TCS – Recent Questions
- 5
1.
If 5+3+2=151022, 9+2+4=183652, then 7+2+5=?
Ans: 143547
Ans: 143547
Let us take the three numbers as x, y, and z.
Thus we have x + y + z = 5 + 3 + 2 =
151022. Similarly
We have x + y + z = 9 + 2 + 4 = 183652.
Looking at the total value in both we
could arrive at a connection among the three digits.
Suppose we split the total value into
three parts then we have |15| |10| |22| and
|18| |36| |52|
A careful look into these split figures
would show the first set is x*y, the second x*z and the third the sum of the
first two sets less the middle digit value -> (x*y) + (x*z) –y
Applying this, we have for 7 + 2 + 5 the
answer as -> |7*2| |7*5| |(7*2) + (7*5) – 2 |
è
143547
2.
1(1!)+2(2!)+3(3!)....2012(2012!) = ?
Ans: 2013!-1
Ans: 2013!-1
We can write 1(1!)=1 ⇒ 2!-1
similarly 1(1!)+2(2!)=1+4=5 ⇒ 3!-1
Proceeding similarly 1(1!)+2(2!)+3(3!)=1+4+18=23 ⇒ 4!-1
similarly 1(1!)+2(2!)=1+4=5 ⇒ 3!-1
Proceeding similarly 1(1!)+2(2!)+3(3!)=1+4+18=23 ⇒ 4!-1
On the same basis
1(1!)+2(2!)+3(3!)+........+2012(2012!)=2013!-1
1(1!)+2(2!)+3(3!)+........+2012(2012!)=2013!-1
3. A
number when successively divided by 5, 3, 2 gives remainder 0, 2, 1
respectively in that order. What will be the remainder when the same
number is divided successively by 2, 3, 5 in that order
a) 4, 3, 2
b) 1, 0,4
c) 2, 1, 3
d) 4, 1, 2
Ans: B
a) 4, 3, 2
b) 1, 0,4
c) 2, 1, 3
d) 4, 1, 2
Ans: B
This question
is a little tricky one. Readers are likely to assume that the number when
divided individually leaves the reminder. This is the crux. The question states
on successful dividing. In other words if A is divided and leaves a quotient B,
then the next division is on B and not A the original digit.
Let
the number be A. When divided by 5 leaves a reminder ‘0’. Let the quotient be B
Then
we have A = 5B.
Now
when B is divided by 3 let the quotient by C and reminder 2
So,
we have B = 3C + 2
Now
when C is divided by 2 let the quotient be D and the reminder 1
So,
we have C = 2D + 1.
We
now have A = 5(3C+2) -> 15C+10 ->
15(2D + 1) + 10 -> 30D + 15 + 10 -> 30D + 25.
Assuming
the value of D as 0, we have the least number as 25.
When
25 is divided by 2, the reminder is 1 and quotient 12.
When
12 si divided by 3, the reminder is 0 and quotient 4
When
4 is divided by 5, the reminder is 4
Hence
the answer is (B) -> 1, 0, 4
4.
Two gears one with 12 teeth and the other one with 14
teeth are engaged with each other. One tooth in smaller and one tooth in bigger
are marked and initially these 2 marked teeth are in contact with each other.
After how many rotations of the smaller gear with the marked teeth in the other
gear will again come into contact for the first time?
a)7
b) 12
c) Data insufficient
d) 84
Ans: (a)
a)7
b) 12
c) Data insufficient
d) 84
Ans: (a)
Let us assume the distance between each tooth in both the
gears as 1cm.
Hence the circumference of each gear is 12 cm and 14cm
respectively.
LCM of 12 and 14 is 84.
The smaller gear with circumference will rotate 7 times
while the larger gear will rotate six times during this period.
Hence both the marked teeth will come in contact for the
first time after 7 rounds.
5.
A owes B Rs.50. He agrees to pay B over a number of
consecutive days from Monday, paying single note of Rs.10 or Rs.20 on each day.
In how many different ways can A repay B.
Ans: 8
ways.
He can pay by all 10 rupee notes = 1 way
He can also pay three ten rupee note and
one twenty rupee note, totally 4 ways.
He can pay through one ten rupee note
and two twenty rupee notes, totally 3 ways.
Total ways = 1 + 4 + 3 = 8
Total ways = 1 + 4 + 3 = 8
6.
Find last two digits of the following expression
(201*202*203*204*246*247*248*249)^2
Ans: 76
The last two digits of a product is obtained by multiplying the last two digits in each number. In this case it is (01*02*03*04**46*47*48*49) This gives a value 122040576 and the last two digits are 76. So we have 76^2 and the value is 5776. Hence the last two digits are 76.
7.
How many two digit numbers are there which when subtracted from
the number formed
by reversing its digits as well as when
added to the number formed by reversing its digits, result in a perfect
square.
Ans:
56.
Let the number be ‘xy’ -> 10x + y.
When the digits are reversed we have yx -> 10y + x
We are given (10y +x) –(10x + y) ->
9(y-x) is a perfect square.
Hence (y-x) should be 1, 4, or 9………………….(i)
Also we are given that
(10x+y) + (10y + x) -> 11(x + y) is a
perfect square.
Hence (x + y) should equal 11
Only (9, 2), (8, 3), (7, 4), (6, 5) …………………….(ii) satisfies this condition.
Comparing (i) and (ii) we find only the
digits 6 and 5 fulfil both the conditions.
Hence the original two digit number is
56.
8.
A property was originally on a 99 years lease and two thirds
of the time passed is equal to
the four fifth of the time to come. How
many years are there to go?
(a) 45 (b) 50 (c) 60 (d) 55
(a) 45 (b) 50 (c) 60 (d) 55
Ans: (a)
Let us assume x years have passed and y years to go. We are given
Let us assume x years have passed and y years to go. We are given
2x/3 = 4y/5 -> 10x = 12y -> x = 6y/5.
But we know x + y = 99. Now substituting the value of x
We have 6y/5 + y = 99. Solving we get the value of ‘y’ as
45.
Hence the number of years to go is 45.
9.
The value of diamond varies
directly as the square of its weight. If a diamond falls and breaks into two
pieces with weights in the ratio 2:3. What is the loss percentage in the value?
Ans: 48%
Let the
original weight be ‘x’. Hence the value is x^2
The diamond
is broken into two pieces of ratio 2 : 3. Let the broken pieces by 2y and 3y.
Now the value
of these two broken pieces are 2y^2 and 3y^2
The value of
broken pieces is 4y2 + 9y2 = 13y2
We know ‘x’
the original weight is equal to 2y + 3y = 5y.
On this basis
the value of the diamond originally was 25y2
Present value
of the two broken pieces 13y2
Hence the
loss percentage is 25y2 – 13y2 = 12 * 100/25 = 48%
Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2
Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2
10.
Cara, a blue whale participated in a weight loss program at
the biggest office. At the end of every month, the decrease in weight from
original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While
Cara made a steadfast effort, the weighing machine showed an erroneous weight
once. What was that?
(a) 2676 (b) 2 (c) 445 (d) 86
(a) 2676 (b) 2 (c) 445 (d) 86
Ans:
(d)
This is a number series problem. The series goes as under:
1 x 1 + 1 = 2
2 x 2 + 2 = 6
6 x 3 + 3 = 21
21 x 4 + 4 = 88
88 x 5 + 5 = 445
445 x 6 + 6 = 2676.
Hence the answer 86 is wrong and that should be 88
