Hi everyone, I am having a few problems on numbers chapter in 'How to prepare for quantitative aptitude for the CAT" book by Arun Sharma. I wanna discuss one by one.
Q) Find the greatest number of 4 digits which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively. a) 9907 b) 9903 c) 9893 d) None of these. Solution: First find the greatest 4 digit multiple of the LCM of 10,11,15 and 22. (In this case it is 9900). Then, Subtract 7 from it to give the answer.
This is the solution provided by the author itself. My Doubt in this is why should we subtract 7 number? why not any other number? What's the logic behind subtracting 7?
Thanks! --------- The only thing that matters is the everlasting present. :banghead::splat:
In each case when the no is being divided by 10,11,15,22 the remainders left is 3,4,8,15. Now the max LCM gives 9900. But 9900 is divisible by 10,11,15,22. To get the remainders as 3,4,8,15, we subtract the no 9900 with 7 then only we will get the remainders as 3,4,8,15.
Note the diff of 10-3 = 7, 11 - 4 = 7 and so on..Hence 7 is used to subtract
Hi everyone, I am having a few problems on numbers chapter in 'How to prepare for quantitative aptitude for the CAT" book by Arun Sharma. I wanna discuss one by one.
Q) Find the greatest number of 4 digits which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively. a) 9907 b) 9903 c) 9893 d) None of these. Solution: First find the greatest 4 digit multiple of the LCM of 10,11,15 and 22. (In this case it is 9900). Then, Subtract 7 from it to give the answer.
This is the solution provided by the author itself. My Doubt in this is why should we subtract 7 number? why not any other number? What's the logic behind subtracting 7?
Thanks! --------- The only thing that matters is the everlasting present. :banghead::splat:
Let the required no. be X ; LCM= found; The remainder can be considered +/- for convenience ::When div . by 10 a +3 remainder is same as -7; similarly when div. by 11 a remainder of +4 is same as -7.Now when all the remainders are same ans:k.LCM-7 Regards, Never Back Down.
In each case when the no is being divided by 10,11,15,22 the remainders left is 3,4,8,15. Now the max LCM gives 9900. But 9900 is divisible by 10,11,15,22. To get the remainders as 3,4,8,15, we subtract the no 9900 with 7 then only we will get the remainders as 3,4,8,15.
Note the diff of 10-3 = 7, 11 - 4 = 7 and so on..Hence 7 is used to subtract
HI, Thanks for clarifying my doubt... well Its a very simple logic.. but I didn't think in that way.
Cheers ------ The only thing that matters is the everlasting present.
Two circles of radii 4cm and 3cm intersect reach other in two points A and B. If the distance between their centres C and D is 5 cm. Find the length of the common chord AB.
Two circles of radii 4cm and 3cm intersect reach other in two points A and B. If the distance between their centres C and D is 5 cm. Find the length of the common chord AB.
Let the intersection point of A and B be called O Now, AO^2 + x^2 = 4^2 ...(i) and AO^2 + (5-x)^2 = 3^2..(ii)
Equate AO^2 = 16 - x^2 in eq(ii) Solving for x we get, x = 2.4 Length of AB = 2x = 4.8
Two circles of radii 4cm and 3cm intersect reach other in two points A and B. If the distance between their centres C and D is 5 cm. Find the length of the common chord AB.
Ans: Let AB and CD intersect at say O . If CO = x then DO = 5-x. Then by geometry AC^2 - x^2 = AD^2 - (5-x)^2 ie 4^2 - x^2 = 3^2 - (5-x)^2. On solving x = 3.2 So AO = sqrt(4^2 - 3.2^2) = 2.4 AB = 2xAO = 4.8 cm .
A boat sails down the river for 10 km and then up the river for 6 km. The speed of the river flow is 1 km/h. What should be the minimum speed of the boat for the boat for the trip from A to B for the trip to take a maximum of 4 hours. 10/x+1+10/x-1=4 16x-4=4x2-4 x=4 am i right?? pls correct if im wrong
A boat sails down the river for 10 km and then up the river for 6 km. The speed of the river flow is 1 km/h. What should be the minimum speed of the boat for the boat for the trip from A to B for the trip to take a maximum of 4 hours. 10/x+1+10/x-1=4 16x-4=4x2-4 x=4 am i right?? pls correct if im wrong
there is a mistake.... 10/X+1+6/X-1=4 now solve this you will get answer as 4km/hr...didn't read your post initially properly
The average of 13 papers is 40. The average of the first 7 papers is 42 and of the last seven papers is 35.find the marks obtained. PLease provide the solution for this type of problem.
he average of 13 papers is 40. The average of the first 7 papers is 42 and of the last seven papers is 35.find the marks obtained in the 7th Paper. Please provide the solution for this type of problem.
The average of 13 papers is 40. The average of the first 7 papers is 42 and of the last seven papers is 35.find the marks obtained. PLease provide the solution for this type of problem.
Hi ; For total:: a1+a2....a13 = (40)(13) For first 7 a1+a2...a7=(42)(7)........A For last 7 a7+a8...a13=(35)(7).........B A+B::: a7+13(40) =7(77) a7=19 Regards, Never Back Down
A contractor undertakes to build a wall in 50 days. He employs 50 people for the same. However after 25 days he finds that only 40% of the work is complete. how many more men will be employed complete the work in time? pls explain with steps
A contractor undertakes to build a wall in 50 days. He employs 50 people for the same. However after 25 days he finds that only 40% of the work is complete. how many more men will be employed complete the work in time? pls explain with steps
my take is 25....more men will be needed... 50*25=4W/10....as 50 people worrk for 25 days so woh itna work khatam kar lete hai now only 6W/10 work is remaning hence (50+N)*25=6W/10 solve both equation you will get your answer edit-calculation mistake forgot to divide it by 2....concept is correct
my take is 50....more men will be needed... 50*25=4W/10....as 50 people worrk for 25 days so woh itna work khatam kar lete hai now only 6W/10 work is remaning hence (50+N)*25=6W/10 solve both equation you will get your answer
1) A B and C had some amount of money . A gives half of his amount to B, who then gives half of what he now has to C. C gives half of what he now has to A, who, now has exactly what he started with. if sum of B's initial amount and twice C's amount is equal to Rs45, what was the amount (in rs) A started with.
2) In a class the ratio of the no of boys to that of girl is 7:3 . Each boy is given only a 50 paise coin and each girl is given a 75 paise coin (assuming 75 paose coin are available). the difference in the amount present with the boys and the girl is Rs 3.75. How many coins should the boys and girls exchange so that hte amount with the boys becomes twice the amount with the girls?
1) A B and C had some amount of money . A gives half of his amount to B, who then gives half of what he now has to C. C gives half of what he now has to A, who, now has exactly what he started with. if sum of B's initial amount and twice C's amount is equal to Rs45, what was the amount (in rs) A started with.
2) In a class the ratio of the no of boys to that of girl is 7:3 . Each boy is given only a 50 paise coin and each girl is given a 75 paise coin (assuming 75 paose coin are available). the difference in the amount present with the boys and the girl is Rs 3.75. How many coins should the boys and girls exchange so that hte amount with the boys becomes twice the amount with the girls?
Ans :
1) People : A B C Amount : x y z 1st iteration :x/2 y+x/2 z 2nd iteration:x/2 y/2+x/4 z+y/2+x/4 3rd iteration: x/2+z/2+y4+x/8 y/2+x/4 z/2+y/4+x/8
Now x=x/2+z/2+y4+x/8 On rearranging 3x = 4z + 2y ---> (1)
Now also y+2z = 45 ----> (2) now combining (1) and (2) we get 3x = 2 x 45 ie x =30 rs
2) Let the no of boys be 7x and girls be 3x. So amounts will be 7x x .5 = 3.5x for boys 3x x 0.75 = 2.25x for girls so difference 1.25x = 3.75 so x=3 ie no of boys = 21 and that of girls = 9. Now as per the exchange criteria, let p be the no of coins to be exchanged. So (21 - p)*0.5 + 0.75*x = 2{(9-p)*0.75 + 0.5*p} On solving p= 4. So no of coins to be exchanged = 4.
1) People : A B C Amount : x y z 1st iteration :x/2 y+x/2 z 2nd iteration:x/2 y/2+x/4 z+y/2+x/4 3rd iteration: x/2+z/2+y4+x/8 y/2+x/4 z/2+y/4+x/8
Now x=x/2+z/2+y4+x/8 On rearranging 3x = 4z + 2y ---> (1)
Now also y+2z = 45 ----> (2) now combining (1) and (2) we get 3x = 2 x 45 ie x =30 rs
2) Let the no of boys be 7x and girls be 3x. So amounts will be 7x x .5 = 3.5x for boys 3x x 0.75 = 2.25x for girls so difference 1.25x = 3.75 so x=3 ie no of boys = 21 and that of girls = 9. Now as per the exchange criteria, let p be the no of coins to be exchanged. So (21 - p)*0.5 + 0.75*x = 2{(9-x)*0.75 + 0.5*x} On solving x = 4. So no of coins to be exchanged = 4.
(21 - p)*0.5 + 0.75*x = 2{(9-x)*0.75 + 0.5*x} how u got (9-x) in this eqn
(21 - p)*0.5 + 0.75*x = 2{(9-x)*0.75 + 0.5*x} how u got (9-x) in this eqn
There are 21 boys with 50p coins each and 9 girls with 75p coins each. Now let them exchange x coins each => (21-x) * 0.5 + 0.75 * x --- 21-x coins of 50p remaining and x coins of 75p after exchanging.. and 2 * --- for girls they gave x of 75p coins..hence 9-x coins of 75p remaining and they recieved 50p coins from the boys hence 0.5x So, => (21-x) * 0.5 + 0.75 * x = 2 * Solve for x, => x = 4
Never Back Down.
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