type 4 difference & equality of si rate & years Practice Questions Answers Test with Solutions & More Shortcuts

Question : 1 [SSC CGL Tier-I 2015]

Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is Rs.1350, then the sum deposited in each case is :

a) Rs.4000

b) Rs.3000

c) Rs.6500

d) Rs.5000

Answer: (d)

Let amount invested in each company be Rs. x.

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

According to the question,

${x × 15 × 5}/100 - {x × 12 × 4}/100$ = 1350

${75x}/100 - {48x}/100$ = 1350

${27x}/100$ = 1350

$x = {1350 × 100}/27$ = Rs. 5000

Using Rule 13

Here, $P_1 = Rs. P, R_1 = 12%, T_1$ = 4 years

$P_2 = Rs. P, R_2 = 15%, T_2$ = 5 years

S.I. = Rs. 1350

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

1350 = ${P × 15 × 5 - P × 12 × 4}/100$

135000 = 75 P - 48P

135000 = 75 P

P = Rs. 5000

Question : 2 [SSC CGL Tier-I 2011]

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%

Question : 3 [SSC CGL Tier-II 2014]

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 : 4 [SSC Delhi Police S.I.2012]

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 [SSC CGL Prelim 2000]

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 = $√{400/9} = 20/3 = 6{2}/3$%

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