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 : 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

The correct answers to the above question in:

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

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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 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 : 2

If x, y, z are three sum of money such that y is the simple interest on x and z is the simple interest on y for the same time and at the same rate of interest, then we have

a) $xyz$ = 1

b) $z^2 = xy$

c) $y^2 = zx$

d) $x^2 = yz$

Answer: (c)

Using Rule 1,

S.I. = ${P × R × T}/100$

$y = {x × T × R}/100$

and $z = {y × T × R}/100$

So, $y/z = x/y = ⇒ y^2 = zx$

Question : 3

The simple interest on a sum of money is $1/16$ of the principal and the number of years is equal to the rate per cent per annum. The rate per annum is

a) 2$1/2$%

b) 1$1/2$%

c) 4$1/2$%

d) 3$1/2$%

Answer: (a)

Let the annual rate of interest = r%

Time = r years

Let the principal be x .

Interest = $x/16$

According to the question,

$x/16 = {x × r × r}/100$ [Since, r = t]

16$r^2$ = 100

$r^2 = 100/16 = 25/4$

r = $√ {25/4} = 5/2 = 2{1}/2$%

Using Rule 5,

Here, n = $1/16$, R = T

RT = n × 100

$R^2 = 100/16$

R = $√{100/16} = 10/4 = 2{1}/2%$

Question : 4

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 : 5

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: (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 = $√{400/9} = 20/3 = 6{2}/3$%

Question : 6

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%

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