Revision Test I

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Questions:

Multiple Choice Question

10 x 1 = 10

Which of the following is neither positive nor a negative rational number?

(a)

(b)

(c)

Such a rational number does not exist

(d)

None of the above

Which of the following is the multiplicative identity for rational numbers?

(a)

(b)

-1

(c)

(d)

None of these

Which of the following is the identity element?

(a)

(b)

-1

(c)

(d)

None of these

What should be subtracted from \(\frac { -3 }{ 4 } \) to get - 1?

(a)

\(\frac { 1 }{ 4 } \)

(b)

\(-\frac { 1 }{ 4 } \)

(c)

(d)

\(-\frac { 3 }{ 4 } \)

What should be added to \(\frac { -3 }{ 4 } \) to get -1?

(a)

\(\frac { 1 }{ 4 } \)

(b)

\(\frac { -1 }{ 4 } \)

(c)

(d)

\(\frac { -3 }{ 4 } \)

The reciprocal of \({-3\over8 } \times {-24\over 13}\) is

(a)

\(9\over 13\)

(b)

\(-9\over 13\)

(c)

\(-13 \over 9\)

(d)

\(13\over9\)

Which of the following statements is always true?

(a)

\(\frac{x-y}{2}\) is a rational number between x and y

(b)

\(\frac{x+y}{2}\) is a rational number between x and y

(c)

\(\frac{x\times y}{2}\) is a rational number between x and y

(d)

\(\frac{x\div y}{2}\) is a rational number between x and y

\(\frac{x+y}{2}\) is a rational number

(a)

between x and y

(b)

less than x and y both

(c)

greater than x and y both

(d)

less than x but greater than y

Three rational numbers lying between\(\frac{-5}{4}\) and \(\frac{1}{2}\)are

(a)

-1, 0 ,\(\frac{4}{3}\)

(b)

\(\frac{-3}{4},\frac{-1}{2},\frac{1}{4}\)

(c)

\(\frac{-3}{4},\frac{4}{3},\frac{1}{4}\)

(d)

\(\frac{-7}{4},-1,0\)

10.

Which of the following is not true?

(a)

\(\frac{10}{11}+\frac{11}{12}=\frac{11}{12}+\frac{10}{11}\)

(b)

\(\frac{10}{11}\times \frac{11}{12}=\frac{11}{12}\times \frac{10}{11}\)

(c)

\(\frac{10}{11}+ \frac{11}{12}=\frac{11}{12}\div \frac{10}{11}\)

(d)

\(\frac{10}{11}\div \frac{11}{12}=\frac{11}{12}\times \frac{10}{11}\)

True or False

10 x 1 = 10

11.

If \({a\over b}\) is a rational number, then b can be any whole number

(a)

True

(b)

False

12.

\({2\over3}-{5\over4}={5\over4}-{2\over3}\)

(a)

True

(b)

False

13.

\(-4\over5\) is greater than \(-5\over4\)

(a)

False

(b)

True

14.

Every rational number has a reciprocal

(a)

True

(b)

False

15.

The product of a non-zero rational number and its reciprocal is 0.

(a)

True

(b)

False

16.

Zero is a rational number

(a)

False

(b)

True

17.

All the fractions are not rational numbers but all the rational numbers are fractions

(a)

True

(b)

False

18.

All the integers are rational numbers.

(a)

False

(b)

True

19.

All the whole numbers are integers

(a)

False

(b)

True

20.

All the natural numbers are whole numbers.

(a)

False

(b)

True

Generic 2

11 x 2 = 22

21.

Find the multiplicative inverse of the following \(-13\over19\)

22.

Find the multiplicative inverse of the following -13

23.

Verify that -(-x) = x,for x = \(-13\over17\)

24.

Verify that -(-x) = x, for x = \(11\over15\)

25.

Write the additive inverse of the following \(19\over-6\)

26.

Write the additive inverse of the following \({2\over-9}\)

27.

Write the additive inverse of the following \(-6\over-5\)

28.

Write the additive inverse of the following \(-5\over9\)

29.

Write the additive inverse of the following \(2\over8\)

30.

Using appropriate properties, find \({2\over5}\times({-3\over7})-{1\over6}\times{3\over2}+{1\over14}\times{2\over5}\)

31.

Is the rational numbers closed under multiplication, Justify?

Generic 3

5 x 3 = 15

32.

If 51x3 is a multiple of 9, where x is a digit, then what is the value of x?

33.

Show that -1728 is a perfect cube. Also, find the number whose cube is -1728.

34.

Write a Pythagorean triplet whose one member is 15.

35.

Simplify: \({ \left[ { \left( \frac { -4 }{ 5 } \right) }^{ -2 } \right] }^{ 2 }\)

36.

Find three rational numbers between -3 and - 4

Match the following

5 x 1 = 5

37.

On a number line the rational number equidistant from a and -1 is:

38.

The product of (-3) and the negative reciprocal of 1^1/2, is:

39.

The reciprocal of (-1) is:

40.

The product of a rational number (other than zero) and its reciprocal is:

41.

The rational number having no reciprocal is

Generic 5

4 x 5 = 20

42.

A machine fills 540 bottles in six hours. How many bottles will it fill in five hours?

43.

A mixture of paint is prepared by mixing 1 part of green pigments with 6 parts of the base. In the following table, find the parts of base needed to be added.

Parts of green pigment	1	4	5	6
Parts of base	6	x₁	x₂	x₃

44.

Complete the following crossword puzzle using the given direction.

Direction:
Across:
(1) The product of number by itself two times, is called its _______
(2) If three numbers a, band c are such that a² + b²= c² then they are called _____ Triplets.
(3) A number is a when it is a product of the same two numbers.
Down: (4) The numbers 2n, n²-1 and n² + 1 where n is a natural number show Pythagorean______.
(5) Finding is the inverse operation of squaring a number.
(6) A number which divides a _______ given number exactly is called a or divisor of that number.

45.

Find the length of the diagonal BD when, the area of the quadrilateral is 32 cm².

Generic 4

2 x 4 = 8

46.

Leela invited some friends for tea on her birthday. Her mother placed some plates and some puris on a table to be served. If Leela places 4 puris in each plate 1 plate would be left empty. But if she places 3 puris in each plate 1puri would be left. Find the number of plates and number of puris on the table.

47.

Take a clock and fix its minute hand at 12.
Record the angle turned through by the minute hand from its original position and the time that has passed, in the following table:

Time Passed (T) (in minutes)	(T₁) 15	(T₂) 15	(T₃) 45	(T₄) 60
Angle turned (A) (in degree)	(A₁) 90	(A₂) __	(A₃) __	(A₄) __
\(T\over A\)	-	-	-	-

What do you observe about T and A? Do they increase together? Is \(T\over A\) same every time?

Is the angle turned through by the minute hand directly proportional to the time that has passed? Yes; From the above table, you can also see
T₁ : T₂ = A₁ A₂, because
T₁ : T₂ = 15 30= 1 :2
A₁ : A₂ = 90 180 = 1 :2
Check if T₂: T₃ = A₂ A₃ and T₃ : T₄ = A₃ : A₄
You can repeat this activity by choosing your own time interval.

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Answers Key:

Multiple Choice Question

10 x 1 = 10

(b)
0

(a)
1

(a)
\(\frac { 1 }{ 4 } \)

(b)
\(\frac { -1 }{ 4 } \)

(d)
\(13\over9\)

(b)
\(\frac{x+y}{2}\) is a rational number between x and y

(a)
between x and y

(b)
\(\frac{-3}{4},\frac{-1}{2},\frac{1}{4}\)

10.

True or False

10 x 1 = 10

11.

(b)
False

12.

(b)
False

13.

(a)
False

14.

(b)
False

15.

(b)
False

16.

(a)
False

17.

(b)
False

18.

(a)
False

19.

(a)
False

20.

(a)
False

Generic 2

11 x 2 = 22

21.

we have, \(-13\over19\)
\(\therefore\)The multiplicative inverse of \(-13\over19\) is\(19\over-13\)

22.

We have, -13
\(\therefore\)The multiplicative inverse of -13 is \(-1\over13\)

23.

We have, x = \(-13\over17\)
\(LHS =-(-x)=-\{-({-13\over17})\}=-\{{13\over17}\}\)
=-\({13\over17}\) = x = RHS
So, - (-x) = x is verified for x = \({13\over17}\).

24.

We have, x =\(11\over15\)
LHS = -(-x) = -\(({11\over15})={11\over15}=X=RHS\)
S0, - (-x) = x is verified for x =\(11\over15\)

25.

we have, \(19\over-6\) so, additive inverse of \(19\over-6\) is \(19\over6\)

26.

we have, \({2\over-9}\) so, additive inverse of \({2\over-9}\) is \({2\over9}\)

27.

we have,\(-6\over-5\)=\({6\over5}\) so, additive inverse of \(({-6\over-5})\) is \(-6\over5\)

28.

we have, \(-5\over9\) so, additive inverse of \(-5\over9\) is \(5\over9\)

29.

we have, \(2\over8\) so, additive inverse of \(2\over8\) is \(-2\over8\)

30.

= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) -\frac { 1 }{ 4 } +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \times \left( \frac { -3 }{ 7 } \right) +\frac { 1 }{ 14 } \times \frac { 2 }{ 5 } -\frac { 1 }{ 4 } \) (Using commutativity)
= \(\frac { 2 }{ 5 } \left[ \frac { -3 }{ 7 } +\frac { 1 }{ 14 } \right] -\frac { 1 }{ 4 } \) (using distributivity)
= \(\frac { 2 }{ 5 } \left[ \frac { -6+1 }{ 14 } \right] -\frac { 1 }{ 4 } =\frac { 2 }{ 5 } \times \frac { -5 }{ 14 } -\frac { 1 }{ 4 } \)
= \(\frac { -1 }{ 7 } -\frac { 1 }{ 4 } =\frac { -4-7 }{ 28 } =\frac { -11 }{ 28 } \)

31.

Rational numbers are closed under multiplication, e.g.\({2\over 12}\times{1\over 12}={2\over 144}\) , which is a rational number.

Generic 3

5 x 3 = 15

32.

We have the sum of the digits of 51x3
= 5+1+x+3=9+x
Since, 51x3 is divisible by 9.
∴ (9 + x) must be divisible by 9.
∴ (9 + x) must be equal to 0 or 9 or 18 or 27
or ... But x is a digit, then
9+x=9 ⇒ x=0
9 + x = 18 ⇒ x = 9
x = 27 ⇒ x = 18, which is not possible.
∴ The required value of x = 0 or 9.

33.

We have

i.e., 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
\(\Rightarrow\) 1728 = 2³ x 2³ X 3³
\(\Rightarrow\) 1728 = (2 x 2 x 3)³
\(\Rightarrow \sqrt[3]{1728}\) = 2 x 2 x 3 = 12
Since 1728 is a perfect cube.
\(\therefore\) -1728 is also a perfect cube.
Also\(\sqrt[3]{1728} =-12\)
\(\Rightarrow\) -1728 is a perfect cube of -12.

34.

Since, a Pythagorean triplet is given by 2n,
n² - 1 and n² + 1.
\(\therefore\) 2n = 15 or n =\(\frac{15}{2}\)s not an integer.
So, let us assume th~t
n² - 1 = 15
or n² = 15 + 1 = 16
or n² = 4², i.e. n = 4
Now, the required Pythagorean triplet is
2n, n² - 1 and n² + 1
or 2(4), 4² - 1 and 4² +
or 8, 15 and 17

35.

Since, \({ \left( \frac { a }{ b } \right) }^{ m }=\frac { { a }^{ m } }{ { b }^{ m } } \quad and\quad { (a }^{ m })^{ n }={ a }^{ mn }\)
\(\therefore \quad { \left[ { \left( \frac { -4 }{ 5 } \right) }^{ -2 } \right] }^{ 2 }={ \left( \frac { -4 }{ 5 } \right) }^{ -2\times 2 }={ \left( \frac { -4 }{ 5 } \right) }^{ -4 }={ \left( \frac { 5 }{ -4 } \right) }^{ 4 }\)
\(=\frac { 5\times 5\times 5\times 5 }{ (-4)\times (-4)\times (-4)\times (-4) } =\frac { 625 }{ 256 } \)
Thuss, \({ \left[ { \left( \frac { -4 }{ 5 } \right) }^{ -2 } \right] }^{ 2 }=\frac { 625 }{ 256 } \)

36.

We have
A rational number between -3 and -4
= \(\frac { (-3)+(-4) }{ 2 } =\frac { -7 }{ 2 } \)
A rational number between (-3) and \(\frac { -7 }{ 2 } \)
\(\left[ \left( -3 \right) +\left( \frac { -7 }{ 2 } \right) \right] \div 2=\left[ \frac { -6+(-7) }{ 2 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -13 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -13 }{ 4 } \)
A rational number between \(\left( \frac { -7 }{ 2 } \right) \) and (-4)
= \(\left[ \frac { -7 }{ 2 } +(-4) \right] \div 2=\left[ \frac { -7+(-8) }{ 2 } \right] \div 2\)
= \(\frac { -15 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -15 }{ 4 } \)
Thus, the three rational numbers
\(\left( \frac { -7 }{ 2 } \right) ,\left( \frac { -13 }{ 4 } \right) \) and \(\left( \frac { -15 }{ 4 } \right) \) are between (-3) and (-4).

Match the following

5 x 1 = 5

37.

()
\(-\frac { 1 }{ 2 } \)

38.

()
2

39.

()
-1

40.

()
1

41.

()
0

Generic 5

4 x 5 = 20

42.

Numbers of bottles filled	Number of hours
540 x	6 5

Let the required number of bottles to be filled in 5 hours be x
Since, more number of bottles, more number of hours would be required.
\(\therefore\) The given quantities very directly
\(\therefore \ \frac { 540 }{ x } =\frac { 6 }{ 5 } \Rightarrow 6\times x=5\times 540\)
\(\Rightarrow \ x=\frac { 5\times 540 }{ 6 } =5\times 90=450\)
Thus, the required number of bottles = 450

43.

Here, as the base increased, the required number of green pigments will also increase.
\(\therefore\) The quantity vary directly:
i.e.,\(\frac { 1 }{ 6 } =\frac { 4 }{ { x }_{ 1 } } =\frac { 5 }{ { x }_{ 2 } } =\frac { 6 }{ { x }_{ 3 } } \)
\(\therefore\) \(\frac { 4 }{ { x }_{ 1 } } =\frac { 1 }{ 6 } \Rightarrow 1\times { x }_{ 1 }=4\times 6\Rightarrow { x }_{ 1 }=24\)
\(\frac { 1 }{ 6 } =\frac { 5 }{ { x }_{ 2 } } \Rightarrow 1\times { x }_{ 2 }=5\times 6\)
\(\Rightarrow { x }_{ 2 }=\frac { 5\times 6 }{ 1 } =30\)
\(\frac { 1 }{ 6 } =\frac { 6 }{ { x }_{ 3 } } \Rightarrow 1\times { x }_{ 3 }=6\times 6\)
\(\Rightarrow \ { x }_{ 3 }=\frac { 6\times 6 }{ 1 } =36\)
Thus, the required unknown quantities are: x₁ = 24, x₂ = 30, and x₃ = 36.

44.

1.SQUARE
2.PYTHAGOREAN
3.PERFECT SQUARE
4.TRIPLET
5.SQUARE ROOT
6.FACTOR

45.

Let the length of the diagonal BD be x cm
\(\because\) Area of a quadrilateral \(=\frac{1}{2}\times(diagonal)\times\) (Sum of the length of perpendiculars on the diagonal from the opposite vertices)
\(\therefore\) Area of the quadrilateral ABCD
\(=\frac{1}{2}\times BD\times(AP+CQ)\)
\(=\frac{1}{2}\times x\ cm\times(4.5\ cm+3.5\ cm)\)
\(=\frac{1}{2}\times x\times 8\ cm^2\)
Since area of the quadrilateral ABCD = 32 cm²
\(\therefore \frac{1}{2}x\times 8=32\Rightarrow x=\frac{32\times2}{8}\ cm=8\ cm\)
Thus, the required length of the diagonal BD = 8 cm.

Generic 4

2 x 4 = 8

46.

Let the number of plates on the table be y and the number of puris be x.
Then, according to the first condition of the problem
4(y - 1) = x ...(1)
and, according to the second condition of the
problem
3y + 1= x ... (2)
From (1) and (2), we have
4(y - 1) = 3y + 1
⇒ 4y - 4 = 3y + 1
⇒ 4y - 3y = 1 + 4
⇒ y=5
Put y = 5 in (1), we get
x = 4(5) - 4 = 20 - 4 = 16
Hence, the number of plates and number of puris on the table are 5 and 16 respectively.

47.

Time Passed (T) (in minutes)	(T₁) 15	(T₂) 15	(T₃) 45	(T₄) 60
Angle turned (A) (in degree)	(A₁) 90	(A₂) __	(A₃) __	(A₄) __
\(T\over A\)	\({15\over 90}={1\over6}\)	\({30\over 180}={1\over6}\)	\({45\over 270}={1\over6}\)	\({60\over 360}={1\over6}\)

We observe about T and A that they increase together and is \(T\over A\) same every time.
Yes; The angle turned by the minute hand is directly proportional to the time that has passed.
On checking, we find that
T₂ : T₃ = A₁ : A₃ = 2: 3
and T₃ : T₄ = A₃ : A₄ = 3 : 4

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