Term
2. If a and b are distinct integers greater than –1,
then what is the value of a^{2} ?
(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/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.



Term
48[(1/2^{4}) + (1/3^{2}) + (1/2^{2})] /3^{2 }= ?
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 reexpress 48 as powers of 2 and 3 and then
distribute. Since 48 = 3 x 16, we can write 48 as
3 x 2^{4} . Then we can multiply that by each of the
fractions in the parentheses and then reduce: ( 3 x 2^{4) }/ 2^{4}= 3 ,
(3 x 2^{4}) / 3^{2 }= 2^{4}/3 and (3 x 2^{4}) /2^{2}= 3 x 2^{2}=12 .
Now add these all together and you get 61/3.
Now, you’re dividing the whole thing by 3^{2} , which is the same as
multipling by 1/3^{2} or 1/9. So  61/3 x 1/9 =61/27 


Term
9. Is m^{2} an integer?
(1). m^{2} 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/a^{3} .
For 1/a^{3} 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) x^{2} + 2xy + y = 16
(2) x^{2} – y^{2} =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 9^{12n+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
Answer: Option B
Explanation:
Let the cost of 1 litre milk be Re. 1
Milk in 1 litre mix. in 1^{st} can = 
3 
litre, C.P. of 1 litre mix. in 1^{st} can Re. 
3 
4 
4 
Milk in 1 litre mix. in 2^{nd} can = 
1 
litre, C.P. of 1 litre mix. in 2^{nd} 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 1^{st} can C.P. of 1 litre mixture in 2^{nd} can 

Mean Price




[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?




Definition
Answer: Option C
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. 

