-please tell the answer asap _/\_ 1 kg of wheat and sugar is the ratio of 5:7 and 1 kg of rice is in the ratio of 3:9. If the price of 30kg of rice is 210. Find the price of 63 kg of sugar. -A B C are rs 50 rs 60 and rs 70. After the shokeeper purchases a given quantity the average price becomes 45. If the total price of C is 900. Find the price of A
Plz help me with this question and the answer is not given!! The demand for a commodity linearly decreases with price and it vanishes when the price is set at rs 60 and equals to the half of the difference of its equilibrium price and this threshold price. The supply of the commodity vanishes when the price is set at Rs 25 and equals the square root of the price in excess of this threshold price. The equilibrium price at which the supply coincides with the demand is (1) Rs 58 (2) Rs 50 (3) Rs 79 (4) Rs 62
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Can anyone answer these two ? Kindly explain the process too
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Five Character Identification code - - - - -
First Two spaces should include 2 of 26 letters. (Repitition of letters not allowed)
Last Three spaces must be one of the ten digits. (Repitition of digits not allowed)
How many identification codes are possible?
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Guys sorry, maybe this is not the place to post, Did XAT officially declared the pattern of the exam?
Divisibilty Methods For checking divisibility by any 'prime/odd' number except for factors of 5', you have the concept of base number.
Number Add Base Number Subtract Base Number
---------------------------------
3: 1 2
7: 5 2
9: 1 8
11: 10 1
13: 4 9
17: 12 5
19: 2 17
21: 19 2
23: 7 ?
27: ? 8
29: 3 ?
...
...
And so on....(i'll describe a method to get these below)
Now for checking divisibility either add the last digit*add base number to the number "formed by removing last digit" or you can subract last digit*subtract base number from the "formed by removing last digit"
for e.g. check for 51/17 either 5- 1*5 =0 or 5 + 1*12 =17 hence divisible.
to check 312/13 we can 31+2*4 =39 hence dvisible or we can 31-2*9 = 13 ence divisible.
to check 61731/19 = 6173 + 1*2 = 6175 = 617 + 5*2 = 627 = 62 + 7*2 = 76 hence divisible.
to check 357976/29 = 35797 + 6*3 = 35815 = 3581 +5*3 = 3596 = 359 + 6*3 = 377 = 37 + 7*3 = 58 hence divisible..
to check 382294/11 = 38229-4*1 =38225 = 3822-5 = 3817 = 381 - 7 = 374 = 37 -4 =33 Hence divisible..
The Subract base Number for a number can be obtained as the {(samllest multiple of number which ends in one)-1}/10
i.e. for 3 or 7 it is (21-1)/10 =2
for 13 it is 91-1/10 = 9.
The AddbaseNumber for a number can be obtained as the {(samllest multiple of number which ends in nine)+1}/10
i.e. for 13 it is (39+1)/10 =4.
for 7 it is 49+1/10 = 5
Proof:
For SubtactBaseNumber say the number abcde...
I want to check divisibility by 17 where subtractbasenumber is 5
I can always write abcde... as 10X+Y (where Y is last digit and X is number formed by removing last digit)
Now X-Y*5 = (10X -50Y)/10 = (10X + Y -51Y)/10 = (OriginalNumber - 51 Y ) / 10
The number '51 Y' is a multiple of 17 so if "OriginalNumber" is divisible by 17 then "OriginalNumber - 51*Y" got to be.. i.e. "10X - 50Y"
as 10 and 17 are co-prime if "10X- 50Y" is divisible the "X-5Y" got to be.....
same theory hold's for addbasenumbers too....
This also defines why it is so easy to check divisibility by 3 or 9 just keep on adding the digits...
And you can check divisibility by 11 just by keeping on subrating digits form previous number.. (which is same as taking sum of even/odd location separately..) Divisibility by 7
Only for those interested in Number theory (Not a Cat short-cut)
say the number is :
38,391,787
Separate into pairs of digits
38 39 17 87
Consider the difference between each pair of digits and the nearest multiple of seven, beginning for the first pair at right, lower (upper) for the first, upper (lower) for the second and so on, alternating for each new pair.
4 -----4 (21-17)
38 39 17 87
---4 ------3 (87-84)
The resulting digits, read from right are 3444 (which is also a number multiple of 7).
Proceed in the same way with 3,444
1
34 44
----2
The final pair 21 is a multiple of seven, so is the original number 38,391,787.
ANOTHER EXAMPLE
Look how fast this method is.
Consider the 15-digit number 531,898,839,909,822
2 ----2--- 3 ----0
5 31 89 88 39 90 98 22
---3 ---4 ----6 ----1
Now we have 10,634,232
4 -----0
10 63 42 32
----0 ----4
And now 4,004
2
40 04
---4
Which gives 42, a multiple of 7.
We only need three steps for a 15-digit number. This is called TOJA's method of divisibility. Incidentally this also works for 11 and 13. Just a little manupulation is required, (in case you get a remainder of more than 9)
Let A = 5,962
7
59 62 77 which is a multiple of 1
--7
EXAMPLE 2
Let A = 5, 971,845
6---- 4
5 97 18 45
--9 ----1
8
14 96 -> 88 ->divisible
----8
EXAMPLE 3
Let A = 80,714,546
8 ----10
80 71 45 46
----5-------2
The resulting numbers ( 2 10 5 8 ) don’t form a decimal number, so proceed in this way: Put the exceeding number 1 from 10, below the 2 and sum.
2 0 5 8 -> 3 0 5 8
1
3
30 58 33
---3
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I know it's late in the day to ask this, but I was wondering if someone could share some useful videos or blogs where they have mentioned ways to quickly convert fractions(especially large fractions) to percentages. I couldn't find anything wothwhile so far.
Any approximation methods with minimum deviation from the actual values will also do.
Thanking you guys in advance.
A car theft occurs and the thief drive away at a constant speed of 70 kmph. By the time the theft is reported the thief is already 50km away from the nearest police station. If the Jeep begin chasing the car at a constant speed of 80kmph, the thief will be caught how many after the police started chasing him? A. 1½hours B. 2 hours C. 4 hours D. 5 hours Plz give me detail solution and answer is not given, thanks
A car theft occurs and the thief drive away at a constant speed of 70 kmph. By the time the theft is reported the thief is already 50km away from the nearest police station. If the Jeep begin chasing the car at a constant speed of 80kmph, the thief will be caught how many after the police started chasing him? A. 1½hours B. 2 hours C. 4 hours D. 5 hours #in the above problem, if the police Jeep constantly increased it speed by 20kmph for every hour starting form the end of the 1st hour of his chase and the thief increased his speed by 10 kmph for every hour starting from the end of the 1st hour of chase, in what time dose the police catch up with the stolen car ? A. 4⅓hr B. 5⅓hr C. 3hr D. 2⅔hr Solution is not given, plz send details Solution, thanks!!
Plz help this question with the detail solution!! And answer isn't given, thanks!
Q1. A container contains mixture of milk and water in which milk is 80%. 75% of mixture is taken out and 10 water is added, now the concentration of milk in the mixture is 60%. Find the quantity of milk initially?
How to solve 37? TIA
I'm a software engnr with 18mths exp, entrances have not been up to the mark. Best option? Please feel free to comment
- Should I continue in software (though I’m finding it not so interseting)?
- XIMB (Global Mgmt)
- XISS Ranchi(Fin/Mark/HR)
- IBS Hyderabad, ICFAI
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