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 5 of 140 pages

Question 41. The number of positive integers which divide

(2^{5})!

are -

213.33.52
28.32.52
211.32.5
28.33.53

Discuss Question

Answer: Option A. -> 213.33.52
:
A

32! = 2^{31} {.3}^{14} . 5^{7} {.7}^{4} {.11}^{2} {.13}^{2} .17 .19 .23 .29 .31

(on prime factorization)
So it has

(31 + 1) (14 + 1) (7 + 1) (4 + 1) (2 + 1) (2 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2^{13} . 3^{3} {.5}^{2}

factors.
Hence, choice (a) is the right answer.

Question 42. The product "n” of three positive integers is 6 times their sum. If one of the numbers is the sum of the other two, then the sum of all possible value(s) of "n” is?

Discuss Question

Answer: Option A. -> 336
:
A
Let the numbers be x,y, x+y.
Therefore xy(x+y) = 12(x+y). Hence, xy= 12.
So, (x,y) = (1,12), (2,6) or (3,4).
Hence "n" = 156, 96 or 84.
Sum "s" (of all possible values of n) = 336.

Question 43. The population of a town was 3600 three years back. It is 4800 right now. What is the rate of growth of population, if it has been constant over the years and has been compounding annually?

Discuss Question

Answer: Option D. -> 10%
:
D
Final Value = Initial Value

{(1 + \frac{r}{100})}^{t}

Thus, Rate

= [{(\frac{F V}{I V})}^{\frac{1}{t}} - 1] \times 100

\Rightarrow \frac{F V}{I V} = \frac{48}{36} = 1.33

Going from answer options -
If 10 is the answer -

{1.1}^{3} = 1.331 \approx 1.33 = \frac{F V}{I V}

Therefore, rate is approximately 10%.

Question 44. Find the complete range of values of "p” for which

(p^{2} - 7 p + 13)^{p} < 1 ?

(−∞,+∞)
0,∞
(−∞,4)
(−∞,0)∪(3,4)
8

Discuss Question

Answer: Option D. -> (−∞,0)∪(3,4)
:
D
Use substitution method,
Put p= 5. Then LHS

>

RHS. Thus, the inequality is not satisfied for this value.
Answer can never be option (a) or (b)
Put p=2. Then LHS>RHS. Thus, the inequality is not satisfied for this value as well. Answer is option (d).

Question 45. ABCDEF is a regular hexagon, with O as the centre. There is another regular hexagon with 2 of the end points PQ inscribed in the above hexagon (again with centre O). Find the area of the unshaded region, given that FPQC lies on a straight line and FP=CQ=2 and FP =

\frac{1}{3}

FO.

ABCDEF Is A Regular Hexagon, With O As The Centre. There Is ...

20√3
24√3
18√3
28√3
118

Discuss Question

Answer: Option B. -> 24√3
:
B
The hexagon can be graphically divided as follows

Hence, there are totally 54 equilateral triangles out of which 24 are unshaded. Each of the triangles have aside 2. Total unshaded area =

\frac{\sqrt{3}}{4}

x 4 x 24 = 24

\sqrt{3}

Question 46. A shopkeeper keeps 100 marbles in two boxes, A and B, such that Box A contains only red and Box B contains only green marbles. Barring an equal number of marbles from each box, he sold all the others and earned Rs. 25 and Rs. 15 from Box A and Box B respectively. Later, he sold half of the remaining marbles. His total earning from all the transactions accumulated to Rs. 45. What is the cost of each marble - red or green.?

40 paise
25 paise
75 paise
50 paise

Discuss Question

Answer: Option D. -> 50 paise
:
D
Let the number of red and green marbles be a and b respectively.
Let the selling price of each marble be Rs.e.
Let the equal number of marbles not sold the first time = c.
Then, e(a - c) = 25 ... (i)
e(b - c) = 15 ... (ii)
The number of marbles that remained that remained with the shopkeeper = 2c
e(a - c) + e(b - c) + ec = 45
ec = 5 ...(iii) (from (i) and (ii))
Also, (100 - c)e - 45

\Rightarrow

100e - 5 = 45 (from (iii)
100e = 50

\Rightarrow

e = 0.5 = 50 Paise.
Hence, [d].

Question 47. Given that a real value function g, is such that g(a+b)=a+g(g(b)) & g(0)=0. Find g(102)?

Discuss Question

Answer: Option D. -> 102
:
D
Assumption
Assume a=0, b=0
g(0)=0+g(g(0))=0
Next assume a=1 and b=0
g(1)=1+g(g(0))=1+0=1
continue in the same way, assume a=2, b=0
g(2)=2+g(g(0))=2
Hence g(x)= x itself. Hence g(102)=102

Question 48. In a school of 150 students, 50 students did not play football, 90 played cricket and 20 neither played football nor cricket
How many students played both, football and cricket?

40
50
60
Cannot be determined

Discuss Question

Answer: Option C. -> 60
:
C

In A School Of 150 Students, 50 Students Did Not Play Footba...

50 students did not play football... i.e. 100 played football.

A \cup B = A + B - A \cap B

130 = 100 + 90 - x

x = 60

Question 49. Find the tens place of the number

7^{281} \times 3^{264}

Discuss Question

Answer: Option C. -> 6
:
C
To find the tens place, we need to find the remainder when this number is divided by 100.

\frac{7^{281}}{100} ∣ r

= \frac{((7^{4})^{70} \times 7^{1})}{100} ∣ r

= \frac{((. .49 \times . .49)^{70} \times 7^{1})}{100} ∣ r = \frac{((. .01)^{70} \times 07)}{100} ∣ r = . .07

Similarly,

\frac{3^{264}}{100} ∣ r

= \frac{(3^{4})^{66}}{100} ∣ r

= \frac{(3^{2} \times 3^{2})^{66}}{100} ∣ r

= \frac{(09 \times 09)^{66}}{100} ∣ r

= \frac{81^{66}}{100} ∣ r = . .81

(66 is of the form 5n+1, and the tens digit of

81^{n}

repeats after a cycle of 5)
Answer is the tens digit of

07 \times 81 = . .6 .

Question 50. The value of a gold bar is directly proportional to the square of its weight. If a gold bar weighing 8 kg breaks into 2 pieces, its total value decreases by