Q:

A:

10 > 2 > -5 > -14 > -17.

Q:

A:

Fraction of games he won = Games he won/Total Games he played

= 8/14

= 4/7

= 8/14

= 4/7

Q:

A:

Arranging terms in ascending order

14,21,22,27,38,48,72

Median = Medial term = 27

Q:

A:

Let the number be 'x'

x- 3 = 4

x- 3 = 4

Q:

A:

No, triangle is not possible as the sum of three angles of a triangle is not equal to 180°.

Q:

Find the supplementary angle of 75°.

A:

Sum of two supplementary angles = 180º

One of them = 75º

Another angle = 180°-75° = 105°

Q:

A:

Q:

A:

{55 + (-38)} – (–128)

= (17) – (–128)

= 17 + 128

= 145.

Q:

A:

(0.3 0.03) 0.003

= 0.009 0.003

= 0.009 0.003

= 0.000027

Q:

23, 35, 40, 49, 49.5, 15, 15.5, 17.5. Find :

1) Highest Marks

2) Lowest Marks

3) Range

A:

1) Highest Marks = 49.5

2) Lowest Marks = 15

3) Range = Maximum marks - Minimum marks

=49.5 – 15

= 34.5

2) Lowest Marks = 15

3) Range = Maximum marks - Minimum marks

=49.5 – 15

= 34.5

Q:

Check whether *x* = -2 is a solution to the equation 7*x *+ 5 = 9 or not.

A:

Here 7*x *+ 5 = 9

Putting *x* = -2

LHS =7 (-2) + 5

or =-14 + 5

or =-9 RHS

* x* = -2 is not a solution of the given equation.

Q:

(i) vertically opposite angles.

(ii) linear pairs.

A:

(i) 1 and 4; 5 and (3 + 2)

(ii) 1 and 5; 4 and 5

(ii) 1 and 5; 4 and 5

Q:

A:

Sum of two interior angles = Opposite exterior angle

50 + x = 110

x = 60

Q:

A:

Q:

Find out the numbers if

i. 10% of number is 600

ii. 12% of number is 1080.

A:

Q:

A:

Q:

A:

Q:

Raghu has 175 sweets out of which he gives one fifth part to his sister and out of the remaining he gives two seventh part to his youngest brother. How many sweets are left with him?

A:

Total sweets = 175

Sweets given to sister = 1/5 of 175

= 1/5 175 = 35

Remaining = 175 – 35 = 140

Sweets given to brother = 2/7 of 140

= 2/7 140 = 40

Now remaining sweets = 140 – 40 = 100

Q:

A:

First five multiples of 5 are 5, 10, 15, 20, 25.

Mean =

Mean = 15

Q:

A:

Thus, the number is 9.

Q:

A:

(i) *l* is parallel to *m*.

(ii)*l* is not parallel to *m*.

(iii)*l* is parallel to *m*.

(ii)

(iii)

Q:

Find angles *x* and *y* in the following figure:

A:

In triangle PQR,

P = x ( vertically opposite angles)

Q = x

R = y = x (vertically opposite angles)

Therefore, P + Q + R = 180^{0}

3x = 180^{0}

x = 60^{0}.

Since, x = y (Verically opposide triangle)

Hence, y = 60^{0}.

Q:

(i) State the three pairs of equal parts in CBD and BCE.

(ii) Is CBD BCE? Why or why not?

(iii) Is DCB = EBC? Why or why not?

A:

(i) In CBD and BCE,

the three pairs of equal parts are as given below:

the three pairs of equal parts are as given below:

BEC = CDB (=90)

BD = CE (Given)

and BC= BC (Common in both)

(ii) From (i) above, CBD BCE (By RHS congruence rule)

(iii) Yes, DCB = EBC (Corresponding parts of congruent triangles)

Q:

A:

C.P of one orange = Rs.

S.P of one orange = Rs. 3/2x3/3 = 9/6 and 4/3x2/2 = 8/6

S.P > C.P

S.P of one orange = Rs. 3/2x3/3 = 9/6 and 4/3x2/2 = 8/6

S.P > C.P

Q:

A:

Q:

(i) Gurpreet attempts all the questions but only 9 of her answers are correct. What is her total score?

(ii) One of her friends gets only 5 answers correct. What will be her score?

A:

Marks given for wrong answer = - 1

(i) Number of correct answers of

Gurpreet = 9

Number of incorrect answers of

Gurpreet = 15 – 9

= 6

Total score of Gurpreet = 9 (3) + 6(-1)

= 27 – 6

= 21

(ii) Number of correct answers of

her friend = 5

Number of incorrect answers of

her friend = 15 – 5 = 10

=15 – 10 = 5

Q:

Weight (in Kg ) | 48 | 50 | 52 | 54 | 58 |

No. of students | 4 | 3 | 2 | 2 | 1 |

Find the Median of the weights.

A:

Weight(x) | No. of students (f) | Cummalative frequency |

48 | 4 | 4 |

50 | 3 | 7 |

52 | 2 | 9 |

54 | 2 | 11 |

58 | 1 | 12 |

N =f = 12 (even number)

Median

Q:

(iii)6z+10=-2, (iv)(a/4)+7=5.

A:

(ii)

(iii)

(iv)

Q:

A:

Length of cloth used in 1 shirts = (49/2)/35 m

= 49/70

= 0.7 m

Length of cloth used for 4 equal shirts = 4×0.7 m

= 2.8 m

Thus, cloth required for 4 equal shirts is 2.8 m.

Q:

A:

ABCD is a rhombus and its diagonals bisect at O.

AC = 12 cm and BD = 10 cm

So, OA = 24/2 = 12 cm, OB = 10/2 = 5 cm

In AOB, AOB = 90°

Using Pythagoras Theorem,

AB^{2 }= AO^{2} + OB^{2 } AB^{2 }= (12)^{2} + (5)^{2 } = 144 + 25

= 169

AB = 13 cm.

So, the perimeter of rhombus = 4× 13 cm

= 52 cm

AC = 12 cm and BD = 10 cm

So, OA = 24/2 = 12 cm, OB = 10/2 = 5 cm

In AOB, AOB = 90°

Using Pythagoras Theorem,

AB

= 169

AB = 13 cm.

So, the perimeter of rhombus = 4× 13 cm

= 52 cm

Q:

In triangle ABC, AB = AC and ADBC. Prove that B = C.

A:

AB = AC and ADBC

To Prove: B = C

Proof: In ABD and ACD,

AB = AC [Given]

ADB = ADC [Each 90°]

AD = AD [Common]

ABD ACD [By RHS]

So, B = C [By CPCT]

Q:

If 1250 is to be divided amongst Raj, Pallavi and Ronit such that Raj gets two parts, Pallavi three parts and Ronit five parts. How much money will each get? What will it be in percentage?

A:

Q:

A:

10 > 2 > -5 > -14 > -17.

Q:

A:

Fraction of games he won = Games he won/Total Games he played

= 8/14

= 4/7

= 8/14

= 4/7

Q:

A:

Arranging terms in ascending order

14,21,22,27,38,48,72

Median = Medial term = 27

Q:

A:

Let the number be 'x'

x- 3 = 4

x- 3 = 4

Q:

A:

No, triangle is not possible as the sum of three angles of a triangle is not equal to 180°.

Q:

Find the supplementary angle of 75°.

A:

Sum of two supplementary angles = 180º

One of them = 75º

Another angle = 180°-75° = 105°

Q:

A:

Q:

A:

{55 + (-38)} – (–128)

= (17) – (–128)

= 17 + 128

= 145.

Q:

A:

(0.3 0.03) 0.003

= 0.009 0.003

= 0.009 0.003

= 0.000027

Q:

23, 35, 40, 49, 49.5, 15, 15.5, 17.5. Find :

1) Highest Marks

2) Lowest Marks

3) Range

A:

1) Highest Marks = 49.5

2) Lowest Marks = 15

3) Range = Maximum marks - Minimum marks

=49.5 – 15

= 34.5

2) Lowest Marks = 15

3) Range = Maximum marks - Minimum marks

=49.5 – 15

= 34.5

Q:

Check whether *x* = -2 is a solution to the equation 7*x *+ 5 = 9 or not.

A:

Here 7*x *+ 5 = 9

Putting *x* = -2

LHS =7 (-2) + 5

or =-14 + 5

or =-9 RHS

* x* = -2 is not a solution of the given equation.

Q:

(i) vertically opposite angles.

(ii) linear pairs.

A:

(i) 1 and 4; 5 and (3 + 2)

(ii) 1 and 5; 4 and 5

(ii) 1 and 5; 4 and 5

Q:

A:

Sum of two interior angles = Opposite exterior angle

50 + x = 110

x = 60

Q:

A:

Q:

Find out the numbers if

i. 10% of number is 600

ii. 12% of number is 1080.

A:

Q:

A:

Q:

A:

Q:

Raghu has 175 sweets out of which he gives one fifth part to his sister and out of the remaining he gives two seventh part to his youngest brother. How many sweets are left with him?

A:

Total sweets = 175

Sweets given to sister = 1/5 of 175

= 1/5 175 = 35

Remaining = 175 – 35 = 140

Sweets given to brother = 2/7 of 140

= 2/7 140 = 40

Now remaining sweets = 140 – 40 = 100

Q:

A:

First five multiples of 5 are 5, 10, 15, 20, 25.

Mean =

Mean = 15

Q:

A:

Thus, the number is 9.

Q:

A:

(i) *l* is parallel to *m*.

(ii)*l* is not parallel to *m*.

(iii)*l* is parallel to *m*.

(ii)

(iii)

Q:

Find angles *x* and *y* in the following figure:

A:

In triangle PQR,

P = x ( vertically opposite angles)

Q = x

R = y = x (vertically opposite angles)

Therefore, P + Q + R = 180^{0}

3x = 180^{0}

x = 60^{0}.

Since, x = y (Verically opposide triangle)

Hence, y = 60^{0}.

Q:

(i) State the three pairs of equal parts in CBD and BCE.

(ii) Is CBD BCE? Why or why not?

(iii) Is DCB = EBC? Why or why not?

A:

(i) In CBD and BCE,

the three pairs of equal parts are as given below:

the three pairs of equal parts are as given below:

BEC = CDB (=90)

BD = CE (Given)

and BC= BC (Common in both)

(ii) From (i) above, CBD BCE (By RHS congruence rule)

(iii) Yes, DCB = EBC (Corresponding parts of congruent triangles)

Q:

A:

C.P of one orange = Rs.

S.P of one orange = Rs. 3/2x3/3 = 9/6 and 4/3x2/2 = 8/6

S.P > C.P

S.P of one orange = Rs. 3/2x3/3 = 9/6 and 4/3x2/2 = 8/6

S.P > C.P

Q:

A:

Q:

(i) Gurpreet attempts all the questions but only 9 of her answers are correct. What is her total score?

(ii) One of her friends gets only 5 answers correct. What will be her score?

A:

Marks given for wrong answer = - 1

(i) Number of correct answers of

Gurpreet = 9

Number of incorrect answers of

Gurpreet = 15 – 9

= 6

Total score of Gurpreet = 9 (3) + 6(-1)

= 27 – 6

= 21

(ii) Number of correct answers of

her friend = 5

Number of incorrect answers of

her friend = 15 – 5 = 10

=15 – 10 = 5

Q:

Weight (in Kg ) | 48 | 50 | 52 | 54 | 58 |

No. of students | 4 | 3 | 2 | 2 | 1 |

Find the Median of the weights.

A:

Weight(x) | No. of students (f) | Cummalative frequency |

48 | 4 | 4 |

50 | 3 | 7 |

52 | 2 | 9 |

54 | 2 | 11 |

58 | 1 | 12 |

N =f = 12 (even number)

Median

Q:

(iii)6z+10=-2, (iv)(a/4)+7=5.

A:

(ii)

(iii)

(iv)

Q:

A:

Length of cloth used in 1 shirts = (49/2)/35 m

= 49/70

= 0.7 m

Length of cloth used for 4 equal shirts = 4×0.7 m

= 2.8 m

Thus, cloth required for 4 equal shirts is 2.8 m.

Q:

A:

ABCD is a rhombus and its diagonals bisect at O.

AC = 12 cm and BD = 10 cm

So, OA = 24/2 = 12 cm, OB = 10/2 = 5 cm

In AOB, AOB = 90°

Using Pythagoras Theorem,

AB^{2 }= AO^{2} + OB^{2 } AB^{2 }= (12)^{2} + (5)^{2 } = 144 + 25

= 169

AB = 13 cm.

So, the perimeter of rhombus = 4× 13 cm

= 52 cm

AC = 12 cm and BD = 10 cm

So, OA = 24/2 = 12 cm, OB = 10/2 = 5 cm

In AOB, AOB = 90°

Using Pythagoras Theorem,

AB

= 169

AB = 13 cm.

So, the perimeter of rhombus = 4× 13 cm

= 52 cm

Q:

In triangle ABC, AB = AC and ADBC. Prove that B = C.

A:

AB = AC and ADBC

To Prove: B = C

Proof: In ABD and ACD,

AB = AC [Given]

ADB = ADC [Each 90°]

AD = AD [Common]

ABD ACD [By RHS]

So, B = C [By CPCT]

Q:

If 1250 is to be divided amongst Raj, Pallavi and Ronit such that Raj gets two parts, Pallavi three parts and Ronit five parts. How much money will each get? What will it be in percentage?

A:

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