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Question 1 of 5
1. Question
What is the sum of all natural numbers between 100 and 200 which are multiples of 3?
Correct
Solution: A
Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = nArithmetic Progression Formula:
an = a1 + (n – 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 – 102 = 3
—> 198 = 102 + (n – 1) * 3
—> 198 – 102 = (n – 1) * 3
—> 96 = (n – 1) * 3
—> (n – 1) = 96/3 = 32
—> n = 32 + 1
—> n = 33Formula:
Sum of n terms = Sn = (n/2) * (a1 + an )
where n = number of elements = 33
a1 = first term = 102
an= last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950Incorrect
Solution: A
Multiples of 3 between 100 and 200 are 102, 105, 108,.. 198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = nArithmetic Progression Formula:
an = a1 + (n – 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 – 102 = 3
—> 198 = 102 + (n – 1) * 3
—> 198 – 102 = (n – 1) * 3
—> 96 = (n – 1) * 3
—> (n – 1) = 96/3 = 32
—> n = 32 + 1
—> n = 33Formula:
Sum of n terms = Sn = (n/2) * (a1 + an )
where n = number of elements = 33
a1 = first term = 102
an= last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950 -
Question 2 of 5
2. Question
One half of a two digit number exceeds its one third by 6. What is the sum of the digits of the number?
Correct
Solution: B
Let the given number be : x
It is given that x/2 – x/3 = 6
Or x/6 =6 or x =36
Therefore, sum of the digits 3+6 = 9
Incorrect
Solution: B
Let the given number be : x
It is given that x/2 – x/3 = 6
Or x/6 =6 or x =36
Therefore, sum of the digits 3+6 = 9
-
Question 3 of 5
3. Question
Mahesh’s age is two didn’t number. If the digits of the age of Mr. Mahesh are reversed then the new age so obtained is the age of his wife. 1/11 of the sum of their ages is equal to the difference between their ages. If Mr. Mahesh is elder than his wife then find the ratio of their ages?
Correct
Solution: D
Let the present age of Mr. Mahesh = (10x + y) yrs.
Present age of his wife = (10y + x) yrs
It is given that,
(1/11)Sum of their ages = Difference of their ages
(1/11)(10x + y + 10y + x) = (10x + y) – (10y + x)
(1/11)(11x + 11y) = 9x – 9y
x + y = 9x – 9y
-8x = -10y
x/y = 10/8
x/y = 5/4
x : y = 5 : 4
Age of Mr. Mahesh = 10(5) + 4 = 54 yrs
Age of his wife = 10(4) + 5 = 45 yrs
Required ratio = 54/45 = 6 : 5.Incorrect
Solution: D
Let the present age of Mr. Mahesh = (10x + y) yrs.
Present age of his wife = (10y + x) yrs
It is given that,
(1/11)Sum of their ages = Difference of their ages
(1/11)(10x + y + 10y + x) = (10x + y) – (10y + x)
(1/11)(11x + 11y) = 9x – 9y
x + y = 9x – 9y
-8x = -10y
x/y = 10/8
x/y = 5/4
x : y = 5 : 4
Age of Mr. Mahesh = 10(5) + 4 = 54 yrs
Age of his wife = 10(4) + 5 = 45 yrs
Required ratio = 54/45 = 6 : 5. -
Question 4 of 5
4. Question
A is twice as good a work man as B and together they finish the work in 14 days. In how many days A alone can finish the work?
Correct
Solution: A
Let A takes x days to finish the work
Then the work done by A for one day = 1/x
Work done by B for one day = 1/2x
It is given that 1/x + 1/2x = 1/14
Thus, 3/2x = 1/14
Or 2x = 42 or x= 21
Days taken by A to complete the work is 21 days
Incorrect
Solution: A
Let A takes x days to finish the work
Then the work done by A for one day = 1/x
Work done by B for one day = 1/2x
It is given that 1/x + 1/2x = 1/14
Thus, 3/2x = 1/14
Or 2x = 42 or x= 21
Days taken by A to complete the work is 21 days
-
Question 5 of 5
5. Question
The distance between Mumbai and Pune is 110 kms. A starts from Mumbai with a speed of 20 kmph at 7 AM for Pune and B starts from Pune with a speed of 25 kmph at 8 AM for Mumbai. When will they meet?
Correct
Solution: B
Since second train starts 8 AM, first train which is leaving from Mumbai at 7 AM had already travelled a distance of 20 Km.
Thus, remaining distance to be covered is D = 110 – 20 = 90
Since, two trains are moving opposite direction, their relative speed (RS) = 20 + 25 = 45 Kmph
Therefore, Time taken= Distance/Speed = 90/45 = 2 hours
Time of Meeting will 2 hrs from 8 AM i.e 10 AM
Incorrect
Solution: B
Since second train starts 8 AM, first train which is leaving from Mumbai at 7 AM had already travelled a distance of 20 Km.
Thus, remaining distance to be covered is D = 110 – 20 = 90
Since, two trains are moving opposite direction, their relative speed (RS) = 20 + 25 = 45 Kmph
Therefore, Time taken= Distance/Speed = 90/45 = 2 hours
Time of Meeting will 2 hrs from 8 AM i.e 10 AM
