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__Solutions__

*Q1 Option D*

*Quantity1:*

*Total weight of 11 persons = 42*11 =462*

*Total weight of first 5 persons = 36*5 = 180*

*Total weight of last 5 persons = 41*5 =205*

*Weight of 6 one = 462- (180+205) = 77kg*

* *

*Quantity 2:*

*Average of five consecutive odd numbers = 45(It should be a middle number) <=(If the difference is equal, then the middle no will be your average value)*

*So, the series must be*

*41, 43,45,47,49*

*Sum of first and last number = 49+41= 90 kg*

*Quantity I < Quantity II*

*Q2 Option C*

*Quantity1:*

*Total cost price = 15000+1000 =16000*

*Selling price = 20000*

*Profit = 20000-16000 =4000*

*Quantity2:*

*Cost price of an article = sp –profit*

*= 2500-250 = 2250*

*Quantity I > Quantity II*

*Q3 Option C
Quantity1:*

*50km in half an hour*

*Speed = 100km/hr*

*60 -100*

*225 – ?*

*Total distance travelled = 375km*

*Quantity2:*

*Quantity I > Quantity II*

*Q4 Option B*

*Quantity1: *

*Quantity 2: *

*Quantity I ≥ Quantity II*

*Q5 Option C*

*Quantity1:*

*Area of circle,*

* = *

*r = 110.25*

*r = 10.5 m*

*Circumference = *

*Total cost = 66*27.5 = Rs.1815*

*Quantity 2:*

*Investment ratio = 7: 8*

*Total investment ratio = 7*9: 8*12 = 63: 96 = 21:32*

*A’s profit = *

*Quantity I > Quantity II*

* *

*Q6.Option D*

*Quantity1:*

*Speed =207 *(5/18) =57.5m/s*

*Time = (650/57.5) =11.3 sec*

*Quantity 2:*

*Speed of Train A = 250/25 = 10m/sec*

*Time taken to cross each other = (250+320) / (25+10) = 16.28sec*

*Quantity I < Quantity II*

* *

*Q7 Option D*

*5 years ago, A_B= 8:5 (we assume this as 8x and 5x)*

*2 years hence, A_B= 5:4*

*32x+28= 25x+35*

*7x=7*

*=>x=1*

*The present age of Suba= 8x+5 =13 years*

*Quantity 2:*

*After 5 years, A_B= 3:4 (we assume this as 3x and 4x)*

*4 years hence, A_B= 29:39*

*3x-1/4x-1 = 29/39*

*117x-39=116x-29*

*=>x= 10*

*The present age of A= 3x-5 = 25 years*

*Quantity I < Quantity II*

*Q8 Option D*

*Quantity 1:*

*Average of first 10 odd number = (1+3+5+7+9+11+13+15+17+19)/10=10*

*(or)*

*If the difference is equal, then the average will be the middle number. Here the middle no is 9 and 11. So, the average is (9+11)/2 = 20/2=10*

*Quantity 2:*

*Average of first 10 prime number = (2+3+5+7+11+13+17+19+23+29) =129/10 =12.9*

*Quantity I < Quantity II*

*Q9 Option C
Quantity 1:*

*1/7 -1/9 = 9-7 / (9*7) = 2/63 = 1/31.5*

*Total time taken = 31.5 hrs*

*Quantity 2:*

*One tap can fill tank in 6 hours and another tap can fill the tank in15 hours*

*The total capacity of the tank is 30. (ie., LCM of 6 and 15)*

*Total hours per hour capacity*

*Tap 1=> 6 5*

*Tap 2=> 15 2*

*Both tap are opened alternatively.*

*The first two hours the capacity of tap 1 and 2 is 7. So, 7*4= 28(8 hour capacity)*

*Remaining 2 will be there. Remaining will be filled by tap 1. So, 2/5 hr*

*The tank will be filled in ** hrs*

*Quantity I > Quantity II*

*Q10 Option D
Quantity1:*

*Probability = 4C _{2} /52C_{2} = 6/(26*51) = 1/221*

*Quantity 2:*

*Total Probability = 6 =36*

*Divisible by 2 = 18 possibilities*

*Probability = 18/36 = 1/2*

*Quantity I < Quantity II*

* *