bunuel maths

10. SUM OF INTEGERS: If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is: A. A+1 inquiry B. A+5 C A+25 D 2A E. 5A

Sum=A, next 5 consecutive will gain additional 5*5=25, so sum of the next five consecutive integers in terms of A is: A+25 Answer: C.

8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures? A. 20 B. 25 C. 40 D. 45 E. 75

Average=50, Sum of temperatures=50*5=250 As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25 Answer: B.

2. THE PRICE OF BUSHEL: The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat? (A) $4.50 (B) $5.10 (C) $5.30 (D) $5.50 (E) $5.60

Note that we are not asked in how many days prices will cost the same. Let y be the # of days when these two bushels will have the same price. First let's simplify the formula given for the rate of decrease of the price of wheat: 2‾√∗x−x=1.41x−x=0.41x, this means that the price of wheat decreases by 0.41x cents per day, in y days it'll decrease by 0.41xy cents; As price of corn increases 5x cents per day, in y days it'll will increase by 5xy cents; Set the equation: 320+5xy=580−0.41xy, solve for xy --> xy=48; The cost of a bushel of corn in y days (the # of days when these two bushels will have the same price) will be 320+5xy=320+5∗48=560 or $5.6. Answer: E.

3. LEAP YEAR: How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year? A. 1 B. 2 C. 3 D. 4 E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4 Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2 So, we are looking for such n (# of people), when 1-(3/4)^n>1/2 n=3 --> 1-27/64=37/64>1/2 Thus min 3 people are needed. Answer: C.

7.. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x? I. -50 II. 25 III. 50 A. I only B. II only C. III only D. I and II only E. I and III only

Re-arrange and factor for x^2: (x2−25)(x2−4)=0. So, we have that x=5, x=−5, x=2, or x=−2. −50=5∗(−5)∗2; 50=5∗(−5)∗(−2). Only 25 is NOT a product of three possible values of x Answer: B.

5. If x^2 + 2x -15 = -m, where x is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero? A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7

Re-arrange the given equation: −x2−2x+15=m. Given that x is an integer from -10 and 10, inclusive (21 values) we need to find the probability that −x2−2x+15 is greater than zero, so the probability that −x2−2x+15>0. Factorize: (x+5)(3−x)>0. This equation holds true for −5<x<3. Since x is an integer then it can take the following 7 values: -4, -3, -2, -1, 0, 1, and 2. So, the probability is 7/21=1/3. Answer: B.

3. If a and b are positive numbers, such that a^2 + b^2 = m and a^2 - b^2 = n, then ab in terms of m and n equals to: A.√2‾‾‾‾‾ m−n B. √2‾‾‾mn C. √2‾‾‾‾‾‾‾m2−n2 D. √2‾‾‾‾‾‾‾n2−m2 E. √2‾‾‾‾‾‾‾m2+n2

Sum the two equations: 2a2=m+n; Subtract the two equations: 2b2=m−n; Multiply: 4a2b2=m2−n2; Solve for ab: ab=(√‾‾‾‾‾‾‾m2−n2)/2 Answer: C.

6. PROBABILITY OF DRAWING: A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn? (A) 2/27 (B) 1/9 (C) 1/3 (D) 4/27 (E) 2/9

This is with replacement case (and was solved incorrectly by some of you): P=2∗39∗29=427 We are multiplying by 2 as there are two possible wining scenarios RW and WR. Answer: D.

8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m? A. -21 B. -20 C. -19 D. -1 E. None of the above

Given m3+380=380m+m. Re-arrange: m3−m=380m−380. m(m+1)(m−1)=380(m−1). Since m is a negative integer, then m−1≠0 and we can safely reduce by m−1 to get m(m+1)=380. So, we have that 380 is the product of two consecutive negative integers: 380=−20∗(−19), hence m=−20. Answer: B.

10. If f(x) = 2x - 1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ? A. -145 B. -24 C. 24 D. 145 E. None of the above

f(x)=2x−1, hence f(n2)=2n2−1. g(x)=x2, hence g(n+12)=(n+12)2=n2+24n+144. Since given that f(n2)=g(n+12), then 2n2−1=n2+24n+144. Re-arranging gives n2−24n−145=0. Next, Viete's theorem states that for the roots x1 and x2 of a quadratic equation ax2+bx+c=0: x1+x2=−b/a AND x1∗x2=c/a. Thus according to the above n1∗n2=−145. Answer: A.

4. ADDITION PROBLEM: AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C? (A) 1 (B) 3 (C) 7 (D) 9 (E) Cannot be determined

AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111. So, A=1. 1B+CD=111 C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111). C=9 Answer: D.

5. RACE: A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B's speed in m/s? (A) 12 (B) 14 (C) 16 (D) 18 (E) 20

Let x be the speed of B. Write the equation: (480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat) (480-48)/x-6=(480-144)/x+2 x=12 First of all thanks for pointing out the typo. Edited the post above. Second it's funny but the equation with typo also gives the correct answer x=12: 392/x-6=296/x+2 --> 96/x=8 x=12 432/x-6=336/x+2 --> 96/x=8 x=12 Answer: A.

9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8? A. 25% B 50% C 62.5% D. 72.5% E. 75%

N=n*(n+1)*(n+2) N is divisible by 8 in two cases: When n is even: No of even numbers (between 1 and 96)=48 AND When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers Total=48+12=60 Probability=60/96=0.62 Answer: C

6. If mn does not equal to zero, and m^2n^2 + mn = 12, then m could be: I. -4/n II. 2/n III. 3/n A. I only B. II only C. III only D. I and II only E. I and III only

Re-arrange: (mn)2+mn−12=0. Factorize for mn: (mn+4)(mn−3)=0. Thus mn=−4 or mn=3. So, we have that m=−4n or m=3n. Answer: E.

2. The equation x^2 + ax - b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b? A. -64 B. -16 C. -15 D. -1/16 E. -1/64

Since one of the roots of the equation x2+ax+15=0 is 3, then substituting we'll get: 32+3a+15=0. Solving for a gives a=−8. Substitute a=−8 in the first equation: x2−8x−b=0. Now, we know that it has equal roots thus its discriminant must equal to zero: d=(−8)2+4b=0. Solving for b gives b=−16. Answer: B.

1. If x=√4‾‾‾‾‾‾‾‾‾(x3+6x2), then the sum of all possible solutions for x is: A. -2 B. 0 C. 1 D. 3 E. 5

Take the given expression to the 4th power: x4=x3+6x2; Re-arrange and factor out x^2: x2(x2−x−6)=0; Factorize: x2(x−3)(x+2)=0; So, the roots are x=0, x=3 and x=−2. But x cannot be negative as it equals to the even (4th) root of some expression (√‾‾‾‾‾‾‾‾‾‾expression≥0), thus only two solution are valid x=0 and x=3. The sum of all possible solutions for x is 0+3=3. Answer: D.

1. THE SUM OF EVEN INTEGERS: The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful: The number of terms in this set would be: n=(k-1)/2 (as k is odd) Last term: k-1 Average would be first term+last term/2=(2+k-1)/2=(k+1)/2 Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1) (k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159 Answer E. MY NOTES ABOUT AP: ARITHMETIC PROGRESSION Sequence a1, a2,...an, so that a(n)=a(n-1)+d (constant) nth term an = a1 + d ( n - 1 ) Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2 Special cases: I. 1+2+...+n=n(1+n)/2 (Sum of n first integers) II. 1+3+5+... (n times)=n^2 (Sum of n first odd numbers). nth term=2n-1 III. 2+4+6+... (n times)=n(n+1) (Sum of n first even numbers) nth term=2n SOLUTION WITH THE AP FORMULA: Sequence of even numbers First term a=2, common difference d=2 since even number Sum to first n numbers of AP: Sn=n*(a1+an)/2=n(2*2+2(n-1))/2=n(n+1)=79*80 n=79 (odd) Number of terms n=(k-1)/2=79 k=159 OR Sum of n even numbers n(n+1)=79*80 n=79 k=2n+1=159

7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE: What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3? A) 1.4 B) sqrt (2) C) 1.7 D) sqrt (3) E) 2.0

This is tough: First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1). Now we can do this by finding the equation of a line perpendicular to given line y=34∗x−3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way. There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it. We know the formula to calculate the distance between two points (x1,y1) and (x2,y2): d=(x1−x2)2+(y1−y2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ BUT there is a formula to calculate the distance between the point (in our case origin) and the line: DISTANCE BETWEEN THE LINE AND POINT: Line: ay+bx+c=0, point (x1,y1) d=|ay1+bx1+c|a2+b2‾‾‾‾‾‾‾√ DISTANCE BETWEEN THE LINE AND ORIGIN: As origin is (0,0) --> d=|c|a2+b2‾‾‾‾‾‾‾√ So in our case it would be: d=|3|12+(34)2‾‾‾‾‾‾‾‾‾√=125=2.4 So the shortest distance would be: 2.4−1(radius)=1.4 Answer: A. OR ANOTHER APPROACH: In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: (x−a)2+(y−b)2=r2 If the circle is centered at the origin (0, 0), then the equation simplifies to: x2+y2=r2 So, the circle represented by the equation x2+y2=1 is centered at the origin and has the radius of r=1‾√=1. Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1). So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line y=34x−3. The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> leg1=4. and the value of y for x=0 (y intercept) --> x=0, y=-3 --> leg2=3. So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: heightleg1=leg2hypotenuse --> height3=45 --> height=2.4. Distance=height−radius=2.4−1=1.4 Answer: A. You can check the link of Coordinate Geometry below for more. links: http://gmatclub.com/forum/bunuel-signature-collection-all-in-one-with-solutions-146628.html http://gmatclub.com/forum/math-coordinate-geometry-87652.html

9. If x=(√‾5−√‾7)^2, then the best approximation of x is: A. 0 B. 1 C. 2 D. 3 E. 4

x=(√‾5−√‾7)2=5−2√‾‾‾35+7=12−2√35. Since √35≈6, then 12−2√35≈12−2∗6=0. Answer: A.

4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ? A. -39 B. -9 C. 0 D. 9 E. 39

−3x2+12x−2y2−12y−39=−3x2+12x−12−2y2−12y−18−9=−3(x2−4x+4)−2(y2+6y+9)−9= =−3(x−2)2−2(y+3)2−9. So, we need to maximize the value of −3(x−2)2−2(y+3)2−9. Since, the maximum value of −3(x−2)2 and −2(y+3)2 is zero, then the maximum value of the whole expression is 0+0−9=−9. Answer: B.