Free Fractions and Decimals Subjective Test 02 Practice Test - 7th grade

Question 1

If a number is multiplied by $\frac{1}{6}$ and then divided by 5 the remainder would be zero. What is the least possible number? [1 MARK]

SOLUTION

Solution :
Answer: 1 Mark

Let the number be x

Given $\frac{x \times \frac{1}{6}}{5}$ = 1

Hence, x = 30

The least value of the number is 30

Question 2

Karan covers 3.2 km in 16 seconds. How far can Karan run in 1 minute? Express it in metres. [2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Answer: 1 Mark
3.2 km = 3200 metres
Given that

Distance covered in 16 seconds = 3200 metres

Distance covered in 1 second = (3200 $\div$ 16)

= 200 metres.

1 minute = 60 seconds

Distance covered in 1 minute $= 60 \times 200$

$= 12000 m$

The distance covered in 1 minute is 12000 metres.

Question 3

Arrange the following in descending order.

(i) $\frac{2}{3}, \frac{4}{21}, \frac{7}{9}$

(ii) $\frac{4}{5}, \frac{6}{7}, \frac{1}{3}$
[2 MARKS]

SOLUTION

Solution : Each option: 1 Mark

Now, $\frac{7}{9}, \frac{2}{3}, \frac{4}{21}$

The LCM of the denominators of the above fractions = 63

$\Rightarrow \frac{7 \times 7}{9 \times 7}, \frac{2 \times 21}{3 \times 21}, \frac{4 \times 3}{21 \times 3}$

$\Rightarrow \frac{49}{63}, \frac{42}{63}, \frac{12}{63}$

Hence, arranging in descending order we have

$\frac{49}{63} > \frac{42}{63} > \frac{12}{63}$

i.e., $\frac{7}{9} > \frac{2}{3} > \frac{4}{21}$

(ii) The LCM of the denominators of the given fractions = 105

Making them like fractions. We have,

$\frac{1 \times 35}{3 \times 35}, \frac{4 \times 21}{5 \times 21}, \frac{6 \times 15}{7 \times 15}$

$\frac{35}{105}, \frac{84}{105}, \frac{90}{105}$

$\Rightarrow \frac{6}{7} > \frac{4}{5} > \frac{1}{3}$

Question 4

Express 3460 grams in kilograms.

Now express $17 \frac{1}{2}$ kg in grams. [2 MARKS]

SOLUTION

Solution :
In kilograms: 1 Mark
In grams: 1 Mark

1 Kilogram = 1000 grams
Or 1 gram= $\frac{1}{1000}$ kilograms

Let x kilograms = 3460 g

$= (\frac{3460}{1000})$ kilograms

= 3.460 kilograms

3460 grams = 3.460 kilograms

Now $17 \frac{1}{2} kg = \frac{17 \times 2 + 1}{2}$

= $\frac{34 + 1}{2}$

= $\frac{35}{2} kg$

= $\frac{35}{2} \times 1000 gms$

= $(35 \times 500) gm$

= $17500 gms$

Question 5

Kartik ate $\frac{3}{5}$ part of an apple and Bharat ate the remaining part of the apple. What fraction of the apple did Bharat eat? Who had the larger share and by how much? [2 MARKS]

SOLUTION

Solution : Each answer: 1 Mark

Given that:
Fraction of the apple Kartik ate $= \frac{3}{5}$

Fraction of the apple uneaten $= 1 - \frac{3}{5}$

$= \frac{5 - 3}{5}$

$= \frac{2}{5}$

This part was eaten by his brother Bharath

$∴$ Bharath ate $\frac{2}{5}$ th part of the apple.

Comparing two fractions we have

$\frac{3}{5} & \frac{2}{5}$

Clearly $\frac{3}{5} > \frac{2}{5}$

$\Rightarrow$ Karthik had the larger share which was larger by $\frac{3}{5} - \frac{2}{5} = \frac{1}{5}$ .

Question 6

How much less is 18km compared to 29000 metres ? [3 MARKS]

SOLUTION

Solution :
Conversion Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
We know that
1 km=1000 m
Converting kilometers into meters
we have

18 km = 18 $\times$ 1000=18000 metres

Difference = 29000-18000

=11000 metres

Or

29000 metres= 29 km

Difference = 29-18

=11 km

Question 7

Given that length of a side of an equilateral triangle is 3.4m. Find its perimeter in cm? [3 MARKS]

SOLUTION

Solution :
Conversion Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

In an equilateral triangle, all the sides are equal.

Let the side of an equilateral triangle be= a cm where a= 3.4 m ( given)

Now Perimeter = a+ a+a=(3a) m

= (3)*(3.4)

=10.2m

Hence perimeter of the equilateral triangle = 10.2m.

Now, 1 m= 100cm

$∴$ 10.2m=10.2 $\times$ 100 = 1020 cm.

Hence, the perimeter of the triangle is 1020 cm.

Question 8

Mohan finished colouring a picture in $\frac{7}{12}$ hour. Keerthi finished colouring the same picture in $\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer? [3 MARKS]

SOLUTION

Solution : Steps: 2 Mark
Answer: 1 Mark
Given that

Mohan finishes the work in $\frac{7}{12}$ of an hour.

Also given that

Keerthi finishes the work in $\frac{3}{4}$ of an hour.

Now, $\frac{7}{12}$ of an hour

$\Rightarrow \frac{7}{12} \times 60 m i n u t e s$

$\Rightarrow 35 m i n s$

Again

$\frac{3}{4}$ of an hour

$\Rightarrow \frac{3}{4} \times 60$

$\Rightarrow 45 m i n u t e s$

Clearly, Keerthi worked more by 45 - 35 = 10 more minutes.

Now, when expressed as a fraction

$\frac{10}{60} = \frac{1}{6}$ of an hour.

Alternative Way

Method II

Now, $\frac{7}{12} a n d \frac{3}{4}$

Making them Like fractions we have

$\frac{7 \times 1}{12 \times 1} & \frac{3 \times 3}{4 \times 3}$

i.e. $\frac{7}{12} < \frac{9}{12}$

Clearly, Keerthi who takes ${\frac{3}{4}}^{t h}$ of an hour takes longer than Mohan who takes ${\frac{7}{12}}^{t h}$ of an hour.

So, subtracting we have

$\frac{9}{12} - \frac{7}{12} = \frac{2}{12} = \frac{1}{6}$ of an hour.

Question 9

In cricket, run rate is defined as the number of runs scored divided by the number of overs bowled. Team 'A' scores 260 runs in 50 overs batting first. The other team batting second scores at half the run rate of the team batting first and plays entire 50 overs. How many runs did the team batting second score? [3 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Given that
Team 'A' scored 260 runs in 50 overs.
So, run rate of the team batting first =(260/50)

= 5.2 runs per over

The team batting second scored at half the run rate of the team batting first

Run rate of team the batting second
= (5.2/2)

= 2.6 runs per over

Runs scored by the team batting second

$= (50) \times (2.6)$

= 130

Hence, the team batting second scored 130 runs.

Question 10

A circle is divided into 8 equal parts. Given that the area of the shaded portion is $\frac{10}{4}$ $m^{2}$ . Find the area of the unshaded part of the circle.

[4 MARKS]

SOLUTION

Solution :
Steps: 2 Marks
Application: 1 Mark
Result: 1 Mark
Given that

Area of the shaded portion = $\frac{10}{4}$ $m^{2}$

The circle is divided into 8 equal parts

Area of the circle = (8) $\times$ $\frac{10}{4}$

= 20 $m^{2}$

Area of unshaded part = Total area - shaded area

   = 20 - $\frac{10}{4}$ $m^{2}$
   = $\frac{80 - 10}{4} m^{2}$
   = $\frac{70}{4}$ = 17.5 $m^{2}$
Hence, area of the unshaded portion is 17.5 $m^{2}$

Question 11

Express $\frac{(0.8 \times 0.7) - (0.65 \times 0.6)}{(0.65 \times 0.6) - (0.62 \times 0.6)}$ in the form of a fraction.
[4 MARKS]

SOLUTION

Solution :
Bodmas operation: 1 Mark
Simplifying the numerator and the denominator: 1 Mark
Answer: 2 Marks

Multiply and divide the equation by 100

= $\frac{(8 \times 7) - (6.5 \times 6)}{(6.5 \times 6) - (6.2 \times 6)}$

= $\frac{(56 - 39)}{(39 - 37.2)}$

= $\frac{17}{1.8}$

= $\frac{170}{18}$

= $\frac{85}{9}$

Question 12

Anubha's father weighs 91 kg. Anubha's mother is 5 times Anubha's dog's weight. If Anubha's dog weighs $\frac{1}{7}$ of Anubha's father's weight, find Anubha's mother's weight. [4 MARKS]

SOLUTION

Solution : Steps: 2 Mark
Application: 1 Mark
Answer: 1 Mark

Given that
Anubha's father weighs 91 kg.
Anubha's dog's weight= $\frac{1}{7}$ of Anubha's father's weight

$\Rightarrow \frac{1}{7} \times 91$

$\Rightarrow \frac{91}{7} \Rightarrow 13 k g$

Also given
Anubha's mother's weight= 5 times Anubha's dog's weight

$= 5 \times 13 = 65 k g$

$∴$ Anubha's mother weighs 65 kg.

Question 13

Add the following decimals and then express it as a fraction.

7.205, 8.012, 3.1, 4.9, 27.3, 3.0009, 8.0102. [4 MARKS]

SOLUTION

Solution : Making them into like decimals: 1 Mark
Mathematical operation: 1 Mark
Fraction expression: 2 Marks

Now converting the decimals given into like decimals and adding it.

$0 - 7.205 0 -$

$0 - 8. 012 0 -$

$0 - 3. 1 0 - 0 - 0 -$

$0 - 4. 9 0 - 0 - 0 -$

$2 7. 3 0 - 0 - 0 -$

$0 - 3. 0009$

$0 - 8. 0102$

We get $6 1. 4281$

(II) Now, 61. 4281 can be rounded up

to 61.428

to 61. 43

to 61. 4

Now
$61.4 = \frac{61.4 \div 2}{100 \div 2} = \frac{307}{50} = 6 \frac{7}{50}$

Question 14

(a) It is said that the weight of a body on the moon is ${\frac{1}{6}}^{t h}$ of its weight on earth. Calculate the weight of a car on the moon if it weighs 420 kg on Earth.
[4 MARKS]

(b) What is the fraction of coloured squares to the total number of squares? State your answer in decimals also.

SOLUTION

Solution : Each part: 2 Marks

(a)

Now, the weight of the car in Moon = $\frac{1}{6}$ of its weight on earth.

$\Rightarrow \frac{1}{6}$ of 420 kg

$\Rightarrow \frac{420}{6}$

$\Rightarrow 70 k g$

$∴$ weight of the car in Moon = 70 kg only.

(b) There are 100 squares. 46 out of them are coloured. This picture shows forty-six hundredths.

To write 46 hundredths as a decimal, we use a place value chart.

The answer is 0.46.

Question 15

Car A covers 5.5 km in 4 litres of petrol and another car B covers 1800 metres in 1 litre of petrol. Calculate the mileage per litre of petrol and now calculate which car travels more and by how much distance compared to the other? [4 MARKS]

SOLUTION

Solution :
Conversion Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given that
Distance covered by car A in 4 liters of petrol is 5.5 km.
Now 1 km=1000 m.
5.5 km=5500 m

Distance covered by car A in 1 litre of petrol = (5500/4) = 1375 metres

Distance covered by car B in 1 liter of petrol = 1800 meters

So car B travels more compared to car A

Difference = 1800-1375 = 425 metres
Car B travels 425 meters more than car A

Free Fractions and Decimals Subjective Test 02 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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