TCS
Questions - 13
1.
The
savings of employee equals income minus expenditure. If the income of A,B,C are
in the ratio 1:2:3 and their expense ratio 3:2:1 then what is the order of the
employees in increasing order of their size of their savings?
Answer:
A, B, C
The
ratio of income of A, B, and C are 1:2:3. Let us assume their Incomes as 10000,
20000 and 30000
The
ratio of their expenditure is 3:2:1. (C spends the least while A spends the maximum). Assuming the
expenditure of A, B and C as 9000, 6000 and 3000, then the
Savings
of A, B and C works out to 1000, 14000 and 27000 the ratio being 1:14:27 The increasing order of employees’ savings is A, B and C.
2. Two full tanks one shaped like the cylinder and the
other like a cone contain liquid fuel. The cylindrical tank held 500 litres
more than the conical tank. After 200 litres of fuel is pumped out from each
tank the cylindrical tank now contains twice the amount of fuel in the conical
tank. How many litres of fuel did the
cylindrical tank had when it was full?
Answer: 1200 litres
Let
the initial quantity of fuel in the conical tank be ‘x’. Then the quantity of fuel in the cylindrical
tank is (x+500). Now 200 litres of fuel are removed from each tank.
The
fuel in the conical tank now is (x-200). The fuel in the cylindrical tank will
be
(x+500)-200=(x+300).
From the question we know (x+300) = 2*(x-200). Solving this we get the value of
‘x’ as 700 litres. Hence the initial
quantity of fuel in the cylindrical tank is (700+500)= 1200 litres.
3.
Eesha
bought two varieties of rice costing 50Rs per kg and 60 Rs per kg and mixed
them in some ratio. Then she sold that mixture at 70 Rs per kg making a profit
of 20 %. What was the ratio of the mixture?
Answer: 1:4
The
sale price of Rs 70/kg includes 20% profit. If the profit element is
eliminated, we get the sale price as 70 * 100/120 = 58.33. (This can be rounded
off to Rs 58.) Now applying Rule of alligation we have
58
X
(B-X) 2
8 (X-A)
Thus the ratio of mixture of the two
varieties Rs 50 and Rs 60 is 2:8 or 1 : 4.
4.
In
a horse racing competition there were 18 horses numbered 1 to 18.The organizers
assigned a probability of winning the race to each horse based on horses health
and training. The probability that horse one would win is 1/7, that horse two
would win is 1/8, and that horse three would win is 1/7. Assuming that tie is
impossible, find the chance that one of these three horses would win the race?
Answer: 15/49
The
chances of Horse No 1 winning 1/7 Probability of horse No 1 not
winning 6/7
The
chances of horse No 2 winning
1/8 Probability of horse No
2 not winning 7/8
The
chances of horse No 3 winning
1/7 Probability of horse No
3 not winning 6/7.
The
chances of any one of these three horses winning the race will be:
(A
winning, B and C not winning) + (B winning and C and A not winning) + (C
winning and A and B not winning)
Now
substituting the relevant information in the above we get the following.
Probability
of any one of the three horses winning is arrived at as
(1/7
+ 7/8 + 6/7) + ( 1/8 + 6/7 + 6/7) + ( 1/7 + 6/7 + 7/8) Reducing this we get the value
Of probability of any one of these three
horses winning as 15/49.
5.
The
missing number in the sequence
5, 6, 15, ?, 89, 170, 291, is
5, 6, 15, ?, 89, 170, 291, is
(a)
50 (b) 40 (c) 42 (d) 32
Answer: (b) 40
Answer: (b) 40
6
– 5 = 1 (1^2) 15 – 6 = 9 (3^2)
170
– 89 = 81 (9^2) 291- 170 = 121 (11^2)
Going by this we observe that between 89 and 15 we should have two digits
namely 5^2 and 7^2. Hence the missing number is
15 5^2 = 40.
6.
Two
trains starting at same time, one from Bangalore to Mysore and other in
opposite direction arrive at their destination 1 hour and 4 hours respectively
after passing each other. How much faster is one train from other?
Answer: One train is four times
faster than the other train.
The distances between the two stations remain the
same.
The ratio of speeds will be inversely proportional to their respective times.
Let speed of train 1 be S1 and that of train 2 be S2.
Let time taken by train 1 be T1=1hr
and time taken by train 2 be T2=4hr
We now have a ratio
S1:S2 = 1/T1:1/T2 = T2:T1 = 4:1
So train 1 is 4 times faster than train 2.
The ratio of speeds will be inversely proportional to their respective times.
Let speed of train 1 be S1 and that of train 2 be S2.
Let time taken by train 1 be T1=1hr
and time taken by train 2 be T2=4hr
We now have a ratio
S1:S2 = 1/T1:1/T2 = T2:T1 = 4:1
So train 1 is 4 times faster than train 2.
7.
If
f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)?
Answer: C
We know f(0) = 1 and f{f(n)} + f(n) =
2n + 3
Now
substituting 0 for n in the above equation, we have f(f(0) + f(0) = 2(0) + 3.
Since
f(0) = 1 substituting this we have f(1) + 1 = 3. Hence f(1) = 2
Similarly
if we put the value of n = 1, then we will get f(2) = 3 and so on .
The
value of f(n) will be n+1
On this basis f(2012) will be 2013. Hence
the answer is ‘c’.
8. A wooden rod AB is 1 meter long. Raju
makes a mark every 3 cm along the rod starting from the end A. Vikram makes a
mark every 7 cm and Aditya every 9 cm both starting from the end B. If the rod
is cut at each of the points where there is a mark, then how many pieces would
it be cut into?
Answer: 51 pieces
Raju will be marking at 33 places starting from the
end A. Thus it will give 33 pieces.
Vikram will be marking at 14 places starting from
the end B. Of these markings five markings at 93 cm, 72 cm, 51, cm, 30 cm, and
9 cm will coincide with the markings of Raju. Also the marking at 37 cm will
coincide with that of Aditya. Thus six markings are common with either Raju or
Aditya. Leaving aside this, Vikrams markings will give 8 pieces.
Aditya’s cutting
starting from the end B will give 11 pieces of which the marking at one place
37 cm is common with that of Vikram. Thus the number of pieces will be 10.
Hence the total number of pieces are
33 + 8 + 10 = 51 pieces.
9.
The
crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,….,
W8), such that there are two rows, each row occupying one of the 2 sides of the
boat and that each side must have 4 members including at least one women.
Further it is also known W1 and M7 must be selected for one of its sides while
M2, M3 and M10 must be selected for other side. What is the number of ways in
which rowing team can be arranged?
Answer: 70 ways.
Side One: Total Number 4. M2, M3 and M10 are must. This leaves one
seat for woman. Out of the eight women, W1 should be on side two. That leaves
remaining 7 women from whom one woman can be selected in 7 ways.
Side Two: Total Number 4. W1 and M7 are must. That
leaves two remaining seats that can be filled exclusively by either women or
one woman and one man. We have 6 remaining women and the two seats can be
filled in 6C2 ways ie 15 ways. Suppose we fill one man and one woman this can
be done in the following manner. Since there are 6 women and 8 men available
the filling can be done in (6C1)*(8C1) = 48 ways.
Thus the remaining seats in
the boat can be filled in 7 + 15 + 48 = 70 ways.
10. At the end of 1994, Rohit, was half as old as his grandmother. The
sum of the years in which they were born is 3844. How old Rohit was at the end of
1999?
A) 48 B)
55 C)
49 D)
53
Answer: (d) 53 years
Let x and y be the years when
the Rohit and grandmother were born.
Thus we have x + y = 3844 …….
(I)
In
1994 Rohit is half the age of his grandmother. We now have another equation
(1994
– y) = 2(1994 – x) this will give a value of 2x – y = 1994………..(II)
Adding
(I) and (II) we get 3x = 5838 and x = 1946 Hence Y = 3844 – 1946 = 1898
In
1994 Rohit’s age was 1994 – 1946 = 48 years.
In 1994
grand-mother’s age was 1994 – 1898 = 96
years (Twice the age of Rohit)
So, Rohit’s age in 1999 will be 48 + 5 = 53
years.
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