Ex. 31. Kiran had 85 currency notes in all, some of which were of Rs. 100 denomination and the remaining of Rs. 50 denomination. The total amount of all these currency notes was Rs. 5000.How much amount did she have in redenomination of Rs.50.
Sol. let the number of 50-rupee notes be X.
Then, the number of 100-rupee notes = (85 – X).
50X+ 100 (85 –X) 5000 X + 2 (85 –X) = 100 X = 70.
So, required amount = Rs.(50 x 70) =Rs. 3500.
Ex.32. when an amount was distributed among14 boys, each of them got Rs.80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount?
Sol. let the total amount be Rs. X. then.
X/14 –x/18 =80 2X/12 = 80 X= 63 x 80 = 5040.
Hence, total amount =Rs. 5040.
Ex.33. Mr. Bhaskar is on tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs.3.for how many days is Mr. Bhaskar on tour?
Sol. supposes Mr. Bhaskar is on tour for X days. Then,
360/X – 360/X + 4 = 3 1 / X – 1/X + 4 = 1/120 X (X + 4) =4 x 120 = 480.
X²+ 4 x – 480 = 0 (X + 24) (X – 20) = 0 X=20.
Hence, Mr.Baskar is on tour for 20 days.
Ex. 34. Two pens and three pencils cost Rs.86. four pens and a pencil cost Rs. 112. Find the cost of a pen and that of a pencil.
Sol. let the cost of a pen and a pencil be Rs. X and Rs. Y respectively.
Then, 2X + 3y = 86. (I) and 4X + y = 112.. (II)
Solving (I) and (II), we get: X= 25 and y =12.
Ex. 35. , Arun and Sajal area friends. Each has some money. If arun gives Rs. 30 to Sajal, then Sajal will have twice the money left with Arun. But, if Sajal given Rs. 10 to Arun, then Arun will have thrice as much as is left with Sajal. How much money does each have?
Sol. suppose arun has Rs. X and Sajal has Rs./ Y. then,
2 (X – 3) = y + 30 2x – y= 90 ... (I)
And X + X+ 10 =3 (y – 10 X – 3y = -40 ..(II)
Solving (I) and (II), we get: X = 62 and y+34.
Arun has Rs. 62 and Sajal has Rs. 34.
Ex. 36. In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers.
Sol. let the number of keepers be X. then,
Total number of heads = (50 +45 + 8 X) = (103 + X).
Total number of feet = (45 +8) x 4 + (50 + X ) x 2 = (312 + 2X).
(312 +2X – (103 + X) = 224 X=15.
Hence, number of keepers = 15.
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