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Consider the simple interest of the following

(A) The simple interest on Rs 8,930 at 6% per annum for 5 years

(B) The simple interest on Rs 3,080 at 13.50% per annum for 4 years

(C) The simple interest on Rs 7,200 at 14.00 % per annum for 7 years

(D) The simple interest on Rs. 3,120 at 4.00 % per annum for 5 years

The simple interests of these in increasing order are:

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

The simple interest formula is:

SI=P×R×T100SI = \dfrac{P \times R \times T}{100}

where PP is the principal, RR is the rate per annum, and TT is the time in years.


For option (A): Principal =Rs 8,930= Rs \space 8,930, Rate =6%= 6\%, Time =5= 5 years

SI=8,930×6×5100SI = \dfrac{8,930 \times 6 \times 5}{100}

SI=267,900100SI = \dfrac{267,900}{100}

SI=Rs 2,679SI = Rs \space 2,679


For option (B): Principal =Rs 3,080= Rs \space 3,080, Rate =13.50%= 13.50\%, Time =4= 4 years

SI=3,080×13.50×4100SI = \dfrac{3,080 \times 13.50 \times 4}{100}

SI=166,320100SI = \dfrac{166,320}{100}

SI=Rs 1,663.20SI = Rs \space 1,663.20


For option (C): Principal =Rs 7,200= Rs \space 7,200, Rate =14%= 14\%, Time =7= 7 years

SI=7,200×14×7100SI = \dfrac{7,200 \times 14 \times 7}{100}

SI=705,600100SI = \dfrac{705,600}{100}

SI=Rs 7,056SI = Rs \space 7,056


For option (D): Principal =Rs 3,120= Rs \space 3,120, Rate =4%= 4\%, Time =5= 5 years

SI=3,120×4×5100SI = \dfrac{3,120 \times 4 \times 5}{100}

SI=62,400100SI = \dfrac{62,400}{100}

SI=Rs 624SI = Rs \space 624


Comparing the simple interests:

(D) Rs 624Rs \space 624

(B) Rs 1,663.20Rs \space 1,663.20

(A) Rs 2,679Rs \space 2,679

(C) Rs 7,056Rs \space 7,056

Therefore, the simple interests in increasing order are: (D), (B), (A), (C)

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