Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

110.

A person distributes his pens among four friends A, B, C and D in the ratio 1 third space colon space 1 fourth space colon space 1 fifth space colon space 1 over 6. What is the minimum number of pens that the person should have?

  • 75

  • 45

  • 57

  • 65


C.

57

A: B : C : D
       equals space space space 1 third space colon space 1 fourth space colon space 1 fifth space colon space 1 over 6
equals space space 60 over 3 space colon space 60 over 4 space colon space 60 over 5 space colon space 60 over 6
equals space space space 20 space colon space 15 space colon space 12 space colon space 10
Hence, minimum number of pen = 20 + 15 + 12 + 10 = 57.

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101.

Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively.

  • 750

  • 150

  • 300

  • 600


A.

750

It can be clearly concluded from the side, that triangle is right angled triangle.
    ∴     Area of the triangular field
                     equals space space 1 half space cross times space AB space cross times space BC
equals space 1 half space cross times space 15 space cross times space 20 space equals space 150 space straight m squared
So, rate of sowing seed per sq m. is ₹ 5
∴    Cost of sowing seed for 150 m2 = 150 x 5 = ₹ 750

    

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106.

If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is

  • 10 cm

  • 8 cm

  • 7.5 cm

  • 6 cm


C.

7.5 cm

Let the sides be 4x, 5x, 6x respectively. As largest side is the base, therefore corresponding altitude (h) is given by,
Now, 
Inradius space left parenthesis straight r right parenthesis space equals space fraction numerator Area space of space triangle over denominator Semi minus Perimeter end fraction
3 space equals space fraction numerator begin display style 1 half end style space cross times space 6 straight x space cross times space straight h over denominator begin display style fraction numerator 15 straight x over denominator 2 end fraction end style end fraction
rightwards double arrow space space space straight h space equals space fraction numerator 3 space cross times space 15 over denominator 6 end fraction space equals space 7.5 space cm

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108.

If a2 + 1 = a,  then the value of a12 + a6 + 1 is

  • 2

  • 3

  • -3

  • 1


B.

3

a2 + 1 = a
  rightwards double arrow space space space space straight a space plus space 1 over straight a space equals space 1
On Squaring both sides, we get,
           straight a squared space plus space 1 over straight a squared space plus space 2 space equals space 1
rightwards double arrow space space space straight a squared space plus space 1 over straight a squared space equals space minus 1
On Cubing both sides, we get,
        space space space open parentheses straight a squared space plus space 1 over straight a squared close parentheses cubed space equals space left parenthesis negative 1 right parenthesis cubed
rightwards double arrow space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 3 straight a squared space cross times space 1 over straight a squared open parentheses straight a squared space plus space 1 over straight a squared close parentheses space equals space minus 1 space space rightwards double arrow space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 3 space cross times space minus 1 space equals space minus 1
rightwards double arrow space space space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 1 space equals space 3 space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
If we put a = 1  in equation (i), it satisfies the equation. i.e. a = 1.
Now,  put a = 1. in  a12 + a6 + 1 i.e (1)12 + (1)6 + 1 = 3

             

27 Views

109.

If straight x space equals space straight a space sec space straight theta space cos space straight ϕ comma space space straight y space equals space straight b space sec space straight theta space sin space straight ϕ and straight z space equals space straight c space tan space straight theta. then the value of straight x squared over straight a squared space plus space straight y squared over straight b squared space equals space straight z squared over straight c squared

  • 9

  • 0

  • 1

  • 4


C.

1

straight x squared over straight a squared space plus space straight y squared over straight b squared space equals space straight z squared over straight c squared
On putting the values of x, y and z, we get
equals space fraction numerator straight a squared sec squared straight theta space cos squared straight ϕ over denominator straight a squared end fraction space plus space fraction numerator straight b squared sec squared straight theta space sin squared straight ϕ over denominator straight b squared end fraction space minus space fraction numerator straight c squared tan squared straight theta over denominator straight c squared end fraction
space equals space sec squared straight theta space open square brackets cos squared straight ϕ space plus space sin squared straight ϕ close square brackets space minus space tan squared straight theta
equals space space sec squared straight theta space minus space tan squared straight theta space equals space 1

28 Views

102.

The value of (tan 1°  tan 2°  tan 3°.... tan 89°) is

  • undefined

  • 0

  • 1

  • 89


C.

1

31 Views

107.

The cost of manufacture of an article was ₹ 900. The trader wants to gain 25% after giving a discount of 10%. The marked price should be

  • ₹ 1000

  • ₹ 1500

  • ₹ 1250

  • ₹ 1200


C.

₹ 1250

Let the marked price be ₹ x.
Cost price of the article  = ₹ 900 (given)
According to the question,
    90 percent sign space of space straight x space equals space 900 space cross times space fraction numerator left parenthesis 100 space plus 25 right parenthesis over denominator 100 end fraction
  rightwards double arrow space space 90 over 100 space cross times space straight x space equals space 900 space cross times space 125 over 100
rightwards double arrow space space straight x space equals space ₹ space 1250

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104.

Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is

  • 25

  • 10

  • 15

  • 20


D.

20

Let the number of sums solved correctly by Mohan be x.
Then, According to the question,
           (3) x (x) - (30 - x) x 2 = 40
 ⟹          3x - 60 + 2x = 40
⟹             5x - 60 = 40
⟹              5x = 60 + 40
⟹                5x = 100
⟹                 x = 100/5 = 20
∴  Mohan attempted 20 questions correctly.

 


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103.

A, B, C and D purchase a gift worth ₹ 60. A pays 1/2 of what others are paying, B pays 1/3rd of what others are paying and C pays 1/4 of what others are paying. What is the amount paid by D?

  • 14

  • 15

  • 16

  • 13


D.

13

A + B + C + D = ₹ 60 (given)
Now, According to the question,
    straight A space equals space 1 half left parenthesis straight B space plus space straight C space plus space straight D right parenthesis
rightwards double arrow space space 2 straight A space equals space straight B space plus space straight C space plus space straight D
rightwards double arrow space space 2 straight A space equals space 60 space minus space straight A
rightwards double arrow space space space 3 straight A space equals space 60
rightwards double arrow space space space straight A space equals space 20
straight B space equals space 1 third left parenthesis straight A space plus space straight C space plus space straight D right parenthesis
rightwards double arrow space space 3 straight B space equals space straight A space plus space straight C space space plus space space straight D
rightwards double arrow space space space 3 straight B space equals space 60 space minus space straight B
rightwards double arrow space space space 4 straight B space equals space 60
rightwards double arrow space space space space straight B space equals space 15

straight C space equals space 1 fourth left parenthesis straight A space plus space straight B space plus space straight D right parenthesis
rightwards double arrow space space 4 straight C space equals space straight A space plus space straight B space plus space straight D
rightwards double arrow space space 4 straight C space equals space 60 space minus space straight C
rightwards double arrow space space 5 straight C space equals space 60
rightwards double arrow space space space straight C space equals space 12

∴    Amount paid by D = ₹ (60 - 20 - 15 - 12) = ₹ 13.


36 Views

105.

The greatest common divisor of
  3 to the power of 3 to the power of 333 end exponent space plus space 1 and 3 to the power of 3 to the power of 334 end exponent space plus space 1 is

  • 3 to the power of 3 to the power of 333 end exponent space plus space 1

  • 20

  • 2

  • 1


D.

1

3 to the power of 3 to the power of 333 end exponent space plus space 1 space space and space 3 to the power of 3 to the power of 334 end exponent space plus space 1
27 to the power of 333 space plus space 1 to the power of 333 space space and space space space 27 to the power of 334 space plus space 1
Now, xm + am is divisible by (x + a) when m is odd
∴  27333 + 1333 is divisible by (27 + 1) = 28
Similarly,
   27334 + 1334 is never divisible by (x + a).
So, the greatest common divisor between 3 to the power of 3 to the power of 333 end exponent space plus space 1 space space and space 3 to the power of 3 to the power of 334 end exponent space plus space 1 is 1.

39 Views