Try some [basic maths] reasoning questions!

for question 1. is the answer 200..:?:

solution:-
three sides are 2,4,5.

for a cube the side will be a lcm of 2,4,5....=>20.

so volume of the large cube is 20*20*20=8000cm^3

and the volume of small cube is 2*4*5=40 cm^3

so the no. of cubes is =8000/40=200.


hope i'm correct.😀

correct answer, nice approach to the solution.

solutions****************************
2>For obtaining identical pieces, cuts must be made parallel to 3 planes of the cube.
x cuts made along one plane results in x+1 pieces
let we made x , y ,z cuts along all three planes
such that (x+1)(y+1)(z+1) = t is maximum & x+y+z =13
different combinations can be possible but t will be maximum in (4,4,5) cuts
so maximum pieces = 5*5*6 = 150

3> similarly as 2 ,
7 cuts can be made (3,2,2) , so maximum pieces = 4*3*3 = 36



Question 4>
Police arrested 3 persons - Charles, Raj & Soban - on suspicion, in a theft case. It was found that one among these three was the thief. During the introgation, their replies were as follows:
Charles : I am not the thief. Raj is the thief.
Raj : I am not the thief. Either Charles or Soban is the thief.
Soban : I am not the thief. Raj is not the thief.
If exactly one person among them always speaks the truth , another always speaks the lies & the third alternates between truth & lies then who is the thief?
a> Charles
b> Soban
c> Raj
d> Cannot be determined

I think Raj is thief assuming that charles speak the truth then it is clear that raj is thief
then soban says I am not the thief. Raj is not the thief.
It also makes clear that raj is thief
Raj says I am not the thief. Either Charles or Soban is the thief.
now assuming that raj speaks alternatively Truth OR Lie then also it is clear that raj is thief

raj is the thief
since one speaks truth,one lies,and other alternates.we can take charles to speak the truth.and the raj to speak lies.and let soban speak truth and lies alternatly.with this we are able to satisfy all the conditions and we get to know that raj is the thief.

thats true, Raj is the thief


question 5>
A teacher distributes some 5 Rs notes, some ten Rs notes & four 20 Rs notes to three of her students. The total amount that each receives is the same. how many maximum number of notes were distributed by the teacher in all?

1> 6
2> 7
3> 9
4> 10
5> 8

Q5. answer >9
Since the notes were distributed , the total amount can be assumed as a multiple of 5 or 10 and ofcourse 3.
there are already 4 - 20 Rs notes implies = 80 Rs.
80 rupees as it is cannot be divided amoung 3 pupil.
1 10 Rs, makes it 90 .. though can be divided equally among 3 pupil, cannot be done as we have 20+20+20+10. There is no way in which equal division can be done without procuring smaller denomination of 20 Rs note.
make it 2 10 notes and we have 100Rs, which doesnot work
put in 1 5 Rupee note which makes it 105Rs. 20+20+20+20+10+10+5. Again cannot be divided unless we have smaller denominations.
when the amount is 120Rs, 20+20+20+20+10+10+10++5+5 , each fo three pupil gets 40 Rs.
20+20
20+10+10
10+20+5+5


so this was to calculate minimum number in which the notes can be divided equally which is 9.

durgaQ5. answer >9
Since the notes were distributed , the total amount can be assumed as a multiple of 5 or 10 and ofcourse 3.
there are already 4 - 20 Rs notes implies = 80 Rs.
80 rupees as it is cannot be divided amoung 3 pupil.
1 10 Rs, makes it 90 .. though can be divided equally among 3 pupil, cannot be done as we have 20+20+20+10. There is no way in which equal division can be done without procuring smaller denomination of 20 Rs note.
make it 2 10 notes and we have 100Rs, which doesnot work
put in 1 5 Rupee note which makes it 105Rs. 20+20+20+20+10+10+5. Again cannot be divided unless we have smaller denominations.
when the amount is 120Rs, 20+20+20+20+10+10+10++5+5 , each fo three pupil gets 40 Rs.
20+20
20+10+10
10+20+5+5


so this was to calculate minimum number in which the notes can be divided equally which is 9.

Nice approach, correct answer!


Proper methodology of Q4.

We will consider that each one could be a truth teller.

1. Charles is truth teller therefore both his statements are true implies Raj is liar & he is the thief. Soban is alternator as his 1st statement is true & his 2nd statement is false.

2. Raj is truth teller therefore both his statements are true implies either charles or Soban is the thief. Since 2nd statement of Soban is true, he must be alternator. So his first statement must be false. Therefore he is thief. But Charles statement are contradicting (1st true & 2nd false). We can't have two alternators out of three. Therefore this case is rejected.
Note : It might be possible that Soban's both statements are true then he is the truth teller which is still not possible (2 truth tellers) So this case will also be rejected.

3. Soban is truth teller implies Charles is the thief. Charle's both statements are false therefore he is liar. Raj's both statements are true, therefore he is truth teller. But we can't have 2 truth tellers out of three. Therefore this case is rejected.

1 is the only valid solution, therefore Raj is the thief.


Question 5.
In a class of 100 students, 50 students passed in Mathematics & 70 passed in English. 5 students failed in both Mathematics & English.

a> How many students passed in both the subjects?
1. 25 2. 50 3. 70 4. 5 5. 10
b> How many students passed in exactly one the two subjects?
1. 5 2. 15 3. 95 4. 70 5. 20
c> How many students failed in at least one subject?
1. 95 2. 50 3. 30 4. 70 5. 75

25 , 70 , 30 !!


Am I correct?

For Question 5.
In a class of 100 students, 50 students passed in Mathematics & 70 passed in English. 5 students failed in both Mathematics & English.

a> How many students passed in both the subjects?
1. 25 2. 50 3. 70 4. 5 5. 10
b> How many students passed in exactly one the two subjects?
1. 5 2. 15 3. 95 4. 70 5. 20
c> How many students failed in at least one subject?
1. 95 2. 50 3. 30 4. 70 5. 75

the answers are
25,70,75

am i right ?

Ya correct, but please post some steps.

correct answer!
explaination of Question 6


Question no. 7
P & Q play volleyball, whereas R & S play football, T & U play handball, but V & W play basketball. Having won their respective games, they sit in a row for a photograph session. T sits to the immediate right of W & P sits at one extreme end. One volleyball player is at fourth place from the left end of the row, W & T don't sit between R & U, who have only 2 persons seated between them, while no one sits to the left of R.
The number of persons seated between P & U is the same as the number of persons sitting between?
1> R & W
2> S & V
3> U & T
4> R & Q
5> Cannot be determined.

First is the question correct , can you please recheck?

And solving with my assumptions gave me the answer as R & Q.

let me know..

Question is cent % correct. There is a tricky part in the question & I think you understood the tricky part, let others solve it, wait from 4-5 days after that you can post your solution.

I think the answer must be R and Q....

the sequence might be R _ _ Q U W T P

(the two blanks may be VS or SV)

I'm not sure... Am I right....

Ya, you right with a slight mistake that the viewer will see them just in a reverse order.
Solution 7
analysing the data
V.B. -- P,Q
B.B. -- V,W
F.B. -- R,S
H.B. -- T,U

All sit for photo session, means that they are facing us. So right implies our left & left implies our right.(this was the tricky part rest is history)
T is the immediate right of W --> TW
P or Q is at the 4th position from left end (w.r.t. the viewer) but P is at one of the extremes --> Q is at fourth position.
None is to the left of R --> R is at postion no. 8 => P is at position no. 1
Hence U is in sit no. 5 as there are 2 people between R & U, but T & W are not between R & U so
they must be in seats 2 & 3 respectively.
So arrangements is as follows
1.P 2.T 3.W 4.Q 5.U 6.S/V 7.V/S 8.R
No. of persons sitting between U & P is same as R & Q i.e. 3
So. 4th option.


Question no. 8

Five pens P,Q,R,S & T are of five different colors, taken from the six colors red, green, blue, yellow, white & orange. P is of green color & T is not a pen of red color. Q is costlier than the pen which is of green color & cheaper than the pen which is of yellow color whereas, yellow colored pen is cheaper than the white coloredpen. S is of white color & T is the costliest of all the pens. Which of the following must be false?
1> The color of pen Q is red or blue or orange.
2> The cheapest pen is P.
3> The color of pen T is yellow
4> Pen R must be costlier than than pen P

Solution 8:
Arranging pens as per their cost
T > S > Pen of yellow color > Q > P

It clearly shows 3rd is false.

Question 9
Kya-Kya is an obscure island in the Pacific Ocean which is inhabited by three tribes namely Tulu, Lulu & Aalu. All inhabitat of this island give two statement as a reply to any question asked to them. Inhabitants of this island belonging to the Tulu tribe always tells the truth, those of lulu always lie & those of aalu alternate between truth & lies but which comes first is not known. One day a visitor to this island meet three persons from different tribes, they are Army, Barmy & Charmy. He asks them which tribe do you belong to? & got following replies?

Army : I belong to lulu. Barmy belong to lulu.
Barmy : Army belong to lulu. Charmy belongs to Aalu
Charmy : Army belongs to aalu. Barmy belongs to Lulu.

Which inhabitant belongs to which tribe?


Question 10
7 girls A,B,C,D,E,F & G sit in a row staisfying the following conditions.
1> D is to the immediate right of G.
2> F is to the right of B & A & is to the immediate left of E.
3> C is in the middle sit between F & D & is to the left of A.
4> G is not to the right of B.

a> Who is sitting at the centre?
1. D 2. B 3. C 4. A

b> Who of the following is sitting next to F?
1. A 2. B 3. C 4. D

c> Without violating any of the conditions in how many ways can we arrange them.
1. 1 2. 2 3. 3 4. 4

It's been more than one year since I posted these questions, people give it a try before I post the solutions.

army is Aalu, barmy is lulu and charmy is tulu. how:
assume army is aalu, let's say he lies about belonging to aalu (which goes with charmy's statement) and tells the truth about barmy belonging to lulu. since barmy is lulu, he always lies. his 1st statement goes with our assumption. he is also lying about charmy belonging to aalu, meaning by process of elimination charmy is tulu, and always tells the truth. as said before, one of his statements is corroborated by army. second of his statements say barmy is a lulu, which goes with our previous reasoning.

as to question 10, I've tried and tried, but I can get only one combination: E F B A C D G. so A sits in center, and E is next to F.

Your 9th question's solution is right but there is a simpler way of doing it.
A : I belong to lulu. Barmy belong to lulu.
B : Army belong to lulu. Charmy belongs to Aalu.
C : Army belongs to aalu. Barmy belongs to Lulu.

From the statements of A, it's evident that he is an alternator.
Explanation : Anybody belonging to Lulu tribe is liar, so none of the Luluians can say that he belongs to Lulu. So, 1st statement is False. As 1st statement is false, he can't belong to Tulu or Lulu. He don't belong to Lulu because that's conclusion from 1st statement of A.
Checkout B's first statement, it's definitely False. So he belongs to Lulu.
Final Solution
Tribe Aalu Lulu Tulu
---------Army Barmy Charmy
1st False False True
2nd True False True


10th solution
As per the rules, the relative positions of people are as follows.
(* denotes, there can be n number of people in between)
GD
B*F
A*F
FE
F[x persons]C[x persons]D OR D[x persons]C[x persons]F
C*A
G*B

Using these conditions, only possible arrangement : - GDBCAFE

Q. 11
There are two tapes in a tank. One alone can empty the tank in 10 hrs, another alone can empty in 15hrs. When both tapes are open, due to some technical defect in the tank, excess of 35 litres leaks out of tank every hr & the tank is empty in 3 hrs. Find the capacity of tank?

vik001indQ. 11
There are two tapes in a tank. One alone can empty the tank in 10 hrs, another alone can empty in 15hrs. When both tapes are open, due to some technical defect in the tank, excess of 35 litres leaks out of tank every hr & the tank is empty in 3 hrs. Find the capacity of tank?

i'm not sure but is it 210 liters?

It is 230 liters..

Let's say the rate of pipe A is x liters/hr and pipe B is y liters/hr.

So, 10x = 15y

and, 10x - 35*3 = 3 * (x + y)

These are the 2 equations.. Solving these, we get x as 23 and y as 15.333.

So, the capacity of the tank is 10*x or 15*y, which is 230 liters..

Please don't ask me for confirmation 😀, so me the procedure!

Q11 Solution
In one hr 10 hr tap empty 1/10 of tank(let its capacity is x)
for 15 hr tap it's 1/15.
Now, both will fill in an hr. 1/10 + 1/15 = 5/30 = 1/6
In one hr. 35 + x/6 litres flow out of the tank.
3(35 + x/6) = x
solve it, x = 210.