Comparing Quantities Test

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Comparing Quantities Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

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?

A
$$650$$

B
$$750$$

C
$$900$$

D
$$800$$

Solution

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. $$750$$.
Question 2
1 / -0

Ajay had purchased a second hand scooter for Rs. $$18000$$ and spent Rs $$1800$$ for repairs. After $$1$$ year he wanted to sell the scooter. At what price should he sell it to gain $$\cfrac{100}{9}$$ $$\%$$ , if $$\cfrac {100}{11}$$ $$\%$$ is to be deducted at the end of every year on account of depreciation?

A
Rs. $$18000$$

B
Rs. $$19800$$

C
Rs. $$20000$$

D
Rs. $$22500$$

Solution

Cost price $$=$$ Rs. $$18000$$
cost on repairing $$=$$ Rs. $$1800$$
So, Total cost $$= 18000 + 1800 = 19800$$
Depreciation $$= \dfrac {100}{11} \%= 9.09 \%$$
Gain $$= \dfrac {100}{9 } \%= 11.11 \%$$
After depreciation, the cost price would be $$= 19800 - 9.09\%$$ of $$19800 =$$ Rs. $$18000$$
S.P. to gain $$11.11 \% = 18000 + 11.11 \%$$ of $$18000 = 18000 + 1999.8 =$$ Rs. $$20000$$.
Question 3
1 / -0

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?

A
$$40$$

B
$$45$$

C
$$60$$

D
$$120$$

E
$$150$$

Solution

$$let,\,C.P=x$$
and $$gain\%=10\%$$
$$S.P=\left(\dfrac{100+gain}{100}\right)\times C.P\Rightarrow S.P=\dfrac{110x}{100}$$

$$new \,C.P=C.P-50$$
$$new \,S.P=S.P-50$$
$$gain\%=5\%\,more \,than\,previous=15\%$$
$$\Rightarrow New\,S.P=\left(\dfrac{100+gain}{100}\right)\times(New\,C.P)$$
$$\Rightarrow (S.P-50)=\left(\dfrac{100+15}{100}\right)\times(C.P-50)$$
$$\Rightarrow (\dfrac{110x}{100}\,-\,50)=\dfrac{115}{100}\times (x-50)$$
$$\Rightarrow \dfrac{110x}{100}\,-50=\dfrac{115x}{100}-\dfrac{115}{2}$$
$$\Rightarrow \dfrac{115}{2}\,-\,\dfrac{100}{2}=\dfrac{115x}{100}-\dfrac{110x}{100}$$
$$\Rightarrow \dfrac{15}{2}=\dfrac{5x}{100}$$
$$\therefore x=150$$
Question 4
1 / -0

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 ?

A
$$ \dfrac{182}{ 3} \%$$

B
$$\dfrac{194}{3}\%$$

C
$$\dfrac{200}{3} \%$$

D
None of these

Solution

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 $$ =$$ $$\dfrac{ (40 \times 100)} { 60} = \dfrac{200}{ 3} \%$$
Question 5
1 / -0

A man sold $$18$$ toys for Rs. $$16800$$, gaining thereby the cost price of $$3$$ toy. find the cost price of a toy.

A
Rs. $$ 600$$

B
Rs. $$ 800$$

C
Rs. $$ 500$$

D
Rs. $$ 900$$

Solution

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. $$800$$.
Question 6
1 / -0

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 $$20\%$$.

A
Rs. $$400$$

B
Rs. $$440$$

C
Rs. $$500$$

D
Rs. $$550$$

Solution

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.
Question 7
1 / -0

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 ?

A
Rs $$30$$

B
Rs $$35$$

C
Rs $$40$$

D
Rs $$45$$

Solution

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 $$40$$.
Question 8
1 / -0

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

A
Rs. $$110$$

B
Rs. $$120$$

C
Rs. $$125$$

D
Rs. $$150$$

Solution

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. $$150$$.
Question 9
1 / -0

A shopkeeper bought 2 dozen of apples at RS. 58 and 4 dozen of oranges at Rs. 48. 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?

A
$$4.1$$

B
$$4.05$$

C
$$4.3$$

D
$$4.0$$

Solution

$$\text{CP of 2 dozen of apples}=Rs.\ 58$$
$$Loss=5\%$$
$$SP=58-0.05\times 58=Rs.\ 55.1$$

$$\text{CP of 4 dozen of oranges}=Rs.\ 48$$
$$Gain=15\%$$
$$SP=48+0.15\times 48=Rs.\ 55.2$$

Total cost price $$=48+58=Rs.\ 106$$
Total selling price $$=55.1+55.2=Rs.\ 110.3$$
Hence, $$SP>CP$$.
So, profit $$=110.3-106=Rs.\ 4.3$$
$$P(\%)=\dfrac{4.3}{106}\times 100=4.05\% $$
Question 10
1 / -0

A trader buys good at $$20\%$$ discount on marked price. If he wants to make a profit of $$25\%$$ after allowing a discount of $$20\%$$. By what percent should his marked price be greater than the original marked price?

A
$$15\%$$

B
$$65\%$$

C
$$25\%$$

D
$$20\%$$

Solution

Let the marked price be Rs. $$100$$.
The trader buys at discount of $$20\%$$.
Hence, his cost price $$=100-20\%$$ of $$100$$ $$=$$ Rs. $$80$$.
He wants to make profit $$25\%$$, hence his selling price is $$=80+25\%$$ of $$80$$ $$=$$ Rs. $$100$$.
However, he wants to get this Rs. $$100$$ after allowing a discount of $$20\%$$, i.e. he will sell at $$80\%$$ of his marked price.
Hence, his marked price $$=\dfrac {100}{0.8}=$$ Rs. $$125$$ which is $$25\%$$ more than the original marked price.

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Re-Attempt Weekly Quiz Competition

Comparing Quantities Test - 45

Result

Comparing Quantities Test - 45

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Rank

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SHARING IS CARING

Question 1

$$650$$

$$750$$

$$900$$

$$800$$

Solution

Question 2

Rs. $$18000$$

Rs. $$19800$$

Rs. $$20000$$

Rs. $$22500$$

Solution

Question 3

$$40$$

$$45$$

$$60$$

$$120$$

$$150$$

Solution

Question 4

$$ \dfrac{182}{ 3} \%$$

$$\dfrac{194}{3}\%$$

$$\dfrac{200}{3} \%$$

None of these

Solution

Question 5

Rs. $$ 600$$

Rs. $$ 800$$

Rs. $$ 500$$

Rs. $$ 900$$

Solution

Question 6

Rs. $$400$$

Rs. $$440$$

Rs. $$500$$

Rs. $$550$$

Solution

Question 7

Rs $$30$$

Rs $$35$$

Rs $$40$$

Rs $$45$$

Solution

Question 8

Rs. $$110$$

Rs. $$120$$

Rs. $$125$$

Rs. $$150$$

Solution

Question 9

$$4.1$$

$$4.05$$

$$4.3$$

$$4.0$$

Solution

Question 10

$$15\%$$

$$65\%$$

$$25\%$$

$$20\%$$

Solution

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