# TCS Ninja, Digital, NQT Latest Placement Questions with solutions - 17

1. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many “different” boxes of chocolates can be made?
Sol:
If n similar articles are to be distributed to r persons, ${x_1} + {x_2} + {x_3}......{x_r} = n$ each person is eligible to take any number of articles then the total ways are ${}^{n + r - 1}{C_{r - 1}}$
In this case ${x_1} + {x_2} + {x_3}......{x_6} = 10$
in such a case the formula for non negative integral solutions is ${}^{n + r - 1}{C_{r - 1}}$
Here n =6 and r=10. So total ways are ${}^{10 + 6 - 1}{C_{6 - 1}}$ = 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
$8x - 5(26 - x) = 0$
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 =$l \times b \times h$ = $6 \times 5 \times 2$  = 60 $c{m^3}$
Now volume is reduced by 19%.
Therefore, new volume = $\displaystyle\frac{{(100 - 19)}}{{100}} \times 60 = 48.6$
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: $\left( {\displaystyle\frac{x}{{100}} \times 6} \right)\left( {\displaystyle\frac{x}{{100}} \times 5} \right)2 = 48.6$
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 $\displaystyle\frac{{n(n + 1)}}{2} \times 10 \le 2888$

For n = 23 we get LHS as 2760.  Remaining terms 128.

Now in the 24th term, we have 24 1's, and next 48 terms are 2's.  So next 72 terms are 3's.
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