Practice Si on n years - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   A and B borrowed Rs. 3000 and Rs. 3200 respectively at the same rate of interest for 2$1/2$ years. If B paid Rs. 40 more interest than A, find the rate of interest.

(a)

(b)

(c)

(d)

Explanation:

Rate of interest = r % per annum

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

According to the question,

${3200 × 5 × r}/{100 × 2} - {3000 × 5 × r}/200$ = 40

80r - 75r = 40

5r = 40 ⇒ r = $40/5$

= 8% per annum

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. = ${P_2R_2T_2 - P_1R_1T_1}/100$

Here, $P_1 = Rs.3000, R_1 = R, T_1 = 5/2$ years

$P_2 = Rs.3200, R_2 = R, T_2 = 5/2$ years

Difference S.I. = Rs.40

40 = ${3200 × R × 5/2 - 3000 × R × 5/2}/100$

4000 = 8000 R - 7500 R

R = 8%


Q-2)   The simple interest on a sum after 4 years is $1/5$ of the sum. The rate of interest per annum is

(a)

(b)

(c)

(d)

Explanation:

Let Prinicpal = Rs.100

S.I. = $100 × 1/5$ = Rs.20

Rate = ${20 × 100}/{100 × 4}$ = 5%

Using Rule 5,

Here, n = $1/5$, T = 4 years.

R = ${n × 100}/T$

R = $1/5 × 100/4$

R = 5%


Q-3)   On a certain sum, the simple interest at the end of 6$1/4$ years becomes $3/8$ of the sum. The rate of interest is

(a)

(b)

(c)

(d)

Explanation:

$\text"Interest"/ \text"Principal" = 3/8$

Rate = ${\text"SI" × 100}/\text"Principal × Time"$

= $3/8 × 100/{25/4}$

= $3/8 × 400/25$ = 6% per annum

Using Rule 5,

Here, n = $3/8$, T = $25/4$ years.

R = ${n × 100}/T$

= $3/8 × 100/{25/4}$ ⇒ R = 6%


Q-4)   The simple interest on a sum for 5 years is one fourth of the sum. The rate of interest per annum is

(a)

(b)

(c)

(d)

Explanation:

$\text"Simple interest"/\text"Principal" = 1/4$

Rate = ${\text"SI" × 100}/\text"Principal × Time"$

= ${1 × 100}/{4 × 5} = 5%$ per annum

Using Rule 5,

Here, n = $1/4$, T = 5 years

R = ${n × 100}/T$

= $1/4 × 100/5$ = R = 5%


Q-5)   Simple interest on a certain sum for 4 years is $12/25$ of the sum. The rate of interest is

(a)

(b)

(c)

(d)

Explanation:

Rate = ${\text"SI" × 100}/\text"Principal × Time"$

= $12/25 × 100/4$ = 12% per annum

Using Rule 5,

Here, n = $12/25$, T = 4 years.

R = ${n × 100}/T$

R = $12/25 × 100/4$

R = 12%


Q-6)   The present worth of a bill due 7 months hence is Rs.1200 and if the bill were due at the end of 2$1/2$ years its present worth would be Rs.1016. The rate per cent is

(a)

(b)

(c)

(d)

Explanation:

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

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

1200 + ${1200 × 7 × r}/{12 × 100}$

= Amount (A)

1200 + 7r = A ...(i)

and, 1016 + ${1016 × 5 × r}/{2 × 100}$ = A

1016 + 25.4r = A ...(ii)

1016 + 25.4r = 1200 + 7r

25.4r - 7r = 1200 - 1016

18.4r = 184 ⇒ r = $184/{18.4}$

= 10% per annum


Q-7)   Simple interest on a certain sum for 6 years is $9/25$ of the sum. The rate of interest is

(a)

(b)

(c)

(d)

Explanation:

Rate = ${\text"SI" × 100}/\text"Principal × Time"$

= $9/25 × 100/6$ = 6% per annum

Using Rule 5,

Here, n = $9/25$, T = 6 years.

R = ${n × 100}/T$

R = $9/25 × 100/6$

R = 6%


Q-8)   In what time will the simple interest be $2/5$ of the principal at 8 per cent per annum?

(a)

(b)

(c)

(d)

Explanation:

Let the principal be x

Interest = $2/5$ x

Rate = 8% per annum

Time = ${\text"Interest" × 100}/\text"Principal × Rate"$

=${{2/5}x × 100}/{x × 8} = 40/8$ = 5 years

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 = $2/5$ and R = 8%

RT = (n × 100)

T = ${n × 100}/R$

T = $2/5 × 100/8$ = 5 years


Q-9)   At the rate of simple interest per annum, the interest on a certain sum of money for 10 years will be $2/5$th part of the amount, then the rate of simple interest is

(a)

(b)

(c)

(d)

Explanation:

Amount after 10 years

= P$(1 + {RT}/100)$ = P$(1 + {R × 10}/100)$

= Rs. P$(1 + R/10)$

Interest = Rs.P$(1 + R/10) × 2/5$

Rate= ${\text"SI" × 100}/\text"Principal × Time"$

R = ${P(1 + R/10) × 2/5 × 100}/{P × 10}$

R = 4$(1 + R/10)$

$R/4 = 1 + R/10$

$R/4 - R/10$ = 1

${5R - 2R}/20$ = 1

3R = 20

R = $20/3 = 6{3}2%$

Using Rule 5,

Here, S.I. = $2/5$ amount

S.I. = $2/5$ (P + S.I.)

S.I. = $2/5$ S.I. + $2/5$ P

$3/5$ S.I. = $2/5$ P

S.I. = $2/3$P

Now, n = $2/3$, T = 10 years.

R= ${n × 100}/T$

= $2/3 × 100/10$

= $20/3 = 6{2}/3%$


Q-10)   The simple interest on a sum at x% for x years is x. What is the sum?

(a)

(b)

(c)

(d)

Explanation:

P = $\text"100 × SI"/\text"R × T"$

= $\text"100 × x"/\text"x × x"$ = $100/x$