Advanced Quant

How many odd integers from 1 to 200 (both inclusive) have odd number of factors? A) 3 B) 4 C) 5 D) 6 E) Greater than 6

In order for a number to have an odd number of factors it must be a perfect square. 36 = 6^2 1*36 2*18 3*12 4*8 6*6 (1,2,3,4,6,8,12,18,36) 9 total odd number of factors Where as factors of 12... 2*6 3*4 12*1 There are 6 factors even. This means that we are going to take all of the off perfect squares from 1 to 200. 1,9,25,49,81,121,169 and the corresponding integers are 1,3,5,7,9,11,13 (more than 6).

Triplets Adam, Bruce, and Charlie enter a triathlon. There are none competitors in the race. If every competitor has an equal chance of winning, and three medals will be awarded, what is the prob that at least two of the triplets will win a medal?

Number of ways awards can be given: 9c3 9!/6!3! = 84 At least two triplets win, so 2 cases (all triplets win or two triplets win) # ways all triplets can win is 1. # ways 2 triplets can win is... 3c2 = 3 now this is times the way that the other 6 athletes can win the remaining award.. 6c1 = 6!/5! = 6 This is 6*3 = 18 Total prob (2 win + 3 win) = 18 + 1 19/84

How many integers are between 1 and 400, inclusive, and are not divisible by 4 and do not contain nay 4s?

Step 1: How many integers are not divisible by 4. Notice that 1,2,3,[4],5,6,7,[8],... 1/4th of all integers from 1 to 100 are divisible by 4. So 400-100 = 300 is how many integers are not divisible by 4. Step 2: Find the number of integers that do not contain 4. Find the integers with 4s. Note: we will not count integers that were eliminated in the previous step. x4y. - tens spot 41,42,43,45,46,47,49 7 integers per 100, so 28 in tens. xy4 - units spot 14,34,54,74, 94 5 integers per 100, so 20 in units. Therefore, there are 400-100-28-20 = 252 integers that satisfy the conditions.

If a and b are consecutive integers, is ab divisible by 30? (1) a^2 is divisible by 25. (2) 63 is a factor of b^2

The prime factorization of 30 is 2*3*5. ab must have these factors. Because a and b are consecutive, one must be even and be divisible by 2. Therefore we just need to see if (1) and/or (2) provide the other factors, 3 and 5.. C

Guessing tactics

The two DS statements are the same (go with D or E) If one statement is inside the other (subset of one another), then eliminate C and/maybe the broader statement. (1) x > 10! (2) x> 50! If one statement adds no info to the other, then eliminate c (1) y^x + y^-x = z (z) y<0 If it is very obvious that the combined statements are sufficient, but you cannot eliminate one of the statements from being individually sufficient, then eliminate c and e. If a yes/no question involves inequalities with variables to cross multiply, then go with c or e. If you have no idea, judge by appearance. Does it have the right variables? Could it likely be manipulated? Good energy?

If x and k are both ints, x>k, and x^-k = 625, what is x? (1) |k| is a prime number. (2) x+k >20

There are several options. 5^-4 25^-2 625^-1 (1) limits it, so suffint (2) insufficient

Is a<0 (1) a^3 <a^2 + 2a (2) a^2 > a^3

Try negatives, positives, and fractions... E

The average of the original six prices for six coats at a clothing tore was 85. After two of the six coats were each discounted by 20% the average price of the six coats was 76. Was the coat with the lowest original price one of the two coats that were discounted? (1) One of the discounted coats was the one with the highest original price (2) Before the discount, none of the coats had a price greater than 180.

Was the coat with the lowest price one of the ones that was discounted? (p1+..P6)/6 = 85 p1 + .... + p6 = 510 with discount p1+...+p6 = 456 so two coats discounted at 20% each led to a decrease in 54 dollars. This means that the sum of the original prices of these coats is 5 *54 = 270. (1) one of the discounted coats was the most expensive. 510 -270 = 240. Case 1 If the most expensive coat cost 269 and the other 1, then the other coats cost 240/4 = 60, so yes. Case 2 If the discounted coats were 130 and 140, then the remaining coats would have an avg of 60, and they would be cheaper, so no.. (2) none of the discounted coats had prices over 180. 270-180 = 90; however, 90 is greater than the new avg. So, no. The lowest price is not in this set. Therefore, the answer is B.

If xy>0, is (5^x)^1/y > 25 (1) 2(xy^4/x^2)^2 = 16y^5/x^2 (2) s>2y

c

What is the sum of 9,999^2 and 10001^2? or 198* 202

recognize that these can be written as quadratic components... (a-b)^2 + (a+b)^2, respectively. This will result in 2(a^2 +b^2). (10000-1)^2 + (10000+1)^2 2(10000^2 +1^2). Units digit will be 2, if several options end in 2 then do math. 198*202... (200-2)(200+2) a-b*(a+b) = a^2 -b^2 200^2 -2^2 = 40000-4 = 39996

If n is an integer and (-3)^4n = 3^(7n-3), then n = ?

since n is an int and 4n is an int, the even exponent will hide the sing. We can treat the bases the same. 3^4n = 3^(7n-3) n = 1.

If A,b,c are ints, is abc divisible by 4? (1) a+b+2c is divisible by 2 (2) a+2b+c is odd

(1) c must even, and a+b must be even, but they could both be even or (odd and odd). This uncertainty could result in abc being odd*odd*even = even only if c is divisible by 4. (2) a +2b + c is odd.. b must be even but a and c must have opposite signs. This means that abc may not be divisible by 4. (1) and (2) a = odd, b = odd, c = even could result in abc not being divisible by 4. E

If x and y are pos ints, what is the remainder when 5^x is divided by y? (1) x = 3 (2) y = 4

(1) insufficent (2) sufficeint Pattern x 5^x remainder 1 5 1 2 25 1 3 125 1 4 625 1 No matter the exponent, 5^x divided by any positive int will always be 1.

If m and n are positive integers, is n a multiple of 24? (1) n = (m+7)!/(m+3)! (2) n is a multiple of (m+4)

(1) sufficient simplifies to (m+7)(M+6)(M+5)(m+4), which is the product of 4 consecutive integers. This will always be a multiple of 24. 4! = 4*3*2*1 = 24. test m = 1, 24 goes in 70 times.

If x is an int, what is the remainder when x is divided by 5? (1) x^2 has a remainder of 4 when divided by 5 (2) x^3 has a remainder of 2 when divided by 5.

(1) x = 2 would have a remainder of a remainder of 4. x = 3 would have a remainder of 5. Insufficient (2) B

f x is a positive integer, what is the remainder when x is divided by 7? (1) The remainder when x is divided by 4 is 3 (2) The remainder when x is divided by 5 is 1

(1) x/4 gives remainder 3, so x = 3 or 7. insufficent x = 4q +3 (2) x/5 gives remainder 1, so x = 6, 11,... insufficient. x = 5q + 1 (1) and (2) 5q + 1 = 1, 6, 11, 16, 21, 26, 31 4q + 3 = 3, 7, 11, 15, 19, 23, 27, 31, 11 and 31.

List A contains 5 positive ints, and the average of the ints is 7. If the inters 6,7,8 are in the list, what is the range of list A? (1) The integer 3 is in List A (2) The largest term in List a is greater than 3 times and less than 4 times the size of the smallest term.

(1) x1 + .. + x5 /5 = 7 x1 + .. +x5 = 35 6+7+8 = 21 + 3 = 24, so missing digit is 11. 11-3 = 8 (2) Since the avg of 6,7,8 is 7, we know that the reaminging digits must also have an avg of 7. x+y /2 = 7, so x+y = 14. Visualize this and see if there are two numbers that satisfy (2) 2 3 4 5 6 (7 mean) 8 9 10 11 12 2 and 12 3 and 11 -- satisfy 4 and 10 ...

Is x a multiple of 12? (1) √(x-3) is odd (2) x is a multiple of 3

(1) √(x-3) = odd = 1,3,5,7,9. x-3 = 1,9,25,49,etc x = 4,12,28,51, etc insufficent (2) x is multiple of 3. 3,9,12,etc insufficent (1) and (2) x = 12 (1) works with (2) C

If √x is a prime number, what is the value of x? (1) -16 < -3x +5 < 22 (2) x^2 is a two digit prime

-16 < -3x+5<22 -21 < -3x < 17 7 > x > -17/3 Let x = 4, whose sqrt is prime. Only number that does this sufficient. (2) insufficient.

Testing cases numbers

-2, neg even -3/2 improper neg frac -1 neg ,-1/2 proper neg frac 0, zero 1/2 proper frac 1 pos 3/2, improper frac 2 pos even pos/neg even/odd improper frac/frac

How many numbers from 1 to 1000, both inclusive, have digits repeated ? A. 9B. 81C. 262D. 648E. 738

1-99➡9 numbers 1000➡1 number100-999➡252 numbers: xxx=9*1*1=9 xxy=9*1*9=81 xyx=9*9*1=81 yxx=9*9*1=81 252+9+1=262 numbersC

a,b,c,d are pos ints. If a/b has a reminder of 9 and c/d has a reminder of 10, what is the min val of bd?

110

The avg of 6 numbers is 18 and the median of the 6 numbers is 16. What is the min possible value for the greatest number?

22 x1 + ..+x6 = 108 x3+ x4 /2 = 16 Only (16,16) and (15, 17) Since we are looking for min max, we want to choose the largest option. 16*4 + x5+x6 = 108 x5+x6/2 = 44 x5 = 22 = x6

A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?

60%

Find sum of An = 1/n(n+1) to n = 100

A1 = 1/2 A2 = 1/6 Sum = 2/3 A3 = 1/12 Sum = 9/12 = 3/4 A4 = 1/20 Sum = 4/5 A5 = 1/30 Sum = 5/6 Recognize that sum is n/(n+1) Therefore, the sum of the terms to n = 100 is 100/101.

Amanda and Todd purchase candy, popcorn, and pretzels at the stadium. If a package of candy costs half as much as a bag of popcorn, how much more money did Amanda and Todd spend on the candy than on the popcorn and pretzels combined? (1) The cost of a bag of popcorn is equal to the cost of a pretzel. (2) Amanda and Todd purchased 24 packages of candy, 6 pretzels, and 6 bags of popcorn.

C Question: ax−(2bx+cz)=?ax−(2bx+cz)=?(1) The cost of a bag of popcorn is equal to the cost of a pretzel --> 2x=z2x=z --> substitute in the question: ax−(2bx+2cx)=x(a−2b−2c)=?ax−(2bx+2cx)=x(a−2b−2c)=?. Not sufficient.(2) Amanda and Todd purchased 24 packages of candy, 6 pretzels, and 6 bags of popcorn --> a=24a=24 and b=c=6b=c=6. Not sufficient.(1)+(2) Substitute the values from (2) into (1): x(a−2b−2c)=x(24−12−12)=0x(a−2b−2c)=x(24−12−12)=0, thus Amanda and Todd spent equal amount of money on the candy AND on the popcorn and pretzels combined. Sufficient.

On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, a close relative, comes up for air on average once every five minutes. The number of times a bottle-nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a beluga whale would come up for air in that same period?A. 50%B. 100%C. 150%D. 200%E. 250%

Dolphin once every 2 min Beluga once every 5 min. times dolphin comes up is what % greater than beluga? 150% (5-2)/2*100. PERCENTAGE GREATER I scored 500 on my gmat, you scored 600. What percent greater did you score? Answer is 20%, however if I said "your score is what percent of my score?" the answer would be 120%.

If mn ≠ 0, is m>n? (1) 1/m < 1/n (2) m^2 >n^2

E This is a positive/negative problem. (1) If m and n are positive, then the scenario holds. 1/6 < 1/4, but 6>4. If m and n are negative, but the scenario does not hold, 1/-6 < 1/-4, but -4 >-6. Insufficient (2) If m and n are positive, it holds. If m and n are both negative, it does not hold. If m or n have different signs it does not hold. (1) + (2) If m and n are both positive, it holds. If m and n are both negative, we get a problem. (1) m = -4, n = -5... 1/-4 < -1/5.. -4 >-5 holds.. (2) 16 < 25 does not hold. Insufficient E