A. -5, 3
B. 3, 5
C. -3, 5
D. -3, -5
Explanation:
x2 + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = -5.
A. 25 hrs
B. 28 hrs
C. 20 hrs
D. 35 hrs
Explanation:
1/x + 1/(x + 10) = 1/12
x = 20
A. Rs.10,840
B. Rs.10,720
C. Rs.10,560
D. Rs.10,280
E. None of these.
Explanation:
Let the rate of interest be R% p.a.
4400{[1 + R/100]2 – 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
A. does not change
B. decreased by 1 %
C. increased by 1 %
D. None of these
Explanation:
Let the given number be x
Increased Number = (110% of x)
= (110/100 × x) = (11x/10)
Finally reduced number = (90 % of 11x/10)
= (90/100 × 11x/10) = 99x/100
Decrease = (x – 99x/100) = x/100
Decrease % = (x/100 × 1/x × 100)% = 1 %
A. 48
B. 34(2/7)
C. 44
D. 45
(1/A)+(1/B) = (1/20) (1/B)+(1/C) = (1/30) (1/A)+(1/C) = (1/40) Solving the three equations,
A=48
A. 0.1
B. 0.2
C. 0.3
D. 0.4
Explanation:
As numerator is of the form a2 – b2 = (a + b) (a – b), so the given expression simplifying becomes
(0.92 – 0.82)/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
A. Rs. 1000
B. Rs. 1200
C. Rs. 1320
D. None of these