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CBSE Questions for Class 8 Maths Comparing Quantities Quiz 7 - MCQExams.com
CBSE
Class 8 Maths
Comparing Quantities
Quiz 7
the customer's price
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0%
$$Rs. 9,620$$
0%
$$Rs. 10,850$$
0%
$$Rs. 12,870$$
0%
$$Rs. 14,240$$
Explanation
M.P=Rs. 11700
Sales tax=10%
$$\therefore $$Customer price$$=11700+10\% of 11700$$
$$\Rightarrow 11700+\frac{10}{100}\times 11700$$
$$\Rightarrow 11700+1170=Rs.12870$$
Madan purchases a compact computer system for Rs. 47,700 which includes 10% rebate on the marked price and then 6% Sales Tax on the remaining price. Find the marked price of the computer.
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0%
$$Rs. 40,000$$
0%
$$Rs. 50,000$$
0%
$$Rs. 60,000$$
0%
$$Rs. 70,000$$
Explanation
Let the marked price of the computer=Rs.x
Rebate=10%
Rebate price=$$\frac{10x}{100}=\frac{x}{10}$$
Remaining price=$$x-\frac{x}{10}=\frac{9x}{10}$$
Rate of sales tax=6%
$$\therefore$$Tax on the remaining price$$=6\% of \frac{9x}{10}$$
$$\Rightarrow \frac{6}{100}\times \frac{9x}{10}=\frac{54x}{1000}$$
Price after the tax$$=\frac{9x}{10}+\frac{54x}{1000}$$
$$\Rightarrow \frac{900x+54x}{1000}=\frac{954x}{1000}=\frac{477x}{500}$$
Then according to the question
$$\therefore \frac{477x}{500}=47700$$
$$\Rightarrow x=\frac{47700\times 500}{477}=50,000$$
The price at which the camera can be bought from the shopkeeper.
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0%
$$Rs. 1,396$$
0%
$$Rs. 1,496$$
0%
$$Rs. 1,596$$
0%
$$Rs. 1,696$$
Explanation
Printed price=Rs.1600
Rate of sales tax=6%
Total S.P$$=1600+6\% of 1600$$
$$\Rightarrow 1600+\frac{6}{100}\times 1600$$
$$\Rightarrow 1600+96=Rs.1696$$
Hence the price at which camera can be bought from the shopkeeper=Rs.1669
Ashopkeeper sells an article at its list price (Rs. 3000) and charges sales-tax at the rate of 12%. If the VAT paid by the shopkeeper is Rs. 72, at what price did the shopkeeper buy the article inclusive of sale-tax?
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0%
$$Rs. 2,590$$
0%
$$Rs. 2,688$$
0%
$$Rs. 2,451$$
0%
$$Rs. 2,100$$
Explanation
Let the profit of the retailer=Rs.x
VAT paid by the shopkeeper is Rs.72 at the rate of sales tax 12%
$$\therefore$$ 12% of x=72
$$\Rightarrow \frac{12}{100}x=72$$
$$\Rightarrow x=\frac{72\times 100}{12}=Rs.600$$
$$\therefore$$ cost price to the retailer$$=3000-600=Rs.2400$$
Sales tax paid by the retailer=12% of 2400
$$\Rightarrow \frac{12}{100}\times 2400=Rs.288$$
$$\therefore$$Price of the article including sales tax paid by the shopkeeper $$=2400+288=Rs.2688$$
An article is sold at a profit of $$14\%$$. If both the cost price and selling price are Rs. $$132$$ less, the profit would be $$12\%$$ more. Find the cost price.
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0%
Rs. $$286$$
0%
Rs. $$268$$
0%
Rs. $$266$$
0%
Rs. $$ 276$$
Explanation
Let cost price be $$x$$
SP $$= x + 14\% $$ of $$ x=x+\dfrac{14x}{100} = 1.14x$$
If CP and SP were Rs. $$132$$ less, the profit would be $$(14 + 12)\%$$
New, CP $$= (x - 132)$$
SP $$ = 1.14x - 132$$
SP $$-$$ CP $$= 26\%$$ of $$(x - 132)$$
$$1.14x - 132 - x + 132 = \dfrac{26\times (x-132)}{100}$$
$$0.14x \times100 = 26x - 3432$$
$$14x - 26x = -3432$$
$$12x = 3432$$
$$x = 286$$
So, CP was Rs. $$286$$.
An article is sold at $$10\%$$ profit . If its cost price and selling price are $$50$$ less, the profit would be $$5\%$$ more. Find the cost price?
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0%
$$40$$
0%
$$45$$
0%
$$60$$
0%
$$120$$
0%
$$150$$
Explanation
$$let,\,C.P=x$$
and $$gain\%=10\%$$
$$S.P=\left(\dfrac{100+gain}{100}\right)\times C.P\Rightarrow S.P=\dfrac{110x}{100}$$
$$new \,C.P=C.P-50$$
$$new \,S.P=S.P-50$$
$$gain\%=5\%\,more \,than\,previous=15\%$$
$$\Rightarrow New\,S.P=\left(\dfrac{100+gain}{100}\right)\times(New\,C.P)$$
$$\Rightarrow (S.P-50)=\left(\dfrac{100+15}{100}\right)\times(C.P-50)$$
$$\Rightarrow (\dfrac{110x}{100}\,-\,50)=\dfrac{115}{100}\times (x-50)$$
$$\Rightarrow \dfrac{110x}{100}\,-50=\dfrac{115x}{100}-\dfrac{115}{2}$$
$$\Rightarrow \dfrac{115}{2}\,-\,\dfrac{100}{2}=\dfrac{115x}{100}-\dfrac{110x}{100}$$
$$\Rightarrow \dfrac{15}{2}=\dfrac{5x}{100}$$
$$\therefore x=150$$
A man bought a chair and sold it at a gain of $$5$$%. If he had bought it at $$5$$% less and sold it for Rs $$45$$ more, he would have gained $$20$$% . Find the cost price of the chair?
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0%
Rs. $$400$$
0%
Rs. $$500$$
0%
Rs. $$450$$
0%
Rs. $$550$$
Explanation
Let CP = x
SP = x + 5% of x = 1.05x
Now,
If he had bought it in 5% less,
CP = x - 5% of x = 0.95x
New SP = 1.05x + 45
New gain = 20% of 0.95x
1.05 + 45 - 0.95x = 20% of 0.95x
0.1x + 45 = 0.19x
0.09x = 45
x = 500
CP = Rs. 500
A man loses $$13 \%$$ by selling a book at a certain price. If he had sold it for Rs. $$9.75$$ more he would have gained $$26\%$$. Find the cost price of the book ?
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0%
$$25$$
0%
$$30$$
0%
$$39$$
0%
$$42$$
Explanation
Let C.P. be $$x$$
S.P. $$= x - 13\%$$ of $$x =x-\dfrac{13}{100}\times x= 0.87 x$$
If he had sold it in Rs. $$9.75$$ more, he had gained,
New S.P. $$= 0.87x + 9.75$$
Then according to the question, we have
$$\Rightarrow 0.87x + 9.75 = x + 26\%$$ of $$ x$$
$$\Rightarrow 0.8x + 9.75 = 1.26x$$
$$\Rightarrow 0.39x = 9.75$$
$$\Rightarrow x = \dfrac{9.75}{0.39}$$
$$\Rightarrow x = 25$$
So, the cost Price was Rs. $$25$$.
A man purchased a bullock and a cart for Rs. $$1800$$. He sold the bullock at a profit of $$20\%$$
and cart at a profit of $$30\%$$. His total profit was $$\dfrac{155}{6} \%$$. Find the cost price of bullock?
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0%
$$650$$
0%
$$750$$
0%
$$900$$
0%
$$800$$
Explanation
Cost Price of bullock and cart $$=$$ Rs. $$1800$$.
Let bullock price was $$X$$ and cart price was $$(1800-X)$$
SP of both $$= 1800+\dfrac {155}{6}\%$$ of $$1800 = 2265$$
Profit $$=$$ Rs. $$465$$
$$X + 20\%$$ of $$X + (1800-X)+30\%$$ of $$(1800-X)=2265$$
$$1.2 X+1800-X +\dfrac { (54000 - 30 X)}{100} = 2265$$
$$1.2 X+1800 - X + 540 - 0.3 X = 2265$$
$$1.2 X - 1.3 X = 2265 - 1800 - 540$$
$$- 0.1 X = -75$$
$$X = 750$$
Cost of bullock is Rs. $$750$$.
A man has to sell pulse at a loss of 10%. If he increases the selling price by Rs 5 per kg, he would make a profit of 15%. Find the cost price and initial selling price per kg of pulse.
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0%
Rs 20, Rs 18
0%
Rs 18, Rs 20
0%
Rs 15, Rs 18
0%
Rs 18, Rs 15
Explanation
Let C.P. $$= x$$
S.P. $$= x - 10\% $$ of $$ x=x-\dfrac{10x}{100}=\dfrac{90x}{100} = 0.9x$$
New S.P. $$= 0.9x + 5$$
Gain $$ =$$ S.P. $$ -$$ C.P.
$$\Rightarrow 15\% $$ of $$ x = 0.9x + 5 - x$$
$$\Rightarrow 0.15x + x - 0.9x=5$$
$$\Rightarrow 0.25x = 5$$
$$\Rightarrow x = \dfrac{5}{0.25 }=$$ Rs. $$20$$
So, initial selling price $$= 20 - 10\% $$ of $$ 20=20-\frac{10}{100}\times 20 =$$ Rs. $$ 18.$$.
A and B invest in a business in the ratio of $$3 : 2$$. If $$5$$ % of total profit goes to charity and A's share is Rs. $$8550$$, then total profit is :
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0%
Rs. $$14000$$
0%
Rs. $$15000$$
0%
Rs. $$16000$$
0%
Rs. $$16500$$
Explanation
Let Total Profit is Rs. 100
After 5 % charity, profit = Rs 95
So, share of A = (95 *3)/5 = Rs. 57
Now, comparing,
A share $$57$$ on profit $$100$$
So, A's share 1 on profit $$= 100/57$$
Thus, A shares Rs. $$8550$$ on =$$ (8550 *100)/57 = Rs. 15000$$
If the duty on an article is reduced by 40% of its present rate, by how much percent must the consumption increase in order that the revenue remains unaltered ?
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0%
$$ \dfrac{182}{ 3} \%$$
0%
$$\dfrac{194}{3}\%$$
0%
$$\dfrac{200}{3} \%$$
0%
None of these
Explanation
Let initial rate was $$100$$
Now, $$40 \%$$ decrement in duty,
rate $$=$$ $$ 100 - 40 = 60$$
To make their revenue $$ 100 $$, consumption must be increase of $$40$$.
So, $$\%$$ increment $$ =$$ $$\dfrac{ (40 \times 100)} { 60} = \dfrac{200}{ 3} \%$$
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?
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0%
$$750$$
0%
$$800$$
0%
$$763$$
0%
$$864$$
0%
None of these
Explanation
Let the C.P. be $$x$$.
Then, Profit $$= 863 - x$$
Loss $$= x - 631$$
Given, p
rofit $$=$$ Loss
$$\Rightarrow 863 - x = x - 631$$
$$\Rightarrow 2x = 863+631 = 1494$$
$$\Rightarrow x = \dfrac {1494}{2} = 747$$
Cost Price $$=$$ Rs. $$747$$
A man sold $$18$$ toys for Rs. $$16800$$, gaining thereby the cost price of $$3$$ toy. find the cost price of a toy.
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0%
Rs. $$ 600$$
0%
Rs. $$ 800$$
0%
Rs. $$ 500$$
0%
Rs. $$ 900$$
Explanation
Let the cost of one toy be $$ x$$
Then, cost of $$18$$ toy $$= 18x$$
Gain $$ = 3x $$
SP of $$18$$ toys $$ =$$ Rs. $$16800$$
Gain $$=$$ S.P. $$-$$ C.P.
$$\Rightarrow 3x = 16800 - 18x$$
$$\Rightarrow 21x = 16800$$
$$\Rightarrow x = \dfrac{16800}{21}=$$ Rs. $$ 800$$
Therefore, cost price of a toy is Rs. $$800$$.
Jacob bought a scooter for a certain sum of money. He spent $$10\%$$ of the cost on repairs and sold the scooter for a profit of Rs. $$1100$$. How much did he spend on repairs if he made a profit of $$20\%$$.
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0%
Rs. $$400$$
0%
Rs. $$440$$
0%
Rs. $$500$$
0%
Rs. $$550$$
Explanation
Total profit $$=$$ Rs. $$1100$$
Percent profit $$= 20$$
$$\Rightarrow 20\% = 1100$$
So, $$100\%=5500$$
Rs. $$5500$$ would be the cost price for Jacob.
He spends $$10\%$$ on repairing. so, the cost price was Rs. $$5000$$.
Thus, he spends Rs. $$500$$ on repairing.
The prices of two articles are in the ratio of $$3 : 4$$. If the price of the first article be increased by $$10$$% and that of the second by Rs $$4$$, the original ratio remains the same. The original price of second article is ?
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0%
Rs $$30$$
0%
Rs $$35$$
0%
Rs $$40$$
0%
Rs $$45$$
Explanation
Let price of two articles are $$3x$$ and $$4x$$
After increment price become $$3.3x$$ and $$(4x + 4)$$
Now, according to question,
$$\dfrac{3.3x}{(4x + 4)} = \dfrac{3}{4}$$
$$13.2x = 12x + 1$$
$$13.2x - 12x = 12$$
$$1.2x = 12$$
$$x = 10$$
So, original price of second article $$= 4x = 4 \times10 = $$Rs $$40$$.
A sells a bicycle to $$B$$ at a profit of $$20\%$$. $$B$$ sells it to $$C$$ at a profit of $$25\%$$. If $$C$$ pays Rs. $$225$$ for it, the cost price of the bicycle for $$A$$ is
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0%
Rs. $$110$$
0%
Rs. $$120$$
0%
Rs. $$125$$
0%
Rs. $$150$$
Explanation
Let the C.P for $$A=$$ Rs. $$ x$$
Profit $$=20\%$$
Then S.P for $$A$$ $$=\dfrac{120}{100}\times x=$$ Rs. $$\dfrac{12x}{10}$$
S.P. for $$A=$$ C.P. for $$B$$ $$=$$ Rs. $$\dfrac{12x}{10}$$
Profit $$=25\%$$
then S.P for $$B$$ $$=\dfrac{25}{100}\times \dfrac{12x}{10}=$$ Rs. $$\dfrac{3x}{2}$$
S.P. for $$B =$$ C.P. for $$C=$$ Rs. $$\dfrac{3x}{2}$$
According to the question,
C.P for $$C=$$ Rs. $$225$$
$$\therefore \dfrac{3x}{2}=225$$
$$\Rightarrow x=\dfrac{225\times 2}{3}=$$ Rs. $$150$$
Hence, cost of the bicycle for $$A$$ is Rs. $$150$$.
A shopkeeper bought 2 dozen of apples at RS. 58 and 4 dozen of oranges at Rs.He sold the oranges to ram at 5% loss and apples to shyam at 15 % gain. What is overall loss or gain percentage in the transaction?
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0%
$$4.1$$
0%
$$4.05$$
0%
$$4.3$$
0%
$$4.0$$
Explanation
$$\text{CP of 2 dozen of apples}=Rs.\ 58$$
$$Loss=5\%$$
$$SP=58-0.05\times 58=Rs.\ 55.1$$
$$\text{CP of 4 dozen of oranges}=Rs.\ 48$$
$$Gain=15\%$$
$$SP=48+0.15\times 48=Rs.\ 55.2$$
Total cost price $$=48+58=Rs.\ 106$$
Total selling price $$=55.1+55.2=Rs.\ 110.3$$
Hence, $$SP>CP$$.
So, profit $$=110.3-106=Rs.\ 4.3$$
$$P(\%)=\dfrac{4.3}{106}\times 100=4.05\% $$
The present population of a city is $$18522000$$. If it has been increased at the rate of $$5\%$$ per annum, find the population(approx) after $$3$$ years.
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0%
$$2522200$$
0%
$$16000000$$
0%
$$21441530$$
0%
$$252200$$
Explanation
Given, Principal amount $$18522000$$, rate of interest $$=5\%$$ and period $$=3$$ years
Total Population $$=18522000\left [1+\dfrac{5}{100}\right]^3$$
$$=18522000\left [1+\dfrac{1}{20}\right]^3$$
$$=21441530$$(approx)
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 ?
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0%
$$56000$$
0%
$$70000$$
0%
$$40000$$
0%
$$60000$$
0%
None of these
Explanation
$$\Rightarrow$$ 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 $$70000$$
Tarun bought an article for Rs. $$8,000$$ and spent Rs. $$1,000$$ for transportation. He marked the article at Rs. $$11,700$$ and sold it to a customer. If the customer had to pay $$10\%$$ sales tax, find:
(i) the customer's price
(ii) Tarun's profit percent
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0%
(i) Rs. $$10,636$$ (ii) $$13\%$$
0%
(i) Rs. $$11,190$$ (ii) $$20\%$$
0%
(i) Rs. $$12,870$$ (ii) $$30\%$$
0%
(i) Rs. $$13,220$$ (ii) $$43\%$$
Explanation
C.P $$=Rs.8000$$
Transportation Cost $$=Rs.1000$$
Total C.P $$=8000+1000=Rs.9000$$
M.P $$=Rs.11700$$
Sales tax Rate$$=10\%$$
(i) Customer's price $$=$$ S.P including sales tax$$=11700+\dfrac{10}{100}\times 11700=Rs.12870$$
(ii) Profit $$=11700-9000=2700$$
Profit $$\%=\dfrac{2700}{9000}\times 100=30$$%
The catalogue price of a colour T.v. is Rs. $$18,000$$. The shopkeeper sells it to a customer at a discount of $$20\%$$ on the catalogue price. He gives a further off-season discount of $$10\%$$ on the balance. But Sales Tax at $$10\%$$ is charged on the remaining amount. Find the Sales Tax amount, the customer has to pay
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0%
Rs. $$1296$$
0%
Rs. $$126$$
0%
Rs. $$196$$
0%
None of these
Explanation
List price $$=Rs.18000$$
Discount$$=20\%\ of\ 18000$$=$$\dfrac{20}{100}\times 18000=Rs.3600$$
T.V price after discount=$$18000-3600=Rs.14400$$
Off season discount$$=10\%$$ of $$14400$$$$=\dfrac{10}{100}\times 14400=1440$$
T.V Price after off season discount=$$14400-1440=Rs.12960$$
(1)The amount of sales tax the customer has to pay$$=10\%$$ of $$12960$$
$$\Rightarrow \dfrac{10}{100}\times 12960=Rs.1296$$
A shopkeeper buys an article at a rebate of $$20\%$$ on the printed price. He spends Rs. $$40$$ on transportation of the article. After charging a sales tax of $$7\%$$ on the printed price, he sells the article for Rs. $$1,070$$. Find his gain as per cent.
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0%
$$4 \displaystyle\frac{1}{16}$$%
0%
$$11 \displaystyle\frac{1}{13}$$%
0%
$$14 \displaystyle\frac{1}{20}$$%
0%
$$19 \displaystyle\frac{1}{21}$$%
Explanation
Let printed price of the article be Rs. $$x$$
According to the given statement:
$$x \left(\displaystyle\frac{100+7}{100}\right) = 1070$$
$$ \Rightarrow x = 1000$$
$$\therefore$$ Printed price of the article $$= Rs. 1,000$$
Given, that the shopkeeper buys the article at $$20\%$$ rebate
$$\therefore$$ For the shopkeeper, C.P. of the article $$= Rs. 1,000 - 20\%$$ of Rs. $$1,000 = Rs. 800$$
Since, he spends Rs. $$40$$ on the transportation of the article
$$\Rightarrow$$ Total (actual) cost price $$= Rs. 800 + Rs. 40 = Rs. 840$$
The selling price (excluding sales tax) $$=$$ Printed price$$ = Rs. 1,000$$
$$\Rightarrow$$ Profit $$= Rs. 1,000 - Rs. 840 = Rs. 160$$
And, Profit $$\% = \displaystyle\frac{160}{840} \times 100 \%$$
$$= 19 \displaystyle\frac{1}{21}$$%
Smith buys an article marked at Rs. $$2,200.$$ The rate of Sales Tax is $$12\%$$. He asks the shopkeeper to reduce the price of the article to such an extent that he does not have to pay anything more than Rs. $$2,240$$ including Sales Tax. Calculate the reduction, as percent, needed in the marked price of the article.
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0%
$$9\displaystyle\frac{2}{13}$$%
0%
$$9\displaystyle\frac{1}{11}$$%
0%
$$7\displaystyle\frac{1}{9}$$%
0%
None of these
Explanation
Let the cost of the article be reduced to Rs. $$x$$.
According to question,
$$Cost\ price+Sales \ tax=Rs.2240$$
$$\therefore x + 12$$% of $$x = 2,240$$
$$\Rightarrow x+\dfrac{12}{100}\times x=2240$$
$$\Rightarrow \dfrac{112x}{100}=2240$$
$$\Rightarrow x=\dfrac{2240\times 100}{112}$$
$$\Rightarrow x = 2,000$$
$$\therefore $$ Reduced price of the article $$= Rs. 2,000$$
Reduction needed $$= Rs. 2,200 - Rs. 2,000 = Rs. 200$$
Hence, reduction as percent of marked-price $$= \displaystyle\frac{Reduction}{Marked \ price} \times 100$$%
$$= \displaystyle\frac{200}{2,200} \times 100$$% $$= 9\displaystyle\frac{1}{11}$$%
A shopkeeper buys an article for Rs. $$1,500$$ and spends $$20\%$$ of the cost on its packing, transportation, etc. Then he marks the article at a certain price. If he sells the article for Rs. $$2,452.50$$ including $$9\%$$ Sales Tax on the price marked, find his profit as per cent.
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0%
$$25\%$$
0%
$$27\%$$
0%
$$31\%$$
0%
$$37\%$$
Explanation
Let marked price of the article be Rs. $$x$$.
$$\therefore$$ Rs. $$x +$$ Rs. $$\displaystyle\frac{9x}{100} = $$ $$Rs. 2,452.50 $$ $$\because$$ Sales-tax $$= 9\%$$
On solving, we get : $$x = 2,250$$
$$\therefore$$ Marked price of the article $$= Rs. 2,250 =$$ Its selling price
Since, the shopkeeper buys the article for Rs. $$1,500$$ and spends $$20\%$$ of the cost as overheads,
$$\therefore$$ Total cost price of the article $$= Rs. 1,500 + 20\%\ of\ Rs. 1,500$$
$$= Rs. 1,500 + Rs. 300 = Rs. 1,800$$
Profit = Selling price - Total cost price
$$= Rs. 2,250 - Rs. 1,800 = Rs. 450$$
Profit % $$= \displaystyle\frac{450}{1800} \times 100$$% $$= 25\%$$
The rate of depreciation
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0%
10%
0%
12%
0%
14%
0%
16%
Explanation
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.
The present population of town is $$12500$$ and it is increasing at the rate of $$8\%$$ per annum. What will be the population of town after two years?
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0%
$$14000$$
0%
$$12500$$
0%
$$14580$$
0%
$$25000$$
Explanation
Given, present population of town $$=12500, R=8\%$$
Therefore, population after $$2$$ years $$=12500\times\begin{Bmatrix}1+\dfrac{8}{100}\end{Bmatrix}^2$$
$$=12500\times\begin{Bmatrix}\dfrac{108}{100}\end{Bmatrix}^2$$
$$=14580$$
A car is valued at Rs. 500000 If sits value depreciates at 2 % p.a. What will be its value after three years?
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0%
Rs. 29440
0%
Rs. 470430
0%
Rs. 470596
0%
Rs. 470596
The price of an article inclusive of $$12\%$$ Sales Tax is Rs. $$2,016$$. Find its marked price. If the Sales Tax is reduced to $$7\%$$, how much less does the customer pay for the article ?
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0%
Rs. $$52$$ more
0%
Rs. $$66$$ more
0%
Rs. $$80$$ less
0%
None of these
Explanation
Let marked price be Rs. $$x$$.
$$\therefore x + 12$$% of $$x = 2,016$$
$$\Rightarrow x+\dfrac{12}{100}x=2016$$
$$\Rightarrow \dfrac{112x}{100}=2016$$
$$\Rightarrow x=\dfrac{2016\times 100}{112}$$
$$\Rightarrow x = 18\times 00$$
$$\Rightarrow x = 1,800$$
$$\therefore$$ Marked price of the article $$= Rs. 1,800$$
Since, new Sales Tax $$= 7\%$$
$$\therefore$$ Now, the customer will pay
$$= Rs. 1,800 +7% $$ of $$1800$$
$$= \displaystyle\frac{107}{100} \times $$ $$Rs. 1,800 = Rs. 1,926$$
$$\therefore$$ Customer will pay for the article $$= Rs. (2,016 - 1,926) less = Rs. 90 less$$
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?
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0%
Rs. $$18050$$
0%
Rs. $$15000$$
0%
Rs. $$1900$$
0%
Rs. $$18000$$
Explanation
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. $$18050$$.
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?
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0%
$$ $ 20,000$$
0%
$$ $ 15,000$$
0%
$$ $ 12,000$$
0%
$$ $ 10,000$$
0%
$$ $ 9,000$$
Explanation
$$\Rightarrow$$ 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 $$\$20,000$$.
By selling a jeans for Rs. $$432$$, John loses $$4\%$$. For how much should John sell it to gain $$6\%$$?
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0%
Rs. $$458$$
0%
Rs. $$428$$
0%
Rs. $$477$$
0%
Rs. $$436$$
Explanation
SP of the shirt $$=432$$
Loss $$=4\%$$
Therefore, $$CP$$ of the shirt $$=\dfrac {100}{100-\text{loss} \%}\times SP$$
$$=\dfrac {100}{100-4}\times 432$$
$$=450$$
Now,$$CP=450$$, desired gain $$=6\%$$
Desired$$,SP=\dfrac {100+\text{gain} \%}{100}\times CP$$
$$=\dfrac {100+6}{100}\times 450$$
$$=477$$
Hence, the desired selling price is Rs. $$477$$.
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?
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0%
$$10350$$
0%
$$10395$$
0%
$$10500$$
0%
$$11500$$
0%
$$12705$$
Explanation
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. $$10,395$$.
The cost of an article is decreased by $$15$$ $$\%$$. If the original cost is Rs. $$80$$, find the new cost.
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0%
$$68$$
0%
$$65$$
0%
$$57$$
0%
$$55$$
Explanation
Original cost of an article $$=$$ Rs. $$80$$
Decrease in cost is $$=15$$ $$\%$$ of Rs. $$80=\dfrac{15}{100}\times80=\dfrac{1200}{100}=$$ Rs. $$12$$
Therefore, new cost of an article $$=$$ Rs. $$80-$$ Rs. $$12=$$ Rs. $$68$$.
The price of a TV is Rs. $$13,000$$. The sales tax on it is $$12\%$$. Find the amount that Vinod will have to pay if he buys it.
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0%
Rs. $$14560$$
0%
Rs. $$2620$$
0%
Rs. $$1456$$
0%
Rs. $$1560$$
Explanation
Sales tax $$=$$ Price $$\times$$ Tax Rate
$$=\dfrac{12}{100} \times 13000=1560$$
Final price $$=13000+1560=14560$$
Therefore, the amount that Vinod will have to pay will be Rs. $$14560$$.
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?
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0%
$$ $ 1$$
0%
$$ $ 2$$
0%
$$ $ 3$$
0%
$$ $ 4$$
0%
$$ $ 12$$
Explanation
$$\Rightarrow$$ 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 = $$\dfrac{120}{40}=\$\,3$$
In a culture, the present bacteria count was found as $$67,60,000$$. It was found that the number was increased by $$4\%$$ per hour. Find the number of hours before which the bacteria count was $$62,50,000$$.
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0%
$$1$$ hr
0%
$$2$$ hr
0%
$$3$$ hr
0%
$$4$$ hr
Explanation
Let the number of years be $$n$$
$$\Rightarrow 67,60,000=62,50,000 \left [1+\dfrac{4}{100}\right]^n$$
$$\Rightarrow \dfrac{676}{625}=[1+\dfrac{1}{25}]^n$$
$$\Rightarrow \left [\dfrac{26}{25}\right]^2=\left [\dfrac{26}{25}\right]^n$$
$$n=2$$ hrs
Thus, the number of hours are $$2$$.
The present population of a city is $$15000$$. If it increases at the rate of $$4\%$$ per annum. Find its population after $$2$$ years.
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0%
$$16224$$
0%
$$26224$$
0%
$$36224$$
0%
$$46224$$
Explanation
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 $$ 16224$$.
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 $$A$$?
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0%
$$66.67$$
0%
$$125$$
0%
$$50$$
0%
$$100$$
Explanation
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 $$A$$.
A person incurs a loss of $$5\%$$ be selling a watch for $$Rs.1140$$. At what price should the watch be sold to earn $$5\%$$ profit ?
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0%
Rs.$$1200$$
0%
Rs.$$1230$$
0%
Rs.$$1260$$
0%
Rs.$$1290$$
Explanation
Formulas required for this sum:
$$Cost\ price(C.P)=\dfrac{100}{100-Loss\%}\times Selling\ price$$
$$Selling\ price(S.P)=\dfrac{100+Profit\%}{100\%}\times Cost\ price$$
$$Loss\%=5\%$$
$$S.P=Rs\ 1140$$
$$C.P=\dfrac{100}{100-5}\times1140$$
$$C.P=\dfrac{100}{95}\times1140$$
$$C.P=1200$$
We found that the $$Cost\ price\ is\ Rs\ 1200$$
$$Profit\%=5\%$$
$$S.P=\dfrac{100+5}{100}\times 1200$$
$$S.P=\dfrac{105}{100}\times 1200$$
$$S.P=Rs\ 1260$$
What is the percentage approximate for the fraction $$\dfrac{12}{35}$$?
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0%
$$34.28\%$$
0%
$$23.78\%$$
0%
$$30.12\%$$
0%
$$29.10\%$$
Explanation
To get the percent you will divide the numerator by the denominator.
Then multiply $$100$$ to the answer and add the percent sign.
therefore, the percentage approximate value if $$\dfrac{12}{35}\times 100 = 34.28$$ $$\%$$.
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?
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0%
$$20000$$
0%
$$22000$$
0%
$$24000$$
0%
$$26000$$
Explanation
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 $$Rs.\ 24000$$.
A merchant marks his goods in such a way that the profit on sale of $$50$$ articles is equal to the selling price of $$25$$ articles. What is his profit margin ?
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0%
$$25\%$$
0%
$$50\%$$
0%
$$100\%$$
0%
$$66.67\%$$
Explanation
Let the selling price per article be $$=$$ Rs. $$1$$
Therefore, selling price of $$50$$ articles $$=$$ Rs. $$50$$
Profit on sale of $$50$$ articles $$=$$ selling price of $$25$$ articles $$=$$ Rs. $$25$$.
S.P $$=$$Rs. $$ 50$$, Profit $$=$$ Rs. $$25$$
Therefore, CP $$=$$ Rs.$$(50-25)=$$ Rs. $$25$$
Profit $$\%=\dfrac{25}{25}\times100\%=100\%$$
Which of the following statement/formulae is correct for Percentage Error?
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0%
$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Exact\quad value } $$
0%
$$\cfrac { \left| Exact\quad value-Approx.value \right| }{ Exact\quad value } $$
0%
$$\cfrac { \left| Approx.value-Exact\quad value \right| }{ Approx.value } \quad $$
0%
None
Explanation
The approximate values is the estimated values and the exact value is the real value.
Mary expected to $$120$$ people for her wedding, but only $$60$$ people appeared. What was the percentage error?
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0%
$$75\%$$
0%
$$25\%$$
0%
$$55\%$$
0%
$$100\%$$
Explanation
Given, absolute value $$=120$$ and exact value $$=60$$
We know, percentage error $$=$$ $$\left | \dfrac{\text{Absolute value} -\text{ exact value}}{\text{exact value}} \right |\times 100$$ $$\%$$
$$=$$ $$\left | \dfrac{120 - 60}{60} \right |\times 100$$ $$\%$$
$$=$$ $$\left | \dfrac{ 60}{60} \right |\times 100$$ $$\%$$
$$=$$ $$\left | 1\right |\times 100$$ $$\%$$
$$= 100\%$$
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?
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0%
$$Rs.16828.45$$
0%
$$Rs.26826.07$$
0%
$$Rs.36828.56$$
0%
$$Rs.46828.80$$
Explanation
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 $$Rs.26,826.07$$.
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.
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0%
$$\text{Rs. }1916$$
0%
$$\text{Rs. }2916$$
0%
$$\text{Rs. }3916$$
0%
$$\text{Rs. }4916$$
Explanation
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 $$\text{Rs. }2916.$$
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?
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0%
$$10497$$
0%
$$20497$$
0%
$$30497$$
0%
$$40497$$
Explanation
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 $$= 20497$$
Martin is planning which crops to plant on his farm for the upcoming season. He has enough seed to plant $$4$$ acres of wheat and $$7$$ acres of soybeans, but the total area of farmland he owns is only $$9$$ acres. He earns $$ $90$$ per acre for every acre of wheat planted and $$ $120$$ for every acre of soybeans planted, and he must pay a $$10$$% tax on all money he earns from selling his crops. Calculate the maximum profit, in dollars that Martin can earn from planting wheat and soybeans this season.
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0%
$$918$$
0%
$$916$$
0%
$$914$$
0%
$$912$$
Explanation
In order to find the greatest profit, maximize the number of acres of soybeans Martin plants, since soybeans bring in more money per acre than does wheat. At most, MartIncan plant $$7$$ acres of soybeans.
Therefore, the most money he can make on soybeans is $$7\times 120=840$$.
He then has $$9 - 7 = 2$$ acres left on which to plant wheat.
The money he makes from this wheat is $$2\times 90=180$$.
The total amount Martin makes before taxes is therefore $$840 + 180 = 1,020$$.
The tax on this money equals $$1020\times 0.10=102$$.
Subtract the amount Martin pays in taxes to get $$1,020 - 102 = 918$$ profit.
The correct answer is $$918$$.
An _______ is a government tax on the taxable profit earned by an individual or corporation.
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0%
income tax
0%
sales tax
0%
property tax
0%
VAT
Explanation
An income tax is a government tax on the taxable profit earned by an individual or corporation.
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