Comparing Quantities Test

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Comparing Quantities Test - 48

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Question 1
1 / -0

Mary expected to $$120$$ people for her wedding, but only $$60$$ people appeared. What was the percentage error?

A
$$75\%$$

B
$$25\%$$

C
$$55\%$$

D
$$100\%$$

Solution

Given, absolute value $$=120$$ and exact value $$=60$$
We know, percentage error $$=$$ $$\left | \dfrac{\text{Absolute value} -\text{ exact value}}{\text{exact value}} \right |\times 100$$ $$\%$$
$$=$$ $$\left | \dfrac{120 - 60}{60} \right |\times 100$$ $$\%$$
$$=$$ $$\left | \dfrac{ 60}{60} \right |\times 100$$ $$\%$$
$$=$$ $$\left | 1\right |\times 100$$ $$\%$$
$$= 100\%$$
Question 2
1 / -0

Which of the following statement/formulae is correct for Percentage Error?

A
$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Exact\quad value } $$

B
$$\cfrac { \left| Exact\quad value-Approx.value \right| }{ Exact\quad value } $$

C
$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Approx.value } \quad $$

D
None

Solution

The approximate values is the estimated values and the exact value is the real value.
Question 3
1 / -0

The present population of a town is $$14000$$. If it increases at the rate of $$10\%$$ per annum, what will be its population after $$4$$ years?

A
$$10497$$

B
$$20497$$

C
$$30497$$

D
$$40497$$

Solution

Population after $$4$$ years $$=$$ $$\text{Present Population}$$ $$\left (1+\dfrac{r}{100}\right)^n$$

Population after $$4$$ years $$=$$ $$14000 \times \left (1+\dfrac{10}{100}\right)^4$$

$$=14000\times1.4641$$

$$=20497$$

Population after $$4$$ years $$= 20497$$
Question 4
1 / -0

The cost of a machine is Rs. $$24500$$. After two year the value of that machine was depreciated by $$6\%$$. Find the value of machine after two year.

A
$$11560$$

B
$$21560$$

C
$$31560$$

D
$$41560$$

Solution

Given: Principal $$= 24500$$, $$r = 6\%, n = 2$$
Reduction $$= 6\%$$ of Rs. $$24500$$ per year
$$= \dfrac{24500\times 6\times 2}{100}=2940$$
Value at the end of $$2$$ years $$= 24500 - 2940 =$$ Rs. $$21560$$
Question 5
1 / -0

A machine was purchased $$3$$ years ago. Its value decreases by $$5\%$$ every year. Its present value is $$Rs.23000$$. For how much money was the machine purchased?

A
$$Rs.16828.45$$

B
$$Rs.26826.07$$

C
$$Rs.36828.56$$

D
$$Rs.46828.80$$

Solution

Given,
Depreciated value $$A=Rs. 23,000$$
Rate of depreciation $$R=5\%$$
Time $$T=3$$ years
So, $$n=3$$
let the machine was purchased in $$Rs. P$$

Depreciated value,$$ A = P\left( 1-\dfrac { R }{ 100 } \right) ^{ n } $$
$$\Rightarrow 23000 = P \times \left( 1-\dfrac { 5 }{ 100 } \right) ^{ 3 } $$

$$ \Rightarrow 23000= P \times \left( 1-\dfrac { 1 }{ 20 } \right) ^{ 3 } $$

$$\Rightarrow 23000 = P \times \left(\dfrac { 19}{ 20 } \right) ^{ 3 } $$

$$\Rightarrow 23000 = P \times \dfrac { 19}{ 20 } \times\dfrac{19}{20}\times\dfrac{19}{20} $$

$$\Rightarrow P=23,000 \times \dfrac { 20}{ 19 } \times\dfrac{20}{19}\times\dfrac{20}{19} $$

$$\Rightarrow P=26,826.07$$

Therefore, the machine was purchased for $$Rs.26,826.07$$.
Question 6
1 / -0

A machine depreciates at the rate of $$10\%$$ of its value at the beginning of a year. If the present value of a machine is $$\text{Rs. }4000$$, find its value after $$3$$ years.

A
$$\text{Rs. }1916$$

B
$$\text{Rs. }2916$$

C
$$\text{Rs. }3916$$

D
$$\text{Rs. }4916$$

Solution

Given:-
$$P = \text{Rs. }4000$$
$$r = 10\%$$
$$n = 3$$ years

Now, as the value of the machine depriciates every year by $$10\%,$$
$$\begin{aligned}{}A& = P{\left( {1 - \frac{r}{{100}}} \right)^n}\\ &= 4000{\left( {1 - \frac{{10}}{{100}}} \right)^3}\\& = 4000{\left( {0.9} \right)^3}\\& = \text{Rs. }2916\end{aligned}$$

Hence, the value of the machine after $$3$$ years is $$\text{Rs. }2916.$$
Question 7
1 / -0

The population of a town increases every year at rate of $$10\%$$. Find the population of the town three years hence if it is $$13000$$ now.

A
$$17303$$

B
$$27303$$

C
$$37303$$

D
$$47303$$

Solution

Given, $$P = 13000, r = 10\%, n = 3$$ years
We know, $$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 13000\left [\left (1+\dfrac{10}{100}\right)^3\right]$$
$$\Rightarrow A= 17303$$
Hence, the population of the town $$3$$ years is $$17303$$.
Question 8
1 / -0

A company bought a car that had a value of Rs. $$10000$$. each year the value of the car depreciates by $$15\%$$. What is the value of the car at the end of $$2$$ years.

A
$$7225$$

B
$$6225$$

C
$$5225$$

D
$$4225$$

Solution

Given, $$P = 10000, r = 15\%$$ depreciation, $$n = 2$$ years
We know $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 10000\left [\left (1-\dfrac{15}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$7225$$
Therefore, the value of car after $$2$$ years is Rs. $$7225$$.
Question 9
1 / -0

Grace bought a new van for Rs. $$15000$$. Each year, the value of her car depreciated by $$10\%$$. Calculate the value of car before $$2$$ years.

A
$$12150$$

B
$$22150$$

C
$$32150$$

D
$$42150$$

Solution

Given, $$P = 15000, r = 10\% $$ depreciated, $$n = 2$$ years
We know, $$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$$
$$\Rightarrow A = 15000\left [\left (1-\dfrac{10}{100}\right)^2\right]$$
$$\Rightarrow A =$$ Rs. $$12150$$
The value of car before $$2$$ years is Rs. $$12150$$.
Question 10
1 / -0

The present population of a city is $$15000$$. If it increases at the rate of $$4\%$$ per annum. Find its population after $$2$$ years.

A
$$16224$$

B
$$26224$$

C
$$36224$$

D
$$46224$$

Solution

Given, $$P=15000, r=4\%, n=2$$ years
Population after $$2$$ years $$=$$ $$P[(1+\frac{R}{100})^2]$$
$$=15000\left [(1+\dfrac{4}{100}\right)^2]$$
$$=15000\times 1.0816$$
$$=16224$$
Population after $$2$$ years is $$ 16224$$.

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 48

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Comparing Quantities Test - 48

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SHARING IS CARING

Question 1

$$75\%$$

$$25\%$$

$$55\%$$

$$100\%$$

Solution

Question 2

$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Exact\quad value } $$

$$\cfrac { \left| Exact\quad value-Approx.value \right| }{ Exact\quad value } $$

$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Approx.value } \quad $$

None

Solution

Question 3

$$10497$$

$$20497$$

$$30497$$

$$40497$$

Solution

Question 4

$$11560$$

$$21560$$

$$31560$$

$$41560$$

Solution

Question 5

$$Rs.16828.45$$

$$Rs.26826.07$$

$$Rs.36828.56$$

$$Rs.46828.80$$

Solution

Question 6

$$\text{Rs. }1916$$

$$\text{Rs. }2916$$

$$\text{Rs. }3916$$

$$\text{Rs. }4916$$

Solution

Question 7

$$17303$$

$$27303$$

$$37303$$

$$47303$$

Solution

Question 8

$$7225$$

$$6225$$

$$5225$$

$$4225$$

Solution

Question 9

$$12150$$

$$22150$$

$$32150$$

$$42150$$

Solution

Question 10

$$16224$$

$$26224$$

$$36224$$

$$46224$$

Solution

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