TCS Questions-7

TCS Questions: 7

1.    The average of four consecutive numbers A, B, C, and D is 49.5. What is the product of B & D?

Answer: 2499

Let the numbers be x, x+1, x+2, and x + 3. Then we have

[x + (x+1) + (x+2) + (x+3)] /4 = 49.5      ie    4x + 6 = 198.  Hence x = 48

B = x+1 = 49.     D = x+3 = 51.   B * D = 49 * 51 = 2499.

2.    Find the unit digit of 25^6251 + 36^528 + 73^54

Answer: 0

As we have to find the unit digit, we can re-write the information as below:

5^6251 + 6^528 + 3^54  ( Unit digit of 5 is always 5 and unit digit of 6 is always 6. Unit digit of 3^54 will be 54/4 reminder 2 ie 3^2 = 9)

Thus the answer is 5 + 6 + 9 = 20

The unit digit being 0  

3.    A number when divided by 5 and 8 leaves remainder 3 and 1 respectively. Find the number.

Answer: 33

Let the quotient be A and B respectively for the two divisions. And let N be the number. Then we have  N = 5A + 3 and N = 8B + 1. Thus we have an equation

5A + 3 = 8B + 1. From here onwards, it will be a trial and error method. Starting from 1 please find for which value of A, we get a natural number for B. The value 6 for A fits in nicely where by the value for B will be 4. Thus the answer is 5*6+3 = 8*4+1 = 33

4.    The last day of a century cannot be which day/s?

Answer: Tuesday / Thursday or Saturday.

100 years contain 5 odd days and the last day of the first century is Friday.

200 years contain 3 odd days and the last day of the second century is Wednesday

300 years contain 1 odd day and the last day of the third century is Monday.

400 years contain no odd days and the last day of the fourth century is Sunday.

As the cycle repeats every four years, the last day of a century cannot be Tuesday/Thursday or Saturday.

5.    A warehouse having a square floor area of 10,000 sq. meters extended a rectangular addition on one side of the floor area that increased the area by 50%. How many meters extension to one entire side was made?

Answer: 50 meters.

Let us assume each side of the square area as 100 meters. Hence the total area is 10,000 sq. meters. Now let X be the additional extension made. Now the new length of one side will extend by 100 + X meters. The new area will be 100 * (100+X).

We are informed that 100 * (100+X) = 15,000 sq. meters.  Ie

10,000 + 100X = 15,000.  100 X =5000.   X = 50 meters.

6.    A digital wristwatch was set right exactly at 8.30 a.m. It started losing 2 seconds, every five minutes. If the watch was operating continuously, then what would be the time indicated in the watch at 6.30 p.m.?

Answer: 6.26 p.m.

Between 8.30 a.m. and 6.30 p.m. the total time difference is 10 hours. The watch loses 2 seconds every 5 minutes. Thus in one hour it will lose 24 seconds. And in 10 hours the watch will lose 240 seconds or 4 minutes. Hence the time by the watch at 6.30 p.m. will be 6.30 – 0.04 = 6.26 p.m.

7.    Sneha’s age is 1/6^th of her father’s age. Sneha’s father will be twice the age of Vimal after 10 years. Vimal’s eighth birthday was celebrated two years ago. What is Sneha’s present age?

Answer: 5 years.

Vimals 8^th birthday was two years ago. Vimal’s present age is thus 10 years. After 10 years Vimal’s age will be 20 years. After 10 years Sneha’s father will be twice the age of Vimal.

Sneha’s father after 10 years will be 2*20 = 40 years

Sneha’s father’s present age is 40 – 10 = 30 year.

Sneha’s present age is 1/6^th of her father’s age. Ie 1/6^th of 30 = 5years.

Hence Sneha’s present age is 5 years.

8.    A batsman by scoring 23 runs improves his average from 15 to 16. If he wants to improve his average to 18, then how many runs he should make in the same match?

Answer: 39 runs.

To answer this question we should find the number of innings played by him prior to this innings. Let us keep it as x. Then we have an equation 15x ( Total runs made by him in x innings) + 23 runs = 16 ( x+1) (Total runs made by him in x+1 innings.

15x + 23 = 16 (x+1) = 16x + 16. Hence the value of x is 7.

Now to improve his average to 18 he must score Y runs. So we have a new equation

7 * 15 + Y = 18 * (7+1) = 18*8 = 144. Hence Y = 144 – 105 = 39 runs.

9.    A man sells apples. He gave half the apples that he had plus half apple to the first person. To the second person he gives half od the remaining apples and a half apple. For the third person he gives half of what he has plus half apple. In this manner he gives apples to seven persons in total and finally left with no apple. How

many apples he had originally?

Answer: 127 apples.

Let the total number of apples he had originally be X.

First he gives   X/2 +1/2 apples   Remaining with him is X – (X/2 + 1/2) mark this as A

Second he gives A/2 +1/2 apple. Remaining with him is A – (A/2 + 1/2) mark this asB

Third he gives B/2 + 1/2  apples  Remaining with him is B – (B/2 + 1/2) mark this asC

Like wise mark the sales as D, E ,F and G. After the last sale no apple remains.

Hence we have an equation F – (F/2 + ½) = 0 This gives the value for F as 1.

Now the next equation will be E – (E/2 + ½) = 1 This will give the value of E as 3

Similarly proceeding further you will have values for D as 7, C as 15, B as 31 and A as 63.

Finally we have the equation X – (X/1 +1/2) = 63. This will give the value of

X as 127 ie the number of apples he had originally.

10.  A club has members of both sexes. If 15 lady members quit, then the number of remaining lady members will be twice the number of gentlemen. If 45 gentlemen quits, then, the number of ladies will be five times the number of gentlemen. How many members of the club are ladies and gentlemen?

Answer: 175 Ladies and 80 Gentlemen.

Let the total ladies be represented as L and total gentlemen as M.

Then we have F – 15 =  2*M       F = 2M + 15   (i)

 Again                                           F = 5 * (M-45) (ii)

Equating (i) and (ii) we have 2M + 15 = 5M – 225  ie 3M = 240  M = 80 members.

 Substituting this is (i) we have F = 2*80+15 = 175 members.

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