Baker's Dozen

5. In how many ways can we select 8 shirts from a closet containing 7 distinct red shirts and 5 distinct blue shirts, such that at least one shirt of each color remains in the closet, if the order of selection does not matter?
A. 460
B. 490
C. 493
D. 455
E. 445

The total number of ways to select 8 shirts out of 12 distinct shirts (7 red and 5 blue) is given by the combination formula: \(C^8_{12}\).

The number of ways to select 8 shirts such that no red shirts are left in the closet is given by the product of the combinations of choosing all 7 red shirts and 1 blue shirts: \(C^7_7 * C^1_5\).

The number of ways to select 8 shirts such that no blue shirts are left in the closet is given by the product of the combinations of choosing all 5 blue shirts and 3 red shirts: \(C^5_5 * C^3_7\).

Therefore, the number of ways to select 8 shirts so that at least one red shirt and at least one blue shirt remain in the closet can be found by subtracting the sum of the two previous cases from the total number of ways to select 8 shirts: \(C^8_{12} - (C^7_7 * C^1_5 + C^5_5 * C^3_7) = 495 - (5 + 35) = 455\).


Answer: D

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.

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.

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

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

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.

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

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.

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.

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.

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.

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

In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?

You figure it out by knowing that there is a 5 as a factor in the final number.

example 8! = 8*7*6*5...*1

Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ...

Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number )
then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure

how many zeros at the end of 312!
answer :- 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence :- 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s .

IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,

Ans 1
Total number of ways = 10 x 10 x 10 x 10 x 10 = 100,000
total number ways with exactly three six = 5C3 * 9*9 = 810
correct Answer B
p.s IMO one more zero is required in the denominator.

Ans 2
resolving the numerator/denominator, we get
3^4 * 5^8 * 2^2 *13^2 * 2^2 *62^2
so A to D is divisible by the above
Correct Answer E
Ans 3
This one is little simple
the equation is 55*k +100 = 60(k + 1)
thus 5k = 40
k = 8
Correct Answer B
Ans 4
3*3*7*2*\sqrt{k} is the square of a positive integer( suppose X^2)
so we need 7 and 2 and put them in square root as 14 * 14= 196
Correct Answer D

x ;x+2;x+2*2;x+2*3....x+2*6

since the set has consecutive odd integers, then mean =median

mean of 5 largest integers=-185/5= -37
median of 5 largest integers is the 3d of the largest or the 5th of the whole set A= x+2*4
x+8=-37 ;x=-45

median of the first 5 small integers =-45+2*2=-41


the sum of the first small odd integers=number of integers*mean (or median)=5*median=5*(-41)=-205
answ is

I am not so certain whether I am right but I would go with D in #9.

IF x<0 and y<0 then sqrt(x^2)/x=|x|/x, then -x/x=-1; because there is a minus in front of x, than -(-x) should be positive.

sqrt two negative y gives square root of y^2, than because y<1, y is negative as well. -y.

From the expression we've got -1+y, thus D should be a correct answer.