A) Statement I alone is sufficient.

B ) Statement II alone is sufficient.

C) Both statement taken together are sufficient to answer this question.

D) Each statement alone is sufficient.

E) Both statements together are not sufficient.

Q1) If a,b and c are natural numbers,is ab^2 + 2bc + a^2c even ?

A ) ab + c^2 is odd.

B ) bc + a is odd.

Solution :-

ab^2 + 2bc + a^2c = 2k.

From statement A, ab + c^2 = odd.

Possibility 1) a/ b = even, c = odd.

Possibility 2) a or b odd, c = even.

Not sufficient.

From B, bc + a = odd.

Possibility 1) a= odd, b/c = even

Possibility 2) a=even, b/c = odd.

Not sufficient.

Hence E

Q2) Ram,Rahul and Raghav are top three rankers of class.The sum of their total marks is 358.Did Rahul get the highest marks ?

A)Ram got 175 marks.

B ) Rahul got 119 marks.

Solution :-

By statement A, Ram got 175 marks.

Marks of Rahul + Marks of Raghav = 183.

We can’t predict Rahul marls is more than 175 or not.

Not sufficient.

By statement B, Rahul got 119 marks.

Ram + Raghav marks = 358 – 119 = 239.

Anyone can get more marks than 119 here.

Sufficient.

Hence B.

Q3) six persons A,B,C,D,E and F are sitting around a circular table, is F sitting opposite D?

A) B is sitting between C & E.

B ) F is sitting to the left to A, who is sitting opposite B.

Solution :-

From statement A, CBE is a group. Then F can be opposite of D.

Sufficient

From statement B, F can be oppposite of C/D/E.

Not sufficient.

Hence 1.

Q4) If 8th march in year X is Thursday, then find X, 2000 < = X < = 2018.

A) year 2018 will have the same calender as that of X.

B ) X+1 is a leap year.

Solution :-

From statement A, there are possibilities 2 or more year can be same as year 2018.

From statement B, X + 1 is a leap year.

X = 2003,2007,2011,2015.

2007 is a year where on 8th March, there is a Thursday.

Q5) The total number of students in class A is 160% more than the total number of students in class B, which in turn, is 20% less than the number of students in class C. In which class among A,B and C is the total number of students who passed the highest ?

A)The total number of students of class A who failed in less than the number of students of class B who failed.

B )The total number of class students of class A who failed in less than the number of students of class C who failed.

Solution :-

Given

A = 1.6 B.

B = 0.8 C.

Suppose C = 400,B = 320, A= 480.

By statement A, Suppose 40 failed in class A. 50 failed in class B.

Pass students in A = 440, B=270.

Not sufficient.

By statement B, Suppose 40 failed in A and 50 failed in C.

Pass students in A=440, C=350.

Not sufficient.

But if we add statement A & B.

Highest number of pass students in A.

Hence C.

Q6) Four students A,B,C and D were among the top four rankers in a class in each of three subjects P,Q and R. No two of them got same rank in any subject and none of the got the same rank in any two subjects. Further,A got in second rank in Q, and C got the fourth rank in P. What are their respective ranks in each of the three subjects ?

A) The sum of the ranks, in all the three subjects put together, of no two persons in the same and the sum of the ranks of D is highest. The rank of D in P is the same as the rank of B in Q.

B ) B got a better rank than D in each of the three sections.

Solutions :- sum of D is highest. It means D’s rank is 2nd ,3rd and 4th in 3 different subjects.

The sum of all 4 individual’s ranks are 6(1+2+3),7(1+2+4),8(1+3+4),9(2+3+4) respectively.

D’s sum of rank is highest. Sum of ranks shall be 9.

C got 4th rank in P.

And as given rank of D in P = rank of B in Q.

Rank of D in P is 3rd, rank of B in Q is 3rd. Rank of D in Q is 4th and in R his rank is 2nd.

Now we can make a table.

Person/Subject | P | Q | R |
---|---|---|---|

A | 1st | 2nd | 4th |

B | 2nd | 3rd | 1st |

C | 4th | 1st | 3rd |

D | 3rd | 4th | 2nd |

Sufficient.

Now let’s take statement B.

B’s rank is better tan D in all three subjects.

It would also be sufficient.

Hence D.

Q7) There are two rods L1 and L2 – one made of metal A and other made of metal B. They are joined end to end to form a third rod L. Due to rise in temparature, the length L1 is increased by 10% and that of L2 of 25%. Find the percentage increase in length of rod L.

A ) The original length of rod L was 120 cm.

B ) The sum of 33 1/3% of the length L1 and 44 4/9% of the length L2 was 44 cm.

Solution :-

From statement A, length of rod was 120 cm. Length L1 and lentgh L2 can’t be determined.

From statement B,

0.33 L1 + 0.44 L2 = 44.

0.03 L1 +0.04 L2 = 4.

Can’t be determined.

If we add both statement A & B then it can be found easily.

Hence D.

Q8) Azim and Kamran sells n mobile handsets each. While Azim makes an overall profit of x%, Kamran incurs an overall loss of x/2 %. If Azim’s total profit is equal to the selling price of m of his handsets and Kamran’s loss is equal to the selling price of y of his handsets,then what is the value of y.

A) mn = 3600

B ) n – m = 900.

Solution :-

Profit x% = m / (n – m).

Loss x/2 % = y / (y + n).

2y / y + n = m / (n – m)

2yn -2ym = my + mn

y = nm / 2n – 3m.

By using both statements together

Answer can be find easily.

Hence C

A) Statement I alone is sufficient.

B ) Statement II alone is sufficient.

C) Both statement taken together are sufficient to answer this question.

D) Each statement alone is sufficient.

E) Both statements together are not sufficient.

Q1) If a,b and c are natural numbers,is ab^2 + 2bc + a^2c even ?

A ) ab + c^2 is odd.

B ) bc + a is odd.

Solution :-

ab^2 + 2bc + a^2c = 2k.

From statement A, ab + c^2 = odd.

Possibility 1) a/ b = even, c = odd.

Possibility 2) a or b odd, c = even.

Not sufficient.

From B, bc + a = odd.

Possibility 1) a= odd, b/c = even

Possibility 2) a=even, b/c = odd.

Not sufficient.

Hence E

Q2) Ram,Rahul and Raghav are top three rankers of class.The sum of their total marks is 358.Did Rahul get the highest marks ?

A)Ram got 175 marks.

B ) Rahul got 119 marks.

Solution :-

By statement A, Ram got 175 marks.

Marks of Rahul + Marks of Raghav = 183.

We can’t predict Rahul marls is more than 175 or not.

Not sufficient.

By statement B, Rahul got 119 marks.

Ram + Raghav marks = 358 – 119 = 239.

Anyone can get more marks than 119 here.

Sufficient.

Hence B.

Q3) six persons A,B,C,D,E and F are sitting around a circular table, is F sitting opposite D?

A) B is sitting between C & E.

B ) F is sitting to the left to A, who is sitting opposite B.

Solution :-

From statement A, CBE is a group. Then F can be opposite of D.

Sufficient

From statement B, F can be oppposite of C/D/E.

Not sufficient.

Hence 1.

Q4) If 8th march in year X is Thursday, then find X, 2000 < = X < = 2018.

A) year 2018 will have the same calender as that of X.

B ) X+1 is a leap year.

Solution :-

From statement A, there are possibilities 2 or more year can be same as year 2018.

From statement B, X + 1 is a leap year.

X = 2003,2007,2011,2015.

2007 is a year where on 8th March, there is a Thursday.

Q5) The total number of students in class A is 160% more than the total number of students in class B, which in turn, is 20% less than the number of students in class C. In which class among A,B and C is the total number of students who passed the highest ?

A)The total number of students of class A who failed in less than the number of students of class B who failed.

B )The total number of class students of class A who failed in less than the number of students of class C who failed.

Solution :-

Given

A = 1.6 B.

B = 0.8 C.

Suppose C = 400,B = 320, A= 480.

By statement A, Suppose 40 failed in class A. 50 failed in class B.

Pass students in A = 440, B=270.

Not sufficient.

By statement B, Suppose 40 failed in A and 50 failed in C.

Pass students in A=440, C=350.

Not sufficient.

But if we add statement A & B.

Highest number of pass students in A.

Hence C.

Q6) Four students A,B,C and D were among the top four rankers in a class in each of three subjects P,Q and R. No two of them got same rank in any subject and none of the got the same rank in any two subjects. Further,A got in second rank in Q, and C got the fourth rank in P. What are their respective ranks in each of the three subjects ?

A) The sum of the ranks, in all the three subjects put together, of no two persons in the same and the sum of the ranks of D is highest. The rank of D in P is the same as the rank of B in Q.

B ) B got a better rank than D in each of the three sections.

Solutions :- sum of D is highest. It means D’s rank is 2nd ,3rd and 4th in 3 different subjects.

The sum of all 4 individual’s ranks are 6(1+2+3),7(1+2+4),8(1+3+4),9(2+3+4) respectively.

D’s sum of rank is highest. Sum of ranks shall be 9.

C got 4th rank in P.

And as given rank of D in P = rank of B in Q.

Rank of D in P is 3rd, rank of B in Q is 3rd. Rank of D in Q is 4th and in R his rank is 2nd.

Now we can make a table.

Person/Subject | P | Q | R |
---|---|---|---|

A | 1st | 2nd | 4th |

B | 2nd | 3rd | 1st |

C | 4th | 1st | 3rd |

D | 3rd | 4th | 2nd |

Sufficient.

Now let’s take statement B.

B’s rank is better tan D in all three subjects.

It would also be sufficient.

Hence D.

Q7) There are two rods L1 and L2 – one made of metal A and other made of metal B. They are joined end to end to form a third rod L. Due to rise in temparature, the length L1 is increased by 10% and that of L2 of 25%. Find the percentage increase in length of rod L.

A ) The original length of rod L was 120 cm.

B ) The sum of 33 1/3% of the length L1 and 44 4/9% of the length L2 was 44 cm.

Solution :-

From statement A, length of rod was 120 cm. Length L1 and lentgh L2 can’t be determined.

From statement B,

0.33 L1 + 0.44 L2 = 44.

0.03 L1 +0.04 L2 = 4.

Can’t be determined.

If we add both statement A & B then it can be found easily.

Hence D.

Q8) Azim and Kamran sells n mobile handsets each. While Azim makes an overall profit of x%, Kamran incurs an overall loss of x/2 %. If Azim’s total profit is equal to the selling price of m of his handsets and Kamran’s loss is equal to the selling price of y of his handsets,then what is the value of y.

A) mn = 3600

B ) n – m = 900.

Solution :-

Profit x% = m / (n – m).

Loss x/2 % = y / (y + n).

2y / y + n = m / (n – m)

2yn -2ym = my + mn

y = nm / 2n – 3m.

By using both statements together

Answer can be find easily.

Hence C