Quantitative Aptitude - Arithmetic Ability Questions

Q:

The calendar of the year 1897 can be used again in the year?

A) 1908 B) 1901
C) 1903 D) 1926
 
Answer & Explanation Answer: C) 1903

Explanation:

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

 

Solution : 

Given Year is 1897, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therfore, The calendar of the year 1897 can be used again in the year 1903

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Filed Under: Calendar

85 28427
Q:

Cosec (90 - θ)

 

A) tan θ B) cot θ
C) sec θ D) cos θ
 
Answer & Explanation Answer: C) sec θ

Explanation:
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Filed Under: Simplification
Exam Prep: Bank Exams

0 28426
Q:

A man went downstream for 28 km in a motor boat and immediately returned. It took the man twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 minutes. Find the speed of the boat in still water and the speed of the river flow.

A) 12 km/hr, 3 km/hr B) 9 km/hr, 3 km/hr
C) 8 km/hr, 2 km/hr D) 9 km/hr, 6 km/hr
 
Answer & Explanation Answer: B) 9 km/hr, 3 km/hr

Explanation:

Let the speed of the boat = p kmph

Let the speed of the river flow = q kmph

From the given data,

2 x 28p + q = 28p - q

 

=> 56p - 56q -28p - 28q = 0

=> 28p = 84q

=> p = 3q.

Now, given that if

283q + 2q + 283q - 2q = 67260=> 285q + 28q = 67260=> q = 3 kmph=> x  =3q = 9 kmph

 

Hence, the speed of the boat = p kmph = 9 kmph and the speed of the river flow = q kmph = 3 kmph.

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Filed Under: Boats and Streams
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

79 28396
Q:

In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

A) 319/420 B) 27/111
C) 114/121 D) 231/420
 
Answer & Explanation Answer: A) 319/420

Explanation:

P(P)=14,P(Q)=15,P(R)=16,P(S)=17
All the events are mutually exclusive hence,

 

Required probability = P(P)+P(Q)+P(R)+P(S)

 

14+15+16+17=319420

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Filed Under: Probability

38 28389
Q:

Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :

A) 20 % B) 40 %
C) 60 % D) 80 %
 
Answer & Explanation Answer: C) 60 %

Explanation:

Let the total original sale be Rs. 100. Then, original number of visitors = 100.

New number of visitors  =  120/0.75 = 160.

Increase % = 60 %.

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Filed Under: Percentage

95 28385
Q:

What is the simplified value of 

125 - 24 - 124-23+123 - 22 - 122 - 21+ 121 - 20 ?

A) 5 + 2√5 B) 5 – 5√2
C) √25 + √21 D) √25 – 4
 
Answer & Explanation Answer: A) 5 + 2√5

Explanation:
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Filed Under: Simplification
Exam Prep: Bank Exams

0 28382
Q:

ΔABC is right angled at B. If m∠C = 45 deg, then find the value of (cosecA - √3).

 

A) (4-√3)/2   B) √2-√3  
C) -√3/2 D) (√6-1)/√3
 
Answer & Explanation Answer: B) √2-√3  

Explanation:
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Filed Under: Simplification
Exam Prep: Bank Exams

0 28367
Q:

In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

A) 1/2 B) 5/12
C) 7/15 D) 3/12
 
Answer & Explanation Answer: B) 5/12

Explanation:

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

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Filed Under: Probability

47 28353