Free Mixed Bag 6 Practice Test - CAT

Question 1

Find the value of (X1-A)(X2-B)(X3-C)……..(X26-Z) where A,B,C…Z take values from 26 in the descending order and X1,X2,X3…..X26 take the square of the values of A,B,C…respectively

A. A(X1.X2.X3…X26)

B. 26

C. X26-(A.B.C..Z)

D. AX1

E. none of these

SOLUTION

Solution : E

option (e) . take note that X26= 1 and Z is also equal to 1. (1-1)=0. the entire product becomes zero. Answer = option (e)

Question 2

Consider the non decreasing sequence of positive integers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5.... in which nth positive number appears n times. Find theremainder when the 2000th term is divided by 4.

A. 0

B. 1

C. 2

D. 3

E. can’t be determined

SOLUTION

Solution : D

Let us see the sequence of the numbers:

Number Last term of the number

1         1

2         3

3         6

4         10

--          --

N   ∑n

We have to find the value of N for the 2000th term. Using iteration we find that if N = 62, the last term that ends with N is (1/2* 62 * 63) = 1953.

Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder is 3. Hence option (d)

Question 3

How many three digit numbers have a digital root = 5?

A. 10

B. 100

C. 20

D. 99

SOLUTION

Solution : B

The numbers will fall in an AP with common difference = 9. Answer will be $\frac{995 - 104}{9} + 1 = 100.$

Question 4

How many three digit numbers have a digital root of 8 with an order 1?

A. 45

B. 36

C. 21

D. 28

SOLUTION

Solution : B

Question is A+B+C=8, where $A > 0$ . So, question changes to A+B+C=7.
Using 1-0 method, answer is based on the arrangement of 7 zeroes and 2 ones $=^{9} C_{2} = 36$

Question 5

Working together, A $&$ B finish a DTP work in 10 days. Had A worked at half his efficiency and B at 5 times her efficiency, then they would have finished the DTP work in half the scheduled time. In how many days can B complete the DTP work on her own?

A. 24

B. 36

C. 15

D. 30

SOLUTION

Solution : D

Let us solve this using answer options

It is better to start with option (c) or option (d) as it is given that B takes ${\frac{1}{5}}^{t h}$ the actual time in the second case. Option (c) and (d) are multiples of 5

Solving for option (c)

$\frac{1}{x} + \frac{1}{15} = \frac{1}{10} \Rightarrow$ A takes 30 days

They would have finished the work in half the time if A takes 60 days and B takes 3 days

But , $\frac{1}{60} + \frac{1}{3} \neq \frac{1}{5}$ . Hence, this is not the answer

Solving for option (d)

$\frac{1}{x} + \frac{1}{30} = \frac{1}{10} \Rightarrow$ A takes 15 days

They would have finished the work in 5 days, if A takes 30 days and B takes 6 days

$\frac{1}{30} + \frac{1}{6} = \frac{1}{5}$ . Hence, this is the answer.

Question 6

How many four-digit numbers divisible by 11 are not palindromes? (A palindrome is a number that reads the same in either direction, e.g. 121)

A. 819

B. 900

C. 625

D. 729

SOLUTION

Solution : D

The number of four digit numbers divisible by $11 = \frac{(9999 - 1001)}{11} + 1 = 818 + 1 = 819$ .

A four-digit palindrome can be written as xyyx then, as (x+y) - (y+x) = 0, it is always divisible by 11. For xyyx, x can be chosen from 1 to 9 and y from 0 to 9.

Hence, the number of four-digit palindrome numbers $= 9 \times 10 = 90$ .

$∴$ The required number = 819 - 90 = 729.

Question 7

If CAT 2017 contains 150 questions and each question contains 5 options, find the total number of ways in which a student can answer atleast two out of 150 questions.

5^{150} - 601

6^{150} - 751

4^{150} - 601

6^{150} - 601

SOLUTION

Solution : B

Option (b)

As a student may answer a question or he may not, So let’s verify the total chances that he can take with a question, i.e. He can take 6 chances with a question namely A,B,C,D,E or not answering it. So with every question he can take 6 chances. So total number of chances is $6^{150}$ .

From this, we have to deduct these chances: - Total number of ways of answering zero questions as well as one question exactly, which is $1 + (150 \times 5) = 751$ . So answer is $6^{150} - 751$ .

Question 8

In the given figure, AD is a median to side BC, and BF is a median to side AD. Given that AC= 30 units, what is the length of AE.

A. 15

B. 10

C. 20

D. 12

SOLUTION

Solution : B

Option (b)

A line DG is drawn || BE.

Then ΔGCD and ΔECB are similar. Also, BD=DC, thus EG=GC.

ΔAFE and ΔADG are similar. Also, AF=FD, thus AE=EG.

Thus, AE=EG=GC= $\frac{30}{3}$ =10 units.

Question 9

What is the remainder of $2000^{1000}$ divided by 13?

A. 3

B. 4

C. 11

D. 8

SOLUTION

Solution : A

$\frac{2000}{13}$ gives remainder 11, now $11^{1000} = 121^{500}$

Also, $\frac{121}{13}$ gives remainder 4. So we have to find the remainder of $\frac{4^{500}}{13} = \frac{2^{1000}}{13}$

We know that $\frac{2^{12}}{13}$ has a remainder 1. Hence, $\frac{2^{1000}}{13} = \frac{(2^{12})^{83} \times 2^{4}}{13} = \frac{2^{4}}{13} = \frac{16}{13}$ , hence remainder is 3.

Alternate Approach - Using Euler's Theorem

$\frac{2000^{1000}}{13} (remainder) = \frac{11^{1000}}{13}$ (remainder)

Euler’s number of a prime number is one less than that number. Hence, Euler’s number for 13 = 12

From Euler's Theorem, remainder of $\frac{N^{E u l e r^{'} s N u m b e r}}{X}$ = 1 (if N and X are co-prime)

Hence $\frac{11^{996}}{13}$ has remainder =1 (as 996 is a multiple of 12, Euler’s number of 13)

The question can be changed to $1 \times \frac{11^{4}}{13}$ .

Remainder of $\frac{14641}{13}$ = 3. $∴$ Answer is 3.

Question 10

Find the number of integral solutions of the equation $p^{2} = 388 + q^{2}$ .

A. 0

B. 1

C. 2

D. more than 3

SOLUTION

Solution : D

The equation can be rewritten as $p^{2} - q^{2} = 388$ or (p+q)(p-q) = 388.
The number of ways in which 388 can be expressed as a product of two numbers such that p and q are integral is 4, i.e. $(- 194) \times (- 2)$ , $194 \times 2$ , $(- 2) \times (- 194)$ and $2 \times 194$ .
Therefore, the number of integral solutions is 4.

Question 11

There are only two machines Alpha and Beta in Limestone Industries. It takes machine Alpha m hours to polish a granite stone that machine Beta can polish in n hours. If machine Alpha operates alone for p hours and is then joined by machine Beta until 100 stones are polished, for how long will the two machines operate simultaneously? Assume that all the granite stones are of identical size

\frac{(100 m n - p)}{(m + n)}

\frac{n (100 m - p)}{m + n}

\frac{100 n (m - p)}{m + n}

\frac{(m + n)}{(100 m n - p)}

SOLUTION

Solution : B

The easiest way to solve this question is by assuming simple values for m,n and p. Let m=1 and n=2 (i.e. Alpha takes 1 hour to polish a stone $&$ Beta takes 2 hours to polish a stone)

Let p=10, Then after 10 hours, 10 stones are completed by Alpha. Together, in 1 hour Alpha and Beta complete $\frac{3}{2}$ stones

Hence to complete the remaining 90 stones, they need 60 hours.Look in the answer options for 60. Answer is option (b)

Question 12

Find the product of all prime numbers between 1 and 20.

A. 9699645

B. 9699690

C. 9699680

D. 9699660

SOLUTION

Solution : B

Going by elimination, option (a) is eliminated because the product of prime numbers 2 and 5 is $2 \times 5 = 10$ , which should yield one zero.

As 3 is a factor of the product, check which of the remaining options are a multiple of 3. Only option (b) and option (d) are both factors of 3.

Checking for 7 and 13, sum of triplets at odd places-sum of triplets at even places=0 in option (b). Hence, option (b) is the answer.

Question 13

The sum of the infinite series 1 + 3x + 4x² + 10x³ + 18x⁴ + ... at x = 1/3 is

A. 28/9

B. 25/8

C. 49/16

D. 3/4

E. none of these

SOLUTION

Solution : E

Option (e) S =1 + 3x + 4x² + 10x³ + 18x⁴ +... ---------- (1)
2x * S= 2x + 6x² + 8x³ + 20x⁴ +..... ----------- (2) --- multiplying by 2x
Subtracting (2) from (1), we get
= (1 - 2x)S= 1 + x - 2x²(1 - x + x² - x³ ......)
= (1 - 2x)S= 1 + x - 2x²/(1+x) [since 0<x <1 and sum of infinite series]
Now substituting x = 1/3 we get S = 7/2
Hence, choice (e) is the correct answer.

Question 14

In the figure given below, the two circles touch externally. AB is a diameter of the larger circle and C is the centre of the smaller circle. If triangle ABC has an area of 289 sq. cm, then which of the following can be the length of the line segment CD?

A. 26 cm

B. 28 cm

C. 30 cm

D. None of these

SOLUTION

Solution : D

Option (d)
Let the radii of the larger and smaller circles be ‘R’ cm and ‘r’ cm.

Area of triangle ABC = $\frac{1}{2}$ * AB * OC = $\frac{1}{2}$ * 2R * (R + r) = R * (r + R) = 289

CD = 2R + r, Using AM>GM, 2R + r > = 2 (R (R + r)) $^{1 / 2}$ = 2 (289) $^{1 / 2}$ = 34 cm

So, minimum length = 34 cm.

Question 15

How many four digit whole numbers ‘n’ are possible such that the last four digits of $n^{2}$ are in fact the original number ‘n’?

A. 0

B. 1

C. 2

D. 3

SOLUTION

Solution : B

Option (b)

Looking at the last digit, the last digit must be either 0, 1, 5 or 6. Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76. Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376. Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376. Out of those, only 9376 is a 4 digit number.

Free Mixed Bag 6 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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