A and B take part in a 200 m race. A run at 5.4 Km/h. A gives B a start of 12 m and still beat him by 6 sec. Speed of B is:
OPtion
1. 5.15 Km/h
2. 4.14 Km/h
3. 4.25 Km/h
4. 4.40 Km/h
5. 4.51 Km/h
6. 5.20 Km/h
7. 4.85 Km/h
8. 4.95 Km/h
9. 5.11 Km/h
10. None of the above
Solution
A’s speed = (5.4x 5/18) m/sec = 27/18 m/sec
Time taken by A to cover 200 m
= (200 x 18/27) sec = 133.33 seconds.
∴ B covers 188 m in 133.33 + 6 = 139.33 seconds.
B’s speed = {(188/139.33) x (18/5)} Km/h
= 4.85 Km/h
The value of a machine is ₹7000. It decreases by 10% during the first year, 15% during the second year and 20% during the third year. What will be the value of the machine after 3 years?
Here, A = 7000, x = –10, y = –15 and z = –20.
∴ Value of the machine after 3 years
= 7000 (1- 10/100)(1-15/100)(1-20/100)
= (7000 x 90 x 85 x 80)/100 x 100 x 100
= 4,28,40,00,000 / 10,00,000
= ₹4284
Here, x = 30 and y = 80.
∴ First number = {(100 + ×)/ (100 + y)} x 100% of the second
= {(100 + 30)/ (100 + 80)} x 100% of the second
= (130 / 180)/100
= 72%
let age of x is a and y is b
now age of x is thrice age of y
a=3b;
now before three years (a-3)=4(b-3)
=>3b-3=4b-12
b=9
=>a=3*9=27
sum of ages is 27+9=36
The distance between two stations is 100 km. 2 cars start from A and B at the same time and return to the original station with speed 48 km/hr and 64 km/hr. The time gap between their meeting twice will be
Total distance covered between their meetings is 200 KM travelled together by 2 cars.
total combined speed = 64+48 = 112
Time take to cover 200KM with 112 KM/H speed is
200/112
=25/ 14 Hour
according to the given conditions
possible values of x, y and z are
x = 2, 3, 4
y = 5, 6, 7
z = 11, 12, 13
so possible combinations are
(2, 5, 12), (2, 6, 11), (3, 5, 11)