Data Sufficiency - 53

Answer: D

From statement (1): Square root of a number less than 1 is also less than 1 - sufficient
From statement (2): Reciprocal of any number less than 1 is greater than the number - sufficient

Data Sufficiency - 52

If x is not equal to zero, is 1/x > 1 ?

1) y/ x > y
2) x^3 > x^2

Answer: B

From statement (1): y/x > y
=> y > xy
=> y(1-x) > 0
y>0 when x is less than 1 or y<0 when x is greater than 1 ---- hence insufficient

From statement (2): x^3 > x^2
=> x^2(x-1) > 0
=> x^2 >0 and x>1
Since x>1, 1/x can not be greater than 1 ---- hence sufficient

Hence B

Problem Solving - 50

If x = -|x|, then which one of the following statements could be true?

I. x=0
II. x < 0
III. x > 0

A). none
B). I only
C). III only
D). I and II
E). II and III

Answer: D

Given x = - |x|
We know that |x| >= 0
=> -|x| = x <=0

Thus inequalities represented by I and II

Data Sufficiency - 33

Is |x - 1| less than 1 ?

1). (x - 1)^2 less than and equal to 1
2). x^2 - 1 greater than 0

Answer: E

|x-1| less than 1 is only true when 0 less than x less than 1

From statement (1): (x-1)^2<=1
True when 0<=x<=2
If x=0.5, then |x-1| less than 1 is true
If x=2, then |x-1| less than 1 is not true
Hence insufficient

From statement (2): x^2>1 means x>1 and x<-1
True when x=1.5, but not when x=3
Hence insufficient

Statement (1) and (2) together: 1 is less than x is less than and equal to 2
Taking x=1.5 and x=2
Hence insufficient

Data Sufficiency - 31

Is |x - 1| less than 1 ?

(1). (x - 1) ^2 less than and equal to 1
(2). (x^2) - 1 greater than 0

Answer: E

From statement (1): (x - 1) ^2 less than and equal to 1
Now this is true only if 0 is less than and equal to x and x is less than and equal to 2. (We know that (x - 1) ^2 less than 1 is only true when 0 is less than x and x is less than 2)
When we take x = 0.5, then |x-1| less than 1 is true
When we take x = 2, then |x-1| less than1 does not holds true
Hence insufficient

From statement (2): x^2 greater than 1 => x is less than -1 and x is greater than 1
When we take x = 1.5, then x^2 greater than 1 is true
When we take x = 3, then x^2 greater than 1 does not holds true
Hence insufficient

Taking both statements (1) and (2) together -- We have 1 is less than x and x is less than and equal to 2
When we take x = 1.5, then 1 is less than x and x is less than and equal to 2 is true
When we take x = 2, then 1 is less than x and x is less than and equal to 2 does not holds true Hence insufficient

DS Question - 30

If q is a multiple of prime numbers, is q a multiple of r?

1) r is less than 4
2) q = 18

Answer: E

From Statement (1) -- q can be positive and r can be negative.
r can also be a real number but not necessarily an integer.
Hence insufficient.

From Statement (2) -- q = 18.
But r can be negative or can be positive.
r can also be a real number not an integer
Hence insufficient

Taking both statements (1) and (2) together -- Again r can be positive or negative. And again r can also be a real number.
Hence insufficient

DS Question - 28

Is the three-digit number n less than 550?

1) The product of the digits in n is 30
2) The sum of the digits in n is 10

Answer: C

From statement (1) : the factors of 30 are 1,2,3,5,6,10,15,30
Now because the number must be digits (single number) we do not need to consider 10,15 and 50

Now if the hundreds digit of the 3-digit number = any digit among the 1,2,3 digits ---> answer to the question is clearly Yes.

But if the hundreds digit = 5 or 6 ---> answer to the question is No.

Hence (1) alone is insufficient.

From statement (2) : There are different combinations where the sum of digits can be equal to 10. e.g 541 and 145.

Hence (2) alone is insufficient.

Combining statements (1) and (2) we get :

If the hundreds digit= 5 , the second digit can only be 1,2,3 because if the digit is equal to 6 it violates statement (2) Hence in this case answer to the question is Yes.

If hundreds digit = 6 then the number n must contain 5 so as to satisfy the first statement but then it will violate statement (2) as the total of digits of n will exceed 10

Thus no 3-digit number exists with the hundreds digit equal to 6 satisfying both the statements together.

Hence the number n will always be less than 550 as it will be the combination of 1, 2, 3 and 5

Hence ans is C

DS Question - 26

Is 5^k less than 1,000?

(1) 5^(k-1) greater than 3,000

(2) 5^(k-1) = 5^k - 500


Answer: D

From statement (1) - 5^(k-1) is greater than 3000

=> 5^k/5 is greater than 3000
=> 5^k is greater than 15000

Hence sufficient

From statement (2) - 5^(k-1) = 5^k - 500

=> 5^(k-1) = 5^k - 500
=> 5^k - 5^(k-1) = 500
=> 5^k(1- 5^-1) = 500
=> 5^k(4/5) = 500
=> 5^k = 2500/4 = 625
Therefore 5^k is less than 1000

Hence sufficient

DS Question - 25

Is 5^k less than 1,000?

(1) 5^(k-1) is less than 3000
(2) 5^(k-1) = 5^k - 500

Answer: B

From statement (1) - It is given that 5^(k-1) less than 3000
=> 5^k/5 less than 3000
=> 5^k less than 15000
=> 5^k could be either more than 1,000 or less than 1,000.
Hence insufficient

From statement (2) - It is given that 5^(k-1) = 5^k - 500
=> 5^k/5 = (5^k) - 500
=> 5^k = 5*(5^k) - 2500
=> (5-1)*5^k = 2500
=> 4*5^k = 2500
=> 5^k = 2500/4 = 625

Hence 5^k less than 1,000.

Hence sufficient.

DS Question - 21

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.

Answer : B

It is given that 9x + 3y <= 20

It is required to prove that 12x + 12y <= 40 or 6x + 6y <= 20

Hence basically the question asks that whether three notebooks can be replaced by 3 pencils in the same price range or not....

Statement (1) - suggests that 2 notebooks can be exchanged for two pencils in the same price range --- well here third can always create some problem in the exchange...hence insufficient to answer the question asked...

Statement (2) - suggests that 4 notebooks can be replaced by 4 pencils in the same price range i.e 20 Swiss francs ----- Hence if 4 notebooks can be exchanged for 4 pencils in the same price range then definitely three notebooks can be exchanged for 3 pencils in the same price range.

Hence statement 2 is sufficient to answer the question.

Thus ans is B

OR

The solution to above problem can be found out graphically:

In the graph above :

Line AB with the axis covers the area representing the information given in the question --- 9x + 3y <= 20

Line ED with the axis covers the area representing the question asked - 6x + 6y <= 20

Statement (1) gives the area covered by line CD with the axis i.e 7x + 5y <= 20.

Clearly we can see the shaded area EFC (covered by line ED) lying outside the area covered by line CD. Hence statement (1) is insufficient to answer the question.

Statement (2) gives the area covered by line drawn from point F to the X - axis. Area given by statement (2) and the area given by line AB together cover the area covered by line ED.

Hence statement (2) is sufficient to answer the question.

DS Question - 19

Is x = square root (x^2) if

(1) x = even
(2) 13 is less than x is less than 17

Answer: B

From Statement (1) -- x = even . Always remember square root (x^2) = mod x.
E.g - squareroot (-4 ^ 2) = 4 ; suareroot (4^ 2) = 4. Hence 1 is insufficient.

From Statement (2) --13 is less than x is less than 17 implies that x can be 14, 15 or 16 ..hence X IS POSITIVE...Hence sufficient.

Note : Is x = square root (x^2) is equivalent to stating Is x = mod x

DS Question - 16

IS x^4 + y^4 > z^4 ?

a) x^2 + y^2 > z^2

b) x + y > z

Answer - E

(1) x^2 + y^2 > z^2, when squared, gives the stated equation.

From this we cannot conclude definitively whether x^4 + y^4 > z^4 because the equation contains (2x^2 * y^2).

If this is removed, then x^4 + y^4 may or may not be > z^4.

Thus insufficient..

e.g - 2+3+4 > 5 --- if we remove 1 number from the left hand side of the inequality then the inequality may or may not hold true..

OR

Statement (1) ---- x^2 + y^2 > z^2

Let x = {(2) ^ 1/2}, y = {(3) ^ 1/2}, z = {(4) ^1/2}

Hence 2+3>4 but at the same time 4+9<16

Let x = 2, y= 4, z=3
4+16>9
And 16 + 256 > 81

Thus (1) is insufficient.

Statement (2) ---- x+y> z
Let x=2, y=3, z=6
Hence 2+3<6

Now let x=2, y=6, z=3
Then 2+6>3

Both 1 and 2 together:insufficient

Hence answer E.

Manhattan Challenge Problem of the week ! - 02/26/07

What is xy?

(1)

(2)

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Answer : A

Simplifying the original expression yields:

Therefore: xy = 0 or y — x = 0. Our two solutions are: xy = 0 or y = x.

Statement (1) says y > x so y cannot be equal to x. Therefore, xy = 0. Statement (1) is sufficient.

Statement (2) says x < x =" y" xy =" 0.">

The correct answer is A.

Manhattan Challenge Problem of the week ! - dec 25

Is n less than 1 ?

(1) nx – n less than 0
(2) x–1 = –2

Answer : C

From stat(1): (n^x) – n less than 0 -- no information about x --- insufficient
From stat(2): x^–1 = –2 -- no information about n --- insufficient

Taking both the statements together we get: From stat 2 we get x = -1/ 2
Substituting value of x in stat 1 we get (n^-1/2) -n less than 0
(n^3/2)>1

=> n will be greater than 1 ---- sufficient

Hence C

Problem Solving - 9

9). AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined


Answer - D

Explanation -- AB + CD = AAA
AB and CD are two digit numbers, hence AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
Hence for all values of B between 3 & 9 ---- C = 9
Thus the answer is D i.e (C = 9)

Manhattan Challenge Problem of the week! - sept 4

The Power of Absolutes

If a and b are integers and a is not equal to b, is ab > 0?

(1) a^b > 0

(2) a^b is a non-zero integer

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Manhattan Challenge Problem of the week ! - Aug 7

Question

If j and k are positive integers where k > j, what is the value of the remainder when k is divided by j?

(1) There exists a positive integer m such that k = jm + 5.

(2) j > 5

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

For the answer click on the link below -
What remains to be seen ?

DS Question - 11

a, b, c all are positive, is a/b > (a+c)/(b+c) ?

1. a > c

2. a > b

Answer - B

statement (1). - insufficient as we do not know the relation between a and b

statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)

a/b > [a+c]/[b+c]

= a(b+c) > b(a+c)

= ab+ac > ab+bc

= ac > bc

=a > b

Problem Solving - 5

If |x| > 3, which of the following must be true?

A). x>3

B). x<3

C). x=3

D). x is not equal to 3

E). x<-3

Answer - D

A,B,E - Ruled out - a modulus sign in the question=>cannot have a single signed (can always find a solution with the opposite sign )

When x is positive ----- x >3

When x is negative ---- x <-3

Thus x can never be equal to 3.

DS Question - 6

Give that n is an integer, is n-1 divisible by 3?

1. n^2 +n is not divisible by 3

2. 3n+5>=k+8, where k is a positive multiple of 3

Answer - A

Sttement (1) Sufficient -- It can be simplified as n(n+1) is not divisible by 3

i.e to satisfy the above condition n can take a value immediately following 3 or multiples of 3.

Such as n = 4, 7, 10 .....

Thus case n-1 is always divisible by 3.

Statement (2) Insufficient -- 3n >= k + 3

This implies n >= (k+3)/3

the possible values of k includes 3,6,9,12 .....

So this can be either multiple of 3 or any other number..