type 4 difference & equality of si rate & years Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 6 EXERCISES
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The following question based on simple interest topic of quantitative aptitude
(a) 10%
(b) $5/2$%
(c) 12.5%
(d) 25%
The correct answers to the above question in:
Answer: (a)
Using Rule 1Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ orS.I. = ${\text"P × R × T"/100$P = ${\text"S.I." × 100}/\text"R × T"$, R = ${\text"S.I." × 100}/\text"P × T"$, T = ${\text"S.I." × 100}/\text"P × R"$ A = P + S.I. or S.I. = A - P
${P × r × 1}/100 = {P × 5 × 2}/100$
[Since,Capital is same in both cases]
r × 1 = 5 × 2 = 10%
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Read more difference equality in si rate years Based Quantitative Aptitude Questions and Answers
Question : 1
The simple interest on a sum of money is $4/9$ of the principal and the number of years is equal to the rate percent per annum. The rate per annum is :
a) 6$2/3$%
b) 5%
c) 7$1/5$%
d) 6%
Answer »Answer: (a)
Using Rule 1Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ orS.I. = ${\text"P × R × T"/100$P = ${\text"S.I." × 100}/\text"R × T"$, R = ${\text"S.I." × 100}/\text"P × T"$, T = ${\text"S.I." × 100}/\text"P × R"$ A = P + S.I. or S.I. = A - P
We know that
S.I. = $\text"PRT"/100$
According to question,
S.I.= $4/9P$ & R = T (numerically)
$4/9P = {P × R × R}/100$
$R^2 = 400/9$
R = $√{400/9} = 20/3 = 6{2}/3$%
Question : 2
A person deposited Rs.500 for 4 years and Rs.600 for 3 years at the same rate of simple interest in a bank. Altogether he received Rs.190 as interest. The rate of simple interest per annum was
a) 5%
b) 4%
c) 3%
d) 2%
Answer »Answer: (a)
Using Rule 1,
Let 'r' be the rate of interest
190 = ${500 × 4 × r}/100 + {600 × 3 × r}/100$
20r + 18r = 190
38r = 190
r = $190/38$ = 5%
Question : 3
If Rs.12,000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for 4$1/2$ years at 16% per annum, the greater part is
a) Rs.6,000
b) Rs.8,000
c) Rs.7,500
d) Rs.7,000
Answer »Answer: (b)
Using Rule 1,
Let the larger part of the sum be x
Smaller part = Rs.(12000 - x)
According to the question,
${x × 3 × 12}/100 = {(12000 - x) × 9 × 16}/{2 × 100}$
36 x = (12000 - x ) 72
x = (12000 - x ) × 2
x + 2x = 24000
3x = 24000
$x = 24000/3$ = Rs.8000
Question : 4
Prakash lends a part of Rs.20,000 at 8% simple interest and remaining at $4/3$% simple interest. His total income after a year was Rs.800. Find the sum lent at 8%.
a) Rs.12,000
b) Rs.8,000
c) Rs.10,000
d) Rs.6,000
Answer »Answer: (b)
Using Rule 1,
Amount lent at 8% rate of interest = Rs.x
Amount lent at $4/3$% rate of interest = Rs.(20,000 - x)
S.I. = ${\text"Principal × Rate × Time"/100$
${x × 8 × 1}/100 + {(20,000 - x) × 4/3 × 1}/100$ = 800
${2x}/25 + {20,000 - x}/75 = 800$
${6x + 20,000 - x}/75 = 800$
5x + 20,000 = 75 × 800 = 60,000
5x = 60,000-20,000 = 40,000
$x = {40,000}/5$ = Rs.8000
Question : 5
Simple interest on a certain sum at a certain annual rate of interest is $16/25$ of the sum. If the number representing rate per cent and time in years be equal, then the rate of interest is
a) 11$1/2$%
b) 8%
c) 12$1/4$%
d) 12$1/2$%
Answer »Answer: (b)
Let the rate of interest be r% and principal be P.
According to the question.
${16P}/25 = {P × r × r}/100$
[Since, r = t numerically]
$r^2 = 1600/25$
r = $40/5$ = 8 %
Using Rule 5If Simple Interest (S.I.) becomes 'n' times of principal i.e.S.I. = P × n then.RT = n × 100
Here, n = $16/25$, R = T
Now R × R = $16/25$ × 100
$R^2 = 1600/25$
R = $√{1600/25} = 40/5$ = 8%
Question : 6
Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is Rs.1350, then the sum deposited in each case is :
a) Rs.4000
b) Rs.3000
c) Rs.6500
d) Rs.5000
Answer »Answer: (d)
Let amount invested in each company be Rs. x.
S.I. = ${\text"Principal × Rate × Time"/100$
According to the question,
${x × 15 × 5}/100 - {x × 12 × 4}/100$ = 1350
${75x}/100 - {48x}/100$ = 1350
${27x}/100$ = 1350
$x = {1350 × 100}/27$ = Rs. 5000
Using Rule 13
Here, $P_1 = Rs. P, R_1 = 12%, T_1$ = 4 years
$P_2 = Rs. P, R_2 = 15%, T_2$ = 5 years
S.I. = Rs. 1350
S.I.= ${P_2R_2T_2 - P_1R_1T_1}/100$
1350 = ${P × 15 × 5 - P × 12 × 4}/100$
135000 = 75 P - 48P
135000 = 75 P
P = Rs. 5000
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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