Quantitaitve Aptitude Clubbed(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394 | Page 139 of 140 pages

The rightmost non-zero digit of the number $30^{2720}$ is?
___

Discuss Question

Answer: Option A. ->
:

30^{2720} = 3^{2720} \times 10^{2720}

The rightmost non-zero digit of

30^{2720}

will be the digit in the unit's place of

3^{2720}

.
3's power cycle is 3, 9, 7, 1 and cyclicity is 4.
2720 = 680

\times

∴

The digit in the unit's place of

3^{2720}

is 1.

∴

The rightmost non-zero digit of

30^{2720}

is 1.

Question 1382.

In how many ways can 7 identical red balls, 6 identical green balls, and 1 yellow ball be put in 14 distinct boxes such that each box contains only one ball?

___.

Discuss Question

Answer: Option A. ->
:
First, the only yellow ball can be placed in one of the 14 boxes in

^{14}

_{1}

= 14 ways.
Out of the remaining 13 boxes, we need to choose 6 places for the 6 identical green balls. This can be done in

^{13}

_{6}

ways.

Therefore, the total number of ways = 14 $\times$ $^{13}$ C $_{6}$ = 24024

Question 1383.

Find the highest power of 5625 that can exactly divide 565!.

Discuss Question

Answer: Option A. -> 34
:
A

$5625 = 5^{4} \times 3^{2}$ .
$∴$ Highest power of 5625 in 565! will depend on the highest power of $5^{4}$ .
So, highest power of $5^{4}$ in 565! is = Integral value of $\frac{113 + 22 + 4}{4}$ = 34.

Question 1384.

Find the greatest number of five digits, which is exactly divisible by 7, 10,14, 15, 21 and 28.

99840
99900
99960
99990

Discuss Question

Answer: Option C. -> 99960
:
C
The number should be exactly divisible by 14(7,2),15 (3, 5), 21 (3, 7), 28 (4, 7).
Hence, it is enough to check the divisibility for 3, 4, 5 and 7.

Only (c) is divisible by all.

Question 1385.

When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?
___

Discuss Question

Answer: Option C. -> 99960
:
Let 10x + y be a two digit number, where x and y are positive single digit integers and

x > 0

.
Its reverse = 10y + x
Now, 10y + x - 10x - y = 18

∴

9(y - x) = 18

∴

y - x = 2
Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴

Other than 13, there are 6 such numbers.

Question 1386.

What digit does $2012^{2012}$ end in?

Discuss Question

Answer: Option B. -> 6
:
B

Since the number in the unit's place in 2012 is 2, we can find the last digit of the expression by considering only the power to which it is raised.
Now, we can write the following: $2012^{2012} =___2^{4 k}$ . Therefore the last digit = 6.

Question 1387.

A train met with an accident 80 km away from old Delhi railway station. It travels at ${\frac{5}{6}}^{t h}$ of the usual speed and reaches the Kanpur central 1 hour 12 min late. Had the accident taken place 120 km further. It would have been only 1 hour late. What is the distance between Old Delhi Railway station and Kanpur Central?

800
720
700
None of these

Discuss Question

Answer: Option B. -> 720
:
B
Since the speed is decreased by 1/6. So, the time will be increased by

\frac{1}{5}

, which is equal to 1 hour 12 minutes. It means the normal time required for this remaining part of journey is 5

\times

72 min = 360 min = 6 hour
Now, accident is supposed to happen 120 km further.
Since, the speed is decreased by

\frac{1}{6}

. So, the time will be increased by

\frac{1}{5}

, which is equal to 1 hour. Hence normal time required for this remaining part of journey is 5

\times

1 = 5 hour.

A Train Met With An Accident 80 Km Away From Old Delhi Railw...

Thus , it is clear that when the trains runs 120 km of its normal speed it takes 1 hour less, which implies that in 1 hour the train can run 120km with its normal speed. Thus the normal speed of train is 120km/h.
Since the train requires 6 hours at its normal speed of 120 km/h for the rest part of journey. So,
AK = 120 $\times$ 6 = 720.
Thus,
The total distance = Distance travelled before accident + Distance travelled after accident
= 80 + 720 = 800.
Option (a).

Question 1388.

An arbelos is the region bordered by three mutually tangent circles with co-linear centers. See shaded region in the figure. If the area of this region is 42π and the radii of the two smaller circles have a ratio of 2 : 3. What is the radius of the largest circle(x)?

2 $\sqrt{21}$
5 $\sqrt{7}$
7 $\sqrt{6}$
35

Discuss Question

Answer: Option B. -> 5

\sqrt{7}

:
B

Let the radii of the two smaller circles be 2y and 3y respectively. This makes the radius of the larger circle 5y. We are told that the area is 42π, so we have
42 $π$ = $\frac{1}{2}$ ( $π (5 y)^{2} - π (3 y)^{2} - π (2 y)^{2}$ ),
y = $\sqrt{7}$ .
But x = 5y; So, x = $5 \sqrt{7}$ .

Question 1389.

In a class 40% of the students enrolled for Math and 70% enrolled for Economics. If 15% of the students enrolled for both Math and Economics, what % of the students of the class did not enroll for either of the two subjects?

Discuss Question

Answer: Option A. -> 5%
:
A
We know that (A

\cup

B) = A + B - (A

\cap

B), where (A

\cup

B) represents the set of people who have enrolled for at least one of the two subjects Math or Economics and (A n B) represents the set of people who have enrolled for both the subjects Math and Economics.
Note: -
(A

\cup

B) = A + B - (A

\cap

\Rightarrow

\cup

B) = 40 + 70 - 15 = 95%
That is 95% of the students have enrolled for at least one of the two subjects Math or Economics. Therefore, the remaining 5% did not enroll for either of the two subjects.
Approach 2- Using S-X Technique
S = 40% + 70% = 110%, X = ? and II = 15%
S - X = II,

\Rightarrow

110% - X = 15%

\Rightarrow

X = 95%. Therefore, the remaining 5% did not enrol for either of the two subjects.

Question 1390.

All the page numbers from a book are added, beginning at page 1. However, one page number was added twice by mistake. The sum obtained was 805. Which page number was added twice?

Discuss Question

Answer: Option D. -> 25
:
D

$- \frac{n (n + 1)}{2} = 805; n (n + 1) = 1610$ , most appropriate value of n= 39.
$\frac{39 \times 40}{2} = 780;$ Therefore,the page number that was counted twice is : 805 -780 =25.