55879.The rates of simple interest in two banks x and y are in the ratio of 10 : 8. Rajini wants to deposit her total savings in two banks in such a way that she receives equal half-yearly interest from both. She should deposit the savings in banks x and y in the ratio of
4 : 5
3 : 5
5 : 4
2 : 1
Explanation:
Let the savings be P and Q and rates of SI be 10x and 8x, respectively.
Then, $P × 10x × 1/2 ×1/100= Q × 8x \times 1/2×1/100$
⇒ 10P = 80
∴ $\dfrac{P}{Q}=\dfrac{8}{10}=\dfrac{4}{5}$
∴ P : Q = 4 : 5.
Let the savings be P and Q and rates of SI be 10x and 8x, respectively.
Then, $P × 10x × 1/2 ×1/100= Q × 8x \times 1/2×1/100$
⇒ 10P = 80
∴ $\dfrac{P}{Q}=\dfrac{8}{10}=\dfrac{4}{5}$
∴ P : Q = 4 : 5.
55880.Geeta borrowed some money at the rate of 6% p.a for the first two years, at the rate of 9% p.a for the next three years, and at the rate of 14% p.a for the period beyond five years. If she pays a total interest of Rs.11400 at the end of nine years, how much did she boorrow?
`Rs 10,000
Rs 11,000
Rs 12,000
Rs 14,000
Explanation:
Let the sum borrowed be x. then,
$\dfrac{(x *6*2)}{100} + \dfrac{(x*9*3)}{100}+\dfrac{(x*14*4)}{100}=11400$
$\dfrac{3x}{25}+\dfrac{27x}{100}+\dfrac{14x}{25}=11400$
$\dfrac{95x}{100}$=11400
$\dfrac{11400 * 100}{95}=12,000$
Hence, sum borrowed = Rs. 12,000.
Let the sum borrowed be x. then,
$\dfrac{(x *6*2)}{100} + \dfrac{(x*9*3)}{100}+\dfrac{(x*14*4)}{100}=11400$
$\dfrac{3x}{25}+\dfrac{27x}{100}+\dfrac{14x}{25}=11400$
$\dfrac{95x}{100}$=11400
$\dfrac{11400 * 100}{95}=12,000$
Hence, sum borrowed = Rs. 12,000.
55881.A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
3.46%
2.5%
6%
5%
Explanation:
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).
[(725 x R x 1)/(100)]+[(362.50 x 2R x 1)/(100 x 3)]= 33.50
→ (2175 + 725) R = 33.50 x 100 x 3
→ (2175 + 725) R = 10050
→ (2900)R = 10050
→ R = 10050/2900 = 3.46
Original rate = 3.46%
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).
[(725 x R x 1)/(100)]+[(362.50 x 2R x 1)/(100 x 3)]= 33.50
→ (2175 + 725) R = 33.50 x 100 x 3
→ (2175 + 725) R = 10050
→ (2900)R = 10050
→ R = 10050/2900 = 3.46
Original rate = 3.46%
55882.An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
10.15%
10.25%
11.35%
None of these
Explanation:
S.I. for first 6 months = Rs.[(100 x 10 x 1)/(100 x 2)] = Rs. 5
S.I. for last 6 months = Rs.[(105 x 10 x 1)/(100 x 2)] = Rs. 5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 - 100) = 10.25%
S.I. for first 6 months = Rs.[(100 x 10 x 1)/(100 x 2)] = Rs. 5
S.I. for last 6 months = Rs.[(105 x 10 x 1)/(100 x 2)] = Rs. 5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 - 100) = 10.25%
55883.A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum.
Rs.3500
Rs.8000
Rs.4500
Rs.6000
Explanation:
[(P x (R+2) x 3)/100] - [ (P x R x 3)/100] = 360
→ 3P x (R+2) - 3PR = 36000
→ 3PR + 6P - 3PR = 36000
→ 6P = 36000
P = 6000
[(P x (R+2) x 3)/100] - [ (P x R x 3)/100] = 360
→ 3P x (R+2) - 3PR = 36000
→ 3PR + 6P - 3PR = 36000
→ 6P = 36000
P = 6000
55884.In which year will the amount on a sum of Rs. 800 at 20% compounded half-yearly exceeds Rs.1000?
1st year
2nd year
4th year
5th year
Explanation:
Let the time taken for this amount to reach Rs.1000 be X.
The important thing to note is that this sum is compounded half-yearly. Hence, we use the formula:
A = P 1 + [$\dfrac{r}{m*100}]^{mT}$
Where, A= Amount,
P = Principal
r = interest rate
m = no. of periods within a year
T = no. of years
We need to obtain T such that the RHS should be greater than the LHS:
⇒A < P$ (1 + [\dfrac{r}{m*100}]^{mT}$
In this case,
A = 1000
P = 800
r = 20%
m = 2 (since it is half-yearly)
Substituting these values, we have
$1000<800(1+20 / [2*100])^{2T}=800(11/10)^{2T}$
$1000/800<(121/100)^T $
⇒ 1.25 < (1.21)T
Now, we need to use trial and error to check for the values of T.
For T = 1, 1.25 > 1.21, hence the condition is not satisfied.
For T = 2, 1.25 < 1.4641, hence the condition is met.
Therefore, it is the second year in which the amount would be greater than Rs.1000.
Let the time taken for this amount to reach Rs.1000 be X.
The important thing to note is that this sum is compounded half-yearly. Hence, we use the formula:
A = P 1 + [$\dfrac{r}{m*100}]^{mT}$
Where, A= Amount,
P = Principal
r = interest rate
m = no. of periods within a year
T = no. of years
We need to obtain T such that the RHS should be greater than the LHS:
⇒A < P$ (1 + [\dfrac{r}{m*100}]^{mT}$
In this case,
A = 1000
P = 800
r = 20%
m = 2 (since it is half-yearly)
Substituting these values, we have
$1000<800(1+20 / [2*100])^{2T}=800(11/10)^{2T}$
$1000/800<(121/100)^T $
⇒ 1.25 < (1.21)T
Now, we need to use trial and error to check for the values of T.
For T = 1, 1.25 > 1.21, hence the condition is not satisfied.
For T = 2, 1.25 < 1.4641, hence the condition is met.
Therefore, it is the second year in which the amount would be greater than Rs.1000.
55885.A certain sum of money lent out at simple interest amount Rs.1380 in 3 years and Rs. 1500 in 5 years. Find the rate percent per annum.
3%
3.5%
4%
5%
Explanation:
To solve this question, we can apply a short trick approach
Rate of interest p.a.
=$\dfrac{100(Amount2-Amount1)}{Amount1Time2-Amount2Time1}$
By the short trick apporach, we get
r%=$\dfrac{100(1500-1380)}{(1380 * 5-1500*3)}$
=$\dfrac{100 * 120}{6900-4500}$
=$\dfrac{120*100}{2400}$
=5%
To solve this question, we can apply a short trick approach
Rate of interest p.a.
=$\dfrac{100(Amount2-Amount1)}{Amount1Time2-Amount2Time1}$
By the short trick apporach, we get
r%=$\dfrac{100(1500-1380)}{(1380 * 5-1500*3)}$
=$\dfrac{100 * 120}{6900-4500}$
=$\dfrac{120*100}{2400}$
=5%
55886.Aman and Raghav are two friends. Aman started a business with an investment of Rs 7200, while Raghav puts 60% of his salary at 40% p.a simple interest for 6 months; Raghav takes the amount received after 6 months and joins Aman in the business. If Aman receives a profit of Rs. 2000 out of a total profit of Rs. 2900 at the end of 1 year, what was the original salary of Raghav?
Rs. 18000
Rs. 9400
Rs. 9000
Rs. 15000
Explanation:
Ratio of profit = 2000 : 900 = 20 : 9
Let amount invested by Raghav be K
Ratio of investment = 12 × 7200 : 6 × K = 14400 : K = 20 : 9
K = Rs 6480
Let original Salary of Raghav = R
60 %( R) at 40% p.a for 6 months
$\dfrac{60}{100} \times R \times \dfrac{40}{200} \times 1=6480-0.6R$
R = Rs. 9000
Hence, option C is correct.
Ratio of profit = 2000 : 900 = 20 : 9
Let amount invested by Raghav be K
Ratio of investment = 12 × 7200 : 6 × K = 14400 : K = 20 : 9
K = Rs 6480
Let original Salary of Raghav = R
60 %( R) at 40% p.a for 6 months
$\dfrac{60}{100} \times R \times \dfrac{40}{200} \times 1=6480-0.6R$
R = Rs. 9000
Hence, option C is correct.
55887.A sum of money becomes 3 times in 5 year. in how many year will be the same sum become 6 times at the same rate of simple interest?
15 year
12.5 year
10 year
7.5 year
Explanation:
To solve this question we can apply a short trick approach,
Required time (t2) = [$\dfrac{(y – 1)}{x – 1} × t1] $years.
Where:
x is the no. of times the sum becomes of itself in the 1st scenario = 3
y is the no. of times the sum becomes of itself in the 2nd scenario = 6
t1 is the time taken in the 1st scenario = 5 yr
t2 is the time taken in the 2nd scenario = ?
By the short trick approach, we get
Required time (t2) = [($\dfrac{6 – 1}{3-1}) × 5]$ =$\dfrac{25}{2} $= 12.5 yrs.
To solve this question we can apply a short trick approach,
Required time (t2) = [$\dfrac{(y – 1)}{x – 1} × t1] $years.
Where:
x is the no. of times the sum becomes of itself in the 1st scenario = 3
y is the no. of times the sum becomes of itself in the 2nd scenario = 6
t1 is the time taken in the 1st scenario = 5 yr
t2 is the time taken in the 2nd scenario = ?
By the short trick approach, we get
Required time (t2) = [($\dfrac{6 – 1}{3-1}) × 5]$ =$\dfrac{25}{2} $= 12.5 yrs.
55888.A sum of Rs.1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest recieved after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is :
5 : 8
8 : 5
16 : 15
31 : 6
Explanation:
Let the sum lent at 5% be Rs. x and that let at 8% be Rs. (1550 – x). Then,
$\dfrac{(x × 5 × 3)}{100}$+$\dfrac{[(1500 – x) × 8 × 3]}{100}$ = 300
⇒ 15x – 24x + (1550 × 24) ⇒ 30000
⇒ 9x = 7200 ⇒ x = 800
ratio= 800 / 750 = 16 : 15
Let the sum lent at 5% be Rs. x and that let at 8% be Rs. (1550 – x). Then,
$\dfrac{(x × 5 × 3)}{100}$+$\dfrac{[(1500 – x) × 8 × 3]}{100}$ = 300
⇒ 15x – 24x + (1550 × 24) ⇒ 30000
⇒ 9x = 7200 ⇒ x = 800
ratio= 800 / 750 = 16 : 15
- Ratio and Proportion Test 4
- Percentage Test 2
- Percentage Test 3
- Percentage Test 4
- Percentage Test 5
- Percentage Test 6
- Ratio and Proportion Test 1
- Ratio and Proportion Test 2
- Ratio and Proportion Test 3
- Percentage Test 1
- Average Test 1
- Average Test 2
- Average Test 3
- Interest Test 1
- Interest Test 2
- Interest Test 3
- Interest Test 4
- Interest Test 5
- Time and Distance Test 3
- Number system Test 2
- Partnership Test 1
- Partnership Test 2
- Profit,Loss and Discount Test 1
- Profit,Loss and Discount Test 2
- Profit,Loss and Discount Test 3
- Time and Distance Test 1
- Time and Distance Test 2
- Number System Test 1
- Time and Work Test 1
- Time and Work Test 2
- Time and Work Test 3
- Time and Work Test 4
- Time and Work Test 5
- Time and Work Test 6
- Time and Work Test 7