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) 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%
Discuss Form
Read more basic problems using formula Based Quantitative Aptitude Questions and Answers
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 »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 »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 onThen 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 »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 »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 12If 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 »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 »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
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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