Directions (1-5):What approximate value should come in place of the question-mark (?) in the following question?
1. 5554.999 ÷50.007 = ?
(1) 110 (2) 150
(3) 200 (4) 50
(5) None of these
2. (18.001)3 =?
(1) 5830 (2) 5500
(3) 6000 (4) 6480
(5) None of these
3. 23.001 ×18.999 × 7.998 = ?
(1) 4200 (2) 3000
(3) 3500 (4) 4000
(5) None of these
4. 9999 ÷99÷9 = ?
(1) 18 (2) 15
(3) 6 (4) 11
(5) None of these
5. 22.005 % of 449.999 = ?
(1) 85 (2) 100
(3) 125 (4) 75
(5) None of these
Directions (6- 10): In each of these questions, two equations are given. You have to solve these equations and find out the values of x and y and give answer—
(1) If x< y (2) If x > y (3) If x≤ y
(4) If x≤ y (5) If x = y
6. I. 16x2 + 20x + 6 = 0
II. 10y2 + 38y + 24=0
7. I. 18x2 + 18x + 4 = 0
II. 12y2 + 29y +14 =0
8. I. 8x2 + 6x = 5
II. 12y2 – 22y + 8 =0
9. I. 17x2 + 48x = 9
II.13y2=32y -12
10. I. 4x + 7y =209
II. 12x -14y = -38
Answers with Solutions(1-5)
1. (1) ? = 5554.999 ÷50.007
= 5555 ÷ 50
= 110 (App.)
2. (1) ? = (18.001)3
= (18)3 = 5832
= 5830 (App.)
3. (3) ? = 23.001 × 18.999 ×7.998
=23 ×19 ×8 = 3496
= 3500(App.)
4. (4) ? (9999 ÷ 99) ÷9
= (101) ÷ 9
= 11.2 = 11(App.)
5. (2) ? = 22.005/100of 449.999
= 22/100 of 450 =99
= 100 (App.)
Solutions (6-10):
6.(2) I.16x2 +20x+6 =0
8x2+10x + 3 =0
(4x+3)(2x+1)= 0
x = – 3/4 or – 1/2
II.10y2+ 38y 24 =0
5y2 +19y +24 =0
(y+ 3) (5y+4) =0
y =− 3 or -4/5
so x > y
7. (4) I. 18x2 +18x + 4
9x2 +9x+ 2 =0
(3x + 2) (3x +1) = 0
x = -2/3 or -1/3
II. 12y2 + 29y + 14 =0
(3y +2) (4y +7) = 0
y = -2/3 or -7/4
so x ≥ y
8. (3) I. 8x2 +6x – 5 =0
(4x +5) (2x – 1) = 0
x = -5/4 or 1/2
II.12y2 −22y +8 =0
6y2 – 11y +4 =0
(2y – 1) (3y −4) =0
y = 1/2 or 4/3
so x ≥ y
9. (1) I. 17x2 + 48x −9 =0
(x+3) (17x −3) =0
x = − 3 or 3/17
II. 13y2−32y +12 =0
(y −2) (13y −6) =0
Y = 2 or 6/13
So X < Y
10.(5) I. 4x + 7y =209
10.(5) I. 4x + 7y =209
II. 12x −14y = − 38
8x + 14y = 418
On addition, 20x =380
x =19
And y = 19
x = y
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