Question 1. H.C.F. Of 3240, 3600 And A Third Number Is 36 And Their L.C.M. Is 24x 35 X fifty two X 72 The Third Number Is?

Answer :

3240 = 23x 34x five; 3600 = 24x32 x 52; H.C.F. = 36 = 22x32

Since H.C.F. Is the manufactured from lowest powers of commonplace factors,

so the 0.33 quantity have to have (22x32) as its elements

Since L.C.M. Is the manufactured from highest powers of not unusual high factors,

so the 0.33 variety ought to have 35and 72 as its elements.

Third number = 22x35x72.

Question 2. The Largest Four ? Digit Number Which When Divided By four, 7 Or thirteen Leaves A Remainder Of three In Each Case, Is?

Answer :

Greatest wide variety of 4 digits is 9999, L.C.M. Of four, 7 and 13 = 364

On dividing 9999 by way of 364, the rest received is 171.

Greatest number of 4 digits divisible by way of 4, 7 and 13 = (9999 ? 171) = 9828

Hence, required quantity = (9828 + 3) = 9831.

Customer Relationship Management Interview Questions

Question 3. The Least Number, Which When Divided By 48, 60, seventy two, 108 And a hundred and forty Leaves 38, 50, sixty two, ninety eight And a hundred thirty As Remainders Respectively, Is?

Answer :

Here (48 - 38) = 10, (60 - 50) = 10, (seventy two - 62) = 10, (108 - 98) = 10 & (a hundred and forty - 130) = 10.

Required variety = (L.C.M. Of 48, 60, seventy two, 108, a hundred and forty) ? 10

= 15120 - 10

= 15110.

Question 4. The Product Of The L.C.M. And H.C.F. Of Two Numbers Is 24. The Difference Of Two Numbers Is 2. Find The Numbers?

Answer :

Let the numbers be x and (x + 2).

Then, x(x + 2) = 24

x2+ 2x - 24 = 0

(x - four) (x + 6) = 0

x = four

So, the numbers are 4 and 6.

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Question 5. Let The Least Number Of Six Digits, Which When Divided By 4, 6, 10 And 15, Leaves In Each Case The Same Remainder Of 2, Be N. The Sum Of The Digits In N Is?

Answer :

Least number of 6 digits is a hundred thousand.

L.C.M. Of four, 6, 10 and 15 = 60.

On dividing one hundred thousand by means of 60, the remainder received is forty.

Least variety of 6 digits divisible by 4, 6, 10 and 15 = a hundred thousand + (60 - forty) = 100020.

N = (100020 + 2) = 100022. Sum of digits in N = (1 + 2 + 2) = five.

Customer Care Interview Questions

Question 6. The Least Number Which Is A Perfect Square And Is Divisible By Each Of The Numbers 16, 20 And 24, Is?

Answer :

The least variety divisible by way of sixteen, 20, 24 = L.C.M. Of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x three x five

To make it a perfect square, it ought to be multiplied via three x 5

Required range = 240 x three x five = 3600.

Question 7. The Least Number Of Five Digits Which Is Exactly Divisible By 12, 15 And 18, Is?

Answer :

Least number of 5 digits is ten thousand. L.C.M. Of 12, 15 and 18 is one hundred eighty.

On dividing 10000 with the aid of a hundred and eighty, the the rest is a hundred.

Required range = ten thousand + (a hundred and eighty - 100) = 10080.

Customer Service Professional Interview Questions

Question eight. The H.C.F. And L.C.M. Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between seventy five And one hundred twenty five, Then That Number Is?

Answer :

Product of numbers = eleven x 385 = 4235.

Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35

Now, co-primes with product 35 are (1, 35) and (five, 7).

So, the numbers are (eleven x 1, 11 x 35) and (11 x 5, 11 x 7).

Since one wide variety lies between 75 and one hundred twenty five, the appropriate pair is (fifty five, 77)

Hence, required variety = seventy seven.

Question nine. The Least Number Which Should Be Added To 2497 So That The Sum Is Exactly Divisible By five, 6, four And 3 Is?

Answer :

L.C.M. Of five, 6, four and 3 = 60.

On dividing 2497 by means of 60, the remainder is 37.

Number to be delivered = (60 – 37) = 23.

HP Application Lifecycle Management Interview Questions

Question 10. Cost Price Of 22 Articles Is Same As The Selling Price Of 18 Articles, Find The Profit Percentage?

Answer :

Let the cost price of one article = Rs. 1

From the given statistics,

Then, the selling rate of one article = 22/18 = eleven/9

Then, Profit = SP - CP = 11/nine - 1 = 2/nine

Required, earnings % = Profit/CP x a hundred

= [(2/9)/1] x a hundred

= two hundred/9

= 22.222%.

Question eleven. Mr. Rajan Invested Rs 1,00,000 In Us Stock Markets When The Gbpinr Rate Was seventy five. After One Year His Investment Appreciated By 20% In Gbp Terms. He Sold Of His Investments And Repatriated The Money To India At The Then Existing Rate Of eighty. What Was Real Returns In Inr?

Answer :

Money invested through Rajan earlier than 1 yr turned into = Rs. 100000

Money in UK pounds @ seventy five is = a hundred thousand/75 = 1333.33 Pounds

Now, after 1 year invested quantity changed into preferred via 20%

=> 20% of 1333.33 = 266.66

Total funding turns into = 1333.33 + 266.66 = 1600 Pounds

This 1600 Pounds @ Indian forex at 80 = 1600 x 80 = Rs. 1,28,000

Hence, Rajan's investment of Rs. 1,00,000 will become Rs. 1,28,000 in 1 yr

Therefore, his earnings % = [(128000 - 100000)/100000] x 100 = 28%.

Desktop Support Interview Questions

Question 12. A Man Gains 20% By Selling An Article For A Certain Price. If He Sells It At Double The Price, The Percentage Of Profit Will Be?

Answer :

Let the C.P. = x,

Then S.P. = (a hundred and twenty/100)x = 6x/five

New S.P. = 2(6x/5) = 12x/five

Profit = 12x/5 - x = 7x/5

Profit% = (Profit/C.P.) * 100

=> (7x/five) * (1/x) * 100 = one hundred forty %.

Customer Relationship Management Interview Questions

Question 13. If The Cost Price Of 12 Items Is Equal To The Selling Price Of 16 Items, The Loss Percent Is?

Answer :

Let the Cost Price of one item = Re. 1

Cost Price of 16 gadgets = 16

Selling Price of 16 items = 12

Loss = 16 - 12 = Rs 4

Loss % = (4/16)* a hundred = 25%.

Question 14. A Producer Of Tea Blends Two Varieties Of Tea From Two Tea Gardens One Costing Rs 18 Per Kg And Another Rs 20 Per Kg In The Ratio five : 3. If He Sells The Blended Variety At Rs 21 Per Kg, Then His Gain Percent Is?

Answer :

Suppose he bought five kg and 3 kg of tea.

Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. A hundred and fifty.

Selling rate = Rs. (8 x 21) = Rs. 168.

Profit = 168 - 150 = 18

So, Profit % = (18/one hundred fifty) * 100 = 12%.

Question 15. A Pair Of Articles Was Bought For Rs. 37.Forty At A Discount Of 15%. What Must Be The Marked Price Of Each Of The Articles ?

Answer :

As question states that rate become of pair of articles,

So rate of One article = 37.40/2 = Rs. 18.70

Let Marked fee = Rs X

then eighty five% of X = 18.70

=> X = 1870/85 = 22.

Call centre Interview Questions

Question sixteen. Naveen Purchased A Cycle At Cost Of ?2000 And It Sold Ram At 20% Profit. Now Ram Spent ?100 For Its Repair And Sold To Jhon At 10% Loss. Find The Cost Price Of Jhon?

Answer :

The Cost Price of Naveen = 2000

Sell price of Naveen = 20% profit on 2000

= 2000 x one hundred twenty / 100 = 2400

Ram fee price = 2400 + 100 = 2500 (because Ram spent ? One hundred for its repair )

Sell fee of Ram = 10% loss on 2500

= 2500 x 90 / 100 = 2250

The Cost rate of Jhon =2250.

Question 17. The Cost Price Of The Two Articles Were Same And They Sold One At 25% Profit And Another Sold At 20% Loss. Now Find The Overall Profit/ Loss In This Transaction?

Answer :

One easy formula at the same time as at same cost rate

Net income / loss percentage = ( x + y ) / 2

Here to be take ” +ve” signal for earnings and “-ve ” sign for Loss & identical rule additionally relevant for final solution.

So x = 25 & y = -20

= ( 25 – 20) / 2 = + 2.5%

Profit = 2.Five%.

TCS Technical Interview Questions

Question 18. A Fruits Vender Cost Price Of 20 Apples Is Equal To Selling Price Of sixteen Apples, What Is The Profit/loss Percentage?

Answer :

Here the usage of the subsequent system

If cost charge of “x “items is identical as promoting price of “y’ objects, earnings / loss percentage

[ ( x – y ) 100 ] / y

Now x = 20 and y = 15

Profit / loss percentage = [ ( 20 – 16) 100 ] / sixteen = four hundred/sixteen = + 25% ( “+ve sig means income and -ve signal means loss)

Profit percentage = 25%.

Customer Care Interview Questions

Question 19. A Girl Bought A Book For Rs.450 And Sold It At 20% Profit. By Using That Amount She Bought Another Book And Sold It At five% Loss. Then Overall Profit Amount Is?

Answer :

Given

Cost Price of 1st ebook = Rs.450.

Profit % on promoting 1st e-book = 20%

Selling charge of 1st e-book = (a hundred + Profit%) / 100 * C.P

= (one hundred + 20) / one hundred * 450

= (one hundred twenty / 100) * 450

= 12 * forty five

= 540 Rs.

Cost fee of 2nd ebook = Rs.540

Loss % on promoting 2nd e-book = 5%

Selling price of 2nd book = (100 - Loss%) / a hundred * C.P

= (a hundred - five) / one hundred * 540

= (ninety five / a hundred) * 540

= 513 Rs.

Overall earnings = Selling fee of 2nd book - Cost Price of 1st e book

= Rs.513 - Rs.450

= Rs. 63.

Question 20. A Shopkeeper Buys Bananas At 15 For Rs 12 And Sells At 12 For Rs.15. Find His Gain Or Loss Percent?

Answer :

Given, the C.P of 15 bananas = Rs. 12

=> C.P of 1 banana = Rs. 12/15

=> C.P of 12 bananas = Rs. (12 /15)* 12

= Rs. A hundred and forty four/15

= Rs. Forty eight/ 5

Thus,C.P of 12 bananas = Rs. Forty eight/ 5

Given,S,P of 12 bananas = Rs. 15

Gain = (S.P - C.P)of 12 bananas

= 15- (48/five)

= (seventy five- forty eight) / 5

= 27/ 5

=>Gain = Rs. 27/ five

Percentage Gain = (Gain/C.P of 12 bananas) * 100%

= (27/5) / (48/5) * a hundred%

= (27/forty eight) * one hundred%

= fifty six.25%

Thus,Gain % = 56.25%.

Desktop Engineer Interview Questions

Question 21. Find The Cost Price Of An Article Which Is Sold At A Loss Of 25% For Rs. 480?

Answer :

We know that the item became bought at lack of 25% it means that best 75% of the value charge is paid.

Which method, 75% of C.P = 480

=> (seventy five/a hundred) x C.P = 480

=> C.P = 480 x 4/three

= a hundred and sixty x four

= 640 Rs

Cost rate of a piece of writing = Rs 640.

Question 22. Arjun Bought A T.V With 20% Discount On The Labelled Price. Had He Bought It With 25% Discount, He Would Have Saved Rs. 500. At What Price Did He Buy The T.V?

Answer :

Let labeled charge became one hundred.

Arjun sold T.V with 20% cut price, then

Cost rate of the T.V =classified price - 20% oflabeled rate

= one hundred - (20% of a hundred)

= eighty Rs.

If Arjunbought it at 25% bargain, then

Cost charge of the T.V =a hundred - 25% of 100

= seventy five Rs.

Given, If Arjun bought T.V with 25% cut price he might have stored Rs.500.

So, Difference among Cost fee for 20% and 25% discount = 500

=> eighty - 75 = 500

=> 5 parts = 500

=> 1 component = 500/5

=>1 element = 100

Then, the real Cost fee of the T.V =eighty components = 100 *eighty = 8000.

Correct Answer is Rs. 8000.

Question 23. In A Box, There Are eight Red, 7 Blue And 6 Green Balls. One Ball Is Picked Up Randomly. What Is The Probability That It Is Neither Red Nor Green?

Answer :

Total variety of balls = (eight + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green

= event that the ball drawn is blue.

N(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/three.

Wipro Aptitude Interview Questions

Question 24. A Bag Contains 2 Red, three Green And 2 Blue Balls. Two Balls Are Drawn At Random. What Is The Probability That None Of The Balls Drawn Is Blue?

Answer :

Total wide variety of balls = (2 + three + 2) = 7.

Let S be the sample area.

Then, n(S) = Number of approaches of drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

N(E) = Number of approaches of drawing 2 balls out of (2 + 3) balls =5C2= 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

Customer Service Professional Interview Questions

Question 25. Three Houses Are Available In A Locality. Three Persons Apply For The Houses. Each Applies For One House Without Consulting Others. The Probability That All The Three Apply For The Same House Is?

Answer :

One individual can choose one residence out of three= 3C1 methods =3.

Hence, 3 persons can pick one residence out of 3 in 3 x three x 3 =9.

Therefore, chance that all thre apply for the equal house is 1/nine.

Question 26. Two Brother X And Y Appeared For An Exam. The Probability Of Selection Of X Is 1/7 And That Of B Is 2/9. Find The Probability That Both Of Them Are Selected?

Answer :

Let A be the event that X is selected and B is the event that Y is selected.

P(A) = 1/7, P(B) = 2/9.

Let C be the event that both are decided on.

P(C) = P(A) × P(B) as A and B are independent events:

= (1/7) × (2/nine) = 2/sixty three.

IBM Aptitude Interview Questions

Question 27. In A Single Throw Of Two Dice , Find The Probability That Neither A Doublet Nor A Total Of 8 Will Appear?

Answer :

n(S) = 36

A = (1, 1), (2, 2), (three, three), (four, 4), (five, five), (6, 6)

B = (2, 6), (three, 5), (four, 4), (5, 3), (6, 2)

n(A)=6, n(B)=5, n(AnB)=1

Required possibility = P(A?B)

= P(A)+P(B)-P(AnB)

= 6/36+5/36-1/36 = 5/18.

HP Application Lifecycle Management Interview Questions

Question 28. An Unbiased Die Is Tossed.Find The Probability Of Getting A Multiple Of 3?

Answer :

Here S = 1,2,three,four,5,6

Let E be the event of getting the multiple of three

Then, E = 3,6

P(E) = n(E)/n(S) = 2/6 = 1/three.

Question 29. What Is The Probability Of Getting At Least One Six In A Single Throw Of Three Unbiased Dice?

Answer :

Find the number of instances in which not one of the digits display a '6'.

I.E. All 3 dice show quite a number apart from '6', five×5×5=125 instances.

Total feasible effects while 3 dice are thrown = 216.

The quantity of results wherein at the least one die indicates a '6' = Total possible effects whilst three dice are thrown - Number of outcomes in which none of them display '6'.

=216-one hundred twenty five=ninety one

The required chance = 91/256.

EXL Service Aptitude Interview Questions

Question 30. A Number X Is Chosen At Random From The Numbers -three, -2, -1, 0, 1, 2, three+2) take X=-1,0,1

=> P( (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

P(E)=n(E)n(S)=6/36=1/6.

Question 32. There Are Four Hotels In A Town. If 3 Men Check Into The Hotels In A Day Then What Is The Probability That Each Checks Into A Different Hotel?

Answer :

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64 = 3/8.

HCL Aptitude Interview Questions

Question 33. In How Many Different Ways Can The Letters Of The Word 'therapy' Be Arranged So That The Vowels Never Come Together?

Answer :

Given word is THERAPY.

Number of letters in the given word = 7

These 7 letters can be arranged in 7! Ways.

Number of vowels in the given word = 2 (E, A)

The number of ways of arrangement in which vowels come together is 6! X 2! Ways

Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! X 2!) ways = 5040 - 1440 = 3600 ways.

Desktop Support Interview Questions

Question 34. If (1 × 2 × 3 × 4 ........ × N) = N!, Then 15! - 14! - 13! Is Equal To ___?

Answer :

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! X 15 (14 - 1)

= 15 × 13 × 13!.

Question 35. In How Many Different Ways Can The Letters Of The Word 'abysmal' Be Arranged ?

Answer :

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! Ways

But the letter is repeated and this can be arranged in 2! Ways

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

Question 36. How Many More Words Can Be Formed By Using The Letters Of The Given Word 'creativity'?

Answer :

The number of letters in the given word CREATIVITY = 10

Here T & I letters are repeated

=> Number of Words that can be formed from CREATIVITY = 10!/2!X2! = 3628800/four = 907200.

Call centre Interview Questions

Question 37. To Fill eight Vacancies There Are 15 Candidates Of Which 5 Are From St. If three Of The Vacancies Are Reserved For St Candidates While The Rest Are Open To All, Find The Number Of Ways In Which The Selection Can Be Done ?

Answer :

ST candidates vacancies can be filled via 5C3 ways = 10

Remaining vacancies are 5 which can be to be stuffed by using 12

=>12C5= (12x11x10x9x8)/(5x4x3x2x1) = 792

Total variety of filling the vacancies = 10 x 792 = 7920.

Question 38. How Many Words Can Be Formed With Or Without Meaning By Using Three Letters Out Of K, L, M, N, O Without Repetition Of Alphabets?

Answer :

Given letters are okay, l, m, n, o = five

wide variety of letters to be inside the words = 3

Total number of phrases that may be fashioned from these 5 letters taken three at a time with out repetation of letters = 5P3 methods.

⇒ 5P3 = 5 x 4 x 3 = 60 phrases.

Question 39. In A Bag, There Are 8 Red, 7 Blue And 6 Green Flowers. One Of The Flower Is Picked Up Randomly. What Is The Probability That It Is Neither Red Nor Green?

Answer :

Total number of flora = (8+7+6) = 21.

Let E = occasion that the flower drawn is neither pink nor green.

= occasion taht the flower drawn is blue.

--> n(E)= 7

--> P(E)= 7/21=1/three.

Question 40. In How Many Ways Can Letter Of The Word Railings Arrange So That R And S Always Come Together?

Answer :

The variety of methods in which the letters of the phrase RAILINGS can be arranged such that R & S usually come collectively is

Count R & S as best 1 area or letter in order that RS or SR can be arranged => 7! X 2!

But in the phrase RAILINGS, I repeated for 2 times => 7! X 2!/2! = 7! Ways = 5040 methods.

TCS Technical Interview Questions

Question 41. In How Many Different Ways Could Couples Be Picked From 6 Men And 9 Women ?

Answer :

Number of mens = 8

Number of womens = 5

Different approaches could couples be picked = 6C1×9C1 = 9 x 6 = 54 ways.

Question forty two. In How Many Different Ways Can The Letters Of The Word 'happyholi' Be Arranged?

Answer :

The given word HAPPYHOLI has nine letters

These nine letters can e organized in nine! Ways.

But here in the given phrase letters H & P are repeated two times every

Therefore, Number of approaches those 9 letters can be organized is

nine!/2! X 2! = 9 x 8 x 7 x 6 x five x 4 x three/2 = ninety,720 approaches.

Desktop Engineer Interview Questions