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Quantitative Aptitude Quiz for SSC(SI and CI): 13 March 2021

13 March, 2021

Source: safalta
1. A sum of money is divided into two parts in the ratio 3 : 5. Both parts are invested in different schemes working on simple interest for 3 years. The rate of interests in first scheme, where smaller part is invested, is 15% while that in second scheme is 12%. The interest earned on second part is Rs. 2250 more than that earned from first part. What was the total sum of money?

a) Rs. 40000
b) Rs. 45000
c) Rs. 50000
d) Rs. 30000

Let the sum of money be 8T. When divided into 2 parts in ratio 3 : 5, the parts will be 3T and 5T.

The rate of interests in first scheme, where smaller part is invested, is 15% while that in second scheme is 12%.

We know, Simple interest = (Principal × Rate × Time)/100

⇒ Interest earned on first part = (3T × 15 × 3)/100 = 1.35T

And, Interest earned on second part = (5T × 12 × 3)/100 = 1.8T

The interest earned on second part is Rs. 2250 more than that earned from first part.

⇒ 1.8T – 1.35T = 2250

⇒ T = 2250/0.45 = 5000

⇒ Total sum of money = 8T = Rs. 40000

2. What will be the simple interest earned on an amount of Rs.16,800 in 9 months at the rate of 6.25% p.a.?

a) Rs.797.50
b) Rs.787.50
c) Rs.757.50
d) Rs.737.50

9 months = 9/12 years = 3/4th year = 0.75 year

We know that simple interest = PRT/100

Where P is the principle amount, R is the rate of interest and T is the number of years.

Simple Interest = (16800×6.25 × 0.75)/100 = Rs.787.50

3. A certain sum of money doubles itself in 5 years at simple interest. In how many years it becomes seven times?

a) 15 years
b) 30 years
c) 20 years
d) 25 years

For simple interest, We know the amount of interest = PRT/100

A certain sum of money doubles itself in 5 years at simple interest.

Hence, interest is same as the sum of money.

Hence, P=(P×R×5)/100

⇒ 5R =100

⇒ R = 20%

The money has to become 7 times, Interest will be 6 times.

Hence, 6P = P×20×t/100

⇒ 20t = 600

⇒ t = 30

Hence, it will become 7 times in 30 years.

4. An amount doubles itself in 10 years with simple interest. What is the rate of interest p.a.?

a) 15
b) 10
c) 20
d) 25

Let the amount be y.

& Rate of interest be R.

The amount becomes 2y in 10 years. Hence, Simple interest = Amount – Principal = 2y – y = y

 We know the Simple interest = PRT/100

Where, P= Principal amount, R = Rate of interest, T = Number of years.

Hence, y = (y×R×10)/100

Or, 10R = 100

Or, R = 10

Hence, the rate of interest = 10% per year.

5. What approximate amount needs to be returned which is due as Rs. 848 at the end of 4 years at 4% per annum simple interest?

a) Rs.953.68
b) Rs.983.68
c) Rs.963.68
d) Rs.973.68

For simple interest, interest = PRT/100

Where P is the principle amount, R is the rate of interest and T is the number of years.

Here, P = 848, R = 4 and T = 4

Hence, Interest = (848 × 4 × 4)/100 = Rs. 135.68

Total amount need to be returned = 848 + 135.68 = Rs.983.68

6. The simple interest for a certain sum of money is half the principal. Find the time and rate of interest if the rate is twice the time.

a) 10 years, 5%
b) 15 years, 10%
c) 5 years, 10%
d) 15 years, 15%

Simple interest = PRT/100 = P/2

R = 2 × T

2T2/100 = 1/2

T = √100/4 = 5 years

R = 2 × 5 = 10%

7. The compound interest on a certain amount for 2 years at the rate of 10 p.a is Rs. 630. What will be the simple interest on the same amount and at the same rate and same time?

a) Rs. 800
b) Rs. 600
c) Rs. 500
d) Rs. 400

⇒ CI = P(1+10/100)^2 – P

⇒ 630 = P× 1.1 × 1.1 – P

⇒ 630 = P(1.21 – 1)

⇒ P = 630/0.21

⇒ P = 3000

Now, we need to find the simple interest.

SI = (P × R× T) / 100

SI = (3000 × 2 × 10) / 100

SI = Rs. 600

8. The simple interest on a sum of money is (1/16)th of principle and the number of years is equal to the rate per cent per annum. The rate per cent annum is _______
a) 2.5%
b) 3.5%
c) 4.5%
d) 1.5%

Let the rate and the number of years be ‘r’ and the sum be ‘x’

∴ According to the given conditions,

⇒ x/16 = (x × r × r)/100

∴ r × r = 100/16

∴ r = 10/4

∴ The required rate is 2.5%

9. If the difference of C.I and S.I on some amount is Rs. 36 and the sum of the C.I and S.I is 160. Find the product of the C.I and S.I? 

a)  Rs. 6276
b)  Rs. 6876
c)  Rs. 6576
d)  Rs. 6076

C.I – S.I = 36

C.I + S.I = 160

(C.I + S.I)2 = (C.I – S.I)2 + 4 × C.I. × S.I

⇒ (160)2 = (36)2 + 4 × C.I. × S.I

⇒ 4 × C.I. × S.I = 25600 – 1296 = 24304

∴ C.I. × S.I = 24304/4 = Rs. 6076

10. A man borrowed Rs. 18750 at a simple interest of 6% per annum. If he cleared his debt by paying an amount of Rs. 22125, after how many years did he cleared his debt?

a)  2 years
b)  4 years
c)  3 years
d)  5 years

Simple interest = Amount - Principal = (Principal × Rate × Time)/100

⇒ 22125 - 18750 = (18750 × 6 × Time)/100

⇒ Time = (3375 × 100)/(18750 × 6)

⇒ Time = 3 years

∴ He cleared his debt after 3 years



 

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