WBBSE Solutions For Class 10 Maths Chapter 2 Simple Interest Exercise 2.1
Question 1:
1. Solution: Interest on Rs. 100 for 1 year is Rs. 5.
∴ Interest on Rs. 600 for 1 year Rs. \(\frac{5}{100} \times 600=30\).
∴ Interest = Rs. 30
2. Solution: Interest on Rs. 100 for 1 year is Rs. \(4 \frac{1}{2}=\text { Rs. } \frac{9}{2}\)
Interest on Rs. 1800 for 1 year \(\text { Rs. } \frac{9}{2} \times \frac{1800}{100}=\text { Rs: } 81\)
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Question 2. In one year, if Shraboni would get Rs. 60 as interest at the rate of simple interest 5% per annum in the bank, then how much amount would she de posit, let us calculate and write it.
Solution: Rs. 5 is the interest for 1 year when the principal is Rs. 100.
Rs. 1 is the interest for 1 year when the principal is Rs. \(\frac{100}{5}\)
Rs. 60 is the interest for 1 year when the principal is Rs. \(\frac{100}{5} \times 60=\text { Rs. } 1200\)
∴ Principal = Rs. 1200.
Question 3:
1. Solution: Rs. 6 is the interest for 1 year when the principle is Rs. 100.
Rs. 1 is the interest for 1 year when the principal is Rs. \(\frac{100}{6}\)
Rs. 90 is the interest for 1 year when the principal is \(\text { Rs. } \frac{100}{6} \times 90=\text { Rs. } 1500\)
∴ Principal = Rs. 1500.
2. Solution: Rs. 3.5 is the interest for 1 year when the principal is Rs. 100
Rs. 59.50 is the interest for 1 year when the principal = Rs. \(\frac{100}{3.5} \times 59.50=\text { Rs. } 1700\)
∴ Principal = Rs. 1700.
Question 4:
1. Solution: Interest (I) = \(\frac{\text { Principal }(P) \times \text { Rate }(r) \times \text { time }(t)}{100}\)
= \(\text { Rs. } \frac{500 \times \frac{25}{4} \times 3}{100}\)
= \(5 \times \frac{25}{4} \times 3 \text { Rs. } 93.75\)
Amount (P + I) = Rs. (500 + 93.75) = Rs. 593.75
2. Solution: Here P = R Rs. 146,
r = \(2 \frac{1}{2} \%=\frac{5}{2} \%\)
t = 1 day
= \(\frac{1}{365} \text { Interest }=\frac{146 \times \frac{5}{2} \times \frac{1}{365}}{100}\)
= Rs. \(\frac{1}{100}\) = Rs. 0.01 Amount
= Rs. (146 + 0.01) = 146.01
3. Solution: Here P = Rs. 4565;
r = 4%;
t = 2 years 6 months
= 2 \(\frac{1}{2}\) years.
= \(\frac{5}{2}\) years.
Interest = Rs. \(\frac{4565 \times 4 \times \frac{5}{2}}{100}\)
= Rs. \(\frac{4565 \times 10}{100}\)
= Rs. 456.50 Amount = Rs. (4565 + 456.50)
= Rs. 5021.50
Question 5: Let us write in the blank calculate it
1. Solution: Here P = \(\frac{\mid \times 100}{r \times t}\)
Here I = Rs. 72
r = \(4 \frac{1}{2} \%=\frac{9}{2} \%\)
t = 4 years.
= \(\text { Rs. } \frac{72 \times 100}{\frac{9}{2} \times 4}\)
= \(\text { Rs. } \frac{72 \times 100}{18}\)
= Rs. 400
2. Solution: Here P = \(\frac{I \times 100}{r \times t}\)
= Here I = Rs. 1
r = 5%
t = 1 day = \(\frac{1}{365}\)
= \(\frac{1 \times 100}{5 \times \frac{1}{365}}\)
= Rs. 73 x 100
= Rs. 7300.
Question 6:
| Principal | Time | Rate of simple interest per annum | Principal with interest |
| Rs.6400 | 4 ½% | Rs.1008 | |
| Rs.500 | 5% | 6% |
Solution: 1. Time = \(\frac{\text { Interest } \times 100}{\text { Principal } \times \text { rate }}\)
Here, Interest = Rs. 1008
Rate = \(4 \frac{1}{2}\)% = \(\frac{9}{2}\)%
Principal = Rs. 6,400
= \(\frac{1008 \times 100}{6400 \times \frac{9}{2}}\)
Time = \(\frac{1008}{32 \times 9}\)
= 3.5 years
= \(3 \frac{1}{2}\) years.
2. Solution: Time = \(\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Rate }}\)
Here, Interest = Rs. 50
Rate = 5%
Principal = Rs. 500
= \(\frac{50 \times 100}{500 \times 5}\)
= 2 years
∴ Time = 2 years.
Question 7. Let us determine the rate of simple interest in percent per annum if the interest of Rs. 500 in 4 yrs. is 100.
Solution:
1st process:
Principal = Rs. 500,
Interest = Rs. 100,
Time = 4 years,
Rate = ?
Interest on Rs. 500 for 4 yrs is 100.
Interest on Rs. 100 for 1 yrs is \(\frac{100}{500} \times 100\)
Interest on Rs. 100 for 1 yr is \(\frac{100 \times 100}{500 \times 4}=5\)
∴ Rate = 5%
Question 8. By calculating let us write the rate of simple interest in percent per annum when the principal with interest of Rs. 910 in 2 yrs. 6 months will be Rs. 955.50.
Solution: Here, Principal = Rs. 910
Interest = Rs. (955.50 – 910)
= Rs. 45.50
Time = 2 years 6 months
= \(2 \frac{1}{2} \mathrm{yrs}\)
= \(\frac{5}{2}\) yrs
∴ Rate = \(\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}\)
= \(\frac{45.50 \times 100}{910 \times \frac{5}{2}}\)
= \(\frac{4550 \times 2}{910 \times 5}=2 \%\)
∴ Rate = 2%.
Question 9. The amount (principal along with interest) of same money becomes Rs. 496 in 3 yrs. and Rs. 560 in 5 yrs. at the same rate of simple interest in percent per annum. By calculating, let us write the principal and the rate of simple interest in percent per annum.
Solution: Principal + Interest for 5 years = Rs. 560
& Principal + Interest for 3 years = Rs. 496
∴ Interest for 2 years = Rs. 64
∴ Interest for 1 year = Rs. \(\frac{64}{2}\) = 32
& Interest for 3 years = Rs. 32 x 3 = Rs. 96
∴ Principal + Interest for 3 years = Rs. 496
Interest for 3 years = Rs. 96
∴ Principal = Rs.(496 – 96) = Rs. 400
Rate = \(\frac{\text { Interest } \times 100}{\text { Principal } \times \text { Time }}=\frac{96 \times 100}{400 \times 3}=8\)
∴ Rate = 8%.
Question 10. Bimalkaku deposited Rs. 1,87,500 for his son of 12 yrs. age and daughter of 14 yrs. age in the bank at the rate of simple interest per annum in such away that both of them will get equal principal along with interest after reaching their 18 yrs. of age. Let us calculate the money he had deposited in the bank for each of his son and daughter.
Solution: Let Bimal babu deposited Rs. x for his 12 yrs old son & Rs. y for his 14 yrs old daughter.
∴ x + y = 187500
When his son will be 18 yrs, i.e., after 6 yrs, h son’s amount = Rs. \(\left(x+\frac{x \times 6 \times 5}{100}\right)\)
= \(\frac{10 x+3 x}{5}=\text { Rs. } \frac{13 x}{10}\)
Again, when his daughter will be 18 years, i.e., after 4 yrs.
His daughter’s amount = Rs \(\left(y+\frac{y \times 4 \times 85}{100}\right)=\frac{5 y+y}{, 5}=\frac{6 y}{5}\)
According to the problem,
\(\frac{13 x}{10}=\frac{6 y}{5}\)or, 13x – 12y = 0
x + y = 187500
or, 12x + 12y = 12 x 187500 = 2250000
& 13x – 12y = 0
Adding, 25x = 2250000
x = \(\frac{2250000}{25}=90000\)
y = 187500 – 90000= 97500
Question 11. Jayanta deposits Rs. 1000 on the first day of every month in monthly savings scheme. In the bank, if the rate of simple interest is per annum, then let us calculate the amount Jayanta will get at the end of 6 months.
Solution: Interest on Rs. 1000 at the rate 5% for each month
= Rs. \(\left[\frac{1000 \times 5 \times 6}{100 \times 12}+\frac{1000 \times 5 \times 5}{100 \times 12}+\frac{1000 \times 5 \times 4}{100 \times 12}+\frac{1000 \times 5 \times 3}{100 \times 12}+\frac{1000 \times 5 \times 2}{100 \times 12}+\frac{1000 \times 5 \times 1}{100 \times 12}\right]\)
= Rs. \(\frac{1000 \times 5}{100 \times 12}[6+5+4+3+2+1]=\text { Rs. } \frac{50}{12} \times 21=\text { Rs. } \frac{175}{2}=87.50\)
Amount = Rs.(1000 x 6 + 87.50) = Rs. (6000 + 87.50)
= Rs. 6087.50
Question 12. Soma aunti deposits Rs. 6,20,000 in such a way in three banks at the rate of simple interest of per annum for 2 yrs. and . respectively so that the total interests in the 3 banks are equal. Let us calculate the money deposited by Soma aunti in each of the three banks.
Solution: Let Soma aunti deposited Rs. p, Rs. x & Rs. y in 3 banks respectively at the rate of 5% S.I.
∴ p + x + y = 620000
Now, the interest from 3 banks are
\(\frac{p \times 5 \times 2}{100}=\frac{x \times 5 \times 3}{100}=\frac{y \times 5 \times 5}{100}\)2p = 3x = 5y = k(let)
∴ \(p=\frac{k}{2} ; x=\frac{k}{3} ; y=\frac{k}{5}\)
As, p + x + y = 620000
or, \(\frac{k}{2}+\frac{k}{3}+\frac{k}{5}=620000\)
or, \(\frac{15 k+10 k+6 k}{30}=62,0000\)
∴ \(\frac{31 k}{30}=620000\)
\(k=\frac{620000 \times 30}{31}=60,0000\)∴ \(p=\frac{k}{2}=\frac{60,0000}{2}=300,000\)
\(x=\frac{k}{3}=\frac{60,0000}{3}=200,000\) \(y=\frac{k}{5}=\frac{60,0000}{5}=120,0000\)∴ She deposited Rs. 300,000; Rs. 200,000 and Rs. 1200,000respectively in 3 banks.