The following question is based on the given data | Arithmetical Reasoning Questions & Answers

Q:

Find by how much percentage the average of first 10 odd numbers (starting from 1) is greater than the last term ?

A) 991/16 %	B) 900/19 %
C) 874/13 %	D) 719/17 %

Answer & Explanation Answer: B) 900/19 %

Explanation:

First ten odd numbers form an arithmetic progression of the form
1,3,5,7,9......
here a = first term = 1
d = common difference = 2
Average of first n numbers = (2a + (n - 1)d)/2
n th term of the AP = a + (n - 1)d
Substituting a = 1, d = 2 and n = 10 in the above formulas
average of first 10 numbers = 10
10 th term of the AP = 19
Therefore average of first 10 terms is 19 - 10 = 9 greater than
the last term. Hence the average is greater than the sum by
9/19 X 100 = 900/19 %

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Q:

On sport’s day, if 12 children were made to stand in each column, 98 columns could be formed. After a certain music being played the plan was that the children have to stand in concentric circles such that the innermost circle will have one student, the next circle will have two students, the third circle will have 3 students and so on. How many circles will be formed in this manner ?

A) 44	B) 46
C) 48	D) 49

Answer & Explanation Answer: C) 48

Explanation:

No. of students will be 98 x 12 = 1176
let us consider n circles then 1 + 2 + 3 + ....+ n = 1176
n(n+1)/2 = 1176
n = 48 circles.

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Q:

An intelligent boy is doing a jigsaw with 275 pieces. Each day that he fits pieces together there are fewer pieces left, and it is reasonable to assume that he fits more pieces each day because the number left to sort out diminishes progressively. Hence he is able to fit an extra piece as each day goes by. On the first day he fits 20 pieces. How many days does it take for him to complete the puzzle ?

A) 11 days	B) 15 days
C) 13 days	D) 16 days

Answer & Explanation Answer: A) 11 days

Explanation:

Let 'n' number of days it take to complete.

we know that sum of n number= n/2[2a+(n-1)d]
where a=20,d=1;
n/2[2x20+(n-1)1]=275
n[n+39]=550
$n^{2}$ + 39n - 550 = 0
$n^{2}$ + 50n - 11n - 550 = 0
(n-11) (n-50) = 0
n-11 = 0;
n = 11(Days).

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Q:

A bird lover was asked how many birds he had in the basket. He replied that there were all parrots but six, all pigeons but six, and all koel but six. How many birds he had in the basket in all ?

A) 8	B) 18
C) 9	D) 10

Answer & Explanation Answer: C) 9

Explanation:

There were all parrots but six means that six birds were not parrots but only pigeons and koels.

Similarly, number of parrots + number of koels = 6 and number of parrots + number of pigeons = 6.

This is possible when there are 3 parrots, 3 pigeons and 3 koels i.e. 9 birds in all.

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Q:

Consider the following statements

P : Suman is brilliant
Q : Suman is rich
R : Suman is honest

The negation of the statement "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as

1) $(P \land ~ R) \leftrightarrow Q$

2) $~ (P \land R) \leftrightarrow Q$

3) $Q \leftrightarrow (P \land ~ R)$

4) $~ (Q \leftrightarrow (P \land ~ R))$

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: D) 4

Explanation:

Suman is brilliant and dishonest is P $\land$ ~ R
Suman is brilliant and dishonest iff suman is rich is
Q $\leftrightarrow$ (P $\land$ ~ R)

Negative of statement is expressed as
~ (Q $\leftrightarrow$ (P $\land$ ~ R)).

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Q:

If 200 cats kill 200 mice in 200 days, then 8 cats would kill 8 mice in how many days ?

A) 8 days	B) 200 days
C) 800 days	D) 1 day

Answer & Explanation Answer: B) 200 days

Explanation:

As given
200 cats kill 200 mice in 200 days
1 cats kill 1 mouse in 200 days
Then
8 cats kill 8 mice in 200 days.

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Q:

There are two examinations rooms P and Q. If 10 students are sent from P to Q, then the number of students in each room is the same. If 20 candidates are sent from Q to P, then the number of students in P is double the number of students in Q. The number of students in room Q is ?

A) 100	B) 70
C) 120	D) 80

Answer & Explanation Answer: D) 80

Explanation:

If 'k' is number of students in P and 'l' is number of students in Q, then
From the given conditions, we have
=> k - 10 = l + 10 .......(1)
=> k + 20 = 2(l - 20) ....(2)

Solving these eqns, we get
k = 100 and l = 80.
Therefore, number of students in class Q is 80.

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Q:

The price of commodity X increases by 60 rupees every year, while the price of commodity Y increases by 20 rupees every year. If in 2002, the price of commodity X was Rs. 320 and that of Y was Rs. 740, in which year commodity X will cost 60 rupees more than the commodity Y ?

A) 2011	B) 2014
C) 2012	D) 2017

Answer & Explanation Answer: B) 2014

Explanation:

If After p years , commodity X will cost 60 rupees more than the commodity Y, then

320 + 60p = 740 + 20p + 60
=> p = 12

So in year (2002 + 12 = 2014), commodity X will cost Rs.60 more than the commodity Y.

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A) 6555	B) 5685
C) 1705	D) 870

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