Class 8: Simple Interest and Compound Interest – …...Class 8: Simple Interest and Compound Interest – Exercise 15A Q1. Find the simple interest and amount on: i. Rs. 4500 for

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Class 8: Simple Interest and Compound Interest – Exercise 15A

Q1. Find the simple interest and amount on:

i. Rs. 4500 for 21

2years at 7

2

3% per annum

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑃 = 𝑅𝑠. 45000, 𝑅 = 72

3% =

23

3%

𝑇 = 2 1

2𝑦𝑒𝑎𝑟 =

5

2𝑦𝑒𝑎𝑟

𝑆. 𝐼. = 45000 ×

233

×52

100

= 𝑅𝑠. 8625 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑃 + 𝑆. 𝐼 = 𝑅𝑠. 53625

ii. Rs. 6360 for 6 years 3 months at 8% per annum

𝑃 = 𝑅𝑠. 6360 𝑅 = 8%

𝑇 = 61

4𝑦𝑒𝑎𝑟 = 25/4 𝑦𝑒𝑎𝑟

𝑆. 𝐼. =(𝑃 × 𝑅 × 𝑇)

100

= (6360 × 8 ×

254

)

100

= 𝑅𝑠. 3180 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 6360 + 3180 = 𝑅𝑠. 9540

iii. Rs. 19200 for 11 months at 93

4% per annum

𝑃 = 𝑅𝑠. 19200, 𝑅 = 93

4% =

39

4%

𝑇 = 11 𝑀𝑜𝑛𝑡ℎ𝑠 = 11

12𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. 𝑅𝑠. = 19200 ×

394

×1112

100

= 𝑅𝑠. 1716 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑃 + 𝑆𝐼

= 𝑅𝑠. (19200 + 1716) = 𝑅𝑠. 20916

2


iv. Rs. 58400 for 75 days at 61

2% per annum

𝑃 = 𝑅𝑠. 58400, 𝑅 = 661

2% =

13

2%

𝑇 = 75 𝑑𝑎𝑦𝑠 = 75


𝑆. 𝐼. = 𝑅𝑠.58400 ×

132

×75

365100

= 𝑅𝑠. 780

𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 59180 Q2. Find the simple interest on Rs.8600 from 18th October, 2006 to 13th march, 2007 at 8% per annum. Also find the amount.

𝑃 = 𝑅𝑠. 8600 𝑅 = 8% 𝑇 = 18 𝑂𝑐𝑡. 𝑡𝑜 13 𝑀𝑎𝑟𝑐ℎ

= (13 + 30 + 31 + 31 + 28 + 13) 𝑑𝑎𝑦𝑠

= 146 𝑑𝑎𝑦𝑠 =146


𝑆. 𝐼 = 𝑃 × 𝑅 × 𝑇

100

= 𝑅𝑠.8600 × 8 ×

146365

100

= 𝑅𝑠. 275.20 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑃 + 𝑆. 𝐼. = 𝑅𝑠. 8600 + 275.2

= 𝑅𝑠. 8875.2 Q3. Ashish lent Rs.10500 to Mark at 7% per annum simple interest. After 5 years, Mark discharged the debt by giving a watch and Rs.3000 in cash. What is the value of the watch?

𝑃 = 𝑅𝑠. 10500, 𝑅 = 7%, 𝑇 = 5 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼 =𝑃 × 𝑅 × 𝑇

100

= 10500 × 7 × 5

100

= 𝑅𝑠. 3675 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑃 + 𝑆. 𝐼.

= 𝑅𝑠. 10500 + 3675 = 𝑅𝑠. 14175

𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑤𝑎𝑡𝑐ℎ = 𝐴𝑚𝑜𝑢𝑛𝑡 – 𝑅𝑠. 13000 = 𝑅𝑠. 14175– 13000 = 𝑅𝑠. 1175

Q4. In what time will the simple interest on Rs.7560 be Rs.1102 at 61

4% per annum?

𝑃 = 𝑅𝑠. 7560, 𝑆. 𝐼. = 1102.5, 𝑅 = 61

4% =

25

4%

𝑊𝑒 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑠 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑦𝑒𝑎𝑟𝑠.

3


𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑂𝑅

𝑇 = 𝑆. 𝐼.× 100

𝑃 × 𝑅

= 1102.5 × 100

7560 ×254

= 2.3 𝑦𝑒𝑎𝑟𝑠 = 2 𝑦𝑒𝑎𝑟𝑠 4 𝑚𝑜𝑛𝑡ℎ𝑠

Q5. In how much time will Rs.25600 amount to Rs.35665, when money is worth 91

4%per

annum simple interest? 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 35664, 𝑃 = 𝑅𝑠. 25600 𝑆. 𝐼. 𝐴𝑚𝑜𝑢𝑛𝑡 − 𝑃

= 𝑅𝑠. (35664 − 25600) = 𝑅𝑠. 10064

𝑅 = 91

4% =

37

4%

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑇 = 𝑆. 𝐼.× 100

𝑃 × 𝑅

= 10064 × 100

25600 ×374

= 4.25 𝑦𝑒𝑎𝑟𝑠 𝑜𝑟 4 𝑦𝑒𝑎𝑟𝑠 3 𝑚𝑜𝑛𝑡ℎ𝑠

Q6. At what rate per cent per annum will Rs.1625 amount to Rs.2080 in 31

2 years?

𝑃 = 𝑅𝑠. 1625, 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 2080, 𝑇 = 31

2𝑦𝑒𝑎𝑟𝑠 =

7


𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑅

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇… … … … … … … … . (1)

𝑆. 𝐼. = 𝐴𝑚𝑜𝑢𝑛𝑡 – 𝑃 = 𝑅𝑠. (2080– 1625) = 𝑅𝑠. 455

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑆. 𝐼. 𝑃 & 𝑇 𝑖𝑛 𝐸𝑞𝑢𝑎𝑙 (1) 𝑤𝑒 𝑔𝑒𝑡

𝑅 =455 × 100

1625 ×72

= 8%

4


Q7. At what rate per cent per annum will the simple interest on Rs.6720 be Rs.1911 in 3 years 3 months?

𝑃 = 𝑅𝑠. 6720, 𝑆. 𝐼. = 𝑅𝑠. 1911 𝑇𝑖𝑚𝑒 = 3 𝑦𝑒𝑎𝑟 3 𝑚𝑜𝑛𝑡ℎ𝑠

= 31


13

4𝑦𝑒𝑎𝑟

𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑅.

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑅 =𝑆. 𝐼.× 100

𝑃 × 𝑇

= 1911 × 100

6720 ×134

= 8.75% = 83

4%

Q8. At what rate per cent of simple interest will a sum of money double itself in 12 years?

𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑃 = 𝑥 𝐴𝑚𝑜𝑢𝑛𝑡 = 2 𝑥 𝑆. 𝐼. = 𝑥 𝑇𝑖𝑚𝑒 = 12 𝑦𝑒𝑎𝑟𝑠 𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑅

𝑆. 𝐼. = .𝑃 × 𝑅 × 𝑇

100

𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇

= 𝑥 × 100

𝑥 × 12=

100

12= 8 𝑦𝑒𝑎𝑟𝑠 4 𝑚𝑜𝑛𝑡ℎ𝑠

Q9. Simple interest will a sum is 9

16 of the sum. Find the rate per cent and the time if both

are numerically equal. 𝐿𝑒𝑡 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑃 = 𝑅𝑠. 𝑥

𝑆. 𝐼. = 𝑅𝑠.9

16𝑥

𝑁𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙 𝑅 = 𝑇

𝑆. 𝐼. = .𝑃 × 𝑅 × 𝑇

100

𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑅.

𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇

𝑇 = 𝑅

𝑅2 = 𝑆. 𝐼.× 100

𝑃

=

916

𝑥 × 100

𝑥=

9

16× 100

𝑅 = 3

4× 10 = 7.5%

5


𝑃𝑒𝑟 𝑎𝑛𝑛𝑢𝑚,

𝑇 = 71


Q10. What sum will yield Rs.406 as simple interest in 1 year 2 months at 61

4% per annum?

𝑆. 𝐼. = 𝑅𝑠. 406

𝑇 = 1𝑦𝑒𝑎𝑟 2 𝑚𝑜𝑛𝑡ℎ𝑠 = 12

12= 1

1

6=

7


𝑅 = 61

4% =

25

4%

𝑊𝑒 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑃

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑃 = 𝑆. 𝐼.× 100

𝑅 × 𝑇

= 406 × 100

254

×76

= 𝑅𝑠. 5568

Q11. What sum will amount to 1748 in 21

2years at 7

1

2% per annum simple interest?

𝐴𝑚𝑜𝑢𝑛𝑡 𝑅𝑠. 1748, 𝑇 = 2.1

2𝑦𝑒𝑎𝑟𝑠, 𝑆 = 7

1

2%

𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑃

𝑆. 𝐼. = .𝑃 ×× 𝑅 × 𝑇

100

𝑃 = 𝑆. 𝐼.× 100

𝑅 × 𝑇

𝑆. 𝐼. = 𝐴𝑚𝑜𝑢𝑛𝑡 – 𝑃 = (1748– 𝑃)

𝑃 = (1748 − 𝑃) × 100

712

× 212

𝑃 ×15

2×

5

2= (1748 − 𝑃)100

𝑂𝑅

𝑃 (100 +75

4) = 1748 × 100

𝑃 (475

4) = 1748 × 100

𝑃 = 1748 × 4 × 100

475

= 𝑅𝑠. 1472

6


Q12. A sum of money becomes 8

5 of itself in 5 years at certain rate of simple interest. Find

the rate of interest.

𝐿𝑒𝑡 𝑃 = 𝑅𝑠. 𝑥, 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 8

5𝑥

𝑇𝑖𝑚𝑒 𝑇 = 5 𝑦𝑒𝑎𝑟𝑠 𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑅.

𝑆. 𝐼. = 𝑅𝑠. (8

5𝑥 − 𝑥)

= 𝑅𝑠. 3

5𝑥

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇

=

35

𝑥 × 100

𝑥 × 5

= 300

25= 12% 𝑝. 𝑎.

Q13. What sum of money lent at 121

2% per annum will produce the same interest in 4 years

as Rs.8560 produces in 5 years at 12% per annum?

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑃 = 𝑅𝑠. 8560, 𝑅 = 12%, 𝑇 = 5 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

= 8560 × 12 × 5

100= 𝑅𝑠. 5136

𝑁𝑜𝑤 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑃 𝑖𝑓 𝑇 = 4 𝑦𝑒𝑎𝑟𝑠, 𝑅 = 121

2% 𝑎𝑛𝑑 𝑆. 𝐼. = 𝑅𝑠. 5316

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

𝑃 = 𝑆. 𝐼.× 100

𝑅 × 𝑇

= 5136 × 100

1212

× 4

= 𝑅𝑠. 10272 Q14. If Rs.1250 amount to 1550 in 3 years at simple interest, what will Rs.3100 amount to in 4 years at the same rate?

𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑅 𝑓𝑜𝑟 𝑃 = 𝑅𝑠. 1250, 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 1550, 𝑇 = 3 𝑌𝑒𝑎𝑟𝑠 𝑆. 𝐼. = 𝐴𝑚𝑜𝑢𝑛𝑡 – 𝑃

= 1550 – 1250

7


= 𝑅𝑠. 300 𝑁𝑜𝑤 𝑓𝑜𝑟 𝑆. 𝐼. 𝑅𝑠. 300

𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇

= 300 × 100

1250 × 3= 8% 𝑝. 𝑎.

𝑁𝑜𝑤 𝑓𝑜𝑟 𝑅 = 8%, 𝑃 = 𝑅𝑠. 3200, 𝑇 = 4 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

= 3200 × 8 × 4

100

= 𝑅𝑠. 1024 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑃 + 𝑆. 𝐼.

= 3200 + 1024 = 𝑅𝑠. 4224

Q15. A sum of money lent at simple interest amounts to Rs.3224 in 2 years and Rs.4160 in 5 years. Find the sum and the rate of interest

𝐺𝑖𝑣𝑒𝑛 𝑖𝑠; (1) 𝐴𝑚𝑜𝑢𝑛𝑡 = 𝑅𝑠. 3224, 𝑇 = 2 𝑦𝑒𝑎𝑟𝑠 (2) 𝐴𝑚𝑜𝑢𝑛𝑡 𝑅𝑠. 4160, 𝑇 = 5𝑦𝑒𝑎𝑟𝑠 𝑃 𝑎𝑛𝑑 𝑅 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑖𝑛 𝑏𝑜𝑡ℎ 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑎𝑟𝑒𝑎 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 (𝑖) 𝑆. 𝐼. = 𝑅𝑠. (3224 − 𝑃)

3224 − 𝑃 = 𝑃 × 𝑅 × 2

100… … … … … … … . . (𝑖)

𝐹𝑟𝑜𝑚 𝑑𝑎𝑡𝑎 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 (𝑖𝑖) 𝑆. 𝐼. = 𝑅𝑠. 4160 − 𝑃

= 4160 − 𝑃 =𝑃 × 𝑅 × 5

100… … … … … … … … … . 𝑖𝑖)

𝐷𝑖𝑣𝑖𝑑𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑖) 𝑏𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑖𝑖)

=3224 − 𝑃

4160 − 𝑃=

𝑃 × 𝑅 × 2100

𝑃 × 𝑅 × 5100

=2

5

(3224 − 𝑃)5 = (4160 − 𝑃)2 𝑂𝑅 3224 × 5 − 5𝑃 = 4160 × 2 − 2𝑃 𝑂𝑅 3𝑃 = 3224 × 5 − 4160 × 2

𝑃 = 3224 × 5 − 4160 × 2

3

= 𝑅𝑠. 2600 𝐹𝑟𝑜𝑚 𝑑𝑎𝑡𝑎 𝑠𝑒𝑡 (𝑖) 𝑆. 𝐼. = 𝑅𝑠. (3224 − 2600)

= 𝑅𝑠. 624

𝑆. 𝐼. = 𝑃 × 𝑅 × 𝑇

100

8


𝑅 = 𝑆. 𝐼.× 100

𝑃 × 𝑇=

624 × 100

2600 × 2= 12% 𝑝. 𝑎.

Q16. A lead Rs.2500 to B and certain sum to C at the same time at 7% per annum simple interest. If after 4 years, a altogether receives Rs.1120 as interest from B and C find the sum lent to C.

𝐹𝑜𝑟 𝐴, 𝑃 = 𝑅𝑠. 2500, 𝑅 = 7% 𝑝. 𝑎. 𝑎𝑛𝑑, 𝑇 = 4𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. 𝑑𝑢𝑒 𝑓𝑟𝑜𝑚 𝐴 = 2500 × 7 × 4

100 = 𝑅𝑠. 700

𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝐵 & 𝐶 = 𝑅𝑠. 1120 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝐶 = 𝑅𝑠. 1120 − 700 = 𝑅𝑠. 420 𝑁𝑜𝑤 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑠𝑢𝑚 𝑟𝑒𝑛𝑡 𝑡𝑜 𝑃

𝑃 =𝑆. 𝐼.× 100

𝑅 × 𝑇

𝑆. 𝐼. = 𝑅𝑠. 420, 𝑇 = 4 𝑦𝑒𝑎𝑟𝑠, 𝑅 = 7% 𝑝. 𝑎.

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑃 = 420 × 100

7 × 4= 𝑅𝑠. 1500

𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑟𝑒𝑛𝑡 𝑏𝑦 𝐴 𝑡𝑜 𝐶 = 𝑅𝑠. 1500 Q17. The simple interest on a certain sum for 3 years at 8% per annum is Rs.96 more than the simple interest on the same sum for 2 years at 9% per annum. Find the sum.

𝑆. 𝐼. (𝑖) 𝑓𝑜𝑟 𝑎 𝑠𝑢𝑚 𝑃, 𝑓𝑜𝑟; 𝑇 = 3𝑦𝑒𝑎𝑟𝑠, 𝑅 = 8% 𝑝. 𝑎.

𝑆. 𝐼. (1) = 𝑃 × 8 × 3

100… … … … … … … (𝑖)

𝑆. 𝐼. (2)𝑓𝑜𝑟 𝑎 𝑠𝑢𝑚 𝑃, 𝑓𝑜𝑟, 𝑇 = 2𝑦𝑒𝑎𝑟𝑠, 𝑅 = 9% 𝑝. 𝑎.

𝑆. 𝐼. (2) = 𝑃 × 9 × 2

100… … … … … … … … … 𝑖𝑖)

𝑆. 𝐼. (1) = 𝑆. 𝐼. (2) + 96 𝑃 × 24

100=

𝑃 × 18

100× 96

𝑂𝑅

𝑃 (24

100−

18

100) = 96

𝑂𝑅

𝑃 ×6

100= 96

𝑂𝑅

𝑃 =96 × 100

6= 𝑅𝑠. 1600

𝑇ℎ𝑒 𝑠𝑢𝑚 𝑅𝑠. 1600

Q18. Two equal sums of money were lent at simple interest at 11% p.a. for 31

2 years and 4

1

2

years respectively. If the difference for two periods was Rs.412.50, find each sum.

9


𝑆. 𝐼. (1) 𝑓𝑜𝑟 𝑠𝑢𝑚 = 𝑃, 𝑅, = 11%, 𝑇 = 31

2𝑦𝑒𝑎𝑟

𝑆. 𝐼. (2 )𝑖𝑛 𝑡ℎ𝑒 𝑆. 𝐼. 𝑓𝑜𝑟 𝑠𝑢𝑚 = 𝑃, 𝑅 = 11%, 𝑇 = 41

2 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. (1) =𝑃 × 11 × 3

12

100

𝑆. 𝐼. (2) =𝑃 × 11 × 4

12

100

𝐺𝑖𝑣𝑒𝑛 𝑆. 𝐼. (2) − 𝑆. 𝐼. (1) = 𝑅𝑠. 412.5 𝑂𝑅

𝑃 × 11 × 4

12

100−

𝑃 × 11 × 312

100= 412.5

𝑂𝑅

𝑃 × 11

100= 412.5

𝑂𝑅

𝑃 =412.5 × 100

11

= 𝑅𝑠. 3750 𝑇ℎ𝑒 𝑒𝑞𝑢𝑎𝑙 𝑠𝑢𝑚𝑠 𝑎𝑟𝑒 𝑅𝑠. 3750 𝑒𝑎𝑐ℎ

Q19. Divide Rs.6000 into two parts so that the simple interest on the first pat for 9 months at 12% per annum is equal to the simple interest on the second part for 1 ½ years at 10% per annum.

𝐿𝑒𝑡 𝑅𝑠. 𝑥 𝑏𝑒 𝑜𝑛𝑒 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑚, 𝑅𝑠. (6000 − 𝑥) 𝑏𝑒 𝑜𝑡ℎ𝑒𝑟 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑠𝑢𝑚.

𝑆. 𝐼. (1)𝑖𝑠 𝑡ℎ𝑒 𝑆. 𝐼. 𝑜𝑛 𝑃 = 𝑅𝑠. 𝑥, 𝑓𝑜𝑟 𝑇 = 9𝑚𝑜𝑛𝑡ℎ𝑠 =3

4 𝑦𝑒𝑎𝑟𝑠, 𝑎𝑡 𝑅 = 12%

𝑆. 𝐼. (1) =𝑥 × 12 ×

34

100… … … … … … … … … 𝑖)

𝑆. 𝐼. (2) 𝑖𝑠 𝑆. 𝐼. 𝑜𝑛 𝑅𝑠. ( 6000 − 𝑥) 𝑓𝑜𝑟 𝑅 = 10%, 𝑇 = 11


𝑆. 𝐼. (2) =(6000 − 𝑥) × 10 × 1

12

100

𝑆. 𝐼. 1 = 𝑆. 𝐼. 2

𝑥 × 12 ×34

100=

(6000 − 𝑥) × 10 × 112

100

𝑂𝑅 𝑥 × 9 = (6000 − 𝑥) × 15

𝑂𝑅 9𝑥 + 15𝑥 = 6000 × 15

𝑂𝑅 24𝑥 = 6000 × 15

10


𝑂𝑅

𝑥 =6000 × 15

24= 𝑅𝑠. 3750

𝑇ℎ𝑒 𝑇𝑤𝑜 𝑃𝑎𝑟𝑡𝑠 𝑎𝑟𝑒; 𝑅𝑠. 3750 𝑎𝑛𝑑 𝑅𝑠. 2250

Q20. Divide a sum of Rs.13500 into two parts such that if one part be lent at 81

3% per

annum for 2 years 9 months and the other at 71

2% per annum for 1 years 8 months, the total

interest received is Rs.2375. 𝐿𝑒𝑡 𝑅𝑠. 𝑥 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑚. 𝑇ℎ𝑒𝑛 𝑅𝑠. (13500 − 𝑥) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑚.

𝑆. 𝐼. (1)𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑜𝑓 𝑃 = 𝑥, 𝑅 = 81

3% ,

𝑇 = 2 𝑦𝑒𝑎𝑟𝑠 9 𝑚𝑜𝑛𝑡ℎ𝑠 = 23


𝑆. 𝐼. (1) =𝑥 × 8

13

× 234

100=

253

×114

100𝑥 =

275

1200𝑥

𝑆. 𝐼. (2) 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑜𝑛 𝑃 = 13500 − 𝑥,

𝑅 = 71

2% 𝑝. 𝑎. 𝑇 = 1 𝑦𝑒𝑎𝑟 8 𝑚𝑜𝑛𝑡ℎ𝑠 = 1

1


𝑆. 𝐼. 2 =(13500 − 𝑥) × 7

12

× 123

100= (13500 − 𝑥)

75

600

𝐺𝑖𝑣𝑒𝑛, 𝑆. 𝐼. (1) + 𝑆. 𝐼. (2) = 2375 275𝑥

1200+ (13500 − 𝑥)

75

600= 2375

𝑂𝑅 275𝑥 + 150(13500 − 𝑥) = 2375 × 1200

𝑂𝑅 275𝑥 − 150𝑥 = 2375 × 1200 − 13500 × 150

𝑂𝑅 125𝑥 = 285000 − 2025000

= 825000 𝑂𝑅

𝑥 = 6600 𝑇ℎ𝑒 𝑜𝑛𝑒 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑠𝑢𝑚 𝑖𝑠 𝑅𝑠. 6600 𝑎𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑 𝑝𝑎𝑟𝑡 𝑅𝑠. 6900

Q21. In what time will a sum of money lent at8 11

3% simple interest become 4 times of

itself. 𝐿𝑒𝑡 𝑃 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑙𝑒𝑛𝑡.

𝐴𝑚𝑜𝑢𝑛𝑡 = 4𝑃, 𝑅 = 81

3%

𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑇. = 𝑆. 𝐼. = 4𝑃 − 𝑃

= 3𝑃

11


3𝑃 =𝑃 × 8

13

× 𝑇

100

𝑂𝑅

3 =8

13

× 𝑇

100

𝑇 =300

813

= 300 × 3

25=

900

5

= 36 𝑦𝑒𝑎𝑟𝑠 Q22. A certain sum of money lent out at 6 2/3% p.a. produced the same simple interest in 6

years as Rs.3200 lent out at 82

5 % p.a. for 7 years. Find the sum

𝐿𝑒𝑡 𝑃 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑙𝑒𝑛𝑡.

𝑅 = 62

3%, 𝑇 = 6 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. (1) 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑖𝑚𝑝𝑙𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑

𝑆. 𝐼. (1) =𝑃 × 6

23

× 6

100=

40

100𝑃 =

4

10𝑃

𝐿𝑒 𝑆. 𝐼. (2) 𝑏𝑒 𝑡ℎ𝑒

𝑆. 𝐼. 𝑓𝑜𝑟 𝑃 = 3200, 𝑅 = 82

5%, 𝑇 = 7% 𝑝. 𝑎.

𝑆. 𝐼. (2) =3200 × 8

25

× 7

100

= 1881.6 𝐺𝑖𝑣𝑒𝑛, 𝑆. 𝐼. (1) = 𝑆. 𝐼. (2) 4

10𝑃 = 1881.6

𝑃 =1881.6 × 10

4= 4704

𝑇ℎ𝑒 𝑠𝑢𝑚 𝑙𝑒𝑛𝑡 𝑖𝑠 = 𝑅𝑠. 4704

Q23. Naveen and Praveen borrowed Rs.42000 and Rs.55000 respectively for 31

2 years at the

same rate of interest. If Praveen has to pay Rs.3640 more than Naveen, find the rate of interest. 𝐿𝑒𝑡 𝑅 𝑏𝑒 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡

𝑆. 𝐼. (1) 𝑏𝑒 𝑡ℎ𝑒 𝑆. 𝐼. 𝑝𝑎𝑖𝑑 𝑏𝑦 𝑁𝑎𝑣𝑒𝑒𝑛, 𝑓𝑜𝑟 𝑃 = 𝑅𝑠. 42000, 𝑇 = 31


7

5 𝑦𝑒𝑎𝑟𝑠,

𝑆. 𝐼. (1) =42000 × 𝑅 ×

72

100= 210 × 7𝑅

𝑆. 𝐼. (2)𝑏𝑒 𝑡ℎ𝑒 𝑆. 𝐼. 𝑝𝑎𝑖𝑑 𝑏𝑦 𝑃𝑟𝑎𝑣𝑒𝑒𝑛 𝑓𝑜𝑟 𝑃 = 𝑅𝑠. 55000, 𝑇 = 31/2 𝑦𝑒𝑎𝑟𝑠 = 7/5 𝑦𝑒𝑎𝑟𝑠,

12


𝐺𝑖𝑣𝑒𝑛 𝑆. 𝐼. (2) − 𝑆. 𝐼. (1) = 3640 𝑖. 𝑒. 275 × 7𝑅 − 210 × 7𝑅 = 3640

𝑂𝑅 (275 − 210) × 7𝑅 = 3640

𝑂𝑅 65 × 7𝑅 = 3640

𝑂𝑅

𝑅 =3640

65 × 7= 8% 𝑝. 𝑎.

Q24. A sum of money was put at simple interest at a certain rate for 2 years. If this sum had been put at 3% higher rate, it would have earned Rs.720 more as interest. Find the sum.

𝐿𝑒𝑡 𝑃 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑙𝑒𝑛𝑡 𝑆. 𝐼. (1) 𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑓𝑜𝑟 𝑠𝑢𝑚 = 𝑃, 𝑅𝑎𝑡𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑅 𝑎𝑛𝑑 𝑇 = 2 𝑦𝑒𝑎𝑟𝑠,

𝑆. 𝐼. (1) =𝑃 × 𝑅 × 2

100=

𝑃 × 𝑅

50

𝐿𝑒𝑡 𝑆. 𝐼. (2) 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑓𝑜𝑟 𝑠𝑢𝑚 𝑃 𝑓𝑜𝑟 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 (𝑅 + 3)

𝑆. 𝐼. (2) =𝑃 × (𝑅 + 3) × 2

100

𝐺𝑖𝑣𝑒𝑛, 𝑆. 𝐼. (2) − 𝑆. 𝐼. (1) = 𝑅𝑠. 720 𝑃 × (𝑅 + 3) × 2

100−

𝑃 × 𝑅 × 2

100= 720

𝑃 × 𝑅 × 2

100+

𝑃 × 3 × 2

100−

𝑃 × 𝑅 × 2

100= 720

𝑂𝑅

6𝑃

100= 720

𝑂𝑅

𝑃 =720 × 100

6

= 12000 𝑆𝑢𝑚 𝑙𝑒𝑛𝑡 𝑖𝑠 𝑅𝑠. 12000

Q25. A sum of money invested at 6% p.a. simple interest for a certain period of time yield Rs.960 as interest. If this sum had been invested for 5 years more, it would have yielded Rs.2160 as interest. Find the sum.

𝐿𝑒𝑡 𝑠𝑢𝑚 𝑙𝑒𝑛𝑡 𝑖𝑠 𝑅𝑠. 𝑃, 𝑅 = 6%, 𝑇 𝑦𝑒𝑎𝑟𝑠

𝑆. 𝐼. (1) =𝑃 × 6 × 𝑇

100

𝐺𝑖𝑣𝑒𝑛, 𝑆. 𝐼. (1) = 𝑅𝑠. 960 𝑃 × 6 × 𝑇

100= 960

13


𝑂𝑅 𝑃 × 𝑇 = 16000

𝐹𝑜𝑟 𝑠𝑢𝑚 = 𝑅𝑠. 𝑃, 𝑅 = 6%, 𝑇 = 5 𝑦𝑒𝑎𝑟𝑠, 𝑆. 𝐼. (2) 𝑖𝑠 𝑡ℎ𝑒 𝑆. 𝐼.

𝑆. 𝐼. (2) =𝑃 × 6 × (𝑇 + 5)

100= 2160

𝑂𝑅 𝑃 × 6 × 𝑇

100+

𝑃 × 6 × 5

100= 2160

𝐹𝑟𝑜𝑚 𝑒𝑞. (1), 𝑃 × 𝑇 = 16000

16000 × 6

100+

𝑃 × 30

100= 2160

𝑃 × 30

100= 2160 − 960 = 1200

𝑃 =1200 × 100

30

= 4000 𝑇ℎ𝑒 𝑠𝑢𝑚 = 𝑅𝑠. 4000

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Class 8: Simple Interest and Compound Interest – …...Class 8: Simple Interest and Compound Interest – Exercise 15A Q1. Find the simple interest and amount on: i. Rs. 4500 for