
Hi,
If f(x) = ax^2 + bx + c
f(5) = 25a + 5b + c (1)
3f(2) = 3(4a+2b+c) (2)
Equating (1) and (2), 37a + 11b + 4c = 0 (3)
3 is a root of f(x), 9a + 3b + c = 0 (4)
Eq (3)  4times Eq(4) gives a = b
Now 3 is a root of f(x), so [b (+/) sqrt (b^2  4ac)]/2a = 3
sqrt (b^2  4ac) = 6a + b = 7a or 7b
since (b + 7b)/2b = 3
So, other root is (b7b)/2b = 4
[B]Answer for Q1. = (b) 4[/B]
Squaring and that b= a gives, a^2  4ac = 49a^2 => c = 12a
Therefore, a+b+c = 2a 12a = 10a
So, infinite solutions, or Cannot be determined.
[B]Answer for Q2. = (e) Cannot be determined.[/B]
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Nice explanation and correct answers mech_guy !!!
Official Correct Answer:
Let f(x) = ax^2 + bx + c where a,b,c are certain constants and a is not equal to 0.. f(5) = 3f(2) and 3 is root of f(x) = 0
Q1. what is the other root of f(x) = 0 ?
a)7
b)4
c)2
d)6
e)cannot be determined
Q2. what is value of a+b+c?
a)9
b)14
c)13
d)37
e)cannot be determined
CB
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Q3. Consider obtuse angled triangles with sides 8 cm,15 cm,x cm. If 'x' is an integer ,then how many such triangles exist?
1.5
2.21
3.10
4.15
5.14
CB
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Well, am not sure about this but lemme try.
I would use 2 properties:
1. Sum of any 2 sides should be greater than third side.
2. Largest angle of a triangle is opposite to largest side.
So largest side could either be 15 or "x".
If largest side is 15, then x can be = 8, 9, 10, 11, 12, 13, 14 (x could be 7 but then triangle would be degenerate, so lets exclude degenerate triangles).
If largest side is x, then x can be = 16, 17, 18, 19, 20, 21, 22 (x could be 23 but then triangle would be degenerate, so lets exclude degenerate triangles).
So total combinations = 7 + 7 = 14
So my answer would be (5) 14.
Am not sure if i have applied any particular property of an obtuse angled triangle.
So, if examination hasn't got any negative marking i would go for 14, else skip it 😀
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Q4). At what time between 23 is the first right angle in that time formed by the hands of the clock??
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Hi ES,
EnglishScared
Q4). At what time between 23 is the first right angle in that time formed by the hands of the clock??
Answer: At
27.27 minutes past 2 O' Clock.
Lets say first right angle between short and long hand is made after t minutes.
In 60 minutes, ShortHand (SH) traverses 30 degrees.
In t minutes, SH will traverse = 0.5t degrees.
In 5 minutes, LongHand(LH) traverses 30 degrees.
In t minutes, LH will traverse = 6t degrees.
Initial angle between SH and LH = 60 degrees.
We thus have following eq : 6t  60  0.5t = 90
which gives, t = 300/11 minutes = 27.2727 minutes.
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Correct Answer!!!
It will be at 90 degrees at 2h.27m.16.362sec
The Explanation from my end is..
Consider the watch to be a 60m road , at 2 PM Hours hand will be 10 m front the minutes hand.
So the minutes hand has to travel 25m to make 90 Degrees.
So in a hour minutes hand travels 55m since hours hand travels 5 m we take the relative.
55m  in 1 hour
25  ??
25/55 *60 = 300/11 in mins.
27.2727 = 27m and 16.262 seconds.
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A five digit number abcde is such that it is divisible by 6 and a<b<c<d<e
Q5) if e<8 then how many such five digit numbers are possible?
a)1 b)2 c)0 d)12 e)16
Q6) How many such five digit numbers are possible in all?
a)13 b)14 c)17 d)8 e)15
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A5 :: a) 1
The only number is 12456.
The reason behind this is that if the number is divisible by 6 than the last digit should be 0,2,4,6,8. But only 6 and 8 can follow the a<b<c<d<e condition. But e<8, so just 6 is left for e. Now there were only 5 numbers following condition a<b<c<d<e. And I just checked them whether they are divisible by 6 or not.
I tried the same method for the 2nd question but it took longer than I thought by giving me 39 combinations. Than I quickly checked the number. I got this answer,
A6 :: a) 13.
Am I right?
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5th is correct 😀
6th wrong .
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Rajdeep,
I think you missed out a number , did you add the number from first answer ? <edited_ES>
So the answer is 13+1 =14.
You have solved it correct but if you put 13 as answer you will be awarded 1 mark in CAT exam.
So careful answering is also required 😀
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EnglishScared
A five digit number abcde is such that it is divisible by 6 and a<b<c<d<e
I just thought it is 5 digit number, if it is 6 digit number than my both answers should be wrong.
Please clarify this.
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Sorry dude, 123456 that was a typo.
I referred to the number from the first answer.
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Ans1 = b and Ans2 = e
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Dipen,
The answers you have posted were already answered ,your answers are correct as well.
If you get any better explanation than the posted solution then post it, or else dont double post the answers once they are declared.
If you have any queries yes you can post your doubts.
Take this suggestion in a positive way 😀
Looking forward for interesting questions and solutions 😀
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Q7) In a Fruit mall certain stall sells only Apples , oranges ,Pears and peaches.
When I asked the fruit vendor about their costs , I was told that any fruit in the stall costs Rs25/.Then I asked him to give a basket full of fruits worth Rs 500/.with at least 2 fruits of each variety.Assuming that there is no additional charge for the basket , how many different combination of fruits can the basket contain?
a) 17c3 b) 19c3 c)15c3 d) 11c3 e)12c3
Q8) Rajesh bought 4 packets of chocolates and distributed all the chocolates among his daughters , It is known that each daughter got exactly one chocolate less than the total number of daughters rajesh has. If each packet contain contains atleast 45 chocolates ,find the minimum number of daughters Rajesh has?
a)16 b)14 c)12 d)13 e)15
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My answers are:
Q7)
c. 15C3
Q8)
b. 14
Are these correct??
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7 is correct , can you please explain the procedure.
8 is wrong 😀
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ms_cs
Member •
Jul 23, 2009
Q8) 16
is it correct?
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Yes, correct .
Explanation guys !!!?
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Hi ES,
EnglishScared
Q8) Rajesh bought 4 packets of chocolates and distributed all the chocolates among his daughters , It is known that each daughter got exactly one chocolate less than the total number of daughters rajesh has. If each packet contain contains atleast 45 chocolates ,find the minimum number of daughters Rajesh has?
a)16 b)14 c)12 d)13 e)15
Number of daughters of Rajesh = N
So each daughter got (N1) chocs.
Minimum number chocs in 4 boxes = 45x4 = 180
Therefore, N(N1) >= 180
N^2  N  180 = 0
N = [1 + sqrt (721) ]/2
721 is not a perfect square, next perfect square is 729 (27)
So, N = 28/2 = 14
So Answer should be (a)14
CrossCheck: Each daughter gets 13 chocs, total chocs = 182 (> 180)Regards
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The answer is not 14 because,
N(n1) should be a multiple of 4.
So in the options the least possible answer is 16.
Got it??
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Got it..
For question 7, here's how I did it.
There are 4 types of fruits and each costs rs.25. We buy for rs.500.
So, we buy 20 fruits in all. Of these, 8 are fixed in that, there should be 2 in every kind.
So, we have to distribute the remaining 12 fruits between the 4 kinds.
That gives 15C3.
Hope it's clear!!
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Ah, one last word.
we have to distribute the remaining 12 fruits between the 4 kinds.
This is similar to asking: "In how many ways can you give 12 chocolates to 4 children?".
The original question was, "In how many ways can you give 20 chocolates to 4 children, such that each one gets at least two".
right ho??
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Yeo thats correct!!!
How does 15c3 came ?? when there are only 12 will be a question again 😀
There is a formula which states this is ..
(12+41)c(41) which results in 15c3
Excellent answer SS.
Thanks for your contribution 😀..
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Q9) A certain quantity of water is evaporated from four pounds of sea water containing y% water to get a pound of wet salt which contains y% of salt.What is the value of y ?
a)60 b)70 c)80 d)81 e)cannot be determined.
Q10) If the digit 5 is appended to the left of the natural number N , it becomes equal to Nl , if the digit 5 is appended to right side of N it becomes Nr. If N is a three digit natural number for how many values of N is Nl<Nr?
a)443 b)444 c)554 d)445 e)555
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Q10)
d) 45
Take any three digit number less than 555 and perform this procedure.
Let's take 456.
Nl is 4565
Nr is 5456.
Nl<Nr.
The first time Nl becomes equal to Nr is at 555.
Nl = Nr = 5555.
So, after 555, Nl>Nr. So, From 555 to 999, there are 445 numbers.
Hence D.
Hope it's correct.
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Q9) No idea, but my guess would be B) 70.
Now, four pounds of water gets converted to one pound of salt. So, b some crude thinking, there must be around 75% water and 25% salt.
Since the whole water doesn't evaporate, 70% of water must have evaporated, leaving behind 70% of salt in the wet mixture.
Am I correct??
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Sorry both the answers are incorrect!!
For Q10 ) You have done very well but the answer is 444 since the 445th number will not satisfy the condition.
Clear dude?
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Q9) Is just a confusing question.
This is the solution.. The amount of salt present in 4 pounds of sea water is = (1y/100)4
The same quantity is present in 1 pound of wet salt and salt content is y/100 pounds.
equating these two equations we get answer as 80.
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EnglishScared
Sorry both the answers are incorrect!!
For Q10 ) You have done very well but the answer is 444 since the 445th number will not satisfy the condition.
Clear dude?
What a silly mistake to make.. Duh!!😔
I got 999  555 = 444,
but I added 1 to it, thinking I've to include the numbers on the border..😔
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EnglishScared
Q9) Is just a confusing question.
This is the solution.. The amount of salt present in 4 pounds of sea water is = (1y/100)4
The same quantity is present in 1 pound of wet salt and salt content is y/100 pounds.
equating these two equations we get answer as 80.
This one is cool. tricky, I'd say..
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Reading the question it makes us confused 😀
Wait for some time will post few more questions tonight.
SS You can also post few questions if possible ok ?
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Q11) N is a natural number. How many values of N exist , such that N2 + 24N +21 (read as N square + 24*N +21) has exactly three factors?
A) 0 B)1 C) 2 D)3 E) More than 3
Q12) What is the sum of the digits of a two digit number , which is 32 less than the square of the product of its digits?
A) 12 B)11 C) 10 D)9 E)8
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EnglishScared
The answer is not 14 because,
N(n1) should be a multiple of 4.
So in the options the least possible answer is 16.
Got it??
Ahhh!! Me a Dumbo!!
Ofcourse i didn't think that each packet will have same number of chocs, it wasn't given but should have been obvious.
Got it ES, thanks mate 😀
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EnglishScared
Q11) N is a natural number. How many values of N exist , such that N2 + 24N +21 (read as N square + 24*N +21) has exactly three factors?
A) 0 B)1 C) 2 D)3 E) More than 3
Q12) What is the sum of the digits of a two digit number , which is 32 less than the square of the product of its digits?
A) 12 B)11 C) 10 D)9 E)8
Ok, Question 12 is easy one :
Lets say N is the number, so
N + 32 = Perfect square
Possible values of N = 17, 32, 49, 68 and 89. No more possibilities because then N will become a three digit number.
We can see that with 17 the given condition satisfies i.e.
17 + 32 = (1x7)^2 = 49
So, sum of digits = 8.
Answer = (E) 8
A doubt regarding Question 11: "Exactly three factors". Three prime factors i suppose question is asking?
I mean if number N^2 + 24N + 21 = 30 (lets say)
30 = 2*3*5; that is 3 prime factors. But it can be said that its factors are 2, 3, 5, 6, 15, 10 and even 1 and 30 itself.
So i assume Question is asking three prime factors. Is it??
Regards
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No dude its not prime factors 😀
I hope you got the answer easily now 😀
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12th answer is correct dude 😀
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EnglishScared
No dude its not prime factors 😀
I hope you got the answer easily now 😀
Oh!! Then i will say there is no value of N possible.
Reason: One of the factors will always be 1. Now if that expression turns out to be a prime number, then there cant be anymore factors. But if that expression is not a prime a number then it can be broken into atleast 2 prime numbers. This will lead to more than 3 factors of the expression.
So my
answer will be (A) 0
Thanks for clarifying ES 😀
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The answer is E,
I don't have the explanation right now, sorry for it..
Will post it tomorrow for sure..
If some one has it please post it.
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EnglishScared
The answer is E,
I don't have the explanation right now, sorry for it..
Will post it tomorrow for sure..
If some one has it please post it.
Hi ES,
In that case, i think three factors are Prime Factors only.
My explanation:
N^2 + 24N + 21 = p
So, N = [24 +/ sqrt(492 + 4p)]/2 = 12 + sqrt(123 + p)
As N is a natural number, (123+p) has to be a perfect square and greater than 12
Possible values of (123 + p) can be = 169, 196, 225, 256.... (Perfect squares of 13 onwards)
In case of 123 + p = 225
p = 102
N^2 + 24N + 21 = 102 => N = 3
Factors of 102 = 2 x 3 x 17
In case of 123 + p = 441 (square of 21)
p = 318 => N = 9
Factors of 318 = 2 x 3 x 53
We can move on like this till infinity and we can be sure to get many many such "N"s.
Thanks
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Yep this looks to be a proper explanation thanks dude..
Check the DI and answer it too 😀
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Few questions on Geometry and Mensuration:
Q13) The minute hand of a clock is 10 cm long , Find the area of the face of the clock described by the minute hand between 9 a.m. and 9.35 a.m.
a) 183.3 sq.cm
b) 366.6 sq.cm
c) 244.4 sq.cm
d) 188.39 sq.cm
Q14) Two circles touch internally . The sum of their areas is 116Pie sq.cm
and distance between their centers is 6cm. Find the radii of the circles.
a) 10 cm , 4 cm
b) 11 cm . 4 cm
c) 9 cm , 5 cm
d) 10 cm , 5 cm
Q15) Iron weights 8 times the weight of oak, Find the diameter of an iron ball whose weight is equal to that of a ball of oak 18 cm diameter.
a) 4.5 cm
b) 9 cm
c) 12 cm
d) cannot be determined
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Q14) A
Out of the answer options, only (A) satisfes the first condition.
ie, pi * {10[sup]2[/sup] + 4[sup]2[/sup]} = 116 pi
Also, if you so want to solve the problem, here's how.
Let the radii be a and b.
so, pi * {a[sup]2[/sup] + b[sup]2[/sup]} = 116 pi
Also, ab=6 (since circles touching internally)
(ab)[sup]2[/sup]=36
a[sup]2[/sup] + b[sup]2[/sup] = 36 + 2*a*b
implies: 36 + 2*a*b = 116
a*b = 40
Now you have two equations
a  b = 6
a * b = 40
Solving these gives you the answer
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Q15) B
"Iron weights 8 times the weight of oak" ==> This means that iron's density is 8 times the density of oak.
Now, weight = volume * density.
Equating weights,
(4/3) * pi * (9)[sup]3[/sup] * X = (4/3) * pi * a[sup]3[/sup] * 8X
where, X is the density of oak, 8X that of iron and a is the radius of the iron ball
Solving, 9 = 2a ==> a = 4.5
So, diameter of the iron ball is 9cm. So, answer is B.
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Both the answers are correct!!.
Well done SS.
13th is still pending.
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Q 13)
PI*(10)^2*(35/60)=183.3 sq.cm
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Perfect!!! Sheeko!!!
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14)abcd is a 4digit number in base of 7 such that 2(abcd) = bcda (a, b 6= 0).
Then a is
a. 1 b. 2 c. 3 d. Data insufficient
15)If a2b3c = 256/27 find min value of a + b + c, given a, b, c are positive
real nos
1.)10 2)11 3)12 4)13 5)14
16) From four positive real nos a, b, c and d, 4 distinct combination of sum
of three numbers are formed S1, S2, S3 and S4. IF abcd = 5. FInd the min
value of S1S2S3S4
1) 40 2.) 135 3.)1080 4.)405 5 .)1280
17)Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at
most 4 apples and 3 oranges with Rs 20 and no more fruit, then the maximum
amount that could be left with Anjli will be about
(1) Rs 1 (2) Rs 1.25 (3) Rs 1.50 (4) Rs 2 (5) Rs 2.25
18)In a classroom there are 14 students seated in 3 rows of 5 chairs. The
place at the centre of the room is unoccupied. A teacher decides to reassign
the seats such that each student will occupy a chair adjacent to his/her
present one (i.e. move one desk forward, backward, right or left). In how
many ways can this reassignment be done?
(1) 1 (2) 13 (3) 0 (4) 27 (5) none of these
19)here is a function f such that f(2) = 60 and −f(1) + f(2) − f(3) +
f(4)−f(5).......+f(n) = nf(n), for all positive integers n > 1... find f(11)..
a) 60 b)12 c)10 d)15 e)11
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Hello somebody answer these!!
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14)1 15)2 16)3 17)2 18)4 19)1
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I got a solution of 10 on some other website. but first of all i dont understand where has that 15 come from. Please explain how did u get that 15 or x assumption??
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