At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.
55 min. are gained in 60 min.
Then 15 min spaces are gained in = min.
Therefore, The hands are coincident at min. past 3 o'clock.
∴ 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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