Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

## Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.

Solve the following:

i. (-16) × (-5)

ii. (72) ÷ (-12)

iii. (-24) × (2)

iv. 125 ÷ 5

v. (-104) ÷ (-13)

vi. 25 × (-4)

Solution:

i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)

= \(\frac{1}{(-1)} \times \frac{72}{12}\)

(-1) × 12

= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)

= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)

= \(\frac { 104 }{ 13 }\)

= 8

vi. 25 × (-4) = -100

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

Solution:

i. 75 = 3 × 25

= 3 × 5 × 5

135 = 3 × 45

= 3 × 3 × 15

= 3 × 3 × 3 × 5

∴ HCF of 75 and 135 = 3 × 5

= 15

LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3

= 675

ii. 114 = 2 × 57

= 2 × 3 × 19

76 = 2 × 38

= 2 × 2 × 19

∴ HCF of 114 and 76 = 2 × 19

= 38

LCM of 114 and 76 = 2 × 19 × 3 × 2

= 228

iii. 153 = 3 × 51

= 3 × 3 × 17

187 = 11 × 17

∴ HCF of 153 and 187 = 17

LCM of 153 and 187 = 17 × 3 × 3 × 11

= 1683

iv. 32 = 2 × 16

= 2 × 2 × 8

= 2 × 2 × 2 × 4

= 2 × 2 × 2 × 2 × 2

24 = 2 × 12

= 2 × 2 × 6

= 2 × 2 × 2 × 3

48 = 2 × 24

= 2 × 2 × 12

= 2 × 2 × 2 × 6

= 2 × 2 × 2 × 2 × 3

∴ HCF of 32, 24 and 48 = 2 × 2 × 2

= 8

LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3

= 96

Question 3.

Simplify:

i. \(\frac { 322 }{ 391 }\)

ii. \(\frac { 247 }{ 209 }\)

iii. \(\frac { 117 }{ 156 }\)

Solution:

i. \(\frac { 322 }{ 391 }\)

ii. \(\frac { 247 }{ 209 }\)

iii. \(\frac { 117 }{ 156 }\)

Question 4.

i. 784

ii. 225

iii. 1296

iv. 2025

v. 256

Solution:

i. 784

∴ 784 = 2 × 2 × 2 × 2 × 7 × 7

∴ √784 = 2 × 2 × 7

= 28

ii. 225

∴ 225 = 3 × 3 × 5 × 5

∴ √225 = 3 × 5

= 15

iii. 1296

∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ √1296 = 2 × 2 × 3 × 3

= 36

iv. 2025

∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5

∴ √2025 = 3 × 3 × 5

= 45

v. 256

∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ √256 = 2 × 2 × 2 × 2

= 16

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 |

Solution:

Question 6.

Simplify the expressions:

i. 45 ÷ 5 + 120 × 4 – 12

ii. (38 – 8) × 2 ÷ 5 + 13

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}

Solution:

i. 45 ÷ 5 + 120 × 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. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)

\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)

\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)

\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}

= 3 × {4[90 – 5] + 2}

= 3 × {4 × 85 + 2}

= 3 × (340 + 2)

= 3 × 342

= 1026

Question 7.

Solve:

i. \(\frac{5}{12}+\frac{7}{16}\)

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)

Solution:

i. \(\frac{5}{12}+\frac{7}{16}\)

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)

= 4 × (-2)

= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)

= \(\frac{7}{4} \times \frac{9}{5}\)

= \(\frac { 63 }{ 20 }\)

Question 8.

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

Solution:

Question 9.

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

Solution:

Question 10.

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

Ans:

In ∆PQR,

m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)

∴ 80 + 70 + m∠R = 180

∴ 150 + m∠R = 180

∴ m∠R = 180 – 150

∴ m∠R = 30°

Question 11.

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

Solution:

Question 12.

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

Solution:

Question 13.

Find the measures of the complementary angles of the following angles:

i. 35°

ii. a°

iii. 22°

iv. (40 – x)°

Solution:

i. Let the measure of the complementary

angle be x°.

35 + x = 90

∴35 + x-35 = 90 – 35

….(Subtracting 35 from both sides)

∴x = 55

∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.

a + x = 90

∴a + x – a = 90 – a

….(Subtracting a from both sides)

∴x = (90 – a)

∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.

22 + x = 90

∴22 + x – 22 = 90 – 22

….(Subtracting 22 from both sides)

∴x = 68

∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.

40 – x + a = 90

∴40 – x + a – 40 + x = 90 – 40 + x

….(Subtracting 40 and adding x on both sides)

∴a = (50 + x)

∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.

Find the measures of the supplements of the following angles:

i. 111°

ii. 47°

iii. 180°

iv. (90 – x)°

Solution:

i. Let the measure of the supplementary

angle be x°.

111 + x = 180

∴ 111 + x – 111 = 180 – 111

…..(Subtracting 111 from both sides)

∴ x = 69

∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.

47 + x = 180

∴47 + x – 47 = 180 – 47

….(Subtracting 47 from both sides)

∴x = 133

∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.

180 + x = 180

∴180 + x – 180 = 180 – 180

….(Subtracting 180 from both sides)

∴x = 0

∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.

90 – x + a = 180

∴90 – x + a – 90 + x = 180 – 90+ x

….(Subtracting 90 and adding x on both sides)

∴a = 180 – 90 + x

∴a = (90 + x)

∴The supplementary angle of (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.

Solution:

i.

ii.

iii.

Question 16.

In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.

- m∠PRT
- m∠P
- m∠Q

Solution:

Here, ∠PRQ and ∠PRT are angles in a linear pair.

m∠PRQ + m∠PRT = 180°

∴70 + m∠PRT = 180

∴m∠PRT = 180 – 70

∴m∠PRT = 110°

Now, ∠PRT is the exterior angle of ∆PQR.

∴m∠P + m∠Q = m∠PRT

∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)

∴2m∠P = 110

∴m∠P = \(\frac { 110 }{ 2 }\)

∴m∠P = 55°

∴m∠Q =

Question 17.

Simplify

i. 5^{4} × 5^{3}

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)

Solution:

Simplify

i. 5^{4} × 5^{3}

= 5^{4+3}

= 5^{7}

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)

Question 18.

Find the value:

i. 17^{16} ÷ 17^{16}

ii. 10^{-3}

iii. (2³)²

iv. 4^{6} × 4^{-4}

Solution:

i. 17^{16} ÷ 17^{16}

= 17^{0}

= 1

ii. 10^{-3}

= \(\frac{1}{10^{3}}\)

= \(\frac{1}{1000}\)

iii. (2³)²

= 2^{3×2}

= 2^{6}

= 2 × 2 × 2 × 2 × 2 × 2

= 64

iv. 4^{6} × 4^{-4}

= 4^{6+(-4)}

= 4^{2}

= 4 × 4

= 16

Question 19.

Solve:

i. (6a – 5b – 8c) + (15b + 2a – 5c)

ii. (3x + 2y) (7x – 8y)

iii. (7m – 5n) – (-4n – 11m)

iv. (11m – 12n + 3p) – (9m + 7n – 8p)

Solution:

i. (6a – 5b – 8c) + (15b + 2a – 5c)

= (6a + 2a) + (-5b + 15b) + (-8c – 5c)

= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)

= 3x × (7x – 8y) + 2yx (7x – 8y)

= 21x² – 24xy + 14xy – 16y²

= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)

= 7m – 5n + 4n + 11m

= (7m + 11m) + (-5n + 4n)

= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)

= 11m – 12n + 3p – 9m – 7n + 8p

= (11m – 9m) + (-12n – 7n) + (3p + 8p)

= 2m – 19n + 11p

Question 20.

Solve the following equations:

i 4(x + 12) = 8

ii. 3y + 4 = 5y – 6

Solution:

i. 4(x + 12) = 8

∴4x + 48 = 8

∴4x + 48 – 48 = 8 – 48

….(Subtracting 48 from both sides)

∴ 4x = -40

∴ x = \(\frac { -40 }{ 4 }\)

∴ x = -10

ii. 3y + 4 = 5y – 6

∴ 3y + 4 + 6 = 5y – 6 + 6

….(Adding 6 on both sides)

∴ 3y + 10 = 5y

∴ 3y + 10 – 3y = 5y – 3y

….(Subtracting 3y from both sides)

∴ 10 = 2y

∴ 2y = 10

∴ y = \(\frac { 10 }{ 2 }\)

∴ y = 5