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Q:

The average of 50 consecutive natural numbers is x. What will be the are average when the next four natural numbers are also included?

A) x + 1 B) x + 2
C) x + 4 D) x + (x/54)
 
Answer & Explanation Answer: B) x + 2

Explanation:
Sum of n consecutive natural numbers = n(n+1)/2
Average of n consecutive natural numbers = (n+1)/2
For first 50 average = 51/2 = x
Last 50 average = 55/2 = x+2
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Filed Under: Average
Exam Prep: Bank Exams

Q:

Consider the following frequency distribution:

x     f

8     6

5     4

6     5

10    8

9     9

4     6

7     4

What is the median for the distribution?

A) 6 B) 7
C) 8 D) 9
 
Answer & Explanation Answer: C) 8

Explanation:
Summation of frequencies = 6+4+5+8+9+6+4 = 42
Median = mid value = average of 21st and 22nd value
Arranging data in increasing order we get

x     f

4     6

5     4

6     5

7     4

8     6

9     9

10    8

 
So mid value i.e 21st and 22nd value = 8
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Filed Under: Average
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:

Consider the following statements:

1) The number of circles that can be drawn through three non-collinear points is infinity.

2) Angle formed in minor segment of a circle is acute.

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: D) Neither 1 nor 2

Explanation:
(1)
Only one circle can be drawn through 3 non collinear points 
Angle in the minor segment is always obtuse
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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

Q:

Consider the following statements:

1) The number of circles that can be drawn through three non-collinear points is infinity.

2) Angle formed in minor segment of a circle is acute.

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: D) Neither 1 nor 2

Explanation:
(1)
Only one circle can be drawn through 3 non collinear points 
Angle in the minor segment is always obtuse
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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