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### Exams > Cat > Quantitaitve Aptitude

## QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394
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**2**of 140 pages**Answer: Option B. ->**42

:

B

n(n+1)2−1+2+32=900:n(n+1)2=903;

n(n+1) = 1806, n = 42.

**Answer: Option B. ->**49{109(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.

**Answer: Option B. ->**25

:

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.

:

Let the number be x.

According to the statement,

290−x6=x2

or x2+x6=290

or 4x6=290

or x = 435.

Therefore, the number is 435.

**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**

__Question 15.__**Answer: Option A. ->**120m

:

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[50−X]

[50+X]×[50−X]=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)

**Answer: Option D. ->**None of these

:

D

Let x=y+1so, x3−y3=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)

**Answer: Option A. ->**2T1

:

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 T

_{2}=1 hour = 2T1. Option (a)

**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, A**

__Question 18.___{n}, 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?

**Answer: Option B. ->**32

:

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

**Answer: Option D. ->**800

:

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.

**Answer: Option B. ->**na1+a2n√a1+√an+1

:

B

Let the AP be 1,2,3,4,5

At n=1, the AP = 1,2

a1=1 and a2=2

The value of the expression =32.4

Only option (b) gives this value