| Column - I | Column - II |
| (i) Gun powder contains 75% nitre and 10% sulphur. The rest of it is charcoal. The amount of charcoal in 9 kg of gun powder (in kg) is | (a) 1.5 |
| (ii) A cycle merchant allows 25% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs. 360 over the sale of one cycle, the marked price of the cycle (in Rs.) is | (b) 1.35 |
| (iii) Time (in years) in which Rs. 64000 will amount of Rs. 68921 at 5% p.a., interest being compounded semi-annually is | (c) 2880 |
A) (i) - (c), (ii) - (b), (iii) - (a)
B) (i) - (b), (ii) - (a), (iii) - (c)
C) (i) - (b), (ii) - (c), (iii) - (a)
D) (i) - (c), (ii) - (a), (iii) - (b)
Correct Answer: C
Solution :
(i) Total quantity of gun powder = 9 kg Quantity of nitre = 75% of 9 kg = 6.75 kg Quantity of sulphur = 10% of 9 kg = 0,9 kg \[\therefore \] Quantity of charcoal \[=[9-(6.75+0.90)]kg=1.35kg\] (ii) Let the marked price be \[\text{Rs}\text{. }x\] Then, \[\text{S}\text{.P}\text{.=75 }\!\!%\!\!\text{ }\] of \[\text{x=Rs}\text{. }\left( \frac{3}{4}x \right)\] According to question, \[\text{C}\text{.P}\text{.=}\frac{\frac{3}{4}x\times 100}{(20+100)}=\frac{5}{8}x\] Now, gain =Rs. 360 \[\Rightarrow \frac{3}{4}\times -\frac{5}{8}x=360=x=2800\] So, marked price = Rs. 2880 (iii) \[\text{P=Rs}\text{. 64000, A=Rs}\text{. 68921}\] \[\text{R=5 }\!\!%\!\!\text{ p}\text{.a}\text{. =}\frac{5}{2}%\] per half year Let time =n years=2n half years \[\therefore 68921=64000{{\left[ 1+\frac{5}{200} \right]}^{2n}}\] \[\Rightarrow \frac{68921}{64000}={{\left( \frac{41}{40} \right)}^{2n}}\Rightarrow {{\left( \frac{41}{40} \right)}^{3}}={{\left( \frac{41}{40} \right)}^{2n}}\] On comparing, \[2n=3\Rightarrow n=1.5\,\,years\]
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