Ex.21. if 1/8 of a pencil is black, 1/2 of the remaining is white and the remaining 31/2 cm is blue, find the total length of the pencil.
Sol. let the total length of the pencil be X cm. Then,
Black part = (X/8) cm. Remaining pat =(x – x/8) cm = (7X/8) cm. white part = (1/2 x 7X/8) .cm
(7x/16) cm. remaining part = (7X / 8 – 7X/16) cm = 7X/16 cm.
Hence, total length of the pencil = 8 cm.
Ex. 22. In a certain office, 1/3 of the workers are women, ½ of the women are married and 1/3 of the married women have children. If ¾ of the men are married and 2/3 of the married men have children, what parts of workers are without children?
Sol. let the total number of workers be X then.
Number of women = X/3 and number of men = (X = X/3) = 2x/3
Number of women having children = 1/3 of 1/2 of e/x = X/18.
Number of men having children. = 2/3 of ¾ of 2X/3 =X/3.
Number of workers having children = (x/18 + X/3 ) =7X 18.
Workers having no children = (x-7X/18)= 11X/18 = 11/18 of all workers.
Ex.23.a crate of manages contains one bruised mango for every 30 mangoes in the crate. If 3 out every 4 bruised manages are considered unsalable, and there are 12 unsalable mangoes in the crate, then how many mangoes are there in the create?
Sol. let the total number of mangoes in the crate be x. then,
Number of bruised mangoes = 1/30 X.
Number of unsalable mangoes = (3/4 x 1/30 X = 1/40 x.
1/40 X = 12 or X = (12 or X = (12 x 40) = 480.
Ex. 24. A train starts full of passengers. At the first station it drops one-third of the passengers and takes 280 more. At the second station, it drops one – half of the new total and takes s12 more. On arriving at the third station, it is found to have 248 passengers. Find the number of passengers in the beginning.
Sol. let the number of passengers in the beginning be X.
After 1st station, number of passengers = 1/2 (2X/3+ 280 ) = (2X/3 + 280).
After 2nd station, number of passengers = ½ (2x/3 + 280) + 12.
1/2(2x/3 + 280) + 12 = 248 2X/3 + 280 = 2 x 236 2X / 3 = 192.
X = (192 x 3/2) = 288.
Ex.25.if a² + b² =117 and a b = 54, then find the value of a+ b/a- b
Sol. (a + b)² =a² + b² + 2ab = 117 +2 x 54 = 225 a + b =15.
(a – b)² = a² + b² -2ab = 117 -2 x 54 = 9 a – b=3.
A+ b/a – b 15/3 = 5.
Ex. 26. Find the value of (75983 x 75983 -45983 x 45983) / 30000)
Sol. Given expression=(75983)² - (45983)²/(75983 – 45983) = (a² - b²/(a –b), where a = 7 75983, b =45983
= (a +b) (a – b) = (a + b) = = (75983 + 45983) =121966.
Ex. 27. Fin the value of (343 x 343 x 33 – 113 x 113 x 113)/343 x 343 +343 x 113 + 113 x 113)
Sol. given expression = (a³ -b³)/(a³ + a b + b²) ,where a = 343, b = 113.
= (a – b) = 343 – (343 -113) = 230.
Ex.28. village X has a population of 688000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year. In how many years will the population of the two villages be equal?
Sol. let the population of villages X and Y be equal after p years.
Then, 68000 – 1200p = 42000 + 800 p 2000 p =26000 p =13.
So, their population will be equal after 13 years.
Ex. 29. From a group of boys and girls, 15 girls leave. There are then left 2 boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. Find the number of girls in the beginning.
Sol. let at present there be X boys. Then, number of girls at present = 5x.
Before the boys had left: number of boys = x + 45 and number of girls = 5X.
X + 45 = 2 x 5X 9X =45 X = 5.
Hence, number of girls in the beginning = 5X 9X = 45 X = 5.
Hence, number of girls in the beginning 5X +15 = 25 + 15 =40.
Ex. 30. An employer pays Rs. 20 for each day a worker works, and forfeits Rs.3 for each day he is ideal. At the end of 60 days, a worker gets Rs. 280.For how many days did the worker remain idle?
Sol. supposes the worker remained idle for x days. Then, he worked for (60 x X) days.
20 (60 – X) – 3X = 280 1200 -23X = 280 23X = 920 X=40.
So, the worker remained idle for 40 days.
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