Suppose the person had deposited ? P at the time of opening the account .
? After one year he had P + (P x 10 x 1)/100 = ? 11P/10
After two years, he had
11P/10 + (11P/10 x 10 x 1)/100 = ? 121P/100 ...(i)
After withdrawn ? 5000 from ? 121P/100, the balance
= ? (121P - 500000)/100
After 3 yr, he had
(121P - 500000)/100 + [(121P - 500000)/100 x 10 x 1]/100
= 11(121P - 500000)/100 ... (ii)
After withdrawn ? 6000 from amount (ii) the balance
= (1331P/1000 - 11500)
? After 4 yr, he had ? (1331P - 5500000)/1000 + 10% of ? (1331P - 5500000)/1000
= ? (11/10) x (1331P/1000 - 11500) ... (iii)
After withdrawn ? 10000 from amount (iii) the balance =0
? 11/10(1331P/1000 - 11500) - 10000 = 0
? P = ? 15470
If A runs 600 m, B runs 600 - 60 or, 540 m.
If A runs 400 m, B runs = 540 x 400 / 600 = 360 m.
Now, when B runs 500 m, C runs 500 - 50 = 450 m.
? When B runs 360 m, C runs = 450 * 360 / 500 = 324 m
? A beats C by 400 - 324 = 76 m.
Let Y = 0.15 / (0.5 / 15)
= (15/100) / [(5/100) / 15)
= 4.5
Putting arbitrary values of a and b.
IIIustration 1:
Let a = 9 and b = 8
HCF (8 + 9, 9 - 8)
HCF (17, 1) = 1
IIIustration 2:
Let a = 23 and b =17
HCF (17 + 23, 23 -17)
HCF ( 40, 6) = 2
Hence, HCF (a + b, a - b) can either be 1 or 2.
5358 x 51 | = 5358 x (50 + 1) |
= 5358 x 50 + 5358 x 1 | |
= 267900 + 5358 | |
= 273258. |
2079 is divisible by each of 3, 7, 9, 11.
Monthly sale = yearly sale / 2
= 132000 / 2
= ? 11000
June's sale = monthly sale / 2
= 11000 / 2
= ? 1000
Given that 0.8 x A = 0.09 x B
? A/B = 0.09/0.8 = 9/80
? A : B = 9 :80
For an income of Rs. 5, investment = Rs. 100
For an income of Rs. 4, investment = Rs. (100/5) x 4 = Rs. 80
For the product of 3.5413 x 2.1
Consider 35413 x 21 = 743673
Total decimal places = 4 + 1
? 3.5413 x 2.1=7.43673
? In 2/3h, 1 article is made.
? In 1h, 3/2 article are made.
? In 71/2 = 15/2h, 3/2 x 15/2 = 45/4 article are made.
? Required articles = 45/4 = 112/4
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.