Cat 2007 Quantitative Questions Day Qqad Problems

Solution to Quantitative Question # 1

Question: 1

The equation |x-1| - |x-2| + |x-4| = m has exactly n real solutions for some real m. Then which among the following relations between m and n can not be true?

(a) m/n = 3/5 (b) m = n (c) m/n = 3/2 (d) m/n = 5/3 (e) m = n-1

Solution:

Since 1, 2, 4 are the critical point, we divide the domain into 4 regions;

1) when x > 4 |x-1| - |x-2| + |x-4| = (x-1)-(x-2)+(x-4) = m => x-3 = m => m>1

2) when 2 < x <=4 |x-1|-|x-2|+|x-4| = (x-1)-(x-2)-(x-4) = m => 5-x = m => 1 <= m < 3

3) when 1 < x <=2 |x-1|-|x-2|+|x-4| = (x-1)+(x-2)-(x-4) = m => x+1 = m => 1 < m <= 3

4) when -Infinity < x <=1 |x-1|-|x-2|+|x-4| = -(x-1)+(x-2)-(x-4) = m => 3-x = m => 2 <= m <= Infinity

Clearly for n=4 we have 2 < m < 3; for n=3 we have m = 2,3; for n=2 we have 1 < m < 2; for n=1 we have m=1; for n=0 we have 1 > m.

Looking at the choices (a) m = 2.4 and n = 4 satisfy (b) m = n = 3 satify (d) m = 10/3 and n = 2 satisy (e) m = -1 and n = 0 satisy

Hence, choice (c) is the right answer.

---------------------------------------------------------------------------Solution to Quantitative Question # 2 (Answered 1st by warrior and best solution by Apple)---------------------------------------------------------------------------

When Katrina get's Swiss chocolcates, she swings in delight, equal to her total chocolates at that time. For instance, if Katrina gets 3 chocolates, then 7 chocolates and then 3 chocolates again, she at first makes 3 swings, then she makes 3+7 = 10 swings and then she makes 3+7+3 = 13 swings, making a total of 3+10+13 = 26 swings. If all of Katrina's Swiss chocolates are in a group of either 3 or 7 and Katrina makes 99 swings during the process, in how many different ways can she get the chocolates in that process ?

(a) 3 (b) 6 (c) 5 (d) 2 (e) none of the foregoing

Solution:

If a1, a2, a3, ... are the numbers in which Katrina gets chocolates in order then Katrina's total swings will be n*a1 + (n-1)*a2 + .... + an where a1, a2 etc. take the values 3 or 7. Given the fact n*a1 + (n-1)*a2 + .... + an = 99; if all a1, a2 etc are 3 then 3n(n+1)/2 = 99 => n(n+1) = 66, so we know n can atmost be 7. Taking a1, a2 as 7 we can see that n is greater than 4.OK, if it's 7a1 + 6a2 + ... + a7 = 99 then because 3 and 7 are each of the form 4k-1 we see 7a1 + 6a2 + ... + a7 = 99 reduces to 4*(7k1 + 6k2 + ...+ k7) = 99 + (1 + 2 +... +7) = 99 + 28, which is not possible. Similarly it's not for n=5. For 6 it's possible as 99 + (1 + 2 +...+ 6) = 120 is div by 4.

Now you are left with 6k1 + 5k2 + ...+ k6 = 30 where k1,k2 ... takes either 1 or 2

(k1, k2, k3, k4, k5, k6) = (1, 2, 1, 2, 1, 2), (2, 1, 1, 1, 2, 2), (1, 1, 2, 2, 2, 1), (1, 2, 2, 1, 1, 1), (2, 1, 1, 2, 1, 1)

Hence, choice (c) is the right answer.

---------------------------------------------------------------------------Solution to Quantitative Question # 3 (Answered 1st by krsh.vik)---------------------------------------------------------------------------

Question:

Let f(x ,y) be a function satisfying f(x ,y)=f(2x + 2y, 2y -2x) for all x , y. Define g(x)=f(2^x , 0). What is the minimum positive integer p if g(x+p)=g(x) for all x?


Solution:

f(x ,y)=f(2x+2y,2y-2) = f(8y ,-8x )

f(8y,-8x) = f (8(-8x), -8(8y)) = f(-64x , -64y) = f(-64(-64x),-64(-64y)) = f(2^12 x ,2^12 y) = f(x,y)

=> f(x,0)=f(2^12 x,0) so , f(2^x,0) = f(2^(x+12) ,0) = > g(x) = g(x+12)

Hence, choice (d) is the right answer.

Solution to Quantitative Question # 4 (Answered 1st by innocent_123 and Apple and best solution by rajan24u)---------------------------------------------------------------------------

Question:

Consider a triangle PQR so that PQ=4, PR=5 and QR=6. Similarly, we have a point S so that QS=5, RS=4, and PQSR is a paralleogram (not an isosceles trapezium). We draw the angle bisector of < P, which hits QS at T, and the angle bisector of <PQT, which intersects PT at U. Then the length of TU is (vx denotes square root of x)

(a) 3 (b) 4v6/ 3 (c) 2v3 (d) 9/2 (e) none of the foregoing

Solution:

Letting <PRQ = x , we have <TPQ=<TPR =x

Let A be the intersection of PT and QR. <PAQ =<APR +<PRA =2x

So , triangle PQR ~ triangle AQP , which gives , PA =4* 5/6 =10/3 Now, since PR||QS , we have PA /AT =PR /QT => PA /PT = PR / PR + QT

<PTQ = <RPT=<QPT ,so triangle PQT is isosceles where PQ=QT=4

Hence , PT = {(PR + QT) /PR} * PA = 9/5 * 10/3 =6

Since the angle bisector of the vertex angle of an isosceles triangle bisects its base,

TU =1/2 *6 =3

Hence, choice (a) is the right answer.

---------------------------------------------------------------------------Solution to Quantitative Question # 5 (only sushane and innocent_123 got this correct)---------------------------------------------------------------------------

Question:

Three positive integers a, b, and c are consecutive terms in an arithmetic progression. Given that n is also a positive integer, for how many values of n below 1000 does the equation a^2 - b^2 - c^2 = n have no solutions?

(a) 458 (b) 493 (c) 524 (d) 559 (e) 596

Solution: (By our member sushan)

Let a = b+d, and c = b-d => n = b(4d - b) put b = 4d - 1, it gives n = (4d - 1)*1 = 4d - 1 = 3,7,11,...,999 for d = 1,2,...,250put b = 4d - 2 it gives n = 8d - 4 = 4,12,...,996 for d = 1,2,...,125put b = 4d -3, n = 12d - 9 which is of the form 4k - 1 and already countedput b = 4d - 4 it gives n = 16d - 16 = 16,32,...,992 for d = 2,3,...,63it can be seen that b = 4d - 3, 4d - 5,... etc. give these already counted values of n.=> total n are 999 - 250 - 125 - 62 = 562. The answer is not in the choices!

For more rigorous mathematical proof please refer innocent_123's post # 214 in the discussion thread.

---------------------------------------------------------------------------Solution to Quantitative Question # 6 (Answered 1st by innocent_123, Apple and sushan)---------------------------------------------------------------------------

Question:

20 teams of five archers each compete in an archery competetion. An archer finishing in kth placecontributes k points to his team, and there are no ties. The team that wins will be the one that has the least score . Given that, the 1st position team's score is not the same as any other team,the number of winning scores that are possible is ?


Solution:

The teams’ scores must sum to 1 + 2 + . . . + ..+99+100 =5050. The winning score must be nolarger than 1/20 *5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15. However, not all scores between15 and 252 inclusively are possible because all teams must have integer scores and no team can tie thewinning team. If the winning score is s, the sum of all teams’ scores is at least s + 19(s + 1) = 20s +19,so solving gives s <= 251. Hence, 251 − 15 + 1 = 237 winning scores are possible.

Hence, choice (b) is the right answer

---------------------------------------------------------------------------Solution to Quantitative Question # 7 (Answered 1st by Apple and sunilnatraj)---------------------------------------------------------------------------

Question:

P(x) is a third degree polynomial and the coefficients of P(x) are rational. If the graph of P(x) touches the x-axis, then how many rational roots does P(x) = 0 have?


Solution:

Since P(x) touches x-axis => P(x)=0 has a repeatable root and is real. If the repeatable root is irrational and is a+vb then the non-repeatable root is k-2vb where k is a rational number. WHY?

=> P(x) = p*(x-a-vb)^2*(x-k+2vb) which means the constant term of P(x) is irrational, a contradiction. Thus, the repeatable root is rational. And so is non-repeatable.

Alternate Method:

Let P(x) be x^3 − ax^2 + bx − c. Let m be a root of P(x) = 0 and P′(x) = 0

m^3 − am^2 + bm − c = 0 … (1) and 3m^2 − 2am + b = 0 … (2) => am^2 − 2bm + 3c = 0 … (3)

=> m = (ab – ac)/2(a^2 – 3b) is rational; also 2m + n = a (rational) => n is also rational

=> all the three roots of P(x) = 0 are rational.

Hence, choice (d) is the right answer

---------------------------------------------------------------------------Solution to Quantitative Question # 8 (Answered 1st by Apple, krsh.vik, Stalwart)---------------------------------------------------------------------------

Question:

QQAD team decides to go on vacation for 8 days trusting their NL software system. The problems are fed into the system and date timer set in advance for each 8 days for the questions to be delivered to the subscribers daily. However, the software follows a weird rule. It doesn't always deliver the NL daily in those 8 days, but never misses 3 consecutive deliveries. How many possible ways are there for the NLs delivery in those 8 days?

(a) 162 (b) 138 (c) 117 (d) 176 (e) 149

Solution:

8 NLs can be delivered in 8C8 = 1 way; 7 NLs can be delivered in 8C7 = 8 ways; 6 NLs can be delivered in 8C6 = 28 ways. 5 NLs can be delivered in 8C5 - (8-3+1) = 50 ways (6 is subtracted as it's no. of ways of 3 consecutive misses). 4 NLs can be delivered in 8C4 - 20 (with 3 consecutive misses) - 5 (with 4 consecutive misses) = 45 ways. 3 NLs can be delivered in 16 ways. WHY? 2 NLs can delivered in 1 way - on 3rd and 6th day. Thus making a total of 149 ways of delivery in those 8 days.

Alternate Method:

On each day NL can take 2 states - it either delivers or it doesn't. Let f(n) be the possible ways for n days with given conditions.=> f(1) = 2, f(2) = 2^2, f(3) = 2^3-1 (1 subtracted as we can't have 3 misses). Let n > 3. Then, NL either delivers on day 1 or it doesn't. When it does it can have f(n-1) ways from there; on day 2 also it either delivers or it doesn't, when it does it can have f(n-2) ways from there. Now on day 3, it has to deliver as we have missed first 2 days, thus after delivering on day 3 it can have f(n-3) ways => f(n) = f(n-1)+f(n-2)+f(n-3) => f(4) = 13, f(5) = 24, f(6) = 44, f(7) = 81 and f(8 ) = 149

Hence, choice (e) is the right answer

---------------------------------------------------------------------------Solution to Quantitative Question # 9 (Answered 1st by Apple and NoopS)---------------------------------------------------------------------------

Question:

Moiz has a machine into which he can put any number of one rupee coins. If he inserts n rupees, the machine returns 2n rupees. Each time he uses the machine, however, he must insert more money than he did on the previous use. If he starts with exactly Rs 1 and use the machine once, he will have Rs 2. On his next use of the machine, he is forced to insert Rs 2 yielding Rs 4, and on his third use of the machine, he can insert either Rs 3 or Rs 4 yielding a total of Rs 7 or Rs 8. The largest 2 digit integer Z such that it is impossible to obtain exactly Z rupees with the machine, starting with Rs 1 is


Solution:

Note that Rs10 is an impossible amount because after four uses of the machine Moiz must have at least Rs 11. Now it is easy to check that all amounts from 11 to 20 rupees can be obtained. Suppose, by way of contradiction, that some value larger than Rs 10 is impossible and let m be the smallest number exceeding 10 such that m rupees cannot be obtained. Then m > 20. If m is even, write m = 2k and note that k > 10. Thus k rupees is possible and he never inserted as much as k rupees to get it. He can thus insert all k rupees to get 2k = m rupees.If m is odd, write m = 2k + 1 with k ¸ 10. Then k + 1 rupees is possible and he never inserted as much as k rupees to get it. He can thus insert k rupees yielding a total of 2k + 1 = m rupees. This shows that m rupees is possible, contradicting the choice of m. Thus no value larger than Rs 10 is impossible to obtain.

Hence, choice (a) is the right answer

---------------------------------------------------------------------------Solution to Quantitative Question # 10 (Answered 1st by innocent_123, krsh.vik, howardroarkrock)---------------------------------------------------------------------------

Question:

Rani draws a square ABCD of side 1 unit. She then draws 10 straight lines connecting A to each of 11 equally spaced points lying internally on CD (including C and D). What is the total area (in unit square) of all the possible triangles that can be formed?

(a) 5 (b) 11/2 (c) 11 (d) 23/2 (e) none of the foregoing

Solution:

Clearly area of each of the smallest triangles is 1/20. Let the 9 internal points be M, N, O ....With AD as one side , the area of all triangles = 1/20 [1+2+..10] = 55/20With AM as one side , the area of all triangles = 1/20[1+2+..9] = 45/20Similarly for others..

SO total area of all such triangles including triangle ABC is 1/40[1*2+2*3+..9*10+10*11] +1/2 =11+1/2 = 23/2 square units.


---------------------------------------------------------------------------Solution to Quantitative Question # 11 (Answered by innocent_123, MBA_007, akangsha, sushane, AC_here)---------------------------------------------------------------------------

Question:

PQR is an acute angled triangle with perimeter 60 cm. S is a point on QR. The circumcircles oftriangles PQS and PSR intersect PR and PQ at T and U respectively such that ST = 8 cm and SU = 7 cm. If <TQR = <QRU, then which of the following represents the value of PT/PU?

(a) 16/19 (b) 3/4 (c) 14/17 (d) 5/6 (e) none of the foregoing

Solution:

Given that <TQR = <URQ = A (suppose)<TQR = <TQS = <TPS (angles made by same arc of circle) = A, <TPS = <RPS = <RUS (angles made by same arc of circle) = A. Similarly <URQ = <URS = <UPS (angles made by same arc of circle) = A, <UPS = <QPS = <QTS (angles made by same arc of circle) = A

Therefore we get PS is the angle bisector of QPR and Triangle USR and Triangle TSQ are isosceles. So, SR = SU = 7 and SQ = ST = 8. Therefore QR = 7 + 8 = 15.

Now using PQ + PR = 60 - QR = 45 and PQ/PR = 8/7 (angle bisector theorem) we get PR = 21 and PQ = 24

Use QU.QP = QS.QR => QU X 24 = 8 X 15 => QU = 5So, UP = 19, similarly TR X 21 = 7 X 15 => TR = 5. So, TP = 16

Therefore PT/PU = 16/19


---------------------------------------------------------------------------Solution to Quantitative Question # 12 (Answered 1st by by Apple)---------------------------------------------------------------------------

Question:

How many two digit or three digit positive integers in base 6 are there such that if 0 in inserted between the units and the tens digit, the multiple of the original number is obtained? Solution:


Solution:

Let AB be the 2 digit number => in base 6 it's 6A+B => 36A+B is divisible by 6A+B; for B=0, A can take 5 values from 1 to 5. 14 and 23 satisfy the given property (check yourself!) => 7 two digit numbers are possible.

Let ABC be the three digit number then 216A+36B+C is divisible by 36A+6B+C; for C=0 we have 5*6 = 30 possibilities. Thus, in all 37 such two or three digit numbers are possible.


---------------------------------------------------------------------------Solution to Quantitative Question # 13 (Answered 1st by by abhijith_69)---------------------------------------------------------------------------

Question:

Apple had to study Mathematics from his teacher Orange. As usual, Apple was sleeping in the class. Immediately he heard a thundering voice. "All roots of this equation are real as well as positive in nature". Apple woke up from deep slumber. He hurried to copy the 10 degree equation written on the board but could copy only the first two terms written on the blackboard before Orange sir wiped it all. Apple however remembered that the constant term was 2. He noted down the equation as2x^10 - 20x^9………. + 2 = 0. Apple was very sure that if someone would tell him the sum of all the coefficients of all the powers of x in the equation, he would solve it anyhow. He asked Mango about the same. The answer which Mango correctly gave was

(a) -1024 (b) 0 (c) 1024 (d) Mango himself was confused (e) none of the foregoing

Solution:

Let the roots be a1 , a2 ,….., a10 then a1 + a2 + …. + a10 = 10 , and (a1.a2….a10)=1 Now since all the roots are real and positive in natureWe can say that (a1+a2+….+a10)/10>=(a1.a2….a10)^1/10 For real numbers, AM >= GM but here we find that AM = GM hence a1 = a2 = … = a10 = 1 So our equation actually is 2(x-1)^10=0 and then the sum of all the coefficients of all the powers of x in the equation is 0.


---------------------------------------------------------------------------Solution to Quantitative Question # 14 (Answered 1st by by amit_vce)---------------------------------------------------------------------------

Question:

The sum of all the possible values of integral n such that (n^2-2n)^(n^2+47)=(n^2-2n)^ (16n-16) is


Solution:

Equate the bases to 0 and 1 and -1, we get n = 0, 1, 2 .. out of which 0 is rejectedEquate the powers to get n=7,9 Thus, we have 1+2+7+9=19 as the answer



Question:

The total number of 7 digit positive integers whose digits are in increasing order (not necessarily strictly) is

(a) 15C7 (b) 6! + 7.4! (c) 14C6 (d) 7! - 4.6! (e) none of the foregoing

Solution:

Consider arranging 7 X and 8 Y in a row. We can do this in 15!/(8!*7!) ways. Replace each X with (1+the number of Ys prior to it), and now remove all Ys. We will have the desired numbers!


---------------------------------------------------------------------------Solution to Quantitative Question # 16 (Answered 1st by by sushane, krsh.vik)---------------------------------------------------------------------------

Question:

The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010?

(a) 93 (b) 48 (c) 119 (d) 38 (e) 140

Solution: (by sushane)

Suppose a number is at nth position. First time n/3, [n/3] + 1, [n/3] + 1 integers less than n are eliminated for n = 3k, 3k+1, 3k+2. So, nth number's new position will be 2n/3, eliminated, [2n/3] for n = 3k, 3k+1, 3k+2.

Working this backwards it's like if a number is at nth position now, earlier it must have been at 3n/2 or [3n/2] + 1. So, what is at 1st position now (winner) was earlier at position [3/2] + 1 = 2. Moving back one more step what was at position 2, even earlier was at position 3*2/2 = 3.Going on like this the only positions that can end up at first position are 1, 2, 3, 5, 8, 12, 18, 27,

41, 62, 93, 140, 210, 315...(These are positions, not actual values)And when I have moved back enough steps to get back the original series (1,2,3,4,5,...), positions become equal to actual values cause in the actual series nth number is at nth position. So these are the actual winning contestant numbers 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315... subject to how many total contestants are there.


---------------------------------------------------------------------------Solution to Quantitative Question # 17 (Answered 1st by by chat.sun)---------------------------------------------------------------------------

Question:

Let 3 statements be made (P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10(Q) The remainder on dividing 16! + 89 by 323 is q(R) 46C23 leaves remainder r on division by 23

Then p+q+r equals


Solution:

The notation a % b = c means that when a is divided by b, c is the remainder that is obtained.

(P) Check for p = 9; 10^12 % 13 = 1 by Fermat's theorem => 10^18 % 13 = 10^6, since 10^3 % 13 = -1 + 13 => 10^18 % 13 = 1 => 10^18 - 10^9 + 1 leaves remainder (1-(-1)+1=3) by p = 9; Check yourself the remainder when p = 8.When p = 7, as 10^12 % 13 = 1 => 10^14 % 13 = 9; also 10^6 % 13 = 1 => 10^7 % 13 = 10. Thus p = 7 satisfies our condition.

(Q) 323 = 17*19. (p-1)! + 1 is divisible by p when p is prime => 16! % 17 = -1 + 17. Also 18! % 19 = -1 + 19If 16! % 19 = x => 18! % 19 = 17*18x => 306x % 19 = 18 => 2x % 19 = 18.

Thus, 16! leaves the remainder 16 by 17 and 9 by 19 => It leaves 237 by 323 => q = 3

(R) 2nCn = (nC0)^2 + (nC1)^2 + ... + (nCn)^2; when n is prime, each of the term in RHS except 1st and last is divisible by n => r = 2


---------------------------------------------------------------------------Solution to Quantitative Question # 18 (Answered 1st by by Apple, krsh.vik )---------------------------------------------------------------------------

Question:

From a cylinder of height 3 cm and radius 2 cm , two identical cones each of height 2 cm are to be cut such that they have the maximum volume. The volume of a cone will be (in cubic cm)

(a) (2/3)*pi (b) (5.12/3)*pi (c) (6.16/3)*pi (d) (7/3)*pi (e) none of the foregoing

Solution:

The two cones must be symmetrically selected with their bases on opposite bases of the cylinder respectively and also the two cones must touch each other.Let the radius of base of each cone be x and let y be the perpendicular distance of the point of tangency of the two cones from the altitude of either cones. Then, 2x +2y = 2*2=4

If the height of cylinder is h, then we have (2h/3 -h/2)/y = (2h/3)/x =>y/x =1/4

Hence, x+x/4 =2 => x = 8/5 . So volume of the cone = pi/3 (8/5)^2 *(2/3)*h = (5.12/3) *pi cubic cm



Question:

Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of

(a) 18% (b) 20% (c) 21 % (d) 23% (e) Cannot be determined

Solution:

Let the CP of two vodkas be Rs 100 and Rs 100x and individual profit in Rs on them being A and B.

=> (A+2B)/3 = 10/100*(100+200x)/3 and (2A+B)/3 = 20/100*(200 + 100x)/3. solving we get A = (70+20x)/3 and B = (20x-20)/3

=> profit percentages on each is (70+20x)/3 and (20x-20)/3x. When they are increased to 4/3 and 5/3 times respectively and mixed in the ratio 1:1 we get total profit % as (4/3*100*(70+20x)/3 + 5/3*100x*(20x-20)/3x)/(100+100x) = 100*(20x+20)/(100+100x) = 20.


---------------------------------------------------------------------------Solution to Quantitative Question # 20 (No one gave the full solution to this) The answer to #19 was given 1st by rahul_4096---------------------------------------------------------------------------

Question:

N girls and 2N boys played a chess tournament. Every player played every other player exactly once. The boys won 7/5 times as many matches as the girls (and there were no draws). Then which among the following is definitely false? (Assume 1 point for a win and 0 for a loss)

(a) Boys pocketed prime number of points against girls(b) Girls always won twice or more matches than boys won against them(c) The sum of the scores of top 3 individual players was not between 25 and 33(d) The sum of the scores of top 3 individual players was 69(e) none of the foregoing

Solution:

Total number of matches among boys were 2nC2, among girls were nC2 and between boys and girls were n*2n. Please note that 2nC2 + nC2 + 2n^2 = 3nC2. Assume 1 point for a win and 0 for a loss.

=> Girls pocketed nC2 points amongst themselves and boys pocketed 2nC2 points among themselves. Let boys take k points from their matches against girls => girls take 2n^2 - k from their matches gainst boys.

=> 2nC2 + k = 7/5*(nC2 + 2n^2 - k), solving we get 8k = n(5n+1). for n = 3, k = 6. For n = 8, k = 41, For n = 11, k = 77.

(a) can be true as for n = 8, k = 41. (b) can be true as can be seen for for n = 3, 8, 11, ... (c) is true as for n = 3, top 3 can score 8+7+6 = 21 points and for n = 11, when 33 matches are played top 3 will always score more than 16+15+14 = 45. (d) is true, For n = 11, we can have the top 3 score as 23+23+23 = 69.


---------------------------------------------------------------------------Solution to Quantitative Question # 21 (Answered 1st by by Doomsayer)---------------------------------------------------------------------------

Question:

Amrutesh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Amrutesh walksaccording to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches thehypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitudeto the hypotenuse of the larger one. He repeats this process forever. What is the total distance thatAmrutesh walks?

(a) 48/25 (b) 12/5 (c) 12 (d) 15 (e) none of the foregoing

Solution:

Let M be the endpoint of the altitude on the hypotenuse. Since we are dealing with right triangles,triangle MAC ~ triangle ABC, so AM = 12/5. Let N be the endpoint he reaches on side AC.Triangle MAC ~ trangle NAM,So , MN/AM =4/5 . This means that each altitude that he walks gets shorter by a factor of 4/5 . The total distance is thus (12/5) /(1- 4/5) =12

Hence, choice (c) is the right answer

---------------------------------------------------------------------------Solution to Quantitative Question # 22 (Answered 1st by by macora and krsh.vik)---------------------------------------------------------------------------

Question:

The minimum possible value of the largest of ab, 1-a-b+ab, and a+b-2ab if 0 <= a <= b <=1 is

(a) 4/9 (b)1/9 (c)5/9 (d)1/3 (e) none of the foregoing

Solution:

Let s = a + b, p = ab, so a and b are (s+/- root(s^2-4p))/2 . Since a and b are real , s^2 - 4p>=0 . If one of the three quantities is less than or equal to 1/9, then at least one of the others is at least 4/9 by the pigeonhole principle since they add up to 1. Assume that s-2p < 4=9, then s^2 - 4p < (4/9 + 2p)^2 - 4p , and since the left side is non-negative we get 0<= p^2 -(5/9) p +4/81 =(p-1/9)(p-4/9).This implies that either p<=1/9 or p>=4/9 , and either way we're done. This minimum is achieved if a and b are both 1/3, so the answer is 4/9


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Solution to Quantitative Question # 23 (Answered 1st by by vivekr)---------------------------------------------------------------------------

Question:

Let f be a factor of 120, then the number of positive integral solutions of xyz = f is


Solution:

Let k be such that k = 120/f. Then, the number of positive integral solutions of xyz = f is same as that of number of positive integral solutions of xyzk=120=2^3.3.5

We can assign 3 and 5 to unknown quantities in 4*4 ways. We can assign all 2 to one unknown in 4C1 ways, to two unknowns in (4C2)(2) and to three unknown in 4C3 ways.

Hence, the number of required solutions

=4*4*[4C1 + (4C2)(2) +4C3] =4*4*20 =320


---------------------------------------------------------------------------Solution to Quantitative Question # 24 (Answered 1st by krsh.vik and Apple)---------------------------------------------------------------------------

Question:

Apple, Bombardier, Chat.sun and Doomsayer are to compile this year's Quant Question A Day. They can finish this work together in a certain number of integer days. However, they work two in a day and it is found that the compilation is completed when (Apple, Bombardier), (Bombardier, Chat.sun) and (Chat.sun, Doomsayer) worked for respectively 5, 9 and 4 days or 7, 6 and 5 days. They could not have all together done the work in

(a) 8 days (b) 9 days (c) 10 days (d) 11 days (No choice (e) in this problem)

Solution:

Let Apple, Bombardier, Chat.sun and Doomsayer do 1/a ,1/b,1/c and 1/d parts of work per day respectively .Then ,

5(1/a +1/b) +9(1/c +1/b) +4(1/c +1/d) =1 ; 7(1/a +1/b) +6(1/c +1/b) +5(1/c +1/d) =1

and 1/a +1/b + 1/c +1/d = 1/n (say)

=>12( 1/c +1/b) = 1-3/n which implies n>3 and 4( 1/a +1/b) = 1- 7/n which implies n>7 .

Consequently, 1/c +1/d = 1/n - ( 1/a +1/b) = 1/n -1/4(1-7/n) =(11-n)/4n

which implies n<11; So, from the given options n=11 is not possible .


---------------------------------------------------------------------------Solution to Quantitative Question # 25 (Answered 1st by macora and tatimatla )---------------------------------------------------------------------------

Question:

Through T, the mid-point of the side QR of a triangle PQR, a straight line is to meet PQ produced to S and PR at U, so that PU = PS. If the length of UR = 2 cm, then the length of QS is

(a) 3/2 cm ( b) 2 cm (c) 5/2 cm (d) 3 cm (e) none of the foregoing

Solution : If <PSU = <PUS =Q , in triangle UTR the sine of < opposite to UR and TR is the same as

sine of < opposite to QS and QT in triangle QST correspondingly.Since , TR=QT =>QS=UR=2

(Alternate Solution by vikasInBlr)

This problem can be solved easily using Menelaus' Theorem.

According to this theorem following relation always holds goodPS.QT.UR = QS.TR.PUas PS = PU and QT = TR we have UR = QS so QS = 2.


---------------------------------------------------------------------------Solution to Quantitative Question # 26 (Answered 1st by tatimatla ,Superstar and vickykansal)---------------------------------------------------------------------------

Question:

A new Ducati is designed for the Indian market such that its mileage at a particular speed follows a certain relationship with that speed. Also , the speed decreases linearly with the mass of the rider while the petrol consumption per km increases linearly with the mass of rider .Ideally , when the mass of the rider is negligible, the speed is 100km/hr and mileage is 100km/l .When the speed of the Ducati is 50 km/hr , the mileage is 50 km/l.When the speed of the Ducati is 75 km/hr , the mileage will be

(a) 60 km/l (b) 67 km/l (c) 72 km/l (d) 75 km/l (e) Cannot be determined

Solution :

(Solution by tatimatla)Speed = 100-k1*massPetrol Consumption per KM = c2+k2*massor Mileage = 100/(1+k2*mass)

When speed = 50KMPH, K1*mass = 50when mileage = 50KMPL, k2*mass = 1k1 = 50*k2

When speed = 75kmph, k1*mass = 25 => k2*mass = 0.5mileage = 100/1.5 = 67km/l


---------------------------------------------------------------------------Solution to Quantitative Question # 27 (Answered 1st by apple ,tatimatla ,siddhesh,doomsayer,xtremelegacy,MG_1806)---------------------------------------------------------------------------

Question:

A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ?

(a) 364 (b) 231 (c) 455 (d) 472 (e) None of the foregoing

Solution :

V(66,14 ) = V(14,66 )=(33/26)* V(14,52) =(33/26) * (26/19)* V(14, 38) = (33/19) *(19/12)* V(14, 24) = (33/12) *(12/5)* V(14,10) = (33/5)* (7/2)*V(10, 4) = (231/10)* (5/3) *V(4, 6) = (77/2)* 3 *V(4, 2) = (231/2)*2* V(2, 2) = 231*2 = 462


---------------------------------------------------------------------------Solution to Quantitative Question # 28 (No one answered correctly)---------------------------------------------------------------------------

Question:

A cargo ship circles a lighthouse at a distance 20 km with speed 1500 km/h. A torpedo launcher fires a missile towards the ship from the lighthouse at the same speed and which moves so that it is always on the line between the lighthouse and the ship. How long does it take to hit?

(a) 37.7 secs (b) 56.57 secs (c) 75.43 secs (d) 94.29 secs (e) 113.14 secs

Solution :

Let M be the position of the ship at the moment the missile is fired. Let V be the point a quarter of the way around the circle from M (in the direction the ship is moving).Take the point at which Lighthouse is situated to be U. Then the missile moves along the semi-circle on diameter UV and hits the ship at V.

To see this take a point T on the quarter circle and let the line UT meet the semi-circle at R. Let Z be the center of the semicircle. The angle VZR is twice the angle VUR, so the arc VT is the same length as the arc VR. Hence also the arc UR is the same length as the arc MT.

So time needed= 10 *pi*3600 /1500 =75.43


---------------------------------------------------------------------------Solution to Quantitative Question # 29 (answered first by doomsayer ,amit_vce)---------------------------------------------------------------------------

Question:

Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

(a) 24 (b) 23 (c ) 22 (d) 20 (e)21

Solution :

Each angle is 180(p-2)/p.

180-{360}/{p}=kSo 360/p has to be an integer.Factors of 360 = 2^3.3^2.5^1So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3So , 24 -2 =22


---------------------------------------------------------------------------Solution to Quantitative Question # 30 (answered first by tatimatla)---------------------------------------------------------------------------

Question:

Consider a pair (x,y) of natural numbers satisfying x + y^2 + g^3 = xyg, where g is the greatest common divisor of x and y .Then , how many such pairs are possible?

(a) 2 (b) 3 (c) 4 (d) 5 (e) 6

Solution :

Put x = Mg, y = Ng, so that M and N are coprime. Then: M + N^2g + g^2 = MNg^2. So g must divide M. Put M = M'g, then M' + N^2 + g = M'Ng^2.

So M' = (N^2 + g)/(Ng^2 - 1). Hence M'g^2 = N + (N+g^3)/(Ng^2 - 1). So (N + g^3)/(Ng^2 - 1) is an integer. If g = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. That gives the solutions (x, y) = (5, 2) and (5, 3). So assume g > 1. Since (N + g^3)/(Ng^2 - 1) is an integer and positive, we must have N + g^3 >= Ng^2 - 1, so N <= (g^3 + 1)/(g^2 - 1). If g = 2, then N <= 3. Then N = 1 gives the solution (x, y) = (4, 2), N = 2 gives (N + g^3)/(Ng^2 - 1) non-integral and hence no solution, N = 3 gives the solution (x, y) = (4, 6).

So assume g > 2. Then (g^3 + 1)/(g^2 - 1) < g + 1. Hence N ≤ g. Hence M' = (N^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(g^2 - 1), since N ≥ 1 and hence M' < 2 (since g > 2). So M' = 1. So N is a root of the quadratic N^2 - g^2N + g+1 = 0. But g^4 - 4(g+1) > g^4 - 4g^2 + 1 = (g^2 - 1)^2 and < g^4, so g^4 - 4(g+1) cannot be a square and hence N cannot be integral. So there are no solutions with g > 2.


------------------------------------------------------------Quantitative Question # 31------------------------------------------------------------

Question:

LMN is a triangle. MO is the angle bisector. The point P on LM is such that <LNP = (2/5) < LNM. MO and NP meet at Z . PO = ON = NZ. Find angle LPO.


Going in detail .. So , quant experts , plz bear with me . As I have found that many newbies find even simple concepts difficult .

Lets frst draw the triangle and assuming <LNM=x , Find out other angles as the conditions given in question are enuf to find out all the angles formed in the Triangle.

<LNP=(2/5)x ; <PNM=(3/5)x ; <NZO=<NOZ={90-(1/5)x} ....... as ON=NZ

<NZO=<PZN={90-(1/5)x} ...... vertical opposite angles

<OPN=<ONP=(2/5)x ..... as PO=ON

<NZO+<OZP=180 ==>> <OZP={90+(1/5)x}

<OZP=<MZN={90+(1/5)x} .... vertical opposite angles

<POZ={90-(3/5)x} .... Sum of angles in a triangle=180

<LOP=(4/5)x ..... Sum of angles in straight line=180

<OMN={90-(4/5)x} ..... Sum of angles in a Triangle=180

<OMN=<LMO={90-(4/5)x} ..... As MO is the angle bisector.

<MPN=x .... Sum of angles in a triangle = 180

<LPO={180-(7/5)x} ..... Sum of angles in straight line=180

<MLN=(3/5)x .... Sum of angles in Triangle=180

Now take triangles LMN and LPO

(LM / MN ) = ( LO / ON ) ..... angle bisector theoremBut ON=OP ... Given==>> ( LM / MN ) = ( LO / OP ) And , <L is common .

==>> Both Triangles are similar .

==>> <LNM=<LPO==>> x=180 - (7/5)x==>> x=<LNM=75 .... (1)

While Required angle = <LPO = 180-(7/5)x = 180-(7/5)*75==>> <LPO = 180 - ( 7*15 )<LPO = 180 - 105 = 75==>> <LPO = 75 deg. .

---------------------------------------------------------------------------Solution to Quantitative Question # 32 (answered first by vivekr ,doomsayer , rahul_4096 )---------------------------------------------------------------------------

Question:

N! is defined for non negative integers as N!=N*(N-1)*(N-2)* ...3*2*1 . The number of positive integers which divide (2^5 )! are

(a) 2^13.3^3.5^2 (b) 2^8.3^2.5^2 (c) 2^11.3^2.5 (d) 2^8.3^3.5^3 (e)None of the foregoing

Solution :

32! is divisible by [32/2] + [32/4] + [32/8] + [32/16] +[32/32] = 31 powers of 2, and

[32/3] + [32/9] + [32/27] = 14 powers of 3. It is divisible by 5^7, 7^4, 11^2, 13^2, 17 , 19, 23, 29, 31.

In other words, 32! = 2^31 .3^14 . 5^7 .7^4 .11^2 ·13 ^2 ·17·19.23 .29.31 So it has (31+1) (14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1) = 2^13 . 3^3 . 5^2


---------------------------------------------------------------------------Solution to Quantitative Question # 33 (answered first by innocent_123 , maxximus )---------------------------------------------------------------------------

Question:

The sum of the infinite series 1/3 + 2/21 + 3/91 + 4/273 + ... is given by which of the following ?

(a) 1/4 (b) 1/2 (c) 3/4 (d) 1 (e) 3/2

Solution :

General term T(n) = n/(n^2-n+1)(n^2+n+1) = 1/2 * (1/(n^2 - n + 1) - 1/(n^2 + n + 1)) Hence sum = 1/2 *(1 - 1/3 + 1/3 - 1/7 + 1/7 - ........) = 1/2 *(1) = 1/2


---------------------------------------------------------------------------Solution to Quantitative Question # 34 (answered first by vivekr )---------------------------------------------------------------------------

Question:

Consider a positive integer Z which when represented in decimal base,does not end in zero. Z and the number obtained by reversing the digits of Z are both multiples of seven. Let the number of such Zs in the set {10 ,11,12... 998,999,1000} be K. Then k is given by which of the following?

(a) 13 (b) 14 (c) 15 (d) 16 (e) None of the foregoing .

Solution :

There is only one 2-digit number, namely 77. Indeed, if 10a + b is (where a, b are integers 1–9), then 7 divides 10a + b and 7 divides 10b + a. Thus,7 divides{ 3(10a + b) − (10b + a)} = 29a − 7b = a + 7(4a − b),so that 7 divides a, and similarly 7 divides b, so we’d better have a = b = 7.

There are 16 3-digit numbers. First consider the 12 palindromic ones (oneswhere the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616,686, 707, 777, 868, and 959. Now consider the general case: suppose 100a + 10b + c is, where a, b, c are integers 1–9. Then 7 divides 100a + 10b + c and 7 divides 100c + 10b + a,so 7 divides (100a+10b+c)−(100c+10b+a) = 99(a−c), and so 7 divides a−c. In order for thisnot to result in a palindromic integer, we must have a − c = +/- 7 and, moreover, both100a + 10b + a and 100c + 10b + c must be palindromic integers. Consultingour list above, we find 4 more integers: 168, 259, 861, and 952.

So K= 1+16 = 17


---------------------------------------------------------------------------Solution to Quantitative Question # 35 (answered first by doomsayer , HarshaRocks)---------------------------------------------------------------------------

Question:

Shravya invests some amount of money in a firm M .This amount grows upto 5000 in 2 years and upto 5500 in 3 years on R% compound inetrest . Then she goes to another firm N and borrows Rs 7000 at a compound interest of R% . At the end of each year she pays back Rs 3000 to firm N . Then , the amount she should pay to firm N at the end of 3 years to clear all the dues is ?


Solution :

CI of 3rd year =5500-5000 =500Rate , R = 500*100/5000 =10%At the end of 3 years she pays = 7000[1+11/100]^3 - 3000[(1+10/100)^2 + (1+10/100)] = 2387


---------------------------------------------------------------------------Solution to Quantitative Question # 36 (answered first by sastry751)---------------------------------------------------------------------------

Question:

If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are (a)36 ( b)33 (c)66 (d)60 (e) None of the foregoing

Solution :

Take b=a+k (where k is any integer)We boil down to |(c^2-k^2)*(2a+k-c)|=15Cases can only be out of the following : (+/-1,+/-15) ,(+/-15,+/-1) , (+/-3,+/-5) , (+/-5,+/-3)

(c,k) can be - > (+/-1,0) ,(0,+/-1),(+/-4,+/-1),(+/-1,+/-4),(+/-8+/-7),(+/-7,+/-8 ),(+/-2,+/-1),(+/-1,+/-2) ,(+/-3,+/-2),(+/-2,+/-3)

And each case for (c,k) will yield 2 values of a and for a given a and c , b is uniquely determined by k

Total c,k from above is (2+2) +(4+4)+(4+4)+(4+4) +(4+4) =36Since for each (c,k) we get two values of a .

Hence total (a,b,c) =2*36 =72

ALTERNATE SOLUTION

2a=(x+y), 2b=(y+z) and 2c=(x+z)

If you take (x,y,z) as (5)(3)(1)=>(a,b,c) is (4,2,3)=>6 cases(-5)(-3)(1)=>(-4,-1,-2)=>6 cases(5)(-3)(-1)=>(1,-2,2)=>6 cases(-5)(3)(-1)=>(-1,1,-3)=>6 cases(15)(1)(1)=>(8,1,8 )=>3 cases(15)(-1)(-1)=>(7,-1,7)=>3 cases(-15)(-1)(1)=>(-8,0,-7)=>6 cases (note that it’s a special one in which -15,-1 and 1 are all distinct, hence it gives 6 cases)

Hence all in all 36 cases for +15 and 36 for -15..so overall 72 triplets are possible.


---------------------------------------------------------------------------Solution to Quantitative Question # 37 (answered first by sourabh phanase , doomsayer)---------------------------------------------------------------------------

Question:

Consider a scalen triangle PQR.Points S ,T and U are selected on sides QR, PR, and PQ respec-tively. The lines PS, QT, and RU meet at point Z. If area(PUZ) = 126, area(UQZ) = 63, andarea(RTZ) = 24, the area of triangle PQR is ?

(a) 324 (b) 351 (c) 360 (d) 364 (e) 378

Solution :

(Solution by doomsayer)

Consider triangle PQT and PQZ and PZTThe area of PQZ = 1/2 QZ *h = 189the area of PZT= 1/2 ZT *h =x so area of PZQ /area of PZT = 189/x = QZ/ZT ----------(1)Then take triangle QTR, similarly area of QZR / area of RTZ = y /24 = QZ/QT ------(2)from 1 and 2---- 189/x = QZ/ZT = y /24

Then take triangle PUR and URQ and apply the same procedure so we get the rel 24 +x /126 = y /63

Solve and get the value of x and y..So get total as 351


---------------------------------------------------------------------------Solution to Quantitative Question # 38 (answered first by rahul_4096 , maxximus , unlimiteddreamz )---------------------------------------------------------------------------

Question:

There are three runners viz , Nishant , Deepak and Mohit who jog on the same path. Nishant goes jogging every two days. Deepak goes jogging every four days. Mohit goes jogging every seven days. If its the first day that they started this routine, what is the total number of days that each person will jog by himself in the next seven weeks?

(a) 12 (b) 13 (c) 14 (d) 15 (e)16

Solution :

Nishant jobs a total of 25 days, Deepak a total of 13 days, and Mohit a total of 7 days. We know that Nishant jogs on days 1,3,5,7,.....49, and Deepak jobs on days 1,5,9,.....49, and Mohit jogs on days 1, 8, 15, 22,...43. Obviously when Deepak jogs, Nishant will always be jogging also, so Deepak jogs 0 days alone. Nishant jogs 25-13=12 days without Deepak also jogging, but Mohit jogs on 4 odd numbered days, 2 of which Deepak also jogs, giving us 12-4+2=10 days that Nishant jogs alone. Nishant jogs on every odd numbered day, so Mohit jogs alone only on even numbered days. Because there are 4 odd numbered days, as stated above, there are 3 even numbered days, so Mohit jogs alone on 3 days.

Answer:Nishant: 10Deepak: 0Mohit: 3

Hence ans=13


---------------------------------------------------------------------------Solution to Quantitative Question # 39 (No one could get this right fully)---------------------------------------------------------------------------

Question:

Consider a polynomial function P(y) =y^3+2y^2+5 and one another polynomial function, Q(y)=y^4 -3y^2+2y+1. Let there be two more functions S(y) and T(y), that satisfy gcd(P(y), Q(y) ) =S(y)*P(y) + T(y)*Q(y). Now consider a function M(y) = T(y) - S(y). Then, the product of all the roots of M(y) is given by


Solution :

First stage division gives ..Q(y) = P(y)(y-2) + y^2 -3y +11y^2 -3y +11 = Q(y) - (y-2)P(y)

And then from the second stage ,

P(y) = (y^2-3y+11)(y+5) + 4y-50= {Q(y) - (y-2)P(y)}(y+5) + 4y-50

From the third stage of division ,4(y^2 -3y +11)= (4y-50)(y+19/2) +519=>519 = 4[Q(y) - (y-2)P(y) ] - [(y^2+3y-9)]P(y) - (y+5) Q(y)] [y+19/2]

=>519 = 4 Q(y) - (4y-8 )P(y) - (y^2+3y-9)[y+19/2] P(y) + (y+5) Q(y) [y+19/2]=-[4y-8+y^3+19y^2/2 +3y^2 +57/2y -9y-171/2]P(y) + [y^2+19/2y +5y+95/2 +4]Q(y)=-P(y)/2 [2y^3 + 25y^2 +47y -187] + Q(y)/2[ 2y^2+29y+103]

=>519 = [-(2y^3 + 25y^2 +47y -187)/2]P(y) + [(2y^2+29y+103)/2]Q(y)

So, T(y) - S(y)= (2y^3 +27y^2 + 76y -84)/2 = y^3 +27/2y^2 + 38y -42

So, product of roots =42

Hence, choice (a) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 40 (Solved first by arnabsadhu )---------------------------------------------------------------------------

Question:

A circle passes through the vertex C of rectangle ABCD and touches its sides AB and AD at P and Q respectively. If the distance from C to the line segment PQ is equal to 4 units, then the area of the rectangle ABCD in sq. units (is)

(a) 20 (b) can not be determined (c) 16 (d) greater than 20 (e) none of the foregoing

Solution :

Let M be the feet of perpendicular from C to PQ. Now, by alternate segment theorem we have=> < CQM = < CPB and < CPM = < CQD. Thus, right triangles CQM and CPB are similar and also CPM and CQD. Thus, CQ/CP = CM/CB and CP/CQ = CM/CD. Thus, CB.CD = CM^2 = 16.

Alternate Solution:

Assume the vertices of A, B, C, D as (0, b), (a, b), (a, 0) and (0,0). Since, AP = AQ, let AQ = m. Thus P = (0, b-m) and Q = (m, b). Now x = 0 is tangent to the circle at P. The eq. of circle is x^2

http://CB.CD/

+ (y - b+m)^2 + kx = 0, here k is variable. But, it's also tangent at Q = (m, b). Thus we get k = -2m.Now, the circle x^2 + (y-b+m)^2 -2mx passes through (a, 0) => a^2 + (b-m)^2 = 2am. Equation of PQ is x-y = b-m. Also |(a-m+b)/v2| = 4. Squaring and using the relation in bold we get 2ab = 32, ab = 16.

Hence, choice (c) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 41 (Solved first by vickykansal, josephalapatt, shyamnaren, rahul_4096)---------------------------------------------------------------------------

Question:

Maxximus needed to calculate the volume of a rectangular room. He multiplied the length and the breadth correctly but the breadth had been incorrectly jotted down, it was one-third larger than what it should have been. To compensate for this, he reduced the height by one-third, then multiplied it on. He figured this was okay since the breadth was equal to the height. He then found his volume was off by 20 m^3. What was the actual volume?

(a) 160 (b) 120 (c) 210 (d) 159 (e) None of the forgoing

Solution :

Actual Volume = L*B*H. Given, L*B*H - L*4/3B*2/3H = 20 => 1/9*LBH = 20

Hence, choice (e) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 42 (Solved first by prateek4u, sunilnatraj, scarletyoke)---------------------------------------------------------------------------

Question:

Which among the following have the same graph? (vx denotes square root of x)

I) y = x-2 II) y = (x^2-4)/(x+2) III) y(x+2) = x^2-4 IV) y = (vx-v2)*(vx+v2)

(a) I and III only (b) II and IV only (c) I, III, IV (d) II and III only (e) none of the forgoing

Solution :

y = x-2, will be a straight-line with slope 45 deg and making intercept of 2 and -2 on X and Y axis respectively. II) Same as I except for a breakpoint at x = -2. III) at x = -2, we will also have a line parallel to the Y axis. IV) Graph is restricted to first and fourth quadrants only.


---------------------------------------------------------------------------Solution to Quantitative Question # 43 (Solved first by vivekr)---------------------------------------------------------------------------

Question:

All the digits of a 50 digit positive number are 4 except for the nth digit. If the number is divisible by 13 for some choice of that nth digit, then how many possible values can n have?

(a) 17 (b) 21 (c) 25 (d) 33 (e) none of them

Solution :

The seed of 13 is 4 which means to check the divisibility by 13 of a n digit number, multiply the last digit by 4 and add the result to the initial (n-1) digits and see if it is div by 13 e.g. 182 -> 18+2*4 = 26. Thus, 182 is div by 13. 1001 -> 100 +1*4 = 104 -> 10 + 4*4 = 26, hence 1001 is div vy 13. It's also the same thing for 182 as saying that 2*(4^2) + 8*4 + 1 is div by 13 or for 1001 1*(4^3) + 1 is div by 13 - this comes from recurrence. Note that we can find the seed of every prime number. The seed of 7 is 5, the seed of 17 is 12, the seed of 19 is 2.

If 444...x...444 is div by 13 => 4*(4^49 + 4^48 + ... + 1) + (x-4)*4^(n-1) is div by 13, where nth digit is x and not 4.

4*(4^49 + 4^48 + ... + 1) = 4/3*(4^50-1).By Fermat's theorem 4^12 = 1 mod (13) => 4^50 = 3 mod (13). Thus, 4*(4^50 - 1) leaves remainder of 8 when div by 13. If 4/3*(4^50-1) leaves x by 13 => 3x%13 = 8 => x = 7. Thus 4/3*(4^50-1) leaves remainder 7 when div by 13.Now, 7 + (x-4)*4^(n-1) is div by 13.when n = 1, 7+(x-4) has to be div by 13, we have no such xwhen n = 2, 7+4*(x-4) has to be div by 13, we have no such xwhen n = 3, 7+3*(x-4) has to be div by 13, we have x = 6when n = 4, 7+12*(x-4) has to be div by 13, we have no such xwhen n = 5, 7+9*(x-4) has to be div by 13, x = 9when n = 6, 7+10*(x-4) has to be div by 13, x = 2when n = 7, 7+(x-4) has to be div by 13, we have no such x and the whole cycle of 6 repeats from here. In each cycle of 6 we have 3 desired n.Till n = 50 we have 48*3/6 + 0 = 24 such numbers.


---------------------------------------------------------------------------Solution to Quantitative Question # 44 (Solved first by Doomsayer)---------------------------------------------------------------------------

Question:

Let a, b, c be positive reals. (I) and (II) are independent statements.

(I) Minimum value of a^3/4b + b/8c^2 + (1+c)/2a is p(II) a + b + 2c = 8 and a^2 + b^2 + 2c^2 = 25. Maximum possible value of c is q.

Then which among the following is |p-q|?

(a) 5/2 (b) 9/4 (c) 1/2 (d) 3/8 (e) none of the foregoing

Solution :

(I) a^3/4b + b/8c^2 + (1+c)/2a = a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a. For positive reals AM >= GM and the equality occurs when all the numbers are equal. Thus, a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a >= 5(a^3/4b*b/8c^2*1/4a*1/4a*c/2a)^1/5 = 5/4. Thus, p = 5/4 when a^3/4b = b/8c^2 = 1/4a = c/2a.

(II) a+b=8-2c; a^2+b^2 >= 1/2*(a+b)^2 = 1/2(8-2c)^2 => 25-2c^2 = 1/2(8-2c)^2 => 1/2 <= c <= 7/2. Thus, q is 7/2 when a=b.

Hence, choice (b) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 45 (Solved first by Apple)---------------------------------------------------------------------------

Question:

ABCD is a convex quadrilateral in which <(BAC) = <(CBD) = 30 deg, <(CAD) = 60 deg, <(CDB) = 15 deg. If E is the point of intersection of AC and BD, then measure of <(BEA) in degrees (is)

(a) can not be determined (b) 75 (c) 105 (d) 120 (e) none of the foregoing

Solution :

Draw a circle passing O through each point B, C, and D. In any circle the angle subtended by an arc at the centre is twice the angle subtended at the perimeter. Since, <(BAC) = 2*(<(CDB)) and <(CAD) = 2*(<(CBD)) => A is the centre of the circle O.

Thus, AB = AD is the radius of circle O => <(ABD) = 1/2*(180-90) = 45 deg => <(BEA) = 105 deg

Hence, choice (c) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 46 (Solved first by buck_was, krsh.vik, v-factor, scarletyoke, dhruv.dingliwal)---------------------------------------------------------------------------

Question:

'All India Pagalguy Meet of year 2008' is in June (which has 30 days), but Allwin forgot which day, so he asked around. Rohit said that the date was an odd number; Apurv claimed it was greater than 13. Divya declared it was not a perfect square, while Sonam swore it was a perfect cube. Finally, Grand-ma tells Allwin the date was less than one-fourth her (Grand-ma's)age, which Allwin knew to be 68. Yesterday Allwin learned that only one of them had told the truth! If the date of the All India PG Meet is D (numerical value), then

(a) D is uniquely determinable(b) D can have exactly 2 values (c) D doesn't exist (d) D has atleast 4 values (e) none of the foregoing

Solution :

The question says that only one of them spoke the truth! Let's see if Rohit spoke the truth or not. If he has then the date D is an odd number <= 13 (since Apurv must have lied), and D > 17 (Grandma must have lied too in this case). Contradiction, and hence we have no such value of D.

If Apurv spoke the truth, then D > 13 and D >= 17 (Grandma must have lied). Since, Rohit and Divya also lie => D is a even perfect square. Not possible as D <= 30.

Please check yourself that if Divya or Sonam speak truth then no value of D exists.

Let's check if Grandma has spoken the truth or not. If she has then D < 17. Rohit lies => D is even, Divya lies => D is a perfect square, Sonam lies => D is not a perfect cube => D can be 4 or 16 but as Apurv also lies => D = 4 is the only possibility.

Hence, choice (a) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 47 (solved correctly only by selective_sloth, kudos to him for that!)---------------------------------------------------------------------------

Question:

Let m be the largest positive term of an harmonic progression whose first two terms are 2/5 and 4/9. A real number r satisfying m/2-1/n < r <= m+1/n, for every positive integer n, is best described by:

(a) 1 < r < 5 (b) 2 < r <= 4 (c) 1 < r <= 5 (d) 2 <= r <= 4 (e) none of the foregoing

Solution :

If 2/5 and 4/9 are the first two terms of the harmonic progression then 5/2 and 9/4 are the terms of the corresponding AP => common difference of the AP is -1/4 => we want 5/2 + (n-1)*(-1/4) as positive as well as minimum. Thus, n = 10 and the AP term is 1/4. m = 4. Please note that our HP consists of 10 terms only.

Now, 2-1/n < r <= 4+1/n for all positive intger n. Putting n = 1 we get 1 < r <= 5. Putting n = 2 we get 3/2 < r < 9/2 etc. If r is more than 4 then r <= 4 + 1/n fails for some n (e.g. if r = 4.01 then n = 101). If r is less than 2 then 2-1/n < r fails for some n (e.g. if r = 1.99 then n = 100)

Note that 1/n > 0 always for any choice of positive integer n. Thus, 2-1/n < 2, and 4 <= 4+1/n is true for all postive integers n.=> the best description of r contains 2 and 4 as well besides the range (2, 4).

Hence, choice (d) is the correct option

---------------------------------------------------------------------------Solution to Quantitative Question # 48 (solved first by vineetvijay, mohit_ranka, lapsy, innocent_123, Apple )---------------------------------------------------------------------------

Question:

The set S has 5 elements. In how many ways can one select two (possibly identical) subsets of S whose union is S?

(a) 32 (b) 63 (c) 64 (d) 93 (e) 122

Solution :

Let the subsets of S be A and B. For each element in S we have three choices (it can belong to either of A, B or both). That gives each pair of subsets twice except for the case A = B = S. Hence, we can select 2 subsets in (3^5 + 1)/2 ways.


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