Question
A 240-litre mixture contains only
milk and honey in the ratio 5:3, respectively. If 'Q' wishes to increase the percentage of milk in the mixture to ___%, then he should add ___ litres of pure milk to it. Which of the following pairs of values, in the given order, will correctly fill the blanks to make the statement true? I. 75, 120 II. 80, 210 III. 70, 60Solution
ATQ, Initial quantity of honey in the mixture = 240 × (3/8) = 90 litres Initial quantity of milk in the mixture = 240 – 90 = 150 litres Statement I: Total quantity of new mixture = 90 ÷ 0.25 = 360 litres Total quantity of milk in the new mixture = 360 × 75% = 270 litres So, quantity of milk to be added = 270 – 150 = 120 litres So, Statement I is true. Statement II: Total quantity of new mixture = 90 ÷ 0.2 = 450 litres Total quantity of milk in the new mixture = 450 × 0.8 = 360 litres So, quantity of milk to be added = 360 – 150 = 210 litres So, Statement II is true. Statement III: Total quantity of new mixture = 90 ÷ 0.3 = 300 litres Total quantity of milk in the new mixture = 300 × 0.7 = 210 litres So, quantity of milk to be added = 210 – 150 = 60 litres So, Statement III is true.
I n the question, assuming the given statements to be true, find which of the conclusion (s) among given three conclusions is /are definitely true and t...
Statements: Z % Y; X # W; U % V; W & V; Y @ X
Conclusions:Â Â Â Â Â
I. U @ X Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ...
In the question, assuming the given statements to be true, find which of the conclusion (s) among given three conclusions is /are definitely true and ...
Statements: G < H  ≤  I, V  ≥ W = G, R  ≥ I = A
Conclusions :
I. R > G
II. A ≥ H  Â
 III. H ≤ R
...Statement: X≤Y<W =Z ≤U<S;S>T ≥V
I. Z≥T
II. Z > X
Statements : I @ L © R * A $ MÂ
Conclusions :Â
I. R * MÂ
II. A % LÂ
III. A % I
Statements: R > S ≥ T = U < V ≤ W; X ≥ Y = Z < U = M ≥ N
Conclusions:
I. S ≥ M
II. T < X
III. W > N
Statements:
Q ≥ J ≤ V; U > J > X; M = U < P
Conclusions:
I. P > X
II. V ≥ M
Statements: G > P > Y ≥ E = N ≥ Q ≥ Z
Conclusions:
I. P > Z
II. Q ≤ YStatements: B ≥ U > P = E ≥ X; X > K > N ≥ J
Conclusions: I. N < P   II. J ≤ X
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