Comparing Quantities Class 8 MCQ Questions

Answer: (c) 29/50

Explanation: In order to convert a Percentage into a Fraction, we divide the Percentage by 100, and remove the sign of Percentage (%).

We have,

58 % = 58/100

On Simplifying, 58/100

HCF of 58 and 100 is 2

Divide both 58 and 100 by their HCF

(58÷2)/(100÷2) = 29/50

Answer: (b) 0.12

Explanation: In order to convert given Percentage into into decimal we have to perform the following steps:

Step 1 – Convert the Percentage into Fraction, by dividing the given Percentage by 100

i.e., 12 % = 12/100

Step 2 – Convert the Fraction obtained at Step 1, into a Decimal form , by dividing numerator by Denominator

i.e., 12/100 = 0.12

Hence, 12 % = 0.12

Answer: (a) 4 %

Explanation: To convert the given ratio into Percentage we have to follow following steps:

Step 1 – Convert the ratio into Fraction

13 : 325 = 13/325

Step 2 – Multiply the Fraction by 100 and put the Percentage sign to the result (%)

( (13/325) x 100 ) %

On solving the given Fraction, we get,

= 4 %

Answer: (a) 32 %

Explanation: To convert a Fraction into Percentage, we need to multiply the Fraction by 100, and assign the Percentage (%) sign to it.

that is, ( (32/100)x 100 ) % = 32 %

Answer: (b) 9 : 25

Explanation: In order to express a Percentage as a ratio, we need to write

First term / Second term = Given Percentage / 100 = 36/100

On Simplifying, 36/100

HCF of 36 and 100 is 4

Divide both 36 and 100 by their HCF

(36÷4)/(100÷4) = 9/25 = 9 : 25

Answer: (a) 2250

Explanation: We can form a linear equation to express the given statement and then arrive at the number

Let the required number be x

We are given that

74 % of x is 1665

or in other words

(74/100) x x = 1665

On cross multiplication, we get

x = (1665×100)/74

x = 2250

Hence, the number is 2250

Answer: (c) 60 %

Explanation: Let a % of 90 litres = 54 litres

Step 1 – Convert the Percentage into Fraction

i.e., a % = a/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ( (a×90)/100 ) litres = ( 54 ) litres

(a×90)/100 = 54

a = (54×100)/90

a = 60

Hence, 54 litres is 60 % of 90 litres

Answer: (a) 50 %

Explanation: Let a % of 600 m = 300 m

Step 1 – Convert the Percentage into Fraction

i.e., a % = a/100

Step 2 – Multiply the Fraction with given quantity.

i.e., ((a×600)/100 ) m = ( 300 ) m

(a×600)/100 = 300

a = (300×100)/600

a = 50

Hence, 300 m is 50 % of 600 m

Answer: (b) 80 %

Explanation: Let a % of 20 kg = 16 kg

Step 1 – Convert the Percentage into Fraction

i.e., a % = a100

Step 2 – Multiply the Fraction with given quantity.

i.e., ( (a×20)/100 ) kg = ( 16 ) kg

(a×20)/100 = 16

a = (16×100)/20

a = 80

Hence, 16 kg is 80 % of 20 kg

Answer: (c) Increased by 19 %

Explanation: We can form a linear equation to express the given statement and then arrive at the number

Let the required number be “a”

Step I:

If the number is increased by 40 % it would be equal to ( Given Number + 40 % of Given Number )

= ( a + (40a/100) )

= ( 1 + (40/100) ) a

= (( 100+40)/100 ) a

= 140a/100

Step II:

If the number at Step 1 is decreased by 15 % it would be equal to ( Increased Number – 15 % of Increased Number )

= ( (140a/100) – (15/100) x (140a/100) )

= ( (140a/100) – (2100a/10000) )

= ( (14000−2100)/10000 ) a

= 11900a/10000

= 119a/100

Net Increase = ( (119a/100) – a )

= ( (119a−100a)/100 )

= 19a/100

Net Increase Percentage = ( (19a/100) x (1/a) x 100 )

= 19 %

Hence, Net Increase Percentage is = 19 %

Answer: (b) 600

Explanation: Here we take the help of linear equation. First we will form the equation.

Let the number be ‘a’

63 % of ‘a’ = (63/100)a

Since, the equation is:

a -(63/100)a = 222

Taking 100 as LCM on LHS

(100a−63a)/100 = 222

(37/100)a = 222

a = (222×100)/37

a = 600

Hence, the number is 600

Answer: (b) 500

Explanation: The problem can be solved using linear equation.

Let the number be ‘a’

125% of a = (125/100)a

We are given that, a + (125/100)a = 1125

Taking 100 as LCM on LHS

(100a+125a)/100 = 1125

(225/100)a = 1125

a = (1125×100)/225

a = 500
Hence, the number is 500

Answer: (c) ₹14500

Explanation: Let Tania’s monthly salary be ₹ a

Expenditure = 12 % of her Salary

Expenditure = 12 % of ₹ a

= (12/100) x ₹ a

= ₹ (12a/100)

Savings = Salary – Expenditure = 100 % – 12 % = 88 %

Savings = ₹ ( a – (12a/100))

= ₹ ( (100a−12a)/100 )

= ₹ (88a/100)

Given,

Savings = 12760 = 88 % of a

₹ 12760 = ₹ (88a/100)

a = ₹ ( (12760×100)/88 )

a = ₹ 14500

Tania’s monthly salary is ₹ 14500

Answer: (a) Gain % = 20 %

Explanation: C.P of 8 Oranges = ₹ 80

C.P of 1 Orange = ₹ 80/8 = 10

S.P of 3 Oranges = ₹ 36

S.P of 1 Orange = ₹ 36/3 = 12

On comparing C.P and S.P

10 , 12

10 < 12

Since,

C.P < S.P

Clearly, some gain has been earned

Gain = S.P – C.P

= 12 – 10

Gain = ₹ 2

Further

Gain% = ( (Gain/C.P) x 100 ) %

= ( (2/10) x 100 ) %

Gain% = 20 %

Answer: (c) 25 %

Explanation: C.P of an article = ₹ 8000

S.P of an article = ₹ 10000

Since,

S.P > C.P

So,

some profit has been earned

We know that,

Gain = S.P – C.P

= 10000 – 8000

= ₹ 2000

Further

Gain% = ( (Gain/C.P) x 100 ) %

= ( (2000/8000) x 100 ) %

Gain% = 25 %

Answer: (b) Gain% = 105 %

Explanation: C.P = ₹ 4600

S.P = ₹ 9430

Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which an Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

Clearly,

S.P > C.P

It means that a profit has been earned on such a Sale

Since,

Gain = S.P – C.P

Gain = 9430 – 4600

= ₹ 4830

Now,

we have to calculate Profit percentage

Gain % = (Gain(in₹)×100)/C.P %

Gain % = ( (4830/4600) X 100 ) %

= 105 %

So, the gain % is 105 %

Answer: (a) 15 %

Explanation: C.P of pair of shoes = ₹ 4000

Transportation charges = ₹ 500

Transportation charge is an overhead expenses and increases the cost of the pair of shoes

Overhead expenses = Overhead expenses are certain other expenses incurred by the Seller, and hence are added to his cost price.

Therefore,

Total Cost Price = Cost price + Transportation charges

Total Cost Price = 4000 + 500

= ₹ 4500

S.P of pair of shoes = ₹ 5175

Clearly,

S.P > C.P

This indicates that some profit has been earned

Therefore,

Gain = S.P – Total Cost Price

= 5175 – 4500

= ₹ 675

Since,

Gain% = ( (Gain/C.P) x 100 ) %

= ( (675/4500) x 100 ) %

= 15 %

Hence, Gain% = 15 %

Answer: (b) 20 %

Explanation: C.P of an article = ₹ 5500

S.P of an article = ₹ 4400

Since,

C.P > S.P

It means that

Some loss has been incurred

We know that,

Loss = C.P – S.P

= 5500 – 4400

= ₹ 1100

Further

Loss% = ( (Loss/C.P.) x 100 ) %

Loss% = ( (1100/5500) x 100 ) %

= 20 %

Answer: (c) 25 %

Explanation: C.P of an article = ₹ 7200

S.P of an article = ₹ 5400

Since,

C.P > S.P

It means that

Some loss has been incurred

We know that,

Loss = C.P – S.P

= 7200 – 5400

= ₹ 1800

Further

Loss% = ( (Loss/C.P) x 100 ) %

Loss% = ( (1800/7200) x 100 ) %

= 25 %

Answer: (b) 1 %

Explanation: C.P of laptop = ₹ 40000

Transportation charges = ₹ 1000

Transportation charge is an overhead expenses

Overhead expenses = Overhead expenses are certain other expenses incurred by the Seller, and hence are added to his cost price.

Therefore,

Total Cost Price = Cost price + Transportation charges

Total Cost Price = 40000 + 1000

= ₹ 41000

S.P of laptop = ₹ 40590

Clearly,

C.P > S.P

This indicates that some loss has been incurred

Therefore,

Loss = Total Cost Price – S.P

= 41000 – 40590

= ₹ 410

Since,

Loss% = ( (Loss/C.P) x 100 ) %

= ( (410/41000) x 100 ) %

= 1 %

Hence, Loss% = 1 %

Answer: (b) ₹ 31500

Explanation: Cost price of refrigerator = ₹ 27000

Discount% = 10 %

M.P = ?

Gain % = 5 %

M.P(marked price) = The price tagged on an article is called it’s marked price.

Discount = The offer of percentage of concession or rebate on the marked price is called Discount.

S.P = ((100+gainpercent)/100)×C.P

S.P = ((100+5)/100) x 27000

= (105/100) x 27000

S.P = ₹ 28350

Since Discount is always given on Marked price

Discount = 10 % of Marked price

Let the M.P be = ₹ x

Discount = (10/100) x x

Or

= 1x/10

S.P = M.P – Discount

= x – (1x/10)

= 10x−(1x/10)

S.P = 9x/10

But, the S.P calculated above = ₹ 28350

So,

9x/10 = 28350

x = (28350×10)/9

x = ₹ 31500

Hence, the Marked price = ₹ 31500

Answer: (c) ₹ 7400

Explanation: Let the original price of bag of cement be = x

GST = 24 %

GST (Goods and Services Tax) = GST is an indirect tax levied on goods and services.

GST = 24 % x original price of bag of cement

= (24/100) x x

= ₹ (6/25) x x

Price including GST = x + (6/25) x x

= (25x+6x)/25

= ₹ 31x/25

But the price included GST is given = ₹ 9176

Therefore,

31x/25 = 9176

x = (9176 × 25)/31

x = 7400

Hence, the original price of bag of cement = ₹ 7400

Answer: (c) ₹ 8610

Explanation: Price of watch = ₹ 8200

Sales tax = 5 %

Sales tax = Sales tax is charged by shopkeeper from the customer on selling price of an item.

Sales tax = 5 % of Price of watch

= (5/100) x 8200

= ₹ 410

Therefore,

The total price charged from the customer for watch = Price of watch + Sales tax

= ₹ ( 8200 + 410 )

= ₹ 8610

Answer: (b) ₹ 12960

Explanation: Price of transistor = ₹ 12000

Sales tax = 8 %

Sales tax = Sales tax is charged by shopkeeper from the customer on selling price of an item.

Sales tax = 8 % of Price of transistor

= (8/100) x 12000

= ₹ 960

Therefore,

The total price charged from the customer for transistor = Price of transistor + Sales tax

= ₹ ( 12000 + 960 )

= ₹ 12960

Answer: (b) ₹ 3000

Explanation: Given

S.P = ₹ 2700

Loss% = 10 %

C.P = ?

Cost Price (CP) – Price at which an Article is purchased.

Selling Price (SP) – Price at which the Article is sold.

Gain – Where the Selling Price is greater than the Cost Price, the difference between the SP and CP is the gain

Using formula to find C.P when Loss% and S.P are given

C.P = (100/(100−loss percent))×S.P

Substituting the value of Loss% and C.P in formula

C.P = (100/((100−10)) x 2700

= (100/90) x 2700

C.P = ₹ 3000

Answer: (b) ₹ 62900

Explanation: Marked price of an article = ₹ 74000

Discount = 15 %

M.P (Marked price) = The price tagged on an article is called it’s marked price.

Discount = The offer of percentage of concession or rebate on the marked price is called Discount.

Since Discount is always given on Marked price

Therefore,

Discount = 15 % of Marked price

= 15 % x 74000

= (15/100) x 74000

= ₹ 11100

So,

S.P = Marked price – Discount

= 74000 – 11100

S.P = ₹ 62900

Answer: (a) ₹ 2400

Explanation: In order to solve this problem, we would need to first find the Cost price of the Shirt given the SP and loss %.

Once we know the cost, we can find the revised Selling Price, given the new profit percentage on Cost

Finding the Cost Price

S.P of radio = ₹ 1400

Loss % = 30 %

Here C.P can be calculated

C.P = (100/(100−loss percent)) x S.P

C.P = (100/(100−30)) x 1400

= (100/70) x 1400

C.P = ₹ 2000

Now we know that the shirt has been sold at Gain = 20 %

Gain % = 20 %

S.P = ?

Since,

C.P = ₹ 2000

S.P = ((100+gain percent))/100)x C.P

= ((100+20)/100) x 2000

= (120/100) x 2000

S.P = ₹ 2400

So, the radio must be sold at ₹ 2400

Answer: (a) Rs. 3993

Explanation: From the given question, we can say

Principal = Rs. 3000

Rate = 10 %

Time (denoted by ‘n’) = 3 years

Computation of Maturity Amount

We know that

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 3000 x (1+(10/100))³

Amount = 3000 x ((100+10)/100)³

Amount = 3000 x(110/100)³

Amount = 3000 x(11/10)³ (Dividing both 110 and 100 by 10 )

Amount = 3000 x (11/10) x (11/10) x (11/10)

Amount = 3993000/1000

Amount = 3993

Hence if An amount of Rs. 3000 is compounded annually at the rate of 10 % p.a. for 3 years , amount to be paid after 3 years would be = 3993.​

Answer: (c) 92400

Explanation: In order to solve this problem, we would need to calculate Compound Interest for given years individually.

The amount for computing interest for second year will change to (Amount of Year 1 + Interest for Year 1)

Computation of Amount at the end of Year 1

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 80000 x (1+(5/100))¹

Amount = 80000 x (105/100)¹

Amount = 80000 x (21/20)

Amount = 84000

Amount at the End of Year 1 = Principal at the beginning of Year 2

Computation of Amount at the end of Year 2

Amount = Principal x (1+(Rate/100))ⁿ

= 84000 x (1+(10/100))¹

= 84000 x (110/100)¹

= 84000 x (11/10)

= 92400

Hence, the Amount = ₹ 92400

Answer: (c) ₹ 1995

Explanation: From the given question, we can say

Principal = ₹ 9500

Rate = 10 %

Time (denoted by ‘n’) = 2 years

We know that

Amount = Principal + Compound Interest

or

Compound Interest = Amount – Principal

In order to solve this problem, we would need to find out Amount, since Principal is already given

Computation of Maturity Amount

Amount = Principal (1+(Rate/100))ⁿ

Amount = 9500 (1+(10/100))²

Amount = 9500 x (110/100)²

Amount = 9500 x (11/10)²

Amount = 9500 x (11/10) x (11/10)

Amount = 1149500/100

Amount = 11495

Compound Interest = Amount – Principal

Compound Interest = 11495 – 9500

= 1995

Therefore, the Compound Interest = ₹ 1995

Answer: (c) ₹ 958728

Explanation: From the given question, we can say

Principal = Rs. 1280000

Rate = 15 % p.a.

(Since we have to compound the interest half-yearly i.e. two times in a year after every 6 months, we will divide the given rate by 2 as the Rate in the Formula for Amount is given Annually)

No. of Times Compounding is done (denoted by ‘n’) = 4 times

(Since the compounding is done half yearly, we will multiply the YEARS i.e. Time by 2)

We will first compute Amount and thereafter take the difference of Principal and Amount to find the Compound Interest.

Computation of Maturity amount

,

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 1280000 x (1+(15/100))⁴

Amount = 1280000 x (115/100)⁴

Amount = 1280000 x (23/20) x (23/20) x (23/20) x (23/20)

Amount = 1280000 x (279841/160000)

Amount = 8 x 279841

Amount = 2238728

Compound Interest = 2238728 – 1280000

= 958728

Answer: (c) ₹ 1261

Explanation: From the given question, we can say

Principal = Rs. 8000

Rate = 5 %

Time (denoted by ‘n’) = 3 years

In order to solve this problem, we would need to find out Amount, since Principal is already given

Computation of Maturity Amount

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 8000 x (1+(5/100))³

= 8000 x (105/100)³

= 8000 x (21/20)³

= 8000 x (21/20) x (21/20) x (21/20)

= 74088000/8000

= 9261

Compound Interest = 9261 – 8000

= 1261

Hence, the Compound Interest = ₹ 1261

Answer: (b) ₹ 1923

Explanation: We would compute both Simple Interest and Compound Interest individually for the given data, and then take their difference.

We are given that the

Principal = ₹ 30000

Rate = 10 % p.a.

(Since we have to compound the interest quarterly i.e. four times in a year after every 3 months, hence we will divide the given rate by 4)

No. of Times Compounding is done (denoted by ‘n’) = 4 times

Computation of Compound Interest

Amount = Principal x (1+(Rate/100))ⁿ

where ‘n’ denotes the number of times compounding is done. Like in this question compounding is done 4 times.

Amount = 30000 x (1+(10/100))⁴

Amount = 30000 x (110/100)⁴

Amount = 30000 x (110/100) x (110/100) x (110/100) x (110/100)

Amount = 30000 x (11/10) x (11/10) x (11/10) x (11/10)

Amount = 30000 x (14641/10000)

Amount = 3 x 14641

Amount = 43923

Compound Interest = 43923 – 30000

= 13923

Computation of Simple Interest

Simple Interest = (Principal × Rate × Time)/100

Simple Interest = (30000×40×4)/100

Simple Interest = 1200000/100

Simple Interest = 12000

Difference between Simple and Compound interest = Compound Interest – Simple Interest

= 13923 – 12000

Hence, the difference between Compound Interest and Simple Interest = ₹ 1923

Answer: (c) 30

Explanation: We would compute both Simple Interest and Compound Interest individually for the given data, and then take their difference.

We are given that the

Principal = Rs. 3000

Rate = 10 % p.a.

(Since we have to compound the interest half-yearly i.e. two times in a year after every 6 months, we will divide the given rate by 2 as the Rate in the Formula for Amount is given Annually)

No. of Times Compounding is done (denoted by ‘n’) = 2 times

(Since the compounding is done half yearly, we will multiply the YEARS i.e. Time by 2)

Computation of Compound Interest

Amount = Principal x (1+(Rate/100))ⁿ

where ‘n’ denotes the number of times the Compounding is required to be done.

Amount = 3000 x (1+(10/100))²

Amount = 3000 x (110/100)²

Amount = 3000 x (110/100) x (110/100)

Amount = 36300000/10000

Amount = 3630

Compound Interest = 3630 – 3000

= 630

Computation of Simple Interest

Simple Interest = (Principal × Rate × Time)/100

Simple Interest = (3000×20×1)/100

Simple Interest = 60000/100

Simple Interest = 600

Difference between Simple and Compound interest = Compound Interest – Simple Interest

630 – 600

Hence the difference between Compound Interest and Simple Interest = 30

Answer: (a) 0

Explanation: We would compute both Simple Interest and Compound Interest individually for the given data, and then take their difference.

We are given that the

Principal = Rs. 5500

Rate = 15 % p.a.

Time (denoted by ‘n’) = 1 year

Computation of Compound Interest

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 5500 x (1+(15/100))¹

Amount = 5500 x (115/100)¹

Amount = 5500 x (115/100)¹

Amount = 5500 x (23/20)

Amount = 275 x 23

Amount = 6325

Compound Interest = 6325 – 5500

= 825

Amount

Simple Interest = (Principal × Rate × Time)/100

= (5500×15×1)/100

= 82500/100

= 825

The difference between Simple Interest and Compound Interest on = Compound Interest – Simple Interest

= 825 – 825

= 0

Answer: (a) Rs. 3388

Explanation: From the given question, we can say

Principal = Rs. 2800

Rate = 10 %

Time (denoted by ‘n’) = 2 years

Computation of Maturity Amount

We know that

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 2800 x (1+(10/100))²

= 2800 x(110/100)² (Dividing both 110 and 100 by 10 )

= 2800 x(11/10)²

= 2800 x (11/10) x (11/10)

= 338800/100

= 3388

Hence, if An amount of Rs. 2800 is compounded annually at the rate of 10 % p.a. for 2 years , The amount to be paid after years would be = Rs. 3388

Answer: (b) ₹ 40000

Explanation: We will compute the Sum by of Amount.

Amount = ₹ 44100

Rate = 5 %

Time (denoted by ‘n’) = 2 years

Let the Principal/Sum be Rs. P

Amount = Principal x (1+(Rate/100))ⁿ

44100 = P x (1+(Rate/100))²

44100 = P x (105/100)²

44100 = P x (21/20)²

44100 = P x (21/20) x (21/20)

44100 = P x (441/400)

44100 x (400/441) = P

17640000/441 = P

40000 = P

Hence, the Sum = ₹ 40000

Answer: (b) 3

Explanation: We will compute TIME by of Amount.

Let us assume TIME to be ‘n’

Amount = Principal x (1+(Rate/100))ⁿ

21952 = 15625 x(1+(12/100))ⁿ

2195215625 = (1+(3/25))ⁿ

2195215625 = (28/25)ⁿ

We know that cube of 28 results in 21952 and cube of 25 results in 15625 .

(28/25)³ = (28/25)ⁿ

Since the Base is common, the index would remain the same

or

n = 3

Answer: (b) 25088

Explanation: Since the increase in population would be based on new population at the end of previous year, this problem can be solved using formula of Compounding of Interest

In this question, instead of AMOUNT we have = Population in the Year 1999

The PRINCIPAL is similar to the beginning population, i.e., Population in the Year 1997

Further the difference of Years will be equal to the time period ‘n’.

Difference of Years = 1999 – 1997 = 2

Population in the year 1997 = 20000

Rate of Increase = 12 % p.a.

Population in 1999 = Population 1997 x (1+(Rate/100))ⁿ

= 20000 x (1+(12/100))²

= 20000 x (112/100)²

= 20000 x (28/25)²

= 20000 x (28/25) x (28/25)

= 20000 x (784/625)

= 32 x 784

= 25088

Hence, the Population at the end of the year 1999 would be = 25088

Answer: (b) 110000

Explanation: Since the increase in population would be based on new population at the end of previous year, this problem can be solved using formula of Compounding of Interest

In this question, instead of AMOUNT we have = Population in the year 2004

The PRINCIPAL is similar to the beginning population, i.e., = Population in the year 2002

Further the difference of Years will be equal to the time period ‘n’.

Difference of Years = 2004 – 2002

= 2

Population in the year = 133100

Rate of Increase = 10 % p.a.

Population in the year 2004 = Population in the year 2002 x (1+(Rate/100))ⁿ

133100 = Population in the year 2002 x (1+(10/100))²

133100 = Population in the year 2002 x (110/100)²

133100 = Population in the year 2002 x (11/10)²

133100 = Population in the year 2002 x (11/10) x (11/10)

133100 = Population in the year 2002 x (121/100)

Population in the year 2002 = 133100 x (100/121)

Population in the year 2002 = 1100 x 100

Population in the year 2002 = 110000

Hence, the Population in the year 2002 was = 110000

Answer: (b) ₹ 951665

Explanation: From the given question, we can say

Principal = Rs. 650000

Rate = 10 % P.a.

(Since we have to compound the interest half-yearly i.e. two times in a year after every 6 months, we will divide the given rate by 2 as the Rate in the Formula for Amount is given Annually)

No. of Times Compounding is done (denoted by ‘n’) = 4 times

(Since the compounding is done half yearly, we will multiply the YEARS i.e. Time by 2)

Computation of Maturity Amount

Amount = Principal x (1+(Rate/100))ⁿ

Amount = 650000 x (1+(10/100))⁴

Amount = 650000 x (110/100)⁴

Amount = 650000 x (11/10)⁴

Amount = 650000 x (14641/10000)

Amount = 65 x 14641

= 951665

Hence, the Amount to be paid at end of 2 years would be = ₹ 951665

Answer: (a) 12 % p.a.

Explanation: We will compute RATE by of Amount.

Amount = Principal x (1+(Rate/100))ⁿ

7840 = 6250 x (1+(Rate/100))²

7840/6250 = (1+(Rate/100))²

784/625 = (1+(Rate/100))²

28/25 = 1 +(Rate/100)

(28/25) – 1 =Rate/100

3/25 = Rate/100

Rate = (3×100)/25

12 = Rate

Hence, the Rate per cent p.a is 12 %