Solution :-
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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%.
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