**Question 1. The H.c.f And L.c.m Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between 75 And 125 , 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 ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 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)

5 letters which can be arranged in 5P

_{5}=5!Vowels A,I = 3!/2!

No.of arrangements = 5! x 3!/2!=360

**Question 3. 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 two prime 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 prime numbers and x >y ,

we have x=23 and y=7.

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

**Question 4. The G.c.d Of 1.08, 0.36 And 0.9 Is?****Answer :**Given numbers are 1.08 , 0.36 and 0.90

H.C.F of 108, 36 and 90 is 18 [ ? G.C.D is nothing but H.C.F]

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

**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 have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).

Required number of ways = (6

_{C1}×4_{C3}) + (6_{C2}×4_{C2}) + (6_{C3}×4_{C1})+(6_{C4})= (24+90+80+15)

= 209.

**Question 6. The Least Number Which When Divided By 5, 6 , 7 And 8 Leaves A Remainder 3, But When Divided By 9 Leaves No Remainder, Is?****Answer :**L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

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

**Question 7. Which Of The Following Has The Most Number Of Divisors?****Answer :**99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

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

Hence, 176 has the most number of divisors.

**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, 4, 6, 8, 10, 12 is 120.

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

In 30 minutes,they will together (30/2)+1=16 times.

**Question 9. 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 by 7 and are of the form 7x and 7y

Difference = 14

=> 7x – 7y = 14

=> x – y = 2

product of numbers = product of their hcf and lcm

=> 7x * 7y = 441 * 7

=> x * y = 63

Now, we have

x * y = 63 , x – y = 2

=> x = 9 , y = 7

The numbers are 7x and 7y

=> 63 and 49.

**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 coins = 18000 , Silver coins = 9600 , Bronze coins = 3600

Find a number which exactly divide all these numbers

That is HCF of 18000, 9600& 3600

All the value has 00 at end so the factor will also have 00.

HCF for 180, 96 & 36.

Factors of

180 = 3 x 3 x 5 x 2 x 2

96 = 2 x 2 x 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

Common factors are 2×2×3=12

Therefore, Actual HCF is 1200

Gold Coins 18000/1200 will be in 15 rooms

Silver Coins 9600/1200 will be in 8 rooms

Bronze Coins 3600/1200 will be in 3 rooms

Total rooms will be (15+8+3) = 26 rooms.

**Question 11. 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 normal speed be x km/h, then

80/x-80/(x+4)=1

?x

^{2}+4x-320=0?x (x + 20) – 16 (x + 20) = 0

(x + 20 ) (x – 16) =0

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore increased speed = 20 km/h.

**Question 12. Akash Leaves Mumbai At 6 Am And Reaches Bangalore At 10 Am . Prakash Leaves Bangalore At 8 Am And Reaches Mumbai At 11:30 Am. At What Time Do They Cross Each Other?****Answer :**Time taken by Akash = 4 h

Time taken by Prakash = 3.5 h

For your convenience take the product of times taken by both as a distance.

Then the distance = 14km

Since, Akash covers half of the distance in 2 hours(i.e at 8 am)

Now, the rest half (i.e 7 km) will be coverd by both prakash and akash

Time taken by them = 7/7.5 = 56 min

Thus , they will cross each other at 8 : 56am.

**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 8.5km 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 speed of the boys = 5.5kmph – 5kmph

= 0.5 kmph

Distance between them is 8.5 km

Time = Distance/Speed

Time= 8.5km / 0.5 kmph = 17 hrs.

**Question 14. A Man Reaches His Office 20 Min Late, If He Walks From His Home At 3 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/speed = x/3 = 20 min late.

time taken at 4 kmph : x/4 = 30 min earlier

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

x/3- x/4 = 50/60

x/12 = 5/6

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 90 Km. How Many Kms Did He Travel By Taxi ?****Answer :**Let x and y be the respective km’s travelled by man via taxi and by his own car.

Given x + y = 90 => x = 90 – y

But according to the question,

7x + 6y = 595

7(90-y) + 6y = 595

=> 630 – 7y + 6y = 595

=> y = 630 – 595 = 35

=> x = 90 – 35 = 55

Therefore, the distance travelled by taxi is 55 kms.

**Question 16. 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 40 Minutes?****Answer :**We are having time and speed given,

so first we will calculate the distance.

Then we can get new speed for given time and distance.

Lets solve it.

Time = 50/60 hr = 5/6 hr

Speed = 48 mph

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

New time will be 40 minutes so,

Time = 40/60 hr = 2/3 hr

Now we know,

Speed = Distance/Time

New speed = 40*3/2 kmph = 60kmph.

**Question 17. A Man Whose Speed Is 4.5 Kmph In Still Water Rows To A Certain Upstream Point And Back To The Starting Point In A River Which Flows At 1.5 Kmph, Find His Average Speed For The Total Journey ?****Answer :**Speed of Man = 4.5 kmph

Speed of stream = 1.5 kmph

Speed in DownStream = 6 kmph

Speed in UpStream = 3 kmph

Average Speed = (2 x 6 x 3)/9 = 4 kmph.

**Question 18. A Person X Started At 3 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 ahead of Y) = (120-30)/20 =4.5h

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

Thus, required difference in time = 3h.

**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 41 poles = 40

So total distance between 41 poles = 40*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 9 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 speed

he walks 15 hours a day(i.e 24 – 9)

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

S = 1200/1050 => 120/105 = 24/21 => 8/7kmph

given 1 1/2 of previous speed

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

New speed = 12/7kmph

Now he rests 10 hrs a day that means he walks 14 hrs a day.

time = 840 x 7 /12 => 490 hrs

=> 490/14 = 35 days

So he will take 35 days to cover 840 km.

**Question 21. A Man Covered A Certain Distance At Some Speed. Had He Moved 3 Kmph Faster, He Would Have Taken 40 Minutes Less. If He Had Moved 2 Kmph Slower, He Would Have Taken 40 Minutes More. The Distance (in Km) Is?****Answer :**Let distance = x km and usual rate = y kmph.

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

Also, x/(y-2) – x/y = 40/60 –> y(y-2) = 3x ——– (ii)

On dividing (i) by (ii), we get:

x = 40 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 original speed be ‘d’ km and ‘k’ kmph respectively.

d/0.8k – d/k = 20/60 => 5d/4k – d/k = 1/3

=> (5d – 4d)/4k = 1/3 => d = 4/3 k

Time taken to cover the distance at original speed

= d/k = 4/3 hours = 1 hour 20 minutes.

**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 around the track in 10 min.

i.e ,Speed of B = 10 min per round

Therefore, A beats B by 1 round

Time taken by A to complete 4 rounds

= Time taken by B to complete 3 rounds

= 30 min

Therefore, A’s speed = 30/4 min per round = 7.5 min per round

Hence, if the race is only of one round A’s time over the course = 7 min 30 sec.

**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 know that

Time = Distance/speed

Required time = (10 1/4)/2 + (10 1/4)/6

= 41/8 + 41/6

= 287/24 = 11.9 hours.

**Question 25. Jagan Went To Another Town Covering 240 Km By Car Moving At 60 Kmph. Then He Covered 400km By Train Moving At 100 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 car 240 km at 60 kmph

Time taken = 240/60 = 4 hr.

By train 240 km at 60 kmph

Time taken = 400/100 = 4 hr.

By bus 240 km at 60 kmph

Time taken = 200/50 = 4 hr.

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

and total speed = 240+400+200 = 840 km

Average speed of the whole journey = 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 by Ramu = 45 x 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu’s speed = 45 – 9 = 36 km/hr and Somu’s speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.

Hence difference in the time taken = 0.5 hours = 30 minutes.

**Question 27. What Should Be The Value Of * In 985*865, If Number Is Divisible By 9?****Answer :**9 + 8 + 5 + * + 8 + 6 + 5 = 9x

41 + * = 9x

Nearest value of 9x must be 45

41 + * = 45

* = 4.

**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 180

Now 180 = 3 x 3 x 2 x 2 x 5

To make it perfect square, it must

be multiplied from 5.

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

**Question 29. How Many Numbers Between 20 And 451 Are Divisible By 9?****Answer :**The required numbers are 27, 36, 45……450.

This is an A.P. with a = 27 and d = 9

Let it has n terms.

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

450 = 27+ 9n – 9

9n = 432

n = 48.

**Question 30. The Largest 4 Digit Number Exactly Divisible By 5, 6 And 7 Is?****Answer :**The required number must be divisible by L.C.M. of 5,6 and 7.

L.C.M. of 5, 6 and 7 = 5 x 6 x 7 = 210

Let us divide 9999 by 210.

210) 9999 (47

840

—-

1599

1470

—-

129

Required number = 9999 – 129 = 9870.

**Question 31. If A Positive Integer N Is Divided By 5, The Remainder Is 3. Which Of The Numbers Below Yields A Remainder Of 0 When It Is Divided By 5?****Answer :**n divided by 5 yields a remainder equal to 3 is written as follows

n = 5 k + 3 , where k is an integer.

add 2 to both sides of the above equation to obtain

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

The above suggests that n + 2 divided by 5 yields a remainder equal to zero.

**Question 32. If An Integer N Is Divisible By 3, 5 And 12, What Is The Next Larger Integer Divisible By All These Numbers?****Answer :**If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and 12 which is 60.

n = 60 k

n + 60 is also divisible by 60 since

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

**Question 33. When The Integer N Is Divided By 8, The Remainder Is 3. What Is The Remainder If 6n Is Divided By 8?****Answer :**When n is divided by 8, the remainder is 3 may be written as

n = 8 k + 3

multiply all terms by 6

6 n = 6(8 k + 3) = 8(6k) + 18

Write 18 as 16 + 2 since 16 = 8 * 2.

= 8(6k) + 16 + 2

Factor 8 out.

= 8(6k + 2) + 2

The above indicates that if 6n is divided by 8, the remainder is 2.

**Question 34. If N Is An Integer, When (2n + 2)^2 Is Divided By 4 The Remainder Is?****Answer :**We first expand (2n + 2)2

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

Factor 4 out.

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

(2n + 2)2 is divisible by 4 and the remainder is equal to 0.

**Question 35. What Is The Smallest Positive 2-digit Whole Number Divisible By 3 And Such That The Sum Of Its Digits Is 9?****Answer :**Let xy be the whole number with x and y the two digits that make up the number.

The number is divisible by 3 may be written as follows

10 x + y = 3 k

The sum of x and y is equal to 9.

x + y = 9

Solve the above equation for y

y = 9 – x Substitute y = 9 – x in the equation 10 x + y = 3 k to obtain.

10 x + 9 – x = 3 k

Solve for x

x = (k – 3) / 3

x is a positive integer smaller than 10

Let k = 1, 2, 3, … and select the first value that gives x as an integer.

k = 6 gives x = 1

Find y using the equation y = 9 – x = 8

The number we are looking for is 18.

It is divisible by 3 and the sum of its digits is equal to 9 and it is the smallest and positive whole number with such properties.

**Question 36. Which Of These Numbers Is Not Divisible By 3?****Answer :**One may answer this question using a calculator and test for divisibility by 3.

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

divisible by3 then the number is divisible by 3.

3 + 3 + 9 = 15 , divisible by 3.

3 + 4 + 2 = 9 , divisible by 3.

5 + 5 + 2 = 12 , divisible by 3.

1 + 1 + 1 + 1 = 4 , not divisible by 3.

**Question 37. The Sum Of Two Number Is 15 And Sum Of Their Square Is 113. The Numbers Are ?****Answer :**Let the number be x and (15 – x)

Then, x

^{2}+ (15 – x)2= 113x

^{2}– 15x + 56 = 0(x-7) (x-8) = 0.

**Question 38. The Number X Is Exactly Divisible By 5 And The Remainder Obtained On Dividing The Number Y By 5 Is 1. What Remainder Will Be Obtained When (x + Y) Is Divided By 5 ?****Answer :**Let x/5 = p and let y when divided by 5 gives 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 100 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 let there be x men.

Less men, More days(Indirect Proportion)

Therefore, (x-10) : x :: 100 :110

=> (x – 10) * 110 = x * 100=> x= 110.

**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 3 Days?****Answer :**Let the required length be x meters

More men, More length built (Direct proportion)

Less days, Less length built (Direct Proportion)

Men20:35

Days6 : 3}?? 56 😡

=> (20 x 6 x X)=(35 x 3 x 56)

=> x = 49

Hence, the required length is 49 m.

**Question 41. 2 Men And 7 Boys Can Do A Piece Of Work In 14 Days; 3 Men And 8 Boys Can Do The Same In 11 Days. Then, 8 Men And 6 Boys Can Do Three Times The Amount Of This Work In?****Answer :**(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys

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

Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys

( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys.

Let the required number of days be x.

More boys , Less days (Indirect proportion)

More work , More days (Direct proportion)

Boys22:11Work1 : 3}?? 14:x

Therefore, (22 * 1 * x) = (11 * 3 * 14)

=> x = 21

Hence, the required number 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(1/3) 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.

**Question 43. A Sum Of Money Amounts To Rs.6690 After 3 Years And To Rs.10,035 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/100)^6=10035…(ii)

On dividing,we get (1+R/100)^3=10025/6690=3/2.

Substituting this value in (i),we get:

P*(3/2)=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+7/100)^n = 34347

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

n = 2years.

**Question 45. 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 3 Years?****Answer :**difference in C.I and S.I in 2years =Rs.32

S.I for 1year =Rs.400

S.I for Rs.400 for one year =Rs.32

rate=[100*32)/(400*1)%=8%

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

=S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56

**Question 46. On A Sum Of Money, The Simple Interest For 2 Years Is Rs. 660,while 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 year =Rs. 36.30

Rate

= (100×36.30/330×1)%

= 11%.

**Question 47. 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% per annum = 5% per quarter.

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

CJ. = Rs. (18522 – 16000) = Rs. 2522.

**Question 48. 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]=525Sum =Rs.2500

S.I.= Rs.(2500*5*4)/100

= Rs. 500.

**Question 49. Find Compound Interest On Rs. 7500 At 4% Per Annum For 2 Years, Compounded Annually?****Answer :**Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25) ) = Rs. 8112.

therefore, 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*4*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. when interest

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

= Rs. 5304.

C.I. when interest is

compounded half-yearly=rs.5000(1+2/100)^3

= Rs. 5306.04

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

**Question 52. A Sum Of Money Invested At Compound Interest Amounts To Rs. 800 In 3 Years And To Rs. 840 In 4 Years. The Rate Of Interest Per Annum Is?****Answer :**S.I. on Rs.800 for 1 year

=Rs[840 – 800]

= Rs.40

Rate

=(100×40/800×1)%

= 5%.

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