Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.
Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.
Answer:
12 boys
Step-by-step explanation:
From the above question:
Number of boys = 3
Number of girls = 2
Boys: Girls
3:2
Let :
a = boys
b = girls
Hence, a : b = 3 : 2
a/b = 3/2
Cross Multiply
2a = 3b .......... Equation 1
a = 3b/2
If four more girls join the class, there will be the same number of boys and girls
Hence,
a: b + 4 = 3 : 3
a/b + 4 = 3/3
Cross Multiply
3a = 3(b + 4)
3a = 3b + 12 ........ Equation (2)
From Equation 1: a = 3b/2
Substitute 3b/2 for a in Equation 2 we have:
3a = 3b + 12 .........Equation 2
3(3b/2) = 3b + 12
9b/2 = 3b + 12
Cross Multiply
9b = 2(3b + 12)
9b= 6b + 24
9b - 6b = 24
3b = 24
b = 8
Substitute 8 for b in Equation 1
a = 3b/2
a = 3 × 8/2
a = 24/2
a = 12
Therefore, the number of boys in the class is 12
Answer:
its 60
Step-by-step explanation:
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