**Solution :-**

*Q11. Option A*

* *

*Let the original price of the jewel be Rs.P and let the profit earned by the third seller*

*be x%*

*Then, (100 + x)% of 125% of 120% of P = 165% of P*

*((100+x))/100 ×125/100×120/100×P= ×[165/100 ×P]*

*(100+x)=[(165 ×100×100)/(125×120)]=110*

*x = 10%*

* *

*Q12. Option D*

*Let C.P be Rs x. then*

*(105% of x) – (80% of x) = 100 *

*i.e 25% of x =100*

*So, x/4 = 100*

*Therefore, CP = 400*

* *

*Q13. Option B*

*10% of x = 15% of y,*

*Where x + y = 30000*

*x/y = 3k/2k*

*Hence the difference = k = 6000*

* *

*Q14. Option C*

*Profit = Rs.(2602.58 – 2090.42) = Rs.512.16*

*Profit % = [512.16/2090.42×100]%=[512160/209042×10]%=24.5%=25%*

* *

*Q15. Option B*

*CP/SP = 2/3*

*So, profile% = ½ x 100 = 50%*

* *

*Q16. Option C*

*Let the MP of 1 kg tea be Rs.1, then CP of 20 kg with discount = 20 × 0.9 = Rs.18*

*Also 1 kg tea is free. So the retailer gets tea worth Rs.21 by paying Rs.18 only.*

*Profit % = goods left/goods sold x 100*

* = (21-18)/18 x 100 = 16.66%*

* *

*Q17. Option A*

*C.P of A = 1818/0.9 = 2020*

*C.P of B = 1818/1.01 = 1800*

*C.P of A/C.P of B = 2020/1800 = 101/90*

* *

*Q18. Option B*

*6.66 % of MP = 25*

*MP = 375*

*SP = MP – 25 = 350*

* *

*Q19. Option C*

*Profit % = 25/100 = (120+k)/880 gives k =100*

*Therefore, net profit% = 100/1000 x 100 = 10%*

* *

*Q20 Option A*

*Charge of 1 call in February = 350/150 = 7/3*

*Charge of 1 call in March = (350 + 50 x 1.4)/250 = 420/250 = 42/25*

*%cheapness of a call in March = (7/3 – 42/25) / 7/3 x 100 = 28%.*