A. Rs. 2000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 20,000
principal = Rs. ( 100 X 5400/12 X 3 ) = Rs. 15000
A. Rs. 2000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 20,000
principal = Rs. ( 100 X 5400/12 X 3 ) = Rs. 15000
A. 3%
B. 4%
C. 5%
D. 6%
S.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = ( 100 X 3000/12500 X 4 )% = 6%.
A. 5%
B. 8%
C. 12%
D. 15%
S.I. for 3 years = Rs. (12005 – 9800) = Rs. 2205. S.I. for 5 years = Rs. ( 2205/3 X 5 ) = Rs. 3675. So Principal = Rs. (9800 – 3675) = Rs. 6125. Hence, rate = ( 100 X 3675/6125 X 5 X 5 )% = 12%.
A. 6 years 3 months
B. 7 years 9 months
C. 8 years 3 months
D. 9 years 6 months
A. 1 : 3
B. 1 : 4
C. 2 : 3
D. Data inadequate
Let the principal be P and rate of interest be R%. So Required ratio = [(P x R x 6/100) / (13 x R x 9/100)] = 6PR/9PR = 6/9 = 2 : 3
A. 10%
B. 10.25%
C. 10.50%
D. None of these
Let the sum be Rs. 100. Then, S.I. for first 6 months = Rs. ( 100 X 10 X 1/100 X 2 ) = Rs. 5. S.I. for last 6 months = Rs. ( 105 X 10 X 1/100 X 2 ) = Rs. 5.25. So, amount at the end of 1 year=Rs.(100 + 5 + 5.25) = Rs. 110.25 Effective rate = (110.25 – 100) = 10.25%.
A. 5 metres
B. 8 metres
C. 10 metres
D. 12 metres
A. 30°
B. 45°
C. 60°
D. 90°
Let AB be the tree and AC be its shadow.
Let ∠ACB = θ.
Then, AC/AB = √3
Cot θ = √3
θ = 30°
A. 149 m
B. 156 m
C. 173 m
D. 200 m
Let AB be the tower. Then, ∠APB = 30° and AB = 100 m, AB/AP = tan 30° = 1/√3 AP = (AB X √3)= 100√3 m. = (100 X 1.73) m = 173 m.
A. 50°
B. 60°
C. 70°
D. 80°
Let AB be the pole and AC be its shadow. Let angle of elevation, ∠ACB = θ. Then, AB = 2√3m, AC = 2 m. tan θ = AB/AC = 2√3/2 = √3 θ = 60°. So,the angle of elevation is 60°