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Problems on Numbers Questions

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You are given 87 matchsticks. They are arranged in rectangular manner.Find the number of triangles formed?


A43

B54

C44

D23

Answer: Option A

Explanation:

1st triangle formed by using 3 sticks.
so remaining sticks are =87-3=84
all other can be formed by just using two sticks =84/2 =42

so total =42(by using two sticks)+1(by using three sticks) =43

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How many Palindrone number are there between 5000 To 67000 . Palindrone no 5335 is palindone no because if u read from first and last the value remains same


A456

B620

C630

D768

Answer: Option B

Explanation:

5000 To 10000 = Per section 10 Total 5 section S/T = 5*10 = 50
Than
10000 To 67000 = Per Section 10 Total 57 Section S/T = 57*100 = 570
Total is = 50+570 = 620

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If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?


A18

B12

C19

D43

Answer: Option C

Explanation:

There are total 9 4's in 14, 24, 34...,94 & total 10 4's in 40,41,42....49

thus, 9+10=19.

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The fourteen digits of a credit card are to be written in the boxes shown above. If the sum of every three consecutive digits is 18, then the value of x is:


fourteen-digits-of-a-credit-card-tcs-mock-test-q4interview

A3

B2

C1

DCannot be determined

Answer: Option A

Explanation:

Let us assume right most two squares are a , b

Then Sum of all the squares = 18 x 4 + a + b ------ (1)
Also Sum of the squares before 7 = 18

Sum of the squares between 7, x = 18 and
sum of the squares between x , 8 = 18

So Sum of the 14 squares = (18 + 7 + 18 + x + 18 + 8 + a + b) ------- (2)
By Solving equation 1 and 2
x = 3

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Given that 0 < a < b < c < d, which of the following the largest?


A(c+d) / (a+b)

B(a+d) / (b+c)

C(b+c) / (a+d)

D(b+d) / (a+c)

Answer: Option A

Explanation:

Take a = 1, b = 2, c = 3, d = 4

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X = 101102103104105106107......146147148149150 (From numbers 101-150). Find out the remainder when this number is divided by 9.


A7

B4

C2

D3

Answer: Option C

Explanation:

The divisibility rule for 9 is sum of the digits is to be divisible by 9. So

We calculate separately, sum of the digits in hundreds place, tenths place, and units place.

Sum of the digits in hundreds place: 1 * 50 = 50

Sum of the digits in tenths place : 0 * 9 + 1 * 10 + 2 * 10 + 3 * 10 + 4 * 10 + 5 * 1 = 105

Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) * 5 = 225

So total = 380

So remainder = 380 / 9 = 2

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A number is 101102103104...150. As 101 102 103 103.... 150. What is reminder when divided by 3?


A5

B2

C7

D1

Answer: Option B

Explanation:

Divisibility rule for 3 also same as 9.
so from the above discussion sum of the digits = 380
and remainder = 380/3 = 2

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7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?


A34

B51

C50

D56

Answer: Option B

Explanation:

Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats.
So for every 4 terms we get one term with 3 in its unit digit.
So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.

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7,77,777,7777,77777,777777....
Find last 3 digits of sum of first 26 terms


A452

B456

C732

D672

Answer: Option C

Explanation:

At unit digit 7 would be coming 26 timing
so, 26*7=182.
2 at unit place
Then 25*7=175+ (18 carry)=193
Then 24*7=187
So, Last digits will be 732.,

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How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?


A65 ways

B45 ways

C12 ways

D60 ways

Answer: Option D

Explanation:

Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6 There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways so total number of ways of arranging all these numbers are 10 * 6 = 60 ways

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