221. Let
number of boys = X. then, number of girls = (60 – X).
X (60 –X ) +
(60 –X ) X = 1600 60X - X² +
60X - X²
= 1600
2X² -
120X + 1600 = 0 X² -
60X + 800 = 0.
(X – 40) ( X – 20)
= 0 x = 40 or X = 20.
So, we are not
definite. Hence, data is inadequate.
222. Let
the distance covered by taxi be X k. then, distance covered by car = (80 – X)
km.
1.5X + 0.5
(80 –X ) = 50 X = 50 – 40 = 10km.
223. Let
the distance covered by taxi be X. number of incorrect answers = (60 –X)
4X =
(60 – X)= 130 5X = 190 X = 38.
224. Let
number of matches lost = X. then, number of matches won = X + 3.
2 – (X + 3 )
– X = 23 X = 17.
Hence, total
number of matches played =X + 9x + 3) =2X = 3= 37.
225. Let
the number of 20-paise coins be X. then, number of 25 –paise coins =(324 –X).
0.20 x X +
0.25 ( 324 – X) = 71 20X + 25 (324 – X)
= 7100
5 X =
1000 X = 200.
Hence, number
of 25-paise coins = (324 – X) = 124.
226. Let
number of notes of each denomination be X.
Then, x + 5X
+ 10X = 480 16X = 480 X = 30.
Hence, total
number of notes = 3X = 90.
227. Original
share of 1 person = 1/8.new share of 1 person = 1/7
Increase =
(1/7 – 18) = 1/56.
Required
fraction = (1 / 56) / (1/8) = (1/56 x 8) = 1/7.
228. Let
total number of sweets be X. Then,
X/140 –X175
= 4 5X – 4X = 3 x 700 X = 2800.
229. Let
the number of persons be x. then,
96/ X – 4 - 96 / X =4 1/Z – 4
- !/z = 4/96 Z – (Z-4) / Z (Z_4) = 1/24.
X²
+ 5X – 150 = 0 ( X + 15) (X – 10) =
0 Z = 10.
230. Let
the number of balls purchased be x.
Then, 450 / X – 450 / X = 5 = 15 1Z – 1/Z + 5 = 15/ 450
X + 5 – X / X (X + 5) = 1/30
X² + 5X – 150 = 0 ( X + 15) ( X – 10) = 0 Z = 10.
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