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Arithmatic-percent

by monirjewel » Sun Nov 14, 2010 10:29 pm
1. A retailer bought a machine at a wholesale price of $90 and later on sold it after a 10% discount of the retail price. If the retailer made a profit equivalent to 20% of the whole price, what is the retail price of the machine?
(A) 81
(B) 100
(C) 120
(D) 135
(E) 160

2. Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have?
(1) Max has fewer than 5 bills worth $5 each
(2) Max has more than 5 bills worth $20 each

3. If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8

4. A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8,12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees?
(A) 5
(B) 7
(C) 8
(D) 10
(E) 12
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by Rahul@gurome » Mon Nov 15, 2010 12:29 am
Question Number 1:
1. A retailer bought a machine at a wholesale price of $90 and later on sold it after a 10% discount of the retail price. If the retailer made a profit equivalent to 20% of the whole price, what is the retail price of the machine?
(A) 81
(B) 100
(C) 120
(D) 135
(E) 160
Say, the retail price is $100x.
After a discount of 10% the machine is sold at a price = $90x

According to the question the machine was sold at a price = Wholesale price + 20% of wholesale price = $90 + ($90)*(0.2) = $90 + $18 = $108

Therefore, 90x = 108 => x = 1.2 => 100x = 120
Thus, retail price is $120.

The correct answer is C.
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by Rahul@gurome » Mon Nov 15, 2010 12:48 am
Question Number 2:
2. Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have?
(1) Max has fewer than 5 bills worth $5 each
(2) Max has more than 5 bills worth $20 each
Given: Max has $125 consisting of bills each worth either $5 or $20.
Say, number of $5 bill = x and number of $20 bill = y.
Thus, (5x + 20y) = 125. We need to solve this equation for integer values of x and y.
Note that, y must be less than or equal to 6. Because if y > 6, x becomes negative.
Therefore, possible combination of x and y are,
  • (1) y = 1, x = 21
    (2) y = 2, x = 17
    (3) y = 3, x = 13
    (4) y = 4, x = 9
    (5) y = 5, x = 5
    (6) y = 6, x = 1
Statement 1: Max has fewer than 5 bills worth $5 each.
Therefore, x < 5. Only possible combination is x = 1 and y = 6

Sufficient.

Statement 2: Max has more than 5 bills worth $20 each.
Therefore, y > 5. Only possible combination is x = 1 and y = 6

Sufficient.

The correct answer is D.
Last edited by Rahul@gurome on Mon Nov 15, 2010 12:49 am, edited 1 time in total.
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by ymach3 » Mon Nov 15, 2010 12:49 am
2. Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have?
(1) Max has fewer than 5 bills worth $5 each
(2) Max has more than 5 bills worth $20 each

Answer is D.

(1) ---> 1*5+6*20=125---> only one combi'n

simillarly, (2)---> 20*6+1*5=125--> One comb'n

Hence , either of the options answers the Qn.


2) If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8

=> (7^4n*243*6^n)/10

We can see that 7 raised to the power 4n always ends with 1
and similarly 6 raised to any power ends with 6

=> (243*1*6)/10

=> (....18)/10

(E )-my answer
=> Remainder=8
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by Rahul@gurome » Mon Nov 15, 2010 1:00 am
Question Number 3:
3. If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8
When any integer is divided by 10, the remainder is nothing but the units digit of the integer.

[7^(4n + 3)]*[6^n] = [(7^4)^n]*[7^3]*[6^n]

Now, units digit of
  • # 7^4 is 1
    # (7^4)^n is 1
    # 7^3 is 3
    # 6^n is 6
Therefore, units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8.

The correct answer is E.
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by Rahul@gurome » Mon Nov 15, 2010 1:15 am
Question Number 4:
4. A club with a total membership of 30 has formed 3 committees, M, S and R, which have 8,12 and 5 members respectively. If no members of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees?
(A) 5
(B) 7
(C) 8
(D) 10
(E) 12
Total member in the club = 30
Total member in committee M = 8
Total member in committee S = 12
Total member in committee R = 5

As no members of committee M is on either of the other 2 committees, total number of people who are in either R or S or both or none of them = 30 - 8 = 22
Using set theoretic notations, 22 = (R U S) + Number of members who are on none of the committees

To maximize number of members who are on none of the committees, we have to minimize (R U S).

Minimum value of (R U S) = Maximum of R and S = S = 12. [This is because number of people in either in R or S or both will be minimum when smaller one of them is a subset of the larger one.)

Thus, greatest possible number of members who are on none of the committees = 22 - 12 = 10

The correct answer is D.
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by AndyB » Mon Nov 15, 2010 2:18 am
Hi Rahul,

3. If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8
When any integer is divided by 10, the remainder is nothing but the units digit of the integer.

[7^(4n + 3)]*[6^n] = [(7^4)^n]*[7^3]*[6^n]

Now, units digit of

# 7^4 is 1
# (7^4)^n is 1
# 7^3 is 3
# 6^n is 6

Therefore, units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8.

How do we make sure that (7^4)^n is 1 and 6^n is 6
Please could you help me in understanding this point.
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by ankur.agrawal » Mon Nov 15, 2010 3:46 am
AndyB wrote:Hi Rahul,

3. If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8
When any integer is divided by 10, the remainder is nothing but the units digit of the integer.

[7^(4n + 3)]*[6^n] = [(7^4)^n]*[7^3]*[6^n]

Now, units digit of

# 7^4 is 1
# (7^4)^n is 1
# 7^3 is 3
# 6^n is 6

Therefore, units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8.

How do we make sure that (7^4)^n is 1 and 6^n is 6
Please could you help me in understanding this point.
Also how do u know that units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8. ?
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by Rahul@gurome » Mon Nov 15, 2010 4:14 am
AndyB wrote:...
How do we make sure that (7^4)^n is 1 and 6^n is 6
Please could you help me in understanding this point.
...
Also how do u know that units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8?
The basic idea is that in case of multiplication, the units digit of the product is nothing but the units digit of the individual product of the units digit of the multiplicands. If this sounds confusing, let's take some examples.

12*7 = 84 => 4 is units digit of 7*2 = 14.
19*14 = 266 => 6 is units digit of 9*4 = 36.
313*72 = 22536 => 6 is units digit of 3*2 = 6.

Now, I assume that you've understood that the units digit of 7^4 is 1. Therefore, if we raise 7^4 to any integer power, its units digit will be 1 as exponentiation with positive integer is nothing but multiplying the number with itself. Thus, units digit of (7^4)^n is always 1.

If we multiply 6 with 6, the result is 36, units digit is again 6. That means if 6 repeatedly multiplied with 6 itself, the units digit of the result will always be 6. Thus, units digit of 6^n is always 6.

Now, units digit of (7^4)^n is 1, units digit of 7^3 is 3 and units digit of 6^n is 6. Therefore units digit of their product will be units digit of (1*3*6) = 18, i.e. 8.

Hope it is clear now.
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by ankur.agrawal » Mon Nov 15, 2010 4:20 am
YOU ARE A LIFE SAVER MAN! THANKS IS A UNDERSTATEMENT. UR AWESOME.
Rahul@gurome wrote:
AndyB wrote:...
How do we make sure that (7^4)^n is 1 and 6^n is 6
Please could you help me in understanding this point.
...
Also how do u know that units digit of the product [(7^4)^n]*[7^3]*[6^n] is 8?
The basic idea is that in case of multiplication, the units digit of the product is nothing but the units digit of the individual product of the units digit of the multiplicands. If this sounds confusing, let's take some examples.

12*7 = 84 => 4 is units digit of 7*2 = 14.
19*14 = 266 => 6 is units digit of 9*4 = 36.
313*72 = 22536 => 6 is units digit of 3*2 = 6.

Now, I assume that you've understood that the units digit of 7^4 is 1. Therefore, if we raise 7^4 to any integer power, its units digit will be 1 as exponentiation with positive integer is nothing but multiplying the number with itself. Thus, units digit of (7^4)^n is always 1.

If we multiply 6 with 6, the result is 36, units digit is again 6. That means if 6 repeatedly multiplied with 6 itself, the units digit of the result will always be 6. Thus, units digit of 6^n is always 6.

Now, units digit of (7^4)^n is 1, units digit of 7^3 is 3 and units digit of 6^n is 6. Therefore units digit of their product will be units digit of (1*3*6) = 18, i.e. 8.

Hope it is clear now.
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by GMATGuruNY » Mon Nov 15, 2010 4:39 am
monirjewel wrote: If n is a positive integer, what is the remainder when ((7^(4n+3)(6^n)) is divided by 10?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8

Two rules we need to know in order to solve this problem:

When an integer is divided by 10, the remainder will be the units digit of the integer:

25/10 = 2 R5 (remainder of 5 = units digit of 25)
137/10 = 13 R7 (remainder of 7 = units digit of 137)

When multiplying positive integers x*y*z, to determine the units digit of the product:

1. Multiply the units digits of x, y and z
2. The units digit of this product will be the units digit of x*y*z

147*356*881:
The product of the units digits is 7*6*1 = 42. Thus, the units digit of 147*356*881 is 2.
584*223*958:
The product of the units digits 4*3*8 = 96. Thus, the units digit of 584*223*958 is 6.

To solve the problem above, we'll need to know the units digit when 7 is raised to a power. Let's examine the pattern when 7 is raised to consecutive powers. We need to determine only the units digits of each result:

7^1 = 7.
7^2 = 9.
7^3 = 3.
7^4 = 1.
7^5 = 7.
etc.

The pattern repeats in a cycle of 4: 7,9,3,1
So when 7 is raised to multiple of 4, the units digit will be 1.

Plug n=1 into the problem above:

((7^(4*1+3)(6^1)
= (7^7)*6

Following the pattern 7,9,3,1, we see that:
7^4 has units digit of 1
7^5 has a units digit of 7
7^6 has a units digit of 9
7^7 has a units digit of 3

Thus, the units digit of (7^7) is 3. The units digit of 6 is 6. Multiplying the units digit of each factor, we get 3*6 = 18. Thus, the units digit of (7^7)*6 is 8.

Since the units digit is 8, when (7^7)*6 is divided by 10, the remainder also will be 8.

The correct answer is D.
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by AndyB » Mon Nov 15, 2010 9:22 pm
Thanks Mitch,

We now have a different way of solving this problem.
But could you please clarify, is the pattern logic applicable to all the questions of this type..???

Regards,
AndyB.
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by AndyB » Mon Nov 15, 2010 9:23 pm
Thanks Rahul,

The solution to the problem is clear now.

Regards,
AndyB.
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by GMATGuruNY » Tue Nov 16, 2010 12:17 pm
AndyB wrote:Thanks Mitch,

We now have a different way of solving this problem.
But could you please clarify, is the pattern logic applicable to all the questions of this type..???

Regards,
AndyB.
To determine the units digit of an integer raised to a large power, we can determine the resulting units digits when the integer is raised to smaller powers until we see the pattern. For example:

What is the units digit of 3^58?

3^1 = 3.
3^2 = 9.
3^3 = 27.
3^4 = 81.
3^5 = 243.

Now we can see that the pattern repeats in a cycle of 4: 3,9,7,1...3,9,7,1...and so on. When 3 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus, 3^56 will have a units digit of 1. Following the pattern, we can see that 3^58 will have a units digit of 9.
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by AndyB » Tue Nov 16, 2010 6:36 pm
Thanks a lot for the explanation.
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