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 135 of 140 pages

3 friends play a game, rules are as follows whenever there is a contest between any 2 of them, the one who has a higher $%$ alcohol should pour 200ml of his wine into the one having lower $%$ alcohol. The game starts as a contest between A and B, then B and C and then C and A. Post this, the game continues in the same cycle on and on. If a player has emptied all his alcohol, then the remaining two play the game with same rules. If 2 players have alcohol of same percentage level, the younger one pours 200 ml of his alcohol into the elder one's glass. All 3 of them start the game with 600 ml of wine. A's mine has $60 %$ alcohol, B's has $48 %$ alcohol, C's has $50 %$ alcohol. They take 3 minutes to play one round of this game. D, a fourth friend leaves the pub immediately after the game begins, returns after an hour and drinks wine from the person who has the highest alcohol percentage. What is the concentration of alcohol that D had ?

$51.5 %$
$52.67 %$
$53 %$
$58.4 %$
Can't be determined

Discuss Question

Answer: Option B. ->

52.67 %

:
B

A v/s B:
A has higher $%$ of alcohol than B, A has to give 200 ml of wine. Considering that A has $60 %$ of strength mixture, B will end up having 800 ml of wine that has $\frac{60 % \times 1 + 48 % \times 3}{4} = 51 %$ of alcohol.
B v/s C:
B has higher $%$ of alcohol than C, B has to give 200 ml of wine to C now. C will end up having wine that has,

Question 1342.

Both the H.C.F and the difference of two numbers is 7. If the L.C.M of the two numbers is a three digit number, then what is the maximum possible value of the smaller number?

Discuss Question

Answer: Option A. -> 77
:
A

Let the smaller number be 7x, then the larger number will be (7x+7) = 7(x+1)

Since x and (x+1) are co-prime, the LCM would be $7 \times x \times (x + 1)$ .

Maximum value of x such that LCM is a three digit number is x=11.

So, smaller number = 77.

Question 1343.

$2 f (x) + 3 f [\frac{(4 x + 5)}{(x - 4)}] = (x + 1)^{2}$
Find the value of f(5) if x is a real number not equal to 4.

-62
1112
442
None of these

Discuss Question

Answer: Option D. -> None of these
:
D

Put x=5, 2f(5)+3f(25)=36.

Put x=25, 2f(25)+3f(5)=676.
f(5) = 391.2

Question 1344.

How many numbers are there less than 100 that cannot be written as a multiple of a perfect square greater than 1?

Discuss Question

Answer: Option A. -> 61
:
A

List all multiples of perfect squares (without repeating any number) and subtract this from 99.

4 - there are 24 multiples of 4 (4,8,12,.....96)

9 - there are 11 multiples, 2 common with 4 (36 and 72) so, add 9 multiples

16 - 0 new multiples

25 - 3 new multiples (25,50,75)

36 - 0 new ones

49 - 2 {49,98}

64 - 0

81 - 0

Total multiples of perfect squares are 38. There are 99 numbers in total. So, there are 61 numbers that are not multiples of perfect squares.

Question 1345.

Vikalp, Nilay and Nilabh and visited a shop to buy LED TV and each one of them bought at least one unit of the TV. The shopkeeper gave a discount of $20 %$ to Vikalp. Nilay, a good bargainer got two successive discounts of $20 %$ and $60 %$ . Nilabh, the hardest bargainer of all got three successive discounts of $20 %$ , $40 %$ and $50 %$ respectively. After selling total of 'x' units to three of them, the shopkeeper calculated that he had given an overall discount of $56 %$ . If
$10 \leq x \leq 20$ , then how many values of 'x' are possible? (Assume every one bought at least one unit).

Discuss Question

Answer: Option B. -> 6
:
B

Discount to Nilay = $1 - 0.8 \times 0.4 = 68 %$
Discount to Nilabh = $1 - 0.8 \times 0.6 \times 0.5 = 76 %$
Let a,b,c be the number of units sold to Vikalp, Nilay, and Nilabh respectively.
So, 0.2a+0.68b+0.76c=0.56(a+b+c)
So, 5c=3.(3a-b),
So,c is multiple of 3.
If,c=3 then (a,b)= (2,1),(3,4), (4,7)
If, c=6 then (a,b)= (4,2), (5,5), (6,8)
Hence 6 solutions.

Question 1346.

If 3sinx + 4cosx + r is always greater than or equal to 0, what is the smallest value that r can take?

5
-5
4
3
Can't be determined

Discuss Question

Answer: Option A. -> 5
:
A

3 sinx +4 cosx + r $\geq$ 0
3 sinx + 4 cosx $\geq$ -r
$5 \times (\frac{3}{5} s i n x + \frac{4}{5} c o s x) \geq - r$
Let $\frac{3}{5}$ = cos A $\Rightarrow$ sin A $\Rightarrow \frac{4}{5}$
5(sin x cos A +sin Acos X) $\geq$ -r
5(sin(X+A)) $\geq$ -r
We have -1 $\leq$ sin (angle) $\leq$ 1
5 sin(X+A) $\geq$ -5
$r_{m i n}$ =5.

Question 1347.

1 unit of $x %$ alcohol is mixed with 3 units of $y %$ alcohol to give $60 %$ alcohol. If $x > y$ , how many integer values can x take?

Discuss Question

Answer: Option D. -> 13
:
D

We know that

$\frac{(x + 3 y)}{4} = 60$ or x+3y =240 and $x > y$

As $0 < x < 100$ , let's start with x=99. When x= 99, y=47.

When x=96, y=48 and so on. x has to be greater than y. Now in the limiting case, when x=60, y=60.

So when x=63, y=59. So x can take the values of 63,66,69...... till 99, a total of 13 values.

Question 1348.

The sum of all the three digit numbers which when divided by 8 give a remainder of 5 is___

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Answer: Option D. -> 13
:

For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{112}{2} [2 \times 109 + (112 - 1) 8]$
=61936.

Question 1349.

Two rugby teams (thirty players) have been selected to play for their school. Five of the players speak Spanish, Afrikaans and English. Nine of them speak only Spanish and English. Twenty speak Afrikaans, of which twelve also speak Spanish. Eighteen speak English. No one speaks only Spanish. How many players speak only English?___

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Answer: Option D. -> 13
:

From the data given in the question, we can draw a Venn diagram as follows

Since N = 30, the sum of the shaded region is 10. Hence 9 + x = 10 $\to$ x = 1.

Question 1350.

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty?___

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Answer: Option D. -> 13
:

Since each box call hold five balls number of ways in which balls; could be distributed so that none is empty are (2. 2, 1) (3, 1, 1)
i.e., ( $^{5}$ C $_{2}$ . $^{3}$ C $_{2}$ . $^{1}$ C $_{1}$ + $^{5}$ C $_{3}$ . $^{2}$ C $_{1}$ . $^{1}$ C $_{1}$ ) $\times$ 3!
=(30+20) $\times$ 6 = 300