Comparing Quantities Test

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Comparing Quantities Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

The profit after selling a pair of trousers for Rs. $$863$$ is the same as the loss incurred after selling the same pair of trousers for Rs. $$631$$. What is the cost price of the pair of trousers?

A
$$750$$

B
$$800$$

C
$$763$$

D
$$864$$

E
None of these

Solution

Let the C.P. be $$x$$.
Then, Profit $$= 863 - x$$
Loss $$= x - 631$$
Given, profit $$=$$ Loss
$$\Rightarrow 863 - x = x - 631$$
$$\Rightarrow 2x = 863+631 = 1494$$
$$\Rightarrow x = \dfrac {1494}{2} = 747$$
Cost Price $$=$$ Rs. $$747$$
Question 2
1 / -0

Mrs. Sharma purchased confectionery goods costing Rs. $$165$$ on which the rate of Sales Tax is $$6\%$$ and some tooth-paste, shaving-cream, soap, etc., costing Rs. $$230$$ on which the rate of Sales Tax is $$10\%$$. If she gives a five-hundred rupee note to the shopkeeper, what money will he return to Mrs. Sharma ?

A
Rs. $$72.10$$

B
Rs. $$12.10$$

C
Rs. $$42.10$$

D
Rs. $$92.10$$

Solution

Price of confectionery goods including Sales Tax $$ = \text{Rs. } 165 + 6\% \text{ of Rs. } 165 $$
$$=165+\cfrac{6}{100}\times 165$$
$$= \text{Rs. } 174.90$$

Price of tooth-paste, shaving-cream, soap, etc., including Sales Tax $$= \text{Rs. } 230 + 10\% \text{of Rs. } 230$$
$$=230+\cfrac{10}{100}\times 230 $$
$$= \text{Rs. } 253$$

$$\therefore$$ Total amount to be paid by Mrs. Sharma $$= \text{Rs. } 174.90 + \text{ Rs. } 253 = \text{Rs. } 427.90$$

Since Mrs. Sharma gave a five-hundred rupee note to the shopkeeper, the money that the shopkeeper will return to Mrs. Sharma $$= \text{Rs. } 500 - \text{Rs. } 427.90 = \text{Rs. } 72.10$$
Question 3
1 / -0

If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college?

A
$$ $ 20,000$$

B
$$ $ 15,000$$

C
$$ $ 12,000$$

D
$$ $ 10,000$$

E
$$ $ 9,000$$

Solution

$$\Rightarrow$$ We have given that $$r$$ = Percent interest.
$$\Rightarrow$$ We are also given that an investment with double in approximately $$70/r$$ years.
$$\Rightarrow$$ We are told to invest $$\$5,000$$ at $$8\%$$ for 18 years.
$$\Rightarrow$$ Put $$r=8$$
$$\Rightarrow$$ So, $$\dfrac{70}{8}$$ is about $$9\,years$$, meaning investment will double in $$9\,years$$.
$$\Rightarrow$$ In the first 9 years, $$\$5,000$$ doubles to $$\$10,000$$
$$\Rightarrow$$ In the next 9 years, $$\$10,000$$ doubles to $$\$20,000$$
$$\therefore$$ The approximate total amount of investment 18 years later, when Pat is ready for college is $$\$20,000$$.
Question 4
1 / -0

Directions For Questions

The cost of a machine depreciated by Rs. 4,752 during the second year and by Rs. 4,181.76 during the third year. Calculate:

...view full instructions

The rate of depreciation

A
10%

B
12%

C
14%

D
16%

Solution

Difference between depreciations of 2nd year and 3rd year
$$= Rs. 4,752 - Rs. 4,181.76 = Rs. 570.24$$
$$\Rightarrow$$ Depreciation of one year on Rs. 4,752 $$=$$ Rs. 570.24
$$\Rightarrow$$ Rate of depreciation $$=\displaystyle \frac{Rs. 570.24}{Rs. 4,752}\times 100\%=12\%$$ Ans.
Question 5
1 / -0

Lucy invested $$10,000$$ in a new mutual fund account exactly three years ago. The value of the account increased by $$10$$ percent during the first year,increased by $$5$$ percent during the second year, and decreased by $$10$$ percent during the third year. What is the value of the account today?

A
$$10350$$

B
$$10395$$

C
$$10500$$

D
$$11500$$

E
$$12705$$

Solution

The first year's increase of $$10$$ percent can be expressed as $$1.10$$, the second year's increase can be expressed as $$1.05$$ and third year's decrease can be expressed as $$0.90$$.
Multiply the original value accounts by each of these years changes,
$$10,000(1.10)(1.05)(0.90)=10,395$$
So, the value of the account today is Rs. $$10,395$$.
Question 6
1 / -0

In a city $$20 \%$$ of the total population is student community which is not employed , if the number of students who are unemployed is $$14000$$, then find the population of the city ?

A
$$56000$$

B
$$70000$$

C
$$40000$$

D
$$60000$$

E
None of these

Solution

$$\Rightarrow$$ In this question we have given, 20% of student are not employed. Total student population is 100%.
$$\Rightarrow$$ Total unemployed student is 14000.
$$\Rightarrow$$ Let $$x$$ be the total population
$$\Rightarrow$$ $$\dfrac{Number\, of\, unemployed\, student}{Total\, number\, of\, student}=\dfrac{Number\, of\, unemployed\, student\, \%}{Total\, number\, of\, student\, \%}$$
$$\Rightarrow$$ $$\dfrac{14000}{x}=\dfrac{20}{100}$$
$$\therefore$$ $$x=70000$$
$$\therefore$$ Population of city is $$70000$$
Question 7
1 / -0

A store currently charges the same price for each towel that it sells. If the current price of each towel were to be increased by $1, 10 fewer of the towels could be bought for $120, excluding sales tax. What is the current price of each towel?

A
$$ $ 1$$

B
$$ $ 2$$

C
$$ $ 3$$

D
$$ $ 4$$

E
$$ $ 12$$

Solution

$$\Rightarrow$$ Let number of towels bought for $$\$120$$ be $$n$$.

$$\Rightarrow$$ So price of single towel $$=\$\dfrac{120}{n}$$

$$\Rightarrow$$ Now price of $$1$$ towel increases by $$\$1$$

$$\Rightarrow$$ So, new price of single towel $$=\$\dfrac{120}{n}+1$$

$$\Rightarrow$$ Number of towel that could be bought at this price = $$n-10$$

$$\Rightarrow$$ So, new price of single towel = $$\dfrac{120}{(n-10)}$$

So, by equating both new price of single towel,

$$\Rightarrow$$ $$\dfrac{120}{n}+1=\dfrac{120}{(n-10)}$$

$$\Rightarrow$$ $$\dfrac{(120+n)}{n}=\dfrac{120}{(n-10)}$$

$$\Rightarrow$$ $$n^2-10n-1200=0$$

$$\Rightarrow$$ $$(n-40)(n+30)=0$$

$$\Rightarrow$$ $$n=40$$ or $$n=-30$$

$$\Rightarrow$$ Number of towels is $$40$$.

$$\Rightarrow$$ Current price per towel = $$\dfrac{120}{40}=\$\,3$$
Question 8
1 / -0

Naman purchased an old bike for Rs. $$20000$$. If the cost of his bike is depreciated at a rate of $$5\%$$ per annum, then find the cost of the bike after $$2$$ years?

A
Rs. $$18050$$

B
Rs. $$15000$$

C
Rs. $$1900$$

D
Rs. $$18000$$

Solution

Initial price of bike is Rs. $$20000$$, time $$=2$$ years, rate od depreciation $$=5\%$$
Cost of bike after $$2$$ years $$=$$ initial price of bike $$\times\left (1-\dfrac {\text{rate of depreciation} }{100}\right)^{\text{time}}$$
Therefore, cost of bike after $$2$$ years $$=20000\left (1-\dfrac {5}{100}\right)^2$$
$$=20000\left (\dfrac {19}{20}\right)^2$$
$$=18050$$
Thus, cost of bike after $$2$$ years is Rs. $$18050$$.
Question 9
1 / -0

Two merchants sell, each an article for Rs. $$1000$$. If merchant $$A$$ computes his profit on cost price, while merchant $$B$$ computes his profit on selling price, they end up making profits of $$25\%$$. By how much is the profit made by Merchant $$B$$ greater than that of Merchant $$A$$?

A
$$66.67$$

B
$$125$$

C
$$50$$

D
$$100$$

Solution

Merchant $$B$$ computes his profit as a percentage of selling price. He makes a profit of $$25\%$$ on selling price of Rs. $$1000$$ i.e. his profit $$=25\%$$ of $$1000=$$ Rs. $$250$$.
Merchant $$A$$ computes his profit as a percentage of cost price.
Therefore, when he makes a profit of $$25\%$$ or $$\left (\dfrac{1 }{4}\right)^{th}$$ of his cost price, then his profit expressed as a percentage of selling price $$=$$ or $$20\%$$ of selling price.
So, merchant $$A$$ makes a profit of $$20\%$$ of Rs. $$1000=$$ Rs. $$200$$.
Merchant B makes a profit of Rs. $$250$$ and merchant A makes a profit of Rs. $$200$$.
Hence, merchant $$B$$ makes Rs. $$50$$ more profit than merchant $$A$$.
Question 10
1 / -0

Sahil purchased a machine at Rs. $$10000$$, then got it repaired at Rs. $$5000$$ and gave its transportation charge Rs. $$1000$$. Then he sold it with $$50\%$$ of profit. At what price he actually sold it?

A
$$20000$$

B
$$22000$$

C
$$24000$$

D
$$26000$$

Solution

Sahil purchased a machine at $$Rs.\ 10000$$.

Its repairing cost is $$Rs.\ 5000$$ and transportation cost is $$Rs.\ 1000$$.

$$\therefore \ $$Total C.P.$$=10000+5000+1000=Rs.\ 16000$$

Given that, on selling the machine, Sahil gained a profit of $$50\%$$.

$$\therefore \ $$ S.P. $$=16000+ \dfrac{50}{100} \times 16000$$

$$=16000+8000$$

$$=Rs.\ 24000$$

Hence, Sahil sold the machine for $$Rs.\ 24000$$.

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Comparing Quantities Test - 46

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Comparing Quantities Test - 46

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SHARING IS CARING

Question 1

$$750$$

$$800$$

$$763$$

$$864$$

None of these

Solution

Question 2

Rs. $$72.10$$

Rs. $$12.10$$

Rs. $$42.10$$

Rs. $$92.10$$

Solution

Question 3

$$ $ 20,000$$

$$ $ 15,000$$

$$ $ 12,000$$

$$ $ 10,000$$

$$ $ 9,000$$

Solution

Question 4

10%

12%

14%

16%

Solution

Question 5

$$10350$$

$$10395$$

$$10500$$

$$11500$$

$$12705$$

Solution

Question 6

$$56000$$

$$70000$$

$$40000$$

$$60000$$

None of these

Solution

Question 7

$$ $ 1$$

$$ $ 2$$

$$ $ 3$$

$$ $ 4$$

$$ $ 12$$

Solution

Question 8

Rs. $$18050$$

Rs. $$15000$$

Rs. $$1900$$

Rs. $$18000$$

Solution

Question 9

$$66.67$$

$$125$$

$$50$$

$$100$$

Solution

Question 10

$$20000$$

$$22000$$

$$24000$$

$$26000$$

Solution

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