Comparing Quantities Test

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Comparing Quantities Test - 35

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Question 1
1 / -0

A tradesman gives 4% discount on his marked price and gives 1 article free for buying every 15 articles and thus gains 35%. By what per cent is the marked price increased above the cost price ?

A
50%

B
40%

C
60%

D
80%

Solution

Let the C.P. of each article be Rs.x
Then C.P. of 16 articles = Rs.16x
S.P. of 15 articles = 135% of Rs.16x = Rs. $$ \left ( \cfrac{135\times 16x}{100} \right )=Rs.\cfrac{108x}{5}$$
S.P. of 1 article $$ =Rs.\left ( \cfrac{108x}{5}\times \cfrac{1}{15} \right )=Rs.\cfrac{36x}{25}$$
Let M.P. = Rs.100 Then S.P. =Rs.96 after a discount of 4%
$$ \therefore$$ If S.P. =Rs.96 then M.P. = Rs.100
If S.P. =Rs.$$ \cfrac{36x}{25}$$ then M.P. $$ =Rs.\left ( \cfrac{100}{96}\times \cfrac{36x}{25} \right )=Rs.\cfrac{3x}{2}$$
$$ \therefore$$ % increase in M.P. over C.P. $$ =\cfrac{\cfrac{3x}{2}-x}{x}\times 100=\left ( \cfrac{x}{2}\times \cfrac{1}{x}\times 100 \right )\%=50\%$$
Question 2
1 / -0

The population of a town increases by 5% annually. If the population in 2009 is 1,38,915, what was it in 2006 ?

A
100,000

B
110,000

C
120,000

D
130,000

Solution

Let the population in 2006 be $$x$$.
Pop. in $$\displaystyle 2009=x\times \left ( 1+\cfrac{5}{100} \right )^{3}$$
$$\Rightarrow 1,38,915=x\times \left ( \cfrac{21}{20} \right )^{3}$$
$$\Rightarrow x=\cfrac{138915\times 8000}{9261}=120000$$
Question 3
1 / -0

A shopkeeper fixes the marked price of an item 35% above its cost price. What is the percentage of discount allowed to gain 8% ?

A
27%

B
25%

C
22%

D
20%

Solution

Let the C.P. = Rs.100 Then M.P. = Rs.135 and S.P. $$\displaystyle =\frac{108\times 100}{100}=Rs.108$$
$$\displaystyle \therefore Discount\: =Rs.135-Rs.108=Rs.27$$ Discount % $$\displaystyle =\frac{27}{135}\times 100=20\%$$
Question 4
1 / -0

The population of a town was 1,60,000 three years ago. If it increased by 3%, 2.5% and 5%, respectively, in the last three years, then what is the present population?

A
166,366

B
177,366

C
166,377

D
177,377

Solution

Present population $$\displaystyle =1,60,000\left ( 1+\frac{3}{100} \right )\left ( 1+\frac{2.5}{100} \right )\left ( 1+\frac{5}{100} \right )$$
$$\displaystyle =160000\times \frac{103}{100}\times \frac{102.2}{100}\times \frac{105}{100}=177366$$
Question 5
1 / -0

The population of a village is 10000. If the population increases by 10% in the first year, by 20% in the second year, and due to mass exodus, it decreases by 5% in the third year, what will be its population after 3 years ?

A
10,540

B
11,540

C
12,540

D
13,540

Solution

Population after 3 years $$\displaystyle =10000\left ( 1+\frac{10}{100} \right )\left ( 1+\frac{20}{100} \right )\left ( 1-\frac{5}{100} \right )$$
$$\displaystyle =10000\times \frac{11}{10}\times \frac{6}{5}\times \frac{19}{20}=12540$$
Question 6
1 / -0

The half life of Uranium - $$233$$ is $$160000$$ years i.e., Uranium $$233$$ decays at a constant rate in such a way that it reduces to $$50\%$$ in $$160000$$ years. in how many years will it reduce to $$25 \%$$?

A
$$80000$$ years

B
$$240000$$ years

C
$$320000$$ years

D
$$40000$$ years

Solution

The half life of $${ U }_{ 233 }$$ $$=160000$$ yrs

$$\therefore $$ It becomes $$50\%$$ of the original amount in $$160000$$ yrs

Now $$25\%=$$ $$\dfrac { 1 }{ 2 } \times$$ $$50\%$$

i.e Another half life is needed to reduce the original amount into $$25\%$$

So, another $$160000$$ yrs. will reduce the original amount into $$25\%$$

$$\therefore $$ The required total number of years

$$=160000+160000=320000$$

i.e $$320000$$ yrs
Question 7
1 / -0

The population of a town increases annually by $$25 \% $$ if the preset population is $$1$$ crore, then what was the difference between the population three years ago and two years ago?

A
$$25,00,000$$

B
$$12,80,000$$

C
$$15,60,000$$

D
None of these

Solution

Given that,
Population increases annually by $$25\%$$

Let population $$3$$ years ago be $$x$$
$$\Rightarrow$$ Population two years ago was $$\dfrac{125}{100}x$$

Population last year was $$\dfrac{125\times125}{10000}x$$

Population this year is $$\dfrac{125\times125\times125}{1000000}x = 10000000$$

$$\Rightarrow$$ Population three years ago was $$x = 5120000$$

$$\therefore$$ Difference between population three years ago and two years ago $$= 25\%$$ of $$x$$
$$= 1280000$$
Question 8
1 / -0

The annual increase in the population of a town is 10%. If the present population of the town is 180,000, then what will be its population after two years ?

A
200,000

B
210,000

C
217,800

D
207,800

Solution

Population after 1 year $$= 180000 + 180000\times \dfrac{10}{100} = 198000$$
Population after 2 years $$ = 198000 + 198000\times \dfrac{10}{100}$$
$$=217800$$
Question 9
1 / -0

How much per cent more than the cost price should a shopkeeper marks his goods so that after allowing a discount of 20% on the marked price he gains 10% ?

A
37.5%

B
40.0%

C
42.5%

D
45.0%

Solution

Let the C.P. = Rs.100 Gain = 10%
$$\displaystyle \therefore$$ S.P. = Rs.110............(i)
Let the M.P. = Rs.x Discount = 20%
$$\displaystyle \therefore$$ S.P. = 80% of $$\displaystyle x=\frac{80x}{100}...........(ii)$$
From (i) and (ii) $$\displaystyle \frac{80x}{100}=110\Rightarrow x=Rs.\frac {100\times 100}{80}=Rs.137.5$$
$$\displaystyle \therefore$$ Marked price =$$\displaystyle37\frac{1}{2}\%$$ above C.P.
Question 10
1 / -0

Ramesh bought a calculator with 20% discount on the tag-price. He obtained 10% profit by selling it for Rs.440. What was the tag-price?

A
Rs.500

B
Rs.400

C
Rs.480

D
Rs.360

Solution

Let the tag price of the calculator be Rs.$$x$$
Then C.P. of Ramesh after 20% discount $$= 80\%$$ of $$\displaystyle Rs.x=Rs.\cfrac{80x}{100}=Rs.\frac{4x}{5}$$ ........(i)
Also, given S.P. = Rs.440 and Profit = 10%
$$\displaystyle \therefore C.P.=Rs.\left ( \frac{440\times 100}{110} \right )Rs.400$$ ............(ii)
From (i) and (ii)
$$\displaystyle \therefore \frac{4x}{5}=400\Rightarrow x=Rs.500$$

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Comparing Quantities Test - 35

Result

Comparing Quantities Test - 35

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SHARING IS CARING

Question 1

50%

40%

60%

80%

Solution

Question 2

100,000

110,000

120,000

130,000

Solution

Question 3

27%

25%

22%

20%

Solution

Question 4

166,366

177,366

166,377

177,377

Solution

Question 5

10,540

11,540

12,540

13,540

Solution

Question 6

$$80000$$ years

$$240000$$ years

$$320000$$ years

$$40000$$ years

Solution

Question 7

$$25,00,000$$

$$12,80,000$$

$$15,60,000$$

None of these

Solution

Question 8

200,000

210,000

217,800

207,800

Solution

Question 9

37.5%

40.0%

42.5%

45.0%

Solution

Question 10

Rs.500

Rs.400

Rs.480

Rs.360

Solution

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