[34] The following distribution gives the daily income of 65 workers of a factory :- Daily income 100-120 120 – 140 140 -160 160- 180 180- 200 (in Rs) No. of worker 14 16 10 16 9 Convert the distribution above to a more than type cumulative frequency distribution and draw its ogive. MARKING SCHEME SUMMATIVE ASSESSMENT – 1 Sl.no marks 1. (b) 1 2. (d) 1 3. (a) 1 4. (a) 1 5. (c) 1 6. (b) 1 7. (b) 1 8. (c) 1 SECTION-B 9. Since opposite sides of the rectangle are equal 1 So. P+3q=13, 3p+q=7 Solving p=1, q=4 1 10. Since α and are the zeros of the polynomial P(x)= 4-2x+k-4 So, = 1 1= k= 8 1 or -x+1)-x³+-3x+5(-x+2 -x³+-x 1 ------------ 2-2x+5 2-2x+2 ----------------- 3 ---------------- 1/2 So, quotient = -x+2 , remainder = 3 11. Here, = Since denominator contains prime factor 7 other than 2 or 5 So, will have a non-terminating repeating decimal expansion. 1 12. we have , + = + = 1 + 2 = 1 K = 6 1 13. 1 So, DN ||BC and DM || AB so, DNBM is a ||grm.
[34] The following distribution gives the daily income of 65 workers of a factory :- Daily income (in Rs) 100-120 120 – 140 140 -160 160- 180 180- 200 No. of worker 14 16 10 16 9 Convert the distribution above to a more than type cumulative frequency distribution and draw its ogive. MARKING SCHEME SUMMATIVE ASSESSMENT – 1 Sl.no marks 1. (b) 1 2. (d) 1 3. (a) 1 4. (a) 1 5. (c) 1 6. (b) 1 7. (b) 1 8. (c) 1 SECTION-B 9. Since opposite sides of the rectangle are equal So. P+3q=13, 3p+q=7 1 Solving p=1, q=4 1 10. Since α and are the zeros of the polynomial P(x)= 4-2x+k-4 So, = 1= k= 8 1 1 or -x+1)-x³+-3x+5(-x+2 -x³+-x ------------ 2-2x+5 2-2x+2 ----------------- 3 ---------------- So, quotient = -x+2 , remainder = 3 1 1/2 11. Here, = Since denominator contains prime factor 7 other than 2 or 5 So, will have a non-terminating repeating decimal expansion. 1 12. we have , + = + = 1 + 2 = K = 6 1 1 13. So, DN ||BC and DM || AB so, DNBM is a ||grm. 1