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

Answer: (d)

Using Rule 1,

True discount

= $\text"Amount × Rate × Time"/ \text"100 +(Rate × Time)"$

= ${2400 × 5 × 4}/{100 + (5 × 4)}$

= ${2400 × 5 × 4}/120$ = Rs.400

S.I. = ${2400 × 5 × 4}/100$ = Rs.480

Required difference

= Rs.(480 - 400) = Rs.80

Answer: (d)

Using Rule 1,

Let the sum lent out at 12.5% be x

Sum lent out at 10% = 1500 - x

Now, ${(1500 - x) × 10 × 5}/100$

= ${x × 12.5 × 4}/100$

50 (1500 - x ) = 50x

2x = 1500

$x = 1500/2$ = Rs.750

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

Answer: (d)

Let the sum lent be x.

${x × 7.5 × 5}/100 - {x × 7.5 × 4}/100 = 150$

${x × 7.5 × 1}/100$ = 150

$x = {150 × 100}/{7.5}$ = Rs.2000

Using Rule 13,

Here, $P_1 = P, R_1 = 7.5%, T_1$ = 4 years.

$P_2 = P, R_2 = 7.5%, T_2$ = 5 years.

S.I. = Rs.150

S.I. = ${P_2R_2T_2 - P_1R_1T_1}/100$

150 = ${P × 7.5 × 5 - P × 7.5 × 4}/100$

15000 = 7.5P

P = $15000/{7.5}$

P = $150000/75$ = Rs.2000

Answer: (b)

Using Rule 1,

Let first part be x and second part be(1750 –x )

According to the question,

x × $8/100 = (1750 - x ) × 6/100$

8x + 6x = 1750 × 6

14x = 1750 × 6

$x = {1750 × 6}/14$ = Rs.750

Interest = 8% of 750

= 750 × $8/100 = Rs.60

Answer: (a)

Let the principal be x and rate be y% per annum.

According to the question,

SI = ${P × R × T}/100$

$x/4 = {x × y × y}/100$

$y^2 = 100/4$ = 25

y = $√{25}$ = 5% per annum

Using Rule 5,

n = $1/5$, R = T

RT = n × 100

$R^2 = 1/4 × 100$ = 25

R = 5%