Aptitude - 2

Total age of 10 students = 13.5 * 10 = 135 years. Total age of 20 students = 14.1 * 20 = 282 years.
Average age of 30 students = (135 + 282)/30 = 13.9 years
Therefore, it is the correct answer.

A number is divisible by 7 if unit's place digit is multiplied by 2 and subtracted from the remaining digits and the number obtained is divisible by 7. 47331-   (22) = 473274732 - (72) = 4718471 - (82) = 45545 - (52) = 3535 is divisible by 7.Therefore, 473312 is divisible by 7. It is the correct answer.

Let the son's age be x years.
So, father's age=5x-4 years
Therefore, x(5x-4)=288
5x²-4x-288=0
5x²-40x+36x-288=0
5x(x-8)+36(x-8)=0
(x-8)(5x+36)=0
x=8 or x=-36/5 therefore x=8
Son's age=8 years and father's age=5x-4=36 years

The net annual increase=5%
Let the initial population be 100.
Then, the population after 2 years =1001.051.05 =110.25
Therefore,% increase in population=110.25-100 =10.25
Thus, this is the correct answer.

Let the one number be x so according to ratio other number will be 4x. Product of two numbers = H.C.F * L.C.Mx * 4x = 84 * 214x² = 1764x² = 441x = 214x = 4 * 21 = 84.Thus, the greater number is 84. It is the correct answer.

If Vishal has 15 more marks, he could have scored 40% marks.
Now, 15 marks more than 185 is 185+15=200
Let the maximum marks be x,
Then, 40% of x=200
40/100 * x=200
x=200* 100/40
x=500
Thus, the maximum marks is 500.
It is the correct answer.

Let quantity of pure aftershave lotion= 6kg
and quantity of pure water=4kg
Therefore, total quantity of the mixture=10kg
Again let x kg of mixture is taken out of 10 kg of mixture.
Now, the amount of aftershave lotion left=(6-(6x/10)) kg
And the amount of water left=(4-(4x/10)) kg
Thus, the amount of water after adding x kg of water becomes
(4-(4x/10)+x)kg=(4+(6x/10)) kg
According to the question,
(6-(6x/10))=(4+(6x/10))
12x/10=2
x=10/6
Therefore, 1/6 of the10 kg mixture is taken out.
It is the correct answer.

In the whole transaction, there is neither gain nor loss.
Therefore,
Amount of gain in one watch=Amount of loss in other watch.
0.20* CP1=0.15*CP2
CP1/CP2=0.15/0.20=3/4
Also CP1+CP2=560
Hence, CP1=3/(3+4) * 630
CP1=270
And CP2=630-270=360
Therefore, cost is Rs. 270 and Rs. 360
It is the correct answer.

Let present worth be Rs. x.
Then, S.I. on Rs. x at 12% for 6 months=Rs. 189
Therefore, x * 12 *6/12 *1/100=189
x=3150
Present worth=Rs 3150
Amount=present worth+true discount
              =3150+189
              =3339
It is the correct answer.

Work done=9/12 Balance work=1-9/12=3/12
Less work,Less days
9/12 --->21 days
3/12--->x days
Then (9/12) x=(3/12) * 21
x=(31221)/(129)
x=7 days
It is the correct answer.

Compound interest (C.I.)=Amount-Principal Simple interest(S.I.)=PrincipalTimeRate%/100. Required difference is (5000(1+(12/100)²-5000)-(5000122/100)
=5000((28/25)*(28/25) -1)-1200
=5000((784-625)/625)-1200
= Rs. 72
It is the correct answer.

Let B's capital is x and C's capital is y
Therefore, A's profit=50001000/(5000+x+y) 400=50001000/(5000+x+y) ---(i)
C's profit=y1000/(5000+x+y)
200=y1000/(5000+x+y) -----(ii)
(i) /(ii) gives
2=5000/y
y=Rs2500
Therefore, x=5,000
It is the correct answer.

Relative speed of the train and the man=(60+6)
=66km/h =66 5/18 m/s
Distance=110m
Therefore,
Time taken in passing the man =Distance/Speed                                                                                                    
=(11018)/(66*5)
=6 s
It is the correct answer.

Let the common base be x m.
Now, area of the triangle=area of parallelogram
1/2 * x * Altitude of the triangle=x * 300
Altitude of the triangle=600 m
It is the correct answer.

Let unit digit be x, then tens digit = 16 - x. Therfore, original number = 10 *(16 - x ) + x = 160-9x. On reversing the digits, we have x at tens place and 16-x at unit place.Therefore, new number = 10x + 16 - x = 9x + 16. Original number-New number = 18 (160-9x)-(9x+16) = 18160 - 9x - 9x - 16 = 18 - 18x + 144 = 1818x = 144-1818x = 126x = 7. In the original number, we have at unit place tens digit = 16 - 7 = 9. Therefore, original number = 97. It is the correct answer.