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 : What sum of money will amount to Rs.520 in 5 years and to Rs.568 in 7 years at simple interest ?

(a) Rs.120

(b) Rs.400

(c) Rs.220

(d) Rs.510

The correct answers to the above question in:

Answer: (b)

Simple interest for 2 years

= Rs.(568 - 520) = Rs.48

Interest for 5 years

= Rs.$48/2 × 5$ = Rs.120

Principal = Rs.(520 - 120) = Rs.400

Using Rule 12,

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

=$({568 × 5 - 520 × 7}/{5 - 7})$

= $({2840 - 3640}/{-2})$

=${- 800}/{- 2}$ = Rs.400

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Read more basic problems using formula Based Quantitative Aptitude Questions and Answers

Question : 1

The simple interest on Rs.7,300 from 11 May, 1987 to 10 September, 1987 (both days included) at 5% per annum is

a) Rs.103

b) Rs.123

c) Rs.223

d) Rs.200

Answer: (b)

Using Rule 1,

Time from 11 May to 10 September, 1987

= 21 + 30 + 31 + 31 + 10

= 123 days

Time = 123 days = $123/365$ year

S.I. = ${7300 × 123 × 5}/{365 × 100}$ = Rs.123

Question : 2

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

Question : 3

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 : 4

Manoj deposited Rs.29400 for 6 years at a simple interest. He got Rs.4200 as interest after 6 years. The annual rate of interest was

a) 2$7/20$%

b) 2$8/21$%

c) 4$8/21$%

d) 3$8/21$%

Answer: (b)

Using Rule 1,

4200 = ${29400 × 6 × R}/100$

R = $4200/{294 × 6} = 50/21 = 2{8}/21$%

Question : 5

A person deposited Rs.400 for 2 years, Rs.550 for 4 years and Rs.1,200 for 6 years. He received the total simple interest of Rs.1,020. The rate of interest per annum is

a) 5%

b) 10%

c) 20%

d) 15%

Answer: (b)

Using Rule 1,

Let the rate of interest be R per cent per annum.

${400 × 2 × R}/100 + {550 × 4 × R}/100$

+ ${1200 × 6 × R}/100 = 1020$

8R + 22 R +72 R = 1020

102 R= 1020

R = $1020/102$ = 10%

Question : 6

If the simple interest on a certain sum of money for 15 months at 7$1/2$% per annum exceeds the simple interest on the same sum for 8 months at 12$1/2$% per annum by Rs.32.50, then the sum of money (in ) is :

a) 312.50

b) 312

c) 3120.50

d) 3120

Answer: (d)

Using Rule 1,

Let the sum be x.

Using formula, I = $\text"PRT"/100$ we have

${x × 15/12 × 15/2}/{100 - x} - {x × 8/12 × 25/2}/100$

= 32.50

${25x}/2400$ = 32.50

$x = {32.50 × 2400}/25$ = 3120

Required sum = Rs.3120

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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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