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13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).

Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

Answer: D.
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9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.
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SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

The total number of 5-digit codes is \(10^5\). It's important to note that it's not \(9*10^4\), as the first digit can be zero in a password.

The number of passwords with three digits as 6 can be calculated as \(9*9*C^3_5 = 810\). We have 9 choices for each of the two remaining digits (not 6), resulting in \(9*9\). The term \(C^3_5\) represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).

The probability is therefore, \(P=\frac{favorable}{total}=\frac{810}{10^5}\).


Answer: B
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12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33


The statement "when \(x\) is divided by \(y\), the remainder is 3" can be expressed as \(x = qy + 3\), where \(q\) is the quotient, an integer that is greater than or equal to 0. This equation implies that the smallest value for \(x\) occurs when \(q\) equals 0, giving us \(x = 3\). Essentially, this shows that the smallest possible value of \(x\) will be less than \(y\). For instance, if we divide 3 by 4, we indeed get a remainder of 3.

Similarly, the condition "when \(y\) is divided by \(z\), the remainder is 8" implies that the smallest possible value for \(y\) is 8, therefore \(y < z\). As a result, the smallest possible value of \(z\) must be one more than 8, or 9. For instance, if we divide 8 by 9, we indeed get a remainder of 8.

Therefore, the smallest possible value of \(x+y+z\) is \(3+8+9=20\).


Answer: B
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6. A swimming pool has two water pumps, A and B, and one drain, C. Pump A can fill the empty pool by itself in \(x\) hours, while pump B can do so in \(y\) hours. The drain C can empty the entire pool in \(z\) hours, with \(z > x\). When both pumps A and B are operating simultaneously and drain C is open until the pool is filled, which of the following represents the fraction of the total pool volume added by pump A until the pool is filled?

A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

When both pumps A and B are running, and the drain is open, the pool fills at a rate of \(\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac{yz + xz - xy}{xyz}\) pools per hour. Therefore, it takes \(\frac{xyz}{yz + xz - xy}\) hours to fill the pool (since time is the reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours, pump A will contribute \(rate * time = \frac{1}{x} * \frac{xyz}{yz+xz-xy} = \frac{yz}{yz+xz-xy}\) of the total pool volume.

For instance, suppose the pool will be filled in 10 hours, and pump A alone can fill the entire pool in 20 hours. In 10 hours, pump A will pump \(\frac{1}{20} * 10 = \frac{1}{2}\) of the total pool volume.

Answer: B
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10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Notice that we are not told that \(x\) and \(y\) are integers.

\(x^2<81\) means that \(-9<x<9\) and \(y^2<25\) means that \(-5<y<5\). Now, since the largest value of \(x\) is almost 9 and the largest value of \(-2y\) is almost 10 (for example if \(y=-4.9\)), then the largest value of \(x-2y\) is almost 9+10=19, so the actual value is less than 19, which means that the largest prime that can be equal to \(x-2y\) is 17. For example: \(x=8\) and \(y=-4.5\).

Answer: D.
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2 If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2} }\), then \(y\) is NOT divisible by which of the following?

A. \(6^4\)
B. \(62^2\)
C. \(65^2\)
D. \(15^4\)
E. \(52^4\)


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2} }=\)

\(=(3^5-3^2)^2*(5^7-5^4)^2=\)

\(=3^4*(3^3-1)^2*5^8*(5^3-1)^2=\)

\(=3^4*26^2*5^8*124^2=\)

\(=2^6*3^4*5^8*13^2*31^2\).

Now, if you examine each option, it becomes clear that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).


Answer: E
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4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=23^27\), so for \(126*\sqrt{k}\) to be the square of an integer, \(\sqrt{k}\) must complete the powers of 2 and 7 to an even number. Therefore, the smallest value of \(\sqrt{k}\) must be \(2*7=14\), which makes the smallest value of \(k\) equal to \(14^2=196\).

Answer: D.
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3. For the past \(k\) days, Liv has baked an average (arithmetic mean) of 55 cupcakes per day. Today, Bibi helped Liv, and together they baked 100 cupcakes, raising the average to 60 cupcakes per day. What is the value of \(k\)?
A. 6
B. 8
C. 9
D. 10
E. 12

The total number of cupcakes baked in \(k\) days was \(55k\), which means that the total number of cupcakes baked in \(k+1\) days was \(55k+100\). The new average is \(\frac{55k+100}{k+1} = 60\).

Thus, \(55k + 100 = 60k + 60\) and \(k = 8\).

Answer: B.
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8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).

Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)

Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185\) --> \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\) --> \((5x+20)+20=-185\) --> \(5x+20=-205\)

Answer: D.
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11. In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?

A. \(10*3^{11}\)
B. \(20*3^{11}\)
C. \(10*3^1\)
D. \(40*3^{11}\)
E. \(20*3^{12}\)

You don't need to use the geometric progression formula to solve this question. Instead, simply recognize the pattern:

\(b_1=1=3^0\);

\(b_2=3=3^1\);

\(b_3=9=3^2\);

\(b_4=27=3^3\);

...

Thus, for any term \(b_n\), its value is \(3^{n-1}\).

The difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms is:

\((b_{13}+b_{15})-(b_{12}+b_{14})=\)

\(=3^{12}+3^{14}-3^{11}-3^{13}=\)

\(=3^{11}(3+3^3-1-3^2)=\)

\(=20*3^{11}\)

Answer: B.
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1) Ans B

The 3 digits that will be '6' can be selected in 5c3 = 10 ways

The remaining two digits can be any digit from 0 to 9 other than 6 i.e 9 digits.

so favorable outcomes = 5c3 *9*9*1*1*1 = 810
total outcome = 10*10*10*10 *10 = 10^5

P(exactly three 6) = 810/10^5


2) Ans E

y = (3^5-3^2)^2/ (5^7 -5^4)^-2

Numerator (3^2(3^3-1))^2 = (3^2(26) )^2 = 3^4*13^2*2^2

Denominator (5^4(5^3-1))^-2 = 1/(5^4(124))^2 = 1/5^8*4^2*31^2 = 1/5^8*2^4*31^2

Numerator/denominator = 3^4 *13^2 *2^2 *5^8 *2^4*31^2 = 3^4*13^2*31^2*2^6

From answer choices we see that 52^4 is not divisble by y


3) Ans B

Old sum/k = 55
old sum = 55 k

old sum +100 /k+1 = 60
old sum+100 = 60k+60

55k+100 = 60k+60

5k = 40 k= 8


4) Ans D

126 * sqrt k = a^2 where a is a +ve integer

7*2*3^2 *sqrt k = a^2

So we need one two and one 7 to make 'a' a perfect square
if k = 196 , 7*2*3^2*sqrt 196 = 2^2*7^2*3^2

5) For selecting 8 marbles from 12 marbles, a maximim of 4 marble can remain in
the jar . The scenarios for atleast 1 red marble and 1 blue marble to remain are:

1r1b 7c1*5c1 = 35
1r2b 7c1*5c2 = 70
2r1b 7c2*5c1 = 105
2r2b 7c2*5c2 = 210
3r1b 7c3*5c1 = 175
1r3b 7c1*5c3 = 70

Total 665 . Not sure if the logic is right..

6) Ans E

Work done by pump a,b,c in 1 hr = 1/x , 1/y, 1/z respectively

Working simultaneously, work done by them in 1 hr is:

1/x+1/y-1/z

yz +xz -xy/xyz job done in 1 hr

So whole job will be completed in xyz/yz+xz-xy hrs

Since pump A work for x hrs, fraction of job completed by pump A is :

x/(xyz/yz+xz-xy) = yz+xz-xy/yz


8) Ans) D

let the 7 consecutive odd integers be:

x, x+2, x+4, x+6, x+8, x+10, x+12 (in ascending order)

Sum of the 5 largest integers is x+12 +x+10 +x+8 +x+6 +x+4 = 5x+40

So 5x +40 = -185
5x = -225
x = -45

Sum of the 5 smallest integers = x+x+2+x+4+x+6+x+8 = 5x+20

5x+20 = (5*-45) +20 = -225+20 = -205


9) Ans c

sqrt x^2 = |x| = -x (since X<0)

sqrt (-y*|y|) = sqrt(-y*-y) sqrt y^2 = y

sqrt x^2/x - sqrt(-y*|y|) = -x/x - y = -1-y


10) Ans c

x^2<81 -9<x<9 y^2<25 -5<y<5

For x -2y to be a maximim prime number, x should be maximim prime number
and y minimum prime number

So x-2y = 7 - (2*-3) =13


11) Ans B

nth term of GP sereis is an = a1*r^n-1 where r =3 common ratio , a1 -=1

a13 = 1*3^12

a15 = 1^3^14

a13+a15 = 3^12 +3^14 = 3^12(1+3^2) = 3^12*10

Similarly a12 = 3^11 a14 = 3^13

a12+14 = 3^11+3^13 = 3^11*10

(a13+a15) - (a12+a14) = 3^12*10 - 3^11*10

3^11*10(3-1) = 3^11 *20


12)Ans B

x = yq1+3 y>3 since divisor should be greater than remainder

y = zq1+8 z>8 since divisor should be greater than remainder

so least value of z is 9 and least value of y 8 since for 8/9 remainder is 8

least value of x is 3 since for 3/8 remainder is 3

Least value of x+y+z = 3+8+9 = 20


13) Ans D

x = (8!)^10 - (8!)^6 / (8!)^5 -(8!)^3

Numerator, taking 8!^6 common ,(8!)^6 ((8!)^4-1) = (8!)^6 [(8!)^2+1)(8!)^2-1)]

Denominator, taking 8!^3 common (8!)^3 [(8!)^2-1]

x = (8!)^6 [(8!)^2+1)(8!)^2-1)] / (8!)^3 [(8!)^2-1]

x = (8!)^3[(8!)^2+1)

x = (8!)^5 +(8!)^3


x/(8!)^3 -39 = [(8!)^5 +(8!)^3/(8!)^3] -39

= [(8!)^2 +1]-39

=(8!)^2-38

8!^2 is a big number and will have 2 trailing zeros..

So some big number ending in 00 - 38 will have
digits 2,6 in units and tens palce respectively

So product is 2*6 =12
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Hi Bunuel,
Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares

Thanks
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7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000

Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders --> Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares;
Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders --> Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares;
Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders --> Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;

Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1-\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).

Answer: D.

It's quite simple: A has $2 and B has 3$ --> A has 2/3rd of B's amount and also 2/(2+3)=2/5th of total amount of $5.

Hope it's clear.
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6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. \(\frac{yz}{x+y+z}\)

B. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

Solution: baker-s-dozen-128782-20.html#p1057508


Reading the discussion on this question made me realize how I end up retracting on the suggested strategies again and again. Since we are talking about Quant, you would expect to have a definite set of rules but no sir, you don't.

I repeat this quite often: When you have a question using variables and the answer is in terms of those variables, your life is easier than if you get a question using numbers. The reason - you can assign your own 'cool' values to the variables for which the relations hold. Then just go on and check the option which gives you the right answer.
I retract it with the following - But if the number of variables is high, say 3 or 4 or more, it might be too cumbersome to plug in values for each variable and keep in mind what stands for what. This could lead to a lot of confusion and errors.
I retract it again - But if you can give some very convenient values to the variables, go ahead and plug it in.

This question has variables and the answer is in terms of the variables so plugging in values is a good option. But the number of variables is 3 which could make it cumbersome. But, you can give the variables such values that you get your answer quickly.

I need to find the rate of A. There are no constraints on the values x, y, and z can take except z > x (drain C empties slower than pipe A fille)

Let's say, x = 4, y = 8, z = 8
What did I do here? I made the rate of B same as the rate of C. This means, whenever both of them are working together, drain C cancels out the work of pipe B. So the entire pool will be filled by pipe A and the amount of water pumped in by A will be the entire pool. Hence, if y = z, pipe A fills the entire pool i.e. the amount of water in terms of fraction of the pool pumped by A is 1.
In the options, put y = z and see which option gives you 1.
Only options (B) and (E) do.

Now let's say, x = 8, y = 4, z = 8.00001 ( z should be greater than x but let's assume it is infinitesimally greater than x such that we can approximate it to 8 only)
Rate of work of C is half the rate of work of B. Rate of work of C is same as rate of work of A. All the work done by pipe A is removed by drain C. So if pipe B fills the pool, drain C empties half of it in that time (this is the water pumped by pipe A)
Put x = z in the options B and E. You get y/z = 4/8 = 1/2 in option (B).

Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. It might give you some ideas of logical solutions in some other questions.
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Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.
Did this way- prob of choosing 6 will be 1/10 on each of he three times (and these choices can be in \(5C3\)=10 ways) and the rest two digits can be any thing from set 0-9 (except 6) so in total we have 9 choices (i..e a prob of 9/10 for the rest of the 2 positions). So total prob will be\(10*(1/10*1/10*1/10*9/10*9/10)= 810/10^5\)(B)
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Bunuel
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).

Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

Answer: D.

Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)--

first I simplified the whole x value as it looks so clumsy/heavy: given
\(x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1] \\
= (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1]\\
so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39\\
= (8!)^2 -38\)
Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since \(8 < 5^2\) so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So \((8!)^2\) has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12...

P.S.- just for refrence # trailing 0's in factorial N is determined by- \(N/5^1+N/5^2+....+N/5^k (where 5^k < N)\)for each division we take integer part (quotient) and then add them up. this should give # trailing 0's
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Bunuel


As for your other question: I don't see a solution for #7, so you can try this one.
ok my answer to the q7 is D , because D is the leader of the day hehe :-D kidding
the answer is really D ,because-

2/3(L+A+W) =F
3/7(F+W+A)=L
4/11(F+W+L)=A
total=360 (I deleted the 0s just to make a number easy)
W=?

2/3(L+A+W) =F means F=2/3(360-F) =>(2/3)*360=5/3F=> F=144
3/7(F+W+A)=L means L=3/7(360-L) => L=108
4/11(F+W+L)=A means A=4/11(360-A)=>A=96

W=360- (L+A+F)=360-144-108-96=12


or W=120 000 (return the 0s which were deleted before)
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