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 Quantitative Aptitude - Arithmetic Ability Questions

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Q:

A boat sails 15 km of a river towards upstream in 5 hours. How long will it take to cover the same distance downstream, if the speed of current is one-fourth the speed of the boat in still water:

A) 1.8h B) 3h
C) 4h D) 5h
 
Answer & Explanation Answer: B) 3h

Explanation:

Upstream speed = B-S

Downstream speed = B+s

B-S = 15/5 = 3 km/h

Again          B= 4S

Therefore    B-S = 3= 3S

=>             S = 1 and B= 4 km/h

Therefore    B+S = 5km/h

Therefore, Time during downstream = 15/5 = 3h

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Filed Under: Boats and Streams

126 26798
Q:

Select the pair that does NOT belong in the following group.

 

(64, 9), (81, 10), (36, 8), (121, 12)

 

A) 121, 12 B) 36, 8
C) 81, 10 D) 64, 9
 
Answer & Explanation Answer: B) 36, 8

Explanation:
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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 26738
Q:

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6 B) b=5; l=6; H=18
C) l=6; b=18; H=15 D) l=5; b=18; H=18
 
Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54⇒x2=54×2⇒x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180⇒H=6

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

33 26731
Q:

A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

A) 0% B) 5%
C) 7.5% D) 10%
 
Answer & Explanation Answer: A) 0%

Explanation:

C.P = Rs.3000 

S.P =Rs. [3600 x 10] / [100+(10 x 2)] = Rs.3000 

Gain =0%

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Filed Under: True Discount

35 26680
Q:

If the ratio of the ages of two friends A and B is in the ratio 3 : 5 and that of B and C is 3 : 5 and the sum of their ages is 147, then how old is B?

A) 27 Years B) 75 Years
C) 45 Years D) 49 Years
 
Answer & Explanation Answer: C) 45 Years

Explanation:

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

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Filed Under: Ratios and Proportions
Exam Prep: GRE

59 26670
Q:

In Arun's opinion, his weight is greater than 65 kg but leas than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all of them are correct in their estimation, what is the average of diferent probable weights of Arun ?

A) 55.5 kg B) 66.5 kg
C) 77.5 kg D) 88.5 kg
 
Answer & Explanation Answer: B) 66.5 kg

Explanation:

Let Arun's weight be X kg.

According to Arun, 65 < X < 72.

According to Arun's brother, 60 < X < 70.

According to Arun's mother, X < 68.

The values satisfying all the above conditions are 66 and 67.

Required average = (66 + 67) / 2 = 66.5 kg

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Filed Under: Average

21 26666
Q:

In the following question, select the odd word pair from the given alternatives.

 

 

 

A) Cup – Tea B) Jug – Water
C) Bowl – Curd D) Pen – Butter
 
Answer & Explanation Answer: D) Pen – Butter

Explanation:
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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 26649
Q:

Consider the following statements:

1) The perimeter of a triangle is greater than the sum of its three medinas.

2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.

Which of the above statements is/are correct?

A) 1 only B) 2 only
C) Both 1 and 2 D) Neither 1 nor 2
 
Answer & Explanation Answer: C) Both 1 and 2

Explanation:
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ΔABD, AD is a median
⇒
AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
∴
AB + BC + AC > AD + BE +CF
 
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD,
AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the
third side]
----
1
In triangle ADC,
AC + DC > AD [because, the sum of any two
sides of a tri
angle is always greater than the
third side]
----
2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved
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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

1 26636