Question 1. 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.

Answer :

Since we recognize that the hobby fee is zero.15, and knowing that the distinction between two years of compound hobby is nothing however hobby on interest, we can find the first yr’s interest as –

45/0.15 = three hundred.

Now if the hobby is 300 at the quit of three hundred and sixty five days, then the main is three hundred / zero.15 = 2,000

Question 2. Look At This Series: eighty, 10, 70, 15, 60, ... What Number Should Come Next ?

Answer :

This is an alternating addition and subtraction collection. In the first pattern, 10 is subtracted from each range to reach at the following. In the second, five is added to every number to arrive at the subsequent

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Question 3. Pointing Out To A Lady, Rajan Said, “she Is The Daughter Of The Woman Who Is The Mother Of The Husband Of My Mother. “who Is The Lady To Rajan ?

Answer :

Mother’s husband – Father: Father’s mom – Grandmother: Grandmother’s daughter – Father’s sister: Father’s sister – Aunt.

So, the lady is Rajan’s Aunt.

Question four. May 6, 1993 Was Thursday. What Day Of The Week Was On May 6, 1992 ?

Answer :

1992 being a bounce 12 months, it has 2 extraordinary days.

So, the day on May, 1993 is two days beyond the day on May 6, 1992.

But, on May 6, 1993 it become Thursday.

So, on May 6, 1992 it turned into Tuesday.

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Question five. Read The Following Information Carefully And Answer The Questions That Follow : A + B Means A Is The Son Of B; A – B Means A Is The Wife Of B; A*b Means A Is The Brother Of B; A/b Means A Is The Mother Of B And A = B Means A Is The Sister Of B. What Does P + R –q Mean ?

Answer :

Clearly , P + R – Q means P is the son of R who's the spouse of Q i.E. Q is the father of P.

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Question 6. A Father Tells His Son, “i Was Of Your Present Age When You Were Born”. If The Father Is 36 Now, How Old Was The Boy five Years Back ?

Answer :

Let the father’s age be x and the son’s age be y. Then, x-y = y or x = 2y

Now, x = 36. So, 2y = 36 or y = 18.

Therefore son’s gift age = 18 years.

So, son’s age 5 years ago = 13 years.

Question 7. Sakshi Can Do A Piece Of Work In 20 Days. Tanya Is 25% More Efficient Than Sakshi. The Number Of Days Taken By Tanya To Do The Same Piece Of Work Is:

Answer :

Ratio of times taken by using Sakshi and Tanya = one hundred twenty five:a hundred = 5:4.

Suppose Tanya takes x days to do the work.

5 : four :: 20 : x ? X =(four x 20)/5

? X = 16 days

Hence, Tanya takes 16 days to complete the work.

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Question eight. A Train Left Station At A Hour B Minutes. It Reached Station Y At B Hour C Minutes On The Same Day, After Travelling C Hours A Minutes (clock Shows Time From zero Hours To 24 Hours). Number Of Possible Value Of A Is?

Answer :

A hours + C ours = B hours .....(i)

A, C and B can not have values more than or same to 24

B minutes + A mins = C minutes .....(ii)

Looking at two equation, we get no fee of A satisfies both equation

Question 9. A Merchant Marks His Goods Up By seventy five% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss?

Answer :

Let us count on that the fee charge of the thing = Rs.A hundred

Therefore, the merchant would have marked it to Rs.One hundred + seventy five% of Rs.100 = a hundred + seventy five = a hundred seventy five.

Now, if he sells it at no earnings or loss, he sells it at the cost rate.I.E. He offers a reduction of Rs.75 on his promoting price of Rs.One hundred seventy five

Therefore, his % cut price = (seventy five/a hundred seventy five) x a hundred= forty two.85%

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Question 10. Two Students Appeared At An Examination. One Of Them Secured 9 Marks More Than The Other And His Marks Was 56% Of The Sum Of Their Marks. The Marks Obtained By Them Are:

Answer :

Let their marks be (x + nine) and x.

Then, x + 9 = (fifty six/a hundred)(x + nine + x)

25(x + nine) = 14(2x + 9)

3x = ninety nine

x = 33

So, their marks are forty two and 33.

Question eleven. Two Pipes Can Fill The Cistern In 10hr And 12 Hr Respectively, While The Third Empty It In 20hr. If All Pipes Are Opened Simultaneously, Then The Cistern Will Be Filled In

Answer :

Work finished by means of all of the tanks working collectively in 1 hour.

=> 1/10 + 1/12 - 1/20 = 2/15

Hence, tank will be crammed in 15/2 = 7.Five hour

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Question 12. A Boat Can Travel With A Speed Of 13 Km / Hr In Still Water. If The Speed Of The Stream Is four Km / Hr. Find The Time Taken By The Boat To Go 68 Km Downstream?

Answer :

Speed Downstream= (13 + four) km/hr

= 17 km/hr.

Time taken to journey 68 km downstream =(68 / 17)hrs

= four hrs.

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Question thirteen. There Are 6 Consecutive Odd Numbers. The Difference Between The Square Of The Average Of The First Three Numbers And The Square Of The Average Of The Last Three Numbers Is 288. What Is The Last Odd Number ?

Answer :

Let the 6 consecutive extraordinary no.’s are:

X, X+2, X+4, X+6, X+eight, X+10

Avg. Of 1st 3 no’s is X+2.

Avg. Of Last 3 no’s is X+8.

Given that (X+8)2-(X+2)2=288

X=19

Last Odd no. Is X+10= 29.

Question 14. A Sum Of Money At Simple Interest Amounts To Rs. 815 In 3 Years And To Rs. 854 In 4 Years. The Sum Is:

Answer :

S.I. For 1 yr = Rs. (854 - 815) = Rs. 39.

S.I. For three years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

Question 15. A Merchant Marks His Goods Up By seventy five% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss ?

Answer :

Let us assume that the fee charge of the article = Rs.One hundred

Therefore,

the service provider might have marked it to Rs.A hundred + seventy five% of Rs.A hundred = one hundred +seventy five = 175.

Now, if he sells it at no profit or loss, he sells it on the price price.I.E., he offers a

bargain of Rs.Seventy five on his selling charge of Rs.A hundred seventy five.

Therefore, his % bargain = (seventy five/175) x a hundred= 42.85%

ABB Group Aptitude Interview Questions

Question sixteen. 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

Answer :

Let the boat pace in nevertheless water be b.

Let the circulate speed be x.

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

5x=b

b/x=5/1

Question 17. How Many Digits Will Be There To The Right Of The Decimal Point In The Product Of 95.75 And .02554?

Answer :

Sum of decimal places = 7.

Since the remaining digit to the extreme proper might be zero (seeing that 5 x 4 = 20), so there will

be 6 giant digits to the right of the decimal factor

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Question 18. Look At This Series: a thousand, 200, forty, …. What Number Should Come Next?

Answer :

eight is the wide variety that comes next.

This is a simple division series. Each quantity is split by way of five.

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Question 19. In A a hundred M Race, A Can Beat B By 25 M And B Can Beat C By four M. In The Same Race, A Can Beat C By ?

Answer :

we can do a by using percentage method

A can beat b by way of

100*75/one hundred=75

25 km

75*ninety six/100=seventy two

one hundred-72=28

Question 20. A Takes three Min 45 Seconds To Complete A Kilometre. B Takes 4 Minutes To Complete The Same 1 Km Track. If A And B Were To Participate In A Race Of 2 Kms, How Much Start Can A Give B In Terms Of Distance ?

Answer :

A can provide B a begin of 15 seconds in a km race.

B takes 4 mins to run a km. I.E a thousand/4= 250 m/min = 250/60 m/sec

Therefore, B will cover a distance of = sixty two.5 meters in 15 seconds.

The start that A can supply B in a km race consequently, is 62.Five meters, the distance that B run in 15 seconds.

Hence in a 2 km race, A can supply B a start of 62.Five * 2 = one hundred twenty five m or 30 seconds.

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Question 21. In A Game Of eighty Points. A Can Give B 5 Points And C 15 Points. Then How Many Points B Can Give C In A Game Of 60 ?

Answer :

A can supply B five factors A / B = 80/75

A can supply C 15 points A / C = eighty/ 65

B can deliver C a points of ( B / A ) * ( A / C)

(75 / eighty) * ( 80 / 65)

(seventy five / 65) = 15/thirteen

Need to be discover in a game of 60 , so multiply numerator and denominator with four such that the ratio turns into 60/52

Therefore B can deliver C of (60-fifty two) = 8 points

Question 22. A Can Run three Kms In three Min 18 Sec And B Can Run Same Distance In 3 Min 40 Sec, Then By How Much Distance A Can Beat B ?

Answer :

Given, 3km or 3000m Difference is 22 2d

three min 40sec = 220 sec

so, 3000*22/220 = 300m

Question 23. In A 200 M Race A Beats B By 35 M Or 7 Seconds. A's Time Over The Cause Is ?

Answer :

B covers 35m in 7 seconds B take time = (2 hundred*7)/35=40

A takes time = (40-7)= 33 Sec.

HCL Aptitude Interview Questions

Question 24. A Can Run 1.5 Km Distance In 2 Min 20 Seconds, While B Can Run This Distance In 2 Min 30 Sec. By How Much Distance Can A Beat B ?

Answer :

A takes time 2.20 min. = a hundred and forty Sec. B takes time 2.30 min. = 150 Sec.

Diffrence = (150-140) = 10 Sec.

Now we're to locate distance blanketed in 10 sec via B

a hundred and fifty Sec. = 1500 m.

1 Sec=10 m.

10 Sec = 10*10=a hundred m

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Question 25. In A Race Of 4 Kms A Beats B By a hundred M Or 25 Seconds, Then Time Taken By A Is

Answer :

B covers 100m in 25 seconds B take time =(4000*25)/one hundred=one thousand sec=sixteen min forty sec.

A takes time =a thousand sec-25sec=975 sec= 16 min 25 sec.

Question 26. In A two hundred M Race, A Beats B By 20 M And C By 38m. In A Race Of three hundred M B Will Beat C By

Answer :

If a runs 200m then b runs two hundred-20=180m

then c runs 200-38=162m

if b runs 300m then c runs 162/a hundred and eighty*300=270

consequently b beats c with the aid of 300-270=30

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Question 27. A Team Of Workers Was Employed By A Contractor Who Undertook To Finish 360 Pieces Of An Article In A Certain Number Of Days. Making Four More Pieces Per Day Than Was Planned, They Could Complete The Job A Day Ahead Of Schedule. How Many Days Did They Take To Complete The Job ?

Answer :

Days taken in the preferred state of affairs = 360/N;

Days taken when 4 articles are prepared more consistent with day = 360/N + four;

The distinction inside the day is one, therefore;

360/n - 360/n+four =1

N2 + 4N – 1440 = 0;

N = 36,

i.E. Number of object organized in trendy situation is 36, and wherein 4 articles organized greater is forty.

Therefore no of days taken to finish the process = 360/forty = 9.

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Question 28. Three Runners A, B And C Run A Race, With Runner A Finishing 12 Meters Ahead Of Runner B And 18 Meters Ahead Of Runner C, While Runner B Finishes eight Meters Ahead Of Runner C. Each Runner Travels The Entire Distance At A Constant Speed. What Was The Length Of The Race?

Answer :

Lets expect distance of race = x mtrs.

Then whilst A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.

=> at this factor B is 6 m beforehand of C. Now to finish race b needs to run every other 12 m,

=> he runs any other 12 m. Whilst B finishes race he is eight m ahead of C.

So remaining 12 m B has run, C has run 10 m.As speeds are steady,

=> x-12/ x-18 = 12/10 => x = forty eight mtrs.

Question 29. Anusha, Banu And Esha Run A Running Race Of one hundred Meters. Anusha Is The Fastest Followed By Banu And Then Esha. Anusha, Banu And Esha Maintain Constant Speeds During The Entire Race. When Anusha Reached The Goal Post, Banu Was 10m Behind. When Banu Reached The Goal Post Esha Was 10m Behind. How Far Was Behind Anusha When The Latter Reached The Goal Post.

Answer :

By that point Anusha included 100m, Bhanu covered 90m.

So ratio in their speeds = 10 : 9 By that time Bhanu reached 100m, Esha protected 90m.

So ratio of their speeds = 10 : 9 Ratio of the rate of all the 3 = 100 : ninety : eighty one

By that time Anusha protected 100m, Esha Covers handiest 81.

Question 30. Alvin, Ben And Clinton Run A Race, With Alvin Finishing forty eight Meters Ahead Of Ben And seventy two Meters Ahead Of Clinton, While Runner Ben Finishes 32 Meters Ahead Of Runner Clinton. Each Runner Travels The Entire Distance At Constant Speed. What Is The Length Of The Race ?

Answer :

Let us expect the speeds of alvin,ben,clinton = A,B,C resp.

And the duration of race = d

let in time t alvin finishes the race

A*t=d

B*t=d-forty eight ------- (1)

C*t=d-72 --------(2)

So,B:C=(d-forty eight):(d-seventy two) ------- (three)

Now Ben covers forty eight mt more to finish the race.

In the same time Clinton covers 40mt (as it's far given that he is crushed through 32mt via ben)

So,B:C=forty eight:40....(4) (speed proportional to distance protected if time is same)

By, evaluating equation (three) and (four)

d=192 mt