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

Answer: (b)

Let the principal be P and rate of interest be r %

According to the question,

${30P}/100 = {P × R × 6}/100$

30 = 6 R

R = 5

Now, let interest be equal to principal in T years.

P = ${P × 5 × T}/100$

T = $100/5$ = 20 years.

Using Rule 5,

Here, n = $30/100 = 3/10$, T = 6 years.

RT = n × 100

R × 6 = $3/10$ × 100

R = 5%

As,S.I. = P

S.I. = ${P × R × T}/100$

100 = RT

100 = 5 × T

This is possible only when T = 20.

Answer: (d)

Using Rule 1,

Let the period of time be T years. Then,

${400 × 5 × T}/100 = {500 × 4 × 6.25}/100$

T = ${500 × 4 × 6.25}/{400 × 5}$

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

Answer: (c)

Using Rule 1,

Let the sum lent in each case be x.

Then, ${x × 9 × 2}/100 + {x × 10 × 2}/100$ = 760

${x × 2}/100$(9 + 10) = 760

${2 × 19x}/100$ = 760

$x = {760 × 100}/{2 × 19}$ = Rs.2000