Mathematics Solutions Solutions for Class 7 Math Chapter 9 Miscellaneous Problems : Set 1 are provided here with simple step-by-step explanations. These solutions for Miscellaneous Problems : Set 1 are extremely popular among class 7 students for Math Miscellaneous Problems : Set 1 Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 7 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.
Page No 61:
Question 1:
Solve the following.
(i) (ii) (iii)
(iv) (v) (vi) 25
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No 61:
Question 2:
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) (ii) (iii)
Answer:
(i)
(ii)
(iii)
Page No 61:
Question 4:
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) (ii)
(iii) (iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 61:
Question 7:
Solve.
(i) (ii) (iii) (iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 61:
Question 8:
Construct âABC 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, â³ABC is the required triangle.
Page No 61:
Question 9:
Construct âXYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Answer:
âSteps 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, â³XYZ is the required triangle.
Page No 61:
Question 10:
Construct âPQR such that, mP = 80°, mQ = 70°, l(QR) = 5.7 cm.
Answer:
In Δ PQR,
∠P + ∠Q + ∠R = 180â (Angle sum property)
⇒ 80â + 70â + ∠R = 180â
⇒ 150â + ∠R = 180â
⇒ ∠R = 180â − 150â
= 30â
âSteps 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, â³PQR is the required triangle.
Page No 61:
Question 11:
Construct âEFG from the given measures. l(FG) = 5 cm, mEFG = 90°, l(EG) = 7 cm.
Answer:
â
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, â³DEF is the required triangle.
Page No 61:
Question 12:
In âLMN, l(LM) = 6.2 cm, mLMN = 60°, l(MN) = 4 cm. Construct âLMN.
Answer:
â
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, â³LMN is the required triangle.
Page No 61:
Question 13:
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.
a + x = 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:
Answer:
(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)°
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
Answer:
(i)
∠PRQ + ∠PRT = 180â (Linear pair angles)
⇒ ∠PRT = 180â − 70â
= 110â
Hence, the measure of ∠PRT is 110ââ.
(ii)
In ΔPQR,
∠P + ∠Q = ∠PRT (Exterior angle property)
⇒ ∠P + ∠P = 110â (∠P = ∠Q)
⇒ 2∠P = 110â
⇒ ∠P = 55â
Hence, the measure of ∠P is 55ââ.
(iii)
In ΔPQR,
∠P + ∠Q = ∠PRT (Exterior angle property)
⇒ ∠Q + ∠Q = 110â (∠P = ∠Q)
⇒ 2∠Q = 110â
⇒ ∠Q = 55â
Hence, the measure of ∠Q is 55ââ.
Page No 62:
Question 17:
Simplify.
(i) (ii) (iii) (iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 62:
Question 18:
Find the value.
(i) (ii) (iii) (iv)
Answer:
(i)
= 1
(ii)
(iii)
(iv)
= 16
Page No 62:
Question 19:
(iii) (7m 5n) ( 4n 11m) (iv) (11m 12n+3p) (9m+7n8p)
Answer:
(i) (6a 5b 8c) + (15b + 2a 5c)
= 6a 5b 8c + 15b + 2a 5c
= 8a + 10b − 13c
(ii) (3x+2y)(7x 8y)
= 3x(7x 8y) + 2y(7x 8y)
= 21x2 − 24xy + 14xy − 16y2
= 21x2 − 10xy − 16y2
(iii) (7m 5n) ( 4n 11m)
= 7m 5n + 4n + 11m
= 18m n
(iv) (11m 12n + 3p) (9m + 7n 8p)
= 11m 12n + 3p 9m 7n + 8p
= 2m 19n + 11p
Page No 62:
Question 20:
Solve the following equations.
(i) 4(âx + 12) = 8 (ii) 3y + 4 = 5y 6
Answer:
(i)
4(âx + 12) = 8
⇒ 4x + 48 = 8
⇒ 4x + 48 − 48 = 8 − 48
â⇒ 4x = − 40
⇒ x = − 10
(ii)
3y + 4 = 5y − 6
⇒ â3y + 4 − 5y = 5y − 6 − 5y
⇒ â4 − 2y = − 6
⇒ â4 − 2y − 4 = − 6 − 4
⇒ â− 2y = − 10
⇒ ây = 5
Page No 62:
Question 1:
The three angle bisectors of a triangle are concurrent. Their point of concurrence is called the ....................... .
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.
= ................
(i) (ii) (iii) (iv)
Answer:
âHence, 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 is ....................
Answer:
= 3
Hence, the correct option is (i).
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Question 4:
Choose the right answer from the options given after every question.
The solution of the equation 3x is ...................
(i) (ii) (iii) 4 (iv)
Answer:
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) (ii)
(iii) (iv)
Answer:
Hence, the correct option is (i).
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