- A contractor had employed 100 labourers for a flyover construction task. He did not allow
any woman to work without her husband. Also, atleast half the men working came with their
wives.He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening. How many men, women and children were working with the constructor?
Answer:- 16 men, 12 women and 72 children were working with the constructor.
Let's assume that there were X men, Y women and Z children working with the constructor.
Hence,
X + Y + Z = 100
5X + 4Y + Z = 200
Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z
As if woman works, her husband also works and atleast half the men working came with their
wives; the value of Y lies between X and X/2. Substituting these limiting values in
equations, we get
if Y = X,
300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428
if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727
But Z must be an integer, hence Z=72. Also, X=16 and Y=12
There were 16 men, 12 women and 72 children working with the constructor.
2. Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they
owned.
It was found that Guran had ten more sheep than Lakha.
If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran,
who then passed on a fifth of his holding to Lakha, they would all have an equal number of
sheep.
How many sheep did each of them possess? Give the minimal possible answer.
Answer:- Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.
Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is
given that at the end each would have an equal number of sheep, comparing the final numbers
from the above table.
Arjan's sheep = Bhuvan's sheep
2A/3 = A/4 + 3B/4
8A = 3A + 9B
5A = 9B
Arjan's sheep = Guran's sheep
2A/3 = A/15 + B/5 + 4G/5
2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)
30A = 3A + 5A + 36G
22A = 36G
11A = 18G
Arjan's sheep = Lakha's sheep
2A/3 = A/60 + B/20 + G/5 + L
2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)
2A/3 = A/6 + L
A/2 = L
A = 2L
Also, it is given that Guran had ten more sheep than Lakha.
G = L + 10
11A/18 = A/2 + 10
A/9 = 10
A = 90 sheep
Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and
Lakha had A/2 i.e. 45 sheep.
3. You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances
only and 3 are Exits only.
One person came in through door F and two minutes later second person came in through door
A. He said, "You will be set free, if you pass through all 6 doors, each door once only and
in correct order. Also, door A must be followed by door B or E, door B by C or E, door C by
D or F, door D by A or F, door E by B or D and door F by C or D."
After saying that they both left through door B and unlocked all doors. In which order must
you pass through the doors?
Answer:- The correct order is CFDABE
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 the Exit as only door D leads to the door A and door A is the
Entrance.
(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 the
Exit and precede door F. Also, the third Exit is B and the W 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.
4. There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the
second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the
last digit. There are 3 pairs whose sum is 11.
Find the number.
Answer :-65292
As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits.
They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)
It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8),
(4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it
must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th
digits is (5, 2, 9)
Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and
(9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is
65292.
5. A person travels on a cycle from home to church on a straight road with wind against him.
He took 4 hours to reach there.
On the way back to the home, he took 3 hours to reach as wind was in the same direction.
If there is no wind, how much time does he take to travel from home to church?
Answer :-Let distance between home and church is D.
A person took 4 hours to reach church. So speed while travelling towards church is D/4.
Similarly, he took 3 hours to reach home. So speed while coming back is D/3.
There is a speed difference of 7*D/12, which is the wind helping person in 1 direction, &
slowing him in the other direction. Average the 2 speeds, & you have the speed that person
can travel in no wind, which is 7*D/24.
Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7 hours.
Answer is 3 hours 25 minutes 42 seconds
