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by ven4gmat » Wed Mar 11, 2009 1:09 am
sureshbala wrote:
sureshbala wrote:Here is the next one....

Find the total number of 4 digit numbers which contain 36 in them and are divisible by 36.

A. 16
B. 15
C. 14
D. 13
E. None of these
A number is divisible by 36, if it is divisible by both 4 and 9.

Divisibility Rule of 4: The number formed by the last digits of the number must be divisible by 4.

Dvisibililty Rule of 9: Sum of the digits of the number must be divisible by 9.

Case 1: Number is of the form 36xy.

It is clear that if the number xy is divisible by 36, then the number 36xy will be divisible by 36.

so xy could be 00, 36 and 72.

Hence 3 numbers are possible in this case

Case 2: Number is of the form x36y.

Now if we take care of the value of y, such that 6y is divisible by 4, we can place the value for x such that the number is dvisible by 9 as well.

So y can take 0, 4 and 8 for which x will take 9, 5 and 1 respectively.

Hence 3 numbers in this case as well

Case 3: Number is of the form xy36.

Since this number is divisible by 4, all we have to see is that this number is divisible by 9. Also, since 3+6 =9, the sum of x and y must be such that it is divisible by 9.

So for x+ y = 9, we have (9,0) (8,1)........(1,8). i.e. a total of 9 cases

Also x+y = 18, we have (9,9)

Hence there are 10 numbers in this case.

So totally 3+3+10 = 16 possibilities. But of this number 3636 is counted in the first case as well as the last case.

Hence total number of possibilities = 15.
This is really a very good and clean explanation Sureshbala. Thanks once again for the efforts.

I am not able to work on BTG for the last week and I know I have missed a lot.

Waiting for the next one...
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by ashaa » Thu Mar 12, 2009 9:28 am
answer to 2nd question in forum

the way i did
draw straight line from angle C on AB at point Q giving two right angles CQB and CQA . As angle CBQ is 45 angle BCQ is 45 too.
Short Guess is In triangle CQA line side AQ is smaller than AC so angle opposite to AQ is 30.
AQC + BQC = 45 + 30= 75
two min solution..... :idea:
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by sureshbala » Fri Mar 13, 2009 7:54 am
ashaa wrote:answer to 2nd question in forum

the way i did
draw straight line from angle C on AB at point Q giving two right angles CQB and CQA . As angle CBQ is 45 angle BCQ is 45 too.
Short Guess is In triangle CQA line side AQ is smaller than AC so angle opposite to AQ is 30.
AQC + BQC = 45 + 30= 75
two min solution..... :idea:
I think there is no way to guess like this and conclude the answer.
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by rs2010 » Fri Mar 13, 2009 9:54 am
Suresh,
When are you going to post the next question? :D
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by sureshbala » Fri Mar 13, 2009 10:10 am
Here is the next question....

A student scored an average of 55 marks in the first 6 tests. If the first test is not counted and the seventh is counted then his average score goes up to 57 marks. If the score in the first test was 50 marks, find his score in the seventh test.

(A) 58
(B) 60
(C) 62
(D) 64
(E) 59
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by DanaJ » Fri Mar 13, 2009 10:25 am
First six tests total: 6*55 = t1 + (t2 + t3 + t4 + t5 + t6).
Average for the last six tests = 57, meaning that total points for last six tests = 6*57 = (t2 + t3 + t4 + t5 + t6) + t7 = 6*55 - t1 + t7.
This is why 6*57 = 6*55 - t1 + t7.
6(57 - 55) = t7 - t1
6*2 = t7 - t1
12 = t7 - t1
t1 = 50, so t7 must be 50 + 12 = 62.
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by rs2010 » Fri Mar 13, 2009 10:34 am
sureshbala wrote:Here is the next question....

A student scored an average of 55 marks in the first 6 tests. If the first test is not counted and the seventh is counted then his average score goes up to 57 marks. If the score in the first test was 50 marks, find his score in the seventh test.

(A) 58
(B) 60
(C) 62
(D) 64
(E) 59
=6*57-6*55+50

you reduced the level :roll:
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by logitech » Fri Mar 13, 2009 1:56 pm
FIRST SET - SECOND SET = (Test # 7 - Test # 1) = 6 x ( 57 - 55 )

This gives us: Test # 7 = Test # 1 + 12 = 50 + 12 = 62
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by sureshbala » Thu Mar 19, 2009 8:03 am
sureshbala wrote:Here is the next question....

A student scored an average of 55 marks in the first 6 tests. If the first test is not counted and the seventh is counted then his average score goes up to 57 marks. If the score in the first test was 50 marks, find his score in the seventh test.

(A) 58
(B) 60
(C) 62
(D) 64
(E) 59
If the score in the seventh test is also 50, the average would not change. But since the average increased by 2 (changed from 55 to 57) upon 6 tests, the increase in the marks is 12. Hence the score in the seventh test = 62

So C is the answer.
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by sureshbala » Mon Mar 23, 2009 2:32 am
Folks, here is the next one...

If sqrt(x+2) + sqrt(x+8) = 10, find the value of [x+(191/100)].

(A) 25

(B) 22

(C) 26

(D) 18

(E) 20
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by rs2010 » Mon Mar 23, 2009 6:50 am
sureshbala wrote:Folks, here is the next one...

If sqrt(x+2) + sqrt(x+8) = 10, find the value of [x+(191/100)].

(A) 25

(B) 22

(C) 26

(D) 18

(E) 20
I am not sure I the shorted procedure or not but my answer is B

(x+2)+(x+8)+2*√(x+2)*(x+8)=100
√(x+2)*(x+8)=45-x

x=49*41/100

x+191/100=2009/100 + 191/100=22

Took me nealy 2 mins. Was there any shortcut?
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by sureshbala » Fri Mar 27, 2009 9:09 pm
hemantsood wrote:
sureshbala wrote:Folks, here is the next one...

If sqrt(x+2) + sqrt(x+8) = 10, find the value of [x+(191/100)].

(A) 25

(B) 22

(C) 26

(D) 18

(E) 20
I am not sure I the shorted procedure or not but my answer is B

(x+2)+(x+8)+2*√(x+2)*(x+8)=100
√(x+2)*(x+8)=45-x

x=49*41/100

x+191/100=2009/100 + 191/100=22

Took me nealy 2 mins. Was there any shortcut?
I guess there is no shortcut for this and my solution to this is quite similar to that of yours...

On squaring both sides, we get

2x+10+2*sqrt(x+2)*sqrt(x+8) = 100
i.e. sqrt(x+2)*sqrt(x+8)=45-x

Again squaring on both the sides, we get

x^2+10x+16 = 45^2+x^2-90x

100x = 45^2 - 16
100x = 45^2- 4^2
x = (49*41)/100


Hence x + 191/100 = 2200/100 = 22.
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by sureshbala » Fri Mar 27, 2009 9:18 pm
Folks, the next question is from geometry.....Have a look at the following figure..

Image
[/img]
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by rs2010 » Sat Mar 28, 2009 7:44 am
sureshbala wrote:Folks, the next question is from geometry.....Have a look at the following figure..

Image
[/img]

Please see attached image.

What we need is OC.


In sqr AEOF

OA^2=AF^2 + OF^2
= x^2 + x^2
= 2* x^2
(AG+GO)^2=2*x^2

AG+GO=√2 * x
AG=x(√2-1)

AG+GC=AC

x(√2-1) + 2*x=16

x=16/(√2+1)
x=16(√2-1)

A.

Next time I will upload scanned image :P
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by ambyclan » Mon Mar 30, 2009 12:28 pm
referring to problem in page 2,can anyone explain why the angle <acb is not 90 degrees
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