Question 1. The H.C.F And L.C.M Of Two Numbers Are eleven And 385 Respectively. If One Number Lies Between 75 And a hundred twenty five , Then That Number Is?

Answer :

Product of numbers = 11 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 (5,7)

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

Since one variety lies seventy five and a hundred twenty five, the precise pair is (fifty five,77)

Hence , required range = 77.

Question 2. In How Many Ways Can The Letters Of The Word 'capital' Be Arranged In Such A Way That All The Vowels Always Come Together?

Answer :

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

five letters which can be organized in 5P5=five!

Vowels A,I = three!/2!

No.Of arrangements = five! X 3!/2!=360

Electrical and Electronics Engineering Interview Questions

Question three. L.C.M Of Two Prime Numbers X And Y (x>y) Is 161. The Value Of 3y-x Is?

Answer :

H. C. F of top numbers is 1.

Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with product 161 are (1, 161) and (7, 23).

Since x and y are top numbers and x >y ,

we've x=23 and y=7.

Therefore, 3y-x = (three x 7)-23 = -2.

Question four. The G.C.D Of 1.08, zero.36 And zero.Nine Is?

Answer :

Given numbers are 1.08 , 0.36 and 0.90

H.C.F of 108, 36 and ninety is eighteen [ ? G.C.D is nothing but H.C.F]

Therefore, H.C.F of given numbers = 0.18.

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Question 5. In A Group Of 6 Boys And 4 Girls, Four Children Are To Be Selected. In How Many Different Ways Can They Be Selected Such That At Least One Boy Should Be There ?

Answer :

We may also have (1 boy and three women)or(2boys and a couple of women)or(3 boys and 1 girl)or(four boys).

Required range of approaches = (6C1×4C3) + (6C2×4C2) + (6C3×4C1)+(6C4)

= (24+ninety+eighty+15)

= 209.

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Question 6. The Least Number Which When Divided By 5, 6 , 7 And 8 Leaves A Remainder three, But When Divided By 9 Leaves No Remainder, Is?

Answer :

L.C.M. Of 5, 6, 7, 8 = 840.

Required range is of the shape 840k + 3

Least price of okay for which (840k + three) is divisible through nine is okay = 2.

Required number = (840 x 2 + 3) = 1683.

Question 7. Which Of The Following Has The Most Number Of Divisors?

Answer :

99 = 1 x three x 3 x eleven

one hundred and one = 1 x one hundred and one

176 = 1 x 2 x 2 x 2 x 2 x eleven

182 = 1 x 2 x 7 x thirteen

So, divisors of ninety nine are 1, three, 9, eleven, 33, .Ninety nine

Divisors of a hundred and one are 1 and one zero one

Divisors of 176 are 1, 2, 4, 8, 11, sixteen, 22, forty four, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the maximum range of divisors.

Electrical Engineering Interview Questions

Question 8. Six Bells Commence Tolling Together And Toll At Intervals Of 2, 4, 6, 8 10 And 12 Seconds Respectively. In 30 Minutes, How Many Times Do They Toll Together?

Answer :

L.C.M. Of 2, four, 6, eight, 10, 12 is a hundred and twenty.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes,they may collectively (30/2)+1=16 times.

Question nine. The Difference Of Two Numbers Is 14. Their Lcm And Hcf Are 441 And 7. Find The Two Numbers ?

Answer :

Since their HCFs are 7, numbers are divisible with the aid of 7 and are of the shape 7x and 7y

Difference = 14

=> 7x - 7y = 14

=> x - y = 2

made of numbers = product of their hcf and lcm

=> 7x * 7y = 441 * 7

=> x * y = 63

Now, we've got

x * y = sixty three , x - y = 2

=> x = nine , y = 7

The numbers are 7x and 7y

=> 63 and forty nine.

Maintenance and Manufacturing Interview Questions

Question 10. In A Palace, Three Different Types Of Coins Are There Namely Gold, Silver And Bronze. The Number Of Gold, Silver And Bronze Coins Is 18000, 9600 And 3600 Respectively. Find The Minimum Number Of Rooms Required If In Each Room Should Give The Same Number Of Coins Of The Same Type ?

Answer :

Gold cash = 18000 , Silver cash = 9600 , Bronze cash = 3600

Find quite a number which precisely divide these kinds of numbers

That is HCF of 18000, 9600& 3600

All the price has 00 at stop so the factor may even have 00.

HCF for a hundred and eighty, ninety six & 36.

Factors of

a hundred and eighty = 3 x 3 x five x 2 x 2

ninety six = 2 x 2 x 2 x 2 x 2 x three

36 = 2 x 2 x three x 3

Common factors are 2x2×3=12

Therefore, Actual HCF is 1200

Gold Coins 18000/1200 will be in 15 rooms

Silver Coins 9600/1200 may be in eight rooms

Bronze Coins 3600/1200 might be in 3 rooms

Total rooms may be (15+eight+three) = 26 rooms.

Question eleven. The Distance Of The College And Home Of Rajeev Is 80km. One Day He Was Late By 1 Hour Than The Normal Time To Leave For The College, So He Increased His Speed By 4km/h And Thus He Reached To College At The Normal Time. What Is The Changed (or Increased) Speed Of Rajeev?

Answer :

Let the everyday velocity be x km/h, then

80/x-eighty/(x+four)=1

?X2+4x-320=zero

?X (x + 20) - sixteen (x + 20) = 0

(x + 20 ) (x - sixteen) =zero

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore expanded velocity = 20 km/h.

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Question 12. Akash Leaves Mumbai At 6 Am And Reaches Bangalore At 10 Am . Prakash Leaves Bangalore At 8 Am And Reaches Mumbai At eleven:30 Am. At What Time Do They Cross Each Other?

Answer :

Time taken with the aid of Akash = 4 h

Time taken through Prakash = three.Five h

For your convenience take the made of instances taken by using each as a distance.

Then the gap = 14km

Since, Akash covers half of the distance in 2 hours(i.E at eight am)

Now, the relaxation 1/2 (i.E 7 km) will be coverd by each prakash and akash

Time taken through them = 7/7.Five = fifty six min

Thus , they'll cross every different at 8 : 56am.

Electrical and Electronics Engineering Interview Questions

Question 13. Two Boys Starting From The Same Place Walk At A Rate Of 5kmph And 5.5kmph Respectively. What Time Will They Take To Be eight.5km Apart, If They Walk In The Same Direction?

Answer :

In this type of questions we need to get the relative speed between them,

The relative pace of the lads = five.5kmph – 5kmph

= zero.5 kmph

Distance between them is 8.Five km

Time = Distance/Speed

Time= eight.5km / zero.Five kmph = 17 hrs.

Question 14. A Man Reaches His Office 20 Min Late, If He Walks From His Home At three Km Per Hour And Reaches 30 Min Early If He Walks 4 Km Per Hour. How Far Is His Office From His House?

Answer :

Let distance = x km.

Time taken at 3 kmph : dist/pace = x/three = 20 min past due.

Time taken at four kmph : x/4 = 30 min earlier

difference between time taken : 30-(-20) = 50 mins = 50/60 hours.

X/three- x/4 = 50/60

x/12 = five/6

x = 10 km.

Question 15. An Employee May Claim Rs. 7.00 For Each Km When He Travels By Taxi And Rs. 6.00 For Each Km If He Drives His Own Car. If In One Week He Claimed Rs. 595 For Traveling ninety Km. How Many Kms Did He Travel By Taxi ?

Answer :

Let x and y be the respective km's travelled by way of guy via taxi and by means of his very own automobile.

Given x + y = ninety => x = 90 - y

But according to the query,

7x + 6y = 595

7(90-y) + 6y = 595

=> 630 - 7y + 6y = 595

=> y = 630 - 595 = 35

=> x = 90 - 35 = fifty five

Therefore, the distance travelled by means of taxi is 55 kms.

Electrician Assistant Interview Questions

Question sixteen. A Train Covers A Distance In 50 Minutes, If It Runs At A Speed Of 48kmph On An Average. Find The Speed At Which The Train Must Run To Reduce The Time Of Journey To forty Minutes?

Answer :

We are having time and speed given,

so first we are able to calculate the space.

Then we will get new velocity for given time and distance.

Lets remedy it.

Time = 50/60 hr = five/6 hr

Speed = 48 mph

Distance = S*T = 48 * 5/6 = 40 km

New time may be 40 minutes so,

Time = 40/60 hr = 2/three hr

Now we know,

Speed = Distance/Time

New pace = forty*three/2 kmph = 60kmph.

Question 17. A Man Whose Speed Is four.Five Kmph In Still Water Rows To A Certain Upstream Point And Back To The Starting Point In A River Which Flows At 1.Five Kmph, Find His Average Speed For The Total Journey ?

Answer :

Speed of Man = four.Five kmph

Speed of circulate = 1.Five kmph

Speed in DownStream = 6 kmph

Speed in UpStream = 3 kmph

Average Speed = (2 x 6 x three)/nine = 4 kmph.

Marketing Interview Questions

Question 18. A Person X Started At three Hours Earlier At 40km/h From A Place P, Then Another Person Y Followed Him At 60km/h. Started His Journey At 3 O'clock, Afternoon. What Is The Diference In Time When X Was30 Km Ahead Of Y And When Y Was 30 Km Ahead Of X?

Answer :

Time ( when X was 30 km beforehand of Y) = (120-30)/20 =four.5h

Time ( when Y turned into 30 km ahead of X) = (120+30)/20 = 7.5 h

Thus, required difference in time = 3h.

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Question 19. A Man In A Train Notices That He Can Count 41 Telephone Posts In One Minute. If They Are Known To Be 50 Metres Apart, Then At What Speed Is The Train Travelling?

Answer :

Number of gaps between forty one poles = forty

So total distance among 41 poles = forty*50

= 2000 meter = 2 km

In 1 minute train is moving 2 km/minute.

Speed in hour = 2*60 = 120 km/hour.

Question 20. A Man Covers A Distance Of 1200 Km In 70 Days Resting nine Hours A Day, If He Rests 10 Hours A Day And Walks With Speed 1½ Times Of The Previous In How Many Days Will He Cover 840 Km ?

Answer :

Distance d = 1200km

let S be the rate

he walks 15 hours an afternoon(i.E 24 - 9)

so absolutely he walks for 70 x 15 = 1050hrs.

S = 1200/1050 => 120/105 = 24/21 => eight/7kmph

given 1 half of of previous speed

so 3/2 * 8/7= 24/14 = 12/7

New velocity = 12/7kmph

Now he rests 10 hrs an afternoon meaning he walks 14 hrs an afternoon.

Time = 840 x 7 /12 => 490 hrs

=> 490/14 = 35 days

So he's going to take 35 days to cowl 840 km.

Personal Protective Equipment (PPE) Interview Questions

Question 21. A Man Covered A Certain Distance At Some Speed. Had He Moved 3 Kmph Faster, He Would Have Taken forty Minutes Less. If He Had Moved 2 Kmph Slower, He Would Have Taken forty Minutes More. The Distance (in Km) Is?

Answer :

Let distance = x km and typical charge = y kmph.

Then, x/y - x/(y+3) = 40/60 --> 2y (y+three) = 9x ----- (i)

Also, x/(y-2) - x/y = forty/60 --> y(y-2) = 3x -------- (ii)

On dividing (i) with the aid of (ii), we get:

x = forty km.

Question 22. A Person Takes 20 Minutes More To Cover A Certain Distance By Decreasing His Speed By 20%. What Is The Time Taken To Cover The Distance At His Original Speed ?

Answer :

Let the distance and unique pace be 'd' km and 'okay' kmph respectively.

D/zero.8k - d/ok = 20/60 => 5d/4k - d/okay = 1/three

=> (5d - 4d)/4k = 1/3 => d = four/three ok

Time taken to cover the gap at authentic speed

= d/okay = four/three hours = 1 hour 20 mins.

Question 23. A And B Runs Around A Circular Track. A Beats B By One Round Or 10 Minutes. In This Race, They Had Completed 4 Rounds. If The Race Was Only Of One Round, Find The A's Time Over The Course?

Answer :

B runs across the music in 10 min.

I.E ,Speed of B = 10 min according to spherical

Therefore, A beats B through 1 round

Time taken by A to complete 4 rounds

= Time taken via B to finish 3 rounds

= 30 min

Therefore, A's pace = 30/4 min in line with round = 7.Five min in step with round

Hence, if the race is best of one spherical A's time over the route = 7 min 30 sec.

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Question 24. A Man Walks At A Speed Of 2 Km/hr And Runs At A Speed Of 6 Km/hr. How Much Time Will The Man Require To Cover A Distance Of 20 1/2 Km, If He Completes Half Of The Distance, I.E., (10 1/4) Km On Foot And The Other Half By Running ?

Answer :

We recognise that

Time = Distance/velocity

Required time = (10 1/four)/2 + (10 1/four)/6

= 41/8 + 41/6

= 287/24 = 11.Nine hours.

Electrical Engineering Interview Questions

Question 25. Jagan Went To Another Town Covering 240 Km By Car Moving At 60 Kmph. Then He Covered 400km By Train Moving At one hundred Kmph And Then Rest 200 Km He Covered By A Bus Moving At 50 Kmph. The Average Speed During The Whole Journey Was ?

Answer :

By vehicle 240 km at 60 kmph

Time taken = 240/60 = 4 hr.

By teach 240 km at 60 kmph

Time taken = four hundred/100 = 4 hr.

By bus 240 km at 60 kmph

Time taken = 200/50 = 4 hr.

So total time = 4 + four + four = 12 hr.

And overall velocity = 240+four hundred+two hundred = 840 km

Average pace of the whole adventure = 840/12 = 70 kmph.

Question 26. Ramu Rides His Bike At An Average Speed Of 45 Km/hr And Reaches His Desitination In Four Hours. Somu Covers The Same Distance In Six Hours. If Ramu Covered His Journey At An Average Speed Which Was 9 Km/hr Less And Somu Covered His Journey At An Average Speed Which Was 10 Km/hr More, Then The Difference In Their Times Taken To Reach The Destination Would Be (in Minutes)?

Answer :

Distance travelled via Ramu = 45 x 4 = 180 km

Somu travelled the equal distance in 6 hours.

His pace = one hundred eighty/6 = 30 km/hr

Hence within the conditional case, Ramu's velocity = 45 - nine = 36 km/hr and Somu's velocity = 30 + 10 = 40km/hr.

Therefore tour time of Ramu and Somu would be 5 hours and 4.Five hours respectively.

Hence difference inside the time taken = zero.5 hours = 30 minutes.

BHEL Aptitude Interview Questions

Question 27. What Should Be The Value Of * In 985*865, If Number Is Divisible By 9?

Answer :

nine + 8 + 5 + * + 8 + 6 + five = 9x

41 + * = 9x

Nearest cost of 9x have to be forty five

forty one + * = forty five

* = four.

Maintenance and Manufacturing Interview Questions

Question 28. The Least Perfect Square, Which Is Divisible By Each Of 15, 20 And 36 Is?

Answer :

LCM of 15, 20 and 36 is one hundred eighty

Now 180 = three x 3 x 2 x 2 x five

To make it perfect square, it have to

be improved from five.

So required no. = 32 x 22 x 52 = 900.

Question 29. How Many Numbers Between 20 And 451 Are Divisible By nine?

Answer :

The required numbers are 27, 36, 45……450.

This is an A.P. With a = 27 and d = nine

Let it has n phrases.

Then Tn = 450 = 27 + (n-1) x9

450 = 27+ 9n - nine

9n = 432

n = 48.

Flipkart Aptitude Interview Questions

Question 30. The Largest four Digit Number Exactly Divisible By 5, 6 And 7 Is?

Answer :

The required number ought to be divisible by L.C.M. Of 5,6 and 7.

L.C.M. Of five, 6 and 7 = five x 6 x 7 = 210

Let us divide 9999 through 210.

210) 9999 (forty seven

840

----

1599

1470

----

129

Required range = 9999 – 129 = 9870.

Question 31. If A Positive Integer N Is Divided By 5, The Remainder Is three. Which Of The Numbers Below Yields A Remainder Of zero When It Is Divided By five?

Answer :

n divided through five yields a remainder equal to 3 is written as follows

n = 5 ok + three , where okay is an integer.

Upload 2 to both aspects of the above equation to achieve

n + 2 = 5 k + five = 5(ok + 1)

The above shows that n + 2 divided by 5 yields a remainder equal to 0.

Question 32. If An Integer N Is Divisible By 3, five And 12, What Is The Next Larger Integer Divisible By All These Numbers?

Answer :

If n is divisible by 3, 5 and 12 it need to a multiple of the lcm of three, five and 12 which is 60.

N = 60 okay

n + 60 is likewise divisible by using 60 for the reason that

n + 60 = 60 ok + 60 = 60(k + 1).

Question 33. When The Integer N Is Divided By 8, The Remainder Is three. What Is The Remainder If 6n Is Divided By 8?

Answer :

When n is split by 8, the the rest is three may be written as

n = eight ok + 3

multiply all terms with the aid of 6

6 n = 6(8 okay + three) = eight(6k) + 18

Write 18 as sixteen + 2 on the grounds that sixteen = 8 * 2.

= eight(6k) + sixteen + 2

Factor 8 out.

= eight(6k + 2) + 2

The above indicates that if 6n is divided by using eight, the the rest is two.

Power Electronics Interview Questions

Question 34. If N Is An Integer, When (2n + 2)^2 Is Divided By 4 The Remainder Is?

Answer :

We first amplify (2n + 2)2

(2n + 2)^2 = 4n^2 + 8n + four

Factor 4 out.

= 4(n^2 + 2n + 1)

(2n + 2)2 is divisible by means of 4 and the remainder is same to zero.

Question 35. What Is The Smallest Positive 2-digit Whole Number Divisible By three And Such That The Sum Of Its Digits Is 9?

Answer :

Let xy be the entire quantity with x and y the two digits that make up the quantity.

The quantity is divisible through three can be written as follows

10 x + y = three okay

The sum of x and y is same to 9.

X + y = nine

Solve the above equation for y

y = nine - x Substitute y = 9 - x within the equation 10 x + y = 3 k to attain.

10 x + 9 - x = three okay

Solve for x

x = (ok - 3) / 3

x is a advantageous integer smaller than 10

Let ok = 1, 2, three, ... And pick the first cost that offers x as an integer.

K = 6 offers x = 1

Find y the use of the equation y = nine - x = eight

The range we're searching out is 18.

It is divisible via three and the sum of its digits is same to nine and it's far the smallest and tremendous entire wide variety with such houses.

Question 36. Which Of These Numbers Is Not Divisible By 3?

Answer :

One may additionally answer this question the usage of a calculator and test for divisibility by way of three.

However we can also test for divisibilty by means of adding the digits and if the end result is

divisible by3 then the number is divisible by using three.

3 + 3 + nine = 15 , divisible through 3.

Three + 4 + 2 = 9 , divisible by using three.

Five + 5 + 2 = 12 , divisible through three.

1 + 1 + 1 + 1 = 4 , no longer divisible with the aid of 3.

Electrician Assistant Interview Questions

Question 37. The Sum Of Two Number Is 15 And Sum Of Their Square Is 113. The Numbers Are ?

Answer :

Let the variety be x and (15 - x)

Then, x2 + (15 - x)2= 113

x2- 15x + fifty six = 0

(x-7) (x-eight) = zero.

Question 38. The Number X Is Exactly Divisible By five And The Remainder Obtained On Dividing The Number Y By 5 Is 1. What Remainder Will Be Obtained When (x + Y) Is Divided By five ?

Answer :

Let x/five = p and allow y while divided by 5 offers q as quotient and 1 as remainder.

Then, y = 5q + 1

Now x = 5p and y = 5q + 1

x + y = 5p + 5q + 1 = 5(p + q) + 1.

Question 39. A Certain Number Of Men Can Finish A Piece Of Work In a hundred Days. If There Were 10 Men Less, It Would Take 10 Days More For The Work To Be Finished. How Many Men Were There Originally?

Answer :

Originally allow there be x men.

Less guys, More days(Indirect Proportion)

Therefore, (x-10) : x :: 100 :a hundred and ten

=> (x - 10) * one hundred ten = x * one hundred=> x= a hundred and ten.

Question 40. If 20 Men Can Build A Wall 56 Meters Long In 6 Days , What Length Of A Similar Wall Can Be Built By 35 Men In three Days?

Answer :

Let the desired length be x meters

More men, More period built (Direct percentage)

Less days, Less period constructed (Direct Proportion)

Men20:35

Days6 : 3?? Fifty six :x

=> (20 x 6 x X)=(35 x three x 56)

=> x = 49

Hence, the required period is forty nine m.

Marketing Interview Questions

Question forty one. 2 Men And 7 Boys Can Do A Piece Of Work In 14 Days; 3 Men And eight Boys Can Do The Same In 11 Days. Then, eight Men And 6 Boys Can Do Three Times The Amount Of This Work In?

Answer :

(2 x 14) men +(7 x 14) boys = (three x 11) guys + (8 x eleven) boys

=>5 guys= 10 boys => 1man= 2 boys

Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = eleven boys

( 8 guys + 6 boys) = (eight x 2 +6) boys = 22 boys.

Let the desired quantity of days be x.

More boys , Less days (Indirect percentage)

More paintings , More days (Direct share)

Boys22:11Work1 : 3?? 14:x

Therefore, (22 * 1 * x) = (11 * three * 14)

=> x = 21

Hence, the desired variety of days = 21.

Question 42. Find Compound Interest On Rs. 8000 At 15% Per Annum For 2 Years 4 Months, Compounded Annually?

Answer :

Time = 2 years 4 months = 2(4/12) years = 2(1/three) years.

Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]

=Rs. [8000 * (23/20) * (23/20) * (21/20)]

= Rs. 11109. .

:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Personal Protective Equipment (PPE) Interview Questions

Question 43. A Sum Of Money Amounts To Rs.6690 After 3 Years And To Rs.10,half After 6 Years On Compound Interest.Find The Sum?

Answer :

Let the sum be Rs.P.Then

P(1+R/100)^3=6690…(i) and P(1+R/a hundred)^6=10035…(ii)

On dividing,we get (1+R/a hundred)^3=10025/6690=3/2.

Substituting this cost in (i),we get:

P*(three/2)=6690 or P=(6690*2/3)=4460

Hence,the sum is rs.4460.

Question 44. The Compound Interest On Rs.30000 At 7% Per Annum Is Rs.4347. The Period Is?

Answer :

Amount = Rs.(30000+4347) = Rs.34347

let the time be n years

Then,30000(1+7/one hundred)^n = 34347

(107/100)^n = 34347/30000 = 11449/10000 = (107/a hundred)^2

n = 2years.

Question forty five. The Compound Interest On A Sum Of Money For 2 Years Is Rs.832 And The Simple Interest On The Same Sum For The Same Period Is Rs.800 .The Difference Between The Compound Interest And Simple Interest For three Years?

Answer :

distinction in C.I and S.I in 2years =Rs.32

S.I for 1year =Rs.Four hundred

S.I for Rs.400 for three hundred and sixty five days =Rs.32

fee=[100*32)/(400*1)%=eight%

difference between in C.I and S.I for 3rd year

=S.I on Rs.832= Rs.(832*eight*1)/a hundred=Rs.66.56

Question forty six. On A Sum Of Money, The Simple Interest For 2 Years Is Rs. 660,even as The Compound Interest Is Rs.696.30,the Rate Of Interest Being The Same In Both The Cases. The Rate Of Interest Is?

Answer :

Difference in C.I and S.I for 2 years

= Rs(696.30-660)

=Rs. 36.30.

S.I for one years = Rs330.

S.I on Rs.330 for 1 yr =Rs. 36.30

Rate

= (100x36.30/330x1)%

= eleven%.

Question forty seven. Find The Compound Interest On Rs. 16,000 At 20% Per Annum For 9 Months, Compounded Quarterly?

Answer :

Principal = Rs. 16000; Time = 9 months =3 quarters;

Rate = 20% according to annum = five% consistent with area.

Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.

CJ. = Rs. (18522 - 16000) = Rs. 2522.

Question forty eight. The Compound Interest On A Certain Sum For 2 Years At 10% Per Annum Is Rs. 525. The Simple Interest On The Same Sum For Double The Time At Half The Rate Percent Per Annum Is?

Answer :

Let the sum be Rs. P.

Then,[p(1+10/100)2-p]=525

Sum =Rs.2500

S.I.= Rs.(2500*five*4)/one hundred

= Rs. 500.

Question forty nine. Find Compound Interest On Rs. 7500 At four% Per Annum For 2 Years, Compounded Annually?

Answer :

Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25) ) = Rs. 8112.

Consequently, C.I. = Rs. (8112 - 7500) = Rs. 612.

Question 50. The Difference Between Simple And Compound Interest Compounded Annually On A Certain Sum Of Money For 2 Years At 4% Per Annum Is Re.1. The Sum Is?

Answer :

sum=Rs.X

C.I=[x(1+4/100)^2-x]=(676/625x-x)=51/625

S.I=(x*four*2)/100=2x/25

x=625.

Question 51. What Is The Difference Between The Compound Interests On Rs. 5000 For 1 1/2 Years At 4% Per Annum Compounded Yearly And Half-yearly?

Answer :

C.I. Whilst interest

compounded yearly=rs.[5000*(1+4/100)(1+1/2*4/100)]

= Rs. 5304.

C.I. While interest is

compounded half of-every year=rs.5000(1+2/a hundred)^three

= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04.

Question fifty two. A Sum Of Money Invested At Compound Interest Amounts To Rs. 800 In 3 Years And To Rs. 840 In four Years. The Rate Of Interest Per Annum Is?

Answer :

S.I. On Rs.800 for 1 yr

=Rs[840 - 800]

= Rs.Forty

Rate

=(100x40/800x1)%

= five%.