Three friends A,B and C went for picnic. A carried 3 chapatis ,B carried 4 chapatis and C did not carry any chapati but was having 14 rupees with him.they sat under a tree and then it was decided that chapati would be equally shared among them,and in return C gave 14 rupees to them.How should they divide 14 rupees between them(i.e A and B).a)A-6,B-8b) A-4,B-10c) A-7,B-7d) None of the above
b)
3 + 4 = 7 3 - 7/3 = 2/3 given by A 4 - 7/3 = 5/3 given by B
Three friends A,B and C went for picnic. A carried 3 chapatis ,B carried 4 chapatis and C did not carry any chapati but was having 14 rupees with him.they sat under a tree and then it was decided that chapati would be equally shared among them,and in return C gave 14 rupees to them.How should they divide 14 rupees between them(i.e A and B).a)A-6,B-8b) A-4,B-10c) A-7,B-7d) None of the above
Option b?
If they divide the chapatis equally then each of them gets 2.33 chapati
Of this A contributes 0.67 chapati and B contributes 1.67 chapati
Little Pika who is five and half years old has just learnt addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add?A. 150 B. 155 C. 156 D. 258 E. None of the above
My take 156.
Explanation :
abc and def are the last 3 digits of the 2 numbers
a+d
Possible solutions are (0,1) (1,2)(2,3)(3,4)(4,5)(9,0)
So 6 solutions forunit's digit .
6 for tens and 6 for hundreth.
SO total 6*6*6 = 216..
So none of these????
Bhai batana sahi hai kya nahi.. aap tho pura XAT level ka la rahe ho .. Lage raho
@grkkrg : thanku bhai.. Just did for unit's digit and applied same for rest should have thought about tens and hundreds bit more carefully
My take None of these.Highly probable that it can be wrong. Explanation :abc and def are the last 3 digits of the 2 numbersa+d Possible solutions are (0,1) (1,2)(2,3)(3,4)(4,5)(9,0)So 6 solutions forunit's digit . 6 for tens and 6 for hundreth. SO total 6*6*6 = 216..So none of these????Bhai batana sahi hai kya nahi.. aap tho pura XAT level ka la rahe ho .. Lage raho
There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8 :32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?A. 0 B. 1/6C. 1/4D. 1/3E. 1
My take 1/3.
Solved recently in quant thread for 2012 ๐
Number of ways of selecting 2 machines = 4C2 = 6
In these selections, there are 2 ways in which he can find out whether the machine is defective or not in 12 minutes.
Two alloys of iron A and B, weighing 6 kg and 12 kg respectively, have different percentage of iron. One piece each of equal weight was cut off from both the alloys and the piece from A was alloyed with B and the piecefrom B was alloyed with A. As a result, the percentage of iron became the same in the resulting two new alloys. What was the weight of each cut-off piece? 4 kg 3 kg 2 kg 1 kg
There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8 :32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?A. 0 B. 1/6C. 1/4D. 1/3E. 1
he has two chances to detect both the defected machines if he wants to catch the last bus.
for first attempt,probability that machine selected is defective=2/4 for second chance,probability that machine selected is defective given that first was defective=1/3
probability that he will catch the last bus=2/4*1/3=1/6
Two alloys of iron A and B, weighing 6 kg and 12 kg respectively, have different percentage of iron. One piece each of equal weight was cut off from both the alloys and the piece from A was alloyed with B and the piecefrom B was alloyed with A. As a result, the percentage of iron became the same in the resulting two new alloys. What was the weight of each cut-off piece? 4 kg3 kg2 kg1 kg
dost its 4 kg.....
lets say there is x nad y % of iron in 6 and 12 kg respectively.
so iron wt= 6x and 12y
now let z part be removed from each..hence zx and zy part iron was removed from each hence remaining iron content in A and B is 6x-zx and 12y-zy respectively
now they are added to each other so total wt remains 6 and 12 kg
and the iron content becomes 6x-zx+zy and 12y-zy+zx respectively
also 6x-zx+zy/6=12y-zy+zx/12
on solving we get z=4....
P.S. i dont like the calc..so a better logical approach would suffice my hunger
A 25 ft long ladder is placed against the wall with its base 7 ft the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:A. (2, 7) B. (5, 8) C. (9, 10)D. (3, 7 ) E. None of the above__/\__ Pranam Bhai @vijay_chandola
My take 9/2Seems to be a CAT question..I had got more or less same one.Had i done this question before CAT i would have got 3 more marks AnywaysThis is based on the property f(x)+f(1-x) = 1so f(1/10)+f(9/10) = 1 = f(2/10)+f(8/10) ...One term is left f(5/10) = 1/2So 4+1/2 = 9/2
bhai f(x)+f(1-x)=1...ye property ko yahan kaise use kara..i mean ye to koi cyclicity pe depend karti hai..plz justify this..
My take 9/2Seems to be a CAT question..I had got more or less same one.Had i done this question before CAT i would have got 3 more marks AnywaysThis is based on the property f(x)+f(1-x) = 1so f(1/10)+f(9/10) = 1 = f(2/10)+f(8/10) ...One term is left f(5/10) = 1/2So 4+1/2 = 9/2 what is the number of posts required to become certified pagal???
P.S. i guess u have to cross few grands for that.....plus it shudn't matter much...leave aside the visual satisfaction..
should have thought about tens and hundreds bit more carefully 
