# TCS latest Questions with Answers – 17

Sol:

If n similar articles are to be distributed to r persons, x1+x2+x3……xr=nx1+x2+x3……xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1n+r−1Cr−1

In this case x1+x2+x3……x6=10×1+x2+x3……x6=10

in such a case the formula for non negative integral solutions is n+r−1Cr−1n+r−1Cr−1

Here n =6 and r=10. So total ways are 10+6−1C6−110+6−1C6−1 = 3003

2. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.

a. 1/3

b. 1/2

c. 5/9

d. 17/36

Sol: Their sum can be 3,4,6,8,9,12

For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 – n) ways.

Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.

So probability is (20/36)=(5/9)

3. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?

a. 2000

b. 4000

c. 5000

d. 3000

Sol:

Time taken by A and B is in the ratio of = 3:2

Ratio of the Work = 2 : 3 (since, time and work are inversely proportional)

Total money is divided in the ratio of 2 : 3 and B gets Rs.3000

4. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero.

a. 10

b. 11

c. 13

d. 12

Sol:

Let x ques were correct. Therefore, (26- x) were wrong

8x−5(26−x)=08x−5(26−x)=0

Solving we get x=10

5. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,

a. 5.5

b. 4.5

c. 7.5

d. 6.5

Sol:

Volume =l×b×hl×b×h = 6×5×26×5×2 = 60 cm3cm3

Now volume is reduced by 19%.

Therefore, new volume = (100−19)100×60=48.6(100−19)100×60=48.6

Now, thickness remains same and let length and breadth be reduced to x%

so, new volume: (x100×6)(x100×5)2=48.6(x100×6)(x100×5)2=48.6

Solving we get x =90

thus length and width is reduced by 10%

New width = 5-(10% of 5)=4.5

6. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?

Sol: We have to consider the number of 4’s in two digit numbers. _ _

If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)

So total 19 ways.

**Alternatively:**

There are total 9 4’s in 14, 24, 34…,94

& total 10 4’s in 40,41,42….49

thus, 9+10=19.

7. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?

Sol: Let man daily wages and woman daily wages be M and W respectively

24M+16W=11600

12M+37W=11600

solving the above equations gives M=350 and W=200

8. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?

Sol:

Profit = 4200

Profit =SP – CP

4200=SP – 300000 therefore SP=304200

x+y = 300000

1.2x + 0.9y = 304200

Solving for x = 114000 = CP of cow.

9. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4……

In the above sequence what is the number of the position 2888 of the sequence.

a) 1

b) 4

c) 3

d) 2

Sol: First if we count 1223334444. they are 10

In the next term they are 20

Next they are 30 and so on

So Using n(n+1)2×10≤2888n(n+1)2×10≤2888

The 2888 term will be “3”.

10. How many 4-digit numbers contain no.2?

Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 )

We try to find the number of numbers not having digit 2 in them.

Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)

Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here ‘0’ cannot be taken

Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832

Total number of numbers having digit 2 in it = 9000-5832 =3168