# Shared Flashcard Set

## Details

GMAT- Math Questions
More difficult questions I missed and whatnot
30
05/01/2013

Term
 2. If a  and b  are distinct integers greater than –1, then what is the value of a2 ?   (1) a–1  is undefined. (2) b  = 2
Definition
 2. A  --> Statement (1) tells us that a  = 0, since only division by 0 is undefined. Since b  is distinct from a,  it cannot be 0, and this statement is sufficient to determine that ab  is 0. Eliminate BCE and keep AD. Statement (2) tells us nothing about a , and so is insufficient; eliminate choice (D).
Term
 3.  Is integer n  greater than 3? (1). (1/100)n  > .0001 (2). (1/100)n - 1 > .001
Definition
 3. D -->  Plug In possible values for n . In both statements, you cannot plug in a value for n  that is greater than 3 without contradicting the statement, so each statement is sufficient, and the answer is (D).
Term
 5.  If a,b, and c, are positive integers, is (a/b)c > 1? (1). b- a=9 (2). c > 1
Definition
 5. A  Statement (1) tells us that a/b is a fraction less than one, and any fraction less than one, when raised to a positive power, will remain a fraction less than one. Statement (1) is sufficient. EliminateBCE. Statement (2) does not tell us anything about a/b; if a/b is an integer greater than one, then (a/b)c will also be greater than one, but if a/b fraction less than one, then (a/b)c  will not be greater than one. Statement (2) is insufficient; eliminate (D).
Term
 7. Is x –y  positive? (1) x  is positive. (2) y  is negative.
Definition
 7. A -->  This question begs to be translated a little bit before even looking at the statements. x-y= 1/xy  The only way 1/xy  can be negative is if x  is negative and y  is odd. Statement (1) tells us that x  is not negative, and since it is not negative, there is no way for x −y  to be negative. Eliminate BCE. Statement (2) doesn’t tell us whether y  is even or odd so that doesn’t help answer the question.
Term
 48[(1/24) + (1/32) + (1/22)] /32 = ? a. 16/27 b. 61/27 c. 61/3 d. 129 e. 183
Definition
 8. B --> This question calls for straight math, covering fractions and exponents. The easiest way to start is to re-express 48 as powers of 2 and 3 and then distribute. Since 48 = 3 x 16, we can write 48 as 3 x 24 . Then we can multiply that by each of the fractions in the parentheses and then reduce: ( 3 x 24) / 24= 3 , (3 x 24) / 32 = 24/3 and (3 x 24) /22= 3 x 22=12 .   Now add these all together and you get 61/3.   Now, you’re dividing the whole thing by 32 , which is the same as multipling by 1/32 or 1/9. So -- 61/3 x 1/9 =61/27
Term
 9.  Is m2 an integer?  (1). m2 is an integer (2). √m is an integer
Definition
 9. B --> Start by translating the question and understanding the pieces of the puzzle given and the pieces needed. To answer this question, we need to know whether m  is an integer. Statement (1) is insufficient because m  could equal 2 or m  could equal √2 . Eliminate AD. Statement (2) tells us that m  must be an integer because it must be the perfect square of an integer, and any integer squared is also an integer.
Term
 11. If a  is not equal to zero, is a–3  a number greater than 1? (1) 0 < a ≤  2 (2) ab  = a
Definition
 11. E -->  This question begs for a little translation and simplification; it is another way to say 1/a3 . For 1/a3 to be greater than 1, a  must be a positive fraction less than 1. Statement (1) does not resolve whether a  is a fraction or not. Eliminate AD. Statement (2) only tells you that b  is 1; it tells us nothing about a . Eliminate (B). When we look at the statements together, we know nothing more about a  than we knew in Statement (1), so together they are still not sufficient -- E
Term
 1. What is the sum of positive integers x  and y ? (1) x2 + 2xy + y  = 16 (2) x2  – y2  =8
Definition
 1. D --  Start with Statement (1). If we factor the equation given, it yields (x + y )2 = 16, so x + y  = 4 (note that we’re told that x  and y  are positive), so Statement (1) is sufficient. Eliminate BCE. Statement (2) can also be factored, and yields (x + y )(x  – y ) = 8. This tells us that x  + y  and x  – y  must be factors of 8. Eight only has four factors, 1, 2, 4, 8. If we consider each possible factor in turn, and if x  and y  are positive integers and must equal one of these factors, there is no way that x + y  can equal 1. If x  + y  must equal 2, then x  and y  must both be 1, but in that instance x  – y  would not equal 4, thus x  + y  cannot be 1. If we continue to try each factor, the only factor of 8 that x  + y  could be is 4, thus this statement is also sufficient.
Term
 5.  What is the value of xyz ? (1) y ! = 6 and x ! > 720 (2) z  is the least even integer greater than –1.
Definition
 5. B --> Statement (1) tells us what the value of y  is, but does not give us the exact value of x  or tell us anything about the value of z , and so is insufficient. Eliminate AD. Statement (2) tells us that z  is 0, and thus we don’t need to know anything about the value of any of the other variables.
Term
 6.  If x  and n  are positive integers, is n ! + x  divisible by x ? (1) n  > x (2) n  is not a prime number.
Definition
 6. A -- This one is tough. To understand the relevance of Statement (1), you have to recognize the following: • A factorial is divisible by all positive integers less than or equal to the integer you are taking a factorial of. For example: x ! is divisible by all positive integers smaller than x . • If b  is a multiple of y , then if you add y  to b,  the result will still be divisible by y . For example, 12 is divisible by 3. If you add 12 + 3 it will still be divisible by 3. Alternatively, plug in values for x  and n  and you will find out the facts mentioned above, but that’s a lot of messy work. Statement (1) is sufficient. Keep AD! - Statement (2) tells us nothing about x nor its relationship to n. Stating that n is NOT prime means it could be a vast number of values. Thus Statement (2) is not sufficient.
Term
 4.  Is m  a multiple of 6? (1) More than 2 of the first 5 positive integer multiples of m  are multiples of 3. (2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.
Definition
 4. B --> Is m 6, 12, 18,24, 30, 36....etc? Start with Statement (1). Multiples of 6 (6, 12, 18, 24, and 30) would yield an answer of “yes.” Multiples of 3 (3, 6, 9, 12, 15) would yield a “no”. Thus Statement (1) is insufficient. Eliminate AD. Approach Statement (2) the same way. The information we are given in this statement doesn’t allow us to use 6 or any multiple of 6 for m , thus answering the question with a definitive “no!”.
Term
 5.  If 6 is a factor of a  and 21 is a factor of b , is ab  a multiple of 70? (1) a  is a multiple of 4. (2) b  is a multiple of 15.
Definition
 5. B -- This question is all about factoring. We need to determine whether 70 is a factor of ab , and the easiest way to do that is to break 70 down into its prime factors. 7 x 5 x  2 = 70. So if ab  is divisible by 7, 5, and 2, then it’s divisible by 70. The question itself lets us know that 70 is divisible by 2 (since 6 is a factor of a ) and by 7 (since 21 is a factor of b ), so all we need is proof that it is divisible by 5. Statement (1) does nothing to help, but Statement (2) shows that b  is divisible by 5, and so is sufficient.
Term
 6. Does s  = t  ? (1) √s  = t (2) s  is both a factor and multiple of t.
Definition
 6. B -- Statement (1) is insufficient because s  and t  could both be 1, which would be equal, or s  could be 4 and t  could be 2. Eliminate AD, down to BCE. What we are given in Statement (2) answers the question because the only number that can be both a factor and a multiple of t is t , thus s  must be equal to t .
Term
 7.  If n  is an integer greater than 0, what is the remainder when 912n+3  is divided by 10? a. 0 b. 1 c. 2 d. 7 e. 9
Definition
 7. E -- When something looks like an insane amount of work, start looking for a shortcut. In this case, the shortcut is a pattern: 91  = 9. 92  = 81. 93  = 81 Å~  9 = 729. Multiply that by another 9? You’ll get a number ending in 1. Then one ending in 9. And so forth and so on. So the bottom line is that whenever 9 is raised to an odd power, the units digit is 9. When it’s raised to an even power, the units digit is 1. Adding 10 to a number won’t do  anything to change the units digit, and when you divide a number by 10, its remainder will always be its units digit. No matter what value you plug in for n , we’re going to be raising 9 to an odd power, so the units digit and the remainder will both be 9.
Term
 14. A man’s working hours a day were increased by 25%, and his wages per hour were increased by 20%. By how much percent were his daily earnings increased ?
Definition
 Sol. Let initially X be number of hours & Y = wages/hour Later these become 1.25 X & 1.2 X respectively. Daily earnings initially = X x Y Now Daily earnings = 1.25X x 1.2Y = 1.5 XY Hence 50% increase.
Term
 15. A tradesman allows a discount of 15% on the marked price. How much above the C.P. must he mark his goods to make a profit of 19 % ?
Definition
 Sol. Let CP = 100, Gain = 19, SP = 100 + 19 = 119 Now Marked price should be such that Marked price reduced by 15% is equal to 119 or 85% of M.P. = 119 or MP = 119 x 100/85 = Rs 140  Answer = 40 % above the C.P.
Term
 6. If a/b > 1 or a > b then (a + X) / (b + X) < a/b a, b, X are natural numbers 7. If a/b < 1 or a < b then (a + X) / (b + X ) > a/b a, b, X are natural numbers
Definition
Term
 2. Find a : b : c if 6a = 9b = 10c.
Definition
 Solution--> a/b = 9/6 = 3 : 2 = 15 : 10, b/c = 10/9 = 10 : 9. Hence a : b : c = 15 : 10 : 9.
Term
 9. A mixture contains milk & water in the ratio 5 : 1. on adding 5 litres of water, the ratio of milk and water becomes 5 : 2. Find the quantity of milk in the original mixture.
Definition
 Sol. Let the quantity of milk be 5X & that of water X. Then 5X / (X + 5) = 5/2 or X = 5  Quantity of milk = 5X = 25 litres
Term
 10.  The ratio of the number of boys to the number of girls in a school of 546 is 4 : 3. If the number of girls increases by 6, what must be the increase in the number of boys to make the new ratio of boys to girls 3 : 2 ?
Definition
 Sol.   Original # of boys = 546 x 4/7 = 312. Original # of girls (312/4=78) --> = 78 x 3 = 234. Final # of girls = 234 + 6 = 240 --> # of boys reqd. to make the new ratio = 240 x 3/2 = 360  The reqd. increase in the # of boys = 360 – 312 = 48 Ans.
Term
 12. Two numbers are in the ratio of 3 : 4. If 5 is subtracted from each, the resulting numbers are in the ratio 2 : 3. Find the numbers.
Definition
 Solution: Let 3X and 4X be the numbers  (3X – 5) / (4X – 5) = 2/3  9X – 15 = 8X – 10  X = 5  The required numbers are 15 and 20 Ans.
Term
 1. If A and B together finish a piece of work in 10 days & B alone can finish it in 20 days. In how many days can A alone finish the work ?
Definition
 Sol. Let X and Y be the number of days required by A and B respectively. By the standard formula, XY / (X + Y) = 10 & Y = 20 X x 20 / (X + 20) = 10 or X = 20 days Ans.
Term
 Translating Word Problems Into Equations 1. Find two consecutive odd numbers the difference of whose squares is 296.
Definition
 Sol. Let the numbers be 2X + 1 and 2X + 3 Then (2X + 3)2 – (2X + 1)2 = 296  X = 36 Hence 2X+ 1 = 2 x 36 + 1 = 73 and 2X + 3 = 75  The required numbers are 73 and 75. [Verification. (75)2 – (73)2= 5625 – 5329 = 296]
Term
 3.  A number consists of three consecutive digits, that in the unit’s place being the greatest of the three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digit. Find the number.
Definition
 Sol. Let the hundred’s digit be X. Then the ten’s digit = X + 1 and the unit’s digit = X + 2  The number = 100 x X + 10(X + 1) + X + 2 = 111X + 12. The number formed by reversing the digits = 100(X + 2) + 10(X + 1) + X= 111X + 210  111X + 210 – 111X – 12 = 22 (X + 2 + X + 1 + X)  X = 2. Hence the required number = 234.
Term

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

 A. 4 litres, 8 litres B. 6 litres, 6 litres C. 5 litres, 7 litres D. 7 litres, 5 litres
Definition

Explanation:

Let the cost of 1 litre milk be Re. 1

 Milk in 1 litre mix. in 1st can = 3 litre, C.P. of 1 litre mix. in 1st can Re. 3 4 4
 Milk in 1 litre mix. in 2nd can = 1 litre, C.P. of 1 litre mix. in 2nd can Re. 1 2 2
 Milk in 1 litre of final mix. = 5 litre, Mean price = Re. 5 8 8

By the rule of alligation, we have:

C.P. of 1 litre mixture in 1st can   C.P. of 1 litre mixture in 2nd can
 3 4
Mean Price
 5 8
 1 2
 1 8
 1 8
 [image] Ratio of two mixtures = 1 : 1 = 1 : 1. 8 8
 So, quantity of mixture taken from each can = [image] 1 x 12 [image] = 6 litres. 2
Term
3.

A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

 A. 10 B. 20 C. 21 D. 25
Definition

Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

 Quantity of A in mixture left = [image] 7x - 7 x 9 [image] litres = [image] 7x - 21 [image] litres. 12 4
 Quantity of B in mixture left = [image] 5x - 5 x 9 [image] litres = [image] 5x - 15 [image] litres. 12 4
[image]
 [image] 7x - 21 [image] 4
= 7
 [image] 5x - 15 [image] + 9 4
9
 [image] 28x - 21 = 7 20x + 21 9

[image] 252x - 189 = 140x + 147

[image] 112x = 336

[image] x = 3.

So, the can contained 21 litres of A.

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