type 1 basic simple interest using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on simple interest topic of quantitative aptitude

Questions : At some rate of simple interest, A lent Rs.6,000 to B for 2 years and Rs.1,500 to C for 4 years and received Rs.9,00 as interest from both of them together. The rate of interest per annum was

(a) 6%

(b) 5%

(c) 10%

(d) 8%

The correct answers to the above question in:

Answer: (b)

Using Rule 1,

If rate of interest be R% p.a. then,

SI = ${\text"Principal × Rate × Time"/100$

${6000 × 2 × R}/100 + {1500 × 4 × R}/100$

= 900

120 R + 60R = 900

180R = 900

R = $900/180$ = 5%

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Question : 1

The time in which Rs.80,000 amounts to Rs.92,610 at 10% p.a. compound interest, interest being compounded semi annually is :

a) 3 years

b) 1$1/2$ years

c) 2$1/2$ years

d) 2 years

Answer: (b)

Using Rule 1 and 2,

Time = t half year

and R = 5% per half year

A = P$(1 + R/100)^T$

$92610/80000 = (1 + 5/100)^T$

$9261/8000 = (21/20)^T$

T = 3 half years or 1$1/2$ years

$(21/20)^3 = (21/20)^T$

Question : 2

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of Rs.10,000 for 3 years will be

a) Rs.2,000

b) Rs.1,600

c) Rs.1,575.20

d) Rs.1,625.80

Answer: (c)

Using Rule 3,
If there are distinct 'rates of interest' for distinct time periods i.e.,
Rate for 1st year → $r_1$%
Rate for 2nd year → $r_2$%
Rate for 3rd year → $r_3$% and so on
Then A = P$(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)$...
C.I. = A - P

Amount = P$(1 + R_1/100)(1 + R_2/100)(1 + R_3/100)$

= 10000$(1+ 4/100)(1 + 5/100)(1 + 6/100)$

= $10000 × 26/25 × 21/20 × 53/50$

A = Rs.11575.2

C.I. = Rs.(11575.2–10000) = Rs.1575.2

Question : 3

In how many years will a sum of Rs.800 at 10% per annum compound interest, compounded semi-annually becomes Rs.926.10 ?

a) 2$1/2$ years

b) 1$1/2$ years

c) 2$1/3$ years

d) 1$2/3$ years

Answer: (b)

Using Rule 1 and 2,

Rate = 10% per annum = 5% half yearly

A = P$(1 + R/100)^T$

926.10 = 800$(1 + 5/100)^T$

$9261/8000 = (21/20)^T$

$(21/20)^3 = (21/20)^T$

Time = 3 half years = 1$1/2$ years

Question : 4

A sum of money lent at simple interest amounts to Rs.880 in 2 years and to Rs.920 in 3 years. The sum of money (in rupees) is

a) 760

b) 700

c) 800

d) 784

Answer: (c)

If the principal be x and rate of interest be r% per annum, then

SI after 1 year = 920 - 880 = Rs.40

SI after 2 years = Rs.80

880 = x + 80

x = Rs.(880 - 80) = Rs.800

Using Rule 12
If certain sum P amounts to Rs. $A_1$ in $t_1$ years at rate of R% and the same sum amounts to Rs. $A_2$ in $t_2$ years at same rate of interest R%. Then,
(i) R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100
(ii) P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$

P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$

= $({920 × 2 - 880 × 3}/{2 - 3})$

= $({1840 - 2640}/{-1})$

= ${- 800}/{- 1}$ = Rs.800

Question : 5

Rs.1,000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will become Rs.2,000 after

a) 18 years

b) 15 years

c) 16$2/3$ years

d) 20 years

Answer: (c)

Using Rule 1,

After 10 years,

SI = ${1000 × 5 × 10}/100$ = Rs.500

Principal for 11th year

= 1000 + 500 = Rs.1500

SI = Rs.(2000 - 1500) = Rs.500

T = ${SI × 100}/{P × R} = {500 × 100}/{1500 × 5}$

= $20/3$ years = 6$2/3$ years

Total time = 10 + 6$2/3$

= 16$2/3$ years

Question : 6

What annual instalment will discharge a debt of Rs.6450 due in 4 years at 5% simple interest ?

a) Rs.1835

b) Rs.1500

c) Rs.1950

d) Rs.1935

Answer: (b)

Let each instalment be x Then,

$(x + {x × 5 × 1}/100) + (x + {x × 5 × 2}/100)$

+ $(x + {x × 5 × 3}/100) + x = 6450$

$(x + x/20) + (x + x/10)$

+ $(x + {3x}/20)$ + x= 6450

${21x}/20 + {11x}/10 + {23x}/20 + x = 6450$

${21x + 22x + 23x + 20x}/20$

= 6450

${86x}/20 = 6450$

$x = {6450 × 20}/86$ = Rs.1500

Using Rule 10
If a sum is to be deposited in equal instalments, then,
Equal instalment = ${A × 200}/{T[200 + (T - 1)r]}$
where T = no. of years, A = amount, r = Rate of Interest.

Equal instalment = ${6450 × 200}/{4[200 + (4 - 1) × 5]}$

= ${6450 × 200}/{4(215)}$

= ${6450 × 50}/215$ = Rs.1500

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GET simple interest PRACTICE TEST EXERCISES

type 1 basic simple interest using formula

type 2 increase or decrease in interest rate

type 3 money multiples in ‘n’ years

type 4 difference & equality of si rate & years

type 5 si on ‘n’ years & ‘x/y ‘of sum

type 6 si with ratios
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