2 Marks Questions - MATHS STD 7 Questions - Vidyadip

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2 Marks Questions

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53 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain.

Answer

Given, children who like getting wet in the rain = 8% of 25
$=\frac{8}{100} \times 25=2$
Hence, 2 children like getting wet in the rain.
Alternate method
Out of 100, 8 children like getting wet in the rain
Therefore, out of 25, the number of children that like
getting wet in the rain $=\frac{8}{100} \times 25=2$
Hence, 2 children like getting wet in the rain.

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Question 22 Marks

What per cent of these figures are shaded?

Answer

(i) It is clear from the figure, that shaded parts are 3 out of 4
Shaded parts in fraction$=\frac{3}{4}$
$\therefore$ Percentage of shaded part $=\left(\frac{3}{4} \times 100\right) \%=75 \%$
Hence, 75% of the figure is shaded.
(ii) From the figure,
fraction of figure which is not shaded
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}=\frac{1+1+2+4}{16}=\frac{8}{16}=\frac{1}{2}$
i.e. half of the figure is shaded.
Percentage of figure which is shaded
$=\left(\frac{1}{2} \times 100\right) \%=50 \%$
Hence, 50% of the figure is shaded.

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Question 32 Marks

If 65% of students in a class have a bicycle, what per cent of the students do not have bicycles?

Answer

Given, percentage of students having bicycle = 65%
This means that if there were 100 students in a class, out of them, 65 would have bicycles and then remaining students would not have bicycles.
$\therefore$ The remaining students would not have bicycle
$=100-65=35$
Hence, 35% of students do not have bicycles.

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Question 42 Marks

We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?

Answer

According to the question, if there are 100 fruits in a basket, out of them 50 are apples and 30 are oranges, then remaining would be mangoes.
Now, mangoes = (100 - (50 + 30))%
= (100 - 80)% = 20%
Hence, there are 20% mangoes in the basket.

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Question 52 Marks

Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?

Answer

Putting the data in the tabular form, we get

Bangels	Number	Fraction	Denominator Hundred	In Percentage
Gold	20	$\frac{20}{30}$	$\frac{20}{30} \times \frac{100}{100}=\frac{200}{3} \times \frac{1}{100}$	$66.67 \%$ or $66 \frac{2}{3} \%$
Silver	10	$\frac{10}{30}$	$\frac{100}{30} \times \frac{100}{100}=\frac{100}{3} \times \frac{1}{100}$	$33.33 \%$ or $33 \frac{1}{3} \%$
Total	30

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Question 62 Marks

A collection of 10 chips with different colours is given

Colour	Number	Fraction	Denominator Hundred	In Percentage
Green
Blue
Red
Total

Complete the table and find the percentage of chips of each colour.

Answer

Given, table is completed as follows

Colour	Number	Fraction	Denominator Hundred	In Percentage
Green	4	$\frac{4}{10}$	$\frac{4}{10} \times \frac{100}{100}=\frac{40}{100}$	40%
Blue	3	$\frac{3}{10}$	$\frac{3}{10} \times \frac{100}{100}=\frac{30}{100}$	30%
Red	3	$\frac{3}{10}$	$\frac{3}{10} \times \frac{100}{100}=\frac{30}{100}$	30%
Total	10

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Question 72 Marks

A shop has the following number of shoe pairs of different sizes.
Size 2:20
Size 3:30
Size 4:28
Size 5:14
Size 6:8
Write this information in tabular form as done earlier and find the percentage of each shoe sizes available In the shop.

Answer

Respective number of shoes of different sizes are out of 100 shoes. Therefore, percentage of each shoe size is shown in the following table

Size of Shoe	Number of Shoes available	In fraction	Percentage of each Shoe size
Size 2	20	$\frac{20}{100}$	20%
Size 3	30	$\frac{30}{100}$	30%
Size 4	28	$\frac{28}{100}$	28%
Size 5	14	$\frac{14}{100}$	14%
Size 6	8	$\frac{8}{100}$	8%
Total	100

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Question 82 Marks

Find the percentage of children of different heights for the following data

Height	Number of Children	In Fraction	In Percentage
110cm	22	-	-
120cm	25	-	-
128cm	32	-	-
130cm	21	-	-
Total	100	-	-

Answer

Respective number of children of different heights are out of 100 children. Therefore, percentage of children of different heights are shown in the following table

Height	Number of Children	In Fraction	In Percentage
110cm	22	$\frac{22}{100}$	22%
120cm	25	$\frac{25}{100}$	25%
128cm	32	$\frac{32}{100}$	32%
130cm	21	$\frac{21}{100}$	21%
Total	100

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Question 92 Marks

Consider the expenditure made on a dress. 20% on embroidery, 50% on cloth, 30% on stitching. Can you think of more such examples?

Answer

Yes, such type of examples are as follow
(i) In class VII, a test of Maths of 100 marks was conducted and observation was as follows 20% students got marks below 50. 36% students got marks between 60 and 75. 44% students got marks above 80 marks.
(ii) An alloy is made up of the following composition i.e. 35% copper, 43% nickel and 22% zinc.

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Question 102 Marks

(i) Can you eat 50% of a cake?
Can you eat 100% of a cake?
Can you eat 150% of a cake?
(ii) Can a price of an item go up by 50%?
Can a price of an item go up by 100%?
Can a price of an item go up by 150%?

Answer

(i) Yes, we can eat 50% of a cake because 50 parts out of 100 parts of given cake are possible.
Yes, we can eat 100% of a cake i.e. whole cake can be eaten.
No, we cannot eat 150% of a cake because 150 parts out of 100 parts of given cake are not possible.
Hence, 150% of a cake cannot be eaten.
(ii) Yes, a price of an item can go up by 50%.
Yes, a price of an item can go up by 100%.
Yes, a price of an item can go up by 150% because price or increment can be more than 100%.

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Question 112 Marks

Look at the examples below and in each of them, discuss which is better for comparison.
(i) In the atmosphere, 1 g of air contains

(ii) A shirt has

Answer

(i) Here, second one (i.e. in per cent) is better for comparison. In the atmosphere, the quantity of air contents in per cent is better for comparison. It means that out of 100 parts of air; 78 parts are of nitrogen, 21 parts are of oxygen and 1 part is of other gas
(ii) Quantity of blending of cotton and polyster in a shirt is easy to understand in terms of percentage. It shows that out of 100 parts of a shirt; 60 parts are of cotton and 40 parts are of polyster.

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Question 122 Marks

The population of a city decreased from 25000 to 24500. Find the percentage decrease.

Answer

Initial population of the city = 25000
Decreased population of the city = 24500
Thus, decrease in population = 25000 - 24500 = 500
$\therefore$ Percentage decrease in population
$=\left(\frac{\text { Decrease in population }}{\text { Initial population }} \times 100\right) \%$
$=\left(\frac{500}{25000} \times 100\right) \%=\left(\frac{50}{25}\right) \%=2 \%$
Hence, the percentage decrease in population is 2%.

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Question 132 Marks

If Meena gives an interest of ₹ 45 for 1 yr at 9% rate p.a.. What is the sun she has borrowed?

Answer

Let the sum borrowed by Meena be ₹ x.
Rate of interest, R = 9%; Time, T = 1yr, Interest, I = ₹ 45
$\therefore$ $I =\frac{P \times R \times T}{100}$
$\Rightarrow$ $45 =\frac{P \times 9 \times 1}{100}$
$\Rightarrow$ $9P = 45 \times 100$
$\Rightarrow$ $P =\frac{4500}{9}$
$\Rightarrow$ $P =$ ₹ $500$
Hence the sum borrowed by Meena is ₹ $500.$

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Question 142 Marks

What rate gives ₹ 280 as interest on a sum of ₹ 56000 in 2 yr?

Answer

Given, Principal, P = ₹ 56000, Interest, f = ₹ 280,
Time, T = 2yr
Let rate of interest be R% p. a..
Now, I=$\frac{P \times R \times T}{100}$
$\Rightarrow\quad$ R=$\left(\frac{I \times 100}{P \times T}\right) \%$
=$\left(\frac{280 \times 100}{56000 \times 2}\right) \%$
=$\left(\frac{28}{56 \times 2}\right) \%$
=$\frac{1}{4}\%~$$=0.25\%$
Hence, the rate of interest is $0.25\%$.

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Question 152 Marks

Find the amount to be paid at the end of 3 yr in each case.
(i) Principal P = ₹ 1200 at 12% p.a.
(ii) Principal P =₹ 7500 at 5% р.а.

Answer

(i) Given, Principal, P = ₹ 1200; Rate of interest, R = 12%'
Time, T = 3yrs
$\therefore$ I = $\frac{P \times R \times T}{100}$
= $\frac{1200 \times 12 \times 3}{100}$
= $144 \times 3$ = ₹ 432
Now, Amount = Principal + Interest = ₹ 1200 + ₹ 432
= ₹ 1632
Hence, the amount paid at the end of 3 yr is ₹ 1632.

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Question 162 Marks

Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?

Answer

Given, CP of a book $=$ ₹ $275$; Loss per cent $=15 \%$
$\therefore$ Loss $=15 \%$ of CP
$=\frac{15}{100} \times 275=\frac{4125}{100}=$ ₹ $41.25$
Now, SP of a book $=C P-$ Loss $=$ ₹ $(275-41.25)$
$=$ ₹ $233.75$
Hence, Amina sells the book for ₹ $233.75$

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Question 172 Marks

(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?

Answer

(i) Given, ratio of calcium, carbon and oxygen in a chalk stick are 10 : 3 : 12.
Sum of the parts = 10 + 3 + 12 = 25
$\therefore\quad$ Percentage of carbon in chalk stick
=$\left(\frac{\text { Part of carbon in chalk stick }}{\text { Total part of chalk }} \times 100\right) \%$
=$\left(\frac{3}{25} \times 100\right) \%$
=$(3 \times 4) \%=12 \%$
Hence, the percentage of carbon in chalk stick is 12%
(ii) Let the weight of the chalk stick be x g.
Now, weight of carbon = 3g (which is 12% of chalk)
$\therefore \quad 12 \%$ of $x=3$
$\Rightarrow \frac{12}{100} \times x=3$
$\Rightarrow\quad$$x=\frac{3 \times 100}{12}=25 g$
Hence, the weight of the chalk stick is 25 g.

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Question 182 Marks

Juhi sells a washing machine for ₹13500. She loses 20% in the bargain. What was the price at which she bought it?

Answer

Given, SP of washing machine = ₹ 13500
Loss per cent 20%
Now, $\quad S P=C P-$ Loss $\Rightarrow S P=C P-20 \%$ of $C P$
$\Rightarrow\quad$13500 = CP -$\frac{20}{100} \times C P$
$\Rightarrow\quad$13500 = CP - $\frac{1}{5} \times C P$
$\Rightarrow\quad$13500 = CP$\left(1-\frac{1}{5}\right)$
$\Rightarrow\quad$13500 = CP$\left(\frac{5-1}{5}\right)$
$\Rightarrow\quad$13500 = $\frac{4}{5}$CP
$\Rightarrow\quad$CP = $\frac{13500 \times 5}{4}$=$3375 \times 5$
$\Rightarrow\quad$CP = ₹ 16875
Hence, Juhi bought a washing machine for ₹ 16875.

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Question 192 Marks

On a certain sum, the interest paid after 3 yr is ₹ 450 at 5% rate of interest per annum. Find the sum

Answer

Let the sum be ₹ P.
Rate of interest, R = 5% Time, T = 3yr Interest, I = ₹ 450
$\therefore\quad$ I = $\frac{P \times R \times T}{100}$ $\quad\Rightarrow\quad$ 450 = $\frac{P \times 5 \times 3}{100}$
$\Rightarrow P \times 15=450 \times 100\quad$ $\Rightarrow\quad$$P=\frac{45000}{15}$ = ₹ 3000
Hence, the required sum is ₹ 3000.

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Question 202 Marks

You have ₹2400 in your account and the interest rate is 5%. After how many years would you earn ₹240 as interest?

Answer

Given, Principal, P = ₹ 2400; Rate of interest, R = 5% Interest, I = ₹ 240; Time, T = ?
$\therefore\quad$ Interest , I = $\frac{P \times R \times T}{100}$
$\Rightarrow\quad$240 = $\frac{2400 \times 5 \times T}{100}$
$\Rightarrow\quad$$240 \times 100$ = $2400 \times 5 \times T$
$\Rightarrow\quad$T=$\frac{240 \times 100}{2400 \times 5}$$\Rightarrow$ T =$\frac{10}{5}$= 2 yr
Hence, the required time is 2 yr.

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Question 212 Marks

₹ 10000 is Invested at 5% interest rate p.a. Find the Interest at the end of 1 yr.

Answer

Given, Principal, P = ₹ 10000; Rate of interest, R = 5%;
Time T = 1yr
$\therefore$ Intrest ,I =$\frac{P \times R \times T}{100}$=$\frac{10000 \times 5 \times 1}{100}$
= 100 ×5 = ₹ 500
Hence, the interest at the end of 1 yr is ₹ 500.

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Question 222 Marks

₹7000 is borrowed at 3.5% rate of interest p.a.. borrowed for 2 yr. Find the amount to be paid at the end of the 2nd yr.

Answer

Given, Principal, P= ₹7000 ; Rate of interest, R = 3.5%;
Time, T = 2yr;
$\therefore\quad$ Interest I = $\frac{P \times R \times T}{100}$= $\frac{7000 \times 35 \times 2}{100}$ = $70 \times 7$ = ₹490.
$\therefore\quad$ Amount = Principal + Interest
$\therefore\quad$ ₹7000 + ₹490 = ₹7490
Hence, the amount paid at the end of 2 yr is ₹7490.

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Question 232 Marks

₹ 6050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 yr.

Answer

Given, Principal, P $=$ ₹ $6050 ;$ Rate of interest, $R=6.5 \%$
Time, $T=3 yr$
$\therefore$ Interest, I =$\frac{P \times R \times T}{100}$= $\frac{6050 \times 6.5 \times 3}{100}$ = $\frac{117975}{100}$
= ₹ 1179.75
$\therefore$ Amount = Principal + Interest
= ₹ 6050 + ₹ 1179.75 = ₹ 7229.75.
Hence, the interest and amount at the end of 3 yr are ₹ 1179.75 and ₹ 7229.75, respectively.

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Question 242 Marks

Cost of an item is ₹ 50. It was sold with a profit of 12%. Find the selling price.

Answer

Given, CP of an item = ₹ 50; Profit % = 12%
$\therefore$ Profit=12% of CP of an item
= 12% of ₹ 50=$\frac{12}{100} \times 50$=$\frac{600}{100}$=₹ 6
$\therefore$ SP of an itam = CP + Profit = ₹ 50 + ₹ 6 = ₹ 56

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Question 252 Marks

A shopkeeper bought a chair for ₹ 375 and sold it for ₹ 400. Find the gain percentage.

Answer

Given, CP (cost price) of a chair = ₹ 375
SP (selling price) of a chair = ₹ 400
Here, SP > CP
So, gain $=S P-C P=$ ₹ $(400-375)=$ ₹ $25$
$\begin{aligned} \text { Now, gain percentage } & =\left(\frac{\text { Gain }}{ CP \text { of a chair }} \times 100\right) \% \\ & =\left(\frac{25}{375} \times 100\right) \%=\frac{100}{15} \% \\ & =\frac{20}{3} \%=6 \frac{2}{3} \%\end{aligned}$
Hence, the gain percentage is $6 \frac{2}{3} \%$

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Question 262 Marks

My mother says, in her childhood petrol was ₹ 1 a litre. It is ₹ 52 per litre today. By what percentage has the price gone up?

Answer

Old price of 1 L petrol = ₹ 1
Present price of 1 L petrol = ₹ 52
Change in price = ₹ (52-1) = ₹ 51
$\begin{aligned} \therefore \text { Percentage increase } & =\left(\frac{\text { Change in price }}{\text { Old price of petrol }} \times 100\right) \% \\ & =\frac{51}{1} \times 100=5100 \%\end{aligned}$
Hence, the price has gone up by 5100%

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Question 272 Marks

Find percentage of increase or decrease.
(i) Price of shirt decreased from ₹ 280 to ₹ 210.
(ii) Marks in a test increased from 20 to 30.

Answer

(i) Decreased price of shirt $=$ ₹ $(280-210)$
$=$ ₹ $70$
$\therefore$ Percentage of decrease
$ \begin{array}{l} =\left(\frac{\text { Amount of change }}{\text { Original amount }} \times 100\right) \% \\ =\left(\frac{70}{280} \times 100\right) \%=25 \% \end{array}$
(ii) Increased marks $=30-20=10$
$\therefore$ Percentage of increase
$\begin{array}{l}=\left(\frac{\text { Increased marks }}{\text { Original marks or Initial marks }} \times 100\right) \% \\ =\left(\frac{10}{20} \times 100\right) \%=50 \%\end{array}$

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Question 282 Marks

Divide 15 sweets between Manu and Sonu so that they get 20% and 80% of them, respectively.

Answer

Total number of sweets = 15
$\begin{aligned} \text { Number of sweets, Manu gets } & =20 \% \text { of } 15=\frac{20}{100} \times 15 \\ & =\frac{15}{5}=3 \text { sweets }\end{aligned}$
and number of sweets, Sonu gets = 80% of 15
$\begin{array}{l}=\frac{80}{100} \times 15=\frac{120}{10} \\ =12 \text { sweets }\end{array}$
Hence, Manu and Sonu get 3 sweets and 12 sweets, respectively.

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Question 292 Marks

Convert the given fractional numbers to per cents.
(i) $\frac{1}{8}$
(ii) $\frac{5}{4}$
(ii) $\frac{3}{40}$
(iv) $\frac{2}{7}$

Answer

(i) $\frac{1}{8}=\left(\frac{1}{8} \times 100\right)\% =\left(\frac{100}{8}\right) \%=\frac{25}{2} \%=12.5 \%$
(ii) $\frac{5}{4}=\left(\frac{5}{4} \times 100\right) \%=(5 \times 25) \%=125 \%$
(iii) $\frac{3}{40}=\left(\frac{3}{40} \times 100\right) \%=\left(\frac{3}{4} \times 10\right) \%=\frac{15}{2} \%=7.5 \%$
(iv) $\frac{2}{7}=\left(\frac{2}{7} \times 100\right) \%=\frac{200}{7} \times=28 \frac{4}{7} \%$

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Question 302 Marks

A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

Answer

Given, total number of matches = 20
Percentage of won matches = 25%
$\begin{aligned} \therefore \text { Total number of winning matches } & =25 \% \text { of } 20 \\ & =\frac{25}{100} \times 20 \\ & =\frac{25}{5}=5\end{aligned}$
Hence, the team won 5 matches.

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Question 312 Marks

Meeta saves ₹ 4000 from her salary. If this is 10% of her salary. What is her salary?

Answer

Let Meeta's salary be ₹ x.
Meeta saves ₹ 4000 which is 10% of her salary
$\therefore 10 \%$ of $x=$ ₹ $4000$
$\Rightarrow \frac{10}{100} \times x=4000$
$\Rightarrow x=\frac{4000 \times 100}{10}=40000$
Hence, Meeta's salary is ₹ 40000.

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Question 322 Marks

Out of 15000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

Answer

Given, 60% voters voted.
This means that, if there were 100 voters, out of them 60 voted and remaining did not vote.
Then, voters who did not vote = (100 - 60)% = 40%
Now, actual number of voters who did not vote
= 40% of 15000
$\begin{array}{l}=\frac{40}{100} \times 15000 \\ =40 \times 150=6000\end{array}$
Hence, 6000 voters did not vote.

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Question 332 Marks

In a city, 30% are females, 40% are males and remaining are childrens. What per cent are childrens

Answer

Given, percentage of females = 30%
Percentage of males 40% and remaining are childrens.
This means that, if the total population of city is 100, out of them 30 are females, 40 are males and remaining childrens.
Then, number of childrens = 100-(30+40)
= 100-70=30
Hence, there are 30% childrens in a city.

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Question 342 Marks

Convert given per cents to decimal fractions and also to fractions in simplest forms.
(i) 25%
(ii) 150%
(iii) 20%
(iv) 5%

Answer

	Per Cents	Fractions	Decimals
(i)	25%	$\frac{25}{100}=\frac{1}{4}$	0.25
(ii)	150%	$\frac{150}{100}=\frac{3}{2}$	1.50
(iii)	20%	$\frac{20}{100}=\frac{1}{5}$	0.20
(iv)	5%	$\frac{5}{100}=\frac{1}{20}$	0.05

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Question 352 Marks

Find
75% of 1 kg

Answer

We have. $75 \%$ of $1 kg=\frac{75}{100} \times 1 kg=0.75 kg$
In terms of gram,
75% of 1kg = 75% of 1000 g $\qquad$[$\because$ 1kg = 1000g]
$=\frac{75}{100} \times 1000 g=750 g$
Hence, 75% of 1 kg is 0.75 kg or 750 g

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Question 362 Marks

Find
20% of ₹ 2500

Answer

We have, $20 \%$ of ₹ $2500=\frac{20}{100} \times$ ₹ $2500=$ ₹ $(20 \times 25)$
$=$ ₹ $500$
Hence, 20% of ₹ 2500 is = ₹ 500.

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Question 372 Marks

Find
1% of 1 hour

Answer

We have, 1% of 1 hour
In terms of $min , 1 \%$ of $1 h=\frac{1}{100} \times 60 min\qquad$[$\therefore$ 1h=60 min]
$=\frac{6}{10} \min =\frac{3}{5}$ min
In terms of second, $1 \%$ of $1 h=\left(\frac{1}{100} \times 3600\right)$ $
\begin{array}{l}
=36 s \\
{[\because 1 h=60 \times 60 s=3600 s]}
\end{array}
$
Hence, $1 \%$ of 1 h is $\frac{3}{5} min$ or 36 s .

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Question 382 Marks

Find
15% of 250

Answer

We have, $15 \%$ of $250=\frac{15}{100} \times 250$
$=\frac{375}{10}=37.5$
Hence, 15% of 250 is 37.5.

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Question 392 Marks

Convert the given decimal fractions to per cents.
(i) 0.65
(ii) 2.1
(iii) 0.02
(iv) 12.35

Answer

(i) $0.65=(0.65 \times 100) \%=\left(\frac{65}{100} \times 100\right) \%=65 \%$
(ii) $2.1=(2.1 \times 100) \%=\left(\frac{21}{10} \times 100\right) \%=210 \%$
(iii) $0.02=(0.02 \times 100) \%=\left(\frac{2}{100} \times 100\right) \%=2 \%$
(iv) $12.35=(12.35 \times 100) \%=\left(\frac{1235}{100} \times 100\right) \%=1235 \%$

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Question 402 Marks

What per cent of 1 day is 1 min?

Answer

Let x % of 1 day is 1 min.
$\because 1 \text { day }=24 h=24 \times 60 min $
$ \Rightarrow 1 \text { day }=1440 min $
$ \therefore \frac{x}{100} \times 1440 min=1 min$
$\Rightarrow x=\frac{100}{1440}=\frac{10}{144} \Rightarrow x=0.069 \%$
Hence, 0.069% of 1 day is 1 min.

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Question 412 Marks

What per cent of 1 h is 30 min?

Answer

Let x% of 1 h is 30 min.
So, $\quad \frac{x}{100} \times 1 h=30 min$
$\therefore \frac{x}{100} \times 60 min=30 min \quad[\because 1 h=60 min]$
$\Rightarrow x=\frac{30 \times 100}{60}$
$\Rightarrow x=\frac{100}{2}=50 \%$
Hence, 50% of 1 h is 30 min.

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Question 422 Marks

Observe and complete the following table.

Per Cent	1%	10%	25%	50%	90%	125%	250%
Fraction	$\frac{1}{100}$	$\frac{10}{100}=\frac{1}{10}$
Decimal	0.01	0.10

Answer

The complete table is shown below

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Question 432 Marks

Rina made a table top of 100 different coloured tiles. She counted yellow, green, red and blue tiles separately and filled the table below. Can you help her complete the table?

Colour	Number of Tiles	Rate per Hundred	Fraction	Written as	Read as
Yellow	14	14	$\frac{14}{100}$	14%	14 per cent
Green	26	26	$\frac{26}{100}$	26%	26 per cent
Red	35	35	-	-	-
Blue	25	-	-	-	-
Total	100	-	-	-	-

Answer

The complete table is shown below

Colour	Number of Tiles	Rate per Hundred	Fraction	Written as	Read as
Yellow	14	14	$\frac{14}{100}$	14%	14 percent
Green	26	26	$\frac{26}{100}$	26%	26 percent
Red	35	35	$\frac{35}{100}$	35%	35 percent
Blue	25	25	$\frac{25}{100}$	25%	25 percent
Total	100

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Question 442 Marks

In an examination, a student must get 40% marks to pass. A student gets 48 marks out of 200. Express his marks as a percentage. How many more marks should he got to pass the examination?

Answer

Pass percentage = 40%
Total marks of examination = 200
Student's marks = 48
Passing marks $=40 \%$ of $200=\frac{40}{100} \times 200$
$=40 \times 2=80\text{ marks}$
$\therefore$ Percentage of marks got by student
$=\frac{48}{200} \times 100=24 \%$
Hence, student needed atleast $=80-48= 32$ marks to pass the examination.

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Question 452 Marks

₹ 5000 amounts to 6000 in 5 yr. What will ₹ 8000 amount be at the same rate of interest in 6 yr.

Answer

Given, amount ₹ $5000$ became ₹ $6000$ in 5 yr.
So, Principal $=$ ₹ $5000$, Time period $=5\text{ yrs}$
$\therefore \text {Interest}=$ ₹ $6000-$ ₹ $5000=$ ₹ $1000$
Let rate of interest be $R \%$ per annum.
$\therefore SI =\frac{P \times R \times T}{100} \Rightarrow 1000=\frac{5000 \times R \times 5}{100}$
$\Rightarrow R=\frac{100000}{5000 \times 5}=\frac{20}{5}=4 \%$
So, SI for the amount ₹ $8000$ will be at same rate for 6 yr.
$\therefore SI =\frac{P \times R \times T}{100}=\frac{8000 \times 4 \times 6}{100}$
$=80 \times 4 \times 6=80 \times 24=$ ₹ $1920$
$\therefore 8000$ amounts to ₹ $9920$ in 6 yr.

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Question 462 Marks

On what principal, the interest is ₹ 1750 for 7 yr at 5% per annum simple interest?

Answer

Let the principal be ₹ $x.$ Simple interest $=$ ₹ $1750$
Time period for principal(T) $=7\text{ yr}$
Rate of Interest $(R)=5 \%$ per annum
$\therefore$ Simple Interest $=\frac{P \times R \times T}{100}$
$\Rightarrow 1750=\frac{x \times 5 \times 7}{100}$
$\Rightarrow 175000=35 x$
$\Rightarrow x=\frac{175000}{35}=5000$
$\therefore$ Principal $=$ ₹ $5000$

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Question 472 Marks

Find the Sl on ₹ 500 for 3 yr at 5% per annum.

Answer

Given, Principal $(P)=$ ₹ $500$
Time $(T)=3\text{ yr};$
Rate of Interest $(R)=5 \%$
Simple interest $( SI )=\frac{P \times R \times T}{100}$
$\therefore SI =\frac{500 \times 5 \times 3}{100}=5 \times 5 \times 3=$ ₹ $75$
Hence, simple interest is ₹ $75.$

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Question 482 Marks

An article is sold for ₹ 1980 at 10% profit. What is the cost price?

Answer

Selling price of the article $=$ ₹ $1980$
Profit $\%=10$
$\because$ Cost price $= SP -\frac{10}{100} CP$
$\Rightarrow C P+\frac{10 CP }{100}= SP$
$\therefore CP =\frac{100}{100+10} \times 1980$
$=\frac{100}{110} \times 1980=$ ₹ $1800$

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Question 492 Marks

The cost price of a scooter is ₹ 20000 and the profit percentage is 12%. What is the selling price?

Answer

Cost price of a scooter $=$ ₹ $20000$, Profit $=12 \%$
$\because$ Selling price $=C P+\frac{12}{100} C P$
$=\left(\frac{100+12}{100}\right) C P$
$\therefore SP =\frac{100+12}{100} \times 20000=\frac{112}{100} \times 20000$
$=112 \times 200=$ ₹ $22400$

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Question 502 Marks

A shopkeeper purchased eggs at the rate of 10 for ₹ 6 and sells them at 6 for ₹ 10. Find his profit or loss per cent.

Answer

Cost price of one egg = ₹$\frac{6}{10}$
Selling price of one egg = ₹$\frac{10}{6}$
Profit on each egg=$\frac{10}{6}-\frac{6}{10}$=$\frac{100-36}{60}=\frac{64}{60}$
$\therefore$ Profit$\%=\frac{\frac{64}{60}}{\frac{6}{10}} \times 100$
$=\frac{64}{60} \times \frac{10}{6} \times 100=\frac{6400}{36} \%$ or $\frac{1600}{9} \%$

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Question 512 Marks

A person sells a watch costing ₹ 750 for ₹ 840. Find his profit or loss per cent.

Answer

$\text {Cost price of watch} =$ ₹ $750$
$\text {Selling price} =$ ₹ $840$
$\because$ Profit $= \text{SP} - \text{CP} =840-750=$ ₹ $90$
$\therefore$ Profit$\%=\frac{\text { Profit }}{\text{CP}} \times 100$
$=\frac{90}{750} \times 100=\frac{900}{75}=12 \%$

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Question 522 Marks

What per cent of ₹ 80 is ₹ 100?

Answer

Let x % of ₹ 80 is ₹ 100.
$\therefore \frac{x}{100} \times 80=100$
$\Rightarrow x=\frac{100 \times 100}{80}=\frac{10000}{80}$ $\Rightarrow x=125 \%$
Hence, 125% of ₹ 80 is ₹ 100.

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Question 532 Marks

Find the whole value/quantities if
(i) 5% of it is 500
(ii) 30% of it is 900 km
(iii) 70% of it is 14 h
(iv) 6% of it is 48 km

Answer

(i) Given, 5% of it is 500.
Let x be the number.
So, 5% of x = 500
$\therefore \frac{5}{100} \times x=500$
$\Rightarrow \quad x=\frac{500 \times 100}{5} \Rightarrow x=10000$
(ii) Given, 30% of it is 900 km.
Let x be the kilometre distance.
So, 30% of x = 900km
$\frac{30}{100} \times x=900$ $\Rightarrow \frac{30 x}{100}=900$
$\Rightarrow \quad 30 x=900 \times 100$
$\Rightarrow \quad x=\frac{900 \times 100}{30}$=$\frac{9000}{3}=3000$
$\Rightarrow \quad x=3000 km$
(iii) Given, 70% of it is 14 h.
Let x be the hours.
So, 70% of x = 14h
$\therefore \quad \frac{70}{100} \times x=14$ $\Rightarrow x=\frac{14}{70} \times 100$
$\Rightarrow x=\frac{14}{7} \times 10=20$ $\Rightarrow x=20 h$
(iv) Given, $6 \%$ of it is 48 km.
Let $x$ be the kilometre distance.
So, $6 \%$ of $x=48 km$
$\therefore \frac{6}{100} \times x=48 \Rightarrow x=\frac{48}{6} \times 100 $
$ \Rightarrow x=8 \times 100=800 $
$ \Rightarrow x=800 km$

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