ICSE Class 7 Mathematics · Selina Solutions Concise Mathematics Class 7 ICSE · 37 solved questions
Find the S.I. and the amount on:
₹ 150 for 4 years at 5% per year.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 150, R = 5% and T = 4 years.
S.I. = $\frac{150 \times 5 \times 4}{100}$
= $\frac{3000}{100}$
= ₹ 30
Amount = P + S.I. = 150 + 30 = ₹ 180.
Hence, S.I. = ₹ 30 and Amount = ₹ 180.
Find the S.I. and the amount on:
₹ 350 for $3\frac{1}{2}$ years at 8% p.a.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 350, R = 8% and T = $3\frac{1}{2} \text{ years } = \frac{7}{2}$ years.
S.I. = $\frac{350 \times 8 \times 7}{100 \times 2}$
= $\frac{19600}{200}$
= ₹ 98
Amount = P + S.I. = 350 + 98 = ₹ 448.
Hence, S.I. = ₹ 98 and Amount = ₹ 448.
Find the S.I. and the amount on:
₹ 620 for 4 months at 8 paise per rupee per month.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 620, R = 8 paise per rupee per month = 8% per month and T = 4 months.
Since the rate is per month, the time is taken in months.
S.I. = $\frac{620 \times 8 \times 4}{100}$
= $\frac{19840}{100}$
= ₹ 198.40
Amount = P + S.I. = 620 + 198.40 = ₹ 818.40.
Hence, S.I. = ₹ 198.40 and Amount = ₹ 818.40.
Find the S.I. and the amount on:
₹ 3,380 for 30 months at $4\frac{1}{2}$% p.a.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 3,380, R = $4\frac{1}{2}$% = $\frac{9}{2}$% and T = 30 months = $\frac{30}{12}$ years = $\frac{5}{2}$ years.
S.I. = $\frac{3380 \times 9 \times 5}{100 \times 2 \times 2}$
= $\frac{152100}{400}$
= ₹ 380.25
Amount = P + S.I. = 3,380 + 380.25 = ₹ 3,760.25.
Hence, S.I. = ₹ 380.25 and Amount = ₹ 3,760.25.
Find the S.I. and the amount on:
₹ 600 from July 12 to Dec. 5 at 10% p.a.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 600, R = 10% p.a.
The starting date (July 12) is not included and the last date (Dec. 5) is included.
Time = (31 − 12) + 31 + 30 + 31 + 30 + 5 days
= 19 + 31 + 30 + 31 + 30 + 5 = 146 days = $\frac{146}{365}$ years = $\frac{2}{5}$ years.
S.I. = $\frac{600 \times 10 \times 2}{100 \times 5}$
= $\frac{12000}{500}$
= ₹ 24
Amount = P + S.I. = 600 + 24 = ₹ 624.
Hence, S.I. = ₹ 24 and Amount = ₹ 624.
Find the S.I. and the amount on:
₹ 850 from 10th March to 3rd August at $2\frac{1}{2}$% p.a.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 850, R = $2\frac{1}{2}$% = $\frac{5}{2}$% p.a.
The starting date (10th March) is not included and the last date (3rd August) is included.
Time = (31 − 10) + 30 + 31 + 30 + 31 + 3 days
= 21 + 30 + 31 + 30 + 31 + 3 = 146 days
= $\frac{146}{365} \text { years }$
= $\frac{2}{5}$ years.
S.I. = $\frac{850 \times 5 \times 2}{100 \times 2 \times 5}$
= $\frac{8500}{1000}$
= ₹ 8.50
Amount = P + S.I. = 850 + 8.50 = ₹ 858.50.
Hence, S.I. = ₹ 8.50 and Amount = ₹ 858.50.
Find the S.I. and the amount on:
₹ 225 for 3 years 9 months at 16% p.a.
By formula,
S.I. = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ and Amount = Principal + S.I.
Given,
P = ₹ 225, R = 16% and T = 3 years 9 months = $3\frac{9}{12} \text { years } = \frac{15}{4}$ years.
S.I. = $\frac{225 \times 16 \times 15}{100 \times 4}$
= $\frac{54000}{400}$
= ₹ 135
Amount = P + S.I. = 225 + 135 = ₹ 360.
Hence, S.I. = ₹ 135 and Amount = ₹ 360.
On what sum of money does the S.I. for 10 years at 5% become ₹ 1,600?
S.I. = ₹ 1,600, T = 10 years and R = 5%.
$$\begin{aligned}\text{ P }= \frac{\text{S.I.} \times 100}{\text{R} \times \text{T}}\\[1em] = \frac{1600 \times 100}{5 \times 10}\\[1em] = \frac{160000}{50}\\[1em] = ₹\ 3,200\end{aligned}$$
Hence, the required sum = ₹ 3,200.
Find the time in which ₹ 2,000 will amount to ₹ 2,330 at 11% p.a.
P = ₹ 2,000, Amount = ₹ 2,330 and R = 11% p.a.
I = Amount − P = 2,330 − 2,000 = ₹ 330.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{330 \times 100}{2000 \times 11}\\[1em] = \frac{33000}{22000}\\[1em] = \frac{3}{2} = 1\frac{1}{2}\ \text{years}\end{aligned}$$
Hence, the required time = $1\frac{1}{2}$ years.
In what time will a sum of money double itself at 8% p.a.?
Let the sum (principal) be ₹ P and R = 8% p.a.
When the money doubles itself, Amount = 2P.
So, I = Amount − P = 2P − P = ₹ P.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{\text{P} \times 100}{\text{P} \times 8}\\[1em] = \frac{100}{8}\\[1em] = \frac{25}{2} = 12\frac{1}{2}\ \text{years}\end{aligned}$$
Hence, the required time = $12\frac{1}{2}$ years.
In how many years will ₹ 870 amount to ₹ 1,044, the rate of interest being $2\frac{1}{2}$% p.a.?
P = ₹ 870, Amount = ₹ 1,044 and R = $2\frac{1}{2}$% = $\frac{5}{2}$% p.a.
I = Amount − P = 1,044 − 870 = ₹ 174.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{174 \times 100 \times 2}{870 \times 5}\\[1em] = \frac{34800}{4350}\\[1em] = 8\ \text{years}\end{aligned}$$
Hence, the required time = 8 years.
Find the rate percent, if the S.I. on ₹ 275 in 2 years is ₹ 22.
P = ₹ 275, T = 2 years and S.I. = ₹ 22.
R = $\frac{\text{S.I.} \times 100}{\text{P} \times \text{T}}$%
$= \frac{22 \times 100}{275 \times 2}$%
$= \frac{2200}{550}$%
= 4%
Hence, the required rate = 4% p.a.
Find the sum which will amount to ₹ 700 in 5 years at 8% p.a.
Let the sum be ₹ P, R = 8% p.a. and T = 5 years.
I = $\frac{\text{P} \times 8 \times 5}{100} = \frac{40\text{P}}{100} = \frac{2\text{P}}{5}$.
Amount = P + I
$$\begin{aligned}\Rightarrow 700 = \text{P} + \frac{2\text{P}}{5}\\[1em] \Rightarrow 700 = \frac{7\text{P}}{5}\\[1em] \Rightarrow \text{P} = \frac{700 \times 5}{7} = ₹\ 500\end{aligned}$$
Hence, the required sum = ₹ 500.
What is the rate of interest, if ₹ 3,750 amounts to ₹ 4,650 in 4 years?
P = ₹ 3,750, Amount = ₹ 4,650 and T = 4 years.
I = Amount − P = 4,650 − 3,750 = ₹ 900.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{900 \times 100}{3750 \times 4}$%
$= \frac{90000}{15000}$%
= 6%
Hence, the required rate = 6% p.a.
In 4 years, ₹ 6,000 amounts to ₹ 8,000. In what time will ₹ 525 amount to ₹ 700 at the same rate?
First, find the rate of interest.
P = ₹ 6,000, Amount = ₹ 8,000 and T = 4 years.
I = Amount − P = 8,000 − 6,000 = ₹ 2,000.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{2000 \times 100}{6000 \times 4}$%
$= \frac{200000}{24000}$%
= $\frac{25}{3}$%
Now, for the second case, P = ₹ 525, Amount = ₹ 700 and R = $\frac{25}{3}$%.
I = Amount − P = 700 − 525 = ₹ 175.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{175 \times 100 \times 3}{525 \times 25}\\[1em] = \frac{52500}{13125}\\[1em] = 4\text{ years}\end{aligned}$$
Hence, the required time = 4 years.
The interest on a sum of money at the end of $2\frac{1}{2}$ years is $\frac{4}{5}$ of the sum. What is the rate percent?
Let the sum (principal) be ₹ P and T = $2\frac{1}{2} \text { years }= \frac{5}{2}$ years.
I = $\frac{4}{5}$ of the sum = ₹ $\frac{4\text{P}}{5}$.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{\frac{4\text{P}}{5} \times 100 \times 2}{\text{P} \times 5}$%
$= \frac{4 \times 100 \times 2}{5 \times 5}$%
$= \frac{800}{25}$%
= 32%
Hence, the required rate = 32% p.a.
What sum of money lent out at 5% for 3 years will produce the same interest as ₹ 900 lent out at 4% for 5 years?
First, find the interest on ₹ 900 at 4% for 5 years.
$$\begin{aligned}\text{ I }= \frac{900 \times 4 \times 5}{100}\\[1em] = \frac{18000}{100}\\[1em] = ₹\ 180\end{aligned}$$
Now, let the required sum be ₹ P, lent at R = 5% for T = 3 years, giving I = ₹ 180.
$$\begin{aligned}\text{ P }= \frac{\text{I} \times 100}{\text{R} \times \text{T}}\\[1em] = \frac{180 \times 100}{5 \times 3}\\[1em] = \frac{18000}{15}\\[1em] = ₹\ 1,200\end{aligned}$$
Hence, the required sum = ₹ 1,200.
A sum of ₹ 1,780 becomes ₹ 2,136 in 4 years. Find:
(i) the rate of interest.
(ii) the sum that will become ₹ 810 in 7 years at the same rate of interest.
(i) P = ₹ 1,780, Amount = ₹ 2,136 and T = 4 years.
I = Amount − P = 2,136 − 1,780 = ₹ 356.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{356 \times 100}{1780 \times 4}$%
$= \frac{35600}{7120}$%
= 5%
Hence, the rate of interest is 5% p.a.
(ii) Let the required sum be ₹ P, R = 5% and T = 7 years.
I = $\frac{\text{P} \times 5 \times 7}{100} = \frac{35\text{P}}{100} = \frac{7\text{P}}{20}$.
Amount = P + I
$$\begin{aligned}\Rightarrow 810 = \text{P} + \frac{7\text{P}}{20}\\[1em] \Rightarrow 810 = \frac{27\text{P}}{20}\\[1em] \Rightarrow \text{P} = \frac{810 \times 20}{27} = ₹\ 600\end{aligned}$$
Hence, the required sum is ₹ 600.
A sum amounts to ₹ 2,652 in 6 years at 5% p.a. simple interest. Find:
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.
(i) Let the sum be ₹ P, R = 5% p.a. and T = 6 years.
I = $\frac{\text{P} \times 5 \times 6}{100} = \frac{30\text{P}}{100} = \frac{3\text{P}}{10}$.
Amount = P + I
$$\begin{aligned}\Rightarrow 2652 = \text{P} + \frac{3\text{P}}{10}\\[1em] \Rightarrow 2652 = \frac{13\text{P}}{10}\\[1em] \Rightarrow \text{P} = \frac{2652 \times 10}{13} = ₹\ 2,040\end{aligned}$$
Hence, the sum is ₹ 2,040.
(ii) When the sum doubles itself, I = Amount − P = 2P − P = ₹ P and R = 5%.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{\text{P} \times 100}{\text{P} \times 5}\\[1em] = \frac{100}{5}\\[1em] = 20\ \text{years}\end{aligned}$$
Hence, the time in which the sum will double itself is 20 years.
P and Q invest ₹ 36,000 and ₹ 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets ₹ 3,080 more interest than Q, find the rate of interest.
Let the rate of interest be R% p.a. and T = 4 years.
For P : Principal = ₹ 36,000, Time = 4 years
I = $\frac{36000 \times \text{R} \times 4}{100} = ₹\ 1,440\text{R}$.
For Q : Principal = ₹ 25,000, Time = 4 years
I = $\frac{25000 \times \text{R} \times 4}{100} = ₹\ 1,000\text{R}$.
Given, P gets ₹ 3,080 more interest than Q.
⇒ 1,440 R - 1,000 R = 3,080
⇒ 440 R = 3,080
$\Rightarrow \text{R} = \frac{3080}{440}$
⇒ R = 7
⇒ R% = 7%
Hence, the required rate of interest = 7% p.a.
A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.
Let the sum (money lent) be ₹ P and T = 5 years.
Interest (I) = one-fourth of the money lent = ₹ $\frac{\text{P}}{4}$.
R% = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
R% $= \frac{\frac{\text{P}}{4} \times 100}{\text{P} \times 5}$%
R% $= \frac{100}{4 \times 5}$%
R% $= \frac{100}{20}$%
R% = 5%
R = 5
Hence, R = 5 and rate of interest = 5%.
The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.
Let the sum be ₹ P and T = 5 years.
I = 30% of the sum = $\frac{30}{100} \times \text{P} = \frac{3\text{P}}{10}$.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{\frac{3\text{P}}{10} \times 100}{\text{P} \times 5}$%
$= \frac{3 \times 100}{10 \times 5}$%
$= \frac{300}{50}$%
= 6%
Hence, the required rate of interest = 6% p.a.
Sangeeta deposited ₹ 1,000 in a bank paying interest at the rate of 10% per year. The amount in 2 years is:
P = ₹ 1,000, R = 10% and T = 2 years.
I = $\frac{1000 \times 10 \times 2}{100}$ = ₹ 200.
Amount = P + I = 1,000 + 200 = ₹ 1,200.
Hence, option 3 is the correct option.
Mercy invested ₹ 2,000 at 12% per year and an equal sum at 10% per year. Thus, total income from interest in one year is:
Interest on ₹ 2,000 at 12% for 1 year = $\frac{2000 \times 12 \times 1}{100}$ = ₹ 240.
Interest on ₹ 2,000 at 10% for 1 year = $\frac{2000 \times 10 \times 1}{100}$ = ₹ 200.
Total income from interest = 240 + 200 = ₹ 440.
Hence, option 1 is the correct option.
Mercy invested ₹ 5,000 at 12% per year and an equal sum at 10% per year. The difference of incomes from two investments in 2 years is:
Interest on ₹ 5,000 at 12% for 2 years = $\frac{5000 \times 12 \times 2}{100}$ = ₹ 1,200.
Interest on ₹ 5,000 at 10% for 2 years = $\frac{5000 \times 10 \times 2}{100}$ = ₹ 1,000.
Difference of incomes = 1,200 − 1,000 = ₹ 200.
Hence, option 3 is the correct option.
A sum of money invested for 10 years earns interest equal to the sum itself. The rate of interest is:
Let the sum be ₹ P and T = 10 years.
I = the sum itself = ₹ P.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{\text{P} \times 100}{\text{P} \times 10}$%
= 10%.
Hence, option 2 is the correct option.
A sum of money invested at 5% per year, earns interest equal to half of the sum invested. The time for which the sum is invested, is:
Let the sum be ₹ P and R = 5%.
I = half of the sum = ₹ $\frac{\text{P}}{2}$.
$$\begin{aligned}\text{ T } = \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{\frac{\text{P}}{2} \times 100}{\text{P} \times 5}\\[1em] = \frac{100}{2 \times 5}\\[1em] = 10 \text{ years }\end{aligned}$$
Hence, option 2 is the correct option.
A sum of money invested at 10% per annum, earns interest which is 1.2 times the sum invested. The time, for which the sum is invested, is:
Let the sum be ₹ P and R = 10%.
I = 1.2 times the sum = ₹ 1.2P.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{1.2\text{P} \times 100}{\text{P} \times 10}\\[1em] = \frac{120}{10}\\[1em] = 12\text{ years}\end{aligned}$$
Hence, option 3 is the correct option.
A sum of money invested at 10% per annum becomes 1.2 times the sum invested. The time, for which the sum is invested, is:
Let the sum be ₹ P and R = 10%.
Amount = 1.2 times the sum = ₹ 1.2P.
I = Amount − P = 1.2P − P = ₹ 0.2P.
$$\begin{aligned}\text{ T }= \frac{\text{I} \times 100}{\text{P} \times \text{R}}\\[1em] = \frac{0.2\text{P} \times 100}{\text{P} \times 10}\\[1em] = \frac{20}{10}\\[1em] = 2\text{ years}\end{aligned}$$
Hence, option 1 is the correct option.
In 8 years, a sum earns interest equal to the sum invested. The rate of interest is:
Let the sum be ₹ P and T = 8 years.
I = the sum invested = ₹ P.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{\text{P} \times 100}{\text{P} \times 8}$%
$= \frac{100}{8}$%
= 12.5%
Hence, option 4 is the correct option.
A sum of money invested at 10% per annum becomes ₹ 3,000 in 5 years. The sum invested is:
Let the sum be ₹ P, R = 10% and T = 5 years.
I = $\frac{\text{P} \times 10 \times 5}{100} = \frac{\text{P}}{2}$.
Amount = P + I
$$\begin{aligned}\Rightarrow 3000 = \text{P} + \frac{\text{P}}{2}\\[1em] \Rightarrow 3000 = \frac{3\text{P}}{2}\\[1em] \Rightarrow \text{P} = \frac{3000 \times 2}{3} = ₹\ 2,000\end{aligned}$$
Hence, option 3 is the correct option.
₹ 4,000 becomes ₹ 5,000 in 5 years. The rate of interest is:
P = ₹ 4,000, Amount = ₹ 5,000 and T = 5 years.
I = Amount − P = 5,000 − 4,000 = ₹ 1,000.
R = $\frac{\text{I} \times 100}{\text{P} \times \text{T}}$%
$= \frac{1000 \times 100}{4000 \times 5}$%
$= \frac{100000}{20000}$%
$= \frac{10}{2}$%
= 5%.
Hence, option 1 is the correct option.
Statement 1: ₹ 400 will yield an interest of ₹ 36 in 3 years at 3% p.a.
Statement 2: Rate percent is the interest paid on ₹ 100 for any period of time.
Which of the following options is correct?
Solving,
$$\begin{aligned}\text{ I }= \frac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] = \frac{400 \times 3 \times 3}{100}\\[1em] = \frac{3600}{100}\\[1em] = ₹\ 36\end{aligned}$$
So, Statement 1 is true.
Rate percent (per annum) is the interest on ₹ 100 for a specific period of time, not for any period of time.
So, Statement 2 is false.
Hence, option 3 is the correct option.
Statement 1: If both rate percent and time are equal, then simple interest on a certain sum is $\frac{16}{25}$ times of the sum.
Statement 2: The value of interest depends on (i) Sum, (ii) Rate of interest p.a. and (iii) time in years.
Which of the following options is correct?
Let the sum be ₹ P and rate = time = n.
Then I = $\frac{\text{P} \times n \times n}{100} = \frac{n^2\text{P}}{100}$.
This equals $\frac{16}{25}$ times the sum only when $\frac{n^2 \text {P}}{100} = \frac{16 \text{P}}{25}$, i.e. n = 8. For any other equal values of rate and time (say n = 5), the interest is not $\frac{16}{25}$ of the sum.
So, Statement 1 is false.
Since I = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$, the interest depends on the Sum, the Rate of interest p.a. and the time in years.
So, Statement 2 is true.
Hence, option 4 is the correct option.
Assertion (A): A sum of money doubles itself at 5% p.a. in 20 years.
Reason (R): Time = $\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Rate}}$
When a sum doubles itself, I = Principal. Taking the sum as ₹ P and R = 5%,
Time = $\frac{100 \times \text{P}}{\text{P} \times 5} = \frac{100}{5}$ = 20 years.
So, Assertion (A) is true.
Time = $\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Rate}}$ is the correct formula.
So, Reason (R) is true.
Hence, option 3 is the correct option.
Assertion (A): ₹ 250 doubles itself in 5 years at simple interest then rate of interest is 20% per month.
Reason (R): Rate % = $\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Time}}$%.
When ₹ 250 doubles itself, I = ₹ 250 and T = 5 years.
Rate = $\frac{100 \times 250}{250 \times 5} = \frac{100}{5}$ = 20% per annum (per year), not 20% per month.
So, Assertion (A) is false.
Rate % = $\frac{100 \times \text{Interest}}{\text{Principal} \times \text{Time}}$% is the correct formula.
So, Reason (R) is true.
Hence, option 2 is the correct option.
Assertion (A): The simple interest on ₹ 850 from 10th March to 2nd August at 2% p.a. is ₹ 6.80.
Reason (R): Amount is the sum of the Principal and interest on it.
The starting date (10th March) is not included and the last date (2nd August) is included.
Time = (31 − 10) + 30 + 31 + 30 + 31 + 2
= 21 + 30 + 31 + 30 + 31 + 2
= 145 days = $\frac{145}{365}$ years.
$$\begin{aligned}\text{ I } = \frac{\text{P} \times \text{R} \times \text{T}}{100} \\[1em] = \frac{850 \times 2 \times \frac{145}{365}}{100} \\[1em] = \frac{850 \times 2 \times 145}{100 \times 365}\\[1em] = \frac{246500}{36500} \\[1em] \approx ₹ 6.75\end{aligned}$$
So, Assertion (A) is false.
Amount = Principal + Interest, which is correct.
So, Reason (R) is true.
Hence, option 2 is the correct option.
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