Qr Practices for Advanced Students

[UCfan]

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First three questions for advanced students. These are some examples of the hardest questions you can see on DAT (It was asked by DAT students who got exceptional DAT scores). Last two are for exceptional students.



1)) How many zeroes will appear if you divide 127! by 10.(Ans: 30) (Corrected. Thank you Street Wolf)
There are so many multiplies of two, so you have to know number of 5 multiplies.
127/5= 25
127/25= 5
127/125= 1
So the answer is 25+5+1=31
31-1=30.
2) How many squares does a chess board have which is 8x8? (Ans:204)
Simply n^2+(n-1)^2+(n-2)^2+………where n=8
which is 64+49+36+25+16+9+4+1=204


3) What is the value of arcsin (74)? (Ans:0.96)
Sin (74) = 2*sin(37)*cos(37)= 2*(3/5)*(4/5)= 24/25
So
Answer is 24/25.

Two more problems for exceptional DAT takers:

4) a+(1/a)= 4. So what is (a^2) + (1/(a^2)) =?

5) (x^3)- 1= 0 and x is not equal to 1. Then what is (x^2)+x=?


If someone find very easy, I can send harder problems.

 

[innocentxgrl]

nice! thanks for this post! post more ques! 🙂

 

[innocentxgrl]

i dont get how you solved the arcsin problem..mind explaining it better?

 

[Marmar tala]

where did you get these questions? I mean which book did you find them in?

 

[Streetwolf]

So questions for advanced students. These are some examples of the hardest questions you can see on DAT.



1)) How many zeroes will appear if you divide 127! by 10.(Ans: 31)
There are so many multiplies of two, so you have to know number of 5 multiplies.
127/5= 25
127/25= 5
127/125= 1
So the answer is 25+5+1=31

2) How many squares does a chess board have which is 8x8? (Ans:204)
Simply n^2+(n-1)^2+(n-2)^2+………where n=8
which is 64+49+36+25+16+9+4+1=204


3) What is the value of arcsin (74)? (Ans:0.96)
Sin (74) = 2*sin(37)*cos(37)= 2*(3/5)*(4/5)= 24/25
So
Answer is 24/25.


If someone find very easy I can send harder problems.

Isn't #1 ans = 30 because you divide by 10?

Other than that plz post more but without answers.

 

[UCfan]

No, the first one is 31. Why are you dividing by 10? U have to know number of 5 and multiple of 5 (which is 5^2=25 and 5^3=125).
For the third one; you have to know that sin 37=3/5, cos 37=4/5 (from special right triangle 3-4-5) and sin(2a)=2(sina)(cosa) = 2(3/5)(4/5)=24/25.

 

[Streetwolf]

1)) How many zeroes will appear if you divide 127! by 10.(Ans: 31)

If you divide by 10, and there are exactly 31 factors of 5 and >31 factors of 2, that makes 31 factors of 10 (as you stated). When you divide by 10 you lose a 0 in the answer, giving you 30 zeros.

On #3 the 3-4-5 triangle is not as well known (in regards to angle degrees). It does have approximately 90-53-37 angles but that means your answer of 24/25 is just an estimate. Close enough for a multiple choice test I'd imagine. I DOUBT that would be on the DAT.

 

[innocentxgrl]

streetwolf, thas exactly wat i was thinking too!! i wonder what the real answer is?

 

[klutzy1987]

Two more problems for exceptional DAT takers:

4) a+(1/a)= 4. So what is (a^2) + (1/(a^2)) =?

5) (x^3)- 1= 0 and x is not equal to 1. Then what is (x^2)+x=?


If someone find very easy, I can send harder problems.

I find these 2 to be the simpler problems here (unless i did something wrong).

4) First you multiply everything by A to get A^2 + 1 = 4A. Therefore A^2 equals 4A-1. Substitute that in and you get (4A-1) + (1/(4A-1)). If you multiply both top and bottom by (4A-1) you can add them together and you get (4A-1)(4A-1)+1/(4a-1). This equals when multiplied out ((16A^2)-8A+2)/(4A-1). Dividing gives (4A-1) and Remainder of 1 (Answer).



5) (x^3)-1=0 Change this around and X^3=1. Divide both sides by x to get x^2=(1/x). Substitute this value into X^2 and you get (1/x)+x=. This equals (X^2+1)/X. dividing out gives an answer of X REMAINDER OF 1.

 

[Streetwolf]

4) a+(1/a)= 4. So what is (a^2) + (1/(a^2)) =?

5) (x^3)- 1= 0 and x is not equal to 1. Then what is (x^2)+x=?

4. Take a + (1/a) and square it and you get a^2 + 2 + 1/a^2, which then equals 4 squared = 16. So a^2 + 1/a^2 = 14.

5. x^3 - 1 = 0 with x not equal to 1.

x^3 -1 is the same as (x^2 + x + 1)(x - 1) which equals 0. If x doesn't equal 1 then the other part of the equation must be true: x^2 + x + 1 = 0. Thus x^2 + x = -1.

 

[toothfairy87]

Thank you!
That was awesome...please keep them coming!!!

 

[klutzy1987]

4. Take a + (1/a) and square it and you get a^2 + 2 + 1/a^2, which then equals 4 squared = 16. So a^2 + 1/a^2 = 14.

5. x^3 - 1 = 0 with x not equal to 1.

x^3 -1 is the same as (x^2 + x + 1)(x - 1) which equals 0. If x doesn't equal 1 then the other part of the equation must be true: x^2 + x + 1 = 0. Thus x^2 + x = -1.

I guess he wanted a numerical answer, huh lol. That was brilliant, i didnt even think of this method. I just did binomial division for some reason. Nice

 

[UCfan]

Good job streetwolf. Do you want some more?
6)(x^4)-(4x^3)+(14x^2)-(20x)+25=0
What are the roots?(advanced)

7) 3^3+5^3+7^3+........+1993^2 is dividing by 990000. What is the remainder? (genius)

8) [(4^2002)+(6^2002)]/25 what is the remainder? (advanced)

9) if x+(1/x)=3, then (x^4)-54x=? (advanced)

 

[Streetwolf]

Okay I did these 2 so far...

(7) I used the summation of i^3 from 1 to n equals [ n(n+1)/2 ]^2 and used n = 1993 to get a really large number. Then I used the summation of (2i)^3 from 1 to n with n = 992 to target the even cubes between 1 and 1993. That was another large number (but not as large). Subtracting the two, I found the sum of 1^3 + 3^3 + ... + 1993^3. Then I subtracted 1^3 = 1 from that to get the sum of 3^3 + 5^3 + ... + 1993^3. Then I divided by 990,000 and found the remainder to be 460,152.

(8) Maybe there's a better way to do this, but what I did was I took 4 and kept multiplying by 4 and keeping track of the last 2 digits. So for example 4x4 = 16. Then 16x4 = 64. Then 64x4 = 256 so I only considered 56. Then 56x4 = 224 so I only considered 24. After doing that I found it cycled back to 4 every 10 times. Thus when done 2002 times, the 2000th time will end the cycle. The 2001st time will end in 04 and the 2002nd time will end in 16.

Do the same with the 6 and it has a cycle of 5 (although the start, 06, is not part of that cycle. Thus the 2001st will end the cycle and the 2002nd will end in 36. When you add 16 + 36 you get 52. Those will be the last 2 digits of the number which is 4^2002 + 6^2002. When you divide that by 25 you will have a remainder of 2 because 25 goes evenly into any number ending in 50.

For (9) I could easily solve for x then plug back into the new equation to get an answer. I assume you want something other than that, like in the problems posted a week ago.