Answers for maths

1. Here it is clear that train engines and the fly start their journeys at the same instant and the fly continues its
journey until the engine collide. So, flight time of the fly is equal to the time taken by the two engines to collide.
Relative speed of the engines = 10 + 10 = 20 km/hr.
So, time taken by the engines to collide with each other =
20
20 = 1 hr.
Hence, distance travelled by fly in 1 hr = 15 km. Ans.(1)
2. Let us suppose S for the two Sharmas and we denote each of the other four couples by M, P, Q and R.
Nobody shook 9 hands, since nobody shook the hand of his/her spouse. Therefore the numbers 0 through 8 are
used in describing the different numbers of handshakes performed by each of the nine people
(other than Mr. Sharma).
Someone shook 8 hands. Say that it is Mr. M. Then how many hands did Mrs. M shake ? Everyone in couples P,
Q, R, S must have shaken Mr. M's hand, in order to account for 8 shakes. So each of the people in couples P, Q,
R, S shook hands at least once. But somebody shook hands zero times. It must be Mrs. M.
Now we eliminate Mr. and Mrs. M from our consideration. Someone shook exactly 7 hands. Say that it is Mrs.
P. We know already that Mrs. P shook the hand of Mr. M. She did not shake the hand of Mrs. M, since nobody
did. To obtain a total of 7 shakes, she must also have shaken the hands of all the people in couples Q, R, S. But
someone had to shake only one hand (the people in Q, R, S have now each shaken at least two hands, since they
each shook Mr. M's hand, as well as Mrs. P's hand). It must be Mr. P who shook only one hand.
Continuing in this fashion, we see that the person who shook 6 hands is betrothed to the person who shook 2.
And the person who shook 5 hands is betrothed to the person who shook 3. That leaves only Mrs. Sharma, who
must have shaken 4 hands - four being the only remaining number.
Note : 4 is the only number that cannot be paired.
The answer to our problem is that Mrs. Sharma shook four hands. Ans.(4)
3. Table (A)
Initial state 2 rupee coin at the top
1 rupee coin
50 paise coin at the bottom
Move (1) 2 rupee coin from table (A) to table (B).
Move (2) 1 rupee coin from table (A) to table (B).
Move (3) 1 rupee & 2 rupee coins from table (B) to table (A).
Move (4) 1 rupee coin, 2 rupee coin & 50 paise coin from table (A) to table (B). Ans.(3)
4. Total number of routes from Bombay to Calicut = (4 × 3 × 2) = 24. Ans.(3)
5. It is given that one person came in through door F and second person came in through door A. It means that door A and
door F are Entrances. Also, they both left through door B. Hence, door B is Exit. As Exit and Entrance should alter each
other and we know two Entrances, let's assume that the third Entrance is W. Thus, there are 6 possibilities with "_"
indicating Exit.
(1) _ W _ A _ F (2) _ W _ F _ A (3) _ F _ W _ A (4) _ F _ A _ W (5) _A _ W _ F (6) _ A _ F _ W
As door A must be followed by door B or E and none of them lead to the door F, (1) and (6) are not possible. Also, door
D must be Exit as only door D leads to the door A and F. Door A and F are Entrances.
Solutions to Maths questions
January 2007 Issue
2 JJanuarryy 2007
(2) _ W _ FDA (3) _ F _ WDA (4) _FDA _ W (5) DA _ W _ F.
Only door D and door C lead to the door F. But door D is used. Hence, door C must be one of the Exits and
should precede door F. Also, the third Exit is B and the W, the assumed Entrance, must be door E. (2) BECFDA
(3) CFBEDA (4) CFDABE (5) DACEBF.
But only door B leads to the door C and both are Exits. Hence, (2) and (5) are not possible. Also, door F does not
lead to door B, discard (3). Hence, the possible order is (4) i.e. CFDABE. Ans.(4)
6. Divide twelve pearls into three groups of 4. Let these are G1, G2, G3. As a first step, weigh G1 against G2.
(i) If they happen to balance, then all eight pearls in G1 and G2 are control pearls. The odd pearl is one of the
pearls in G3.
(ii) If they do not balance, then all the pearls in G3 are control pearls. The odd pearl is either in G1 or G2 but we
do not know which.
We first consider Case 1. Take any three pearls from G1 and weigh them against any three pearls from G3.
If they balance, then the odd pearl is the remaining pearl from G3. Weighing that last pearl against one of
the pearls from G1 will tell whether the odd pearl is heavy or light. If they do not balance, then the odd pearl
will be among the three selected from G3 and we will know whether it is lighter or heavier (since the pearls
from G1 are control pearls). Now a third weighing, as usual, will pin down the odd pearl from among those
three that we selected from G3.
For Case 2, we suppose for specificity that G1 is heavier and G2 lighter. Give the pearls in G1 names a, b, c,
d and give the pearls in G2 the names a', b', c', d'. For the second weighing, we weigh {a, b, a'} against {c,
d, b'}.
(a) If they balance, then the odd pearl is one of c', d' (the two pearls from G1 and G2 that we omitted from this
second trial). Of course c', d' come from the light side, so we know that the odd pearl is light. For the third
weighing, we weigh c' against d'. The odd pearl is the lighter of the two.
(b) If they do not balance, then say that {a, b, a'} is heavier. This must mean that c, d are control pearls, and
so is a', or else the balance would be the other way. Thus the odd pearl is either a, b or b'. Finally weigh a
against b. If they balance, then the odd pearl is b' and it is light. If they do not balance then the odd pearl
is the heavier of the two (since a and b come from G).
(c) The case that {c, d, b'} is heavier is handled just as in subcase (b). Ans.(2)
7. Here it is given that two statements of each student must be true and one statement must be false. So we can
check by each statement.
From Anita’s statements : if third statement is true, then first and second statement also will be true. But it
contradicts the mentioned condition.
So, third statement is false and remaining are true, i.e., Anita and Sameer did not steel the purse. From
Sunita’s statements : If third statement is true, then first statement is false then second statement must be
true and if third statement is false, then first and second seconds statements are true.
From Dev’s statement : The third statement is false so first and second statements must be true.
From Manmohan’s statement : Third statement is false, so first and second must be true. So, from Sunita’s &
Manmohan’s statements we conclude that Sunita stole the purse. Ans.(1)
8. Let the number of pebbles in first, second and third heap be x, y and z respectively.
In the first operation, ‘y’ pebbles are taken from the first heap and added to the second. Next, we take ‘z’
pebbles from the second heap and add them to the third. Finally, we see that the first heap will contain “x–y”
pebbles after the first operation. We take an equal number of pebbles from the third heap and add them to the
first one.
We can tabulate the same :
Heap-1 Heap-2 Heap-3
x y z
x – y 2y z
x – y 2y – z 2z
2(x –y) 2y – z 2z – (x – y)
3 JJanuarryy 2007
Now, 2(x – y) = 2y – z Þ 2x – 4y + z = 0 .......(1)
2(x – y) = 2z – (x – y) Þ 3x – 3y + 2z = 0 .......(2)
2y – z = 2z – (x – y) Þ x + y – 3z = 0 .......(3)
From (2) and (3) i we get;
x - y = z and x + y = z 2
3
3
\x = z and y = z
11
6
7
6
.
\ Total numbers of pebbles in the 3 heaps = 11 + + =
6
7
6
z z z 4z.
We have been that the number of pebbles is a multiple of 4.
Options (1); 4z = 12 Þ z = 3
\x = Not possible
11
2
( )
Option (2); 4z = 24 Þ z = 6
\ x = 11 and y = 7.
Options (3) and (4) are greater than 24. Ans.(2)
9. Let R is the radius of the Earth at equator in cm.
Circumference = C = 2pR cm.
To increase the radius by 1 cm, we will have to increase the circumference also. Let the new circumference be
C1, where
C1 = 2p (R + 1) cm.
So, the required change in length of steel band
= C1 – C = 2 p (R + 1 – R)= 2p cm = 2 × 3.14 cm = 6.28 cm. Ans.(4)
10. Here it is clear that train engines and the fly start their journies at the same instant and the fly continues it
journey until the engine collide. So, flight time of the fly is equal to the time taken by the two engines to collide.
Relative speed of the engines = 10 + 10 = 20 km/hr.
So, time taken by the engines to collide with each other =
20
20 = 1 hr.
Hence, distance travelled by fly in 1 hr. = 15 km. Ans.(3)
11. We can take a maximum of 2 black, 2 white, 1 red, 1 yellow and either 1 violet or 1 green wire. Thus the
maximum number of wires is seven. Ans.(2)
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12. Asmita, Babita, Chandramukhi, Deekshaand Elina are in one weight-group. Their weight-group cannot have
any more people and their instructor is Aparna. We can tabulated the information gives in the question as :
Instructor Weight-group Members
Aparna W1
Asmita, Beena, Chandramukhi,
Deeksha, Elina
Firoza, Gauri
Hillary
Indira
Jaya
The other constraints can be represented as :
Kanika – Jaya Manisha – Jaya
Lalita – Hillary Manisha – Kanika
Kanika – Indira (in W4) Manisha – Hillary
Also, Naina – Firoza Deeksha – W1
and Oxana – Firoza Naina – Oxana
Since both Naina and Oxana belong to a weight-group with four members, they have to be with Indira and
Kanika in W4 and their instructor is Beena.
Also, since Manisha cannot be with either of Hillary, Kanika or Jaya, she is in the same weight-group as Firoza.
Also, since Lalita cannot be in the same group as Hillary, she has to be in the same group with Jaya. Hence the
weight-groups can new be tabulated as :
Instructor Weight-group Members
Aparna W1
Asmita, Beena, Chandramukhi,
Deeksha, Elina
Chandrakanta Firoza, Gauri, Manisha
Beena Hillary
Deepali W4 Indira, Kanika, Naina, Oxana
Ekta Jaya, Lalita
Ans.(1)
13. One cube is cut into 8 cubes and the other into 27 cubes. These two cubes are now joined together and painted
red. The cubes at the edges would be painted on three sides and there are 8 such cubes. The cube which has
been divided into 27 cubes would have 1 cube painted red on two sides on each of its 4 edges on the extreme
end. There are 4 edges which have 2 cubes painted red on 2 sides. Thus there are 4 + 8 = 12 cubes. Also the
cube which is divided into 8 cubes will have 4 cubes painted red on two sides. Thus there are a total of 16 cubes
which are painted red on two sides. For the cube divided into 27 cubes, there are 2 cubes on the inner side
which are not painted red. Ans.(3)