# Free Rational Numbers Subjective Test 01 Practice Test - 7th grade

### Question 1

Define Rational Number. [1 MARK]

#### SOLUTION

Solution :Any number which can be represented in the form of pq where p and q are integers and q≠0 is a rational number.

### Question 2

Which of these is not a rational number and why? [2 MARKS]

39,−411,√2,π

#### SOLUTION

Solution :Answer: 1 Mark

Reason: 1 Mark

A rational number is one which has an integer as its numerator and denominator but the denominator cannot be equal to zero. Hence, the rational numbers here are: 39,−411

As the numbers √2,π cannot be written in simple fractions therefore thay are irrational ( Note: π=227 is a popular approximation, but it is not accurate.)

### Question 3

The diagram shows a circle with radius 3 and centre at O.

The circumference and the area of this circle are rational or irrational? [2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

The circumference is given by the formula C=2×π×r

So, C=2×π×3=6π

The area is given by the formula A=π×r2

So, A=π×32=9×π

As, π is irrational, so the circumference and area are both irrational.

### Question 4

What is the value of additive inverse of [2 MARKS]

(i) 78

(ii) 12

#### SOLUTION

Solution :Each part: 1 Mark

(i) Additive Inverse of 78 will be -78 since the sum of both is 0.

(ii) Additive Inverse of 12 will be -12, since their sum will be 0.

### Question 5

Give the additive inverse of the reciprocal of: [2 MARKS]

(i) −25

(ii) −1716

#### SOLUTION

Solution :Each part: 1 Mark

(i) Reciprocal of −25=−52

Additive Inverse of −52=52

(ii) Reciprocal of −1716=−1617

Additive Inverse of −1617=1617

### Question 6

(a) How many rational numbers lie between 25 and 65? Mention atleast two.

(b) If x=110, y=−38. Evaluate: (x+y),(x−y),(x×y),(x÷y)

[3 MARKS]

#### SOLUTION

Solution :a) Each rational number: 0.5 Marks

b) Each operation: 0.5 Marks

(a) There are infinite numbers of rational numbers that lie between two points on a number line.Consider the rational numbers 25 and 65.

Take the average of two numbers 2+65×2 = 810810 lies in the middle of 25 and 65.

Now take the average of 25 and 810. It is (25+810)×12 = 4+810×12 = 1220

So, as we proceed in this, we will get an infinite number of rational number between two points.

(b) (i) x+y=110+−38

⇒110−38=8−3080

⇒−2280=−1140

(ii) x−y=110−−38

⇒110+38=8+3080

⇒3880=1940

(iii) x×y=110×−38=−380

(iv) x÷y=110÷−38

⇒110×8−3=8−30

⇒−830=−415

### Question 7

(a) Find the sum of:

17,(23×52),−512

(b) If −57=x28. Find x

(c) Simplify 37÷(21−55)

[3 MARKS]

#### SOLUTION

Solution :Each part: 1 Mark

(a) 17+(23×52)+−512

17+(53)−512

12+140−3584 (LCM of 7, 3, 12 = 84)

11784

(b) −57=x28

x28=−57

x=28×(−5)7=4×(−5)=−20

(c) 37÷21−55

⇒37×−5521

⇒17×−557

⇒−5549

### Question 8

(a) Amit used to be on the playground 412 times as much as his brother. Now he cut down 113 of the time he used to be on the playground as much as his brother. How much time does he spend on the playground when compared to his brother?

(b) The sum of two rational numbers is −58. If one of the numbers is −611, find the other number.

(c) Find x such that −38 and x−24 are equivalent fractions.

[3 MARKS]

#### SOLUTION

Solution :Each part: 1 Mark

(a) Amit played 412=92 times of his brother.

He cut down his play by 113=43

The duration Amit plays now

= 92−43=196=316.So, now Amit plays 316 times as his brother.

(b) Let the number be x.

−611+x=−58

x=−58−−611

x=−58+611

x=−788

(c) For −38 and x−24 to be equivalent.

⇒−38=x−24

⇒x=−3×−248

⇒x=−3×−31=9

### Question 9

(a) Find wether the following numbers are equal or greater or less than 1926. The numbers are 247338, 1921 and 2632.

(b) Find a rational number which lies between 12 and 14.

[3 MARKS]

#### SOLUTION

Solution :(a) Answer: 2 Marks

(b) Answer: 1 Mark

(a) When we multiply by 13 in denominator and the numerator of 1926 we get 247338, therefore 1926=247338.

By inspection, when the numerator is same and the denominator is different then the fraction having a large value of denominator is smaller than the fraction having less value of denominator.

Hence, 1926 < 1921.For comparing 1926 and 1316 we have to make the denominator of both the fraction by taking LCM of 26 and 16 = 416.

(19×16)(26×16)=304416, (13×26)(16×26)=338416

338>304. Hence, 2632 >1926.

(b) A rational number which lies between 12 and 14 is:

⇒12+142

⇒342

⇒38

### Question 10

(a)

A, B, C and D are rectangles.

The length of the diagonal of which rectangles are not irrational and which are irrational?

(b) Solve: 513−−826

[4 MARKS]

#### SOLUTION

Solution :(a) Formula: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

(b) Answer: 1 Mark

(a) To find the length of the diagonal, d, of each rectangle, use Pythagoras Theorem.

For A, d =√(12+22)=√(1+4)=√(5)

For B, d =√(22+32)=√(4+9)=√(13)

For C, d =√(32+42)=√(9+16)=√(25)

For D, d =√(42+52)=√(16+25)=√(41)

Of these, only √(25)=5 is rational.

The length of the diagonal of rectangle C is rational.

The length of the diagonal of rectangle A, B and D are irrational

(b) 513−−826

⇒513+826

⇒10+826

⇒1826=913

### Question 11

Arrange the following rational number in increasing order. [4 MARKS]

(i) 43, 32, 79, 54.

(ii) 85, 127, 92, 2320.

#### SOLUTION

Solution :Each part: 2 Marks

(i) Find the LCM of denominators, i.e. the LCM of 3, 2, 9 and 4 = 36.Now make the denominators common to the above fraction.

So, 4×123×12 = 4836 ; 3×182×18 = 5436 ; 7×49×4 = 2836 ; 5×94×9 = 4536;

Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude.

So, correct sequence is 2836, 4536, 4836, 5436.

⇒79<54<43<32

(ii) Find the LCM of denominators, i.e. the LCM of 5, 7, 2 and 20 = 140.Now make the denominators common to the above fraction.

So, 8×285×28 = 224140;

12×207×20 = 240140;

9×702×70 = 630140;

23×720×7 = 161140;Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude.

So, correct sequence is 161140, 224140, 240140, 630140.

⇒2320<85<127<92

### Question 12

(a) An investor invested 20 cr in a company for the development of a mobile application. The company distributed this money among various departments to cover up the needs. One-fourth of invested money is allocated to advertisement team. Two third of the remaining money was given to development team. In development team, there are two sub-teams namely Graphics and content team. Among them, it is distributed in such a way that content team will get 32 times more money than that of graphics team. Find the money allocated to each team separately.

(b) Sean is helping his dad build a tiled walkway in their backyard. The walkway will be 60 feet long and 2 feet wide. The local hardware store sells tiles which are 2 by 2 feet and come in boxes of 6.

How many boxes of tiles do they need?

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) Total money invested = 20 crAmount allocated to advertisement team = 20×14 = 5 cr.

Remaining amount: 20 - 5 = 15cr; Two third of 15 cr is allocated for development team =15×23=10 cr

Let graphics team got x ,then content team got 32x.

By solving, x + 32x =10; x=4

Hence, amount allocated for graphics and content team were 4cr and 6cr respectively.

(b) Area of the walkway = (60 × 2)feet2 =120feet2

Area of 1 tile = (2 × 2)feet2 = 4feet2

Total tiles required = 1204=30

1 box contains 6 tiles.

∴ No. of boxes for 30 tiles

=306=5

### Question 13

(a) A ball is dropped from a height of 30m. After striking the surface it rises to 23 of its height. Again it falls on the surface and this time it covered only 25 of its previous height. It continued for next two times and it only covered half of its previous height. Find the distance covered by the ball.

(b) Which of the following inequalities are correct?

(i) −15<−35

(ii) −35<−15

(iii) 35>15

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) After striking to the surface, the height gained by ball is 30m×23=20m.Now it attained the height of 20m, height attained by ball after 2nd

^{ }strike is 20×25=8m.After 2nd strike, height attained by ball is 8×12=4.

So for 3rd

^{ }strike, it falls from a height of 4m.Again he attained half of the height of previous so this time it covers 2m.

So total distance covered by ball = 30+ 20+8+4+2 = 64.

(b) (i) −15<−35 is incorrect

The following inequalities are correct:

(ii) −35<−15

(iii) 35>15

### Question 14

(a) Ram is training for his school's annual sports meet. He earlier used to run at 8 m/s, now after training, he has increased his speed by 12 times his original speed. Find how much time will Ram take to complete a 100m race.

(b) List four rational numbers that lie between 57 and 78

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) Original speed of Ram =8m/s

After training, his speed has increased to = 8+12×8=12m/s

Time taken to complete the 100m race =10012=253 seconds.

(b) One of the methods is taking the average of the two rational numbers. Let's do it with LCM method here.

Taking LCM of 7 and 8 = 56

⇒57=5×87×8=4056

⇒78=7×78×7=4956

∴ Four rational numbers between 4056 and 4956 can be 4356,4456,4556,4656

In standard form the rational numbers are 4356,1114,4556,2328

### Question 15

(a) Rahul's school is 10 km from his house. To cut the distance short he takes a shortcut. He first takes a route that is 45 of the original route but, he finds the road blocked at the end. The distance of his school from the blocked path is 13 of the original route. What was the difference in distance that Rahul covered by this new path and his original route?

(b) From a rope 68m long, pieces of equal size are cut. If the length of one piece is 414m, find the number of such pieces.

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) Distance of the first path taken for short cut =45×10=8km

Distance of the second path taken for short cut =13×10=103km

Total distance covered by Ram = 103+8=343km

Difference in Distance =343−10=43km

Ram ended up covering a larger distance of 1.3 km.

(b) Let the number of pieces be y.

Length of one piece =414m

Total length of y pieces = Total length of the rope

414×y=68m

174×y=68

y=68×417

=4×4

=16

∴ 16 equal pieces of the given size can be cut.