## QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394 | Page 2 of 140 pages
Question 11. On every birthday of my life, my mother has seen to it that my cake contains my age in candles. Starting on my fourth birthday, I have always blown out all my candles. Before that age, I averaged a 50% total blowout rate. So far, I have blown out exactly 900 candles. How old am I
1.    43
2.    42
3.    44
4.    45
:
B
n(n+1)21+2+32=900:n(n+1)2=903;
n(n+1) = 1806, n = 42.
Question 12. Find the sum of given series Sn = 4 + 44 + 444 + 4444 + ...................n terms
1.    49{109(10(n−1)−1)−n}
2.    49{109(10n−1)−n}
3.    49{10(10n−1)−n}
4.    49{(10n−1)−n}
:
B
Reverse Gear Approach
Assume n= 1, then S1=4
Now put n=1in options.Eliminate those options , where you are not getting 4.
Only option (b) will give 4.
Question 13. Find the remainder when 755 is divided by 29.
1.    4
2.    25
3.    1
4.    0
5.    3q+ r
:
B
We need to use reverse of Euler's theorem as the Euler's number for 29=28.
And the question reduces to (72729) (Rem).
The right answer multiplied by 7, should give a remainder of 1 when divided by 29, as (72829) (Rem) = 1.
Thus, the answer is option (b).
Verification -
25 or (4)×7= -28 remainder of 1.
Question 14. If 16th of a number is subtracted from 290, the result is equal to 50% of the number. Find the number.___

:
Let the number be x.
According to the statement,
290x6=x2
or x2+x6=290
or 4x6=290
or x = 435.
Therefore, the number is 435.
Question 15. A 50m long platoon is marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position. In the mean time the whole platoon has moved ahead by 50m. How much distance (approximately) did the last person cover in that time. Assuming that he ran the whole distance with uniform speed
1.    120m
2.    100m
3.    102m
4.    97m
5.    111m
:
A
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backward - are equal. Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
[50+X]X=X[50X]
[50+X]×[50X]=X×X
Solving, X=35.355 meters
Thus, total distance covered by the last person
=[50+X]+X
=2×X+50
=2×[35.355]+50
=120.71 metres.
=120 m (Approximately)
Question 16. Let x and y be consecutive integers. Suppose m be the number of solutions of x3y3=3k2 and n be the number of solutions of  x3y3=2t2 , where k, t are integers. Then m + n equals:
1.    13
2.    5
3.    6
4.    None of these
5.    111m
Answer: Option D. -> None of these
:
D
Let x=y+1so, x3y3=1+3y(y+1)clearly this is not divisible by either 2 or 3 so m = 0 and n = 0 so m+n=0.
Hence option (d)
Question 17. Consider two points from which two persons start walking towards each other. They cross each other after time T1. They went further to the opposite points and return instantaneously to cross each other again after time T2. Then T2 =
1.    2T1
2.    T1
3.    3T1
4.    32×T1
5.    Data Insufficient
:
A
Suppose the distance between the two points is 'd'.
When the persons meet for the first time, total distance covered = d
When persons meet for the second time, total distance covered = d + 2d
Distance = d, time taken = T1; distance = 2d, time taken = 2T1 = T2
Hence Option (a)
Alternative Approach: Using Assumption
You can also solve this using numbers. If the distance is 100 km, and the speed is 50 kmph for each person, they will meet after 12 an hour. They will reach the opposite corners in 1 hour and will meet again in 112 hours; thus T2=1 hour = 2T1. Option (a)
Question 18. Four concentric circles share the same centre. The smallest circle has a radius of 1 inch. For n ≥ 1, the area of the nth smallest circle in square inches, An, is given by An+1=An+(2n+1)π. What is the ratio of the sum of the areas of the four circles to the sum of their circumferences?
1.    1
2.    32
3.    2
4.    3
5.    3q+ r
:
B
Following the pattern
A1=π
A2=4π
A3=9π
A4=16π
Hencer = 1,2,3 4
Thus
C1= 2π
C2= 4π
C3= 6π
C4= 8π
Ratio = 1.5
Question 19. The integers from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked (that is 1, 16, 31, etc.). This process is continued until a number is reached which has already been marked. How many unmarked numbers remain?
1.    732
2.    136
3.    635
4.    800
5.    123
:
D
Consider the first round: - All the numbers which leave a remainder of 1 when divided by 15 will be marked. (1, 16, 31, 46 ... 991)
Consider the second round: - The first number to be marked is 991 + 15 - 1000 = 6. Thereafter all the numbers which leave a remainder 6 when divided by 15 will be marked. (6, 21, 36, 51 ... 996)
Consider the third round: - The first number to be marked is 996 + 15 - 1000 = 11. And all the numbers which leave a remainder 11 when divided by 15 will be marked. (11, 26, 41, 56 ... 986)
The first number to be marked in the fourth round is 986 + 15 - 1000 = 1. Now the cycle will repeat.
If we see the numbers which are getting marked are: 1, 6, 11, 16, 21, 26, 31, 36 ... 996
i.e. all the numbers which divided by 5 leave a remainder 1. So there are 10005=200 numbers. So, 1000 - 200 = 800 numbers remain unmarked.
Question 20. Given that a> 0 and i belongs to a set of natural numbers. If a1,a2,a3.....a2n are in AP, then find the value of a1+a2na1+a2+a2+a2n1a2+a3............+an+an+1an+an+1
1.    n-1
2.    na1+a2n√a1+√an+1
3.    n−1√a1+√an+1
4.    n+1√a1+√an+1
5.    3q+ r