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 134 of 140 pages

Question 1331.

How much milk should be added to 800 dl of a 15% milk solution to make the strength 32%?

Discuss Question

Answer: Option C. -> 200
:
C

Let x dl of pure milk be added. The concentration of pure milk is 100%

Total milk content in x dl of milk = x * 100% = x dl

Total milk content in 800 dl of 15% mixture = 800 * 15% = 120 dl

Therefore, total milk content in final mixture = x + 120

Total volume of mixture = 800 + x

Hence concentration of milk in final mixture = $\frac{x + 120}{800 + x}$

This is given to be 32%.

Hence $\frac{x + 120}{800 + x}$ = 32 %

Solving, we get x = 200 dl.

Shortcut- Using Alligation

There may be different ways to solve this question, but one of the fastest approaches is using the concept of Alligation. Pure milk has 100% concentration, and it is mixed with a solution with 15% milk to make it 32%. This information can be used to get the ratio of the mixture as follows

For every 4 parts of 15% milk, 1 parts of 100% milk needs to be added.

Hence to 800 dl of the 15% solution, $\frac{800}{4}$ = 200 dl of 100% milk needs to be added

Question 1332.

If in a $Δ$ ABC, $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ , then what can be said about the triangle?

Isoceles triangle
Right angled triangle
Equilateral triangle
Acute triangle
1

Discuss Question

Answer: Option C. -> Equilateral triangle
:
C

Since rule says that = $\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = k$
Therefore, a=k(sinA), b=k(sinB) and c=k(sinC)
We can write $\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}$ as
$\frac{c o s A}{k s i n A} = \frac{c o s B}{k s i n B} = \frac{c o s C}{k s i n C}$ = cot A=cot B=cot C
$\Rightarrow$ A=B=C(Equilateral Triangle).

Question 1333.

For the T-20 World Cup, Indian team of 15 players is to be formed from 20 players. Out of 20 players, 7 are batsmen, 7 are bowlers, 5 all-rounders and 1 wicketkeeper. In how many ways this can be done by taking exactly 5 bowlers, 6 batsmen, 1 wicketkeeper and at most 3 all-rounders.

1470
1890
6280
10
1

Discuss Question

Answer: Option A. -> 1470
:
A

To select a team of 15, you have to select almost 3 all rounders.
Therefore the required number of ways
= $^{7}$ C $_{5}$ $\times$ $^{7}$ C $_{6}$ $\times$ $^{1}$ C $_{1}$ $\times$ $^{5}$ C $_{3}$
=1470.

Question 1334.

How many numbers are there between 1 and 100 that are not divisible by 2,3 and 5?

Discuss Question

Answer: Option A. -> 26
:
A

Numbers divisible by 2 = 50

Numbers divisible by 3 = 33

Numbers divisible by 5 = 20
Numbers divisible by 6 (LCM of 2 & 3) = 16
Numbers divisible by 10 (LCM of 2 & 5) = 10
Numbers divisible by 15 (LCM of 3 & 5) = 6
Numbers divisible by 30 (LCM of 2, 3 & 5) = 3

Numbers divisible by 2, 3 & 5 combined = 50 + 33 + 20 - (16 + 10 + 6) + 3 = 103 - 32 + 3 = 106 - 32 = 74
Numbers not divisible by 2, 3 & 5 = 100 - 74 = 26

Question 1335.

A tank has two pipes A and B. Pipe A is for filling the tank and Pipe B is for emptying the tank. Pipe A can fill the tank in 20 hours and Pipe B can empty the tank in 25 hours. How many hours will it take to completely fill a one-fourth empty tank?

10 hours
25 hours
50 hours
100 hours
5 hours

Discuss Question

Answer: Option B. -> 25 hours
:
B

In one hour pipe A can fill $\frac{100}{20}$ = 5% of the tank

In one hour pipe B can empty $\frac{100}{25}$ = 4% of the tank

(A+B)'s work = 5-4 = 1%

To fill a one fourth of empty tank, it will take = $\frac{25}{1}$ = 25 hours

Question 1336.

Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP=6 units PB=4 units and DP=3 units. What is the area of the circle?

$\frac{125 π}{4}$
$\frac{100 π}{7}$
$\frac{125 π}{8}$
$\frac{52 π}{43}$
1

Discuss Question

Answer: Option A. ->

\frac{125 π}{4}

:
A

Two Mutually Perpendicular Chords AB And CD Meet At A Point ...

As AB and CD are two chords that intersect at P, AP

*

PB = CP

*

6 * 4 = C P * 3 \Rightarrow C P = 8

From centre O draw OM

⊥

AB and ON

⊥

CD
We get AM = MB = 5 (perpendicular from center bisects the chord)
Therefore MP = 1, ON = 1, CD = 11
CN = ND = 5.5 (perpendicular from centre bisects the chord)

O N^{2} + C N^{2} + O C^{2}

1 + (5.5)^{2} = r^{2}

Area of circle =

π r^{2} = π (1 + (5.5)^{2}) = \frac{125 π}{4}

Question 1337.

Two circles are placed in an equilateral triangle as shown in Fig. What is the ratio of the area of smaller circle to that of equilateral triangle?

Discuss Question

Answer: Option C. ->

:
C

Question 1338.

If a 3 digit number 'abc' has 3 factors, how many factors does the 6-digit number 'abcabc' have?

16 factors
24 factors
16 or 24 factors
20 factors
5.12 am

Discuss Question

Answer: Option C. -> 16 or 24 factors
:
C

'abc' has exactly 3 factors, so 'abc' should be square of prime number.
Let the square root of 'abc' be 'p' (where 'p' is prime).

'abcabc' = abc $\times$ 1001 = $p^{2} \times 13 \times 11 \times 7$
If 'p' is any prime number other than 13, 11 or 7, then no. of factors = 3 x 2 x 2 x 2 = 24 (no. will be of the form $p^{2} \times 13 \times 11 \times 7$ )
If 'p' is any prime number among 13, 11 or 7, then no. of factors = 4 x 2 x 2 = 16 (no. will be of the form $7^{3} \times 13 \times 11, i f p = 7$ )
$∴$ No. of factors can be 16 or 24.

Question 1339.

Consider three numbers a, b and c Max(a,b,c)+Min(a,b,c)=13. Median (a,b,c)-Mean(a,b,c)=2. Find the median of a, b and c.

11.5
9
9.5
12.5
5.12 am

Discuss Question

Answer: Option C. -> 9.5
:
C

The numbers can be arranged in ascending order min $<$ median $<$ max
We have min + max =13
$M e a n = \frac{m i n + m e d + m a x}{3} = m e d i a n - \frac{12 + m e d}{3} = 2$
On solving we get median =9.5.

Question 1340.

The Karnataka express started from Bangalore to Bhopal at 7 pm at a speed of 60 Kmph. Another train, MP express started from Bhopal to Bangalore at 4 am next morning at a speed of 90 Kmph. Find the time at which the two trains will meet, if the distance between Bangalore and Bhopal is 800 km.