TCS - Recent Questions - 5

TCS – Recent Questions - 5

1.       If 5+3+2=151022, 9+2+4=183652, then 7+2+5=?
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

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

On the same basis
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

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)

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

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

Ans: (a)
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

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

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