Arithmatical - True Discount

DIRECTIONS : Problems based on True discounts.
1. If Rs.10 be allowed as true discount on a bill of Rs.110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is
  A.  Rs.21.81
  B.  Rs.18.33
  C.  Rs.21
  D.  Rs.22
Solution
S.I on Rs.(110 - 10) for a certain time = Rs.10.
S.I on Rs. 100 double the time= Rs.20.
T.D on Rs.120= Rs.(120 - 100)
= Rs.20.
T.D on Rs.110= Rs.(20/120×110)
= Rs.18.33.
2. A man buys a watch for Rs.1950 in cash and sells it for Rs.2200 at a credit of 1 year. If the rate of interests is 10% per annum, the man
  A.  Gain Rs.55
  B.  Gain Rs.30
  C.  Loses Rs.30
  D.  Gains Rs.50
Solution
S.P=P.W. of Rs.2200 due 1 year
= Rs.[2200×100/100+(10×1)]
= Rs.2000.
Gain=Rs.(2000 - 1950)
= Rs.50.
3. Rs.20 is the true discount on Rs.260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same?
  A.  Rs.10
  B.  Rs.10.40
  C.  Rs.15.20
  D.  Rs.13
Solution
S.I.on Rs.(260-20) for a given time = Rs. 20
S.I. on Rs. 240 for half the time= Rs. 10.
T.D. on Rs. 250= Rs. 10.
T.D. on Rs. 260= Rs.(10/250×260)
= Rs. 10.40
4. A trader owens a merchant Rs.10,028 due 1 year hence. The trader wants to settle the account after 3 months. If the rate of interest is 12% per annum, how much cash should he pay?
  A.  Rs.9025.20
  B.  Rs.9200
  C.  Rs.9600
  D.  Rs.9560
Solution
Required money = P.W of Rs.10028 due 9 months
Rs.[10028x100/100+(12x9/12)]
= Rs.9200.
5. The true discount on Rs.2562 due 4 months hence is Rs.122. The rate percent is
  A.  12 %
  B.  13 %
  C.  15 %
  D.  14 %
Solution
P.W= Rs.(2562 - 122)
= Rs. 2440.
S.I on Rs.2440 for 4 months is Rs.12
Rate= (100 x 122/2440x1/3)%
= 15%.
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