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
(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 »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 »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 10If 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 »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 »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 »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 »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
simple interest Shortcuts and Techniques with Examples
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type 1 basic simple interest using formula
Defination & Shortcuts … -
type 2 increase or decrease in interest rate
Defination & Shortcuts … -
type 3 money multiples in ‘n’ years
Defination & Shortcuts … -
type 4 difference & equality of si rate & years
Defination & Shortcuts … -
type 5 si on ‘n’ years & ‘x/y ‘of sum
Defination & Shortcuts … -
type 6 si with ratios
Defination & Shortcuts …
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