A. Rs.17684
B. Rs.1684
C. Rs.2522
D. Rs.3408
Explanation:
A = 1600(21/20)3 = 2522
A. 3 hours 20 minutes
B. 3 hours 15 minutes
C. 3 hours 30 minutes
D. 3 hours
Let his usual speed be x kmph, the distance to his office be y km and his usual travel time be t hrs. y = xt = (5/6)x * (t+(40/60)) Solving the equation,
t = 3.33 hrs = 3 hrs 20 minutes
A. 3000m
B. 7500m
C. 3750m
D. 7000m
Average speed for Raj is: v = (2*9*10)/(9+10) = 180/19 km/hr = 50/19 m/s
Average speed for Rohit = 12 km/hr = 10/3 m/s
Now, v = d/t
As d is same, v*t = constant.
Let, t be the time taken by Rohit in seconds. Hence, time taken by Raj is (t +600)s
50/19 * (t + 600) = 10/3 * t
150*(t + 600) = 190t
t = 2250s
Let d be the distance between the house and the office
2d = 2250 * 10/3
2d = 7500 m
d = 3750m
A. 7
B. 6
C. 3
D. 4
E. None of these
Explanation:
Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.
A. 50m
B. 100m
C. 150m
D. 200m
Explanation:
Let length = 5x meters and breadth = 3x meters
Perimeter = 2 x (5x + 3x) m = 16x metres.
Perimeter = Total cost/Rate = (3000/7.50) m = 400 m.
16x = 400 or x = 25
(Length) – (breadth) = (5 x 25 – 3 x 25)m
= (2 x 25) = 50 m
A. 9
B. 8
C. 7
D. 6
1 to 9 = 9 * 9 = 81
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5th = 50 + 40 = 90 – 81 = 9
A. Rs 100000
B. Rs 11000
C. Rs 120000
D. Rs 170000
Explanation:
Let the sum be Rs x.
Then S. I = Rs (x × 8/100 × 2) = Rs 4x/25
C. I = Rs [x × (1 + 8/100)2 – x]
= Rs (x × 27/25 × 27/25 – x)
=Rs 104x/625 (C.I) – (S.I)
= Rs(104x/625 – 4x /25)
= Rs 4x /625
Therefore 4x/625 = 768
=> x = ((768 ×625)/4) = 120000
Therefore sum = Rs 120000