CAT Questions

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Monosaccharides, also called simple sugars, are the most basic units of carbohydrates. They are fundamental units of carbohydrates and cannot be further hydrolyzed to simpler compounds.

Hence, Monosaccharides can be classified according to the spatial arrangement of their atoms is true about Monosaccharides.

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton 

i.e. 3 engines of latter type consumes 1 ton for one hour

Hence 9 engines consumes 3 tons for one hour

For 15 hours it is 15 x 3 = 45 metric tonnes.

By trial and error method from the given options,

as 54 - 45 =  (1/11) x (45+54)  = 99/11  = 9

Hence this satisfies the given conditions.

Then the woman's husband age is 54.

3300 ----- 3% ↑ (1st time interval, 99) → 3399.

Here, time interval is given as half-yearly i.e. 6 months.

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.

The given series follows the rule,
+(11x1), +(11x3), +(11x5), +(11x7), +(11x9),....

So starting from 7, 7 + 11 = 18, 18 + 33 = 51, 51 + 55 = 106, ...
So wrong number is 40 which is to be 51.

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.

The molar mass of H2O can be calculated as the following.

M(H2O)= [2 x M(H)] + [M(O)].

And we know that M(H) = 1g/mol, M(O) = 16g/mol.

So, M(H2O)= 2x1 + 16 = 18g/mol,

That means 1 mole of H2O contains 18g.

Hence, one mole of H2O corresponds to 18 grams.