Four Problem Solving Questions from OG 12

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Four Problem Solving Questions from OG 12

by gmat_mba » Wed Apr 08, 2009 2:52 pm
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!
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1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12

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2) If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?
a)Three
b)Four
c)Five
d)Six
e)Nine

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3) If x, y, and k are positive numbers such that (x / x+y)(10) + (y / x+y)(20) = k and if x<y, which of the following could be the value of k?

a)10
b)12
c)15
d)18
e)30

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4)For any positive integer n, the sum of the fist n positive integers equals (n(n+1)) / 2. What is the sum of all even integers between 99 and 301?
a)10,100
b)20,200
c)22,650
d)40,200
e)45,150
Source: — Problem Solving |

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Re: Four Problem Solving Questions from OG 12

by dtweah » Wed Apr 08, 2009 5:00 pm
gmat_mba wrote:If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!
----------------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12

--------------------------------------------------------------------------------------------------------
2) If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?
a)Three
b)Four
c)Five
d)Six
e)Nine

--------------------------------------------------------------------------------------------------------
3) If x, y, and k are positive numbers such that (x / x+y)(10) + (y / x+y)(20) = k and if x<y, which of the following could be the value of k?

a)10
b)12
c)15
d)18
e)30

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4)For any positive integer n, the sum of the fist n positive integers equals (n(n+1)) / 2. What is the sum of all even integers between 99 and 301?
a)10,100
b)20,200
c)22,650
d)40,200
e)45,150
1)

96.12= 96 12/100= 96 3/25= 96 3/25 (3/3) = 96 9/75
Choose B.

Sume of first 301 numbers =
301

2. 1/2^6 x2^3 x 125 = 1/64 x 1000 = .016/1000 = .000016 =

Choose B


3 (x/x+y +2y/x+y )10
( x+2y/x+y)10
IF x=y
( 3y/2y) x10- 15

Since x<y demonitor is smaller result must be > 15. That leaves 30 and 18. 30 means the fractional term has to equal 3, which is not possible. Choose D.

4)

I don't need the formula provided to solve.

There are 202 integers between 100 and (301 301 -100 +1) inclusive and since the sequence begins with even and ends with odd, There are 101 even numbers. I go to the sum of an arithmetic sequence

S_101 = 101/2 ( 2 x100 +(101-1)2)
101/2 (200+200)
101 x 200

20200

Choose B

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by krisraam » Wed Apr 08, 2009 5:15 pm
1. x/y = 96.12

The remainder when divided by y gives the decimals .12

9/y = .12 ==> y = 75.


2. 1/(2^3 * 5^3 * 2^6)

1/64000

No of zeros is 4.

3. simplifying we get 10 + 10(y/x+y) = k

We know that x<y

0< y/(x+y) < 1 and x < y.

substitute the values for k.
K = 18 satisfies the equation with x = 1 and y = 4. Rest of the values have
x > y.

4. No of even numbers between 99 and 301. ie between 100 and 300 included is 101

Sum Sn = 101/2(2*100 + (101 -1)2) = 20200

Thanks
Raama