Number Set I(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

NUMBER SET I MCQs

Total Questions : 90 | Page 5 of 9 pages

Question 41. Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, and 16 is 4?
(CAT 1994)

137
1361
140
172

Discuss Question

Answer: Option D. -> 172
:
D
Ans: (d)
LCM of (7, 12, 16) = 336. If we divide 1856 by 336 then remainder is 176. Since it is given that remainder in this condition is 4. Hence the least number to be subtracted = (176 - 4) = 172.
Alternate Approach:
In case you don't remember this method: go from answer options by subtracting the answer option from 1856 and divide by each of 7,12, and 16 to get a reminder of 4.

Question 42. If Roopa leaves home with 30 flowers, the number of flowers she offers to each deity is : (CAT 1999)

Discuss Question

Answer: Option C. -> 32
:
C
Ans: (c)
If

x = 30,

y = 16 \times 2 = 32

Let Roopa have x flowers with her; then

\begin{matrix} B a l a n c e b e f o r e o f f e r i n g & F l o w e r s o f f e r e d & B a l a n c e a f t e r o f f e r i n g I^{s t} P l a c e & 2 x & y & 2 x - y I I^{n d} P l a c e & 4 x - 2 y & y & 4 x - 3 y I I I^{r d} P l a c e & 8 x - 6 y & y & 8 x - 7 y I V^{t h} P l a c e & 16 x - 14 y & y & 16 x - 15 y \end{matrix}

16 x - 15 y = 0 \Rightarrow 16 x = 15 y \Rightarrow y = \frac{16 x}{15}

Question 43. The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is : (CAT 1993)

26
18
31
None of these

Discuss Question

Answer: Option A. -> 26
:
A
Ans:(a)
There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 numbers which are divisible by 3. Out of remaining there are 7 numbers which are divisible by 5. Hence numbers which are not divisible by 2, 3, 5 = (50- 17 - 7) = 26
Alternate Approach:
Taking the Euler's number approach:
If we take the prime factors of a number then we get all the numbers that are co-prime to that number, which means all the multiples of the prime factors are removed, using the same formula we can find out the numbers that are not divisible by a particular set of numbers.
100=29
As we don't want the number 2, 3, and 5 also to in the list remove these 3 numbers, hence option will be 26

Question 44. Elements of S1 are in ascending order, and those of S2 are in descending order. a24 and a35 are interchanged, then which of the following statements is true?

S1 continues to be in ascending order.
S2 continues to be in descending order.
S1 continues to be in ascending order and s2 in descending order.
None of the above

Discuss Question

Answer: Option A. -> S1 continues to be in ascending order.
:
A
Ans: (a)
Let S1 = 1, 2, 3, 4,......., 24, S2 = 50, 49,........., 25;
New series after interchange S1 = 1, 2, 3, 4........, 25, S2 = 50, 49,.......24
It is therefore clear that S1 continues to be in ascending order.
As this is a variable based question: the word "ANY" can be used
Let the series of integers

a_{1}, a_{2}, . . . . . . ., a_{50}

1, 2, 3, 4, 5, . . . . . ., 50.

S_{1} = 1, 2, 3, 4, . . . . . . 24, S_{2} = 25, 26, 27, . . . . . . . . .50

Question 45. The remainder obtained when a prime number greater than 6 is divided by 6 is :
(CAT 1995)

1 or 3
1 or 5
3 or5
4 or 5

Discuss Question

Answer: Option B. -> 1 or 5
:
B
Ans: (b)
This is a generalized question: the important word "ANY" can be used here. Let us solve the question for some prime numbers greater than 6 i.e., 7, 11, 13 and 17. If these numbers are divided by 6, the remainder is always either 1 or 5.

Question 46.

Let $N = 55^{3} + 17^{3} - 72^{3} .$ N is divisible by : (CAT 2000)

both 7 and 13
both 3 and 13
both 17 and 7
both 3 and 17

Discuss Question

Answer: Option D. -> both 3 and 17
:
D

Ans: (d).

$N = 55^{3} + 17^{3} - 72^{3} = (54 + 1)^{3} + (18 - 1)^{3} - 72^{3}$

$= (51 + 4)^{3} + 17^{3} - (68 + 4)^{3} .$

These two different forms of the expression are divisible by 3 and 17 both.

Elimination strategy:

Check the divisibility by 13, it is seen that it is not divisible, now check for 3, it is

divisible, hence the only option will be (d)

Question 47.

L Let D be a recurring decimal of the form $D = 0. a_{1} a_{2} a_{1} a_{2} a_{1} a_{2} . . . . . . . . . . . .;$ where digits $a_{1}$ and $a_{2}$ lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? (CAT 2000)

Discuss Question

Answer: Option C. -> 198
:
C

Ans: (c)

$D = 0. a_{1} a_{2}; 100 D = a_{1} a_{2} . a_{1} a_{2}$ (recurring); Thus, $99 D = a_{1} a_{2}; \Rightarrow D = (\frac{a_{1} a_{2}}{99})$

Required number should thus be a multiple of 99. Hence 198 is the required number.

Question 48.

Three pieces of cakes of weights $4 \frac{1}{2}$ lbs, $6 \frac{3}{4}$ lbs and $7 \frac{1}{5}$ lbs respectively are to be divided into parts of equal weights. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained? (CAT 2001)

54
72
20
None of these

Discuss Question

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

Ans: (d)

Total weight of three pieces: $= (\frac{9}{2} + \frac{27}{4} + \frac{36}{5}) = \frac{359}{20} = 18.45 l b .$

Required weight of a single piece is HCF of $(\frac{9}{2} + \frac{27}{4} + \frac{36}{5}) = \frac{H C F o f (9, 7, 36)}{L C M o f (2, 4, 5)} = \frac{9}{20} l b .$

Number of guests = $\frac{18.45}{(\frac{9}{20})} = 14$

Question 49.

Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (CAT 2001)

1050
540
1040
1590

Discuss Question

Answer: Option D. -> 1590
:
D

Ans: (d)

53x - 35x = 540; $\Rightarrow$ 18x= 540 or x = 30; Therefore, new product $= 53 \times 30 = 1590.$

Question 50.

Lety = $\frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . . . . .}}}}$ What is the value of y?

-2
2
$\frac{(- \sqrt{15} - 3)}{2}$
$\frac{(\sqrt{15} - 3)}{2}$

Discuss Question

Answer: Option D. ->

\frac{(\sqrt{15} - 3)}{2}

:
D

Ans: (d)

Given:y=

y = $\frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . . . . .}}}}$

Thus y = $\frac{1}{2 + \frac{1}{3 + y}}$

$y = \frac{(3 + y)}{(6 + 2 y + 1)} \Rightarrow 2 y^{2} + 7 y = 3 + y;$ On solving, $y = \frac{(- 3 \pm \sqrt{15})}{2};$ since the given fraction is positive, the value of y is