Std 7 Mathematics · Part One

Miscellaneous Problems: Set 1

Step-by-step solutions to Miscellaneous Problems Set 1, covering Chapters 1 to 8: integers, HCF and LCM, square roots, joint bar graph, simplification, constructions, pairs of angles, indices, algebraic expressions and equations. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

Questions 1 to 7

1Solve the following. (i) (−16) × (−5) (ii) (72) ÷ (−12) (iii) (−24) × (2) (iv) 125 ÷ 5 (v) (−104) ÷ (−13) (vi) 25 × (−4)

(i) (−16) × (−5) = 80
negative × negative = positive
(ii) (72) ÷ (−12) = −6
positive ÷ negative = negative
(iii) (−24) × (2) = −48
negative × positive = negative
(iv) 125 ÷ 5 = 25
positive ÷ positive = positive
(v) (−104) ÷ (−13) = 8
negative ÷ negative = positive
(vi) 25 × (−4) = −100
positive × negative = negative

2Find 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

75 = 3 × 5 × 5

135 = 3 × 3 × 3 × 5

HCF = product of the common prime factors = 3 × 5 = 15
LCM = product of every prime factor, each taken the greatest number of times it occurs = 3 × 3 × 3 × 5 × 5 = 675

(ii) 114, 76

114 = 2 × 3 × 19

76 = 2 × 2 × 19

HCF = product of the common prime factors = 2 × 19 = 38
LCM = product of every prime factor, each taken the greatest number of times it occurs = 2 × 2 × 3 × 19 = 228

(iii) 153, 187

153 = 3 × 3 × 17

187 = 11 × 17

HCF = product of the common prime factors = 17 = 17
LCM = product of every prime factor, each taken the greatest number of times it occurs = 3 × 3 × 11 × 17 = 1683

(iv) 32, 24, 48

32 = 2 × 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

48 = 2 × 2 × 2 × 2 × 3

HCF = product of the common prime factors = 2 × 2 × 2 = 8
LCM = product of every prime factor, each taken the greatest number of times it occurs = 2 × 2 × 2 × 2 × 2 × 3 = 96

Check for two numbers: HCF × LCM = product of the numbers. For example, 15 × 675 = 10125 = 75 × 135, and 38 × 228 = 8664 = 114 × 76.

3*Simplify. (i) 322391 (ii) 247209 (iii) 117156

(i) 322391
322 = 2 × 7 × 23, 391 = 17 × 23
HCF of 322 and 391 = 23
322391 = 322 ÷ 23391 ÷ 23 = 1417

(ii) 247209
247 = 13 × 19, 209 = 11 × 19
HCF of 247 and 209 = 19
247209 = 247 ÷ 19209 ÷ 19 = 1311

(iii) 117156
117 = 3 × 3 × 13, 156 = 2 × 2 × 3 × 13
HCF of 117 and 156 = 39
117156 = 117 ÷ 39156 ÷ 39 = 34

To reduce a fraction to its simplest form, divide the numerator and the denominator by their HCF.

4Find the square root of the following numbers. (i) 784 (ii) 225 (iii) 1296 (iv) 2025 (v) 256

(i) 784 = 2 × 2 × 2 × 2 × 7 × 7
= (2 × 2) × (2 × 2) × (7 × 7)
√784 = 2 × 2 × 7 = 28

(ii) 225 = 3 × 3 × 5 × 5
= (3 × 3) × (5 × 5)
√225 = 3 × 5 = 15

(iii) 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3)
√1296 = 2 × 2 × 3 × 3 = 36

(iv) 2025 = 3 × 3 × 3 × 3 × 5 × 5
= (3 × 3) × (3 × 3) × (5 × 5)
√2025 = 3 × 3 × 5 = 45

(v) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)
√256 = 2 × 2 × 2 × 2 = 16

Write the number as a product of prime factors and pair equal factors. Take one factor from each pair; their product is the square root.

5There 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. (Navodaya Vidyalaya: women 500, men 440; Vidyaniketan School: 520, 640; City High School: 680, 760; Eklavya School: 800, 600)

Step 1: Draw the X-axis and Y-axis on graph paper. Show the polling booths on the X-axis and the number of voters on the Y-axis.

Step 2: Choose a scale. The greatest number is 800, so take 1 cm = 100 voters on the Y-axis.

Step 3: For each booth draw two bars of equal width side by side, with no gap between them: one for women and one for men. Keep equal gaps between the pairs.

Step 4: Colour or shade the bars for women and for men differently, and show this in a key (legend).

Joint bar graph: voters at the four polling booths (scale: 1 cm = 100 voters on the Y-axis)

Polling booth	Women (bar height)	Men (bar height)
Navodaya Vidyalaya	500 → 5 cm	440 → 4.4 cm
Vidyaniketan School	520 → 5.2 cm	640 → 6.4 cm
City High School	680 → 6.8 cm	760 → 7.6 cm
Eklavya School	800 → 8 cm	600 → 6 cm

6Simplify the expressions. (i) 45 ÷ 5 + 20 × 4 − 12 (ii) (38 − 8) × 2 ÷ 5 + 13 (iii) 53 + 47 ÷ 3221 (iv) 3 × { 4 [ 85 + 5 − (15 ÷ 3) ] + 2 }

(i) 45 ÷ 5 + 20 × 4 − 12

= 9 + 80 − 12 (division and multiplication first)
= 89 − 12
= 77

(ii) (38 − 8) × 2 ÷ 5 + 13

= 30 × 2 ÷ 5 + 13 (bracket first)
= 60 ÷ 5 + 13
= 12 + 13
= 25

(iii) 53 + 47 ÷ 3221

= 53 + 47 × 2132 (division first)
= 53 + 4 × 217 × 32 = 53 + 38
= 4024 + 924
= 4924 = 2124

(iv) 3 × { 4 [ 85 + 5 − (15 ÷ 3) ] + 2 }

= 3 × { 4 [ 85 + 5 − 5 ] + 2 } (round bracket first)
= 3 × { 4 × 85 + 2 } (then the square bracket)
= 3 × { 340 + 2 }
= 3 × 342
= 1026

Order of operations: brackets first — ( ), then [ ], then { } — then division and multiplication from left to right, and finally addition and subtraction.

7Solve. (i) 512 + 716 (ii) 325 − 214 (iii) 125 × (−10)3 (iv*) 438 ÷ 2518

(i) 512 + 716

LCM of 12 and 16 = 48
= 5 × 412 × 4 + 7 × 316 × 3 = 2048 + 2148
= 4148

(ii) 325 − 214

= 175 − 94
= 6820 − 4520
= 2320 = 1320

(iii) 125 × (−10)3

= 12 × (−10)5 × 3 = −12015
= −8

(iv) 438 ÷ 2518

= 358 × 1825 (multiply by the reciprocal)
= 35 × 188 × 25 = 7 × 94 × 5
= 6320 = 3320

Questions 8 to 12: Constructing Triangles

8Construct ΔABC such that m∠A = 55°, m∠B = 60°, and l(AB) = 5.9 cm.

Step 1: Draw a rough figure and show the given measures in it.

Step 2: Draw seg AB of length 5.9 cm.

Step 3: At point A, draw ray AP such that m∠PAB = 55°.

Step 4: On the same side of seg AB, draw ray BQ at point B such that m∠QBA = 60°.

Step 5: Name the point where the two rays intersect C. ΔABC is the required triangle.

ΔABC: AB = 5.9 cm, m∠A = 55°, m∠B = 60°

Check: m∠C = 180° − (55° + 60°) = 65°; on measuring, l(AC) ≈ 5.6 cm and l(BC) ≈ 5.3 cm.

9Construct ΔXYZ such that l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.

Step 1: Draw a rough figure and mark the given lengths on it.

Step 2: Draw seg YZ of length 7.7 cm (the longest side) as the base.

Step 3: Open the compass to 3.7 cm. With the point on Y, draw an arc above seg YZ.

Step 4: Open the compass to 6.3 cm. With the point on Z, draw an arc that cuts the first arc. Name the point of intersection X.

Step 5: Draw seg XY and seg XZ. ΔXYZ is the required triangle.

ΔXYZ: YZ = 7.7 cm, XY = 3.7 cm, XZ = 6.3 cm

10Construct ΔPQR such that m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.

The given side is QR, but the given angles are ∠P and ∠Q. ∠P is not at an end of seg QR, so first find ∠R using the angle sum property of a triangle.

m∠P + m∠Q + m∠R = 180°
80° + 70° + m∠R = 180°
m∠R = 180° − 150° = 30°

Now the angles at both ends of seg QR are known: m∠Q = 70° and m∠R = 30°.

Step 1: Draw seg QR of length 5.7 cm.

Step 2: At point Q, draw ray QA such that m∠AQR = 70°.

Step 3: On the same side of seg QR, draw ray RB at point R such that m∠BRQ = 30°.

Step 4: Name the point where the two rays intersect P. ΔPQR is the required triangle.

ΔPQR: QR = 5.7 cm, m∠Q = 70°, m∠R = 30° (so m∠P = 80°)

Check: on measuring, m∠P = 80°, l(PQ) ≈ 2.9 cm and l(PR) ≈ 5.4 cm.

11Construct ΔEFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.

m∠EFG = 90°, so EG (opposite ∠F) is the hypotenuse.

Step 1: Draw a rough figure and mark the right angle at F.

Step 2: Draw seg FG of length 5 cm as the base.

Step 3: At point F, draw ray FT perpendicular to seg FG (m∠TFG = 90°).

Step 4: Open the compass to 7 cm (the hypotenuse). With the point on G, draw an arc cutting ray FT at E.

Step 5: Join G and E. ΔEFG is the required triangle.

ΔEFG: m∠F = 90°, FG = 5 cm, hypotenuse EG = 7 cm

Check: on measuring, l(FE) ≈ 4.9 cm.

12In ΔLMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) = 4 cm. Construct ΔLMN.

Step 1: Draw a rough figure and show the given measures in it. The angle ∠LMN is included between sides LM and MN.

Step 2: Draw seg ML of length 6.2 cm as the base.

Step 3: At point M, draw ray MT such that m∠TML = 60°.

Step 4: Open the compass to 4 cm. With the point on M, draw an arc cutting ray MT at N.

Step 5: Join L and N. ΔLMN is the required triangle.

ΔLMN: LM = 6.2 cm, m∠M = 60°, MN = 4 cm

Check: on measuring, l(LN) ≈ 5.4 cm.

Questions 13 to 15: Pairs of Angles

13Find the measures of the complementary angles of the following angles. (i) 35° (ii) a° (iii) 22° (iv) (40 − x)°

Two angles are complementary if the sum of their measures is 90°. So, complement = 90° − the given angle.

(i) Complement of 35° = 90° − 35° = 55°
(ii) Complement of a° = (90 − a)° = (90 − a)°
(iii) Complement of 22° = 90° − 22° = 68°
(iv) Complement of (40 − x)° = 90° − (40 − x)° = (90 − 40 + x)° = (50 + x)°

14Find the measures of the supplements of the following angles. (i) 111° (ii) 47° (iii) 180° (iv) (90 − x)°

Two angles are supplementary if the sum of their measures is 180°. So, supplement = 180° − the given angle.

(i) Supplement of 111° = 180° − 111° = 69°
(ii) Supplement of 47° = 180° − 47° = 133°
(iii) Supplement of 180° = 180° − 180° = 0°
(iv) Supplement of (90 − x)° = 180° − (90 − x)° = (180 − 90 + x)° = (90 + x)°

In (iii), 180° is a straight angle, so the arithmetic gives 0°. A 0° angle has no opening, so in practice a straight angle does not have a supplementary angle that can be drawn.

15Construct 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) A pair of adjacent angles

m∠CBA = 50°, m∠ABD = 65°

∠CBA and ∠ABD have the common vertex B and the common arm BA, and their interiors do not overlap. So they are a pair of adjacent angles.

(ii) Two supplementary angles which are not adjacent angles

m∠PQR = 120°

m∠XYZ = 60°

120° + 60° = 180°, so ∠PQR and ∠XYZ are supplementary. They have no common vertex or arm, so they are not adjacent.

(iii) A pair of adjacent complementary angles

m∠NML = 35°, m∠LMO = 55°

∠NML and ∠LMO have the common vertex M and the common arm ML, so they are adjacent. 35° + 55° = 90°, so they are also complementary. Together they form the right angle ∠NMO.

Any suitable measures may be used, as long as the conditions of each part are met.

Questions 16 to 20

16In ΔPQR, the measures of ∠P and ∠Q are equal and m∠PRQ = 70°. Find the measures of the following angles. (i) m∠PRT (ii) m∠P (iii) m∠Q

ΔPQR with side QR extended to T

(i) m∠PRT

∠PRQ and ∠PRT form a linear pair, so their sum is 180°.
m∠PRT = 180° − m∠PRQ = 180° − 70° = 110°

(ii) and (iii) m∠P and m∠Q

An exterior angle of a triangle equals the sum of its two remote interior angles.
m∠P + m∠Q = m∠PRT = 110°
Since m∠P = m∠Q, 2 × m∠P = 110°
m∠P = 55° and m∠Q = 55°

Check: 55° + 55° + 70° = 180°, the angle sum of a triangle.

17Simplify. (i) 5⁴ × 5³ (ii) (23)⁶ ÷ (23)⁹ (iii) (72)⁸ × (72)⁻⁶ (iv) (45)² ÷ (54)

a^m × aⁿ = a^m+n a^m ÷ aⁿ = a^m−n (a^m)ⁿ = a^mn a^−m = 1a^m a⁰ = 1

(i) 5⁴ × 5³

= 5⁴⁺³ = 5⁷ = 78125

(ii) (23)⁶ ÷ (23)⁹

= (23)⁶⁻⁹ = (23)⁻³
= (32)³ = 278

(iii) (72)⁸ × (72)⁻⁶

= (72)⁸⁺⁽⁻⁶⁾ = (72)²
= 494

(iv) (45)² ÷ (54)

= (45)² × (45) (multiply by the reciprocal)
= (45)³ = 64125

18Find the value. (i) 17¹⁶ ÷ 17¹⁶ (ii) 10⁻³ (iii) (2³)² (iv) 4⁶ × 4⁻⁴

(i) 17¹⁶ ÷ 17¹⁶

= 17¹⁶⁻¹⁶ = 17⁰ = 1

(ii) 10⁻³

= 110³ = 11000 = 0.001

(iii) (2³)²

= 2^3×2 = 2⁶ = 64

(iv) 4⁶ × 4⁻⁴

= 4⁶⁺⁽⁻⁴⁾ = 4² = 16

19Solve. (i) (6a − 5b − 8c) + (15b + 2a − 5c) (ii) (3x + 2y)(7x − 8y) (iii) (7m − 5n) − (−4n − 11m) (iv) (11m − 12n + 3p) − (9m + 7n − 8p)

(i) (6a − 5b − 8c) + (15b + 2a − 5c)

= (6a + 2a) + (−5b + 15b) + (−8c − 5c) (group like terms)
= 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 (change the sign of every term in the second bracket)
= 18m − n

(iv) (11m − 12n + 3p) − (9m + 7n − 8p)

= 11m − 12n + 3p − 9m − 7n + 8p
= 2m − 19n + 11p

20Solve the following equations. (i) 4(x + 12) = 8 (ii) 3y + 4 = 5y − 6

(i) 4(x + 12) = 8

Divide both sides by 4: x + 12 = 2
Subtract 12 from both sides: x = 2 − 12
x = −10

Check: 4(−10 + 12) = 4 × 2 = 8.

(ii) 3y + 4 = 5y − 6

Add 6 to both sides: 3y + 10 = 5y
Subtract 3y from both sides: 10 = 2y
y = 5

Check: 3 × 5 + 4 = 19 and 5 × 5 − 6 = 19.

Multiple Choice Questions

1The 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 angle bisectors meet at the incentre. (The perpendicular bisectors of the sides meet at the circumcentre.)

Answer: (iii) incentre

2[(37)⁻³]⁴ = ________. (i) (37)⁻⁷ (ii) (37)⁻¹⁰ (iii) (73)¹² (iv) (37)²⁰

= (37)^−3×4 = (37)⁻¹² = (73)¹²

Answer: (iii) (73)¹²

3The simplest form of 5 ÷ (32) − 13 is ________. (i) 3 (ii) 5 (iii) 0 (iv) 13

= 5 × 23 − 13 (division first)
= 103 − 13 = 93 = 3

Answer: (i) 3

4The solution of the equation 3x − 12 = 52 + x is ________. (i) 53 (ii) 72 (iii) 4 (iv) 32

3x − x = 52 + 12
2x = 3
x = 32

Answer: (iv) 32

5*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

(i) 30 + 7 = 37 (ii) 40 + 2 = 42 (iii) 32 + 3 = 35 (iv) 27 + 2 = 29

Answer: (i) 10 × 3 + (5 + 2)

← 8. Algebraic Expressions and Operations on them 9. Direct Proportion and Inverse Proportion →