Free Rational Numbers Subjective Test 02 Practice Test - 7th grade 

There are infinite rational numbers between two points hence we cannot practically calculate the number of rational numbers skipped by the frog.

Each part: 1 Mark

(a) Four irrational numbers are $\sqrt{2},\sqrt{3},\sqrt{5}$ and $\pi$

(b) Let the other number be $x$.

 $\Rightarrow \frac{3}{4}=\frac{5}{6}\times x$

 $\Rightarrow x=\frac{3\times 6}{4\times 5}=\frac{9}{10}$

(a) Solution: 1 Mark
(b) Steps: 1 Mark
     Result: 1 Mark

(a) Firstly we simplify the fraction $\frac{9}{24}$ = $\frac{3}{8}$

Now the LCM of denominator 8 and 16 is 16.

$\frac{9}{16}+\frac{3}{8}=\frac{9+6}{16}=\frac{15}{16}$


(b) 
  

Arun walks $\frac{1}{3}$km towards East from place A

Then he walks $1\frac{2}{3}=\frac{5}{3}$km towards West, ie. in the opposite direction.

Arun's new position is $\frac{5}{3}-\frac{1}{3}=\frac{4}{3}=1\frac{1}{3}$

Arun is $1\frac{1}{3}$km towards West from place A.

(a) Explanation: 2 Marks
(b) Answer: 1 Mark

(a) $\frac{22}{7}$ is a rational number because we can write it in a form of $\frac{p}{q}$.

The actual value of $\pi$ is 3.141516........which implies that there is no end to the values after the decimal, the decimal goes on forever without repeating.
Hence we are not able to find the actual value of $\pi$ therefore we can't represent it on a number line and moreover, we can't represent it in the form of $\frac{p}{q}$, that's why it is an irrational number.


(b) Yes, the numbers $\frac{1}{3}$ and $\frac{-5}{2}$ lie on the opposite sides of 0 on the number line as one of them is positive and the other is negative.


 

(a) Answer: 1 Mark
(b) Steps: 1 Mark
     Result: 1 Mark

(a) Each corner has three mutually perpendicular edges, hence forming three right angles in each corner. A room has 8 corners. Hence, there are 24 right angles in the room.

(b) Let the number to be subtracted be x.

 $\Rightarrow \frac{5}{12}+\frac{3}{2}-x=2$

 $\Rightarrow \frac{5+18}{12}-x=2$

 $\Rightarrow \frac{23}{12}-x=2$

 $\Rightarrow x=\frac{23}{12}-2$

$\Rightarrow x=\frac{23-24}{12}$
$\Rightarrow x=\frac{-1}{12}$

$\Rightarrow x=\frac{-1}{12}$

Each part: 2 Marks

(a) First write the rational numbers with positive denominator = $\frac{-7}{10}$, $\frac{5}{-8}$, $\frac{2}{-3}$.

LCM of denominators (LCM of 10, 8, 3 is 120)

$\frac{-7}{10}$ = $\frac{(-7\times 12)}{(10\times 12)}=\frac{-84}{120}$

$\frac{-5}{8}$ = $\frac{(-5\times 15)}{(8\times 15)}=\frac{-75}{120}$

$\frac{-2}{3}$ = $\frac{(2\times 40)}{(3\times 40)}=\frac{-80}{120}$

Comparing the numerators of these numbers, we get:

- 84 < - 80 < - 75

∴ $\frac{-84}{120}$ < $\frac{-80}{120}$ < $\frac{-75}{120}$;
$\frac{-7}{10}$ < $\frac{-2}{3}$ < $\frac{-5}{8}$.

(b)  $\frac{4}{3}$, $\frac{17}{2}$, $\frac{5}{18}$

LCM of denominators (LCM of 3, 2, 18 is 18)

$\frac{4}{3}$ = $\frac{(4\times 6)}{(3\times 6)}=\frac{24}{18}$

$\frac{17}{2}$ = $\frac{(17\times 9)}{(2\times 9)}=\frac{153}{18}$

$\frac{5}{18}$ = $\frac{(5\times 1)}{(18\times 1)}=\frac{5}{18}$

Comparing the numerators of these numbers, we get:

5 < 24 < 153

∴ $\frac{5}{18}$ < $\frac{24}{18}$ < $\frac{153}{18}$;
$\frac{5}{18}$ < $\frac{4}{3}$ < $\frac{17}{2}$.

a) Steps: 1 Mark
    Answer: 1 Mark
b) Each sub-part: 0.5 Marks


(a) Let the total number of students $=x$.

Then, number of candy bars each student has $=\frac{x}{8}$

Now, no. of student has been half  $=\frac{x}{2}$,

Number of candies each student would have got = 16

$\frac{x}{2}\times 16=x\times \frac{x}{8}$

Or No. of initial students = 64 ; $\frac{x}{2}=32.$

So, number of candies distributed = 32 x 16 = 512 


(b) 
(i) $\frac{6}{3}+\frac{-1}{6}=2-\frac{1}{6}=\frac{11}{6}$

(ii) $\frac{-3}{2}-\frac{3}{8}=\frac{-12-3}{8}=\frac{-15}{8}$

(iii) $\frac{-9}{6}+\frac{-3}{5}=\frac{-3}{2}-\frac{3}{5}$

$\frac{-15-6}{10}=\frac{-21}{10}$

(iv) $\frac{-1}{3}-\left(\frac{-3}{5}\right)=\frac{-1}{3}+\frac{3}{5}$

$\frac{-5+9}{15}=\frac{4}{15}$

Each part: 2 Marks

(a) As given in the question AP = PQ = QB = $\frac{1}{3}$, because interval of one between AB is divided by into three equal parts.
Similarly, US = SR = RT = $\frac{1}{3}$, since the interval of one between U and T is divided into three equal parts.  

So, value of P $=2+\frac{1}{3}=\frac{7}{3}$;

Q $=\frac{7}{3}+\frac{1}{3}=\frac{8}{3}$

Similarly for R $=-1-\frac{1}{3}=-\frac{4}{3}$;

S $=-\frac{4}{3}-\frac{1}{3}=-\frac{5}{3}$

(b) Time taken by 1 person to prepare 1 bag = 2 minutes

Time taken by 6 people to prepare 1 bag = $\frac{2}{6}$minutes

Time taken by 6 people (Ashley and her 5 friends) to prepare 900 bags

= $\frac{2}{6}\times 900=300minutes=5hrs$

Each part: 2 Marks

(a) In normal jump frog covers 5m in first jump but this time his jump covered $=5+(5\times \frac{1}{5})=6m$

In the second jump it covered $=4+(4\times \frac{1}{2})=6m$.

So, distance covered in second jump = 6m.

Total distance covered by frog = 6m + 6m = 12m


(b) (i) The statement is false as $\frac{-21}{5}$ and $\frac{7}{-31}$ both are negative numbers and lie to the left side of  0 on the number line i.e. they are at the same side to that of 0 which is left.

(ii) This statement is true as the number $\frac{-3}{-5}$ is actually positive $\frac{3}{5}$  and  $\frac{-4}{7}$ is negative. Therefore, $\frac{3}{5}$ is  is on the right of  $\frac{-4}{7}$ on the number line.

Each part: 2 Marks

(a) The LCM of denominator -5 and 2 = -10. Converting these rational numbers to equivalent rational numbers having common denominator.

$\frac{-2\times 2}{5\times 2}=\frac{-4}{10}=\frac{-4\times 2}{10\times 2}=\frac{-8}{20}$

$\frac{1\times 5}{2\times 5}=\frac{5}{10}=\frac{5\times 2}{10\times 2}=\frac{10}{20}$

Clearly -7, -6, -5.............8, 9 are integers between numerators -8 and 10 of these equivalent rational number. Thus we can take any 10 rational numbers and the numbers are $\frac{-7}{20}$, $\frac{-6}{20}$, $\frac{-5}{20}$,$\frac{-3}{20}$,$\frac{-1}{20}$, 0,$\frac{1}{20}$, $\frac{3}{20}$, $\frac{4}{20}$, $\frac{5}{20}$.


(b) $\sqrt{2}$​ = 1.414 (approx)

$\sqrt{3}$​​​​ = 1.732 (approx)

5 rational numbers between  $\sqrt{2}$​ and $\sqrt{3}$​​​​ are 1.42, 1.43, 1.44, 1.45, 1.50. (there are many more)

2 irrational numbers between  $\sqrt{2}$​ and $\sqrt{3}$​​​​ are  $\sqrt{2}.1$ and $\sqrt{2}.9$ (there are many more)