GATE Questions

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Now as per question, Karan invested for 12 months and Satish invested for 7 months.

So, Karan:Satish = (29000 x 12):(18000 x 7)

= 348:126

= 58:21

Satish Ratio in profit will be= 16970×2179 = Rs. 4511

Here the given numbers are related in such a manner that,

$$\begin{gathered} 52 = 25 \\ 5+12 + 1 = 37 \\ Similarly, \\ 72 = 49 \\ 7+12 + 1 = 64 + 1 = 65 \end{gathered}$$

Hence, 25 : 37 :: 49 : 65.

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.

Given 40% minimum passing marks for boys = 483 + 117 = 600
=> 1% = 600/40
=> 100% = 600 x 100/40 = 1500
Minimum passing marks for girls = 35% of 1500 = 35 x 1500/100 = 525.

NH3, Ammonia is a neutral compound as the individual oxidation numbers elements that make up the compound NH3 are Nitrogen (N) and Hydrogen (H) sum to zero. 

We do not speak of the oxidation number of a molecule. We can speak of the oxidation numbers of the individual elements that make up the compound.

Here the oxidation number of Nirogen is -3 and that of Hydrogen is +1.

Now, NH3 = 1(-3) + 3(+1) = -3 + 3 = 0.

Dates on which Tuesday fall are 3,10,17, and 24

Hence, the 27th day of the month will be (Tuesday + 3 days=)Friday

Required day of the month = Friday - 4 = Monday

Ratio of Milk and water in a vessel A is 4 : 1

Ratio of Milk and water in a vessel B is 3 : 2

Ratio of only milk in vessel A = 4 : 5

Ratio of only milk in vessel B = 3 : 5

Let 'x' be the quantity of milk in vessel C

Now as equal quantities are taken out from both vessels A & B

=> 4/5     :     3/5
               x 
  3/5-x          x - 4/5

=> 35-xx-45 = 11 (equal quantities)

=> x = 7/10

Therefore, quantity of milk in vessel C  = 7

=> Water quantity = 10 - 7 = 3

Hence the ratio of milk & water in vessel 3 is 7 : 3 

Let C1 be the cost price of the first article and C2 be the cost price of the second article.
Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%.

We know, C1 + C2 = 600.
The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1
The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2.

The total selling price of the first and second article = 1.22C1 + 0.92C2.

As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2.

Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600
As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get
1.22C1 + 0.92(600 - C1) = 600
or 1.22C1 - 0.92C1 = 600 - 0.92*600
or 0.3C1 = 0.08*600 = 48
or C1 = 48/(0.3) = 160.
If C1 = 160, then C2 = 600 - 160 = 440.
The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.