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Page No 61:

Question 1:

Solve the following.


(i) - 16 × - 5   (ii) 72 ÷ - 12        (iii)  - 24 × 2

(iv) 125 ÷ 5            (v)  - 104 ÷ - 13  (vi)   25 × - 4

Answer:

(i) - 16 × - 5=80   

(ii) 72 ÷ - 12=-6       

(iii)  - 24 × 2=-48

(iv) 125 ÷ 5=25           

(v)  - 104 ÷ - 13=8 

(vi)   25× - 4=-100

Page No 61:

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

Answer:

(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

Page No 61:

Question 3:

Simplify.

(i) 322391   (ii)  247209    (iii)  117156

Answer:

(i) 

322391=322÷23391÷23=1417

(ii)

247209=247÷19209÷19=1311

(iii)  

117156=117÷3156÷3=39÷352÷3=1314

Page No 61:

Question 4:

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

Answer:

(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.

Page No 61:

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

Answer:

The joint bar graph for the given data is :

Page No 61:

Question 6:

Simplify the expression.

(i) 45 ÷ 5 + 20 × 4 - 12      (ii) 38 - 8 × 2 ÷ 5 + 13

(iii) 53 + 47 ÷ 3221                (iv) 3 ×  4  85 + 5 - 15 ÷ 3  + 2

Answer:

(i) 
45÷5+20×4-12=9+80-12=89-12=77     
(ii) 
38-8×2÷5+13=30×2÷5+13=60÷5+13=12+13=25

(iii) 
53+47÷3221=53+47×2132=53+47×21 332 8=53+38       
=5×83×8+3×38×3=4024+924=40+924=4924
(iv) 
3×485+5-15÷3+2=3×485+5-5+2=3×485+2=3×340+2=3×342=1026

Page No 61:

Question 7:

Solve.

(i) 512 + 716      (ii)   325 - 214       (iii)  125 × -103    (iv)  438 ÷ 2518

Answer:

(i) 
512+716=5×412×4+7×316×3=2048+2148=20+2148=4148     
(ii)   
325-214=175-94=17×45×4-9×54×5=6820-4520
=68-4520=2320       
(iii)  
125 × -103=4 12×-10 -25×3=4×-2=-8   
(iv)  
438 ÷ 2518=7 354 8×9 185 25=7×94×5=6320

Page No 61:

Question 8:

Construct ∆ABC such that mA = 55°, mB = 60°, and l(AB) = 5.9 cm.

Answer:



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.

Page No 61:

Question 9:

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

Answer:

 

​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.

Page No 61:

Question 10:

Construct ∆PQR such that, mP = 80°, mQ = 70°, l(QR) = 5.7 cm.

Answer:

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.

Page No 61:

Question 11:

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

Answer:


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.

Page No 61:

Question 12:

In ∆LMN, l(LM) = 6.2 cm, mLMN = 60°, l(MN) = 4 cm. Construct ∆LMN.

Answer:

​ 

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.

Page No 61:

Question 13:

Find the measures of the complementary angles of the following angles.
(i) 35° (ii) a° (iii) 22° (iv)  40 - x°

Answer:

(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

Page No 61:

Question 14:

Find the measures of the supplements of the following angles.
(i) 111° (ii) 47° (iii) 180° (iv) 90 - x°

Answer:

(i) 111° (ii) 47° (iii) 180°  (iv) 90 - x°
(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

Page No 61:

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.

Answer:

(i)


(ii)


(iii)



Page No 62:

Question 16:

figure

In ∆PQR, the measures of P and Q are equal and mPRQ = 70°. Find the measures of the following angles.
(i) mPRT (ii) mP (iii) mQ

 

Answer:

(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​.
 

Page No 62:

Question 17:

Simplify.

(i) 54 × 53    (ii)  236 ÷ 239    (iii) 728 × 72-6    (iv)  452 ÷ 54

Answer:

(i) 

54 × 53=57                               am×an=am+n   

(ii)

  236 ÷ 239=236-9                              am÷an=am-n=23-3 =323                                 a-m=1am   

(iii) 

728 × 72-6=728+-6                        am×an=am+n=722   


(iv)  

452 ÷ 54=452 × 45  =452+1             am×an=am+n=453 

Page No 62:

Question 18:

Find the value.

(i) 1716 ÷ 1716     (ii) 10-3    (iii)  232     (iv)  46 × 4-4

Answer:

(i) 
1716 ÷ 1716=1716-16                      am÷an=am-n=170 
= 1
(ii) 
10-3=1103                       a-m=1am=11000   
(iii)  
232=26                 amn=amn=64     
(iv)  
46 × 4-4=46+-4                           am×an=am+n=46-4=42
= 16

Page No 62:

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)

Answer:

(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
 

Page No 62:

Question 20:

Solve the following equations.

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

Answer:

(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

Page No 62:

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.

Answer:

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

Page No 62:

Question 2:

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

37-34 = ................

(i) 37-7   (ii) 37-10    (iii)  7312   (iv)  3720

Answer:

37-34=37-34              amn=amn=37-12=7312                           a-n=1an
​Hence, the correct option is (iii).

Page No 62:

Question 3:

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

The simplest form of 5 ÷32 - 13 is .................... 

(i) 3 (ii) 5 (iii) 0 (iv)  13

Answer:

5÷32 - 13=5×23-13=103-13=10-13=93
= 3
Hence, the correct option is (i).

Page No 62:

Question 4:


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

The solution of the equation 3x - 12 = 52 + x  is ...................

(i) 53 (ii) 72  (iii) 4  (iv) 32

Answer:

3x-12=52+x3x-x-12=52+x-x2x-12+12=52+12
2x=52+122x=3x=32
Hence, the correct option is (iv).

Page No 62:

Question 5:

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

(i) 10 × 3 + 5 + 2     (ii)     10 × 4 +  5 - 3  

(iii) 8 × 4 + 3               (iv)     9 × 3 + 2

Answer:

10 × 3 + 5 + 2=10×3+7=30+7=37  
Hence, the correct option is (i).



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