The sum of the areas of the 10 squares, the lengths of whose sid

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41.

The sum of the areas of the 10 squares, the lengths of whose sides are 20 cm, 21 cm,....29 cm respectively is

  • 6085 cm2

  • 8555 cm2

  • 2470 cm2

  • 2470 cm2


A.

6085 cm2

Required sum = 202 + 212 + ......292
 = (12 + 22 +.......+292) - (12 + 22+........+192)

equals space fraction numerator 29 left parenthesis 29 plus 1 right parenthesis thin space left parenthesis 2 space cross times 29 space plus space 1 right parenthesis over denominator 6 end fraction space minus space fraction numerator 19 left parenthesis 19 plus 1 right parenthesis space left parenthesis 2 cross times 19 plus 1 right parenthesis over denominator 6 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space 1 squared space plus space 2 squared space plus.... straight n squared space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis thin space left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction close square brackets
equals space 8555 space minus space 2470 space equals space 6085 space cm squared space

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42.

The missing term in the sequence
2, 3, 5, 7, 11, .......17, 19 is

  • 16

  • 15

  • 14

  • 14

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43.

The wrong number in the sequence
8, 13, 21, 32, 47, 63, 83 is

  • 32

  • 47

  • 63

  • 63

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44.

A number, when divided successively by 4, 5 and 6 leaves remainders 2, 3 and 4 respectively. The least such number is

  • 50

  • 53

  • 56

  • 58

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45.

A number when divided by 361 gives a remainder 47. If the same number is divided by 19, the remainder obtained is

  • 3

  • 8

  • 9

  • 9

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46.

The unit digit in the product (2467)153 x (341)172 is

  • 7

  • 8

  • 4

  • 4

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47.

If the sum of the digits of a three-digit number is subtracted from that number, then it will always be divisible by

  • 3 only

  • 9 only

  • Both 3 and 9

  • Both 3 and 9

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48.

If straight U subscript straight n space equals space 1 over straight n space minus space fraction numerator 1 over denominator straight n plus 1 end fraction, then the value of U1 + U2 + U3 + U4 + U5

  • 1 fourth
  • 5 over 6
  • 1 over 6
  • 1 over 6
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49.

The sum of the cubes of two numbers in the ratio 3 : 4 is 5824. The sum of the number is 

  • (5824)1/3

  • 28

  • 24

  • 24

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50.

What least value must be assigned to '*' so that the number 451*603 is exactly divisible by 9?

  • 7

  • 8

  • 5

  • 5

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