A +10 = 3(B - 10)
? A - 3B = -20 ....(i)
And ( A - 10) = (B + 10)
? A - B = 20 ...(ii)
on subtracting Eq.(ii) from Eq.(i), we get
(A - 3B) - (A - B) = -20 - 20
? - 2B = - 40
there4; B = 20
Now, by using Eq. ...(ii),
there4; Ratio of the numbers of students of A and B = 40 : 20 = 2 : 1
Couldn't be determined, since the total amount of money is not given in either of the case
Total of maximum marks of all subjects = 105 x 5 = 525
Total marks obtained by Nandita = 80% of 525 = 525 x 80/100 = 420
Obtained marks of four subjects (Hindi, Sanskrit, Mathematics and English)
= 89 + 92 + 98 + 81 = 360
So, the obtained marks in Science = 420 - 360 = 60
Let total number of employees in beginning = 8K and total number of employees in present = 7K
Let salary of an employee in beginning = 5L and salary of an employee in present = 6L
? Total salary in beginning = 8K x 5L = 40KL
and total salary in present = 7K x 6L = 42 KL
Required ratio = 40KL : 42KL = 20 : 21
Clearly, salary is increased.
Since there are 70 males out of 120 applicants, there must be 50 females out must have For the minimum number of males to have a driver's licence all 50 females must have a driver's license Thus the number of males having a driver's licence will be 80 - 50 = 30. The maximum possible number of males having a driver's license is 70. The ratio between the minimum and the maximum is 30 : 70 or 3 : 7 .
Quantity of acid in the mixture = (2/3) x 60 = 40 L
Quantity of water in the mixture = (1/3) x 60= 20 L
Let the required quantity of water be x L.
According to the question,
40/(20+ x) = 1/2
? 80 = 20 + x
? x = 60 L
Let the cost of third variety tea will be ? N per kg
According to the question,
(126 + 135 + 2N)/4 = 153
? (261 + 2N) = 4 x 153
? 2N = 612 - 261 = 351
? N = 351/2 = 175.50
? N = ? 175.5
Hence, the cost of third variety tea is ? 175.50 per kg .
Required compound ratio = (2 x 5 x 4) / (7 x 3 x 7)
= 40/147
= 40 : 147
Given that
A : B = 3 : 4 = ( 3 x 2 ) : (4 x 2 ) = 6 : 8
B : C = 8 : 9
? A : B : C = 6 : 8 : 9
As consequent of the first ratio is equal to the antecedent of second ratio.
(a + b) / (a - b) = 5/3
? 3a + 3b = 5a - 5b
? 2a = 8b
? a = 4b,
? a/b = 4/1
Now, ( a2 + b2 ) / (a2 - b2 ) = [a2 / (b2+1)] / [a2 / (b2 -1)]
= [( a/b )2 + 1] / [( a/b )2 - 1] = [( 4/1 )2 + 1] / [( 4/1 )2 - 1]
= [16 + 1] / [16 - 1] = 17/15
? ( a2 + b2 ) : ( a2 - b2 ) = 17 : 15
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