some tricky PS and DS questions

Question

I had some difficulties solving these problems:


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I. A decorator bought a bolt of defective cloth that he judged to be 3/4 usable, in which case the cost would be 0.80 USD per usable yard. If it was later found that only 2/3 of the bolt could be used, what was the actual cost per usable yard?

0.60; 0.90; 1.00; 1.20; 1.70


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II. If x, y, and z are single-digit integers and 100(x) + 1,000 (y) + 10(z) = N, what is the units' digit of the number N?

0; 1; x; y; z


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III. Three stacks containing equal numbers of chips are to be made from 9 red chips, 7 blue chips, and 5 green chips. If all of these chips are used and each stack contains at least 1 chip of each color, what is the maximum number of red chips in any one stack?

7; 6; 5; 4; 3


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IV. What is the remainder when the positive integer x is divided by 2?

 (1) x is an odd integer.
 (2) x is a multiple of 3.


Guys. please advise/

pi10t · Answer

The answer is (A).

Let's suppose that $X is the price of the bolt. The bolt size is Z yards.

Then (1) : X / (Z*3/4) = $0.8. it was expected by the decorator.

In reality he got (2) : X / (Z*2/3) = $Y. We need to find out what is the actual cost $Y per usable yard?

substitute X from (1) to (2) : 
  /   = $Y. Z is canceled.

3/4*0.8*(3/2) = 0.6 = Y.

So (A).

pi10t · Answer

(A) : 0.

In any set of (x,y,z) we get the number xyz0 or y000 + x00 + z0 = xyz0.
For example 9870 : 9000 + 800 + 70 = 9870. 

Every time the units' digit is 0.

pi10t · Answer

(C) : 5.

Every stack consists of the 7 chips ( 21 for the 3 stacks).

Let's assume 1 stack consists of the one blue chip, one green chip and 5 of the red ones:

BG(5R). 

5 red is the maximum number. 

it is impossible to insert more than 5 red chips in one stack. because the rule will be broken "each stack contains at least 1 chip of each color".

pi10t · Answer

(A).

x is divided by 2? x = 2*k + R, where k- integer and R- remainder. R = ?

(1) x is odd. it means that x = 2*k +1 , i.e. x = 1,3,5,7,9, ..... 
then R = 1. so (A)

(2) x = 3*k = 2*k + k, so R=k . R could have any value k. insuff.

Thus, only (A).

==================

by the way I still can't figure out where are any tricks here?

and what is the source of the questions?

nick_sun · Answer

Thank you, pi10t. Your answers are correct. Now I can see that there are no tricks there

The source is G+. :read

Update: Sorry, I have overlooked it :lol:

OA for the first problem is B (0.90)

Mishari · Answer

I. A decorator bought a bolt of defective cloth that he judged to be 3/4 usable, in which case the cost would be 0.80 USD per usable yard. If it was later found that only 2/3 of the bolt could be used, what was the actual cost per usable yard? 

X: the judged cost by the decorator
Y: the actual cost

 3/4 x = 2/3 y 
x = 0.8 --> y = 3/4 x 3/2 x 0.8 = 9/8 x 0.8 = 0.9

the actual cost is  0.9  USD per usavle yard 

NOTE:  Pi10t : you got your equation correct
3/4*0.8*(3/2) = 0.6 = Y. This should yield to 0.9 = Y.
I think you had an arithmatic error here.
the answer is  B 


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II. If x, y, and z are single-digit integers and 100(x) + 1,000 (y) + 10(z) = N, what is the units' digit of the number N? 
 
No matter what x,y,and z are, each one of them is multiplied by 10, 100, or 1,000. Therefore, the units' digit is 0

A

If you can't visualize it, try to pick simple single-digit integers for x , y , and z and see what happens:  
N = 200 + 4000 + 70 = 4270 --> units' digit of N is  zero  


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III. Three stacks containing equal numbers of chips are to be made from 9 red chips, 7 blue chips, and 5 green chips. If all of these chips are used and each stack contains at least 1 chip of each color, what is the maximum number of red chips in any one stack? 

 First, we have 9+7+5 = 21 chips for 3 stacks --> each stack contains 21/3 = 7 chips.

Now, since we're looking for maximum # of red chips in one stack, yet each stack must contain at least one red chip, let's use one chip only to minimize the number of red chips in two of the stacks. This will allow to maximize the number of red chips in the third stack.

Stack 1   :1 green , 1 blue , 5 red
There are enough red chips such that each other stack would contain at least one red chip. 

 Answer is C   



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IV. What is the remainder when the positive integer x is divided by 2? 

(1) x is an odd integer. 
(2) x is a multiple of 3. 


 x/y = qy + r where q is the qoutient and r is the remainder

x/2 = 2q + r

Statement 1: x is odd 
if x is odd, then x is not divisible by 2 and the remainder r = 1
You can pick numbers to confirm this fact.

--> Statement 1 is sufficient

Statement 2: x is a multiple of 3 
x could be 6 and thus r = 0 or x could be 9 and thus r = 1

--> Statement 2 is insufficient 

 Answer is A

Summer3 · Answer

My answers: I 0.90 II Zero III 5 IV A For sum no I, I used 12 yard for the total (since 3/4 and 2/3 are given and denominators 4*3 = 12).. to get the whole numbers for 3/4 and 2/3. 3/4 of 12 = 9 yard * 0.80 = $ 7.2 2/3 of 12 = 8 yard => 7.2/8 = $ 0.90 per yard <----Answer Other sums were very easy

some tricky PS and DS questions

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