∴ C.P. of 1 kg of mixture = Rs. | ❨ | 100 | x 9.24 | ❩ | = Rs. 8.40 |
110 |
By the rule of allilation, we have:
C.P. of 1 kg sugar of 1st kind Cost of 1 kg sugar of 2nd kind | ||
Rs. 9 | Mean Price Rs. 8.40 |
Rs. 7 |
1.40 | 0.60 |
∴ Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
⟹ x = | ❨ | 7 x 27 | ❩ | = 63 kg. |
3 |
39.8% of 400 + ?% of 350 = 230
? 159.2 + (350 x ?)/100 = 230
&Arr; ? = (230 - 159.2) x 100/360 = 20.23 = 20
3x + y = 243 = 35 ⟺ x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.
Three days after Saturday is Tuesday and Tuesday is a day before a day before yesterday So, yesterday is Tuesday.
Hence, today is Friday.
Let salary be Rs. A, then A - 15 % of A - 30% of 85% of A = 2380
? A - 15A/100 - (30 x 85 x A)/(100 x 100) = 2380
? 200A - 30A - 51A = 2380 x 200
? 119A = 2380 x 200
? A = (2380 x 200)/199 = 4000
Let the number be x.
? (x - 4)/6 = 9
? x = 58
Again (x - 3)/5 = (58 - 3)/5 = 11
Given Expression = (1.04)2 + 1.04 x 0.04 + (0.04)2 / (1.04)3 - (0.04)3
= (a2 + ab + b2) / (a3 - b3)
= (a2 + ab + b2) / [(a-b) (a2 + ab + b2)]
= 1 / (a - b)
= 1/ (1.04 - 0.04)
= 1
Given expression
= [(0.47)3 - (0.33)3] / [(0.47)2 + 0.47 x 0.33 + (0.33)2]
= [a3 - b3 ] / [ a2 + ab + b2]
= [(a-b) (a2 + ab + b2)] / [(a2 + ab + b2)]
= (a - b)
= (0.47 - 0.33)
= 0.14
According to question ,
No. of students who passed in one or more subjects = 11 + 9 + 13 + 17 + 15 + 19 + 7 = 91.
No. of students who failed in all the subjects = 100 ? 91 = 9.
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