700+ GMAT Data Sufficiency Questions With Explanations

700+ GMAT Data Sufficiency Questions With Explanations

Collected by Bunuel

Solutions by Bunuel

1. Probability / Combinatorics

If 2 different representatives are to be selected at random from a group of 10 employees

and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women.

(2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people

we have

-->

So basically question asks is

(1)

not sufficient.

this is true only when

?

and this should be

. (w # of women

)

. So

(2)

-->

(1)+(2)

,

not sufficient

Answer E.

You can use Combinations, to solve as well:

# of selections of 2 women out of

-->

, not sufficient

employees;

total # of selections of 2 representatives out of 10 employees.

Q is

(1)

-->

, not sufficient.

--> -->

-->

?

(2)

# of selections of 2 men out of

employees -->

--

>

-->

-->

, not sufficient

(1)+(2)

,

not sufficient

Answer E.

Discussed at:

/forum/

2. Coordinate Geometry

If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater

than 15?

(1) A < 3

(2) The triangle is right

First of all right triangle with hypotenuse 5,

doesn't mean

that we have (3, 4, 5) right triangle. If we

are told that values of all sides are integers, then yes: the only integer solution for right triangle with

hypotenuse 5 would be (3, 4, 5).

To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right

triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also

true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So ANY point on circumference of a circle with diameter

would make the right triangle with diameter.

Not necessarily sides to be

and

. For example we can have isosceles right triangle, which would be

45-45-90: and the sides would be

. OR if we have 30-60-90 triangle and hypotenuse is

, sides would

be

and

. Of course there could be many other combinations.

Back to the original question:

If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater

than 15?

(1) A < 3 --> two vertices are on the X-axis and the third vertex is on the Y-axis, below the point (0,3).

The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at

(0,-100) and the area would be more than 15. So not sufficient.

(2) The triangle is right. --> Obviously as the third vertex is on the Y-axis, the right angle must be at

the third vertex. Which means the hypotenuse is on X-axis and equals to 5. Again if we consider the

circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of

two intersections of the circle with Y-axis. We'll get the two specific symmetric points for the third

vertex, hence the area would be fixed and defined. Which means that it's possible to answer the

question whether the area is more than 15, even not calculating actual value. Sufficient.

Answer: B.

If we want to know how the area could be calculated with the help of statement 2, here you go:

One of the approaches:

The equation of a circle is

We know:

, as the hypotenuse is 5.

and

0) and (4, 0).

, where

is the center and

is the radius.

, as the center is on the X-axis, at the point

, half the way between the (-1,

for the circle.

We need to determine intersection of the circle with Y-axis, or the point

So we'll have

-->

and

. The third vertex is either at the point

OR

. In any

case

.

Discussed at:

/forum/urgent-help-required-

?hilit=only%20integer%20solution%20triangle#p65662 8

3. Geometry

Is the perimeter of triangle ABC greater than 20?

(1) BC- AC=10.

(2) The area of the triangle is 20.

This problem could be solved knowing the following properties:

For (1):

The length of any side of a triangle must be larger than the positive difference of the other two

sides, but smaller than the sum of the other two sides.

Now, as BC-AC=10 then BC>10 and also according to the above property the third side AB is also more

than 10, so the perimeter is more than 20. Sufficient.

For (2):

A. For a given perimeter equilateral triangle has the largest area.

B. For a given area equilateral triangle has the smallest perimeter.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral

triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that

perimeter must be more than 20.

, (

.)

?

hence area is more than 20.

Sufficient.

Answer: D.

Discussed at:

/forum/

4. Modulus / Inequalities

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

, is

(1)

Two cases:

A.

-->

-->

. But remember that

, so

? Which means is

? (

)

B.

-->

-->

.

Two ranges

or

. Which says that

either in the first range or in the second. Not

sufficient to answer whether

. (For instance

can be

or

)

OR:

Either

multiply both sides of inequality by

and

, so

or

-->

or

.

would be

(side note: we can safely do this as absolute value is

-->

:

-->

.

;

non-negative and in this case we know it's not zero too) -->

Or

and

, so

The same two ranges:

(2)

. Well this basically tells that

is negative, as if x were positive or zero then

equal to

. Only one range:

, but still insufficient to say whether

. (For

instance

can be

or

)

Or two cases again:

-->

-->

.

-->

: never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range

(

(from 2)

and

or

(from 1), hence

). Every

from this range is definitely in the

range

. Sufficient.

Answer: C.

To demonstrate on diagram:

Range from (1):

-----(-1)

----

(0)- ---(1)

----

Range from (2):

-----(-1)--- -

(0)----(1)----

or

, blue area;

, green area;

From (1) and (2):

----(-1)

----

(0)----(1)----

, common range of

from (1) and (2)

(intersection of ranges from (1) and (2)), red area.

Discussed at:

/forum/

5. Coordinate Geometry / Modulus

On the number line shown, is zero halfway between r and s

(1) s is to the right of zero

(2) The distance between t and r is the same as the distance between t and -s

NOTE:

In GMAT we can often see such statement:

is halfway between

and

. Remember this

statement can ALWAYS be expressed as:

.

Also in GMAT we can often see another statement: The distance between

and

is the same as the

distance between

and

. Remember this statement can ALWAYS be expressed

as:

.

Back to original question: is 0 halfway between r and s?

OR is

(1)

? --> Basically the question asks is

, clearly not sufficient.

.

?

(2) The distance between

and

is the same as the distance between

and -

:

is always positive as

is to the left of the

, hence

;

BUT

can be positive (when

, meaning

is to the right of -

) or negative (when

,

meaning

is to the left of -

, note that even in this case

would be to the left of

and relative

position of the points shown on the diagram

still

will be the same). So we get

either

OR

.

In another words:

is the sum of two numbers from which one

, is greater than

. Their sum

clearly can be positive as well as negative. Knowing that one is greater than another doesn't help to

determine the sign of their sum.

Hence:

-->

OR

-->

So the only thing we can determine from (2) is:

Not sufficient.

(1)+(2)

and

.

-->

.

;

(as

is to the right of

) hence

.

Hence

. -->

-->

. Sufficient.

Answer: C.

About the relative position of the points on diagram:

In general, you should not trust the scale of GMAT diagrams, either in Problem Solving or Data

Sufficiency.

It used to be true that Problem Solving diagrams were drawn to scale unless mentioned

otherwise, but I've seen recent questions where that is clearly not the case. So I'd only trust a diagram

I'd drawn myself. ...

Here I'm referring only to the scale of diagrams; the relative lengths of line segments in a triangle, for

example. ...

You can accept the relative ordering of points and their relative locations as given

(if

the vertices of a pentagon are labeled ABCDE clockwise around the shape, then you can take it as given

that AB, BC, CD, DE and EA are the edges of the pentagon;

if a line is labeled with four points in A, B,

C, D in sequence, you can take it as given that AC is longer than both AB and BC;

if a point C is

drawn inside a circle, unless the question tells you otherwise, you can assume that C is actually within

the circle;

if what appears to be a straight line is labeled with three points A, B, C, you can assume

the line is actually straight, and that B is a point on the line

-- the GMAT would

never

include as a

trick the possibility that ABC actually form a 179 degree angle that is imperceptible to the eye, to give

a few examples).

So don't trust the lengths of lines, but do trust the sequence of points on a line, or the location of

points within or outside figures in a drawing.

Discussed at:

/forum/

6. Inequalities

Are x and y both positive?

(1) 2x-2y=1

(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the

following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we

know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line)

is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both

negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less

time-consuming and comfortable for you personally.

One of the approaches:

-->

-->

Sufficient.

--> substitute x -->

-->

is positive, and as

,

is positive too.

Answer: C.

NOTE:

does not mean that

then

.

. If both x and y are positive, then

, BUT if both are negative,

From (2)

, we can only deduce that x and y have the same sigh (either both positive or both

negative).

Discussed at:

/forum/

7. Inequalities / Word Problem

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12

notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.

(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Given

, question is

true? Or is

true? So basically we

are asked whether we can substitute 3 notebooks with 3 pencils. Now if

we can easily substitute

notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But

if

we won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both

cases (

or

) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil,

but we won't be sure for 3 (more than 2).

(1)

. We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.

(2)

. We can substitute 5 notebooks with 5 pencils, so in any case (

or

) we

can substitute 3 notebooks with 3 pencils. Sufficient.

Answer: B.

Discussed at:

/forum/?hilit=swiss#p674919

8. Word Problem

Ten years ago, scientists predicted that the animal z would become extinct in t years. What is t?

(1) Animal z became extinct 4 years ago.

(2) If the scientists had extended their extinction prediction for animal z by 3 years, their prediction

would have been incorrect by 2 years.

(1) The only thing we can get from this statement is when animal z

actually extincted

: 4 years ago or 6

years after the prediction. Not sufficient.

(2) Also not sufficient: t+3=actual extinction +/- 2.

(1)+(2) Animals extincted 6 years after the prediction: t+3=6-2 --> t=1 OR t+3=6+2 --> t=5. Two answers,

not sufficient.

Answer: E.

Discussed at:

/forum/

9. Modulus

If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) -4x - 12y = 0

(2) |x| - |y| = 16

(1)

-->

-->

and

have opposite signs --> so

either

either

:

Sufficient.

(2)

. Sum this one with the equation given in the stem -->

--

>

,

.

(x and y have opposite sign) or

(x and y have the

same sign). Multiple choices. Not sufficient.

Answer: A.

Discussed at:

/forum/

10. Number properties

If k is an integer greater than 1, is k equal to 2^r for some positive integer r?

(1) k is divisible by 2^6

(2) k is not divisible by any odd integer greater than 1

Given:

, question is

.

Basically we are asked to determine whether

has only 2 as prime factor in its prime factorization.

(1)

, if

is a power of 2 then the answer is YES and if

is the integer other than 2 in any

power (eg 3, 5, 12...) then the answer is NO.

(2)

is not divisible by any odd integers greater then 1. Hence

has only power of 2 in its prime

factorization. Sufficient.

Answer: B.

Discussed at:

/forum/

11. Inequalities

What is the value of integer x?

(1) 4 < (x-1)*(x-1) < 16

(2) 4 < (x+1)*(x-1) < 16

Note:

is an integer.

(1)

-->

-->

is a perfect square between 4 and 16 -->

or

--

and

OR

and

:

,

-->

,

,

,

OR

, same answer.

there is only one perfect square: 9 -->

-->

>

or

. Two answers, not sufficient.

(2)

-->

-->

-->

-->

is

a perfect square between 5 and 17 --> there are two perfect squares : 9 and 16 --

>

or

-->

or

or

or

. Four answers, not sufficient.

(1)+(2) Intersection of values from (1) and (2) is

. Sufficient.

Answer: C.

Discussed at:

/forum/tough-inequation-ds-what-is-the- value-of-integer-x-

12. Inequalities / Modulus

Is

x>y>z?

?

(

1) x-y = |x-z|+|z-y|

(2) x > y

(1)

First of all as RHS is the sum of two non-negative values LHS also must be non-negative. So

.

Now, we are told that the distance between two points

and

, on the number line, equals to the

sum of the distances between

and

AND

and

.

The question is: can the points placed on the number line as follows ---z---y---x---. If you look at the

number line you'll see that it's just not possible. Sufficient.

OR algebraic approach:

If

is true, then

, will become

-->

,

which contradicts our assumption

. So

is not possible. Sufficient.

(2)

--> no info about

. Not sufficient.

Answer: A.

Discussed at:

/forum/

13. Coordinate Geometry / Algebra

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1)

(2)

Distance between the point A (x,y) and the origin can be found by the formula:

So we are asked is

(1)

(2)

(1)+(2)

From (1)

>

as

and

and

and

-->

and

. Not sufficient.

-->

.

, for some non-zero

-->

, then

. Substitute in (2)

) --> so

and

. Not sufficient.

? Or is

?

.

--

(another solution

is not possible as

in (1) given in

denominator and can not be zero, so

Now square

-->

--

>

-->

cancels out -->

. Sufficient.

Answer: C.

Discussed at:

/forum/

14. Remainders

If w, x, y, and z are the digits of the four-digit number N, a positive integer, what is the remainder

when N is divided by 9?

(1) w + x + y + z = 13

(2) N + 5 is divisible by 9

Remainder when a number is divided by 9 is the same as remainder when the sum of its digits is divided

by 9:

Let's show this on our example:

Our 4 digit number is

When 1000w is divided by 9 the remainder is

remainder

,

remainder

.

The same with

and

So, the remainder when

.

is divided by 9 would be:

. what is the remainder when it's divided by 9?

:

(1) w + x + y + z = 13 --> remainder 13/9=4, remainder N/9=4. Sufficient.

(2) N+5 is divisible by 9 --> N+5=9k --> N=9k-5=4, 13, 22, ... --> remainder upon dividing this numbers

by 9 is 4. Sufficient.

Answer: D.

Discussed at:

/forum/

15. Word Problem (800 level question)

Laura sells encyclopaedias, and her monthly income has two components, a fixed component of

$$1000, and a variable component of $$C for each set of encyclopaedias that she sells in that month

over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $$600 lower

than it was.

(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been

over $$4000.

Laura's income

(

-->

(1) Three cases:

, where

is number of sets she sold and

).

is target number

-->

-->

-->

(surplus of 600$$ was generated by 1 set);

(surplus of 600$$ was generated by 2 set);

(surplus of 600$$ was generated by 3 set).

If

equals to 1, 2, or 3, then income for March will be 1600$$, BUT if

, then income will be

more then 1600. Or another way: if we knew that c=300 or 600, then we couldd definitely say

that

BUT if

, then

(for

)

or more

(for

). Not

sufficient.

(2)

then

-->

. Not sufficient.

-->

--> as the lowest value of

,

(1)+(2) as from (2)

, then from (1)

-->

. Sufficient.

Answer: C.

ADDITIONAL NOTES FOR (1) :

had sold

three fewer sets in March...

and

had

, but if she

had sold

then her income would have been

.

Now:

If

is 1 more than

(or as I wrote

), then it would mean that 600$$ was generated by only 1

set;

If

is 2 more than

(or as I wrote

), then it would mean that 600$$ was generated by 2 sets;

If

is more than 3 more than

(or as I wrote

), then it would mean that 600$$ was

generated by all 3 sets;

In first two cases income for March will be 1600$$, BUT for third case income can be 1600$$ or more. So

this statement is not sufficient.

Discussed at:

/forum/?hilit=laura

20. Number properties

If p is a positive integer, what is a remainder when p^2 is divided by 12?

(1) p>3.

(2) p is a prime.

(1) not sufficient

(2) not sufficient

For (1) and (2), just plug two different integers: for (1) 4 an 5 (>3) and for (2) 2 and 3 (primes) to see

that you'll get two different answers for remainders.

(1)+(2) Any prime >3 when divide by 6 can only give remainder 1 or 5 (remainder can not be 2 or 4

because than p would be even, it can not be 3 because p would be divisible by 3). So any prime more

than 3 can be expressed as

or

(

)

So either:

which gives remainder 1 when divided by 12,;

OR:

which also gives remainder 1 when divided by 12.

Answer: C.

Discussed at:

/forum/

21. Coordinate Geometry

In the xy-plane, if line k has negative slope and passes through the point (-5,r) , is the x-intercept

of line k positive?

(1) The slope of line k is -5.

(2) r> 0

Let the

points

Question: is

(1)

whether

-->

or not. Not sufficient.

) but we

intercept be the point

and

would be

?

?

? We can not determine

. Slope

is rise over run and for two

-->

.

? --> is

(2)

and

-->

?

is some positive value (as

don't know whether it's more than

or not. Not sufficient.

(1)+(2)

Not sufficient.

Answer: E.

and

-->

-->

.

can be positive as well as negative.

This can be done by visualizing the question. Statement (2) tells us that the point

, as

, is

in the II quadrant. Line with negative slope through the point in the II quadrant can have

intercept

positive as well as negative.

Taken together: as we don't know the exact location of the point

in II quadrant we can not say

even knowing the slope whether the

intercept would be positive or negative.

Discussed at:

/forum/og-12-ds-question-line-concept- very-good-one-key-

22. Inequalities /Modulus

Is |x| + |x -1| = 1?

(1) x ≥ 0

(2) x ≤ 1

Q is

. Let's check when this equation holds true. We should consider three ranges (as

there are two check points

and

):

A.

-->

-->

, but this solution is not valid as we are checking the

range

;

B.

-->

-->

, which is true. That means that for ANY value from the

range

C.

-->

, equation

-->

holds true.

, but this solution is not valid as we are checking the range

.

.

So we get that equation

holds true ONLY in the range

Statements:

(1)

. Not sufficient, as

must be also

;

(2)

. Not sufficient, as

must be also

;

(1)+(2)

, exactly the range we needed. Sufficient.

Answer: C.

Discussion of this question at:

/forum/

23. Number properties

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.

(2) The sum of all distinct factors of N is even.

Probably the best way of solving would be making the chart of perfect squares and its factors to check

both statements, but below is the algebraic approach if needed.

Couple of things:

1. Note that if

is a perfect square powers of its prime factors must be even, for

instance:

, powers of prime factors of 2 and 3 are even.

2. There is a formula for

Finding the Number of Factors of an Integer

:

First make prime factorization of an integer

of

and

,

, and

are their powers.

, where

,

, and

are prime factors

The number of factors of

will be expressed by the formula

include 1 and n itself.

Example:

Finding the number of all factors of 450:

Total number of factors of 450 including 1 and 450 itself

.

NOTE:

this will

is

factors.

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even- factors. For

instance odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6

even factors).

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say

factors of

is even -->

then powers of its primes

, given that the number of

. But as we concluded if n is a perfect square

,

, and

must be even, and in this case number of factors would

be

.

Hence

can not be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if n is a perfect square then (according to 3) sum of

odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect

square would be

. Hence

can not be a perfect square. Sufficient.

Answer: D.

There are some tips about the perfect square:

? The number of distinct factors of a per

fect square is ALWAYS ODD.

? The sum of distinct factors of a perfect square is ALWAYS ODD.

? A perfect square ALWAYS has an ODD number of Odd

-factors, and EVEN number of Even- factors.

? Perfect square always has even number of powers of prime factors.

Discussed at:

/forum/

24. Statistic

Lists S and T consist of the same number of positive integers. Is the median of the integers in S

greater than the average (arithmetic mean) of the integers in T?

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.

(2) The sum of the integers in S is greater than the sum of the integers in T.

Q: is

? Given: {# of terms in S}={# of terms in T}, let's say N.

(1) From this statement we can derive that

as set S and set T are evenly spaced their medians equal

to their means

. So from this statement question becomes is

? But this

statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25}

OR

S{20,22,24} and

T{1,3,5}.

(2)

. Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set

T{3,3,3} (Mean{T}=3)

OR

S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1).

(1)+(2) From (1) question became is

? --> As there are equal # of term in sets

and mean(average)=(Sum of terms)/(# of terms), then we have: is

Is

? This is exactly what is said in statement (2) that

Hence sufficient.

Answer: C.

Discussed at:

/forum/

25. Number Properties

If n is an integer >1, is 3^n-2^n divisible by 35?

(1) n is divisible by 15.

(2) n is divisible by 18.

RULE: for x^n-y^n:

is ALWAYS divisible by

.

is divisible by

when n is even.

If n is an integer >1, is

divisible by 35?

true? -->

.

(1) n is divisible by 15. -->

--> 5m may or my not be even, so

insufficient to answer, whether it's divisible by 27+8=35.

(2) n is divisible by 18. -->

--> 6m is even, so 3^(18m)-

2^(18m)=27^(6m)-8^(6m) is divisible by 27+8=35. Sufficient.

Answer: B.

Discussed at:

/forum/

26. Geometry

The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the

hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.

(2) LM is 6 inches.

Properties of Similar Triangles:

? Corresponding angles are the same.

? Corresponding sides are all in the same

proportion

.

? It is only necessary to determine that two sets of angles are identical in order to conclude tha

t two

triangles are similar; the third set will be identical because all of the angles of a triangle always sum to

180?

.

? In similar triangles, the sides of the triangles are in some proportion to one another. For example, a

triangle with lengths 3, 4, and 5 has the same angle measures as a triangle with lengths 6, 8, and 10.

The two triangles are similar, and all of the sides of the larger triangle are twice the size of the

corresponding legs on the smaller triangle.

? If two similar triangles have sides

in the ratio

, then their areas are in the ratio

.

OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their

sides:

.

For more on triangles please check Triangles chapter of the Math Book (link in my signature).

Back to original question:

The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the

hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees --> ABC and KLM are similar triangles --

>

Sufficient.

, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20.

(2) LM is 6 inches --> KM=8 -->

-->

. But just knowing the

are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48

and 4 and you'll get different values for hypotenuse. Not sufficient.

Answer: A.

Discussed at:

/forum/mgmat-ds-help-

?hilit= different%20values%20for%20hypotenuse%20sufficient .#p723237

27. Number properties

If N is a positive integer, what is the last digit of 1! + 2! + ... +N!?

(1) N is divisible by 4

(2) (N^2 + 1)/5 is an odd integer.

Generally the last digit of

can take ONLY 3 values:

A. N=1 --> last digit 1;

B. N=3 --> last digit 9;

C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33,

the terms

after N=4 will end by 0 thus not affect last digit and it'll remain 3).

So basically question asks whether we can determine which of three cases we have.

(1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.

Answer: D.

Discussed at:

/forum/

28. Remainders

The integers m and p are such that 2

is divided by m, is r>1?

(1) the greatest common factor of m and p is 2

(2) the least common multiple of m and p is 30

Given:

and

.

. q:

(1) the greatest common factor of m and p is 2 --> p and m are even (as both have 2 as a factor) -->

even divided by even can give only even remainder (0, 2, 4, ...), since remainder is not zero

(as

), then remainder must be more than 1. Sufficient.

(2) the least common multiple of m and p is 30 --> if m=5 and p=6, remainder=1

=1

, answer to the

question would be NO. BUT if m=10 and p=15 remainder=5

>1

answer to the question would be YES. Two

different answers. Not sufficient.

Answer: A.

Discussed at:

/forum/

29. Inequalities

Is

(1)

(2)

?

Is

? --> is

? Now, nominator is non-negative, thus the fraction to be positive

nominator must not be zero (thus it'll be positive) and denominator mut be positive --

>

and

.

Statement (1) satisfies first requirement and statement (2) satisfies second requirement, so taken

together they are sufficient.

Answer: C.

NOTE:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its

sign since you do not know whether you must flip the sign of the inequality.

So you CAN NOT multiply

What you CAN DO is:

-->

by

since you don't know the sign of

--> common denominator is

.

-->

-->

multiply be 3 -->

-->

.

Discussed at:

/forum/

30. Coordinate Geometry

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their

slopes negative?

(1) The product of the x-intersects of lines L and K is positive.

(2) The product of the y-intersects of lines L and K is negative.

We have two lines:

and

. The question: is

?