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A student attempts x number of questions. He answers 15 correctly out of first 20 questions and of the remaining questions, he answers 1/3 correctly. If all questions have same credit and the student gets 50 % marks, then find the value of x.

A.

30

B.

35

C.

45

D.

50

Answer with explanation

Answer: Option DExplanation

Student attempts x questions.

2) Out of 20 questions he answers 15 correctly and of (x – 20) questions he answered 1/3 correctly.

3) The student gets 50 % marks.

Therefore,

15 + | 1 | (x – 20) = 50% of x |

3 |

15 + | 1 | (x – 20) = | 50 | × x |

3 | 100 |

15 + | 1 | (x – 20) = | x |

3 | 2 |

90+2 (x-20)=3x

Solving this equation, we get

x=50

Hence, the number of question attempted by the students = 50

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A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in the calculation ?

A.

54 %

B.

64 %

C.

74 %

D.

84 %

Answer with explanation

Answer: Option BExplanation

Let the number be x.

Then, ideally he should have multiplied by x by 5/3. Hence Correct result was x * (5/3)= 5x/3.

By mistake he multiplied x by 3/5 . Hence the result with error = 3x/5

Then, error = (5x/3 – 3x/5) = 16x/15

Error % = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

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405 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. How many sweets did each child recieve ?

A.

9

B.

10

C.

11

D.

12

Answer with explanation

Answer: Option AExplanation

Workspace

1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of the girls pass. The percentage of the total who failed is :

A.

58

B.

62 *2/3

C.

64

D.

67

Answer with explanation

Answer: Option BExplanation

Total number of students = 1100 + 700 = 1800.

Number of students passed = (42% of 1100 + 30% of 700) – (462 + 210) = 672.

Number of failues = 1800-672 = 1128.

Percentage failure = (1128/1800 * 100 )% = 62 * 2/3 %.

Workspace

If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 350%, the resultant fraction is 25/51. What is the original fraction ?

A.

31/25

B.

15/17

C.

14/25

D.

11/16

Answer with explanation

Answer: Option BExplanation

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If the pice of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumpion of sugar by

A.

15

B.

20

C.

25

D.

30

Answer with explanation

Answer: Option BExplanation

Let original consumption = 100 kg and new consumption = x kg,

So, 100 * 6 = x * 7.50 => x = 80kg

Reduction in consumption = 20%.

Workspace

Of the 1000 inhabitants of a town, 60 % are males of whom 120 % are literate. If, of all the inhabitants, 25% are literate, then what percent of the females of the town are literate ?

A.

32.5 %

B.

43 %

C.

46.6 %

D.

53.2 %

Answer with explanation

Answer: Option AExplanation

Number of males = 60% of 1000 = 600. Number of females = (1000 – 600) = 400.

Number of literates = 25% of 1000 = 250.

Number of literate males = 20% of 600 = 120.

Number of literate females = (250 – 120) = 130.

Required pecentage = (130/400 * 100 ) % = 32.5 %.

Workspace

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