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#### Question 1:

Solve the following.

(i)    (ii)         (iii)

(iv)             (v)    (vi)   25

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Question 2:

Find the prime factors of the following numbers and find their LCM and HCF.
(i) 75, 135 (ii) 114, 76 (iii) 153, 187 (iv) 32, 24, 48

(i) 75, 135

HCF = 3 × 5 = 15
LCM = 3 × 5 × 5 × 9 = 675

(ii) 114, 76

HCF = 2 × 19 = 38
LCM = 2 × 19 × 2 × 3 = 228

(iii) 153, 187

HCF = 17
LCM = 17 × 9 × 11 = 1683

(iv) 32, 24, 48

HCF = 2 × 2 × 2 = 8
LCM = 2 × 2 × 2 × 2 × 2 × 3 = 96

#### Question 3:

Simplify.

(i) $\frac{322}{391}$   (ii)  $\frac{247}{209}$    (iii)  $\frac{117}{156}$

(i)

$\frac{322}{391}=\frac{322÷23}{391÷23}\phantom{\rule{0ex}{0ex}}=\frac{14}{17}$

(ii)

$\frac{247}{209}=\frac{247÷19}{209÷19}\phantom{\rule{0ex}{0ex}}=\frac{13}{11}$

(iii)

$\frac{117}{156}=\frac{117÷3}{156÷3}\phantom{\rule{0ex}{0ex}}=\frac{39÷3}{52÷3}\phantom{\rule{0ex}{0ex}}=\frac{13}{14}$

#### Question 4:

Find the square root of the following numbers.
(i) 784 (ii) 225 (iii) 1296 (iv) 2025   (v)  256

(i)

The square root of 784 is 28.

(ii)

The square root of 225 is 15.

(iii)

The square root of 1296 is 36.

(iv)

The square root of 2025 is 45.

(v)

The square root of 256 is 15.

#### Question 5:

There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

 Polling         Booths Navodaya       Vidyalaya Vidyaniketan       School City High              School Eklavya          School Women 500 520 680 800 Men 440 640 760 600

The joint bar graph for the given data is :

#### Question 6:

Simplify the expression.

(i)       (ii)

(iii)                 (iv)

(i)
$45÷5+20×4-12=9+80-12\phantom{\rule{0ex}{0ex}}=89-12\phantom{\rule{0ex}{0ex}}=77$
(ii)
$\left(38-8\right)×2÷5+13=30×2÷5+13\phantom{\rule{0ex}{0ex}}=60÷5+13\phantom{\rule{0ex}{0ex}}=12+13\phantom{\rule{0ex}{0ex}}=25$

(iii)

$=\frac{5×8}{3×8}+\frac{3×3}{8×3}\phantom{\rule{0ex}{0ex}}=\frac{40}{24}+\frac{9}{24}\phantom{\rule{0ex}{0ex}}=\frac{40+9}{24}\phantom{\rule{0ex}{0ex}}=\frac{49}{24}$
(iv)
$3×\left\{4\left[85+5-\left(15÷3\right)\right]+2\right\}=3×\left\{4\left[85+5-5\right]+2\right\}\phantom{\rule{0ex}{0ex}}=3×\left\{4\left[85\right]+2\right\}\phantom{\rule{0ex}{0ex}}=3×\left\{340+2\right\}\phantom{\rule{0ex}{0ex}}=3×342\phantom{\rule{0ex}{0ex}}=1026$

#### Question 7:

Solve.

(i)       (ii)          (iii)      (iv)

(i)
$\frac{5}{12}+\frac{7}{16}=\frac{5×4}{12×4}+\frac{7×3}{16×3}\phantom{\rule{0ex}{0ex}}=\frac{20}{48}+\frac{21}{48}\phantom{\rule{0ex}{0ex}}=\frac{20+21}{48}\phantom{\rule{0ex}{0ex}}=\frac{41}{48}$
(ii)
$3\frac{2}{5}-2\frac{1}{4}=\frac{17}{5}-\frac{9}{4}\phantom{\rule{0ex}{0ex}}=\frac{17×4}{5×4}-\frac{9×5}{4×5}\phantom{\rule{0ex}{0ex}}=\frac{68}{20}-\frac{45}{20}$
$=\frac{68-45}{20}\phantom{\rule{0ex}{0ex}}=\frac{23}{20}$
(iii)

(iv)

#### Question 8:

Construct ∆ABC such that m$\angle$A = 55°, m$\angle$B = 60°, and l(AB) = 5.9 cm.

Steps of constructions:

(1) Draw seg AB of length 5.9 cm.
(2) Draw ray AD such that ∠BAD = 55°.
(3) Draw ray BE such that ∠ABE = 60°.
(4) Name the point of intersection of ray AD and BE as C.
Therefore, △ABC is the required triangle.

#### Question 9:

Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.

​Steps of constructions:

(1) Draw seg XZ of length 6.3 cm.
(2) Draw an arc of 3.7 cm from the vertex X.
(3) Draw another arc of 7.7 cm from the vertex Z, cutting the previously drawn arc at Y.

Therefore, △XYZ is the required triangle.

#### Question 10:

Construct ∆PQR such that, m$\angle$P = 80°, m$\angle$Q = 70°, l(QR) = 5.7 cm.

In Δ PQR,
∠P + ∠Q + ∠R = 180   (Angle sum property)
⇒ 80 + 70 + ∠R = 180
⇒ 150 + ∠R = 180
⇒ ∠R = 180 − 150
= 30

​Steps of constructions:

(1) Draw seg QR of length 5.7 cm.
(2) Draw ray QA such that ∠RQA = 70°.
(3) Draw ray RB such that ∠QRB = 30°.
(4) Name the point of intersection of ray RB and QA as P.

Therefore, △PQR is the required triangle.

#### Question 11:

Construct ∆EFG from the given measures. l(FG) = 5 cm, m$\angle$EFG = 90°, l(EG) = 7 cm.

Steps of constructions:

(1) Draw seg FG of length 5 cm.
(2) Draw ray FA such that ∠GFA = 90°.
(3) Draw an arc of 7 cm from the vertex G, cutting the ray FA at E.

Therefore, △DEF is the required triangle.

#### Question 12:

In ∆LMN, l(LM) = 6.2 cm, m$\angle$LMN = 60°, l(MN) = 4 cm. Construct ∆LMN.

​

Steps of constructions:

(1) Draw seg LM of length 6.2 cm.
(2) Draw ray MA such that ∠LMA = 60°.
(3) Draw an arc of 4 cm from the vertex M, cutting the ray MA at N.

Therefore, △LMN is the required triangle.

#### Question 13:

Find the measures of the complementary angles of the following angles.
(i) 35° (ii) a° (iii) 22° (iv)

(i)
Let the measure of the complementary angle be a.
35 + a = 90
∴ a = 55°
Hence, the measure of the complement of an angle of measure 35° is 55°
(ii)
Let the measure of the complementary angle be x.
ax = 90
∴ x = (90 − a
Hence, the measure of the complement of an angle of measure a° is (90 − a
(iii)
Let the measure of the complementary angle be a.
22 + a = 90
∴ a = 68°
Hence, the measure of the complement of an angle of measure 22° is 68°
(iv)
Let the measure of the complementary angle be a.
(40 − x) + a = 90
∴ a = (50 + x
Hence, the measure of the complement of an angle of measure (40 − x)° is  (50 + x

#### Question 14:

Find the measures of the supplements of the following angles.
(i) 111° (ii) 47° (iii) 180° (iv)

(i) 111° (ii) 47° (iii) 180°  (iv)
(i)
Let the measure of the complementary angle be a.
111 + a = 180
∴ a = 69°
Hence, the measure of the complement of an angle of measure 111° is 69°
(ii)
Let the measure of the complementary angle be x.
47 + x = 180
∴ x = 133°
Hence, the measure of the complement of an angle of measure 47° is 133°
(iii)
Let the measure of the complementary angle be a.
180 + a = 180
∴ a = 0°
Hence, the measure of the complement of an angle of measure 180° is 0°
(iv)
Let the measure of the complementary angle be a.
(90 − x) + a = 180
∴ a = (90 + x
Hence, the measure of the complement of an angle of measure (90 − x)° is  (90 + x

#### Question 15:

Construct the following figures.
(i) A pair of adjacent angles
(ii) Two supplementary angles which are not adjacent angles.
(iii) A pair of adjacent complementary angles.

(i)

(ii)

(iii)

#### Question 16:

figure

In ∆PQR, the measures of $\angle$P and $\angle$Q are equal and m$\angle$PRQ = 70°. Find the measures of the following angles.
(i) m$\angle$PRT (ii) m$\angle$P (iii) m$\angle$Q

(i)
∠PRQ + ∠PRT = 180   (Linear pair angles)
⇒ ∠PRT = 180 −  70
= 110
Hence, the measure of  ∠PRT is 110​.

(ii)

In ΔPQR,
∠P + ∠Q = ∠PRT   (Exterior angle property)
⇒ ∠P + ∠P = 110              (∠P = ∠Q)
⇒ 2∠P = 110
⇒ ∠P = 55
Hence, the measure of  ∠P is 55​.

(iii)

In ΔPQR,
∠P + ∠Q = ∠PRT   (Exterior angle property)
⇒ ∠Q + ∠Q = 110              (∠P = ∠Q)
⇒ 2∠Q = 110
⇒ ∠Q = 55
Hence, the measure of  ∠Q is 55​.

#### Question 17:

Simplify.

(i)     (ii)      (iii)     (iv)

(i)

(ii)

(iii)

(iv)

#### Question 18:

Find the value.

(i)      (ii) ${10}^{-3}$    (iii)  ${\left({2}^{3}\right)}^{2}$     (iv)

(i)

= 1
(ii)

(iii)

(iv)

= 16

#### Question 19:

Solve.

(i) (6a $-$5b $-$ 8c) + (15+ 2a $-$ 5c)  (ii) (3x+2y)(7x $-$ 8y
(iii) (7m $-$ 5n) $-$ ( $-$ 4n $-$ 11m)        (iv) (11m $-$ 12n+3p) $-$ (9m+7n$-$8p)

(i) (6a $-$5b $-$ 8c) + (15+ 2$-$ 5c)
= 6a $-$5b $-$ 8c + 15+ 2$-$ 5c
= 8a + 10b − 13c
(ii) (3x+2y)(7x $-$ 8y
= 3x(7x $-$ 8y) + 2y(7x $-$ 8y
= 21x− 24xy + 14xy − 16y2
= 21x− 10xy − 16y2
(iii) (7m $-$ 5n$-$ ( $-$ 4n $-$ 11m)
= 7m $-$ 5n + 4n + 11m
= 18m $-$ n
(iv) (11m $-$ 12+ 3p$-$ (9+ 7$-$ 8p)
= 11m $-$ 12+ 3p $-$ 9$-$ 7n + 8p
= 2m $-$ 19+ 11p

#### Question 20:

Solve the following equations.

(i) 4(​x + 12)  =  8        (ii)  3y + 4 =  5y $-$ 6

(i)
4(​x + 12)  =  8
⇒ 4x + 48 = 8
⇒ 4x + 48 − 48 = 8 − 48
​⇒ 4x = − 40
x = − 10
(ii)
3y + 4 =  5y − 6
⇒ ​3y + 4 − 5=  5y − 6 − 5y
⇒ ​4 − 2=  − 6
⇒ ​4 − 2y − 4 =  − 6 − 4
⇒ ​− 2y =  − 10
⇒ ​y =  5

#### Question 1:

Choose the right answer from the options given after every question.

The three angle bisectors of a triangle are concurrent. Their point of concurrence is called the ....................... .
(i) circumcentre (ii) apex (iii) incentre (iv) point of intersection.

The three angle bisectors of a triangle are concurrent. Their point of concurrence is called the incentre.
Hence, the correct option is (iii).

#### Question 2:

Choose the right answer from the options given after every question.

${\left[{\left(\frac{3}{7}\right)}^{-3}\right]}^{4}$ = ................

(i) ${\left(\frac{3}{7}\right)}^{-7}$   (ii) ${\left(\frac{3}{7}\right)}^{-10}$    (iii)  ${\left(\frac{7}{3}\right)}^{12}$   (iv)  ${\left(\frac{3}{7}\right)}^{20}$

​Hence, the correct option is (iii).

#### Question 3:

Choose the right answer from the options given after every question.

The simplest form of 5  is ....................

(i) 3 (ii) 5 (iii) 0 (iv)  $\frac{1}{3}$

= 3
Hence, the correct option is (i).

#### Question 4:

Choose the right answer from the options given after every question.

The solution of the equation 3x   is ...................

(i) $\frac{5}{3}$ (ii) $\frac{7}{2}$  (iii) 4  (iv) $\frac{3}{2}$

$3x-\frac{1}{2}=\frac{5}{2}+x\phantom{\rule{0ex}{0ex}}⇒3x-x-\frac{1}{2}=\frac{5}{2}+x-x\phantom{\rule{0ex}{0ex}}⇒2x-\frac{1}{2}+\frac{1}{2}=\frac{5}{2}+\frac{1}{2}$
$⇒2x=\frac{5}{2}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒2x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}$
Hence, the correct option is (iv).

#### Question 5:

Choose the right answer from the options given after every question.
Which of the following expressions has the value 37 ?

(i)      (ii)

(iii)                (iv)