How many 2’s are there between the terms 112 to 375?

Answer: c

Explanation:

Let us calculate total 2’s in the units place. (122, 132, 142 … 192), (201, 212, 222, … 292), (302, 312, … 372) = 8 + 10 + 8 = 26

Total 2’s in tenth’s place, (120, 121, 122, …, 129) + (220, 221, …, 229) + (320, 321, …, 329) = 30

Total 2’s in hundred’s place = (200, 201, … 299) = 100.

Total 2’s between 112 and 375 = 26 + 30 + 100 = 156

Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?

Answer:

Explanation:

Let Sandra, David and Mary each has s, d and 11 respectively.

After the first distribution,

David has d + d = 2d, Mary has 11 + 11 = 22 and Sandra has s – d – 11.

After the second distribution,

Sandra has 2×(s – d – 11) , mary has 2×22 = 44 and david has 2d – (s – d – 11) – 22=3d – s –11.

After the third distribution,

Sandra has 2×2(s – d – 11), david has 2×(3d – s – 11) and mary has 44 – 2(s – d – 11) – (3d – s – 11) = 77 – s – d

It is given that finally Mary has Rs.17. So, 77 – s – d=17

⇒ s + d = 60

⇒ s + d + 11 = 60 + 11 = 71.

Apples cost L rupees per kilogram for the first 30 kilograms and Q per kilogram for each additional kilogram. If the price paid for 33 kilograms of Apples is Rs.1167 and for 36 kilograms of apples if Rs.1284, then the cost of the first 10 kgs of apples is:

a. Rs.117

b. Rs.350

c. Rs.281

d. Rs.1053

Answer: b

Explanation:

Given that

30L + 3Q = 1167

30L + 6Q = 1284

Solving we get Q = 39, L = 35

So cost of first 10 kgs of apples = 35 × 10 = 350

A bag contains 1100 tickets numbered 1, 2, 3, … 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?

a. 291/1100

b. 292/1100

c. 290/1100

d. 301/1100

Answer: c

Explanation:

Numbers which dont have 2 from 1 to 9 = 8

Numbers which don’t have 2 from 10 to 99:

Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbers.

Numbers which don’t have 2 from 100 to 999 =_ _ _ = 8 × 9 × 9 = 648

Numbers which don’t have 2 from 1000 to 1099 =10_ _ = 9 × 9 = 81

Finally 1100 does not have 2. So 1.

Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810

Tickets with 2 in them = 1100 – 810 = 290

Required probability = 290 / 1100

Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth?

a)David

b)Querishi

c)Chitra

d)Thara

Sol: As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.

Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.

What is the 32nd word of “WAITING” in a dictionary?

Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W

Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways

so can’t be arranged starting with A alone as it is asking for 32nd word so it is out of range

AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.

AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24

AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways

so,24+12 =36th word so out of range. So we should not consider all the words start with AGN

now AGNI_ _ _can be arranged in 3! ways =6 ways

so 24+6=30 within range

Now only two word left so, arrange in alphabetical order.

AGNTIIW – 31st word

AGNTIWI – 32nd word

George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete?

a) 11.30 am

b) 12 noon

c) 12.30 pm

d) 1 pm

Let the total work = 120 units.

As George completes this entire work in 8 hours, his capacity is 15 units /hour

Similarly, the capacity of paul is 12 units / hour

the capacity of Hari is 10 units / hour

All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74

Remaining work = 120 – 74 = 46

Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx)

So work gets completed at 1 pm

3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs.144. What is the combined price of 1 apple, 1 peach, and 1 mango.

a) 37

b) 39

c) 35

d) 36

Sol: Note: It is 114 not 144.

3m + 4a = 85 ..(1)

5a + 6p = 122 ..(2)

6m + 2p = 114 ..(3)

(1) x 2 => 6m + 8a = 170

(3) => 6m + 2p = 114

Solving we get 8a – 2p = 56 ..(4)

(2) => 5a + 6p = 122

3 x (4) = 24a – 6p = 168

Solving we get a = 10, p = 12, m = 15

So a + p + m = 37

If you have any doubts regarding TCS Ninja Written Test, please comment here in the comment section and we will help you out.

In TCS Ninja it consists 4 section.. but I have doubt in coding section because I have did the code at last the time has been completed later it shown me that not attempted .. will they call for me interview or not