7th Maths Term 1 Exam 2022 Thiruvannamalai District - PDF

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ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in N Neett N Neett lalai.i. l a l i a.i. l a l i a assaa aaddaassaa aaddaassaa ww. P. P w w.P.P w www w www N Nett e N Neett lalai.i. l a l i a.i. l a l i a et assaa aaddas a aa s ad a a d s a aa s. P.P. P. P.N w w w www www www l a l i a N.i. Neett lai l a l i a N.i. Neett l a l i a sa assaa aaddaasaa s aaddaassaa ww. P. P w w. P. P w www w www da Pa N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa ad adaassaa aaddaasaa s w. ww. P. P w w. P. P w www w www ww N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa aaddaasaa s ad a a d s a aa s w w.P. P w w. P. P w www w www N Neett N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593  ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in N Neett N Neett lalai.i. l a l i a.i. l a l i a assaa aaddaassaa aaddaassaa ww. P. P w w.P.P w www w www N Nett e N Neett lalai.i. l a l i a.i. l a l i a et assaa aaddas a aa s ad a a d s a aa s. P.P. P. P.N w w w www www www l a l i a N.i. Neett lai l a l i a N.i. Neett l a l i a sa assaa aaddaasaa s aaddaassaa ww. P. P w w. P. P w www w www da Pa N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa ad adaassaa aaddaasaa s w. ww. P. P w w. P. P w www w www ww N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa aaddaasaa s ad a a d s a aa s w w.P. P w w. P. P w www w www N Neett N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593  ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in Orange International Matric Hr Sec School #4/4, Shiva Gardens, Chetpet Road, Aagaram Village , N Neett N Neett Vinnamangalam Post , Arni -632316.Tiruvannamalai district. lalai.i. l a l i a.i. l a l i a assaa aa aa ANSWER KEY Q.No ddaass ANSWER a ddaass MARK 1 c)0. P. Paa.P.Paa 1 2 w d) (-6) x (+5) w w w w w 1 3 4 a) 70 sq.m www All options www [The correct answer is 3.6 m , but these are not given in the options to be selected(90 1 1 sq.m/25 m is 3.6 m)] N Ne 5 tt e b) -7 N Neett 1 lalai.i. 6 a) 0 l a l i a.i. l a l i a 1 et assaa aa aa 7 b) -1 1 8 d) 147 ddas a s d a d s a s 1 9 a) 360°. P.Paa. P. Paa 1.N 10 b) 0 1 w w w www www www 11 -40 1 12 x 1 13 32 1 lai 14 15 1 15 ₹ 84 1 N16 Neett 2c N Neett 1 l a l i a.i.17 180° l a l i a.i. l a l i a 1 sa assaa saa s ssaa 18 -44 1 19 Unlike aaddaa aaddaa 1 20 80 ww. P. P w w. P. P 1 21 w www w www 2 da Pa N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa ad adaassaa 22 (−15) × 13 × (−7) = [(−15) × (−7)] × 13 =[(105)] × 13 aaddaasaa s 2 w. ww. P. P = 1365 w w. P. P 23 w www Base of a parallelogram (b) = 16 cm, w www Height of a parallelogram (h) = 6 cm(Assume length is height) 2 ww Area of a parallelogram = b × h sq.units. Therefore, Area = 16 × 6 = 96 sq. cm. N Neett N Neett Thus, area of the parallelogram is 96 sq. cm. l a l i a.i.24 Amount with Geetha = ₹ 150 l a lai.i. l a l i a 2 assaa aa aa Cost of bag = ₹ 275 ddaass Amount to be borrowed = 275 – 150 d a d s a s.P. Paa = ₹ 125. P. Paa w w 25 Parallel sides a = 24 cm, b = 20 cm w w w w 2 www Height h = 15 cm www Area of the trapezium = 1/2 h (a + b) sq.units Area of the trapezium = 1/2 × 15 (24 + 20) sq. units = 1/2 × 15 × 44 sq.cm = 330 sq.cm N Neett Area of the trapezium = 330 sq.cm N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593  ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in 26 (29x + 4y– 40) –(11x + 8y – 3)= (18x - 4y –37) 2 27 Like terms : 2 N Neett x terms : 7x, –8x, –12x N Neett lalai.i. y terms : 5y, 12y, –9y l a l i a.i. l a l i a assaa aa aa z terms : 6z, z, 11z ddaass ddaass. P. Paa.P.Paa www w w w www www 28 Let x be the number of pencils required for 20 children. As the number of children increases, number of pencils also increases. 2 Number of children 7 20 Number of pencils 28 x N Nett e N 𝑥 Neett 𝑥 In the case of direct proportion we take 𝑦1 = 𝑦2.i..i. 1 2 lalai 7 ⇒ 28 = 20 l a l i a l a l i a et assaa aa aa 𝑥 ⇒ 𝑥= 20×28 7 ddas a s d a d s a s ⇒ 𝑥 = 20 × 4 ⇒ 𝑥 = 80. P.Paa. P. Paa.N w w w www www www Hence, 72 pencils are required for 80 children. 29 12x + 10 = 70 2 ⇒12x = 70 – 10 lai ⇒12 x= 60 ⇒ x= 60/12 N Neett ⇒ x= 5 N Neett l a l i a.i. l a l i a 30 Let x be the required number of days..i. l a l i a 2 sa assaa saa s aa Number of hens 144 112 Number of days 28 x ddaa s ddaa s. P. Paa. P. Paa When number of hens decrease food last for days will be increased So, it is in inverse proportion. www w w w www www da Hence x1 y1 = x2 y2 144 × 28 = 112 × x x = ( 144 × 28) / 112 = 36 days The food will last for 36 days. Pa N 31 Neett N Neett 2 l a l i a.i. l a lai.i. l a l i a assaa ad adaassaa aaddaasaa s w. ww. P. P w w. P. P 32 w x = 65° (corresponding angles) www w www 2 ww N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa aa aa ∠ AOC + ∠ BOC = 180° 72° + x° = 180° ddaass d a d s a s x° = 180° – 72° = 108°.P. Paa. P. Paa 33 w w The area of the rhombus = 100 sq.cm w w w w 2 www www The length of one diagonal d1 = 8 cm The length of other diagonal = d2 Area of the rhombus = 1/2 × d1 × d2 = 100 = 1/2 × 8 × d2 = 100 d2 = 100/4 = 25 cm N Neett The length of the other diagonal = 25 cm N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593  ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in 34 [(−6)×4]×(−3) = [−24] × (−3) = + 72 3 (−6)×[4×(−3)] = (−6)× [−12] = + 72 [(−6)×4]×(−3) =(−6)×[4×(−3)] N Neett Hence it is proved N Neett lalai.i.35 a i a.i. Height of the parallelogram h = 14 m l l l a l i a 3 assaa s aa s aa base b = 8m longer than its height= (8+ 14) m = 22 m ddaa s Area of the parallelogram = b × h sq. units ddaa s. P. Paa = 22 × 14 sq. m.P.Paa www = 308 sq. m w w w www Cost of levelling 1 sq.m = ₹ 15 Cost of levelling 308 sq.m = ₹ 308 × 15 = ₹ 4620 www Cost of levelling the ground = ₹ 4620 N Nett e N Neett 36 (3x + 2y– z)+(6x - 5y+7z)-(7x - 7y- 6z)= (3+6-7)x+(2-5+7)y+(-1+7+6)z 3 lalai.i. l a l i a.i. = (9-7)x+(9-5)y+(-1+13)z = 2x+4y+12z l a l i a et assaa s aa s s aa s 37 x = 2, y = 3 3 d aa i) 2x – 3y = 2(2) – 3(3) = 4 – 9 = –5 aa d ad a a d a. P. ii) x + y = 2 + 3 = 5 P. P. P.N w w iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10 w www www iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0 www 38 Using unitary method we can solve this as follows: 3 The cost of 4 notebooks = ₹ 32 lai The cost of 1 notebook = 32/4 = ₹8 N Neett N Neett Therefore, the cost of 10 notebooks = 10×₹8 = ₹ 80 Hence, Sheela has to pay ₹80 for 10 notebooks. l a l i a.i. l a l i a.i. l a l i a sa 39 Given l is parallel to m and n is transversal to l and m. 3 assaa aaddaasaa s aaddaassaa ww. P. P w w. P. P w www w www da Pa We get, y = 2x [Vertically opposite angles are equal] N Neett N Neett y + 4x = 180° [sum of interior angles that lie on the same side of the transversal] l a l i a.i. 2x + 4x = 180° [since y = 2x] l a lai.i. l a l i a assaa aa aa 6x = 180° ddaa Dividing by 6 on both sidesss ddaass aa aa w. x/6 = 180°/6 gives, x = 30°. w. P. P Now, y = 2(30°) = 60°. w w w. P. P w www w www 40 Let x be the number of letters sorted in 9 hours. Number of letters sorted 838 x 3 ww Time (hrs) 6 9 As the time increases the number of letters sorted also increases N Neett So, it is in direct proportion Hence, x1 / y1 = x2 / y2 N Neett l a l i a.i. 838 / 6 = x / 9 l a lai.i. l a l i a assaa aa aa x = [ 838 × 9 ] / 6 = 7542/6 ddaass d a d s a s x = 1257.P. Paa. P. Paa w w 1257 letters can be sorted in 9 hours. w w w w www www N Neett N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593  ww.. w w.. wwww wwww www.Padasalai.Net www.CBSEtips.in 41 5 N Neett N Neett lalai.i. l a l i a.i. l a l i a assaa aaddaassaa aaddaassaa ww. P. P w w.P.P w www w www N Nett e N Neett lalai.i. l a l i a.i. l a l i Step 1 : Draw a line. Mark two points A and B on it so that AB = 9 cm a et assaa s aa s s aa s Step 2 : Using compass, A as center and radius more than half of the length of AB draw two aaddaa arcs one above AB and one below AB. ad a a d a. P.P. P. P Step 3 : With the same radius and B as center draw two arcs above and below AB. They cut.N w w w www www www the previous arcs at C and D. Step 4 : Join C and D. CD intersects AB. Mark the point of intersection as O. CD is the required perpendicular bisector of AB. lai 42 5 N Neett N Neett l a l i a.i. l a l i a.i. l a l i a sa assaa aaddaasaa s aaddaassaa ww. P. P w w. P. P w www w www da Pa N Neett N Neett l a l i a.i. l a lai.i. l a l i a assaa addaassaa Step 1 : Draw a line. Mark a point A on it. a aaddaasaa s w. ww. P. P w w. P. P Step 2 : With A as center, draw an arc of convenient radius to the line at a point B. w www w www Step 3 : With the same radius and B as center draw an arc to cut the previous arc at C. Step 4 : With the same radius and C as center draw an arc to cut the previous arc at D. ww Step 5 : Join AD. ∠ BAD = 120° Step 6 : With C as center, draw an arc of convenient radius in the interior of ∠CAD N Neett N Neett Step 7 : With the same radius and D as center draw an arc to cut the previous arc at E. l a l i a.i. l lai.i. Step 8 : Join AE. Now ∠BAE = 90° is the required angle. a l a l i a assaa aaddaasaa s aa d a assaa __________________________________________________________________________________________________ d by : w w.P. P w w. P. P Prepared w www ww E.VIKRAM,w w M.Sc.,B.Ed(MATHEMATICS) BT Assistant in Maths N Neett N Neett lalai.i. l a l ai.i. l a l i a assaa dda a s saa d aassaa PP a a Kindly send me your district question PP d aa papers to our whatsapp number: 7358965593