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 138 of 140 pages

$Δ$ ABC has a circle centred on vertex C that passes through points A and E. If $∠$ ABC $\approx$ $∠$ DEC, AC = 1 and CD = x, what is the distance between E and B?

2x
$\frac{1 - x}{x}$
1 - x
1 + x

Discuss Question

Answer: Option B. ->

\frac{1 - x}{x}

:
B

Since $∠$ DEC $≅$ $∠$ ABC, it follows that $Δ$ DEC $\approx$ $Δ$ ABC.
So, $\frac{1}{x} = \frac{B E + 1}{1}$
$\Rightarrow x (B E + 1) = 1$
$\Rightarrow (B E + 1) = \frac{1}{x}$
$\Rightarrow B E = \frac{1 - x}{x}$

Question 1372.

At a new index of 12 shares, the shares of HBL, Niposys and Brilliance have a weightage of $7 %$ , $13 %$ and $1 %$ respectively. What is the increase in the price of the other shares, if the rise in these three is $9 %$ , $10 %$ and $4 %$ , when the index rises by $6 %$ ?

$4 %$
$3.4 %$
$6 %$
None of these

Discuss Question

Answer: Option C. ->

6 %

:
C
Lets say the value of the index = 100
HBL = 7
Niposys = 13
Brilliance = 1
Others = 100 - 21 = 79
Increased index value = 106
Increased individual values
HBL = 1.09

\times

7=7.63
Niposys = 1.1

\times

13 =14.3
Brilliance = 1.04

\times

1 = 1.04
Total = 22.97

Others = 106 - 22.97 = 83.03

Increase $= \frac{83.03 - 79}{79} \times 100 = 5.1 %$ (approx)

Question 1373.

A mixture of 125 litres of milk and water contains 20% water. What amount of water needs to be added to this milk-water mixture in order to increase the percentage of water to 25% of the new mixture?

7 L
5.66 L
8.33 L
None of these

Discuss Question

Answer: Option C. -> 8.33 L
:
C

We need to find out how much of a solution of 100% water needs to be added to a solution containing 20% water to attain a dilution of 25%.
This can be found out as follows:

Ratio = 75:5 = 15:1

i.e. for 15 parts of a 20% water solution, one part of 100% water solution needs to be added. Therefore, for a solution of 125 litres, $(\frac{125}{15})$ litres of water needs to be added.

Question 1374.

If $a_{n} = x^{n} + \frac{1}{x^{n}},$ n being a natural number, then which of the following is equal to $a_{n + 3}$ ?

$a_{1} . a_{n + 2} - a_{n + 1}$
$a_{1} . a_{n + 2} - a_{n + 1} + a_{n}$
$a_{1} . a_{n + 2} - a_{n + 1} - a_{n}$
$a_{1} . a_{n + 2} + a_{n + 1} - a_{n}$

Discuss Question

Answer: Option A. ->

a_{1} . a_{n + 2} - a_{n + 1}

:
A

Approach :1 Using the options
$a_{1} . a_{n + 2}$ exists in all options.
We will calculate it first.
$a_{1} . a_{n + 2}$ exists in all options.
We will calculate it first.
$a_{1} . a_{n + 2} = (x + \frac{1}{x}) (x^{n + 2} + \frac{1}{x^{n + 2}})$
$= x^{n + 3} + \frac{1}{x^{n + 3}} + x^{n + 1} + \frac{1}{x^{n + 1}}$
$\Rightarrow a_{1} . a_{n + 2} = a_{n + 3} + a_{n + 1}$
$\Rightarrow a_{n + 3} = a_{1} a_{n + 2} - a_{n + 1}$
Approach 2: Assumption
Since the question is of the form "Variable to Variable”, assume any value for "n” and "x”
Assuming n=1 $&$ x=1 . Then $a_{1} = 2$ . Similarly $a_{2} = 2$ , $a_{3} = 2$ $& a_{4} = 2$
Eliminating answer options
Option (a) =2
Option (b) $\neq$ 2

Option (c) $\neq$ 2 $&$ option (d) $\neq$ 2 . Answer is option (a)

Question 1375.

In the figure, EAF is a common tangent to the circles at the point A. Chords AC and BC of the smaller circles are produced to meet the larger circle at G and D respectively . Which of the following must be true?

1. $∠$ ADG = $∠$ EAG
2. $∠$ ABD = $∠$ AG D
3. $∠$ BAE = $∠$ ADB

1 only
2 only
1 and 3 only
2 and 3 only

Discuss Question

Answer: Option A. -> 1 only
:
A

From the tangent secant theorem, $∠$ ADG= $∠$ EAG. Other equalities don't hold. Option(a)

Question 1376.

The number of positive integers n in the range $12 < n < 40$ such that the product (n-1)(n-2)...3.2.1 is not divisible by n is:

Discuss Question

Answer: Option B. -> 7
:
B

The question is asking you for all prime numbers between 12 and 40. There are 7 prime numbers.

Question 1377.

In a certain examination paper, there are n questions. For j = 1,2 ...n, there are $2^{n - j}$ students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is:

Discuss Question

Answer: Option A. -> 12
:
A
Let us say there are only 3 questions.
Thus there are

2^{3 - 1} = 4

students who have answered 1 or more questions wrongly,

2^{3 - 2} = 2

students who have answered 2 or more questions wrongly and

2^{3 - 3} = 1

student who must have answered all 3 wrongly.
Thus total number of wrong answers

= 4 + 2 + 1 = 7 = 2^{3} - 1 = 2^{n} - 1

In our question, the total number of wrong answers $= 4095 = 2^{12} - 1.$ Thus n = 12.

Question 1378.

What are the last two digits of $3844^{47}$ ?

Discuss Question

Answer: Option D. -> 64
:
D

(3844)^{47} = (. . .44)^{47} = (11 \times 4)^{47} = (11)^{47} \times (2)^{94} = . .71 \times (2)^{90} \times 2^{4} = . .71 \times . .24 \times 16

(

∵ (2^{10})^{o d d n u m b e r}

always ends in 24)

The last two digits of the above product = 64.

Question 1379.

Two circles centered at points A and B, respectively, intersect at points E and F as shown. These circles intersect segment AB at points C and D. If AC=1 and CD = DB = 2, determine EF.

2.4
4.8
2 $\sqrt{5}$
2 $\sqrt{7}$

Discuss Question

Answer: Option B. -> 4.8
:
B

In triangle AFB, $\frac{1}{2} \times A B \times F P = \frac{1}{2} \times A F \times F B$
$\Rightarrow F P = \frac{12}{5}$
Hence, EF= $\frac{24}{5}$ = 4.8.

Question 1380.

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

11 $\leq$ e $\leq$ 66
10 $\leq$ e $\leq$ 66
11 $\leq$ e $\leq$ 65
0 $\leq$ e $\leq$ 11

Discuss Question

Answer: Option A. -> 11

\leq

\leq

66
:
A
the question is basically asking us to find out the minimum number and maximum number of lines that can be drawn using 12 points.
We can draw a minimum of 11 lines. We can visualize this as one central point and all lines connecting to this point i.e. one point has a degree of 11 and the other 11 have a degree of 1.
There can be a maximum of