type 3 money multiples in ‘n’ 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) 6$1/2$ years
(b) 6$1/4$ years
(c) 6$2/3$ years
(d) 6$1/3$ years
The correct answers to the above question in:
Answer: (c)
If the principal be x, the amount = 2x
SI = x
Time = ${SI × 100}/\text"Principal × Rate"$
= ${x × 100}/{x × 15} = 20/3 = 6{2}/3$ years
Using Rule 3,
T = ${(n - 1)}/R × 100%$
= $({2 - 1}/15) × 100$
= $100/15 = 20/3$ Years
= 6$2/3$ years
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Read more money multiples in n years Based Quantitative Aptitude Questions and Answers
Question : 1
In how many years will a sum of money double itself at 12% per annum?
a) 6 yrs. 9 months
b) 8 yrs. 6 months
c) 7 yrs. 6 months
d) 8 yrs. 4 months
Answer »Answer: (d)
If the principal be Rs.100 then S.I. = Rs.100.
Time = ${SI × 100}/\text"Principal × Rate"$
= ${100 × 100}/{100 × 12} = 25/3$ years
= 8 years 4 months
Using Rule 3,
T = ${(n - 1)}/R × 100%$
= ${(2 - 1)}/12 × 100%$
= $100/12 = 25/3$ years.
= 8$1/3$ years
= 8 years, 4 months.
Question : 2
At what rate per cent per annum will the simple interest on a sum of money be $2/5$ of the amount in 10 years ?
a) 6%
b) 4%
c) 6$2/3$%
d) 5$2/3$%
Answer »Answer: (c)
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
Let P be the principal and R% rate of interest.
S.I. = ${\text"PR" × 10}/100 = \text"PR"/10$
According to the question,
${PR}/10 = (P + {PR}/10) × 2/5$
$R/10 = (1 + R/10) × 2/5$
$R/10 = 2/5 + R/25$
$R/10 - R/25 = 2/5$
${5R - 2R}/50 = 2/5$
${3R}/50 = 2/5$
R = ${50 × 2}/{3 × 5} = 20/3 = 6{2}/3$%
Question : 3
In how many years will a sum of money double itself at 6$1/4$% simple interest per annum ?
a) 20 years
b) 24 years
c) 12 years
d) 16 years
Answer »Answer: (d)
TIme = ${SI × 100}/\text"Principal × Rate"$
= ${x × 100}/{x × 25/4}$ = 16 years
Using Rule 3,
T = ${(n - 1)}/{R%} years$
= ${(2 - 1)}/{25/4} × 100$ years = 16 years.
Question : 4
If a sum of money deposited in a bank at simple interest is doubled in 6 years, then after 12 years, the amount will be
a) 3 times the original amount
b) $5/2$ times the original amount
c) 4 times the original amount
d) $7/2$ times the original amount
Answer »Answer: (a)
Case I,
Principal = Rs.x
Interest = Rs.x
Time = 6 years
Rate = ${Interest × 100}/\text"Principal × Time"$
= ${x × 100}/{x × 16} = 50/3%$ per annum
Case II,
Interest = ${x × 12 × 50}/{100 × 3}$ = Rs.2x
i.e., Amount is thrice the principal.
Question : 5
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest ?
a) 10 years
b) 8 years
c) 14 years
d) 12 years
Answer »Answer: (a)
Case-I
Let the principal be x
Amount = 3x
Interest = 2x
Time = 20 years
I = $\text"PRT"/100$
2x = ${x × R × 20}/100$ ⇒ R = 10%
Case-II
I = x, P = x, R = 10, T = ?
I = $\text"PRT"/100$
x = ${x × 10 × T}/100$ ⇒ T = 10 years.
Using Rule 3,
R% = ${(3 - 1)}/20$ × 100% = 10%
Now, T = ${(n - 1)}/R$ years
T = ${2 - 1}/10 × 100$ = 10 years
Question : 6
The rate of simple interest for which a sum of money becomes 5 times of itself in 8 years is :
a) 40%
b) 30%
c) 55%
d) 50%
Answer »Answer: (d)
Principal = Rs.x (let)
Amount = Rs.5x
Interest = Rs.(5x - x) = Rs.4x
Rate = ${S.I. × 100}/\text"Principal × Time"$
= ${4x × 100}/{x × 8}$ = 50% per annum
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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