## 3i Infotech Interview Questions

### Twice The Speed Of A Boat Downstream Is Equal To Thrice The Speed Upstream. The Ratio Of Its Speed In Still Water To The Speed Of Current Is

**Let the boat pace in nonetheless water be b.**

Let the stream pace be x.

2(b+ x) = 3(b-x)

5x=b

b/x=5/1

### A Person Has A Chemical Of Rs. 25 Per Litre. In What Ratio Should Water Be Mixed With That Chemical So That After Selling The Mixture At Rs. 20/litre He May Get A Profit Of 25% ?

This may be solved utilizing alligation.

What is required on the finish of blending is a worth of 20/1.25 = 16.

So the alligation would appear like this:

Water/0 Mixture/16 Chemical/25

Hence the ratio could be (25 – 16) : 16 = 9 : 16

Hence required ratio of Water : Chemical is 9:16.

### The Difference Between The Simple Interest And Compound Interest On A Certain Sum Of Money For 2 Years At 15% P. A. Is Rs. 45. Find The Sum ?

Since we all know that the rate of interest is 0.15, and understanding that the distinction between two years of compound curiosity is nothing however curiosity on curiosity, we will discover the primary 12 months’s curiosity as –

45/0.15 = 300.

Now if the curiosity is 300 on the finish of 1 12 months, then the principal is 300 / 0.15 = 2,000

### How Many Terms Are There In An A.p. Whose First And Fifth Terms Are -14 And 2, Respectively, And The Sum Of Terms Is 40 ?

Now the frequent distinction of this AP is 16/4 = 4.

The sum of an AP is n/2 {2a + (n – 1)d}

Substituting we get, 40 = n/2 {2×-14 + (n – 1)4}

The finest method to clear up that is by plugging choices. Put in n = 10 and get the RHS as 40.

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### In A Class, 50 Students Play Cricket, 20 Students Play Football And 10 Play Both Cricket And Football. How Many Play At Least One Of These Two Games ?

The required reply is 50 + 20 – 10 = 60.

### A Bottle Is Full Of Dettol. One-third Of It Is Taken Out And Then An Equal Amount Of Water Is Poured Into The Bottle To Fill It. This Operation Is Done Four Times. Find The Final Ratio Of Dettol And Water In The Bottle ?

As in denominator we have now to take 1/Three 4 occasions so, we begin by assuming 81 ml of dettol within the bottle. After the primary iteration you may be left with

2/3 × 81 = 54 ml. After the second iteration you may be left with

2/3 × 54 = 36 ml. After the third iteration you may be left with

2/3 × 36 = 24 ml. After the fourth iteration you may be left with

2/3 × 24 = 16 ml. So the required ratio will probably be 16 : (81 – 16) = 16 : 65

### In A Survey Of Defaulted Payments Of Electrical Bills Of A Residential Complex Of 125 Houses, It Is Found That 50 Houses Defaulted On Their Payment Of Electrical Bills In January, 60 In February And 40 In March. Houses Can Default In Consecutive Months Only. 20 Defaulted In January And February. 10 Defaulted In February And March. How Many Houses Defaulted In All The 3 Months?

We use formulation for intersection of three units, preserving in thoughts that Jan ∩ Mar doesn’t exist, since they don’t seem to be consecutive months.

Let x be the variety of folks defaulting in all Three months.

We get the equation as : 125 = 50 + 60 + 40 – 20 – 10 + x. Solving we get x = 5.

### India Plays Two Matches Each With West Indies And Australia. In Any Match The Probabilities Of India Getting Points 0, 1, 2 Are 0.45, 0.05 And 0.50 Respectively. Assuming That Outcomes Are Independent, The Probability Of India Getting At Least 7 Points Is

Getting 7 factors is feasible in 2 instances.

**Case 1:** India wins all Four matches.

Probability: (.5)4 = .0625.

**Case 2:** India wins any of the three matches and attracts the remaining match. This can occur in complete Four methods. Probability: Four x (.50)Three x (.05) = .025.

So, required chance: .0625 + .025 = .0875

### In Order To Obtain An Income Of Rs. 650 From 10% Stock At Rs. 96, One Must Make An Investment Of

To acquire Rs. 10, funding = Rs. 96.

To acquire Rs. 650, funding = Rs.96/10x 650 = Rs. 6240.

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### Express A Speed Of 36 Kmph In Meters Per Second ?

36 * 5/18 = 10 mps

### Express 25 Mps In Kmph?

25 * 18/5 = 90 kmph

### A Boy Has Nine Trousers And 12 Shirts. In How Many Different Ways Can He Select A Trouser And A Shirt?

- The boy can choose one trouser in 9 methods.
- The boy can choose one shirt in 12 methods.
- The variety of methods wherein he can choose one trouser and one shirt is 9 * 12 = 108 methods.

### How Many Three Letter Words Are Formed Using The Letters Of The Word Time ?

The variety of letters within the given phrase is 4.

The variety of three letter phrases that may be shaped utilizing these 4 letters is ⁴P₃ = 4 * 3 * 2 = 24.

### Using All The Letters Of The Word “thursday”, How Many Different Words Can Be Formed ?

Total variety of letters = 8

Using these letters the variety of Eight letters phrases shaped is ⁸P₈ = 8!.

### Using All The Letters Of The Word “nokia”, How Many Words Can Be Formed, Which Begin With N And End With A ?

**There are 5 letters within the given phrase.**

Consider 5 blanks .

The first clean and final clean should be stuffed with N and A all of the remaining three blanks may be stuffed with the remaining Three letters in 3! methods.

The variety of phrases = 3! = 6.

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### The Number Of Arrangements That Can Be Made With The Letters Of The Word Meadows So That The Vowels Occupy The Even Places ?

The phrase MEADOWS has 7 letters of which Three are vowels.

-V-V-V-

As the vowels must occupy even locations, they are often organized within the Three even locations in 3! i.e., 6 methods. While the consonants may be organized amongst themselves within the remaining Four locations in 4! i.e., 24 methods.

Hence the whole methods are 24 * 6 = 144.

### The Number Of Permutations Of The Letters Of The Word ‘mesmerise’ Is_.

n objects of which p are alike of 1 variety, q alike of the opposite, r alike of one other variety and the remaining are distinct may be organized in a row in n!/p!q!r! methods.

The letter sample ‘MESMERISE’ consists of 10 letters of which there are 2M’s, 3E’s, 2S’s and 1I and 1R.

Number of preparations = 9!/(2!)2 3!

Question 18. A Committee Has 5 Men And 6 Women. What Are The Number Of Ways Of Selecting 2 Men And 3 Women From The Given Committee ?

The variety of methods to pick two males and three ladies = ⁵C₂ * ⁶C₃

= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)

= 200