(3 + 1) interesting math problems

[dat_student]

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#1)
The perimeter of a sector of a circle is 12 units. Find the radius so that the area of the sector is a maximum. 🙂

Answer: r = 3 units

#2)
Find the last two digits of N = 11^10 - 1 without using a calculator & without multiplying 11 ten times 😉

Answer: N ends in 00.

#3)
Find the least value of x^4 + y^4 subject to the restriction x^2 + y^2 = c^2

Answer = (C^4) / 2

#4)
A person, starting with 64 cents,making 6 bets, winning 3 times and losing 3 times. The wins and losses come in random order, and each wager is for half the money remaining at the time the wager is made. If the chance for a win equals the chance for a loss, find the final result

Answer: There is a loss of 37 cents although number of wins = number of losses AND the chance for a win = the chance for a loss - easy problem but interesting answer)

#5)
What is the maximum value of ab / 4(a + b)^2 when a & b are positive numbers

Answer = 1/16

Everyone, please add your math questions to this thread...

 

[organichemistry]

#1)
The perimeter of a sector of a circle is 12 units. Find the radius so that the area of the sector is a maximum. 🙂

Answer: r = 3 units

#2)
Find the last two digits of N = 11^10 - 1 without using a calculator & without multiplying 11 ten times 😉

Answer: N ends in 00.

#3)
Find the least value of x^4 + y^4 subject to the restriction x^2 + y^2 = c^2

Answer = (C^4) / 2

#4)
A person, starting with 64 cents,making 6 bets, winning 3 times and losing 3 times. The wins and losses come in random order, and each wager is for half the money remaining at the time the wager is made. If the chance for a win equals the chance for a loss, find the final result







Answer: There is a loss of 37 cents although number of wins = number of losses AND the chance for a win = the chance for a loss - easy problem but interesting answer)

definitely harder than the QR

 

[mutual]

#1)
The perimeter of a sector of a circle is 12 units. Find the radius so that the area of the sector is a maximum. 🙂

Answer: r = 3 units

Hey guys, this is a good question. I've seen similar but easier question of this type both in the DAT and Q14 of QRT 2 in DAT Achiever. (A word problem on 3-sided corral fronting a river mouth...)

Solution:

Perimeter of a sector of any angle, q rad, P:

r + r + rq = 2r + rq = 12

q = (12 -2r)/r ---------- (1)

Area of a sector:

A = (pi * r^2) * (q/2pi) = (r^2)/2 * q ---------- (2)

Substitute (1) into (2),

A = (r^2)/2 * (12 - 2r)/r

A = 6r - r^2

dA/dr = (6 - 2r) = 0 when A is at its maximum

r = 3 units

(Note: Knowing how to apply basic calculus rule(s) is one testing edge to judge how well you're going to perform in the QRT.) 👍 👍 😉

 

[mitshi]

Hey guys, this is a good question. I've seen similar but easier question of this type both in the DAT and Q14 of QRT 2 in DAT Achiever. (A word problem on 3-sided corral fronting a river mouth...)

Solution:

Perimeter of a sector of any angle, q rad, P:

r + r + rq = 2r + rq = 12

q = (12 -2r)/r ---------- (1)

Area of a sector:

A = (pi * r^2) * (q/2pi) = (r^2)/2 * q ---------- (2)

Substitute (1) into (2),

A = (r^2)/2 * (12 - 2r)/r

A = 6r - r^2

dA/dr = (6 - 2r) = 0 when A is at its maximum

r = 3 units

(Note: Knowing how to apply basic calculus rule(s) is one testing edge to judge how well you're going to perform in the QRT.) 👍 👍 😉

The corral word question appearing in the DAT and DAT Achiever is of equal amount of difficulty (not easier like mutual said). The key point here is to differentate equation with respect to one varible instead of two via substitution of (1) into (2).

This is one question that has trapped many inexperienced candidates working backward (a long and clumsy way) from answers instead of applying a simple calculus rule here.

dat_student, if you need to get 30 like you mentioned in an earlier post from other thread, you may want to pay attention to this. 😍

 

[luder98]

Hey guys, this is a good question. I've seen similar but easier question of this type both in the DAT and Q14 of QRT 2 in DAT Achiever. (A word problem on 3-sided corral fronting a river mouth...)

Solution:

Perimeter of a sector of any angle, q rad, P:

r + r + rq = 2r + rq = 12

q = (12 -2r)/r ---------- (1)

Area of a sector:

A = (pi * r^2) * (q/2pi) = (r^2)/2 * q ---------- (2)

Substitute (1) into (2),

A = (r^2)/2 * (12 - 2r)/r

A = 6r - r^2

dA/dr = (6 - 2r) = 0 when A is at its maximum

r = 3 units

(Note: Knowing how to apply basic calculus rule(s) is one testing edge to judge how well you're going to perform in the QRT.) 👍 👍 😉

A = 6r - r^2 = -(r^2 - 6r) = -(r^2 - 6r + 9 -9) = -(r^2 - 6r + 9) + 9
= -(r-3)^2 + 9 <= 9
A is at max (9) when r = 3

Note: Be careful with 1st derivative, it must change sign when crossing its root. Otherwise, it is not a max nor a min point. In this question, it changes from + to -. Thus, it's a max.

 

[luder98]

#3)
Find the least value of x^4 + y^4 subject to the restriction x^2 + y^2 = c^2

Answer = (C^4) / 2

x^4 + y^4 = (x^2)^2 + (y^2)^2 = (x^2)^2 + (c^2 - x^2)^2

= (x^2)^2 + c^4 - 2*c^2*x^2 + (x^2)^2

= 2*(x^2)^2 - 2*c^2*x^2 + c^4

= 2[ (x^2)^2 - c^2*x^2 + (c^4)/2]

= 2[ (x^2)^2 - c^2*x^2 + (c^4)/4 + (c^4)/4 ]

= 2[ (x^2)^2 - c^2*x^2 + (c^2/2)^2] + (c^4)/2

= 2(x^2 - c^2/2)^2 + (c^4)/2 is greater or equal to (c^4)/2

The least value is (c^4)/2 when x^2 = c^2/2

Plug into the restriction equation you'll get y = c^2/2

If you can't see it, rewrite the problem by replacing x^2 with X, y^2 with Y, and c^2 with C.

 

[dat_student]

I just added question #5 (see above)...Everyone, please post your math questions...

 

[luder98]

#5)
What is the maximum value of ab / 4(a + b)^2 when a & b are positive numbers

Answer = 1/16

The key point to solving your max/min problems is to know that (x+/-y)^2 is always positive or equal to zero. Apply to your prob: a^2 + b^2 >= 2ab. That said, your problem can be re-written

ab/[4(a^2+b^2+2ab)] is less than or equal to ab/[4(2ab + 2ab)] = 1/16

Thus, max is 1/16 when a^2 + b^2 = 2ab or a = b

 

[dat_student]

#2)
Find the last two digits of N = 11^10 - 1 without using a calculator & without multiplying 11 ten times 😉

Answer: N ends in 00.

Nobody attempted this problem:
Here is the solution:
N = 11^10 - 1 = 10^10 + 10 * 10 ^ 9 + ... + 10 * 10 + 1 - 1
Each of these terms ends in at least 2 zeros....

 

[dat_student]

Ok, here is a true or false question:

(T/F) 1 is equal to 0.9999999....(infinite number of 9s)

Let's see how many people get this right 😉

 

[mahme]

Ok, here is a true or false question:

(T/F) 1 is equal to 0.9999999....(infinite number of 9s)

Let's see how many people get this right 😉

True.

 

[organichemistry]

Ok, here is a true or false question:

(T/F) 1 is equal to 0.9999999....(infinite number of 9s)

Let's see how many people get this right 😉

it is true. there is no number between 0.9999... and 1; therefore, the values are equal.

 

[dat_student]

Here is the proof:

x = 0.999....
10x = 9.999....
10x - x = 9.9999... - 0.9999... (because x = 0.999...)
9x = 9
x = 1

x = 1 = 0.999....

 

[dat_student]

If we define (2n + 1)! to mean the product (1)(2)(3)...(2n + 1) and (2n + 1)!! to mean the product (1)(3)(5)...(2n + 1) express (2n + 1)!! in terms of (2n + 1)!

 

[luder98]

If we define (2n + 1)! to mean the product (1)(2)(3)...(2n + 1) and (2n + 1)!! to mean the product (1)(3)(5)...(2n + 1) express (2n + 1)!! in terms of (2n + 1)!

(2n + 1)!! = (1)(3)(5)...(2n + 1) * 2*4....*(2n)/[2*4....*(2n)]
= (2n+1)! / [2^n(1*2*3....*n)] = (2n+1)!/[2^n(n!)]

Is it?

 

[dat_student]

(2n + 1)!! = (1)(3)(5)...(2n + 1) * 2*4....*(2n)/[2*4....*(2n)]
= (2n+1)! / [2(1*2*3....*n)] = (2n+1)!/[2(n!)]

Is it?

Hint: [2(1)][2(2)][2(3)].....[2👎]

 

[dat_student]

(2n + 1)!! = (1)(3)(5)...(2n + 1) * 2*4....*(2n)/[2*4....*(2n)]
= (2n+1)! / [2^n(1*2*3....*n)] = (2n+1)!/[2^n(n!)]

Is it?

Now, that is the right answer.

 

[luder98]

Everyone, please post your math questions...

What is the max of 5sin(x) + 12cos(x)?

 

[dat_student]

What is the max of 5sin(x) + 12cos(x)?

I'd guess 13 😉

 

[luder98]

I'd guess 13 😉

Well educated guess. Do you have proof? 😉

 

[dat_student]

Well educated guess. Do you have proof? 😉

5sin(x) + 12cos(x) = sq root of (25 + 144) * sin ( x + inv of sin 12/13)

the max for the sin is 1

so 13 * 1 = 13

 

[luder98]

5sin(x) + 12cos(x) = sq root of (25 + 144) * sin ( x + inv of sin 12/13)

the max for the sin is 1

so 13 * 1 = 13

I guess someone knows his/her trig 👍

 

[dat_student]

I guess someone knows his/her trig 👍

I guess that someone is luder98

 

[luder98]

I guess that someone is luder98

No, I meant you. I'd solve it differently 😉

 

[dat_student]

No, I meant you. I'd solve it differently 😉

how would you solve it?

 

[luder98]

how would you solve it?

By applying Cauchy-Schwarz inequality.

5sinx + 12cosx <= |5sinx + 12cosx |<= sqrt(5^2+12^2)(sin^2(x)+cos^2(x))=13

 

[dat_student]

By applying Cauchy-Schwarz inequality.

5sinx + 12cosx <= |5sinx + 12cosx |<= sqrt(5^2+12^2)(sin^2(x)+cos^2(x))=13

That's trig. 😉

 

[dat_student]

Find the sum of n terms of the arithmetic series whose first term is the sum of the first n natural numbers and whose commen difference is n.

 

[luder98]

Find the sum of n terms of the arithmetic series whose first term is the sum of the first n natural numbers and whose commen difference is n.

Sn = (a1 + an)n/2
a1 = 1 + 2 + ... + n = (1+n)n/2
an = a1 + (n-1)d = 1 + ... + n + (n-1)n = (1+n)n/2 + (n-1)n

Sn = [ (1+n)n/2 + (1+n)n/2 + (n-1)n]n/2 = n^3

Is it?

Edited: My bad. I had n in place of d in the formula.

 

[dat_student]

n^3

 

[dat_student]

Here is another problem:

1/4 + 1/16 + 1/64 + ... is exactly equal to ....

 

[luder98]

Here is another problem:

1/4 + 1/16 + 1/64 + ... is exactly equal to ....

Usual way: This is a geometric series whose u1 = 1/4 , r=1/4, just use the formula Sn = u1(1-r^n)/(1-r)

Sn = 1/4(1-1/4^n)/(1-1/4)

When n approaches infinity, 1/4^n approaches 0.

Thus, Sn = 1/4/(3/4) = 1/3

Short cut:

Sn = 1/4 + 1/16 + 1/64 + 1/256 + ....
= 1/4 (1 + 1/4 + 1/16 + 1/64 + 1/256 + ...) = 1/4 (1 + Sn)

Or 4Sn = 1 + Sn or Sn = 1/3

Is it?

Here is a more interesting series problem. Find sum of 1/2 + 1/6 + 1/12 + ...

 

[dat_student]

Or 4Sn = 1 + Sn or Sn = 1/3

Is it?

Here is a more interesting series problem. Find sum of 1/2 + 1/6 + 1/12 + ...

it's 1/3
----------------------
1/2 * (1/3) = 1/6
1/6 * (1/2) = 1/12
is the pattern: *1/3, *1/2, *1/3, *1/2.... ?

 

[luder98]

it's 1/3
----------------------
1/2 * (1/3) = 1/6
1/6 * (1/2) = 1/12
is the pattern: *1/3, *1/2, *1/3, *1/2.... ?

The pattern is 1/[n(n+1)]. The answer is 1.

 

[dat_student]

The pattern is 1/[n(n+1)]. The answer is 1.

oops! thanks..Now, I am predicting that I'll get QR=10

 

[luder98]

oops! thanks..Now, I am predicting that I'll get QR=10

I'm sure you'll score in high 20s. Unlike you, I didn't practice as much. I ended up with a 20. I should do better this time around.

 

[dat_student]

I'm sure you'll score in high 20s. Unlike you, I didn't practice as much. I ended up with a 20. I should do better this time around.

When are you planning to take the DAT?
I haven't practiced much for QR. Just a few random problems and I doubt the problems, that I have worked on, will appear on the DAT. Honestly, at this time, I think I'll get single digit scores across the board. I have no confidence...

 

[luder98]

When are you planning to take the DAT?
I haven't practiced much for QR. Just a few random problems and I doubt the problems, that I have worked on, will appear on the DAT. Honestly, at this time, I think I'll get single digit scores across the board. I have no confidence...

Your posted problems are harder than the real thing itself. I didn't see any of those on my test. But it doesn't hurt to overprepare. Where did you get them from? I'm gonna purchase Ace the Dat for QR and PAT. I ran out of time on the real thing last time. I heard good things about Ace the Dat on the two subjects. I plan to re-take it in August. That's when my app will arrive at the schools.

 

[dat_student]

Your posted problems are harder than the real thing itself. I didn't see any of those on my test. But it doesn't hurt to overprepare. Where did you get them from? I'm gonna purchase Ace the Dat for QR and PAT. I ran out of time on the real thing last time. I heard good things about Ace the Dat on the two subjects. I plan to re-take it in August. That's when my app will arrive at the schools.

I am not overpreparing. I just grabbed some questions that I thought were interesting from different textbooks. Isn't August too late to take the DAT?