type 3 money multiples in ‘n’ years 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 : Rs.6,000 becomes Rs.7,200 in 4 years at a certain rate of simple interest. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be

(a) Rs.8,250

(b) Rs.8,000

(c) Rs.9,000

(d) Rs.9,250

The correct answers to the above question in:

Answer: (a)

Using Rule 1,

SI = Rs.(7200–6000) = Rs.1200

SI = $\text"PRT"/100$

1200 = ${6000 × R × 4}/100$

R = ${1200 × 100}/{6000 × 4}$ = 5%

New rate of R = 5×1.5 = 7.5%

Then, SI = ${6000 × 7.5 × 5}/100$ = Rs.2250

Amount = Rs.(6000 + 2250) = Rs.8250

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Read more money multiples in n years Based Quantitative Aptitude Questions and Answers

Question : 1

A sum of money becomes $41/40$ of itself in $1/4$ years at a certain rate of simple interest. The rate of interest per annum is

a) 1%

b) 10%

c) 5%

d) 2.5%

Answer: (b)

Let the principal be Re.1

S.I. = $41/40 - 1 = 1/40$

Now, rate = ${\text"Interest" ×100}/{\text"Principal × Time"}$

= ${1/40 × 100}/{1 × 1/4} = {100 × 4}/40$ = 10%

Using Rule 3,

R = ${(41/40 - 1) × 100%}/{1/4}$

= $1/40 × 4 × 100%$ = 10%

Question : 2

At what rate of simple interest per annum will a sum become $7/4$ of itself in 4 years ?

a) 18$1/4$%

b) 18%

c) 18$1/2$%

d) 18$3/4$%

Answer: (d)

Let the principal be x

Principal + SI = ${7x}/4$

SI = ${7x}/4 - x = {3x}/4$

Rate = ${SI × 100}/\text"Principal × Time"$

= ${3x × 100}/{4 × x × 4} = 18{3}/4%$

Using Rule 3,

R = ${(7/4 - 1)}/4 × 100%$

= $3/16$ × 100%

= $75/4% = 18{3}/4%$

Question : 3

A sum doubles itself in 16 years, then in how many years will it triple itself; rate of interest being simple

a) 32 years

b) 25 years

c) 64 years

d) 48 years

Answer: (a)

Case I

Principal = Rs. x

Interest = Rs. x

Rate = ${SI × 100}/\text"Principal × Time"$

= ${x × 100}/{x × 16} = 25/4%$ per annum

Case II

Interest = Rs. 2x

Time = ${SI × 100}/\text"Principal × Rate"$

= ${2x × 100 × 4}/{x × 25}$ = 32 years

Using Rule 3,

R = ${(n - 1)}/T × 100%$

= ${(2 - 1)}/16 × 100%$

= $25/4 % = 6{1}/4%$

Now, T = ${(n - 1)}/R × 100%$

= ${(3 - 1)}/{25/4} × 100$

= $800/25$ = 32 years.

Question : 4

The rate of simple interest per annum at which a sum of money doubles itself in 16$2/3$ years is

a) 5%

b) 4%

c) 6$2/3$%

d) 6%

Answer: (d)

According to the question,

Principal = Rs.x.

Interest = Rs.x.

Time = $50/3$ years

Rate = ${Interest × 100}/\text"Principal × Time"$

= ${x × 100}/{x × {50/3}}$

= ${100 × 3}/50$ = 6% per annum

Question : 5

A certain sum of money amounts to Rs. 2200 at 5% p.a. rate of interest, Rs. 2320 at 8% interest in the same period of time. The period of time is :

a) 4 years

b) 3 years

c) 2 years

d) 5 years

Answer: (c)

Principal = Rs.P and time = T years

S.I. = $\text"Principal × Time × Rate"/100$

According to the question,

P + ${PT × 5}/100$ = 2200

P + ${PT}/20$ = 2200 ...(i)

Again, ${PT × 8}/100 - {PT × 5}/100$

= 2320 - 2200

${3PT}/100$ = 120

PT = ${120 × 100}/3$ = 4000 ...(ii)

From equation (i),

P + $4000/20$ = 2200

P = 2200 - 200 = Rs. 2000

From equation (ii),

PT = 4000

T = $4000/2000$ = 2 years

Alternative Method

Difference in rates

= 8 - 5 = 3%

Since,3% ≡ 2320 - 2200 = 120

5% ≡ $120/3$ × 5 = 200

Principal = Rs.(2200 - 200) = Rs.2000

Time = ${200 × 100}/{2000 × 5}$ = 2 years

Question : 6

A sum of money becomes $7/6$ of itself in 3 years at a certain rate of simple interest. The rate per annum is :

a) 6$5/9$%

b) 5$5/9$%

c) 25%

d) 18%

Answer: (b)

Principal = P

Amount = ${7p}/6$

S.I. = ${7p}/6 - P = P/6$

R = ${S.I × 100}/{P × T} = {P × 100}/{6 × p × 3}$

= $50/9 = 5{5}/9%$

Using Rule 3
If a certain sum becomes 'n' times of itself in T years on Simple Interest, then the rate per cent per annum is.
R% = ${(n - 1)}/T × 100%$ and
T = ${(n - 1)}/R × 100%$

R% = ${(7/6 - 1) × 100%}/3$

= $1/18 × 100% = 50/9%$

= 5$5/9$%

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