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Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.1

Question 1.
In ∆PQR, D is the mid point of QR.
\(\overline{\mathrm{PM}}\) is …………
PD is ………….
Is QM = MR ?

Solution:
\(\overline{\mathrm{PM}}\) is the altitude of ∆PQR. PD i Since median is the segment.
hence, QD = DR
But, QM ≠ MR.

Question 2.
Draw rough sketches for the folowing;
(a) In ∆ABC, BE is a median.
(b) In ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude in the exterior of th triangle.
Answer:
(a) Hence BE is a median[fig (a)]

(b) In a ∆PQR, PQ and PR are sides of ∆PQR. [Fig. (b)]

(c) YL is the altitude of the triangle. ∆XYZ [Fig. (c)]

Question 3.
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Solution:
The median and altitude of an isosceles triangle can not be same.
In a ∆ABC AD, BE, CF are median but AL, BM and CN are altitude.
Hence O and O’ are the mid point of median and altitude respectively.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 5 Lines and Angles InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 5 Lines and Angles InText Questions

Try These (Page No. 94):

Question 1.
List ten figures around you and identify the acute, obtuse and right angles found in them.
Solution:

Think, Discuss & Write (Page No. 95):

Question 1.
Can two acute angles be complement of each other ?
Solution:
Yes complementary 30° is and acute angle and 60° is also an acute angle.
30° + 60° = 90° (Complementary angle)

Question 2.
Can two obtuse angles be complement of each other ?
Solution:
No obtuse angle > 90°

Question 3.
Can two right angles be complement of each other ?
Solution:
No, one right angle = 90°

Try These (Page No. 95):

Question 1.
Which pairs of following angles are complementary ?
Solution:
(i) and (iv), because complementary angle = 90°.

Question 2.
What is the measure of the complement each of following given angles ?
(i) 45° (ii) 65° (iii) 41° (iv) 54°
Solution:
(i) Let the complement of 45° is re.
∴ x + 45° = 90°
or x = 90° -45° = 45°
∴ Complement of 45° is 45°.

(ii) Let the complement of 65° is x.
∴ x + 65° = 90°
or re = 90° -65° = 25°
∴ The complement of 65° is 25°.

(iii) Let the complement of 41° be m.
∴ m + 41° = 90°
or m = 90° – 41° = 49°
or m = 49°
Thus, the complement of 41° is 49°.

(iv) Let the complement of 54° be x.
∴ x + 54° = 90°
or re = 90° – 54° = 36°
Thus, the complement of 54° is 36°.

Question 1.
The difference in the measures of two complementary angles is 12°. Find the measures of the angles.
Solution:

90° -x = 12°
90° – 12° = x
or 78° = x
∴  x = 78°

Think, Discuss & Write (Page No. 96):

Question 1.
Can two obtuse angles be supplementary ?
Solution:
No, supplementary angle = 180°
obtuse > 90°.

Question 2.
Can two acute angles be supplementary ?
Solution:
Yes, acute angle < 90°

Question 3.
Can two right angles be supplementary ?
Solution:
Yes, right angle
= 90°

Try These (Page No. 97):

Question 1.
Find the supplementary angles in the following given pairs :
Solution:
(i) Measures of the given angles are 110° and 50°.
∵ 110°+ 50° = 160°.
and 160° ≠ 180°
∴ 110° and 50° are not a pair of supplementary angles.

(ii) Measures of the given angles are 105° and 65°.
∵ 105°+ 65° = 170°.
and 170° ≠ 180°
∴ 105° and 65° are not a pair of supplementary angles.

(iii) Measures of the given angles are 50° and 150°.
∵ 50° + 130° = 180°.
∴ 50° and 130° are a pair of supplementary angles.

(iv) Measures of the given angles are 45° and 45°.
∵ 45° + 45° = 90°. and 90° ≠ 180°
∴ 45° and 45° are not a pair of supplementary angles.

Question 1.
What will be the measure of the supplement of each one of the following angles ?
(i) 100° (ii) 90° (iii) 55° (iv) 125°
Solution:
(i) Let the supplementary of 100° be a.
∴ 100° + x = 180°
or x = 180° – 100° = 80°
∴ The measure of the supplement of 100° is 80°.

(ii) Let the supplementary of 90° be x.
∴ x + 90° = 180°
or x = 180°- 90° = 90°
The measure of the supplement of 90° is 90°.

(iii) Let the supplementary of 55° be x.
∴ 55° + x = 180°
or x = 180° -55° = 125°
The supplement of 55° is 125°.

(iv) Let the supplementary of 125° bey.
∴ y + 125° = 180°
or y = 180°-125° = 55°
∴ The supplement of 125° is 55°.

Question 2.
Among two supplementary angles the measure of the larger angle is 44° more than the measure of the smaller. Find their measures.
Solution:
A-t-q, x + x + 44 = 180°
or, 2x + 44 = 180°

2x = 180° – 44°
x = \(\frac{136}{2}\) = 68°
∴ 68°, 112°

Try These (Page No. 97-98):

Question 1.
Are the angles marked 1 and 2 adjacent ? If they are not adjacent.


Solution:
(i) Yes, (ii) Yes, (iii) No, (iv) Yes,
Yes.
(iii) No, because two angles in a plane are said to be adjacent angles, if they have a common vertex, a common arm and the other two arms on opposite sides of the common arm.

Question 2.
In the given figure are the following adjacent angles ?
(a) ∠AOB and ∠BOC
(b) ∠BOD and ∠BOC

Justify your answer.
Solution:
(a) Yes, because they have a common vertex O and a common arm OB.
(b) No, because they are not placed next to each other.

Think, Discuss & Write (Page No. 98):

Question 1.
Can two adjacent angles be supplementary ?
Solution:
Yes, because two adjacent angles form a supplementary.

Question 2.
Can two adjacent angles be complementary ?
Solution:
Y es, because two adjacent angles form a complementary.

Question 3.
Can two obtuse angles be adjacent angles ?
Solution:
Yes, in the figure, ∠BOC and ∠COD are obtuse angles and they are adjacent angles.

Question 4.
Can an acute angle be adjacent to an obtuse angle ?
Solution:
Yes, in the figure, ∠1 and ∠2 are adjacent angles. ∠1 is acute and ∠2 is an obtuse angle.

Think, Discuss & Write (Page No. 99):

Question 1.
Can two acute angles form a linear pair ?
Solution:
No, because acute angle < 90° and linear pair = 180°. Question 2. Can two obtuse angles form a linear pair ? Solution: No, because obtuse angle > 90°
and linear pair = 180°.

Question 2.
Can two right angles form a linear pair ?
Solution:
Yes, because right angle = 90°
and linear pair = 180°.

Try These (Page No. 99):

Question 1.
Check which of the following pairs of angles form a linear pair:
Solution:
(i) Yes,
∵ 140°+ 40° = 180°
∴ The given pair of angles form a linear pair.

(ii) No,
∵ 60° + 60° = 120°
and 120° ≠ 180°.

(iii) No,
∵ 90° + 80° = 170°
and 170° ≠ 180°.

(iv) Yes,
∵ 115°+ 65° = 180°.

Try These (Page No. 101):

Question 1.
In the given figure, if ∠1 = 30°. find ∠2 and ∠3.

Solution:
Since ∠1 + ∠2 = Linear pair = 180°
or, 30° + ∠2 = 180°
or, ∠2 = 180°-30°
∴ ∠2 = 150°
Now, ∠2 + ∠3 = 180°
or, 150° + ∠3 = 180°
or, ∠3 = 180° -150°
∴ ∠3 = 30°
or, ∠1 = ∠2 = vertically opposite angle
∴ 30° = ∠2
∴ ∠2 = 30°

Question 2.
Give an example for vertically opposite angles in your surroundings.

Solution:
Two angles formed by two intersecting lines having common arm are said to be vertically opposite angles.
Hence, ∠COB = ∠AOD = 60°
= vertically opposite angle
and ∠BOD = ∠AOC = 120°
= vertically opposite angles

Try These (Page No. 104):

Question 1.
Find examples from your surround-ings where lines intersect at right angles.
Solution:
(i) A black board, (ii) A table, (iii) A television (iv) A computer.

Question 2.
Find the measures of the angles made by the intersecting lines at the vertices of an equilateral triangle.

Solution:
Vertices, A, B and C. (Fig.)

Question 3.
Draw any rectangle and find the measures of angles at the four vertices made by the intersecting lines.
Solution:

Question 4.
If two lines intersect, do they always intersect at right angles ?
Solution:

Try These (Page No. 105):

Question 1.
Suppose Two lines are given. How many transversals can you draw for these lines.

Solution:
Many

Question 2.
If three lines have a transversal, how many points of intersections are there ?
Solution:
Three.

Question 3.
Try to identify a few transversals in your surroundings.
Solution:
Road, Rail, ladder etc.

Try These (Page No. 1.6):

Question 1.
Name the pairs of angles in each Figure.


Solution:
(ii) i.e. < 1 and < 2 are called pairs corresponding angles. [Fig. (i)]
(it) i.e. <3 and < 4 are called pairs of alternate interior angles. [Fig. (ii)]
(iii) i.e. < 5 and < 6 are called pairs of interior angles on the same side of the transversal. [Fig. (iii)]
(iv) i.e. < 7 and < 8 are called pairs of corresponding angles. [Fig.(iv)]
(v) i.e. < 9 and < 8 are called pairs of alternate interior angles. [Fig. (v)]
(vi) i.e. < 11 and < 12 are called linear pairs. [Fig. (vi)]

Try These (Page No. 109):

Question 1.


Solution:
(i) x = 60°
[∵ x and 60° are alternate interior angles]

(ii) ∠y = 55°
[∵ y and 55° are alternate interior angles]

(iii) If two non-parallel lines are intersected by a transversal, then the angle of pairs of alternate interior angles are not equal.
Hence ∠1 = ∠2

(iv) 60° + z = 180°
⇒ z = 180° – 60°
z = 120°
x = 120°
[ ∵ and 60° are interior angles on the same side of the transversal]
∴ z = 120°.

(v) x = 120°
[∵ x and 120° are corresponding angles]

(vi) a + 60° = 180°
⇒ a = 180° – 60° = 120°
a – c
⇒ c = 120° [alternate angles]
c = b
[vertically opp. angles]
⇒ b = 120° [linear pair]
⇒ b+d = 180°
⇒ 120° + d = 180°
∴ d = 180° – 120° = 60°.

Try These (Page No. 100):

Question 1.


Solution:
(i) Yes, l || m, because alternate interior angles are equal and 50°.
(ii) Yes, l || m, because alternate interior angles are equal and 50°.
(iii) If two parallel lines are intersected by a transversal, then the sum of the interior angles on the same side of the transversal is 180° or supplementary.
Hence, 180° + x = 180°
or, x = 180° -180°
x = 0°
But in this figure one angle is 180°. Hence it is not possible.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 5 Lines and Angles Ex 5.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 5 Lines and Angles Exercise 5.2

Question 1.
State the property that is used in each of the following statements?
(i) If a // b, then ∠1 = ∠5.
(ii) If ∠4 – ∠6, then a //b.
(iii) If ∠4 + ∠5 = 180°, then a // b.
Solution:

If two lines are intersected by a transversal and if we have any of the following three conditions.
(a) The angles of any pair of corresponding angles are equal.
(b) The angles of any pair of alternate angles are equal.
(c) The sum of the interior angles on the same side of the transversal is 180°, then the lines are parallel.
(i) In the above figure, if a || b, then ∠1 = ∠5 = corresponding angles.
(ii) If ∠4 = ∠8 – corresponding angles, hence a || b.
(iii) If ∠4 + ∠5 = 180° = supplementary, hence a||b.

Question 2.
In the given figure, identify :
(i) The pairs of corresponding angles.
(ii) The pairs of alternate interior angles.
(iii) The pairs of interior angles on the same side of the transversal.
(iv) The vertically opposite angles.

Solution:
(i) ∠1 and ∠5, ∠4 and ∠8, ∠3 and ∠7, ∠2 and ∠6.
(ii) ∠3 and ∠5, ∠2 and ∠8
(iii) ∠2 and ∠5, ∠3 and ∠8
(iv) ∠1 and ∠3, ∠4 and ∠2, ∠8 and ∠6, ∠5 and ∠7.

Question 3.
In the adjoining figure, p || q. Find the unknown angles.

Solution:
∠e + ∠125° = 180°(Linearpair)
∴ ∠e = 180°-125° = 55°
∠e = ∠f = 55° (vertically opposite angles)
∠d = 125° (corresponding angles)
∠d = ∠b =125° (vertically opposite angles)
∠e = ∠a = 55° (corresponding angles)
∠a = ∠c = 55° (vertically opposite angles)
i.e., a = 55°, b = 125°, c = 55°, d = 125°, e = 55° and f = 55°.

Question 4.
Find the value of x in each of the following figures if l || m.

Solution:
(i) ∠x = ∠x = (alternate angles)
New ∠x + ∠100° = 180° (Linear pair)
=> Lx = 180°—110°
∴ x = 70°.(Fig.i)
(ii) x + 2x = 1800 (supplementary angles)
3x = 180°
x = \(\frac{180}{3}\)
∴ x = 60°.
x = 100 (Corresponding angles)
or, y + 80° = 180° (Supplemcnatary angles)
∴ y = 100°
s = y = 100° (alternate angles)
(Fig. iii)

Question 5.
In the given figure, the arms of two angles are parallel. If ∠DGC = 70° then find s
(i) ∠DGC
(ii) ∠DEF

Solution:
∠ABC = ∠DGC = corresponding angles
= 70°
∴ ∠DGC = 70°
∠DGC = ∠DEF = 70°
= corresponding angles
hence, (i) ∠DGC = 70°
(ii) ∠DEF = 70°

Question 6.
In the given figures below, decide whether l is parallel to m ?


Solution:
(i) ∠126° + ∠44° = 170° ≠ 180° (Corresponding angles)
Hence l is not parallel to m.
(ii) l is not parallel to m because alternate angles are not equal.
(iii) ∠57° + x = 180° (Linear pair)
∴ x = 180° -57° = 123°
Now x = 123° = alternate angle
Hence l || m because alternate angles are equal.
(iv) l is not parallel to m because alternate
angles are not equal. .

Haryana State Board HBSE 7th Class Maths Solutions Chapter 5 Lines and Angles Ex 5.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 5 Lines and Angles Exercise 5.1

Question 1.
Find the complement of each of the following angles:

Solution:
If the sum of two angles is 180°, then they are said to be supplementary angles and each of them is called the supplement of the other.
(i) 180° -105° = 75°
(ii) 180°-87° = 93°
(iii) 180°-105° = 75°.

Question 2.
Find the supplement of each of the following angles:

Solution:
If the sum of two angles is 180°, then they are said to be supplementary angles and each of them is called the supplement of the other.
(i) 180° — 105° = 750
(ii) 180° — 87° = 93°
(iii) 180° — 105° = 75°.

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°
Solution:
Complementary angles = 90° Supplementary angles = 180°
(i) 65° + 115° = 180° = Supplementary
(ii) 63° + 27° = 90° = Complementary
(iii) 112° + 68° = 180° = Supplementary
(iv) 130° + 50° = 180° = Supplementary
(v) 45° + 45° = 90° = Complementary
(vi) 80° + 10° = 90° » Complementary.

Question 4.
Find the angle which is equal to its complement.
Solution:
∵ Let the required angle be x. v It is equal to its complement,
∴ x = 90° – x
⇒ x + x = 90°
⇒ 2x = 90°
∴ x = \(\frac{90^{\circ}}{2}\) = 45
Thus, 45° is equal to its complement.

Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the required angle be x and supplement of x = 180 – x
∴ x is equal to its supplement .
∴ x = 180° – x
x + x = 180°
2x = 180°
∴ x = \(\frac{180^{\circ}}{2}\) = 90°
Thus, 90° is equal to its supplement.

Question 6.
In the given figure ∠1, and ∠2 are supplementary angles. If E ∠1 is decreased, what changes should take place in ∠2 so that both the angles supplementary.

Solution:
∠1 + ∠2 = 180° .
Supplementary angles if ∠1 is decreased, then 2 is increased but ∠1 + ∠2 = 180°
Hence, 2 is increased so that the sum of the two angles remains the same.

Question 7.
Can two angles be supplementary if both of them are :
(i) acute (ii) obtuse (iii) right ? Sol: Supplementary = 180°
acute angle = < 90° acute angle = > 90°
acute angle = 90°
(i) No, (ii) No, (iii) Yes [because 90° + 90° = 180°]

Question 8.
An angle is greater than 45°. Is its complementary angle grater than 45° or equal to 45° or less than 45° ?

Solution:
A-t-q,
∠AOC < 45°
And BOC < 45°
Because ∠AOC + ∠BOC = 90°
= Complementary angles
Hence, It is complementary and one angle is greater than 45° and other is less than 45″.

Question 9.
In the given figure
(i) Is ∠1 adjacent to ∠2 ?
(ii) Is ∠AOC adjacent to ∠AOE?
(iii) Do ∠COE and ∠COD form a linear pair ?
(iv) Are ∠BOD and ∠DOA supplementary ?
(v) Is ∠1 vertically opposite to ∠4 ?
(vi) Wliat is the vertically opposite angle of ∠5 ?

Solution:
(i) Yes
(ii) No, Because OC and OA are not on the opposite side of OE.
(iii) No, Because ∠COE + ∠COD 180°.
(iv) Yes, Because ∠BOD + ∠DOA = 180° Supplementary.
(v) Yes.
(vi) (∠2 + ∠3) is the vertically opposite angle of ∠5.

Question 10.
Indicate which pairs of angles are
(i) Vertically opposite angles.
(ii) Linear pairs.

Solution:
(i) ∠1 = ∠4
i.e. ∠l, ∠4
(ii) ∠4 + ∠5 = 180′ i.e., ∠4, ∠5
and ∠1 + ∠2 + ∠3= 180°,i.c., ∠1, ∠2, ∠3
and ∠1 + ∠5 = 180″, i.e., ∠1,∠5

Question 11.
In the given Fig. is ∠1 adjacent to ∠2 ? Given reasons.

Solution:
No, ∠1 and ∠2 are not adjacent angles because they do not have a common vertex.

Question 12.
Find the values of the angles X, Y and Z in each of the following:

Solution:
(i) ∠X = ∠55°
= Vertically opposite angles and
∠X + ∠Y = 180° = Linear pairs
or, ∠55° + ∠Y = 180°
or, ∠Y = 180°-55°
∴ ∠Y = 125°
and ∠Y = ∠Z = Vertically opposite
angles
125° = ∠Z Fig.
∴ ∠Z = 125°
i.e., X = 55°, Y = 125°, Z = 125°.

(ii) ∠Z = 40
= Vertically opposite angles and
∠Y + 40° = Linear pairs = 180°
or, ∠Y = 180° – 40° = 140°
∴ ∠Y = 140°
Now, ∠Y = ∠X + ∠25°
= Vertically opposite angles
or, 140° = ∠X + ∠25°
or, 140°-25° = ∠X
or, 115° = ∠X
∴ ∠X = 115°
i.e., X = 115°, Y = 140°, Z = 40°.

Question 13.
Fill in the blanks :
(i) If two angles are complementary, then the sum of their measures is ……………..
(ii) If two angles are supplementary, then the sum of their measures is …………
(Hi) Two angles forming a linear pair are …………….
(iv) If two angles are supplementary, they form a ………………
(v) If two lines intersect at a point, then the vertically opposite angles are always ………….
(vi) If two lies intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of
vertically opposite angles are ……………………
Solution:
(i) 90°. (it) 180°. (iii) adjacent angles. (iv) linear pair (u) equal, (vi) obtuse angles.

Question 14.
In the Fig., name the following pairs of angles.

(i) Obtuse
vertically opposite angles
(ii) Adjacent complementary angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair.
Solution:
(i) ∠BOC = ∠AOE + ∠EOD, i.e., ∠BOC, (∠AOE + ∠EOD)
(ii) ∠AOB + ∠AOE = 90°, i.e., ∠AOB, ∠AOE
(iii) ∠BOE + ∠EOD = 180°, i.e., ∠BOE, ∠EOD
(iv) ∠AOB + ∠BOC = ∠180°, i.e., ∠AOB, ∠BOC
(v) ∠COD and ∠DOE, ∠DOE and ∠EOA, ∠BOA and ∠AOE.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations InText Questions

Try These (Page No. 78):

Question 1.
The value of the expression (lOy – 20) depends on the value of y. Verify this by giving 5 different values to y and finding the value of (lOy – 20) for each. From the different values of (lOy – 20) you obtain, do you see a solution to 10y – 20 = 50 ? If there
is no solution, try giving more values of y and find whether the condition 10y – 20 = 50 is met.
Solution:
The value of an expression depends upon the value of the variable from which the expression is formed.
10y – 20 = 50?
L.H.S. = 10y – 20
If y = 0, 10y – -20 = 10 ×  0 – 20 = 0 – 20 = -20
If y = 1, 10y – -20 = 10 × 1 – 20  = 10 – 20 = -10
If y = 2, 10y – -20 = 10 × 2 – 20 = 20 – 20 = 0
If y = 3, 10y – -20 = 10 × 3 – 20 = 30 – 20 = 10
If y = 4, 10y – -20 = 10 × 4 – 20 = 40 – 20 = 20
If y = 5, 10y – -20 = 10 × 5 – 20 = 50 – 20 = 30
If y = 6, 10y – -20 = 10 × 6 – 20 = 60 – 20 = 40
If y = 7, 10y – -20 = 10 × 7 – 20 = 70 – 20 = 50
if, y = 7, 10y – 20 = 50

Try These (Page No. 80):

Question 1.
Write at least one other from each equation.
(i) 5P = 20, (ii) 3n + 7 = 1, (iii) \(\frac{m}{5}\) – 2 = 6
Solution:
(i) (a) five times a number p is 20. (b) multiple of p is 20.
(ii) (a) Add 7 to three times n to get 1. (b)
Three times n plus 7 gets you 1.
(iii) (a) You get 6, when you subtract 2 from one fifth of a number m.
(b) One fifth of a number m mumus 2 leaves 6.

Try These (Page No. 88):

Question 1.
Start with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5.
Solution:
x = 5
Multiply both sides by 2, we get,
2x = 10
Multiply both sides by 2, we get
4x = 20
Check: If x = 5, 2 x 5 = 10
10 = 10
If x = 5, 4 x 5 = 20
20 = 20

Try These (Page No. 88):

Question 1.
Try to make two number puzzles, one with the sulution 11 and another with 100.
Solution:
(i) If x = 11
Multiply both sides by 2.2x = 22
Add 4 to both sides 2x + 4 = 26
(ii) If y = 100
Multiply both sides by 2.3y = 300
Subtract 100 from
both sides 3y -100 = 200

Try These (Page No. 90):

Question 1.
(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is ?
(ii) What is that number one third of which added to 5 gives 8 ?
Solution:
(i) Let x be the number.
hence, 6x – 5 =7
or, Qx = 7 + 5 = 12
or, x = \(\frac{12}{6}\) = 2
Hence number be 2.
(ii)Let y be the number.
hence, \(\frac{y}{3}\) + 5 = 8
\(\frac{y+15}{3}\)
or, y + 15 = 24
∴ y = 24 – 15 = 9
Hence number be 9.

Try These (Page No. 90):

Question 1.
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of managoes contained in the smaller box ?
Solution:
Let the number of mangoes contained in the smaller box be x.
According to questions, 8x + 4 = 100
or, 8x = 100 – 4
or, 8x = 96
or, x = \(\)\frac{96}{8}\(\) = 12
Hence, 8 mangoes contained in the smaller box.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.4

Question 1.
Set up equations and solve them to find the undnown numbers in the following cases :
(a) Add 4 to eight times a number; you get 60.
(b) One fifth of a number minus 4 gives 3.
(c) If 1 take three fourths of a number and count up 3 more, 1 get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the rusult is \(\frac{7}{11}\).
Solution:
(a) Let a number be x.
According to question, + 4 = 60
⇒ 8x = 60 – 4
⇒ x = \(\frac{56}{8}\) =7
Required number is 7.

(b) \(\frac{x}{5}\) – 4 = 3
⇒ \(\frac{x-20}{5}\) = 3
⇒ x – 20 = 15
⇒ x = 15 + 20
⇒ x = 35
Required number is 35.

(c) \(\frac{3 x}{4}\) + 3 = 21
\(\frac{3 x+12}{4}\) = 21
⇒ 3x + 12 = 84
⇒ 3x = 84-12
⇒ x = \(\frac{72}{3}\)
∴ x = 24
Required number is 24.

(d) 2x -11 = 15
⇒ 2x = 15 + 11
⇒ x = \(\frac{26}{2}\)
∴ x = 13
Required number is 13.

(e) 50 – 3x = 8
⇒ 50 – 8 = 3x
⇒ 3x = 42
x = \(\frac{42}{3}\)
∴ x = 14
Required number is 14.

(f) \(\frac{x+19}{5}\) = 8
⇒ x + 19 = 8 x 5
⇒ x = 40 – 19
⇒ x = 21
Required number is 21.

(g) \(\frac{5 x}{2}-7=\frac{11}{2}\)
⇒ \(\frac{5 x+14}{2}=\frac{11}{2}\)
⇒ 2(5x -14) = 11 x 2
⇒ 10x – 28 = 22
⇒ 10x = 22 + 28
⇒ x = \(\)
∴ x = 5
Required number is 5.

Question 2.
Solve the following :
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score ?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°).
(c) Smita’s mother is 34 years old. Two years from now mother’s age will be 4 times Smita’s present age. ?What is Smita’s present age ?
(d) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
(a) Let lowest score be x
According to question, 2x + 7 = 87
⇒ 2x = 87 – 7
⇒ x = \(\frac{80}{2}\)
∴ x = 40
Required lowest score is 40.

(b) Let base angle be x now base angle are equal We know that the sum of three angles of a triangle is 180°.

x = x
hence, 40° + x + x = 180°
⇒ 2x = 180°-40°
⇒ x = \(\frac{140}{2}\)
∴ x = 70°
Required base angles are 70° and 70°.

(c) Let Smita presentage be x.
According to question 4x – 2 = 34
⇒ 4x = 34 + 2
⇒ x = \(\frac{36}{4}\)
∴ x = 9
Smita’s present age is 9 years.

(d) Let Rahul’s scores be
hence sachin’s scores = 2x
A.t.Q 2x + x = 200-2
⇒ 3x = 198
⇒ x = \(\frac{198}{3}\)
∴ x = 66
Hence, required Rahul’s scores be 99 runs.
Sachin’s scores be 2 x 66 = 132 runs.

Question 3.
Solve the following :
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have ?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age ?
(iii) Maya, Madhra and Mohsina are friends studying in the same class. In a call test in geography, Maya got 16 out of 25. Madhura got 20. Their average score was 19. How much did Mohsina score ?
(iv) People ofSutidargram planted a total of 102 trees in the village garden Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted ?
Solution:
(i) Let pannit’s marbles be m.
A.t.Q, 5m + 7 = 37
⇒ 5m = 37 – 7
⇒ x = \(\frac{30}{5}\)
∴ x = 6
Hence, parmit have 6 marbles. (ii) Let Laxmi’s age be years.
A.t.Q, 3y + 4 = 49
3y = 49 – 4
⇒ y = \(\frac{45}{3}\)
⇒ y = 15
Hence, Laxmi’s age is 15 years.

(iii) Let Mohsina score be x.
According to question, \(\frac{16+20+x}{3}\) = 19
⇒ 36 + x = 19 x 3
⇒  x = 57 — 36
∴ x = 21
Hence, Mohsina score is 21.

(iv) Let the number of fruit trees planted be x.
According to questions^: + (3x + 2) = 102
⇒ 4x = 102-2
⇒ x = \(\frac{100}{4}\)
∴ x = 25
Hence, the number of fruit trees planted be 25.

Question 4.
Solve the following riddle :
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Solution:
Let I be denoted by x.
According to question, (7x + 50) + 40 = 300
⇒ 7x + 90 = 300
⇒ 7x = 300-90
⇒ x = \(\frac{210}{7}\)
∴ x = 30
Thus I am 30.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.3

Question 1.
Solve the following equations :
(a) 2y + \(\frac{5}{2}=\frac{37}{2}\)
(b) 5t + 28 = 10
(c) \(\frac{a}{5}\) + 3 = 2
(d) \(\frac{q}{4}\) + 7 = 5
(e) \(\frac{5}{2}\)x = -10
(f) \(\frac{5}{2} x=\frac{25}{4}\)
(g) 7m + \(\frac{19}{2}\) = 13
(h) 6z + 10 = -2
(i) \(\frac{3 \cdots}{2}=\frac{2}{3}\)
(j) \(\frac{2b}{3}\) – 5 = 3
Solution:
(a) 2y + \(\frac{5}{2}=\frac{37}{2}\)
\(2 y=\frac{37-5}{2}=\frac{32}{2}=\frac{37}{2}-\frac{5}{2}\)
2y = 16
y = \(\frac{16}{2}\)
∴ y = 8

(b) 5t + 28 = 10
or, 5t = 10 – 28 = -18
∴ t = \(\frac{-18}{5}\)

(c) \(\frac{a}{5}\) + 3 = 2
or \(\frac{a}{5}\) = 1 – 3 = -1
a = -1 × 5 = -5
∴ a = -5

(d) \(\frac{a}{4}\) + 7 = 5
or, \(\frac{a}{4}\) = 5 – 7 = -2
a = -2 × 4 = -8
∴ a = -8

(e) \(\frac{5}{2}\)x = -10
5x = 2 × (-10) = -20
x = \(\frac{-20}{5}\)
∴ x = -4

(f) \(\frac{5}{2} x=\frac{25}{4}\)
5x × 4 = 25 × 2
20x = 50
x = \(\frac{50}{20}=\frac{5}{2}\)
∴ x = \(\frac{5}{2}\)

(g) 7m + \(\frac{19}{2}\) = 13
or, 7m = 13 – \(\frac{19}{2}\)
= \(\frac{29-19}{2}=\frac{7}{2}\)
or, 7m × 2 = 7
or, m = \(\frac{7}{7 \times 2}=\frac{1}{2}\)
∴ m = \(\frac{1}{2}\)

(h) 6z + 10 = -2
or, 6z = -2 – 10 = -12
or z = \(\frac{-12}{6}\) = -2
∴ z = -2

(i) \(\frac{3 \cdots}{2}=\frac{2}{3}\)
3l × 3 = 2 × 2 = 4
∴ l = \(\frac{4}{9}\)

(j) \(\frac{2b}{3}\) – 5 = 3
\(\frac{2b}{3}\) = 3 + 5 = 8
2b = 3 × 8 = 24
b = \(\frac{24}{2}\) = 12
∴ b = 12

Question 2.
Solve the following equations :
(a) 2(x + 4) = 12
(b) 3(n-5)=21
(c) 3(n-5) = -21
(d) 3 – 2(2 – y)-7
(e) – 4(2 – x) = 9
(f) 4(2 – x) = 9
(g) 4 + 5 (p – 1) = 34
(h) 34 – 5(p – 1) = 4
Solution:
(a) 2(x + 4) = 12
or, x + 4 = \(\frac{12}{2}\) = 6
or, x = 6 – 4 = 2
∴ x = 2

(b) 3(n – 5) = 21
or, (n – 5) = \(\frac{21}{3}\) = 7
n = 7 + 5 = 12
∴ n = 12

(c) 3(n – 5) = -21
or, (n – 5) = \(\frac{-21}{3}\)= – 7
n = – 7 + 5 = – 2
∴ n = -2

(d) 3 – 2(2 – y) = 7
or, -2(2 – y) = 7 – 3 = 4
or, (2 – y) = \(\frac{4}{-2}\) = -2
-y = – 2 – 2 = – 4
∴ y = 4

(e) -4(2 – x) = 9
or, 2-x = \(\frac{9}{-4}\)
or, -x = \(\frac{-9-8}{4}=\frac{-17}{4}\)
∴ x = \(\frac{17}{4}\)

(f) 4(2 – x) = 9
or, 2 – x = \(\frac{9}{4}\)
or – x = \(\frac{9}{4}\) – 2
= \(\frac{9-8}{4}=\frac{1}{4}\)
∴ x = \(-\frac{1}{4}\)

(h) 34 – 5(p – 1) = 4
or, – 5(p – 1) = 4 – 34 = – 30
or, (p-1) = \(\frac{-30}{-5}\) = 6
or, P = 6 + 1
∴  P = 7.

Question 3.
Solve the following equations :
(a) 4 = 5(p-2)
(b) – 4 = 5(p – 2)
(c) -16 = -5(2 – p)
(d) 10 = 4 + 3(t + 2)
(e) 28 = 4 + 3(t + 5)
(f) 0 = 16 + 4(m -6)
Solution:
(a) 4 = 6(p – 2)
or, 5(p – 2) = 4
or, (p – 2) = \(\frac{4}{5}\)
or, P = \(\frac{4}{5}+2=\frac{4+10}{5}=\frac{14}{5}\)
∴ P = \(\frac{14}{5}\)

(b) -4 = 5(p – 2)
or, 5 (p – 2) = _4
(p-2) = \(\frac{-4}{5}\)
p = \(\frac{-4}{5}+2=\frac{-4+10}{5}=\frac{6}{5}\)
∴ p = \(\frac{6}{5}\)

(c) -16 = -5(2-p)
or, 5(2-p) = 16
2-p = \(\frac{16}{5}\)
-p = \(\frac{16}{5}-2=\frac{16-10}{5}=\frac{6}{5}\)
∴ p = \(\frac{6}{5}\)

(d) 10 = 4 + 3(t + 2)
or, 4 + 3(t + 2) = 10
3 (t + 2) = 10 – 4 = 6
t + 2 = \(\frac{6}{3}\) = 2
t = 2 – 2 = 0
∴ t = 0

(e) 28 = 4 + 3(t + 5)
or, 4 + 3(t + 5) = 28
3(t + 5) = 28-4 = 24
(t + 5) = \(\frac{24}{3}\) = 8
t = 8-5 = 3
∴ t = 3

(f) 0 = 16 + 4(m – 6)
or, 16 + 4(m – 6) = 0
or, 4 (m -6) = -16
(m – 6) = \(\frac{-16}{4}\) = – 4
m = -4 + 6 = 2
∴ m = 2.

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x=- 2.
Solution:
(a) x = 2
2x = 2 × 2 = 4 ⇒ 2x = 4
3x = 2 × 3 = 6 ⇒ 3x= 6
4x = 2 × 4 = 8 ⇒ 4x = 8

(b) x =-2
2x = -2 × 2 = -4 ⇒ 2x = -4
3x = – 2 × 3 = – 6 ⇒ 3x = – 6
5x = – 2 × 5 = – 10 ⇒ 5x = -10.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.2

Question 1.
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y -4 = – 7
(f) y -4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4.
Solution:
(a) x – 1 = 0
L.H.S. = x- R.H.S.
1 + 1 = x = 0 + 1 = 1
∴ x = 1
Which is the required solution.

(b) x + 1 = 0
L.H.S. = x + 1 – 1 = x
R.H.S. = 0 – 1 = -1
∴ x = -1
Which is the required solution.

(c) x – 1 = 5
L.H.S. = x – 1 + 1 = x
R.H.S. = 5 + 1 = 6
∴ x = 6
Which is the required solution.

(d) x + 6 =2
L.H.S. = x + 6 – 6 = x
R.H.S. = 2 – 6 = – 4
∴ x = -4
Which is the required solution.

(e) y- 4 = -7
L.H.S. = y – 4 + 4 = y
R.H.S. = – 7 + 4 = – 3
∴ y = -3
Which is the required solution.

(f) y – 4 = 4
L.H.S. = y – 4 + 4 = y
R.H.S. = 4 + 4 = 8
∴ y = 8
Which is the required solution.

(g) y + 4 = 4
L.H.S. = y + 4 + 4= y
R.H.S. = 4 – 4 = 0
∴ y = 0
Which is the required solution.

(h) y + 4 = – 4
L.H.S. = y + 4 — 4 = y
R.H.S. = – 4 – 4 = – S
∴ y = -8
Which is the required solution.

Question 2.
Give first the step you will use to separate the variable and then solve the equation.
(a) 3l = 42
(b) \(\frac{b}{2}\) = 6
(c) \(\frac{p}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = – 10.
Solution:
(a) 3l = 42
L.H.S. = 3l = \(\frac{3}{3}\) x l = l
R.H.S. = 42 = \(\frac{42}{3}\) = 14
∴ l = 14
Which is the required solution.

b) \(\frac{b}{2}\) = 6
L.H.S = \(\frac{b}{2}=\frac{b}{2}\) x 2 = b
R.H.S = 6 = 6 x 2 = 12
∴ b = 12.
Which is the required solution.

(c) \(\frac{p}{7}\) = 4
L.H.S. = \(\frac{p}{7}=\frac{p}{7}\) x 7 = p
R.H.S. = 4 = 4 x 7 = 28
∴ p = 28
Which is the required solution.

(d) 4 x = 25
L.H.S = 8y = \(\frac{8y}{8}\) = y
R.H.S = 36 = \(\frac{36}{8}=\frac{9}{2}\)
∴ y = \(\frac{9}{2}\)
Which is the required solution.

(f) \(\frac{z}{3}=\frac{5}{4}\)
L.H.S = \(\frac{z}{3}=\frac{z}{3}\) x 3 = z
R.H.S = \(\frac{5}{4}=\frac{5}{4} \times 3=\frac{15}{4}\)
∴ y = \(\frac{15}{4}\)
Which is the required solution.

(f) \(\frac{a}{5}=\frac{7}{15}\)
L.H.S. = \(\frac{a}{5}=\frac{a}{5}\) x 5 = a
R.H.S = \(\frac{7}{15}=\frac{7}{15}\) x 5 = \(\frac{7}{3}\)
∴ a = \(\frac{7}{3}\)
Which is the required solution.

(h) 20t = -10
L.H.S. = 20t = \(\frac{20 t}{20}\) = t
R.H.S = -10 = \(\frac{-10}{20}=-\frac{1}{2}\)
∴ t = \(-\frac{1}{2}\)
Which is the required solution.

Question 3.
Give the step you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac{20 p}{3}\) = 40
(d) \(\frac{3 p}{10}\) = 6
Solution:
3n – 2 = 46
L.H.S. = 3n – 2 + 2
= 3n = \(\frac{3n}{3}\) = n
R.H.S. = 46 + 2 = 48 = \(\frac{48}{3}\) = 16
∴ n = 16
Which is the required solution.

(b) 5m + 7 = 17
L.H.S. = 5 m + 7 = 5m + 7-7
= 5m = \(\frac{5m}{5}\) = m
R.H.S = 17 = 17 – 7 = 10
= \(\frac{10}{5}\) = 2
∴ m = 2
Which is the required solution.

(c) \(\frac{20 p}{3}\) = 40
L.H.S = \(\frac{20 p}{3}=\frac{20 p}{3} \times 3\)
= 20 p = \(\frac{20 p}{20}\) = p
R.H.S. = 40 = 40 x 3
= 120 = \(\frac{120}{20}\) = 6
∴ P = 6
Which is the required solution.

(d) \(\frac{3 p}{10}\) = 6
L.H.S.
= \(\frac{3 p}{10}=\frac{3 p}{10} \times 10=3 p=\frac{3 p}{3}=p\)
R.H.S. = 6 x 10 = 60 = \(\frac{60}{3}\) = 20
∴ p = 20
Which is the required solution.

Question 4.
Solve the following equations
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac{p}{4}\) = 5
(d) \(\frac{-p}{3}\) = 5
(e) \(\frac{3 p}{4}\) = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:
(a) 10p = 100
\(\frac{10 p}{10}=\frac{100}{10}\)
∴ p = 10

(b) 10p + 10 = 100
10p = 100 – 10
10p = 90
\(\frac{10 p}{10}=\frac{90}{10}\)
∴ p = 9

(c) \(\frac{p}{4}\) = 5
\(\frac{p}{4}\) x 4 = 5 x 4
∴ p = 20

(d) \(\frac{-p}{3}\) x 3 = 5 x 3
-p = 15
∴ p = -15

(e) \(\frac{3 p}{4}\) = 6
\(\frac{3 p}{4}\) x 4 = 6 x 4
3p = 24
\(\frac{3 p}{3}=\frac{24}{3}\)
∴ p = 8

(f) 3s = -9
\(\frac{3 s}{3}=\frac{-9}{3}\)
∴ s = -3

(g) 3s+ 12 = 0
3s = -12
\(\frac{3 s}{3}=\frac{-12}{3}\)
∴ s = -4

(h) 3s = 0
or \(\frac{3 s}{3}=\frac{0}{3}\)
∴ s = 0

(i) 2q = 6
or \(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3

(j) 2q – 6 = 6
or, 2q = +6
\(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3

(k) 2q + 6 = 0
or 2q = -6
\(\frac{2 q}{2}=\frac{-6}{2}\)
∴ q = -3

(l) 2q + 6 = 12
or 2q = 12 – 6
\(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.1

Question 1.
Complete the last column of the table.

Solution:
1. Put x = 3, x + 3 = 0++
3 + 3 = 0
⇒ 6 ≠ 0 → No.

2. Put x = 0, x + 3 = 0
0 + 3 = 0
⇒ 3 ≠ 0 → No.

3. Put x = – 3, x + 3 = 0
-3 + 3 = 0
⇒ 0 ≠ 0 → Yes.

4. Put x = 7, x – 7 = 1
7 – 7 = 1
⇒ 0 ≠ 1 → No.

5. Put x = 8, x – 7 = 1
8 – 7 = 1
⇒ 1 = 1 → Yes.

6. Put x = 0, 5x = 25
5×0 = 25
⇒ 0 ≠ 25 → No.

7. Put x = 5, 5x = 25
5 x 5 = 25
⇒ 25 = 25 → Yes.

8. Put x = – 5, 5x = 25
5 x (- 5) = 25
⇒ -25 ≠ 25 → No.

9. Put m = – 6, \(\frac{m}{3}\) = 2
\(\frac{-6}{3}\) = 2
⇒ -2 ≠ 2 → No.

10. Put m = 0, \(\frac{m}{3}\) = 2
\(\frac{0}{3}\) = 2
⇒ 0 ≠ 2→ No.

11. Put m = 6, \(\frac{m}{3}\) = 2
\(\frac{6}{3}\) = 2
⇒ 2 = 2→ Yes.

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not.
(a) n + 5 = : 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (P = 1)
(e) 4p – 3 = 13 (P = -4)
4p – 3 = 13 (P = 0)
Solution:
(a) Put n = 1, n + 5 = 19
1 + 5 = 19
1 + 5 = 19
6 ≠ 19
So, the value given in the brackets is not a solution of equation.

(b) Put n = – 2, In+ 5 =19
7 x (- 2) + 5 = 19 -14 + 5 =19
-11 ≠ 19
So, the value given in the brackets is not a solution of equation.

(c) Put n = 2, 7n + 5 =19
7 x 2 + 5 =19
14 + 5 = 19
19 = 19
So, the value given in the brackets is a solution of equation.

(d) Put p = 1, 4p – 3 =13
4 x 1 – 3 = 13
4 – 3 =13
⇒ 1 ≠ 13
So, the value given in the brackets is a not solution of equation.

(e) Put p = – 4, 4p – 3 =13
4 x (- 4) – 3 =13
-16 – 3 =13
-19 ≠ 13
So, the value given in the brackets is not a solution of equation.

(f) Put p = 0, 4p – 3 =13
4 x 0 – 3 = 13
0 – 3 = 13
-3 ≠ 13
So, the value given in the brackets is not a solution of equation.

Question 3.
Solve the following equations by a trial and error method.
(i) 5p + 2 = 17 (ii) 3m -14 = 4
Solution:
(i) 5p + 2 = 17

When p = 3, then L.H.S. = R.H.S.
∴ p = 3 is the solution of the equation.

(ii) 3m – 14 = 4

When m = 6, then L.H.S. = R.H.S.
∴ m = 6 is the solution of equation.

Question 4.
Write equations for the following statements :
(i) The sum of numbers x and 4 is 9.
(ii) The difference ofy and 2 is 8.
(Ui) Ten times a is 70.
(iv) The number b divided by 5 gives 6*.
(v) Three fourth oft is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One fourth of a number x minus 4 leaves 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one third of z, you get 30.
Solution:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10a = 70
(iv) b/5 = 6
(v) \(\frac{3}{4}\)t = 15
(vi) 7m+ 7 = 77
(vii) \(\frac{1}{4}\)n – 4 = 4
(viii) 6y – 6 = 60
(ix) \(z + 3

Question 5.
Write the following equations in statement forms :
(i) p + 4 = 15
(ii) m-7 = 3
(iii) 2m = 7
(iv) [latex]\frac{m}{5}\) = 3
(v) \(\frac{3m}{5}\) = 6
(vi) 3p + 4= 25
(vii) 4p-2 = 18
(viii) \(\frac{p}{2}\) + 2 = 8
Solution:
(i) The sum of numbers p and 4 is 15.
(ii) Taking away 7 from m gives 3.
(iii) Two times a number m is 7.
(iv) One fifths of m is 3.
(v) Three fifth of m is 6.
(ut) Add 4 to three times p to get 25.
(vii) Taking away 2 from four times ofp gives 18.
(viii) Half of a number p plus 2 is 8.

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles parmit has. Irfan has 37 marbles. (Take m to be the number of Permit’s marbles).
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’ age to be y years).
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l).
(iv) In an isosceles triangle, the vertex angle is twice either base angel. (Let the base angle be b in degrees. Remember that the sum of angles of a triangles is 180 degrees).
Solution:
(i) Let us take m to be the number of parmit’s marbles.
Hence the required equation, 5m + 7 = 37
(ii) Let us take y to be the Laxmi’s age.
Hence the required equation, 3y + 4 = 49
(iii) Let us take the lowest score to be l.
Hence the required equation, 2l + 7 = 87
(iv) Let the base angle be b°,
Hence the required equation, 2 b° + b° + b°= 180 or, 4 b° =180°

Haryana State Board HBSE 7th Class Maths Solutions Chapter 3 Data Handling InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 3 Data Handling InText Questions

Try These (Page No. 59):

Question 1.
Weigh (in kg.) atleast 20 children (girls and boys) of your class. Organise the data and answer the following questions using this data.
(i) Who is the heaviest of all ?
(ii) What is the most common weight ?
(iii) What is the difference between your weight and that of your best friend ?
Solution:
Sameer reads in class VII. Sameer’s colleage helped his organise the data in the following way:
Five conclusions:

(a) My name is Sameer and my weight is 48 kg.
(b) My best friend is Rohan.
(c) In my class minimum weight of Manisha is 32 kg.
(d) In my class maximum weight of Rohan is 50 kg.
(e) Average weight of class VII students are 38.5 kg.
(i) Rohan is the heaviest of all.
(ii) 38.5 kg. is the most common weight. (Hi) Difference between Sameer and
Rohan
= 50 – 48 = 2 kg.

Try These (Page No. 61):

Question 2.
How would you find the average of your study hours for the whole week ?
Solution:
One a week = 7 days.
Ashish study hours for the whole week = 5, 7, 6, 8, 9, 10, 4.
Total study hours = 49 hours
∴ Mean
= \(\frac{Sum of all observations}{Number of observations}\)
= \(\frac{5+7+6+8+9+10+4}{7}=\frac{49}{7}\) = 7
∴ The average hours for study = 7 hours.

Try These (Page No. 61):

Question 1.
Find the mean of your sleeping hours during one week.
Solution:
My sleeping hours during one week = 4, 3, 5, 6, 7, 5, 6.
Total sleeping hours during one week = 36 hours.
= \(\frac{Sum of all observations}{Number of observations}\)
= \(\frac{4+3+5+6+7+5+6}{7}\)
= \(\frac{36}{7}\) = 5.14 hours.

Question 2.
Find atleast 5 numbers between \(\frac{1}{2}\) and \(\frac{1}{3}\)
Solution:

Try These (Page No. 65):

Question 1.
Find the mode of:
(i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4.
(ii) 12, 14, 16, 12, 14, 14, 16, 14, 10, 14, IS, 14.
Solution:
(i) Arranging the numbers with same values together, we get,
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6.
Mode of this data is 2, 3 and 4 because it occurs three times frequently than other observations.
(ii) Arranging the number with same values together, we get,
10, 12,12,14,14, 14, 14, 14,14, 16, 16,18.
Mode of this data is 14 because it occurs more frequently than other observations.

Try These (Page No. 65-66):

Question 1.
Find the mode of the following data :
12, 14, 12, 16, 15, 13, 14, 18, 19, 12,
14, 15, 16, 15, 16, 16,15, 17, 13, 16,
16, 15, 15, 13, 15, 17, 15, 14, 15,
13, 15, 14.
Solution:
Observations in ascending order are:
12, 12, 12, 13, 13, 13, 13, 14, 14, 14,
14, 14, 15, 15, 15, 15, 15, 15, 15,
15, 15, 15, 16, 16, 16, 16, 16, 16,
17, 17, 18, 19.
We observe that 15 is occuring ten times i.e., maximum number of times.
∴ Mode = 15

Question 1.
Heights (in cm) of 25 children are given below:
168, 165, 163, 160, 163, 161, 162,
164, 163, 162, 164, 163, 160, 163,
163, 165, 163, 162, 163, 164, 163,
160, 165, 163, 162.
What is the mode of the heights ?
What do we understand by Mode here ?
Solution:
Observations in ascending order are:
160, 160, 160, 161, 162, 162, 162,
162, 163, 163, 163, 163, 163, 163,
163, 163, 163, 163, 164, 164, 164,
165, 165, 165, 168.
We observe that 163 is occuring ten times, i.e., maximum number of times.
Hence, Mode = 163
The mode is the most frequently occuring observation, i.e., the- observation having maximum number of frequencies.

Try These (Page No. 66):

Question 1.
Discuss with your friends and give
(a) Two situations when mean would be an appropriate representative value to use and
(b) Two situations where mode would be and appropriate representative value to use.
Solution:
(a) (i) The height of 5 boys in a group are :
152 cm, 170 cm, 156 cm, 164 cm and 158 cm.
Mean = \(\frac{Sum of the given observations}{Number of observation}\)
= \(\) = 160 cm
hence, It is clear that mean is the appropriate representive value.

(ii) Mean of the first 5 whole number \(\frac{0+1+2+3+4}{5}=\frac{10}{5}\) = 2
Hence, This mean is the appropriate representive value.

(b) The mode of
(i) 5, 5, 7, 8, 8, 9, 7, 9, 10, 10, 11, 9,13.
(ii) 110, 120, 130, 120, 110, 140, 130, 120, 140,120.
Solution:
(i) In the given data, we see that the observation 9 occurs maximum number of times i.e., 3 times
Hence, mode is 9.
(ii) Since, the value 120 occurs maximum number of times, i.e. , 4 times.
Hence, mode is 120.
Hence two sutations, mode is the appropriate representative data.

Try These (Page No. 67):

Question 1.
Your friend found the median and the mode of a given data. Describe and correct your friends error if any:
35, 32, 35, 42, 38, 32, 34
Median = 42, Mode = 32.
Solution:
We arrange the data in ascending order, we get,
32,32,34, 35, 35, 38, 42.
Median is the middle observation. Therefore 35 is the median.
Here, 32 and 35 both occur two times. Therefore, they are both modes of the data.

Try These (Page No. 71-72):

Question 1.
The bar chart shows the result of a survey to test water resistant watches made by different companies.
Each of these companies claimed that their watches were water resistant. After a test the above results were revealed.

(a) Can you work a fraction of the number of watches that leaked to the number tested for each company ?
(b) Could you tell on this basis which company has better matches ?
Solution:
(a) In company A,
Number that leaked = 20,
Number that tested = 40
In company B, No. that leaked = 10,
Number that tested = 40.
In company C, No. that leaked = 12,
Number that tested = 40
In company D, No. that leaked = 22,
Number that tested =S ‘ 40
(b) B company has better matches.

Question 2.
Sale of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below. Draw a double bar graph and answer the following questions :

Draw a double bar graph and answer the following questions:
(i) In which year was the difference in the sale of the two languages books least ?
(ii) Can you say that the demand for English books rose faster ? Justify.
Solution:

(i) 1998 (650 books – 620 books = 30 books)
(ii) Yes, 1995 to 1998 English books increases to = 50, 50, 170.
1995 to 1998 Hindi books increases to = 25, 75, 50.

Try These (Page No. 74):

Question 1.
Think of some situations, atleast 3 examples of each, that are certain to happen, some that are impossible and some that may or may not happen or Le., situations that have some chance of happening.
Solution:
(i) The sun coming up from the east. This is possible.
(ii) An elephant growing to two ft. height. This is not possible.
(iii) It will be Wednesday on 6th January. If the day on 5 th January is Monday, then on 6th January it will be Tuesday. Thus, it is impossible that the day on 6th January will be Wednesday.

Try These (Page No. 75):

Question 2.
Toss a coin 100 times and record the data. Find the number of times heads and tails occur in it.
Solution:

Question 3.
Aftaab threw a die 250 times and got the following table. Draw a bargraph for this data.

Solution:

Question 4.
Throw a dice 100 times and record the data. Find the number of times 1, 2,3, 4, 5, 6 occur.
Solution:

Try These (Page No. 76):

Question 1.
Construct or think of five situations where outcomes do not have equal chances.
Solution: