Arrange 9 integers from 1 to 9 without repetition in a 3*3 size square box (total 9 small boxes) in such a manner that sum of 3 integers (S) of 3 horizontal , 3 vertical and 2 diagonal lines are same. S will be
A father said to his son that his (father) age was 7 times his (son) age and his mothers age was 6 times his age 15 years before. If age of father is 5 years more than that of mother, then sum of their ages (all the three) at present will be
let age of father = x
that of mother = x-5
and that of son = y
now according to the given conditions
x-15 = 7(y-15) or x-7y = -90
x-5-15 = 6(y-15) or x-6y = -70
on solving we will get y = 20 and x = 50
so sum = 50 + 45 + 20 = 115
A Milk dealer has 125 liters, 150 liters and 275 liters of three different kinds of milk. Find the least number of containers of equal size required to store all the milk without mixing.
Since minimum number of containers are required, the size of the containers is greatest. Also, the containers in three cases are of equal size.
The size of the containers is the H.C.F. of 125, 150 and 275 which is 25.
Now, the number of containers required for storing the milk = (125+150+275) ÷ 25 = 22
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%