8th Maths : Chapter 4 : Life Mathematics : Compound Interest : Exercise 4.3 : Text Book Back Exercises Questions with Answers, Solution

**Exercise
4.3**

** **

**1. Fill in the blanks:**

(i) The compound interest on ₹5000 at
12 % p.a for 2 years, compounded annually is ____________. **[Answer: ₹ 1272]**

**Solution:**

Solution 2:

Compound Interest (CI) formula is

CI = Amount − Principal

Amount = A (1 + [r/100] )^{n}

= 5000 ( 1 + 12/100 )^{2} = 5000 × (112 / 100)^{2}
= 6272

∴ CI = 6272 – 5000 = **₹
1272**

(ii) The compound interest on ₹8000 at
10 % p.a for 1 year, compounded half yearly is ____________. **[Answer: ₹ 820 ]**

**Solution:**

Solution 2:

Compound Interest (CI) = Amount − Principal

Amount = P (1 + [r/100] )^{2n} [2n as it is compounded half yearly]

*r* = 10% p.a, for half yearly *r*
= 10/2 = 5

∴ A = 8000 (1 + [5/100])^{2x1} = 8000 × (105 / 100)^{2}
= 8820

CI = Amount – Principal = 8820 − 8000 = ₹ 820

(iii) The annual rate of growth in population
of a town is 10 % . If its present population is 26620, then the population 3 years
ago was_________. **[Answer: ₹ 20,000]**

**Solution:**

Solution 2:

Rate of growth of population *r *= 10%

Present population = 26620

Let population 3 years ago be *x*

∴ Applying the formula for population growth which is similar to
compound interest,

26620 = *x* (1 + [r/100])^{3}

∴ *x* = 26620 = *x* (1 + [10/100])^{3
}= *x* (110/100)^{3}

∴ *x* = 26620 × (110/100)^{3}
= **₹ 20,000**

The population 3 years ago was
₹ 20,000

(iv) If the compound interest is calculated
quarterly, the amount is found using the formula __________. **[Answer: **[or] **A = P ( 1 + r/400 ) ^{4n}]**

**Solution:**

Quarterly means 4 times in a year.

∴ The formula for compound interest is

A = P ( 1 + r/400 )^{4n}

(v) The difference between the C.I and
S.I for 2 years for a principal of ₹5000 at the rate of interest 8 % p.a is ___________.
**[Answer: ₹ 32]**

**Solution:**

Solution 2:

Difference between S.I & C.I is given by the formula

CI − SI = P(*r*/100)^{2}

Principal (P) = 5000, *r *= 8% p.a

∴ CI − SI = 5000 (8/100)^{2}
= 5000 × 8/100 × 8/100 **= ₹ 32**

2. **Say True or False.**

(i) Depreciation value is calculated
by the formula, **[Answer: True]**

**Solution:**

Depreciation formula is P ( 1 – *r*/100 )^{n}

(ii) If the present population of a city
is P and it increases at the rate of r % p.a, then the population n years ago would
be . **[Answer: False]**

**Solution:**

Solution 2:

Let the population '*n*’
yrs ago be '*x*'

∴ Present population (P) = *x* × ( 1 + [*r*/100] )^{n}** **

∴ *x* = P / (1 + [*r*/100])^{n}** **

(iii) The present value of a machine
is ₹16800. It depreciates at 25 % p.a. Its worth after 2 years is ₹9450. **[Answer: True]**

**Solution:**

Solution 2:

Present value of machine = ₹**
**16800

Depreciation rate (*n*) =
25%

Value after 2 years = P ( 1 + *r*/100
)^{n}** = **16800 (1 –
25/100)^{2}

= 16800 × ( 1 – 1/4 )^{2} = 16800 × 3/4 × 3/4 = 9450

(iv) The time taken for ₹1000 to become
₹1331 at 20 % p.a, compounded annually is 3 years. **[Answer: False]**

**Solution:**

Solution 2:

Principal money = 1000

rate of interest = 20%

Amount (A) = 1331, applying in formula we get

A = (1 + [r/100] )^{n}

∴1331 = 1000 (1 + [20/100] )^{n}

1331/1000 = (1 + 1/5)^{n}

1331/1000 = (6/5)^{n}^{ }∴ *n* ≠ 3 (False)

(v) The compound interest on ₹16000 for
9 months at 20 % p.a, compounded quarterly is ₹2522. **[Answer: True]**

**Solution: **

Solution 2:

Principal (P) = 16000

*n *= 9 months = 9/12 years

*r* = 20% p.a

For compounding quarterly, we have to use below formula.

Amount (A) = P × ( 1 + [*r*/100] )^{4n}

Since quarterly we have to divide ‘*r*’ by 4

*r* = 20/4 = 5%

A = 16000 ( 1 + [5/100] )^{9/12x4}

= 16000 (105/100)^{ 9/12x4 }=^{ }16000^{ }(105/100)^{
9/3}

= 16000 × (21/20)^{3 }= 16000 × 21/20^{ }×^{ }21/20
× 21/20^{}

∴ Interest = A − P = 18522
− 16000 = 2522 (True)

** **

**3. Find the compound interest on ₹3200
at 2.5 % p.a for 2 years, compounded annually.**

**Solution:**

Solution 2:

Principal (P) = ₹ 3200

*r* = 2.5% p.a

*n* = 2 years comp, annually

∴ Amount (A) = (1 + [*r*/100] )^{n}

= 3200 ( 1 + [2.5/100] )^{2}

= 3200 × (1.025)^{2} = 3362

Compound Interest (CI) = Amount − Principal

= 3362 – 3200 =** ₹ 162**

** **

**4. Find the compound interest for 2 1/2
years on ₹4000 at 10 % p.a, if the interest is compounded yearly. **

**Solution:**

Solution 2:

Principal (P) = ₹ 4000

*r* = 10% p.a

Compounded yearly

*n* = 2 1/2 years. Since it is of
the form *a* *b*/*c* years

Amount (A) = (1 + [*r*/100]
)^{a} ( 1 + ([*b*/*c* × *r*] / 100) )^{1}

= 4000 × (1 + 10/100)^{2}
( 1 + [ (1/2 × 10) / 100] )^{1}

= 4000 × (110 / 100)^{2}
× (105 / 100)^{1}

= 4000 × 1.1 × 1.1 × 1.05 =
5082

∴ CI = Amount − Principal = 5082 – 4000 = 1082

** **

**5. A principal becomes ₹2028 in 2 years
at 4 % p.a compound interest. Find the principal.**

**Solution:**

Solution 2:

*n* = 2 years

*r* = rate of interest = 4% p.a

Amount A= ₹ 2028

Amount (A) = P ( 1 + [*r*/100]
)* ^{n}* (Applying in formala]

2028 = P ( 1 + [4/100] )^{2}

2028 = P (104/100)^{2}

∴ P = [ 2028 × 100 × 100] / [104 × 104] =
**₹ 1875**

** **

**6. In how many years will ₹3375 become
₹4096 at 13 1/3 % p.a if the interest is compounded half-yearly?**

**Solution:**

Solution 2:

Principal = ₹ 3375

Amount = ₹ 4096

*r* = 13 1/3 % p.a = 40/3 % p.a

Compounded half yearly *r*
= [40/3] / 2 = 20 / 3 % half yearly

Let no. of years be *n*
for compounding half yearly, formula is

A = P ( 1 + [*r*/100] )^{2n}

∴ 4096 = 3375 ( 1 + [20/3 / 100] )^{2n}

∴ 4096 / 3375 = (1 + [ 20 / (3×100) ] )^{2n} = (1 + 1/15)^{2n}

( [15 + 1] / 15)^{2n}
= (16/15)^{2n} ⇒ ∴ 4096 / 3375 = (16/15)^{2n}

Taking cubic root on both
sides.

(16/15)^{2n/3} = ^{3}√4096
/ ^{3}√3375 = 16/15

∴ 2*n*/3 = 1

∴ *n* = 3 / 2 = 1.5 years or 1 ½ years

** **

**7. Find the C.I on ₹15000 for 3 years
if the rates of interest are 15 % , 20 % and 25 % for the I, II and III years respectively.**

**Solution:**

Solution 2:

Principal (P) = ₹ 15000

rate of interest 1 (*a*) = 15% for year I

rate of interest 2 (*b*) = 20% for year II

rate of interest 3 (*c*) = 25% for year III

Formula for amount when rate of interest is different for
different years is

A = (1 + *a*/100)^{1} (1 + *b*/100)^{1}(1
+ *c*/100)^{1}

Substituting in the above formula, we get

A = 15000 (1 + 15/100) (1 + 20/100) ( 1 + 25/100)

= 15000 × 115/100 × 120/100 × 125/100 = 25,875

Compound Interest (CI) = A – P = 25,875 – 15,000 = ₹ 10,875

CI = **₹ 10,875**

** **

**8. Find the difference between C.I and
S.I on ₹5000 for 1 year at 2 % p.a, if the interest is compounded half yearly.**

**Solution:**

Solution 2:

Principal (P) = ₹ 5000

time period (*n*) = 1 yr.

Rate of interest (*r*) = 2% p.a

for half yearly *r* = 1%

Difference between CI & SI is given by the formula.

CI − SI = P (r/100)^{2n
}[for half yearly compounding]

∴ CI − SI = 5000 (1/100)^{2x1
}= 5000 × 1/100 × 1/100 = ₹ **0.50**

** **

**9. Find the rate of interest if the difference
between C.I and S.I on ₹8000 compounded annually for 2 years is ₹20.**

**Solution:**

Solution 2:

Principal (P) = ₹ 8000

time period (*n*) = 2 yrs.

rate of interest (*r*) = ?

Difference between CI & SI is given by the formula

CI − SI = P ^{(1+r/100)n}

Difference between CI & SI is given as 20

∴ 20 = 8000 × (*r*/100)^{2}

∴ (*r*/100)^{2} = 20/8000 = 1/400

Taking square root on both sides

*r*/100 = √[1/400] = 1/20

∴ *r *= 100 / 20 = **5%**

** **

**10. Find the principal if the difference
between C.I and S.I on it at 15 % p.a for 3 years is ₹1134.**

**Solution:**

Solution 2:

Rate of interest (*r*) = 15% p.a

time period (*n*) =
3 years

Difference between CI & SI is given as 1134

Principal = ? → required to find

Using formula for difference

C.I − S.I = P(*r*/100)^{2} (3 + *r*/100)

1134 = P(15/100)^{2} (3 + 15/100)

1134 = P(0.15)^{2} ([300 + 15]/100)

1134 = P (0.0225 × [315/100])

1134 = P × 0.070875

P = 1134/0.070875

P = ₹ 16,000

** **

**Objective
Type Questions**

**11. The number of conversion periods
in a year, if the interest on a principal is compounded every two months is___________.**

(A) 2

(B) 4

(C) 6

(D) 12

**[Answer: (C) 6]**

**Solution:**

Conversion period is the time period after which the interest is
added to the principal. If principal is compounded every two months then in a
year, there will be 6 (12/2) conversion periods.

**12. The time taken for ₹4400 to become
₹4851 at 10 % , compounded half yearly is _______.**

(A) 6 months

(B) 1 year

(C) 1 years

(D) 2 years

**[Answer: (B) 1 year]**

**Solution:**

Solution 2:

Principal = ₹ 4400

Amount = ₹ 4851

Rate of interest = 10% p.a

for half yearly, divide by 2,

*r* = 10/2 = 5%

Compounded half yearly, so the formula is

A = P (1 + [*r*/100])^{2n}

Substituting in the above formula, we get

4851 = 4400 (1 + [5/100])^{2n} = 4400 ( [100 + 5]
/ 100 )^{2n}

∴ 4851/4400 = (105/100)^{2n} = (21/20)^{2n}

(21/20)^{2n} = 4851/4400 = [11 × 441] / [11 × 400]
= 441/400

Taking square root on both sides, we get

(21/20)^{2n} = (21/20)^{2}

Equating power on both sides

∴ 2*n* = 2, *n* = **1**

**13. The cost of a machine is ₹18000 and
it depreciates at 16 2/3 % annually. Its value after 2 years will be___________.
**

(A) ₹12000

(B) ₹12500

(C) ₹15000

(D) ₹16500

**[Answer: (B) ₹ 12500]**

**Solution:**

Solution 2:

Cost of machine = ₹ 18000

Depreciation'rate = 16 2/3 % = 50/3 % p.a

time period = 2 years

∴ As per depreciation
formula,

Depriciated value = Original value (1 – *r*/100]^{n}

Substituting in above formula, we get

Depreciated value after 2 years = 18000 (1 – [(50/3) / 100] )^{2}
= 18000 (1 – [50/ (100 × 3)] )^{2}

= 18000 (1 − 1/6)^{2} = 18000 × (5/6)^{2}

= 18000 × 5/6 × 5/6 = **₹ 12,500**

**14. The sum which amounts to ₹2662 at
10 % p.a in 3 years, compounded yearly is_______.**

(A) ₹2000

(B) ₹1800

(C) ₹1500

(D) ₹2500

**[Answer: (A) ₹ 2000]**

**Solution:**

Solution 2:

Amount = ₹ 2662

Rate of interest = 10% p.a

Time period = 3 yrs. Compounded yearly

Principal (P) → required to find?

Applying formula A = P (1 + [*r*/100])^{n}

2662 = P ( 1 + [10/100] )^{3} = P ( [100 + 10] / 100 )^{3}

2662 = P (110/100)^{3} = 2662 = P(11/10)^{3}

∴ P = [ 2662 × 10 × 10
× 10 ] / [ 11 × 11 × 11 ] = 2 × 10 × 10 × 10 =** ₹ 2000**

**15. The difference between compound and
simple interest on a certain sum of money for 2 years at 2 % p.a is ₹1. The sum
of money is __________ .**

(A) ₹2000

(B) ₹1500

(C) ₹3000

(D) ₹2500

**[Answer: (D) ₹ 2500]**

**Solution:**

Solution 2:

Difference between CI and SI is given as Re 1

Time period (*n*) = 2 yrs.

Rate of interest (*r*) = 2% p.a

Formula for difference is

CI − SI = P × (1 + *r*/100)^{n}

Substituting the values in above formula, we get

1 = P × (2/100)^{2}

∴ P = 1 × (100/2)^{2}
= 1 × (50)^{2} = **₹ 2500**

**Answer:**

**Exercise 4.3 **

**1. (i) ₹1272 (ii) ₹820
(iii) ₹20,000 (iv) A = P (1 + [r/400]) ^{4n} (v) ₹32**

**2. (i) True (ii) False
(iii) True (iv) False (v) True **

**3. ₹162 **

**4. ₹936.80 **

**5. ₹1875 **

**6. 1 1/2 years **

**7. ₹10,875 **

**8. ₹0.50**

**9. 5% **

**10. ₹ 16000 **

**11. (C) 6 **

**12. (B) 1 year **

**13. (B) ₹ 12500**

**14. (A) ₹2000 **

**15. (D) ₹2500**

** **

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8th Maths : Chapter 4 : Life Mathematics : Exercise 4.3 (Compound Interest) | Questions with Answers, Solution | Life Mathematics | Chapter 4 | 8th Maths

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