CAT Questions

Q:

A bird lover was asked how many birds he had in the basket. He replied that there were all parrots but six, all pigeons but six, and all koel but six. How many birds he had in the basket in all ?

A) 8 B) 18
C) 9 D) 10
 
Answer & Explanation Answer: C) 9

Explanation:

There were all parrots but six means that six birds were not parrots but only pigeons and koels.

Similarly, number of parrots + number of koels = 6 and number of parrots + number of pigeons = 6.

This is possible when there are 3 parrots, 3 pigeons and 3 koels i.e. 9 birds in all.

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Filed Under: Arithmetical Reasoning
Exam Prep: AIEEE , Bank Exams , CAT
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17 13950
Q:

Refer the below data table and answer the following question.

What is the average bonus in Rs. ?

 

A) 606154 B) 7880002
C) 253333 D) 224000
 
Answer & Explanation Answer: A) 606154

Explanation:
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Filed Under: Table Charts
Exam Prep: Bank Exams , CAT

0 13942
Q:

A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1½ times of the previous in how many days will he cover 840 km ?

A) 39 days B) 37 days
C) 35 days D) 33 days
 
Answer & Explanation Answer: C) 35 days

Explanation:

Distance d = 1200km
let S be the speed

he walks 15 hours a day(i.e 24 - 9)
so totally he walks for 70 x 15 = 1050hrs.

S = 1200/1050 => 120/105 = 24/21 => 8/7kmph


given 1 1/2 of previous speed
so 3/2 * 8/7= 24/14 = 12/7

New speed = 12/7kmph

Now he rests 10 hrs a day that means he walks 14 hrs a day.

time = 840 x 7 /12 => 490 hrs
=> 490/14 = 35 days

So he will take 35 days to cover 840 km.

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Filed Under: Time and Distance
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
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20 13939
Q:

Can you find the mistake 1 2 3 4 5?

can_you_find_the_mistake_1_2_3_4_51539067038.jpg image

Answer

Here in the above given question, the mistake is "the is reapeated twice" before mistake.

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Subject: Logic Puzzles Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
Job Role: Analyst , Bank Clerk , Bank PO

53 13937
Q:

If 5 men and 2 boys working together, can do four times as much work per hour as a man and a boy together. Find the ratio of the work done by a man and that of a boy for a given time  ?

A) 1:2 B) 1:3
C) 3:1 D) 2:1
 
Answer & Explanation Answer: D) 2:1

Explanation:

5M + 2B = 4(1M + 1B)

5M + 2B = 4M + 4B

1M = 2B

The required ratio of work done by a man and a boy = 2:1

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Filed Under: Time and Work
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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8 13929
Q:

On multiplying a number by 7, the product is a number each of whose digits is 3. The smallest such number is  ?

A) 476190476 B) 48617
C) 47619 D) 4587962
 
Answer & Explanation Answer: C) 47619

Explanation:

By hit and trail, we find that

47619 x 7 = 333333.

7) 333333 (47619
    333333
----------------
        0

And 476190476 x 7 = 3333333333 but smallest number is 47619.

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Filed Under: Problems on Numbers
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16 13924
Q:

If a number 72k23l is divisible by 88. Then find value of k and l ?

A) k=8 & l=2 B) k=7 & l=2
C) k=8 & l=3 D) k=7 & l=1
 
Answer & Explanation Answer: B) k=7 & l=2

Explanation:

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

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Filed Under: Problems on Numbers
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10 13924
Q:

Which device is required for the internet connection?

A) cd drive B) modem
C) nic card D) joystick
 
Answer & Explanation Answer: B) modem

Explanation:

The device required for the internet connection is a MODEM.

which_device_is_required_for_the_internet_connection1537861312.png image

A modem (modulator–demodulator) is a network hardware device that modulates one or more carrier wave signals to encode digital information for transmission.

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Filed Under: Computer
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13 13919