Arey bhai, you are in a better state. Mera, last year ka ek sim diya sectional main toh 30 aaya. 22A 13C. Fir aaj SIM8 diya, 24A 15C. 100%ile aane ke baad se aise hi scores aa rahe hai.. I swear I wasn't or am not complacent... LR main tagda katta ho raha hai. Suddenly ban hi nahi rahe LR. Aur VA main mistakes badh gayi hai all of a sudden. RC main mera stock number hai 3 mistakes ka..So 3RC + 4VA + 2LR -> 9 mistakes, aur main saare LR kar nahi paaya toh less than 25 attempts. I have a plan to get myself out of this, but it is just a plan until it works.
mere toh haal aur bekar hota jaa rha h din pe din...i hv started panicking in sectn 2...LR nhi banta toh heavy faat leti h...SIM 7 and 8 mein aisa hi kuch hua...LR nhi hua initially toh heavy pressure aa gya aur baaki paper barbaad kar dia...jaldi jaldi ke chakkar mein naaya duub rahi h...wats ur plan maine bhi ek banaya toh h...lets c if it wrks...
agar kisi ne advanced sectionals diye hain VA/LR ke to please yeh query solve karo... LR set jisme 5 students ke number pata karne thay woh yeh kyun nahi ho sakte?? S1- 170, S2- 170, S3- 152, S4- 182, S5- 176 OA is s1-170,s2-150,s3-160,s4-190,s5-180 [image was removed because it was too large]
@ankit3007 said:agar kisi ne advanced sectionals diye hain VA/LR ke to please yeh query solve karo...LR set jisme 5 students ke number pata karne thay woh yeh kyun nahi ho sakte??S1- 170, S2- 170, S3- 152, S4- 182, S5- 176OA is s1-170,s2-150,s3-160,s4-190,s5-180
exactly.. yahi dout aa ya tha mujhe.. time is crazy...
mere toh haal aur bekar hota jaa rha h din pe din...i hv started panicking in sectn 2...LR nhi banta toh heavy faat leti h...SIM 7 and 8 mein aisa hi kuch hua...LR nhi hua initially toh heavy pressure aa gya aur baaki paper barbaad kar dia...jaldi jaldi ke chakkar mein naaya duub rahi h...wats ur plan maine bhi ek banaya toh h...lets c if it wrks...
A and B are mother and father of C respectively.C has 5 aunts and four uncles.A has 3 siblings.How many sisters does B have? A:A has 2 brothers B:B has 1 brother.
@CBZ123 said: A and B are mother and father of C respectively.C has 5 aunts and four uncles.A has 3 siblings.How many sisters does B have?A:A has 2 brothersB:B has 1 brother.
questn can be answered using either of the statement.....
ok.but I have a doubt ..nothing has been given of "married" uncles/aunts.cos lets suppose in first condition both the brothers are married then their respective spouse too will be C's aunt.where m i going wrong?.
@CBZ123 said: ok.but I have a doubt ..nothing has been given of "married" uncles/aunts.cos lets suppose in first condition both the brothers are married then their respective spouse too will be C's aunt.where m i going wrong?.
@CBZ123 exactly yr, even I got this doubt while solving it bt since thr is no detail mentioned abt them gettn married so I ignored it...I hv seen such a type f questn in some mock, nd they hv followed the same thng....
A triangle ABC has 2 points marked on side BC, 5 points marked on side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have at least one vertex common with the triangle ABC?
A triangle ABC has 2 points marked on side BC, 5 points marked on side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have at least one vertex common with the triangle ABC?
that will be 237--> 13C3-(7c3+4c3+5c3)------------------- (1)
now subtract from this which have none of ABC as it's vertex...
for that the following cases are:
take two vertices on AB and one on either AC or BC--> 3c2*(5c1+2c1)
take two vertices on AC and one on either BC or AB--> 5c2*(2c1+3c1) take two vertices on BC and one on either AB or AC--> 2c2*(3c1+5c1)
take one vertex on AB and one vertex on BC and one vertex on AC--> 3c1*2c1*5c1
add all the four u'll get 109--------------------------- (2)
(the reason for the bold terms is whenever u have 'and' in any sentence it means multiplication, and if u have 'or' it means addition.. plzz dont cram it it's a concept of boolean algebra.. it's just for simplification)
now subtract (2) from (1) we get 128.. as the question is from the new triangles atleast 1 vertex should be A or B or C.. therefore we have to subtract 1 more triangle which is ABC(it was included in 13C3)
