I bought 5 Rasgullas, 7 Ladoos and 4 Samosas. Preeti bought 6 Rasgullas, 8 Samosas and 14 Ladoos for an amount which was half more than what I had paid.
What percent of the total amount paid by me was paid for the Rasgullas?
Assume R is Cost of one Rasgulla, L is cost of one Ladoo& S is cost of one Samosa
Let total cost of 5 Rasgullas, 7 Ladoos and 4 Samosas be X
5R+7L+4S=X -> Equation (i)
Preeti has bought 6 Rasgullas, 8 Samosas and 14 Ladoos at 1.5X cost (half more than mine)
So, 6R+14L+8S=1.5X -> Equation (ii)
Multiplying Equation (i) by 2 and subtracting 2nd equation from it:
10R+14L+8S=2X
06R+14L+8S=1.5X
--------------------------------
4R = 0.5X
=> R = 0.5X/4
Total amount paid by me for Rasgulla = 5R= (0.5X/4) * 5 = 2.5X/4
Percent of the total amount paid by me for the Rasgulla = {(2.5X/4)/X}* 100 = 62.5%
2 solid spheres of radius 2 m and a solid cone of radius 2 m, height 2 m are melted to form a bigger solid sphere. What is the radius of this bigger solid sphere?
Seema bought more mangoes than guavas. She sells mangoes at Rs.23 apiece and makes 15% profit. She sells guavas at Rs.10 apiece and makes 25% profit. If she gets Rs.653 after selling all the mangoes and guavas, find her profit percentage.
Assume Seema Sells X mangoes and Y guavas, where X > Y
Then, 23X + 10Y = 653 ---(i)
Try putting X = 1, 11, 21, 31â€¦
(Why are we trying X =1,11,21? because as per equation (i) last digit of 23*X needs to be3 )
X=1 cannot be true as X>Y
When X=11, Y = 40 so this also cannot be true as X >Y
Next we find X = 21 and Y = 17 is the only solution where X > Y
Cost of aMango = 23/1.15 and cost of aGuava = 10/1.25
Overall cost = 483/1.15 + 170/1.25 = 556
Then profit = 653 â€“ 556=97
Profit percentage = (97/556)Ã—100=17.4%
Option 5)
Two numbers are randomly selected from following set {3, 4,7,8,and 9}. What is the probability that product of these 2 selected numbers is an even number?
For faster solution letâ€™s find out probability of Product of 2 number being odd
Product of 2 numbers a and b would be odd only if both a and b are odd
Odd numbers in given set {3,7,9}
Number of such numbers where product of 2 selected number is odd = 3C2 = 3*2/2 = 3
Total Number of possible ways to select 2 numbers from set {3,4,7,8,9} = 5C2 = 5*4/2 = 10
Probability that product of these 2 selected numbers is odd = 3/10
Probability that product of these 2 selected numbers is even = 1-
(3/10) = 7/10
5 kg of adulterated rice has 2.72% stones in it and the rest is rice. Half of the stone content was removed. Now the percentage of stone content in it is
Weight of stones in 5 KG of adulterated rice = (2.72/100)*5000 = 136 gm
Weight of rice = 5000 - 136 = 4864 gm
Weight of Half of the stones which are removed = 136/2 = 68 gm
Total weight of remaining adulterated rice = 4932 gm
Percentage of Stones = 68/(4932) = 1.3787%
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