Tuesday, 1 September 2015

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Daily quiz- 01 sep 2015

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    Daily quiz- 01 sep 2015      

1. If a shopkeeper sales a cow at 16(2/3) % profit and a buffalo at 12(1/2)% loss, he gains Rs. 30 in whole transaction. If he sales a cow at 12.5 % loss and the buffalo at 16(2/3)% profit he neither gains nor losses. Find sum of their prices ?
(a)  620
(b)  720
(c)  725
(d)  625
(e)  None of these
 solution:- (b) Rs. 720
Let the price of Cow and Buffalo be X and Y respectively.
Now, ATQ
7X/6 + 7Y/8 = X+Y+30
== > 28X + 21Y = 24X+24Y+720
== > 4X – 3Y = 720…(i)
Also, 7X/8 + 7Y/6 = X+Y
== > -3X + 4Y = 0 …(ii)
Adding (i) and (ii), we get
X+Y = 720

2. When a train is running at a speed of 7/11 of its usual speed, it takes 22 hours to cover a certain distance. Find the actual time to cover the certain distance ?
(a)  14 hours
(b)  20 hours
(c)  8 hours
(d)  12 hours
(e)  None of these
solution:- (a) 14 Hours
New Speed = 7/11 of usual speed
New Time Taken = 11/7 of usual time.
22 = 11/7 of usual time
Usual Time = 22*7/11 = 14 hours.

3. Three taps A, B and C can file a tank in 12 hours, 15 hours and 20 hours respectively. First tap opened at 7 am., 2nd tap opened at 10 am. and 3rd tap opened at 11 am. At what time the tank will fill completely ?
(a)  3PM
(b)  2PM
(c)  6PM
(d)  1PM
(e)  None of these


solution:-. (b) 2PM
Let the total work be 180
Efficiency of A = 180/12 = 15/hr.
Efficiency of B = 180/15 = 12/hr.
Efficiency of C = 180/20 = 9/hr.
Now, A worked from 7AM to 10 AM alone before Tap B opened.
Work done in 3 hrs(7-10AM) = 15/hr*3 = 45
Remaining work = 180-45 = 135.
Now, Tap A and B worked together until Tap C opened at 11AM.
Efficiency of A and B together = 27/hr.
Work done in 1 hr (10AM-11AM)= 27.
Remaining Work after 11AM= 108.
Now, all 3 taps are opened after 11AM
Hence, Efficiency of A+B+C = 36/hr
Time taken = 108/36 = 3 hrs.
Therefore, 11AM+3hrs = 14 Hrs= 2PM.

Also can be done in the following way:-
A opened for (7-11)= 4 hrs.
B opened for (10-11) = 1 hr
4/12 + 1/15 + 4t/20 = 1.
20+4+12t = 60
12t = 36
== > t=3hrs.
Hence, 3 hrs after 11AM i.e, at 2PM.

4. Two trains running in opposite directions cross a man standing on platform in 28 sec and 15 sec respectively, and they cross to each other in 22 sec. Find the ratio of their speed ?
(a)  6:7
(b)  7:6
(c)  4:3
(d)  Cannot be determined
(e)  None of these
solution:-. (b) 6:7
Let Speed of the train be X and Ym/s
Length of First train = 28x
Length of Second train = 15y
Also, (28x + 15y) / (x+y) = 22
== > 6x-7y = 0
6x = 7y
x/y = 7/6.

5. Three partners A, B and C invested money in a business in the ratio of 5:3.5 : 5.5 and ratio of their duration of investment is 2:8/3:1.33. Then what % of total profit will A get?
(a)  30%
(b)  35%
(c)  37.5%
(d)  40%
(e)  None of these
solution:-. (c) 37.5%
Money Invested in the ratio = 5:7/2:11/2 = 10:7:11
The Ratio of Duration of Investment = 2:8/3:4/3= 6:8:4= 3:4:2
Ratio of their Profit = 10x3:7x4:11x2 = 30:28:22 = 15:14:11
Profit % of A = 15x100/40 = 37.5%.

6. The ratio of present ages of son and daughter of a lady is in the ratio of 7:8 and the age of lady is sum total of ages of her daughter and son. If the ratio of ages of son after 3 years and the age of daughter 6 years ago is 4 : 3 then what will be the ratio of the present age of lady and the age of her son after four years?
(a)  4:5
(b)  5:3
(c)  6:5
(d)  9:5
(e)  None of these
solution:-. (d) 9:5
Let the present age of Son and Daughter be 7x and 8x respectively.
Then, Age of Lady = 15x
Now, (7x+3)/(8x-6) = 4/3
== > 21x + 9 = 32x-24
== > 11x = 33 == > x= 3
Son’s Age = 21 years
Daughter = 24 years
Lady’s Age = 45 years
Required Ratio = 45/25 = 9/5 = 9:5

7. In a two digit number the digit at unit's place is less than twice the digit at ten's place by 1. And the difference between the number obtained by interchanging its digits and the original number is 50% of the number obtained by adding 1 in original number. What is the original number?
(a)  46
(b)  57
(c)  35
(d)  68
(e)  None of these

solution:-. (c) 35
Let the number be 10x+y
Now, ATQ, we have following conditions
(i) y = 2x-1
(ii) 9y-9x = 50%(10x+y+1) = > 9y-9x = 1/2(10x+y+1)
== > 18y-18x = 10x+y+1
== > 17y-28x = 1
Solving (i) and (ii), we get x=3 and y=5
Thus, Number = 10x3+5 = 35

8. An alloy contains mixture of two metals A and B in the ratio of 3:5, and the second alloy contains, mixture of same metals in the ratio of 2.33 : 3. In what ratio should be both alloy mixed to make a new alloy containing 42(6/7)% metal A?
(a)  1:2
(b)  3:2
(c)  2:3
(d)  1:6
(e)  None of these
solution:-. (d) 1:6
Quantity of A in first alloy = 3/8.
Quantity of A in second Alloy= 7/16.
Let two alloys are mixed in the ratio x:y
Therefore, % of A in new mixture = [(3x/8)+(7x/16)]x100 / [x+y]
But, Given % of A in new mixture = 42(6/7)= (300/7)%
== > [(3x/8)+(7x/16)]x100 / [x+y] = 300/7
== > (6x+7y)/16(x+y) = 3/7
== > 42x+49y = 48x+48y
== > 6x = y  == > x=y = 1/6
Ratio = x:y= 1:6

9. The profit earned by a compnay is to be distributed among the managers and officers in the ratio of 7 : 5. If the number of managers and officers is in the ratio of 7 : 12 the amount recieved by each officers is 30,000 what was the total profit earned if the number of managers was 28?
(a)  35 Lakh
(b)  17.5 Lakh
(c)  34.56 Lakh
(d)  19 Lakh
(e)  None of these
solution:-. (c) 34.56 Lakh
Let the Managers get Rs.X
7X/12*3000 = 7/5 == > X= Rs. 72000
Total Money = 72000*28*12/7 = 34.56 Lakh


10. The profit earned by a compnay is to be distributed among the managers and officers in the ratio of 7 : 5. If the number of managers and officers is in the ratio of 7 : 12 the amount recieved by each officers is 30,000 what was the total profit earned if the number of managers was 28?
(a)  70%
(b)  60%
(c)  65%
(d)  80%
(e)  None of these
solution:-. (c) 60%
Total Discount Given = -10-15+150/100 = -25+1.5 = -23.5
Now, Let CP be Rs. 100
SP = Rs. 153
MP = 153x100/(100-23.5)= 153x100/76.5 = 200
Therefore, Marked Price = 200
On Giving 20% discount , SP = 200x80/100 = 160.
SP=160
CP=100
Profit = 60%




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