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Linear Equations ( 21 to 25 ):
APTITUDE QUESTIONS (LINEAR EQUATIONS):
21. If x + y = 7 and 3x – 2y = 11, then
A) x = 2, y = 5
B) x = 5, y = 5
C) x = 5, y = 2
D) x = 0, y = 3
E) None of these
View Answer
Answer is C
Explanation:
Multiply eq. (i) by 2 and adding
2 * (x + y = 7) …(i)
3x – 2y = 11 …(ii)
2x + 2y = 14
3x – 2y = 11
————–
5x = 25 => x = 5
Put in eq (i) 5 + y = 7 => y = 2
So, x = 5 and y = 2
22. The solution of the system of linear equations 0.4x + 0.3y = 1.7 and 0.7x – 0.2y = 0.8 is
A) x = 3, y = 2
B) x = 2, y = -3
C) x = 2, y = 3
D) x = -2, y = 3
E) None of these
View Answer
Answer is C
Explanation:
Here, (4x/10) + (3y/10) = (17/10) and (7x/10) – (2y/10) = (8/10)
Or 4x + 3y = 17 …(i)
7x – 2y = 8 …(ii)
Solving eq (i) and (ii)
We get x = 2 and y = 3
23. A horse and two cows together cost Rs.680. If a horse costs Rs.80 more than a cow, then the cost of horse is
A) Rs.170
B) Rs.280
C) Rs.200
D) Rs.220
E) None of these
View Answer
Answer is B
Explanation:
Let cost of one horse be Rs. x and Cost of one cow be Rs. Y
So, x + 2y = 680 …(i)
x – y = 80 …(ii)
subtracting eq. (i) and (ii)
3y = 600 => x = 200
∴ cost of one horse = 200 + 80 = 280
24. Sunita has 10 paise and 50 paise coins in her purse. If the total number of coins is 17 and their total value is Rs. 4.50, then the number of 10 paise coins is
A) 9
B) 7
C) 10
D) 5
E) None of these
View Answer
Answer is C
Explanation:
Let number of 10 paise coins be x and number of 50 paise coins be y
Then, x + y = 17 …(i)
and 10x + 50y = 450 …(ii)
from eq (ii) x + 5y = 45 …(iii)
subtracting eq (i) and (iii), we get
4y = 28
y = (28/4) = 7
Number of 10 paise coins = y = 17 – x = 17 – 7 = 10
25. Korna has pens and pencils which together are 40 in number. If she had 5 more pencils and 5 less pens, the number of pencils would have become 4 times the number of pens. Then, the original number of pencils Korna had:
A) 19
B) 27
C) 13
D) 17
E) None of these
View Answer
Answer is B
Explanation:
Let the original number of pens be x and original number of pencils be y.
As, x + y = 40 …(i)
and (y + 5) = 4(x – 5) …(ii)
From eq.(i) y = 40 – x
Put in eq. (ii) (40 – x) + 5 = 4(x – 5)
=> 45 – x = 4x – 20
=> -5x = -65
=> x = 13
∴Original number of pencils = 40 – 13 = 27