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Q:

If the mean of m observations out of n observations is n and the mean of remaining observations is m, then what is the mean of all n observations?

A) 2m –m^2/n B) 2m –m/n
C) 2m D) 2m –m
 
Answer & Explanation Answer: A) 2m –m^2/n

Explanation:

Mean of m observations is n

Mean of n-m observations is m

So total = nm+ (n-m)m

Total observations = n

Mean = Total / Total observations = (2mn-m^2)/n = 2m –m^2/n

 

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Filed Under: Simplification
Exam Prep: Bank Exams

Q:

The monthly expenditure of a person is Rs. 6000. The distribution of expenditure on various items is as follows:

Item of expenditure            Amount(in Rs.)
Food                                      2000
Clothing                                   660
Fuel and rent                          1200
Education                                 480
Miscellaneous                         1660

If the above data is represented by a percentage bar diagram of height 15 cm, then what are the lengths of the two segments of the bar diagram corresponding to education and miscellaneous respectively?

 

A) 1.25 cm and 5 cm B) 1.2 cm and 4.15 cm
C) 1.2 cm and 3.5 cm D) 4.15 cm and 6 cm
 
Answer & Explanation Answer: B) 1.2 cm and 4.15 cm

Explanation:
15 cm corresponds to 6000 rs
Education = 480/6000 * 15 cm = 1.2 cm
Miscellaneous = 1660/6000 * 15 cm = 4.15 cm
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Filed Under: Bar Charts
Exam Prep: Bank Exams

Q:

Consider the following grouped frequency distribution:

  x           f

0-10       8

10-20     12

20-30     10

30-40      p

40-50      9

 

If the mean of the above data is 25.2, then what is the value of p?

 

A) 9 B) 10
C) 11 D) 12
 
Answer & Explanation Answer: C) 11

Explanation:
Mean = (sum of fixi/ (sum of f) = (8*5 + 12*15 + 10*25 + P*35 +9*45) / (8+12+10+P+9) = 25.2
(875 + 35P)/(39+P) = 25.2
P = 11
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Filed Under: Average
Exam Prep: Bank Exams

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

Q:

Two equal circles intersect such that each passes through the centre of the other. If the length of the common chord of the circles is 10√3 cm, then what is the diameter of the circle?

A) 10 cm B) 15 cm
C) 20 cm D) 30 cm
 
Answer & Explanation Answer: C) 20 cm

Explanation:

Let there be 2 circles with centre O1 and OAB is the common chord

Since both passes through the center of each other as shown in figure

So O1O is the radius of both

Let O1O = r = AO1= AO

AX = AB / 2 = 5√3 cm (since OX perpendicular to chord bisects it)

AOO1 forms an equilateral triangle with on side = radius = r

Sin 60 = √3/2 = AX / AO = 5√3/r

So r = 10 cm

So diameter = 20 cm

 

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Filed Under: Simplification
Exam Prep: Bank Exams

Q:

A ladder is resting against a vertical wall and its bottom is 2.5 m away from the wall. If it slips 0.8 m down the wall, then its bottom will move away from the wall by 1.4 m. What is the length of the ladder?

A) 6.2 m B) 6.5 m
C) 6.8 m D)  7.5 m
 
Answer & Explanation Answer: B) 6.5 m

Explanation:
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Filed Under: Height and Distance
Exam Prep: Bank Exams

Q:

The hypotenuse of a right-angled triangle 10 cm and its area is 24 cm^2. If the shorts side is halved and the longer side is double, the new hypotenuse becomes

 

A) √245 cm B) √255 cm
C) √265 cm D) √275 cm
 
Answer & Explanation Answer: C) √265 cm

Explanation:
Hypotenuse = 10cm
Let the other 2 perpendicular sides be a and b
Area ½ a*b = 24
So a*b = 48 cm^2
Also using Pythagoras
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Filed Under: Area
Exam Prep: Bank Exams

Q:

The angles of a triangle are in the ratio 1 : 1 : 4. If the perimeter of the triangle is k times its largest side, then what is the value of k?

A) 1 + 2/√3 B) 1 - 2/√3
C) 2 + 2/√3 D) 2
 
Answer & Explanation Answer: A) 1 + 2/√3

Explanation:
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Filed Under: Area
Exam Prep: Bank Exams