Solve the following equations:- (2×2=4) a) 2 (y + 7) = 3 (y –10) b) 3 – 4 5 x = 2 4 2 x 2 In a school the monthly Fee collection of 350 students is Rs. 3,85,000. What will be the collection fee of 58 students for a month? (2) 3 Circumference of a circle is 88 cm. Find the diameter of the circle. (2) 4 The base and height of a triangle are in the ratio of 3:2 and it’s area is 108 cm2 . Find it’s base and height. (2 ) 5 A wire when bent in the form of a square encloses an area of 484 sq. cm. If the same wire is re-bent in the form of a circle find the radius of the circle. (2 ) 6 The length of a rectangle is 4 more than twice of its breadth. If the perimeter of the rectangle is 2012. Find the dimensions of the rectangle. (3) 7 At what rate percent per annum will Rs. 7000 produce Rs. 350 as simple interest in 2 years (3) 8 Find the area of the shaded portion. (3x2) a) S R b) D C 36cm T 28cm P Q A 40cm B 36cm 9 Calculate the area of quadrilateral PQRS shown in the figure given below. (3) 20m S R 8m P Q 12m
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Answer:
1) monthly fees collection of 350 students=3,85, 000
monthly fees of 1 student=3,85,000÷350
=1,100
monthly fees of 58 student=1,100×58
=63,800
Step-by-step explanation:
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Answer:
fees collected from 350 students=rs.385000
fee collected from 1 student=385000/350=rs.1100
so,for fees for 58 students=rs.1100*58=rs.63800
circumference of a circle =πr^2=88 cm
22/7*r^2=88
r^2=88*7/22
=4*7
so, r=2✓7 cm
therefore. d=2*r=2*2✓7=4✓7 cm