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Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that:

Sum due = Rs. 156 due 4 years hence;

Present Worth (P.W.) = Rs. 100;

True Discount (T.D.) = Rs. (156 - 100) = Rs. 56 = (Sum due) - (P.W.)

We define: T.D. = Interest on Present Worth;Amount = Present Worth + True Discount

Interest is reckoned on P.W. and true discount is reckoned on the amount.

Important Formulae

Let rate = R% per annum and Time = T years. Then,

P.W. =	(100 x Amount) / (100 + (R x T))    = (100 x T.D.)/ (R x T)
T.D. = (P.W. x R x T) / 100   = (Amount x R x T) / (100 + (R x T))	
Sum = (S.I. x T.D.) / (S.I. - T.D.)
S.I. - T.D.  = S.I. on T.D.
When the sum is put at compound interest, then   P.W. = Amount/ (1+R/100)T  

Solved Examples

Q 1 - Find the P.W. of Rs. 9920 due 3 years at 8% p.a.

A - Rs 7000

B - Rs 8000

C - Rs 6000

D - Rs 7000

Answer - B

Explanation

P.W.= ((100 x Amount))/(100+(R x T))  = Rs {(100 x 9920)/(100 (3 x 8))}  = Rs  (100 x 9920)/124   = Rs 8000  

Q 2 - The genuine rebate on a bill due 9 months consequently at 6% for each annum is Rs. 180. Discover its present worth.

A - Rs 4000

B - Rs 5000

C - Rs 3000

D - Rs 6000

Answer - A

Explanation

P.W. = ((100 x T.D.))/((R x T))  = Rs {(100 x 180)/(6 x 3/4)}  = Rs 4000  

Q 3 - The genuine markdown on a sure total of cash due 3 years consequently is Rs 200 and the straightforward enthusiasm on the some aggregate for the same time and at the same guideline is Rs. 240. Discover the aggregate and the rate percent.

A - Rs 1200

B - Rs 1000

C - Rs 1200

D - Rs 1000

Answer - A

Explanation

T.D = Rs 200 and S.I. = Rs 240   Sum due = (S .I. x T.D.)/(S.I.-T.D.)   = Rs ((240 x 200))/((240-200))  = Rs 1200  

Q 4 - The genuine rebate on Rs 1860 due after a sure time at 5% p.a. is Rs. 60. Discover the time after which it is expected.

A - 10 months

B - 8 months

C - 7 months

D - 6 months

Answer - B

Explanation

P.W. = (Amount)-(T.D.)   = Rs (1860-60)   = Rs 1800   T.D. is S.I. on P.W.   Rs. 60 is S.I. on Rs 1800 at 5% p.a.   Time = ((100 x 60))/((5 x 1800)) years   = 2/3 years   = 2/3 x 12 months   = 8 months  

Q 5 - Find the rebate on Rs. 9920 due 3 years at 8% p.a.

A - Rs 1720

B - Rs 1820

C - Rs 1920

D - Rs 1220

Answer - C

Explanation

P.W.= ((100 x Amount))/(100+(R x T))  = Rs {(100 x 9920)/(100 (3 x 8))}  = Rs  (100 x 9920)/124   = Rs 8000  T.D. = (Amount)-(P.W.)  = Rs (9920-8000)  = Rs 1920  

Q 6 - The genuine rebate on a bill due 9 months consequently at 6% for each annum is Rs. 180. Discover the charge's measure.

A - Rs 4180

B - Rs 3180

C - Rs 2180

D - Rs 1180

Answer - A

Explanation

P.W. = ((100 x T.D.))/((R x T))  = Rs {(100 x 180)/(6 x 3/4)}  = Rs 4000  Sum = (P.W.+T.D.)   = Rs (4000+180)   = Rs 4180   

Q 7 - The genuine markdown on a sure total of cash due 3 years consequently is Rs 200 and the straightforward enthusiasm on the some aggregate for the same time and at the same guideline is Rs. 240. Discover the rate percent.

A - 50/3% p.a

B - 40/3% p.a

C - 10/3% p.a

D - 20/3% p.a

Answer - D

Explanation

T.D = Rs 200 and S.I. = Rs 240   Sum due = (S .I. x T.D.)/(S.I.-T.D.)   = Rs ((240 x 200))/((240-200))  = Rs 1200  T.D is S.I. on the sum.   Rs. 240 is S.I. on Rs 1200 for a long time.   R= ((100 x 240))/((1200 x 3))% p.a.  = 20/3% p.a  

Q 8 - The genuine markdown on Rs 2575 due 4 months thus is Rs. 75. Discover the rate of hobby.

A - 6% p.a.

B - 7% p.a.

C - 8% p.a.

D - 9% p.a.

Answer - D

Explanation

Amount = Rs 2575, T=4/12 years = 1/3 years, T.D = Rs. 75   P.W. = (Amount) - (T.D.) = Rs (2575-75) = 2500.   T.D. Is S.I. on P.W.   R.s 75 is S.I. on Rs. 2500 or 1/3 years  Rate = ((100x75)/(2500x1/3))% p.a. = 9% p.a.  

Q 9 - The genuine rebate on a bill due 10 months consequently at 6% p.a. is Rs 26.25. Discover the charge's measure.

A - Rs 451.25

B - Rs 551.25

C - Rs 351.25

D - Rs 251.25

Answer - B

Explanation

T.D. = Rs 26.25, T = 10/12  year=  5/6  year,R=6% p.a.  Let P.W. be Rs x. Then,  S.I. on Rs x at 6% p.a. for 5/6 year is Rs. 26.25  ∴  (x * 6 * 5/6)/100 = 25.25   => x= (26.25 * 20) = 525  ∴  (P.W.) + (T.D.) = Rs. (525+26.25) = Rs 551.25  

Q 10 - The contrast between the S.I. what's more, T.D. on a sure whole of cash for 6 months at 6% p.a. is Rs. 27. Discover the total.

A - 32900

B - 30900

C - 31900

D - 33900

Answer - B

Explanation

S.I. = Rs. (x * 6 * 1/2)/100   = Rs. 3x/100 and T.D.   = Rs {(x * 78/12)/(100+(6 * 6/12) )}  = Rs. 3x/103  ∴ 3x/100-3x/103 = 27   => (309x-300x) = (27 * 100 * 103)  => x = ((27*100*103))/9   = 30900

Factors and Multiples

If a number P divides another number Q exactly, we say that P is a factor of Q i.e. Q is a multiple of P.

H.C.F

The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly.

L.C.M

The least number which is exactly divisible by each one of the given numbers is called their L.C.M

Product of two numbers

Product of their H.C.F and L.C.M

Co-primes

Two numbers are co-primes if their H.C.F is 1.

H.C.F and L.C.M of fractions

  1. H.C.F = H.C.F of Numerators⁄L.C.M of Denominators
  2. L.C.M = L.C.M of Numerators⁄H.C.F of Denominators

Solved Examples

Q 1 - Compute H.C.F of (22* 23*5*74), (23*32*52*73) and (22*53*75).

A - 6760

B - 6860

C - 6960

D - 7060

Answer - B

Explanation

Prime numbers which are common to all the given numbers are 2,5 ,7.  ∴ H.C.F = (22*5*73)= (4*5*343) = 6860  

Q 2 - Find the H.C.F of 108, 360 and 600.

A - 12

B - 13

C - 14

D - 15

Answer - A

Explanation

108   = (22*33) , 360 = (23*32*5)  and 600 =  (23*52*3)  ∴ H.C.F =  (22* 3) = (4* 3)=12  

Q 3 - Find the H.C.F of 148 and 185.

A - 37

B - 38

C - 39

D - 40

Answer - A

Explanation

Remainder of 185/148 = 37  Remainder of 148/37 = 0  ∴ H.C.F. = 37  

Q 4 - Find the H.C.F of 204, 1190 and 1445.

A - 16

B - 17

C - 18

D - 19

Answer - B

Explanation

Remainder of 1190/204 = 170  Remainder of 204/170 = 34  Remainder of 170/34 = 0    ∴ H.C.F. of 204, 1190 = 34    Remainder of 1145/34 = 17  Remainder of 34/17 = 0    ∴ H.C.F. of 204, 1190 and 1145 = 17  

Q 5 - Reduce 391/667 to lowest terms.

A - 7/29

B - 27/29

C - 17/29

D - 37/29

Answer - C

Explanation

First we find the H.C.F of 391 and 667.  Remainder of 667/391 = 276  Remainder of 391/276 =115  Remainder of 276/115 = 46  Remainder of 115/46 = 23  Remainder of 46/23= 0    ∴ H.C.F. of 391, 667 = 23  ∴    391/667   =( 391/23)/ (667/23)=  17/29  

Q 6 - Find the L.C.M of (22*32*5*7) , (2³*3*52*72) and (2*3*7*11).

A - 970200

B - 97020

C - 9702

D - 970

Answer - A

Explanation

We have  L.C.M = product of terms containing highest powers of (2,3,5,7,11)  = (2³* 32* 52*72*11) = (8*9*25*11*49)= 970200  

Q 7 - Find the L.C.M of 15, 18, 24, 27, 56.

A - 7260

B - 7360

C - 7460

D - 7560

Answer - D

Explanation

15 = 3 * 5  18 = 2* 3 * 3 = 2 * 3 2  24 = 2 *2 * 2 * 3   =  2 3 * 3  27 = 3 * 3 * 3 =  3 3  56 = 2 * 2 * 2 * 7=  2 3 * 7  L.C.M  = product of terms containing highest powers of (2,3,5,7) =2 3 * 3 3 * 5 * 7 = 7560  

Q 8 - Find the H.C.F and L.C.M of 2/3 , 8/9 , 10/27 and 16/81.

A - 45

B - 55

C - 65

D - 75

Answer - D

Explanation

H.C.F of 2,8,10,16     = 2  L.C.M of  3,9,27,81 = 81                        H.C.f =  H.C.F of 2,8,10,16/L.C.M of 3,9,27,81 = 2/81            L.C.M =  L.C.M of 2,8,10,16/H.C.F of  3,9, 27,81 = 80/3  

Q 9 - Two numbers are in the ratio 8:11 . Considering their H.C.f as 6, find the numbers.

A - 58.79

B - 48,66

C - 38,56

D - 28,33

Answer - B

Explanation

Let the numbers be 8x and 11 x. then, their H.C.F = x   So, the numbers are (8*6), (11*6) i.e 48 and 66.   

Q 10 - Given the H.C. F of two numbers as 7 and their L.C.M as 210. If one of the numbers is 35, find the other.

A - 32

B - 42

C - 52

D - 62

Answer - B

Explanation

Let the Other number be X. then,   Product of numbers = product of their H.C .F and L.C.M  35*x=   7* 210  ⇒ x=  7*210/35  = 42  Hence, the other number is 42.  

Q 11 - Three big drums contain 36 liters, 45 liters and 72 liters of oil. What is the biggest measure which can measure all the different quantities exactly?

A - 9 liters

B - 10 liters

C - 11 liters

D - 12 liters

Answer - A

Explanation

Required measure =   H.C.F  of 36 L, 45 L, and 72 L  = (32) liters = 9 liters  [As 36 = 22*32, 45 = 32*5 and 72 = 24* 34]  

Q 12 - Four electronic devices make a beep after duration of 30 minutes, 1 hour, 3/2 hours and 1 hour 45 min. respectively. If all the devices beeped together at 12 noon at what time will they beep together again?

A - 9 am

B - 10 am

C - 11 am

D - 11:30 am

Answer - A

Explanation

Intervals of beeping 30  min, 60 min, 90 min, 105 min.  Interval of beeping together= L.C.M of 30 min. 60 min. 90 min. 105 min  = (3*5*2*2*3*7) min. = 1260 min = 21 hrs.  So, they will beep together again next morning at 9 am.  

Q 13 - Find the largest number which can exactly divide 513, 783 and 1107.

A - 22

B - 23

C - 24

D - 25

Answer - B

Explanation

Remainder of 783/513 = 270  Remainder of 513/270 =243  Remainder of 270/243 = 27  Remainder of 243/27 = 0    Remainder of 46/23= 0    ∴ H.C.F. of 513, 783 = 23    Remainder of 1107/23 = 0    ∴ H.C.F. of 513, 783 and 1107= 23  

Q 14 - Find the smallest number which is exactly divisible by each one of the numbers 12, 15, 20 and 27.

A - 540

B -530

C - 520

D - 510

Answer - A

Explanation

Required no. = L.C.M of 12,15, 20 and 27  = (3*2*2*5*9) = 540  

Q 15 - Find the least number which if divided by 6, 7, 8, 9, 12 leaves the same remainder 2 in each case.

A - 506

B - 504

C - 502

D - 500

Answer - A

Explanation

Required number =  (L.C.M  of 6,7,8,9,12)+2   =  (2*3*2*7*2*3)+ 2 = (504+2)= 506.  

Q 16 - Find the largest natural number which can divide the product of any 4 consecutive natural numbers.

A - 23

B - 24

C - 25

D - 26

Answer - B

Explanation

(1*2*3*4) =  24  ∴ Required number = 24  

Q 17 - Find the least number which if divided by 35, 45 and 55 leaves the remainder 18, 28 and 38 respectively.

A - 3448

B - 3458

C - 3468

D - 3478

Answer - A

Explanation

Here (35-18) =  17 , (45-28)=  17  and (55- 38) =  17  Required number =  (L.C.M of 35,45, 55)- 17  = (3465 -17) = 3448  

Q 18 - The H.C.F of 1/2 , 2/3 , 3/4 , 4/5 is

A - 1/120

B - 12/5

C - 100/3

D - 10/3

Answer - A

Explanation

H.C.F =   H.C.F of 1,2,3,4/ L.C.M  of  2,3,4,5   = 1/120                                                   

Q 19 - The H.C.F of 2/3, 8/9 , 10/27 , 32/81.

A - 160/81

B - 160/3

C - 2/81

D - 2/3

Answer - C

Explanation

H.C.F  = H.C.F of 2, 8,10, 32/  L.C.M of 3,9, 27, 81   = 2/81  

Q 20 - Which of the following is a pair of Co-primes?

A - (14, 35)

B - (18, 25)

C - (31, 93)

D - (32,62)

Answer - B

Explanation

H.C.F of  18 and 25 is 1.

S Naga Malleswara Reddy

S NAGA MALLESWARA REDDY

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