Solutions Test 2

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QQAD, Practice test 2: CAT 2007 (Solutions)

Instructions:

1) The duration of this test is 50 minutes and the test is meant to be taken in one-go without any break(s).

2) This test has 2 sections, A and B with 12 and 20 questions respectively. Each question in section A and B carries 5 marks and 2 marks respectively and has 5 options. However, for section A the 6th unmentioned choice (f) is none of the foregoing. Mark (f) in the answer sheet for section A if you believe that the answer is not among the choices. For wrong answers, penalty per question for section A and B is -1 and -1/2 mark respectively.

3) Use of slide rule, log tables and calculators is not permitted.

4) Use the blank space in the question paper for the rough work.

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Keys to Practice Test 2

1) D2) F3) F4) C5) B6) D7) E8 C9) D10)A11)B12)B13)A14)C15)D16)E17)A18 A19)A20)B21)D22)B23)E24)D25)D26)A27)A28 B29)E30)D31)A32)D

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Part A

1) Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

Solution: Area of square = x^2, let area of the triangle AED be m*(x^2).1/2*y*x =m*(x^2) => y= 2*m*x.let the base of the other triangle be k then slope will be same,hence x/k=y/x=>x/k=(2*m*x)/x or, k= x/(2*m)hence area of the other triangle is (1/2)*x*x/(2*m) = (x^2)/4m

The ratio of the area of the other small right triangle to the area of the square is[(x^2)/4m]/x^2 = 1/4m

Hence, option(d) is the right answer

2) Call a 7-digit telephone number MEMORABLE if the prefix sequence

is exactly the same as either of the sequences or (possibly both). Assuming that each can be any of the ten decimal digits 0,1,2,...9, the number of different memorable telephone numbers is

(A) 19810 (B) 19910 (C) 20000 (D) 20010 (E) 20100

Solution:There are 10000 ways to write the last 4 digits and among these there are 9990 whose all the digits are not same. For each of these there are 2 ways to adjoin three digits d1d2d3 to obtain a memorable number. There are ten

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memorable numbers for which the last 4 digits are the same. Thus, in all 19990 numbers.


3) Consider the triangular array of numbers with 0,1,2,3, ... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.

Let denote the sum of the numbers in row . What is the remainder when is divided by 100?

(A) 12 (b) 30 (c) 50 (d) 54 (e) 62

Solution: f(100) = 2^1+2^2+2^3+.......................+2^99 = 2*(2^99-1)/(2-1) = 2*(2^99-1)2^100 = 1 mod (25) by fermat’s theorem => 2^100 – 2 leaves remainder of 24 on division by 25. Also 2^100 – 2 leaves remainder 2 on division by 4. Since 25 and 4 are co-prime, remainder of 2^100 – 2 on division by 100 is 74.

Hence, option(f) is the right answer

4)

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Solution:Roll the side of the cylinder out, the strip becomes a parallelogram with base 3 and height 80.

Hence, option(c) is the right answer

5)

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Solution:Area of shaded region = (8/13)* area of unshaded region.=> let the acute angle be 'x'.now (2*x/2*(1^2)) + ( 2*((pi-x)/2)*(2^2 - 1^2)) + (2*x/2*(3^2 - 2^2)) = 8/13*[ {2*((pi-x)/2)*(1^2)} + {2*x/2*(2^2 - 1^2)} + {2*((pi-x)/2)*(3^2 - 2^2)}solving this we get, x = pi/7

Hence, option (b) is the right answer

6) Let denote the greatest integer not exceeding . Let and satisfy the simultaneous equations

If is not an integer, then is

Solution:y = 2[x]+3 and y = 3[x-2]+5equating both we get; 2[x]+3 = 3[x-2]+5

This is only satisfied if x lies between 4 and 5, and at that time y = 11.Hence, x+y is between 15 and 16.


7) In a narrow alley of width a ladder of length is placed with its foot at point between

the walls. Resting against one wall at , the distance above the ground makes a angle with the ground. Resting against the other wall at , a distance above the ground, the ladder makes a angle with the ground. The width is equal to

Solution:tan(75 degree) = tan (45 degree + 30 degree) = (tan45 + tan 30)/(1- tan45*tan30)

= [1+1/v3] / [ 1-1/v3] = 2+v3.

now, w = k+[h/(2+v3)] = k+(2-v3)*h

also a =[2*v2*h]/[v3+1]

k = a/v2 = 2h/(v3+1) = (v3-1)h

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hence w = k+(2-v3)*h = (v3-1)h + (2-v3)*h = h

Hence, option (E) is the right answer

8) The negation of the proposition "For all pairs of real numbers , , if , then "

is: There are real numbers such that

Solution:The negation of if A then B is A and not B.

Hence, option (C) is the right answer

9) Given a circle of radius 2, there are many line segments of length 2 that are tangent to the circle at their midpoints. The area of the region consisting of all such line segments is

Solution:

The area enclosed will be a ring between the circles with radii 2 and ((2)^2 + (2/2)^2)^1/2 = v5

Hence, option (D) is the right answer

10) Consider the sequence defined recursively by (any positive integer), and

For which of the following values of must ?

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(A) 16 (b) 18 (c) 15 (d) 14 (e) 17

u1 = a; u2 = (-1)/(a+1); u3 = (-1-1/a); u4 = a.

Hence the pattern u1 = u4 =u7 = u10 = u13 = u16 = u19= ...............= a


11) Let be the set of points in the coordinate plane, where each of and may be , , or . How many distinct lines pass through at least two members of ?

Solution:Total number of points = 9. These 9 points are arranged three in a row. i.e. 3 rows of 3 points each.Total number of straight lines segment= 9C2 = 36

We have to subtract two straight line segment corresponding to each such combination in which three points are in a row i.e. three horizontal, three vertical lines and two diagonals = total 8 such lines hence subtract 16 from the above calculated total number of lines to get the distinct straight line.

= 36 - 16 = 20.

Hence, option (B) is the right answer

12) An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(I) the selection of four red marbles; (II) the selection of one white and three red marbles; (III) the selection of one white, one blue, and two red marbles; and (IV the selection of one marble of each color.

What is the smallest number of marbles satisfying the given condition?

Solution:

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Rc4=W*Rc3=W*B*Rc2=R*W*B*Gwhere R=no. of red balls, W=no. of white balls, B=no. of blue balls, G=no. of green ballstherefore, R=3B+2, R=2G+1, R=4W+3

for W=1 will not satisfy(B=5/3 which is not an integer)for W=2 we get smallest no. of marbles(R=11,B=3,G=5)


Part B

13) A right circular cone of volume , a right circular cylinder of volume , and a sphere of volume all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then

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Solution:

A = (2/3)*pi*(r^3); M = 2*pi*(r^3); C = (4/3)*pi*(r^3).=> A-M+C =0

Hence, option (A) is the right answer

14) Three cards, each with a positive integer written on it, are lying face-down on a table. Twinkle, Raveena, and Akshay are told that

(a) the numbers are all different, (b) they sum to 13, and (c) they are in increasing order, left to right

First, Twinkle looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Akshay looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Raveena looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?

(A) 2 (B) 3 (C) 4 (D) 5 (E) can not be determined

Solution:Options are: (1, 2, 10); (1, 3, 9); (1, 4, 8); (1, 5, 7);(2, 3, 8); (2, 4, 7); (2, 5, 6);and (3, 4, 6);Now eliminate, 3 in card seen by Twinkle, Otherise she'll definiely know other digits.Now, eliminate 6, 9, 10 in the card seen by Akshay. Otherwise he'll definiely know other digits.Now, eliminate 3, 5 in the card seen by Raveena. Otherwise she'll definiely know other digits. So, the card seen by Raveena must be 4.


15) The midpoints of the sides of a regular hexagon are joined to form a smaller hexagon. What fraction of the area of is enclosed by the smaller

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hexagon?

Solution:In regular hexagon joining the centre to the vertex, 6 equilateral triangles we can get with sides = hexagon sides(a, say).

Interior angle = 120 deg. The next hexagon sides to be equal to x(say) = a*sqrt(3)/2. So, we get the ratio = 3/4.


16) When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?

Solution:When we add another no. it is the median if it is 3,else any other no. 4 will be the medianCase1) x=3

Median=3,Mode=2,Mean=28/7=4Median=Mean-1Mode=Median-1Therefore, it is an arithmetic progression

Case2) x is not equal to 3

Median=4,Mode=2,using AP formula mean=6.Therefore (25+x)/7=6therefore x=17Hence, sum of all possible real values of x is 17 + 3= 20


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17) A sum of money gets doubled in little over 4 years at X% SI. Had the rate been Y% the sum would have been thrice in little less than 7 years. Then the least possible value of (Y-X), given that it is an integer will be

(A) 4 (B) 5 (C) 2 (D) 6 (E) 3

Solution:S > S*4*(X)/100 and 2S < S*7*(Y)/100


18) If x and y are integers then the equation 5x+19y = 64 has:

(A) a solution for 250 < x < 260 (B) no solution for x > 250 and y > -100(C) no solution for x < 300 and y < 0 (D) a solution for -59 < y < -56 (E) Exactly two of the foregoing

Solution:19y = (64 – 5x); x = 9 satisfies this. Thus, x = 19K + 9 must be a solution. When K = 13, x = 256


19) Let and be distinct elements in the set

What is the minimum possible value of

(A) 34 (B) 50 (C) 30 (D) 40 (E) 32

Solution:(-2-5+4+6)^2 + (-7-3+13+2)^2 = 9+25 =34.


20) The sum of the lengths of the twelve edges (all integral numbers) of a rectangular box is , and the distance from one corner of the box to the farthest corner is . The total

surface area of the box is

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Solution:4*(l+b+h) = 140 => (l+b+h) = 35 also l^2+b^2+h^2 = 21^2

Total surface area = 2*(lb+bh+lh) = (35^2) - (21^2) = 784.


21)

Solution:Let AB = a. <EBA = 15 and cos(15) = (v3+1)/2v2.

Hence, option (D) is the right answer 22) A positive integer has 60 divisors and has 80 divisors. What is the greatest integer such that divides ?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Solution:N must have a divisor as 49 and not 343.


23) For each real number , let be the minimum of the numbers , ,

and . Then the maximum value of is

Solution:plot the graph of f(x)= 4x+1; f(x) = x+2; f(x) = -2x+4

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f(x) = x+2; f(x) = -2x+4 intersect at (2/3,8/3) and f(x)= 4x+1; f(x) = x+2 intersect at (1/3,7/3) hence max f(x) is at 8/3.


Directions: (24-25) is based on the information below

A, B, and C are three persons on a circular track. They start moving simultaneously with constant speeds such that after some time T, A, B, C are at initial positions of B, C and A respectively (none of them having completed a full revolution during the time).

24) If A, B and C can complete a full rotation of a circular track in 4, 6 and 12 minutes respectively, then T (in seconds) is

(A) 60 (B) 75 (C) 90 (D) 120 (E) 150

Solution:Let n1, n2, n3 be the number of revolutions made per second => (n1+n2+n3)T = 1 => (1/4+1/6+1/12)*T/60 = 1 => T = 120 seconds


25) If A, B and C come again to their initial positions simultaneously next time after 1 hour; then T can not be equal to

(a) 10 minutes (b) 12 minutes (c) 15 minutes (d) 18 minutes (e) exactly two of the foregoing

Solution:They again come to their initial position after 60 minutes => they must make an integral number of revolutions in 60 minutes. => (m+n+p)T = 60


Directions (26-27) is based on the information below

A group of students are asked to take atleast one subjects from Physics, Chemistry, and Mathematics for their courses.

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26) The number of students taking exclusively one subject is 50 while the number of students taking exclusively two subjects is 30. The number of students taking physics, chemistry, and mathematics is 45, 44, and 48 respectively. The number of students taking all subjects is

(A) 9 (B) 10 (C) 11 (D) 13 (E) 15

Solution:Let the number of students taking all the three subjects be x;

then 3*x + 50 + 2*30 = 137 => 3x = 27x =9


27) The number of students taking physics, chemistry and mathematics is 42, 45 and 49 respectively while the number of students taking physics or chemistry is 65, that taking chemistry or mathematics is 70 and that taking physics or mathematics is 72. What is the number of students not taking all 3 subjects?

(A) 71 (B) 77 (C) 58 (D) 66 (E) can not be determined

Solution:A+B – A&B = 65; B+C – B&C = 70; C+A – C&A = 722(A+B+C) – (A&B + B&C + C&A) = 207Thus, AUBUC - (A&B + B&C + C&A) = 71


28) If the base representation of a perfect square is , where , then equals

Solution:ab3c = 512a + 64b + 24 + c. Any odd perfect square is of the form 8n+1. Thus, c can’t be 3.Any even perfect square is of the form 16k or 16k+4. Thus, c can’t be 0 or 4.Looking for c = 1 we find 729 and 3481 indeed satisfy this.


29) How many non-congruent right triangles are there such that the perimeter in and the area in are numerically equal?

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Solution:Let a, b be the non-hypotenuse sides of the triangle => ab/2 = a + b + (a^2 + b^2)^1/2

(ab/2 – a – b)^2 = a^2 + b^2 => (ab)^2 = 4ab(a+b) - 8ab => ab = 4(a+b–2)

There are infinitely many non-congruent right angles triangles.


Directions: (30-31) is based on the information below

A shopkeeper sells 3 eggs per rupee. He noted that if he sells 4 eggs per rupee, his rate of sales get double i.e. during the same amount of time his sale in later case would be of twice the amount of the sale in former case. If he further increases an egg a rupee, his rate of sales gets double. Let x, y, z be the number of hours during which he sells 3, 4 and 5 eggs per rupee respectively.

30) If his profit in all the three cases remain same, then which among the following holds true?

(A) 2/y = 1/x + 2/z (B) 2/y = 2/x + 1/z (C) 4/y = 1/x + 4/z (D) 4/y = 4/x + 1/z (E) 4/y = 1/x + 1/y

Solution:Let the shopkeeper bought n eggs per rupee and let he sold eggs worth Rs k per hour when he offered 3 eggs per rupee. Then, the profits when eggs are sold 3/rupee, 4/rupee, 5/rupee are 3xk(1/3-1/n), 4y(2k)(1/4-1/n), 5z(4k)(1/5-1/n) respectively.

3xk(1/3-1/n) = 4y(2k)(1/4-1/n) = 5z(4k)(1/5-1/n)


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31) During the same amount of time, his profits when he sells the egg at 3 or 4 per rupee will be same if he buys per rupee

(A) 5 eggs (B) 6 eggs (C) 12 eggs (D) 15 eggs (E) 18 eggs

Solution:3xk(1/3-1/n) = 4y(2k)(1/4-1/n) and x = y => n = 5


32)

Solution:CF = CB as length of tangents from same point on a circle will be same.=> CF = 2.

Let EA = EF = x, in rt. angled triangle CED;ED^2+CD^2 = EC^2or (2-x)^2+2^2= (2+x)^2=>x=1/2.CE=2+1/2=5/2


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Solutions Test 2