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

 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).

 3.  Is integer n  greater than 3?

(1). (1/100)ⁿ  > .0001

(2). (1/100)^{n - 1} > .001

 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).

5.  If a,b, and c, are positive integers, is (a/b)^c > 1?

(1). b- a=9

(2). c > 1

 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).

 7. Is x ^–y  positive?

(1) x  is positive.

(2) y  is negative.

 7. A -->  This question begs to be translated a little bit

before even looking at the statements. x^-y= 1/x^y 

The only way 1/x^y  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.

48[(1/2⁴) + (1/3²) + (1/2²)] /3² = ?

a. 16/27

b. 61/27

c. 61/3

d. 129

e. 183

 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

9.  Is m² an integer? 

(1). m² is an integer

(2). √m is an integer

 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.

11. If a  is not equal to zero, is a^–3  a number greater

than 1?

(1) 0 < a ≤  2

(2) ab  = a

 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

 1. What is the sum of positive integers x  and y ?

(1) x² + 2xy + y  = 16

(2) x²  – y²  =8

 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.

5.  What is the value of xyz ?

(1) y ! = 6 and x ! > 720

(2) z  is the least even integer greater than –1.

 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.

 6.  If x  and n  are positive integers, is n ! + x  divisible

by x ?

(1) n  > x

(2) n  is not a prime number.

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.

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.

 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!”.

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.

 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.

 6. Does s  = t  ?

(1) √s  = t

(2) s  is both a factor and multiple of t.

 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 .

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

 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.

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 ?

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.

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 % ?

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.

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

2. Find a : b : c if 6a = 9b = 10c.

Solution--> a/b = 9/6 = 3 : 2 = 15 : 10, b/c = 10/9 = 10 : 9.

Hence a : b : c = 15 : 10 : 9.

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.

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

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 ?

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.

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.

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.

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 ?

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.

Translating Word Problems Into Equations

1. Find two consecutive odd numbers the difference of whose squares is 296.

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]

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.

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.

Answer: Option B

Explanation:

Let the cost of 1 litre milk be Re. 1

By the rule of alligation, we have:

Answer: Option C

Explanation:

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

[image] 252x - 189 = 140x + 147

[image] 112x = 336

[image] x = 3.

So, the can contained 21 litres of A.