## QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394 | Page 1 of 140 pages
Question 1. A Swedish watch company is assigning ID cards for its employees. The ID is made of 7 digits. How many ID numbers can exist such that at least one of the digits repeat?
1.    9456745
2.    7898760
3.    8455680
4.    None of these
5.    none of these
:
C
Total number of combinations = No repeats + 1 number repeat + 2 number repeat +.....7 number repeat
Total number of combinations - No repeats = 1 number repeat + 2 number repeat +...... 7 number repeat
Total number of combinations - No repeats = Atleast one number repeat
106×9(9.9.8.7.6.5.4)=9000000544320=8455680
Question 2.  The diameter of a circle X is given by ([1p2]+p2), which of the following best approximates the circumference of circle X?
(Given that [1p] + p = 3)
1.    27
2.    22
3.    30
4.    12
5.    118
:
B
([1p]+p)2 = 9
([1p2]+p2+2) = 9
([1p2]+p2) = 7
Circumference = 2πr 22
Question 3. The sum of 4 consecutive two-digit odd numbers, when divided by 10 becomes a perfect square. Which of the following can possibly be one of these 4 numbers?
1.    23
2.    49
3.    87
4.    75
5.    118
:
C
The sum has to end with a zero. The numbers have to end with 7,9,1,3. Option (d) is out. Go from answer options, answer is option (c).
The numbers are 87, 89, 91, 93 which add up to 360. 36 is a perfect square.
Question 4. Given that: Side of square ABCD = 4cm, E, F, G and H are midpoints of the sides AD, AB, BC and CD respectively. Also, the line passing through ON would bisect the line XF. Similar is the case for all other pairs: PO, RS and TM. Lastly, the lengths TM = NO = PQ = RS = 14 side of square (Note: The figure given below is a symmetric figure). X is the centre point of the square. The area of the shaded region is ___

:
Using graphical division, we can divide the figure to look like the following:

Looking at the figure, we can easily make out that the shaded region is made out of 16 triangles,
Each being 1/64th the size of the larger square. (Since, area of each triangle =14 * 116 is area of square)
Therefore, area of shaded region = (14×116×16×16)cm2
Thus, required area of shaded region = 4cm2.
Question 5. Find the sum of areas of POQ and ORS
1.    pq2
2.    p2q22
3.    p2+q22
4.    p2+q28
:
A
Since the only condition we need to maintain is that the diagonals need to intersect at right angles, assume the simplest quadrilateral which satisfies this condition- a square. Now the question can be solved by simple graphical division.
Take the side of the square =p=q=2. Area = 4.

We need the area of the shaded region, which is half of that of the square= 2. Look in the answer options for 2 on substituting p=2 and q=2. Answer is option (a)
Question 6. What is the remainder when 15987797 is divided by 10?
1.    3
2.    1
3.    7
4.    9
:
C
Express the power in the form of 4K + X. In this case 797 = 4K + 1.
159871 = Unit digit is 7. Hence, remainder is 7 when divided by 10.
Question 7. If a, b, c >0, a + b + c =6 and f(x) = (6x) - 1, find the minimum value of f(a).f(b).f(c)?
1.    6
2.    12
3.    24
4.    8
5.    Cannot be determined.
:
D
f(a).f(b).f(c) = (6a)(6b)(6c)abc = 21636(a+b+c)+6(ab+bc+ca)abcabc = (ab+bc+ca)abc - 1 = 6(1a+1b+1c) - 1
As AMHM
(1a+1b+1c)3 = 6a+6b+6c3 > 3
6a+6b+6c > 9
f(a).f(b).f(c)> 9 - 1 = 8
Question 8. A, b, c and d are real positive numbers such that a + b = 5, c + d = 2 and ad = bc = 1. Then ac+bd+(ac) is:
1.    11.5
2.    12.5
3.    10.5
4.    13
5.    Cannot be determined.
:
C
Multiply both equations - (a+b)(c +d) = 10
a(c+d)=2a (c+d=2)
ac=52...(ii)
Hence (i)+(ii) = ac+bd+(ac) = 8+52=10.5
Question 9. Given that N= abcdefghij is a ten-digit number . All of the digits (a, b, c, d, ...) are different from one another. If 1 1 1 1 1 divides it evenly, how many different possibilities are there for abcdefghij?
1.    3024
2.    3456
3.    5076
4.    1692
:
B
If all the digits are different, and there are 10 of them, then all the digits from 0 to 9 must appear.
That implies that the digits of abcdefghij add up to 45 =0+1+2+3+...+9.
That implies that abcdefgh is definitely divisible by 9.
Since 9 and 1 1 1 1 1 are relatively prime, their least common multiple is 9×11111=99999,
Thus abcdefgh must be divisible by 99999.
We have to find the condition such that abcdefghij is divisible by 99999.
abcdefghij =abcde×105+ fghij and, this number is exactly divisible by(105-1).
Hence by remainder theorem, (abcde×105+ fghij) must be 99999.
Then the following equations must all hold:
f =9-a, g =9-b,h =9-c, I =9-d, j =9-c
Now there are 9 Options for a (it can't be 0) and then f is known.That leaves 8 options for b, and
then'g' is known. That leaves 6 Options for c, and then h is known.
That leaves 4 Options for d and then I is known.
That leaves 2 Options for c, and then j is known.
Thus, the total number of such number abcdefghij is 9×8×6×4×2=3456.
Question 10. A box has a ball and everyday starting with today two balls are added to the box for every ball present in it. After 4 weeks the balls are distributed equally among five boxes. The remaining balls are thrown away. How many were thrown away?
1.    0
2.    1
3.    2
4.    3
5.    4