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

Questions : 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%

The correct answers to the above question in:

Answer: (a)

Using Rule 1
Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ or
S.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 = $√{400/9} = 20/3 = 6{2}/3$%

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Read more difference equality in si rate years Based Quantitative Aptitude Questions and Answers

Question : 1

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: (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 : 2

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: (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 : 3

The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by Rs.42. The sum is :

a) Rs.280

b) Rs.210

c) Rs.840

d) Rs.750

Answer: (c)

According to question,

Interest of one year = Rs.42

Rate = 5% and Time = 1 year

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

= ${42 × 100}/{5 × 1}$ = Rs.840

Using Rule 13
The difference between the S.I. for a certain sum $P_1$ deposited for time $T_1$ at $R_1$ rate of interest and another sum $P_2$ deposited for time $T_2$ at $R_2$ rate of interest is
S.I. = ${P_2R_2T_2 - P_1R_1T_1}/100$

$P_1 = P, R_1 = 5%, T_1$ = 3years.

$P_2 = P, R_2 = 5%, T_2$ = 4 years.

S.I.= 42

42 = ${20P - 15P}/100$

P = 42 × 20 = Rs.840

Question : 4

The rate of interest per annum at which the total simple interest of a certain capital for 1 year is equal to the total simple interest of the same capital at the rate of 5% per annum for 2 years, is

a) 10%

b) $5/2$%

c) 12.5%

d) 25%

Answer: (a)

Using Rule 1
Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ or
S.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%

Question : 5

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: (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 : 6

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: (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 5
If 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%

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GET simple interest PRACTICE TEST EXERCISES

type 1 basic simple interest using formula

type 2 increase or decrease in interest rate

type 3 money multiples in ‘n’ years

type 4 difference & equality of si rate & years

type 5 si on ‘n’ years & ‘x/y ‘of sum

type 6 si with ratios
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