When a cylindrical tank is 3/7 full, an inlet pipe with a capacity of 4928 liters per hour fills it completely in 4 hours. If the height of a tank is two times its diameter, then the radius of a tank is:
OPtion
1) 2.8 m
2) 1.9 m
3) 2.1 m
4) 2.4 m
5) 1.4 m
6) 2.0 m
7) 1.8 m
8) 3.5 m
9) 1.2 m
10) None of these
Solution
4/7 volume of a tank = 4928x4 = 19712 liters
.'. Total capacity of tank = (7/4)x19712 = 34496 liters
⇒ Volume of a cylindrical tank = πr²h = 34496 litres = 34496000 cm³
Given, h=2xDiameter=2x(2r)=4r
.'. πr²(4r) = 34496000
(22/7)x4xr³ = 34496000
r³ = 2744000
r = 140 cm = 1.4 m
Correct Option 5)
20 women work for 4 days on a certain project and complete one-tenth of it. 15 men work for the next 'X' days and complete one-fifth of it, If after (4+X) days, 20 women and 15 men work together and complete the remaining work in 12 days, then how many days will 15 men take to complete the whole work ?
20 Women's 1 day work=1/(4*10)=1/40
15 men's 1 days work=1/(X*5)=1/(5X)
After total (4+X) days, work comleted=1/10 + 1/5=3/10 and remaining work=7/10
As 20 women and 15 men complete the remaining work in 12 days, so their 1 day work=(1/12)*(7/10)=7/120
Comparing 1 day's work of 20 women and 15 men=1/40 + 1/(5X)=7/120
(X+8)/(40X)=7/120
Solving, we get X=6
.'. 1 day's work of 15 men=1/(5*6)=1/30
Hence, 15 men can complete the whole work in 30 days.
Correct Option 2)
A boy started for a school which is 12 km from his home by bicycle at a speed of 8 km/hr but after 15 minutes he remembers that he forgot one of the book at home, so he increases his speed and returns back to home and also reaches to school in time. The percentage increase in his speed is
Usual time to reach school=12/8=1.5 hrs
Distance covered in 15 min. at a speed of 8 km/hr=(15/60)*8=2 km
So, he comes back 2 km to home and again return to school at a speed of say 'x' km/hr.
Then, usual time of 1.5 hrs = 15 min. + Time taken for the journey of (2+12) km with increased speed.
1.5 = (1/4) + (14/x)
x=11.2
Therefore % increase in the speed = [(11.2 - 8)/8]*100 = 40%
Correct Option 7)
Six friends, A, B, C, D, E and F are sitting in two rows, three in each. D is not at the end of any row. E is a neighbour of C. C and F are sitting diagonally opposite to each other. A is second to the left of F. Who among the following three friends are in the same row?
According to the information given, the sitting arrange of A,B,C,D,E and F row-wise is:
C - E - B
A - D - F
So, from the given options Only B, E and C will be in one row.
Correct option 8)
When letters from A to Z are numbered from 1 to 26 respectively.
The sequence of position of first letter Z(26), V(22), R(18), N(14),..decrease by 4, so next letter at position 10=J
The second letter position C(3), E(5), I(9), O(15),... Increasing difference of 2, 4, 6,.., so next difference will be 8 and position 23=W
Two-digit number after letters is obtained by the sum of positions of first two letters i.e. ZC(26+3=29), VE(22+5=27), RJ(18+9=27), NO(14+15=29), so next number will be JW(10+23=33)
Hence correct cluster is JW33
Option 8)
A bus can travel 66.66% faster than a car. Both start from point X at the same time and reach point Y which is 114 kms away from X at the same time. On the way, however, the bus lost about 48 minutes while stopping at the stations. The speed of the bus (in km/hr) is:
Let speed of the car=x km/hr, then speed of the bus=(166.66/100)*x=1.6666x=5x/3
Time taken by bus while travelling + Time of stoppages = Time taken by car
114/(5x/3) + 48/60 = 114/x
(342 + 4x) / 5x = 114/x
4x = 228
x = 57
.'. Speed of the bus = 5x/3 = 5*57/3 = 95 km/hr
Correct Option 6)
Chirag and Bhavesh working together on certain job takes 7½ days to complete it. Chirag alone takes 8 days less than Bhavesh alone takes. In how many days Akash alone can complete the same work, if he is 60% more efficient than Chirag?
In the Meena Bazar ladies purchases with a fair bargaining, so the shopkeepers markup the prices too much. One of the shopkeeper marked up an article at Rs. X expected huge profit if it is sold on the marked price. But a lady purchased it at X/2 with his fine bargaining skills, so the expected profit of the shopkeeper diminished by 66.66%. What is the percentage discount fetched by the customer through bargaining?
Let CP=C, Given MP=X .'. Profit without any discount=(X-C)
With discount, when SP=X/2, then Profit=X/2 - C
Given, profit after discount is 66.66 less or is 33.33 of the original profit.
⇒ (33.33/100)(X-C) = (X/2 - C)
(1/3)*(X-C) = (X-2C)/2
X=4C
⇒ MP is 4 times the CP.
When CP=100, MP=400 and after bargain SP=200
.'. % Discount fetched by customer = [(400-200)/400]*100 = 50%
Correct Option 6)
For an Arithmetic Progression (AP), the sum of first four terms and first eight terms is 288. If the sum of first twelve terms is 480, then sum of first 15 terms of this AP is:
Length, breadth and height of a room are in the ratio 6:5:3. If a new room is constructed where length is increased by 40%, breadth by 20% but the area of all four wall is kept same as original room, then change in volume of a new room compared to original is
For the original room, let Length=6, Breadth=5 and Height=3, then its area of 4 walls=2(6*3 + 5*3)=66 unit² and Volume=6*5*3=90 unit³
For a new room, Length=1.4*6=8.4, Breadth=1.2*5=6. If height is 'x' then area of 4 walls=2(8.4x + 6x)=28.8x unit²
As the area of walls of both rooms is same .'. 28.8x=66 ⇒ x=165/72
So, the volume of new room=8.4*6*(165/72)=115.5 unit³
Hence percentage increase in volume=[(115.5-90)/90]*100=28.33%
Correct Option 9)