Ex. 3. The true discount on a certain sum of money due 3 years hence is Rb. 250 and the simple interest on the same sum for the same time and at the same rate is Rs. 375. Find the sum and the rate percent.

Sol. T.D. = Rs. 250 and S.I. = Rs. 375. Sum due =S.I. xT.D./ S.I. -T.D.

Ex. 4. The difference between the simple interest and true discount on a certain sum of money for 6 months at 12â??% per annum is Rs. 25. Find the sum.

Sol. Let the sum be Rs. x. Then, T.D. = (x*25/2*1/2)/(100+(25/2*1/2))=x*25/4*4/425=x/17 S.I=x*25/2*1/2*1/100=x/16 x/16-x/17=25 ⇒17x-16x=25*16*17 ⇒x=6800 Hence, sum due = Rs. 6800.

Ex. 5. A bill falls due in 1 year. The creditor agrees to accept immediate payment of the half and to defer the payment of the other half for 2 years. By this arrangement ins Rb. 40. What is the amount of the bill, if the money be worth 12-z% ? Sol. Let the sum be Rs. x. Then, [x/2+(x/2*100)/100+(25/2*2)]-[(x*100)/(100+25/2*1]

=40 ⇒x/2+2x/5-8x/9=40 ⇒x=3600 Amount of the bill - Rs. 3600.

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Ex. 1. Find the present worth of Rs. 930 due 3 years hence at 8% per annum. Also find the discount.

Sol.P.W=100 x Amount /[100 + (R x T)]

=Rs.100 x 930/100+ (8x3)

= (100x930)/124

= Rs. 750,

T.D. = (Amount) - (P.W.) = Rs. (930 - 750) = Rs. 180.

Ex. 2. The true discount on a bill due 9 months hence at 12% per annum is Rs. Find the amount of the bill and its present worth.

Sol. Let amount be Rs. x. Then,x*R*T/100 + (R x T)

=T.D.

⇒x * 12*3/ 4/[100+[12*3/4]]

=540

x= 540x109 = Rs.6540

Amount - Rs. 6540. P.W. = Rs. (6540 - 540) - Rs. 6000.

Ex. 3. The true discount on a certain sum of money due 3 years hence is Rb. 250 and the simple interest on the same sum for the same time and at the same rate is Rs. 375. Find the sum and the rate percent.

Sol. T.D. = Rs. 250 and S.I. = Rs. 375.

Sum due =S.I. xT.D./ S.I. -T.D.

=375x250/375- 250

=Rs.750.

Rate=[100*375/750*3]%=16 2/3%

Ex. 4. The difference between the simple interest and true discount on a certain sum

of money for 6 months at 12â??% per annum is Rs. 25. Find the sum.

Sol. Let the sum be Rs. x. Then,

T.D. = (x*25/2*1/2)/(100+(25/2*1/2))=x*25/4*4/425=x/17

S.I=x*25/2*1/2*1/100=x/16

x/16-x/17=25

⇒17x-16x=25*16*17

⇒x=6800

Hence, sum due = Rs. 6800.

Ex. 5. A bill falls due in 1 year. The creditor agrees to accept immediate payment of the half and to defer the payment of the other half for 2 years. By this arrangement

Sol. Let the sum be Rs. x. Then,ins Rb. 40. What is the amount of the bill, if the money be worth 12-z% ?

[x/2+(x/2*100)/100+(25/2*2)]-[(x*100)/(100+25/2*1]

=40

⇒x/2+2x/5-8x/9=40

⇒x=3600

Amount of the bill - Rs. 3600.