbers a and b(a < b) is such that LCM = 144 × HCF.

If the difference between a and b is the minimum possible and a has 7 factors, how many factors does b have?

a. 5 b. 7 c. 15 d. 25 Sol. dSol. dSol. dSol. d Let the HCF be h. Thus the numbers can be assumed to be N = ℎ × and � = ℎ × O where m and n are co-prime. Further m and n should be as close to each other so that the difference between the numbers is minimum Since m and n are co-prime, the LCM of ℎ × and ℎ × O will be ℎ × × O which is given to be equal to 2I × 3M × ℎ. Thus, × O = 144 w� 2I × 3M and we need two separate them into two co-prime numbers that are as close as possible. This is when m and n are 9 and 16 respectively (a is less than b). Thus N = 3M × ℎ and has 7 factors. Since the number of factors is odd, the only way this is possible is when a is the 6th power of a prime number i.e. when ℎ = 3I.


Hence � = 2I × 3I and would have 25 factors. Direction Direction Direction Direction for 301for 301for 301for 301----302:302:302:302: If ABC is a circumscribed triangle such that measurements of lengths AF, FC,CE,EB,BD and DA are prime numbers and l(AC)=12cm and l(BC)=9cmI in incentre of the inscribed circle

301.301.301.301. Area of △ABC = ? a. 14 cm^2

302.302.302.302. Inradius of △ABC = ? a. 7/8 cm

Sol: 301 [b]Sol: 301 [b]Sol: 301 [b]Sol: 301 [b],,,, 302 [c302 [c302 [c302 [c]]]]

Since small line segments are prime and AC=12 has two segments 5 and 7, Since tangents from same points are equal in length so, AF = AD = 5 cm, FC=CE=7cm & EB=BD= 2 cm.SO sides are AC = 12cm , BC=9cm & AB= 7 cm , s(semiBy Heron's Formula area (△) = & In-radius (r) = △/s = KI√�

KI 303.303.303.303. Let S be a set of all natural numbers less than 100 that are the

natural number, n > 1. Let elements, each element being an odd number. How many distinct sets

a. 11 Sol .c


and would have 25 factors.

If ABC is a circumscribed triangle such that measurements of lengths AF, FC,CE,EB,BD and DA are prime numbers and l(AC)=12cm and l(BC)=9cmI in incentre of the inscribed circle, then

b. 14√5 cm^2 c. 16√5 cm^2

△ABC = ? b. ^z √5 cm^2 c. √5 cm^2

Since small line segments are prime and BC = 9 cm so it has segments of 7 or 2 & AC=12 has two segments 5 and 7,

e tangents from same points are equal in length so, AF = AD = 5 cm, FC=CE=7cm & EB=BD= 2 cm. SO sides are AC = 12cm , BC=9cm & AB= 7 cm , s(semi-perimeter)=14 cm

△) = 14√5 cm^2

� cm= √5 d

Let S be a set of all natural numbers less than 100 that are the Onatural number, n > 1. Let ¢K be a subset of S such that it has an odd number of elements, each element being an odd number. How many distinct sets

b. 20 c. 32


If ABC is a circumscribed triangle such that measurements of lengths AF, FC,CE,EB,BD and DA are prime numbers and l(AC)=12cm and l(BC)=9cm &

d. None of these

d. None of these

BC = 9 cm so it has segments of 7 or 2 &

perimeter)=14 cm

Oi¯ power of some be a subset of S such that it has an odd number of

elements, each element being an odd number. How many distinct sets ¢K are there? d. 16


Since all elements in ¢K are odd, let’s just focus on the odd elements of S. These will be any powers of odd numbers but have to be less than 100. Thus the set S will have the following odd numbers: 1², 3², 3³, 3I, 5M NOP 7M Now ¢K could have 1 or 3 or all 5 of these odd elements. Thus, ¢K can be formed in 6]K + 6 ]_ +6]�= 6 + 20 + 6 = 32 ways. 304.304.304.304. An Arithmetic Progression of 3 terms is such that each term is a perfect square.

If d is the common difference, which of the following can be a value of d? a. 45 b. 63 c. 96 d. 198 Sol: [c]Sol: [c]Sol: [c]Sol: [c] Let the three terms in AP be NM, �MNOP dM. Thus, dM − NM = 2P, where d is a natural number. In number systems we have seen that only odd numbers or multiples of 4 can be written as difference of squares. If you do not remember, then read on …. (c – a) = p and (c + a) = q will result in natural values of c and a only when both p and q are odd or when both are even. Since (d − N) × (d + N) = 2P i.e. an even number, both of them cannot be odd. Thus, both of them have to be even and 2d has to be a multiple of 4 i.e. d has to be a multiple of 2. Thus, among the given options, d = 45, d = 63 are ruled out. Only possible values from the options are d = 96 or d = 198. Also�M − NM = P. By the same logic as explained earlier d has to be odd or a multiple of 4. But the odd possible values are ruled out earlier. And d = 198 is not a multiple of 4. Thus, there would be no natural number solution to �M − NM = 198. Hence from the given choices only d = 96 can be a possibility. The question is solved. The following is only for those interested in knowing the AP. We need natural number solutions to �M − NM=96 and dM − NM = 192 such that a is common to the solutions. Writing 96 and 192 as a product of two even numbers, the possible ways and possible values of b and a and c and a are (� − N) × (� + N) = 96 (d − N) × (d + N) = 192 2 × 28: � = 15, N = 13 2 × 96: d = 49, N = 47 4 × 24: � = 14, N = 10 4 × 48: d = 26, N = 42 6 × 16: � = 11, N = 5 6 × 32: d = 19, N = 13 8 × 12: � = 10, N = 2 8 × 24: d = 16, N = 8 12 × 16: d = 14, N = 2


The only value of a that is common to the two sets is 2. Thus, a = 2, b = 10, c = 14 and the AP will be 4, 100, 196. Directions for questions 305 to 307:Directions for questions 305 to 307:Directions for questions 305 to 307:Directions for questions 305 to 307: In each questions, the word at the top is used in four different ways, numbered a to d. Choose the option in which the usage of the word is incorrect or inappropriate. 305.305.305.305. Kiss a. Good rating gave the program the kiss of life. b. It would be the kiss of death for the company if it could be proved that the food was unsafe. c. He can kiss off that promotion. d. I could kiss up my career good-bye. 306.306.306.306. Face a. He’ll have to face the music, sooner or later. b. He asked her to shut up and get out of his face. c. She slammed the door on my face. d. The code of conduct required that he strike back or lose face. 307.307.307.307. Mark a. I’d have bought it had he marked it down a bit. b. They mark up the price of imported wines by 66 percent. c. He had been a man of mark. d. Johnny’s not feeling to the mark at the moment. Sol 305Sol 305Sol 305Sol 305.d.d.d.d The correct form should be: I could kiss my career good-bye, not kiss up.up.up.up. To kisskisskisskiss something goodbyegoodbyegoodbyegoodbye means to accept the certain loss of something. Kiss upKiss upKiss upKiss up means to behave obsequiously or sycophantically toward (someone) in order to obtain something: e.g. She’s always kissing up to the senior management. Kiss of lifeKiss of lifeKiss of lifeKiss of life means an action or event that revives a failing enterprise. Kiss of deathKiss of deathKiss of deathKiss of death is an action or event that causes certain failure for an enterprise. Kiss offKiss offKiss offKiss off means to give up or regard as lost. Sol 306Sol 306Sol 306Sol 306.... cccc The correct form should be: She slammed the door in my face, not onononon my face.


In one’s faceIn one’s faceIn one’s faceIn one’s face means directly at or against one. Face the musicFace the musicFace the musicFace the music means to be confronted with the unpleasant consequences of one’s actions. Get out of someone’sGet out of someone’sGet out of someone’sGet out of someone’s facefacefaceface means to stop harassing or annoying someone. To lose facelose facelose facelose face means to suffer a loss of respect. Sol 307Sol 307Sol 307Sol 307.... dddd The correct form should be: Johnny’s not feeling up to the mark at the moment, not to the mark. Mark somethingMark somethingMark somethingMark something down is to reduce the indicated price of an item. Mark upMark upMark upMark up means to mark for sale at a higher price. Of markOf markOf markOf mark means having importance and distinction. 308.308.308.308. If W� + 1 is a factor of W�b + 1, then what is the summation of all possible values

of b if range of b : 0<b<100 ? a. 100 b. 1000 c. 2500 d. None of these Sol: [c] Plugging W� = °, we have to question as “is y + 1 a factor of °b + 1?” Substituting y = 1 in °b + 1, the remainder is (−1)b + 1 which will be zero only if b is odd. So b is odd between 1-99, and summation of all odds starting from 1 = OM Here there are 50 consecutive odd starting from 1 , so summation of all = 50M = 2500 309.309.309.309. The average marks of 17 students in a class is A. The marks of students when

arranged in ascending or descending order were found to be in A.P. The class teacher found that the students who were ranked 4, 6, 9, 11, 15 had copied in the exam and hence got all of them rusticated. Rank 1 has maximum marks and so on. The average of the remainder in the class was B. Then

a. A = B b. A > B c. A < B d. Data Insufficient Sol.aSol.aSol.aSol.a Let the descending order of marks (ranks being 1, 2, 3, …) be a, a – d, a – 2d, a – 3d, …


Thus average will be the marks of the 9th ranked student i.e. a – 8d. Now the sum of marks of the 4th, 6th, 9th, 11th and 15th ranked students will be (a – 3d) + (a – 5d) + (a – 8d) + (a – 10d) + (a – 14d) = 5a – 40d. The average of these 5 students is a – 8d i.e. same as the average of the entire class. Thus, even after their leaving, the average of the remainder of the class will remain the same i.e. B = A. Directions for questions Directions for questions Directions for questions Directions for questions 333310 to 10 to 10 to 10 to 333312121212:::: When Mr and Mrs Thadani gave a dinner-party, they invited Mr and Mrs Joshi, Mr and Mrs Goyal, Mr and Mrs Patel and Mr and Mrs Shroff. The ten people were seated at a circular table, and no man sat next to his own wife. Mr Thadani sat next but one to Mrs Shroff. Mr Goyal sat between two ladies, and so did Mr Patel. Mrs Joshi sat next to her sister, while Mr Joshi sat next but one to Mr Shroff, who was immediately to the left of his father-in-law. Mrs Goyal sat next but two to her husband, and Mrs Joshi sat next but two to Mrs Patel. Three of the ladies each sat between two men. 310.310.310.310. Which of the following could exchange places and still no man would sit next to

his own wife? a. Mr. Patel & Mr. Thadani b. Mrs Goyal & Mrs Shroff c. Mrs Patel & Mrs Joshi d. Mrs Patel & Mr Goyal 311.311.311.311. How many men sat between two ladies i.e. had ladies as both their neighbours? a. 1 b. 2 c. 3 d. 4 312.312.312.312. Who is Mr. Shroff’s father-in-law? a. Mr. Goyal b. Mr. Thadani c. Mr. Patel d. Mr. Joshi For For For For 333310 to 10 to 10 to 10 to 333312:12:12:12: Point 1: Since two ladies are sitting together and each of the other three are sitting between two men, it necessarily has to be the case that two men are also sitting together and each of the other three men are sitting between two women. Further Mr. Shroff is sitting immediately left to his father in law and thus, Mr. Joshi should be the third man who is sitting between two ladies (other two being explicitly given as Mr. Goyal and Mr. Patel. This also concludes that the 5th man i.e. Mr. Thadani is Mr. Shroff’s father in law. Point 2: Forming smaller units of being seated (a man is denoted by a box and a woman by a circle)…. S T S


Left RightThe person between Mr. Thadani and Mrs Shroff has to be a woman (else three men would become together) and of tJoshi. Thus…

(In the above figure, left is antithe people sitting facing inwards, all that matters is that you be consistent in your approach to the direction. If you still are confused that logically left should clockjust draw the mirror-image of the following figures and see that you arrive at the same answer) Point 3: ‘Mrs Goyal sat next but two to her husband’ couple. Of which one position (Mrs. Goyal between Mr. Shroff & Mr. Joshi) is not possible because then Mr. & Mrs. Patel have to sit together. Thus, the only possible arrangement is

310.d; 310.d; 310.d; 310.d; 311.c; 311.c; 311.c; 311.c; 313.313.313.313. N is the total number of 3 d

M is the total number of 3 digit numbers whose sum of the digits is divisible by 5. What is N – M?

a. 60


Left Right The person between Mr. Thadani and Mrs Shroff has to be a woman (else three men would become together) and of the only two ladies sitting together, one has to be Mrs.

(In the above figure, left is anti-clockwise. It does not matter whether it is in sync with the people sitting facing inwards, all that matters is that you be consistent in your

ach to the direction. If you still are confused that logically left should clockimage of the following figures and see that you arrive at the same

Point 3: ‘Mrs Goyal sat next but two to her husband’ – defines only two couple. Of which one position (Mrs. Goyal between Mr. Shroff & Mr. Joshi) is not possible because then Mr. & Mrs. Patel have to sit together. Thus, the only possible

312.b312.b312.b312.b

number of 3 digit numbers whose sum of the digits is divisible by 3. number of 3 digit numbers whose sum of the digits is divisible by 5.

b. 80 c. 120


The person between Mr. Thadani and Mrs Shroff has to be a woman (else three men he only two ladies sitting together, one has to be Mrs.

clockwise. It does not matter whether it is in sync with the people sitting facing inwards, all that matters is that you be consistent in your

ach to the direction. If you still are confused that logically left should clock-wise, image of the following figures and see that you arrive at the same

defines only two positions for the couple. Of which one position (Mrs. Goyal between Mr. Shroff & Mr. Joshi) is not possible because then Mr. & Mrs. Patel have to sit together. Thus, the only possible

igit numbers whose sum of the digits is divisible by 3. number of 3 digit numbers whose sum of the digits is divisible by 5.

d. 180


Sol .cSol .cSol .cSol .c Starting with the smallest three digit number, 100 and checking, every next three digit number for their sum of digits being divisible by 3, we get the following numbers: 102, 105, 108, 111, 114, 117, 120, 123… i.e. an AP with common difference being 3. The last number in this series will be 999. These are a total of ��LKpM

_ + 1 = z�^_ + 1 = 300.

Similarly, starting with the smallest three digit number, 100 and checking every next digit number for their sum of digits being divisible by 5, we get the following numbers: 104, 109, 113, 118, 122, 127, 131, 136, 140, 145, 154, 159, 163, 168, …. Thus, in every 10 numbers, there are two such numbers. Since there a 900 three digit numbers, they can be divided into 90 non-overlapping sets of 10 numbers each and each set will have 2 numbers whose sum of digits is divisible by 5. Hence there will be 180 such numbers. N – M = 300 – 180 = 120. 314.314.314.314. What is the minimum value of 9 cotM ± + 4 tanM ±

a. 12 b. 13 c. 12.5 d. None of these Sol : [a] Let tanM ± = W , �

GR + 4WM = � by solving this you will get minimum of K=12 315. A 900 digit number is written by writing the digits 1 to 9 side by side 100 times, in

order (12345678912345…….89) but while writing, one digit is written incorrectly hence

distorting the required number. It was found that when this number was divided by 9,

remainder was 5. How many such distorted numbers are possible?

a. 1024 b. 1000 c. 900 d. 800

Sol [b] The required 900 digit number would be divisible by 9 since 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 =45 is repeated 100 times. If the remainder was 5, it means that total sum was either increased by 5 or decreased by 44. Any of the hundred 1’s or 2’s or 3’s or 4’s could be written as 6 or 7 or 8 or 9 respectively to increase the sum by 5. Each of these ways writing a number at a different position incrementing it by 5 would result in 400distinct distorted numbers.


Also any of the hundred 4’s or 5’s or 6’s or 7’s or 8’s or 9’s could be written as 0 or 1 or 2 or 3 or 4 or 5 respectively to give a deficit of 4. Hence we have another 600 distinct distorted numbers.

Directions for questions 3Directions for questions 3Directions for questions 3Directions for questions 316161616 to to to to 138 138 138 138 :::: Each of the following questions has a paragraph with one capitalized word that does not make sense. Choose the most appropriate replacement for that word from the options given below the paragraph. 316.316.316.316. Over the previous decades, Britain’s fascination with the East had mingled with a

sense of dread and fear; myths and misconceptions had competed with forms of positive or at least quasi-scientific knowledge; genuine admiration had contended with rapacious and exploitative MELFN.

a. veneration b. plaudits c. scorn d. hostility 317.317.317.317. When students are confronted with a problem, they often resort to rather

primitive ways of thinking. Sometimes well-trained students will consciously think of BENKUL problems previously solved to see if there is anything that these previous experiences can bring to the current problem. When primitive methods are used, a solution is unlikely, and if it emerges it will have taken considerably more time than an elegant solution that may result from thoughtful reasoning.

a. analogous b. unrelated c. extraneous d. superfluous 318.318.318.318. Emily Dickinson, a 19th-century poet, thought that poets, at their best, “rinsed the

language”. They made language KUTZEN by cleansing it of the accumulated crud of cliché, sloppiness and general woolliness.

a. purge b. avant-garde c. anew d. odious 316 316 316 316 .d.d.d.d There are three clauses in this paragraph and each has contrasting elements: fascination vs. dread and fear, myth and misconceptions vs. positive or quasi-scientific knowledge, genuine admiration vs.? ‘a’ and ‘b’ can be eliminated because they don’t contrast with admiration but are in fact synonymous. Between ‘c’ and‘d’ hostility is better because it goes with rapacious and exploitative. Exploitative scorn is not the correct phrase.


317317317317.... aaaa “previously solved” is the hint here. Unrelated, extraneous and superfluous are all synonymous and totally absurd in the context. How can an unrelated or external or an unnecessary problem help us solve another one? Analogous means comparable. 318318318318.... cccc Purge means to purify, cleanse. …..make the language ‘purge’ …. Here purge is used in the verb form and is not grammatically correct. Also the cleansing part is repetitive – cleanse the language by cleansing ….. Hence this option is ruled out. Avant-garde is used to denote ‘experimental’experimental’experimental’experimental’ treatment/advancement, which is not what Emily is talking about. Anew means ‘afresh’ and is the best among the given options. Odious is ‘offensive/repugnant’ and would in no way be a result of ‘rinsing’ or ‘cleansing’. 319.319.319.319. How many ways we can select more than 2 letters from word INDEPENDENCE a. 32 b. 300 c. 476 d. None of These Sol. [d] I NNN DD EEEE P C No of ways to Select No letter = 1 No of ways to Select 1 letter = 6C1 = 6 No of ways to Select 2 letter = 6C2 + 3C2 = 15+3 = 18 [All different + Both Similar] Select any number of letters = 2_ ∗ 3 ∗ 4 ∗ 5 =480 Selection of more than 2 letters = 480 - (1+6+18) = 455 320.320.320.320. Krishna has 5 roses (similar), 4 Lilies(similar) and 5 different types of flower. In

how many different ways he can give minimum one flower to Radha. Ans -959 = 6*5*2^5 - 1


321.321.321.321. 322.322.322.322. 323.323.323.323. Answer: 321- 560, 322-520, 323Check the discussion thread https://www.facebook.com/photo.php?fbid=595098523875601 Directions for questions Directions for questions Directions for questions Directions for questions 323232324 to 4 to 4 to 4 to B. Both the products have to be processed on the three machines, X, Y and Z, in any order. Further AB Ltd has 3 machines of type X, 2 machines of type Y and 4 machines of type Z. A shift consists of 8 hours.The following table gives the processing time required on each of the two products:

(Time in mins)

Product A

Product B

324.324.324.324. If either A or b but not both have to be produced in a shift, what are the

maximum number of products that can be made? a. 36 325.325.325.325. If for every A that the firm produces, it necessarily has to produce atleast two B’s,

find the maximum number of products that the firm can produce in one shift. a. 45 326.326.326.326. What are the maximum numbers of products that can be produced in one shift? a. 56


520, 323-40 will upload the short cut ASAP.

https://www.facebook.com/photo.php?fbid=595098523875601

4 to 4 to 4 to 4 to 327327327327:::: Company AB Ltd. Produces only two products A and roducts have to be processed on the three machines, X, Y and Z, in any

order. Further AB Ltd has 3 machines of type X, 2 machines of type Y and 4 machines of type Z. A shift consists of 8 hours. The following table gives the processing time required on each type of machines for

M/c X M/c Y M/c Z

15 20 30

40 10 20

If either A or b but not both have to be produced in a shift, what are the maximum number of products that can be made?

b. 46 c. 47 If for every A that the firm produces, it necessarily has to produce atleast two B’s,

find the maximum number of products that the firm can produce in one shift.b. 46 c. 47

What are the maximum numbers of products that can be produced in one shift?b. 57 c. 58


Company AB Ltd. Produces only two products A and roducts have to be processed on the three machines, X, Y and Z, in any

order. Further AB Ltd has 3 machines of type X, 2 machines of type Y and 4 machines of

each type of machines for

If either A or b but not both have to be produced in a shift, what are the

d. 48 If for every A that the firm produces, it necessarily has to produce atleast two B’s,

find the maximum number of products that the firm can produce in one shift. d. 48

What are the maximum numbers of products that can be produced in one shift? d. 59


327.327.327.327. If 20 products of A are already produced in a shift, how many products of B can be produced in the remaining time in the same shift?

a. 26 b. 27 c. 28 d. 29 Solutions: For Qs. 324 to 327:

Each shift is of 480 minutes and based on the number of machines of each type that are

available, the available time on Mc X, Y and Z is 1440, 960 and 1920 minutes in each shift.

324.d

The number of A’s that can be produced is min~KIIpK� , �hp

Mp , K�Mp_p � i.e. min (96,48, 64) i.e. 48.

The number of B’s that can be produced is min ~KIIpIp , �hp

Kp , K�MpMp � i.e. min (36, 96, 96) i.e. 36.

Thus, the maximum number of products of one variety that can be produced is 48.

325.a

Consider each ‘packet’ of one A and two B. Such a packet will require 95, 40 and 70 mins

respectively on M/c X, Y and Z. Thus, the number of such packets that can be produced is

min ~KIIp�� , zhp

Ip , K�Mp^p � i.e. min (15, 24, 27) i.e. 15.

The questions say that for every A produced. ATLEAST two B has to be produced. So we

can produce more number of B’s also if possible.

M/c X has the least amount of time left because it was the limiting constraint in deciding the

maximum number of packets of 1 A and 2 B to be produced. So we just need to check, it any

B could be produced int eh time left available on this machine. There is ample time available

on other machines.

The amount of time left on M/c X is 1440 – 15 X 95 = 1440 – 1425 = 15 minutes. In 15

minutes, we would not be able to make any B since each B requires 40 minutes on M/c X.

Thus, the maximum number of products that can be made is 15 X 3 = 45.

326.d

In Q. 41 we found that we could produce 48 products of A. This limit was imposed by M/c

Y.

For every 1 unit of A produced less, we would free 20 minutes of time on machine Y and

hence produce 2 units of B, thereby increasing the total number of products produced.

Then, is it not possible to NOT produce any of the 48 units of A and instead produce 96 units

of B? Not really, because as B is produced at the expense of A, M/C X will become the

constraining factor, because B uses far too much time on M/c X.


With 48 units of A produced, time left on M/c X is 1440 – 48 X 15 = 1440 – 720 =

720 mins.

For each 1 unit of A not produced and instead 2 units of B being produced, focusing on M/c

X, we would free up 15 minutes of time, but consume 80 minutes of time i.e. we would use

up 65 minutes of the free time. In 720 mins of free time, we could thus do ^Mph� =11 instances

of the above.

Thus, we could increase the total production by 11 i.e. 48 + 11 = 59. (This is possible with

producing 48 – 11 = 37 units of A and 22 units of B)

327.c

Solving the set so far, we would have realized that M/c X is the constraining factor to

produce B. So we could save our time by focusing only on M/c X.

Having produced 20 units of A, free time available on M/c X is 1440 – 20 X 15 = 1440 – 300

= 1140 mins.

With this available time, we can produce KKIp

Ip = 28 units of B.

328.328.328.328. Let gi¯ term of series 2, 6, 12, 20,be denoted by Ni and gi¯ terms of series 0, 6, 24, 60, 120, 210,…………..be denoted by �i What is the value of NKp + �Kp?

a. 1100 b. 900 c. 660 d. 880 Sol: [a] For the first series S10 = 2+6+12+20+..............+a10 - [S10 = 2+6+12+20+.............+a10 ] 0 = 2 + 4+6+..........................(till 10 terms) - a10 So, a10 = 110 For the first series , 0, 6, 24, 60, 120, 210,…

�y = 2(O − 1) ∗ ~y(yJK)M � = (O − 1)O(O + 1)

So, b10 = 9*10*11 = 990

¢w, NKp + �Kp = 110 + 990 = 1100

329.329.329.329. If x, y and z are two-digit natural numbers, then the following equation has how many solutions?

5WM + 20°M + 2µM = 4W° + 8°µ + 4µW a. 2 b. 15 c. 24 d. 60


Sol: [a]Sol: [a]Sol: [a]Sol: [a] 5WM + 20°M + 2µM =After solving above equation in the form of (W − 2°)M + (4° − µ)M + (2WIt is only possible if, x-2y =0 ; 4y-z = 0 ; 2x-z = 0 => x:y:z = 2:1:4Since x, y and z are double digit integers then we have only (x,y,z) = (20,10,40) to (48, 24, 330.330.330.330.

Answer [c] easy try your self 331.331.331.331. Aarti and Anamika run around a circular track starting from the same point and

in the same direction. However Aarti starts 15 sec after Anamika starts. And they meet for the first time after Aarti has run exactly half a round. If the ratio of speeds of Anamika to that of Aarti is 4:2, after how much time, from Anamika’s start, will they meet for the second time?

a. 99 sec b. 1 min Sol: [d] Let distance of the track be x m and speed of Aarti be S m/s and that of Anamika be 2S m/s.


= 4W° + 8°µ + 4µW After solving above equation in the form of (N − �)M, we will get

( W − µ)M = 0

z = 0 => x:y:z = 2:1:4 Since x, y and z are double digit integers then we have only fifteen possible values

(20,10,40) to (48, 24, 96) for z= 10 to 24

ABCD is a concave quadrilateral EA and EC are internal angle bisector of internal angle A and C of quadrilateral. α and β are internal and external angles at vertex B and D respectively. X is angle AEC then X = ?X = ?X = ?X = ? a. 2β b. (β- α)/2 c. (β+ α)/2 d. None of These

Answer [c] easy try your self

Aarti and Anamika run around a circular track starting from the same point and in the same direction. However Aarti starts 15 sec after Anamika starts. And they

et for the first time after Aarti has run exactly half a round. If the ratio of speeds of Anamika to that of Aarti is 4:2, after how much time, from Anamika’s start, will they meet for the second time?

b. 1 min c. 1.5 min d. 105 sec

Let distance of the track be x m and speed of Aarti be S m/s and that of Anamika be 2S


possible values of

ABCD is a concave quadrilateral EA and EC are internal angle bisector of internal angle A and C of quadrilateral. α and β are internal and external angles at vertex B and D

d. None of These

Aarti and Anamika run around a circular track starting from the same point and in the same direction. However Aarti starts 15 sec after Anamika starts. And they

et for the first time after Aarti has run exactly half a round. If the ratio of speeds of Anamika to that of Aarti is 4:2, after how much time, from Anamika’s start, will

Let distance of the track be x m and speed of Aarti be S m/s and that of Anamika be 2S


According to question they have started from same point and Anamika whose speed is more than Aarti started first so they would met when Anamika completed her one round, so in the first meeting Anamika covered 1.5 D and Aarti covered 0.5 D.

So, K.�¸M¹ − p.�¸

¹ = 15 ⇒ ¸º = 60 (glm gN�mO �° \N�gl gw dw»vmgm wOm �w¼OP)

For the second meeting time (since they are starting from same place and time) = ¸

º�xxe ½x¾�i¿Àx = ¸¹ = 60 �md

Total time from Anamika's start = 1st Meeting Time of Anamika + 2nd Meeting Time

= K.�¸M¹ + ¸

¹ = 1.75 ~¸¹� = 1.75 ∗ 60 = 105 �md

Directions for questions 332 to 334: Each of the questions is followed by two statements, I and II. Answer each question using the following instructions. Choose (a) if the question can be answered by using statement I alone but cannot be answered by using II alone Choose (b) if the question can be answered by using statement II alone but cannot be answered by using I alone Choose (c) if the question can be answered by using both the statements together but not by either statement alone Choose (d) if the question cannot be answered on the basis of the two statements 332.332.332.332. The world cup takes place every four years. Will it be held in a leap year this

time? I. Last time the world cup was not held in a leap year. II. Next time, the world cup will be held in a leap year. 333.333.333.333. Five people, A, B, C, D and E speak one after the other, not necessarily in the

given order. It is known that A speaks before C and B speaks before D. Who speaks the last?

I. D speaks after C and A is not the first to speak II. E is not the first to speak 334.334.334.334. The flowers in a basket double every minute. After how many minutes, from

start, will at least 5/6th of the basket be filled with flowers? I. At start the basket had 2 flowers II. At start, the basket was 1/8th full.


332.c332.c332.c332.c A leap year is found if the year is divisible by 4. But in case the year is divisible by 100, then the condition for the year to be leap is that it should be divisible by 400. Thus, each of 1700, 1800, 1900, 2100, etc, though divisible by 4 is not a leap year. Statement I: Even if the last time the world cup was not held in a leap year, it is quite possible that this year it could be held in a leap year. E.g. Last time the world cup was held in 1900 (not a leap year) and this time it will be held in 1904 (a leap year). Needless to say, this time the world cup could also be held in a non-leap year. E.g. last time it was held in 2007, this time it will be held in 2011. Thus, no firm answer can be found for the questions. Statement II: By the same using this statement independently, the question can still not be answered. Next time the world cup will be held in 1904 means this time the world cup is held in 1900, a non-leap year. Whereas if next time the world cup is held in 2004 means this time the world cup will be held in 2000, a leap year. Using both the statement: Last time it was not held in a leap year and next time it will be held in a leap year results only in possibilities like 1900, 1904, 1908 or 2100, 2104, 2108 or similar years. Thus, this time the world cup will surely be held in a leap year.

333.d

A and B are not the speakers who speak last.

Using statement I: Additionally we just know that C is not the last person to speak. The last

to speak could be any of D or E as seen in the following two arrangements: B A C D E or B

A C E D

Using statement II: Nothing can be said about the last person. The two examples cited above

still remain a possibility and thus, the last speaker cannot be ascertained uniquely.

Using both statement: If you paid careful attention to the use of the two cases cited above for

both statements independently, then you should have realized that the two cases satisfy the

conditions given in both the statements simultaneously. Thus, even after using both the

statements together, we cannot identify the last speaker uniquely.

334.b

Using statement I: Since we do not know the capacity of the basket we cannot know what

does 5/6th

of the basket mean.


Using statement II: Even if we do not know the capacity of the basket, after 1 minute

the basket will be 2 ×1/8th

i.e. 1/4th

filled. After another minute the basket will be ½ filled.

And so on. Thus, we can find out after how many minutes the baskets will atleast 5/6th be

filled. (In-fact after 3 minutes, the basket will be filed completely.)

Directions for questions Directions for questions Directions for questions Directions for questions 335335335335 to to to to 338338338338:::: Six people with varying lifestyles had a certain breakfast time routine. They each had a particular breakfast, at a certain time whilst enjoying a breakfast pastime. Breakfast times are (all AM): 6:30, 6:50, 7:15, 7:35, 7:50 and 8:30. Further following clues about their breakfast routine are known: A. The person who read the newspaper had breakfast later than the person who liked to listen to the news on the radio, but earlier than the person who enjoyed silence at breakfast time. B. Jagmohan had his breakfast directly before the person who had toast, who had breakfast directly before the person who listened to music on the radio. C. Cereal was eaten 20 minutes after the person who watched the Breakfast News on TV. Shivinder ate breakfast at 7:15. Lalit had breakfast sometime after Madhuban. A banana was not eaten at half past the hour. D. Chintamani’s corn-flaks were not the breakfast taken last. Sausages & eggs were eaten directly after the toast. E. Joginder had his breakfast directly after the person who ate sausage & eggs, but directly before the person whose idea of a fun breakfast was shouting at the children. 335.335.335.335. What breakfast did the person listening to news on the radio have? a. Corn-flakes b. Sausages and Eggs c. Banana d. Cereal 336.336.336.336. Which of the following two persons had a gap of 45 minutes between their

breakfasts? a. Chintamani and Jagmohan b. Jagmohan and Madhuban c. Madhuban and Lalit d. Joginder and Lalit 337.337.337.337. What is the time duration between the breakfast timings of the person having

cornflakes and the person listening to music while having breakfast? a. 40 minutes b. 55 minutes c. 35 minutes d. 65 minutes 338.338.338.338. For which person can we not ascertain which breakfast item is he having?


a. Lalit b. Joginder c. Madhuban d. Shivinder For Qs. 335 to 338:

We have four variables – name, break-fast, past-time and time of break-fast. Since time has a

fixed order, let’s keep that as the header row of the arrangement. And the next rows would

be in order of name, break-fast and past-time.

From clue B:

Jagmohan

Toast

Music on Radio

Clue D adds to the above block:

Jagmohan

Toast Sausages & Eggs

Music on Radio

Clue E also adds to the above block:

Jagmohan Joginder


Music on Radio Shout at Children

Since these are 5 columns, they could either be the columns 1 – 5 or column 2 – 6 in the

overall table.

If we above are columns 1 – 5, adding other obvious clues from A and C….

6:30 6:50 7:15 7:35 7:50 8:30

Jagmohan Shivinder Joginder


Music on Radio Shout at Children


Clue C states that cereal was eaten 20 minutes after the person who watched news on

TV. This is not possible with above possibility because the only two time intervals with 20

minutes gap is 6:30 and 6:50 (breakfast at 6:50 is already toast) and 7:15 and 7:35 (past-time

at 7:15 is listening to music on radio).

Thus, the above 5 column HAVE to necessarily be columns 2 – 6:

6:30 6:50 7:15 7:35 7:50 8:30

Jagmohan Shivinder Joginder

Cereal Toast Sausages & Eggs

News on TV Music on Radio Shout at Children

Clue A tell about past time: Listening to news on radio is before reading newspaper, which in

turn is before enjoying silence. This will help us fill the last row completely.

Clue C tells us that banana is not eaten at half-past an hour. So banana has to be eaten at

7:50. Since Chintamani’s cornflakes were not the last break-fast taken, it has to be the first,

as there is no other space left.

Thus, entire table cane be filled up.

335.d; 336.b; 337.c; 338.a

339.339.339.339. √1 − W_= x - 1 For how many real value of x is the above equation satisfied? a. 3 b. 2 c. 1 d. 0 Ans [c] only one x=1 340. Let ƒ(x) = ax² + bx + c have two roots p and q such that p > 0 and q < 0 and p > |q|.

Then how many of them would be necessarily false?

I. a×b<0, a×c<0 II. a×b>0,a×c<0 III. a×c<0,b×c<0

IV. a×c>0,b×c>0

a. Only one b. only Two c. Only Three d. All four

Sol. [C] only I is correct Sum of Roots = p + q = - b/a > 0 since p > |q|, i.e. either one of a or b are positive and

other is negative

Product of roots = p*q = c/a < 0 since p > 0 and q < 0 i.e. either one of a or c are positive

and other is negative


aaaa BBBB cccc a*ba*ba*ba*b a*ca*ca*ca*c b*cb*cb*cb*c

++++ ---- ---- ---- ---- ++++

---- ++++ ++++ ---- ---- ++++

341.341.341.341. In triangle ABC, E divides AC in the ration 2: 3, F divides AB in ration 4 : 3. ED is

drawn parallel to AB and FG is drawn parallel to BC, where D and G lie on BC and respectively. If area of triangle ABC = 1225 sq. units, what is the area of triangle EPG? (P is intersection of ED and FG)

a. 100 sq. units b. 60 sq. units c. 36 sq. units d. 25 sq. units Ans [c] by ratio formula 342.342.342.342. A, B and C decide to play 100 games of chess themselves. Two players play a

game and next game is played between winner of previous game and the third person. If a game results in a draw then another game is played between the same players till one of them loses and is replaced by the third person. B and C play the 1st game and A plays against C in 7th, 17th, 26th, 38th, 52nd, 73rd and 98th game ONLY. If B won n games then which of the following best describes n?

a. 7 ≤ O ≤ 93 b. 7 ≤ O ≤ 86 c. 0 ≤ O ≤ 93 d. 0 ≤ O ≤ 86 Sol: [d]Sol: [d]Sol: [d]Sol: [d] Case I : All matches in between B & C results draw except matches just before A Vs C, in which C wins so no win for B Case II : All matches in between B & C results B's wins except matches just before A Vs C, C wins and in matches between A Vs C B cant win so B wins 86 matches. 343.343.343.343. Z(W + 2°, W − 2°) = W ∗ ° for all value of x and y then f(x,y) =? for all value of x

and y

a. ÂGRLÃRÄR

z(GRLÃR) b. ÅRJÆR

z c. ÅRLÆR

z d. (a) and (c)

Ans : [c] Best is Solve by options [a] can not be answer since it does not hold when x^2 - y^2 =0 and we want for any value of x and y


344.344.344.344. The highest power of 5 that can divide 100! is 24. This can also be written as

100!=5MI × O, where n cannot be divided by 5. The highest power of how many other natural numbers that can divide 100! = WMI × O such that n cannot divided by x valid? a. 0 b. 15 c. 22 d. None of these Sol: [d] 21 Number of 2s in 100! = 97 Number of 2^2 s in 100! = 48 Number of 2^3 s in 100! = 32 Number of 2^4 s in 100! = 24 Number of 3s in 100! = 48 Number of 3^2 s in 100! = 24 Number of 5s in 100! = 24 So x should be combination of above values If x is factor of 2^4*3^2*5 such that it multiple of either 2^4 or 3^2 or 5 Options are (2^0,2^1,2^2,2^3,2^4)(3^0,3^1,3^2)(5^0,5^1) Multiple of 2^4 = 6 Multiple of 3^2 = 10 Multiple of 5 = 15 Multiple of 2^4 *3^2= 2 Multiple of 2^4 *5= 3 Multiple of 3^2 *5= 5 Multiple of 2^4 *3^2*5= 1 Hence required value = of either 2^4 or 3^2 or 5 = 6+10+15-2-3-5+1 = 22 But questions is how many other values (other than 5) so its 21 345.345.345.345. A change in price has an effect on the quantity sold. For a product consider that

if price is changed by a%, the quantity sold changes by b%. If revenue is product of


price and quantity then for which of the following values of a and b will the revenue have the maximum change?

a. a = -10, b = 25 b. a = -12.5, b = 30 c. a = 0, b = 12 d. a = -20, b = 40 Sol: [b] Final Effect in Revenue a. a = -10, b = 25 => New Revenue = Revenue*(0.9)*(1.25) = Revenue*1.125 b. a = -12.5, b = 30 => New Revenue = Revenue*(0.875)*(1.3) = Revenue*1.1375 c. a = 0, b = 12 => New Revenue = Revenue*(1)*(1.12) = Revenue*1.12 d. a = -20, b = 40 => New Revenue = Revenue*(0.8)*(1.4) = Revenue*1.12

346.346.346.346. How many numbers greater than 999 can be formed using any of the digits 1, 2, 3, 4, 5, 6, 7 exactly once such that digits are in ascending or descending order?

a. 64 b. 214 c. 128 d. 256 Sol [c] Number having more than three digits are greater than 999. So total such combinations = 7C4 + 7C5 + 7C6 + 7C7 = 64 And in each combination we get one 2 cases in which number is either in ascending or descending order. So required answer = 64*2 = 128 347.347.347.347. What is the average % change which is equivalent to 10% and 30% growth in

two equal time frame ? a. 20 % b. 15% c. 19.58% d. None of these Sol [c] Average is a single value which shows same effect if applied in all the data (number of data or phases or frame) Let a% be average change of 10% and 30% growth in two equal time frame. So we have two phases (time frame)


So effect or a5 comes in two time frame which is equivalent to 10% and 30% growth in two equal time frame, So, (1.a)*(1.a) = (1.1)*(1.3) (1.a)^2 = 1.43 1.a=1.1958 (approx) So a= 19.58%

348.348.348.348. x ∗ (W + 2) ∗ (W + 4) = 2� × 3y has how many integral solutions for x where m

and n are whole numbers. a. 0 b. 2 c. 3 d. More than 5 e. None of these Ans [c] x=2,4, -3 Let x =2n then x *(x+2)*(x+4)=2n*(2n+2)*(2n+4) =2^3 *n(n+1)(n+2) For only multiple of 2 and 3, n(n+1)(n+2) 349.349.349.349. x ∗ (W + 2) ∗ (W + 4) ∗ (W + 6) = 2� × 3y has how many integral solutions for x

where m and n are whole numbers. a. 0 b. 2 c. 3 d. More than 5 e. None of these Ans : [c] x=2,-8,-3 350. If

GÃ + Ã

G = 1 ( x, y ≠ 0) , the value of Wh – °h is

a. 1 b. -1 c. 2 d. 0

Sol: [d]

Wh – °h = (W_ + °_)(W_ – °_) = (W + °)(WM − W° + °M)(W − °)(WM + W° + °M) = 0

Since GÃ + ÃG = 1 ⇒ WM − W° + °M = 0

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