1) a person on an excercise regime decides to loose 500 cal. in a day. For every 30 minute of excercise he loses 80 cal. , but gains 25 cal. per cake that he eats. In the morning he excercise for a certain time and he eats a certain no of cakes. in the afternoon he excercise for 1.5 times the times he spends in the morning session ans eats 4 times the no of cakes he eats in the morning. Had he not eaten any cake, he would have exceeded his targate by 220 cal. if he actually falls by 180 cal. ,then, what is the no of cakes he has eaten in the evening?
2)A shopkeeper sold a certain no of toys. the no of toys as well as the price of each toy was a two digit no. By mistake, he reversed the digits of both the no of toys he sold and the price of each toy. As a result he found that his stock account at the end of the day showed 81 items more than it was actally was
a)find the actual no of toys sold b) if the faulty calculations show a total sale of Rs 882, find the actual selling price of each toy. c)using the info in the previous Question find his actual sales revenue in Rs
1) Hours Cakes Morning : x y Afternoon: 1.5x 4y
Total no of hrs = 2.5 x Without eating any cake he burns 220 calories extra, ie 720 calories. But when he eats the cakes also he burns only 380 calories instead of the 720. so 720 - 25x5y = 380 solving this you get a fraction. But it can't be. So vbhv..u sure its 4 times the cake?...say if its 3 then u can arrive at a whole number which would be 4 and so the answer will be 16. Correct me if I am wrong :)
2) The question is errenous because of the figure 81. If its any other multiple of 9 (below 81 of course) then it can be solved. Please check it.
Total no of hrs = 2.5 x Without eating any cake he burns 220 calories extra, ie 720 calories. But when he eats the cakes also he burns only 380 calories instead of the 720. so 720 - 25x5y = 380 solving this you get a fraction. But it can't be. So vbhv..u sure its 4 times the cake?...say if its 3 then u can arrive at a whole number which would be 4 and so the answer will be 16. Correct me if I am wrong :)
2) The question is errenous because of the figure 81. If its any other multiple of 9 (below 81 of course) then it can be solved. Please check it.
Sry Buddy i posted wrong Q actually i forgot to add one line...... 1) a person on an excercise regime decides to loose 500 cal. in a day. For every 30 minute of excercise he loses 80 cal. , but gains 25 cal. per cake that he eats. In the morning he excercise for a certain time and he eats a certain no of cakes. in the afternoon he excercise for double the time but eats thrice the number of cakes.In the evening he excercise for 1.5 times the times he spends in the morning session and eats 4 times the no of cakes he eats in the morning. Had he not eaten any cake, he would have exceeded his targate by 220 cal. if he actually falls by 180 cal. ,then, what is the no of cakes he has eaten in the evening? ans for this is 8 cake
Sry Buddy i posted wrong Q actually i forgot to add one line...... 1) a person on an excercise regime decides to loose 500 cal. in a day. For every 30 minute of excercise he loses 80 cal. , but gains 25 cal. per cake that he eats. In the morning he excercise for a certain time and he eats a certain no of cakes. in the afternoon he excercise for double the time but eats thrice the number of cakes.In the evening he excercise for 1.5 times the times he spends in the morning session and eats 4 times the no of cakes he eats in the morning. Had he not eaten any cake, he would have exceeded his targate by 220 cal. if he actually falls by 180 cal. ,then, what is the no of cakes he has eaten in the evening? ans for this is 8 cake
and that 2nd Q is right that is 81 only........
Ans)
1)
Hours cakes Morning x y AN 2x 3y Evening 1.5x 4y
Without eating cakes he burns 220 calories extra meaning he burns 720 calories total. When he eats cakes he burns 180 calories short. So 720 - 8y*25 = 500-180 =320 So 400 = 200y or y =2 So he eats 4y = 8 cakes in the evening
2)About this question i still have a doubt. I have solved a similar question before where it was 72 instead of 81. Anyways the approach is this: Let the no of toys be 10a+b and cost/toy = 10p+q. But due to reversal he ends up wid 81 extra as provided. so (10b+a) - (10a-b) = 9(b-a) = 81 or b-a = 9 NOw the problem comes. Since its a two digit number and the difference is 9 the only possible solution set for (a,b) is (0.9). which is not possible because in that case the nimber will be 09 which is not possible. Say its 72 then b-a = 8 ie (a.b) = (1,9) or say if its 63 then (b-a) = 7 ie (a,b) = { (8,1) , (9.2) } . Correct me if I am wrong 😃
1) a person on an excercise regime decides to loose 500 cal. in a day. For every 30 minute of excercise he loses 80 cal. , but gains 25 cal. per cake that he eats. In the morning he excercise for a certain time and he eats a certain no of cakes. in the afternoon he excercise for 1.5 times the times he spends in the morning session ans eats 4 times the no of cakes he eats in the morning. Had he not eaten any cake, he would have exceeded his targate by 220 cal. if he actually falls by 180 cal. ,then, what is the no of cakes he has eaten in the evening?
2)A shopkeeper sold a certain no of toys. the no of toys as well as the price of each toy was a two digit no. By mistake, he reversed the digits of both the no of toys he sold and the price of each toy. As a result he found that his stock account at the end of the day showed 81 items more than it was actally was
a)find the actual no of toys sold b) if the faulty calculations show a total sale of Rs 882, find the actual selling price of each toy. c)using the info in the previous Question find his actual sales revenue in Rs
2)a grocer uses a weighing balance in which one pan weights 0.5 kg and the other 0.75 kg. He puts a certain quantity of food grains in 0.5 kg pan and finds the weights(in kg) as a two digit no. however as the customer insists, he puts it in 0.75 kgs pan. Now the indicated weight is 9.5 kg more than the weights which is obtained by reserving the digits of previous weight. Which of the following cannot be the actual weight(in kg) of the food grains?
2)a grocer uses a weighing balance in which one pan weights 0.5 kg and the other 0.75 kg. He puts a certain quantity of food grains in 0.5 kg pan and finds the weights(in kg) as a two digit no. however as the customer insists, he puts it in 0.75 kgs pan. Now the indicated weight is 9.5 kg more than the weights which is obtained by reserving the digits of previous weight. Which of the following cannot be the actual weight(in kg) of the food grains?
multiplying the eqn 1) with 2 and eqn 2) with 3 we get
Two products are manufactured sequentially on two m/cs. The time available on each m/c is 8 hrs per day and may be increased by 4 hrs of overtime, if necessary, at an additional cost of Rs 1000 per hour. suggest Optimum production shedule.... production rate from machine A 5units/hr & 5 units /hr & from machine B 8units/hr & 4 units /hr price per unit 1200 & 1275 for product A & B receptively. pls help me out
Two products are manufactured sequentially on two m/cs. The time available on each m/c is 8 hrs per day and may be increased by 4 hrs of overtime, if necessary, at an additional cost of Rs 1000 per hour. suggest Optimum production shedule.... production rate from machine A 5units/hr & 5 units /hr & from machine B 8units/hr & 4 units /hr price per unit 1200 & 1275 for product A & B receptively. pls help me out
kindly provide options...with options it can be solved quite easily...
2)a grocer uses a weighing balance in which one pan weights 0.5 kg and the other 0.75 kg. He puts a certain quantity of food grains in 0.5 kg pan and finds the weights(in kg) as a two digit no. however as the customer insists, he puts it in 0.75 kgs pan. Now the indicated weight is 9.5 kg more than the weights which is obtained by reserving the digits of previous weight. Which of the following cannot be the actual weight(in kg) of the food grains?
hi kindly provide options because without options i am not able to finalise my answer my approach- let X kg weight is been put now X+.5kg=ab(assume it as ab because it said it is an 2 digit number) and by 2nd procedure X+.75kg=ba+8.75 now we can solve both the above equation then i got ab-ba=9.25 this means the options in which difference between the obtained weight and reverasal of digit is not equal to 9.25 that won't be the answer... i hope this will help...kindly correct me if i am wrong somewhere
2)a grocer uses a weighing balance in which one pan weights 0.5 kg and the other 0.75 kg. He puts a certain quantity of food grains in 0.5 kg pan and finds the weights(in kg) as a two digit no. however as the customer insists, he puts it in 0.75 kgs pan. Now the indicated weight is 9.5 kg more than the weights which is obtained by reserving the digits of previous weight. Which of the following cannot be the actual weight(in kg) of the food grains?
a)43.25 b)36.25 c)41.5 d)more than one of the above. ANS d)more than one of the above.
1)A man wins a bet and the amount gets doubled. Thrice the numerical value of the sum of his age and the numerical value of the original amount with him is 200 more than the numerical value of the sum of his age and that of the money after the winning. He bets the new amount and it doubles again. Now twice the sum of the numerical value of his amount before the first bet and the and the numerical value of the age in years is 100 less than the sum of the numerical value of his age in years and that of the final amount after the second winning. Find the original amount with the man.(all accounts are counted in rupees and age in years). Ans 80
2)Seven men and 12 women are divided into two groups each containing at least one man and one women . If the number of men in the first group is more than the no of women in the group, what is the least no by which the no of women exceed the no of men in the second group? Ans 6
1)A man wins a bet and the amount gets doubled. Thrice the numerical value of the sum of his age and the numerical value of the original amount with him is 200 more than the numerical value of the sum of his age and that of the money after the winning. He bets the new amount and it doubles again. Now twice the sum of the numerical value of his amount before the first bet and the and the numerical value of the age in years is 100 less than the sum of the numerical value of his age in years and that of the final amount after the second winning. Find the original amount with the man.(all accounts are counted in rupees and age in years). Ans 80
2)Seven men and 12 women are divided into two groups each containing at least one man and one women . If the number of men in the first group is more than the no of women in the group, what is the least no by which the no of women exceed the no of men in the second group? Ans 6
1. Let the age is a and the initial money is x then by the given conditions... 3(a+x) - (a+2x) = 200 2a + x = 200 ---- (1)
(a+4x) - 2(a+x) = 100 2x - a = 100 ---- (2) solving (1) & (2) we get x = 80
2. Let groups be A and B In A: 6M and 5W In B: 1M and 7W Difference in B is 6
(2n!) / (n!)^2 = (2n)Cn => Number of ways in which 'n' things can be chosen out of '2n' things => Always going to be an integer
=> (2n)! is divisible by (n!)^2
An alternative:
1*2*3*........(n-1)*n is divisible by n! now we have the remaining series as (n+1)*(n+2)*........*(2n) Note that there are 'n' numbers left behind and product of 'n' consecutive numbers is divisible by n! therefore the series is divisible by n!*n! 😃
