type 4 difference & equality of si rate & years Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Let the sum be x

${x × 5 × 15}/{100 × 12} - {x × 4 × 8}/{100 × 12}$ = 129

$x/{100 × 12}$(75 - 32) = 129

$x = {129 × 1200}/43$ = Rs.3600

Using Rule 13,

$P_1 = P, R_1 = 4%$

$T_1 = 8 months = 8/12$ years

$P_2 = P, R_2$ = 5%

$T_2 = 15 month = 15/12$ years

S.I. = Rs.129

129 = ${P × 5 × 15/12 - P × {4 × 8}/12}/100$

12900 = ${75P - 32P}/12$

12900 = ${43P}/12$

P = Rs.3600

Answer: (d)

Let $r_1$, and $r_2$ be the required rate of interest

Then, ${13.50} = {1500 × 3 × r_1}/100$

– ${1500 × 3 × r_2}/100 = 4500/100(r_1 - r_2)$

$r_1 - r_2 = 135/450 = 27/90$

= $3/10$ = 0.3%

Using Rule 13,

$P_1 = Rs.1500, R_1 , T_1$ = 3 years.

$P_2 = Rs.1500, R_2 , T_2$ = 3 years.

S.I. = Rs.13.50

13.50 = ${1500 × R_2 × 3 - 1500 × R_1 × 3}/100$

$1350/100 = {4500(R_2 - R_1)}/100$

$R_2 - R_1 = 1350/4500 = 27/90$

= $3/10$ = 0.3%

Answer: (d)

$\text"Simple interest"/ \text"Principal" = 1/9$

If the annual rate of interest be r%, then

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

$r = 1/9 × 100/r$

$r^2 = 100/9$

r = $√{100/9} = 10/3 = 3{1}/3$%

Using Rule 5,

Here, n = $1/9$, R = T

RT = n × 100

$R^2 = 1/9 × 100 = 100/9$

R = $√{100/9} = 10/3 = 3{1}/3$%

Answer: (b)

Using Rule 1,

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

${4000 × 3 × x}/100$

= ${5000 × 2 × 12}/100$

$x = {5 × 2 × 12}/{4 × 3}$

= 10% per annum

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 = $√{100/16} = 10/4 = 2{1}/2%$

Answer: (c)

Using Rule 1,

S.I. = ${P × R × T}/100$

$y = {x × T × R}/100$

and $z = {y × T × R}/100$

So, $y/z = x/y = ⇒ y^2 = zx$