Free Mixed Bag 8 Practice Test - CAT

Question 1

Find the number of integral solutions of the equation p²=388+q²

A. 0

B. 1

C. 2

D. 3

E. more than 3

SOLUTION

Solution : E

option (e)
the equation can be rewritten as p²-q²=388. the number of ways in which 388 can be expressed as difference of squares in one way as 388 is a prime multiple of 4 ( 4 x 97). However, since the question is about integral solutions, the total number of ways in which it can be expressed will be 4 (2 cases for p, positive and negative and 2 cases for q)

Question 2

(x² – 1) is a factor of f(x) = (x⁵ + ax⁴ + bx³ + cx² + x + d). The graph of f(x) intersects the Y axis at (0, –3). Find the value of (a + c).

A. 1

B. 4

C. 2

D. 5

E. 3

SOLUTION

Solution : E

(x² -1) = (x – 1) (x + 1) is a factor of f(x).

So, f(1) = f(-1) = 0

1 + a + b + c + 1 + d = –1 + a + b + c – 1 + d

b = -2

(1 + a + b + c + 1 + d) = 0

(–1 + a + b + c – 1 + d) = 0

i.e., a + b + c + d = 2

a – b + c + d = 2.

Since graph passes through (0, –3)

f(0) = –3 or d = –3

∴a + b + c = 1 and a – b + c = 5

∴a + c = 3 Hence option (e)

Question 3

Pebbles can buy eggs from two stores: a new Reliance supermarket that sells eggs only in crates of 4, and a road side shop that sells single eggs without a crate. If Pebbles wants to ensure that the total number of eggs she buys is a multiple of 5, what is the minimum number of eggs she must buy from the roadside shop?

A. 0

B. 1

C. 2

D. 3

SOLUTION

Solution : A

Pebbles can buy 0 eggs from the roadside shop and 20 eggs from the supermarket.

Question 4

Let S be the set of all four-digit positive integers whose digits are 3, 5, 7 and 9, with no digit repeated in the same integer. Calculate the remainder when the sum of all of the integers in S is divided by 9.

A. 6

B. 3

C. 0

D. None of these

SOLUTION

Solution : C

There are 24 numbers formed with the four given numbers.

Six of these numbers have a 3 in the 1000s position, six have a 5 in the 1000s position, six have a 7 in the 1000s position and six have a 9 in the 1000s position.

The same can be said about the distribution of numbers in the 100s, 10s and units positions.

Therefore, the sum of the 24 numbers is

6(3 + 5 + 7 + 9)(1000) + 6(3 + 5 + 7 +9)(100) + 6(3 + 5 + 7 + 9)(10) + 6(3 + 5 + 7 + 9) = 159984

The remainder is 0 when 159984 is divided by 9.

Shortcut -

Sum of the digits will be 24 in each case, which will leave a remainder 6 when divided by 9. there are 4! numbers possible. Remainder when $4! \times 6$ is divided by 9 is zero. Hence option (c).

Question 5

A closed cylinder is made out of paper whose base has a circumference of meters and an equilateral triangle is painted on the interior side of the base. A drop of paint is dropped into the tank, which has an equal probability of landing on any particular point on the base. If the probability of the drop of paint landing on the portion of the base outside the triangle is 3/4, what is the length of the side of the triangle?

A. 2√6

B. 6√6/2

C. √6

D. √3

E. 2

SOLUTION

Solution : E

Option E is the correct answer.

Question 6

PQRS is a square of length x, a natural number >1. Let $L_{1}, L_{2}, L_{3}, L_{4} \dots$ be points on QR such that Q $L_{1} = L_{1} L_{2} = L_{2} L_{3} = L_{3} L_{4} \dots . = 1$ and $M_{1}, M_{2}, M_{3}$ be points on RS such that R$M_1=M_1M_2=M_2M_3. _{. . . . . .}=1$. Then, $_{n - 1}^{a - 1} \sum (P L_{n}^{2} + L_{n} M_{n}^{2})$ is equal to?

\frac{1}{2}

\times

\times

(a-1)

^{2}

\frac{1}{2}

\times

\times

(a-1)

\times

(4a-1)

\frac{1}{2}

\times

(a-1)

\times

(2a-1)

\times

(4a-1)

\frac{1}{2}

\times

(a+1)

\times

(2a-1)

\times

(4a-1)

SOLUTION

Solution : B

Option (b)

Assume a square of side 2 as follows with L $_{1}$ as the midpoint of side QR and M $_{1}$ as the mid-point of side RS

At n=1, the expression yields a value 5+2=7

Look in the answer options for 7, when n=1. Eliminate those answer options where this is not obtained

$\frac{1}{2}$ *a*(a-1) $^{2}$ = 1

$\frac{1}{2}$ *a*(a-1)*(4a-1)= 7

$\frac{1}{2}$ *(a-1)*(2a-1)*(4a-1)= $\frac{21}{2}$

$\frac{1}{2}$ *(a+1)*(2a-1)*(4a-1)= $\frac{63}{2}$

a $^{2}$ *2n $^{2}$ = 8

Question 7

In a sequence of (4n+1) terms, the first (2n+1) terms are in Arithmetic Progression whose common difference is 2 and the last (2n+1) terms are in Geometric Progression whose common ratio is 0.5. If the middle terms of the AP & GP are equal, then the middle term of the sequence is

A. (nx2ⁿ⁺¹)/(2ⁿ-1)

C. nx2ⁿ

D. n

E. n²-n

SOLUTION

Solution : A

Assumption

At n=1, there are 5 terms in this sequence. The 2^nd and the 5^th term are equal. One such sequence which satisfies the given conditions is

02421

Based on this, the middle term is 4

Look in the answer options for a value 4, on substitution of n=1. Answer is option (a)

Question 8

k = 2 ^{-{| x + 8| - | x + 1|}} . Find the maximum value of k?

A. 234

B. 2⁷

C. 2^1/7

D. 2³

E. 256

SOLUTION

Solution : B

Let f(x)= (|x + 8| - | x + 1|)
the expression changes to k= 2^-f(x)
To maximize k, we need to minimize f(x)
f(x) will be minimum at x=-8 and or x=-1
at x=-8, f(x)=-7
at x=-1, f(x)= 7

hence the minimum value of f(x)=-7 and so the value of k maximum will be 2⁷

Question 9

There is a game which has a board which looks just like a chess board, with n rows and n columns, only n is any odd integer greater than 7. Numbers are filled in this board subject to the following rules-
(I) Any number is the negative of the number directly above it, if there is a number directly above it.
(II) Any number is double the number directly to its left if there is a number directly to its left.
If the number is in the upper left corner (first row and first column) is 1, what is the sum of all the numbers in the table?

A. 0

n^{2} - 1

2^{n} - 1

2^{n} + 1

SOLUTION

Solution : C

The point to note here is that although it is mentioned that $n > 7$ , the pattern will hold good for all odd values of n.
At n=1, the answer will be 1, which is obtained only at option (c).

Hence, the sum = $2^{n} - 1$ . Answer is option (c).

Question 10

The set M consists of p consecutive integers with sum 2p. The set N consists of 2p consecutive integers with sum p. The difference between the largest elements of M and N is 9. Then p is

A. -15

B. 36

C. 9

D. 21

SOLUTION

Solution : D

Let the first term of series M and N be 'a' and 'b' respectively

For the series M we have, (a+1)+...+(a+p) = p(2a+p+1)/2 = 2p

For the series N we have (b+1)+...+(b+2p) = 2p(2b+2p+1)/2 = p

Also- b+p-a=9 or a-b-p=9

Solving we get, p = 21 or - 15 (not possible)

Hence choice (d) is right option.

Question 11

Bangalore dairy sales Milk in two kinds of packets: Plastic Bag $&$ Tetra Packs. Each of these packets goes through three steps: Filling, Sealing and Labelling. The company has three filling machines, 6 sealing machines and 3 labelling machines. Each of these 12 machines can work for exactly 8 hours in a day. One box of Tetra Packs requires 20 minutes for filling, 1 hour for sealing and 20 minutes for Labelling. The box of Plastic Bags requires 15 minutes for filling, 1 hour for sealing and 10 minutes for labelling. If the dairy has to manufacture only 40 boxes of Tetra Packs then how many machine hours are utilised on that day?

A. 2 hrs

66 \frac{2}{3}

hours

C. 30 hrs

D. 72 hrs

SOLUTION

Solution : B

Each box of Tetra Pack takes 20 minutes for filling, 1 hour for sealing and 20 minutes for Labelling.
Total Machine Hours Utilised = 800 + 2400 + 800 = 4000 = $66 \frac{2}{3}$ hours.

Question 12

Bangalore dairy sales Milk in two kinds of packets: Plastic Bag $&$ Tetra Packs. Each of these packets goes through three steps: Filling, Sealing and Labelling. The company has three filing machines, 6 sealing machines and 3 labelling machines. Each of these 12 machines can work for exactly 8 hours in a day. One box of Tetra Packs requires 20 minutes for filling, 1 hour for sealing and 20 minutes for Labelling. The box of Plastic Bags requires 15 minutes for filling, 1 hour for sealing and 10 minutes for labelling.If the number of sealing machines is increased by $50 %$ then at a maximum, how many boxes of Tetra Packs can be manufactured in one day?

A. 60

B. 82

C. 48

D. 86

E. 72

SOLUTION

Solution : E

So, 72 boxes of Tetra Packs can be manufactured. Hence option (e)

Question 13

There are 7 couples in a dance party. How many dance pairs can be there such that exactly 2 pairs are the original couples?

A. 765

B. 924

C. 1222

D. 904

E. none of these

SOLUTION

Solution : B

Number of ways of selecting 2 couples out of $7 =^{7} C_{2}$

Derangement of 5 couples can be done in 44 ways.

Total number of ways = 924

Question 14

In a triangle ABC, AB = AC = 20 cm. D, E, F are the midpoints of sides AB, AC and BC respectively. Find the ratio of area of quadrilateral ADFE to the area of triangle ABC .

A. 1 : 2

B. 1 : 1

C. 2 : 3

D. 1 : 3

SOLUTION

Solution : A

AD = BD = 10

ABC is isosceles triangle. Median is same as perpendicular bisector. So angle AFC = Angle AFB = 90

BC is tangent to the circle; AF is perpendicular so AF must be the diameter. (Radius is perpendicular to tangent)

So angle ADF and angle AEF = 90 degrees. Let DF = x.

Area ADF = Area AFE = $\frac{1}{2}$ *10 * x = 5x, Area ADFE = 10x

Area ABF = Area AFC = $\frac{1}{2}$ * 20 * x = 10x, Area ABC = 20x

So the required ratio is 1:2

Question 15

If x₁,x₂,x₃,x₄….x_2n+1are in Arithmetic Progression, then find the value of [(x_2n+1–x₁)/ (x_2n+1+x₁)] + [(x_2n–x₂)/ (x_2n–x₂)]………….. [(x_n+2–x_n)/ (x_n+2–x_n)]

A. n-1

C. n(n+1)/2

D. [n(n+1)/2]/[x₂-(x₁/x_n)+1]

E. (n+1)(x₂-x₁)

SOLUTION

Solution : D

Assumption assume an AP 1,2,3 substituting in the equation, we get the value as ½ at n=1 look in the answer options and eliminate those options, where you do not get ½ answer is option (d)

Free Mixed Bag 8 Practice Test - CAT

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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