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A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260.

If the zoo had either pigeons or horses, how many horses were there in the zoo?

A.30

B.40

✔ C.50

D.60

Solution:Option(C) is correct

Let the number of horses =x=x

Then the number of pigeons =80−x.=80−x.

Each pigeon has 2 legs and each horse has 4 legs.

Therefore, total number of legs =4x+2(80−x)=260=4x+2(80−x)=260

⇒4x+160−2x=2604x+160−2x=260

⇒2x=1002x=100

⇒x=50x=50

A girl wrote all the numbers from 100 to 200. Then she started counting the number of one's that has been used while writing all these numbers. What is the number that she got?

A.111

B.119

✔ C.120

D.121

Solution:

Option(C) is correct

From 100 to 200 there are 101 numbers.

There are 100, 1's in the hundred place.

10, 1's in tens place and10, 1's in unit place.

Thus, the answer is 100+10+10=100+10+10= 120.

If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261....491492259260261....491492.

How many 8's will be used to write this large natural number?

A.32

B.43

C.52

✔ D.53

Solution:

Option(D) is correct

From 259 to 458, there are 200 natural numbers so there will be 2×20=402×20=40 8's

From 459 to 492 we have 13 more 8's and so answer is 40+13=40+13= 53

IBM-Daksh observes that it gets a call at an interval of very 10 minutes from Seattle, at every 12 minutes from Arizonia, at the interval of 20 minutes from New York and after every 25 minutes it gets the call from Newark.

If in the early morning at 5:00 a.m. it has received the calls simultaneously from all the four destinations, then at which time it will receive the calls at a time from all places on the same day?

✔ A.10:00 am

✖ B.3:00 am

C.5:00 am

D.Both (A) and (B)

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4.0/4

Discuss Toggle Dropdown

Solution:

Option(A) is correct

IBM – Daksh gets simultaneous calls from all the placers after an interval of time given by the LCM of 10, 12, 20 and 25 which is 300.

So, the next simultaneous calls are received after 300 minutes or after 5 hours or at 10:00 a.m.

Hence option(A) is the correct choice

(AA)2=DCBA(AA)2=DCBA, where AA, BB, CC and DD are distinct digits with BB being odd. Find the value of DD.

✔ A.4

B.6

✖ C.1

D.1 or 4

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4.0/5

(11) Discuss Toggle Dropdown

Solution:

Option(A) is correct

Check the digits A=1,2,3,4,......9A=1,2,3,4,......9

11×11=12111×11=121 (Not valid, only 3 digit number)

22×22=48422×22=484 (Not valid, only 3 digit number)

33×33=108933×33=1089 (Not valid, BB is NOT odd)

44×44=193644×44=1936 (NOT valid since digit AA is not 44 on the ridht hand side - unit's place)

55×55=302555×55=3025 (Not valid, BB is NOT odd)

66×66=435666×66=4356 (Valid choice, keep note of it)

77×77=592977×77=5929 (Not valid, digits are NOT distinct)

88×88=774488×88=7744 (Not valid, digits are NOT distinct)

99×99=980199×99=9801 (Not valid, BB is NOT odd)

So we see that, only 66×66=435666×66=4356 satisfy the criteria.

Hence, D=4

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The number obtained by interchanging the two digits of a two digit number is less than the original number by 27. If the difference between the two digits of the number is 3, what is the original number?

A.74

B.63

C.85

✔ D.Cannot be determined

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0/0

Discuss Toggle Dropdown

Solution:

Option(D) is correct

Let the number be xyxy.

The number =10x+y=10x+y

On interchanging the digits of the number =10y+x=10y+x

⇒ 10x+y−10y−x=2710x+y−10y−x=27

⇒ x−y=3x−y=3

Now, yy is not equal to zero and the set of digits satisfying the condition are :

(9,6),(8,5),(7,4),(6,3),(5,2),(4,1)(9,6),(8,5),(7,4),(6,3),(5,2),(4,1)

⇒ We can't reach on the distinct answer.

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How many zeroes will be there in the expansion of the expression

11×22×33×44……..×10010011×22×33×44……..×100100

A.1200

B.1232

✔ C.1300

D.1296

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5.0/4

(3) Discuss Toggle Dropdown

Solution:

Option(C) is correct

Number of zeroes will be decided by the power of 2 and 5 in the product.

Since, the power of 5 is less than the power of 2 hence number of zero will be equal to power of 5.

Power of 5=55×1010×1515×……×1001005=55×1010×1515×……×100100

⇒ (5+10+15+20+(25×2)+30+40+.....)(5+10+15+20+(25×2)+30+40+.....)

⇒ (5+10+15+....100)+(25+50+75+100)(5+10+15+....100)+(25+50+75+100)

⇒ 202[2×5+19×5]+250202[2×5+19×5]+250

⇒ 1050+250=1050+250= 1300

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How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?

A.3001

✔ B.2893

C.2704

D.2890

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4.2/5

Discuss Toggle Dropdown

Solution:

Option(B) is correct

While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 key strokes.

There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers.

There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.

Then 1000 is a four-digit number which requires 4 keystrokes.

Totally, therefore, one requires 9+180+2700+4=9+180+2700+4= 2893 keystrokes.

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When writing numbers from 1 to 10,000, how many times is the digit 9 written?

A.3200

B.3600

✔ C.4000

D.4200

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How do you find this question?

4.5/8

(3) Discuss Toggle Dropdown

Solution:

Option(C) is correct

The digits 9 occurs in the thousands place in 1000 numbers.

It occurs in the hundreds place in 1000 numbers and so on

The digit occurs 4000 times.

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zz4CVh3SPhE

When writing numbers from 1 to 10,000, how many times is the digit 9 written?

A.3200

B.3600

✔ C.4000

D.4200

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How do you find this question?

4.5/8

(3) Discuss Toggle Dropdown

Solution:

Option(C) is correct

The digits 9 occurs in the thousands place in 1000 numbers.

It occurs in the hundreds place in 1000 numbers and so on

The digit occurs 4000 times.

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