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Question 1 of 5
1. Question
Direction : Read the following information carefully and answer the questions below it.
Three small children Sonu, Monu and Tonu went on a picnic with their dog Jhony. They carried with them few chocolates, which none of them incidentally counted on their way. They rested under a tree and slept for a while. After some time Sonu woke up, gave one chocolate from the total to Jony and distributed the remaining into three equal parts, ate his share and slept. After Some time, Monu woke up, gave one chocolate to Jony and distributed the remaining into three equal parts, ate his share and slept. After some time Tonu woke up and repeated the same. A little later all of them woke up together, gave one chocolate from the total to Jhony and divided the remaining chocolates among them and each one ate his share. By chance we know that the total number of chocolates were less than 150 in the beginning and they didn’t break any chocolate
How many chocolates were in the beginning?
Correct
Ans-d
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
Incorrect
Ans-d
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
-
Question 2 of 5
2. Question
Direction : Read the following information carefully and answer the questions below it.
Three small children Sonu, Monu and Tonu went on a picnic with their dog Jhony. They carried with them few chocolates, which none of them incidentally counted on their way. They rested under a tree and slept for a while. After some time Sonu woke up, gave one chocolate from the total to Jony and distributed the remaining into three equal parts, ate his share and slept. After Some time, Monu woke up, gave one chocolate to Jony and distributed the remaining into three equal parts, ate his share and slept. After some time Tonu woke up and repeated the same. A little later all of them woke up together, gave one chocolate from the total to Jhony and divided the remaining chocolates among them and each one ate his share. By chance we know that the total number of chocolates were less than 150 in the beginning and they didn’t break any chocolate
What is the difference in the number of chocolates eaten by Monu and Tonu?
Correct
Ans-a
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
Incorrect
Ans-a
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
-
Question 3 of 5
3. Question
Direction : Read the following information carefully and answer the questions below it.
Three small children Sonu, Monu and Tonu went on a picnic with their dog Jhony. They carried with them few chocolates, which none of them incidentally counted on their way. They rested under a tree and slept for a while. After some time Sonu woke up, gave one chocolate from the total to Jony and distributed the remaining into three equal parts, ate his share and slept. After Some time, Monu woke up, gave one chocolate to Jony and distributed the remaining into three equal parts, ate his share and slept. After some time Tonu woke up and repeated the same. A little later all of them woke up together, gave one chocolate from the total to Jhony and divided the remaining chocolates among them and each one ate his share. By chance we know that the total number of chocolates were less than 150 in the beginning and they didn’t break any chocolate
What is the difference in the number of chocolates eaten by Sonu and Tonu?
Correct
Ans-b
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
Incorrect
Ans-b
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
-
Question 4 of 5
4. Question
Direction : Read the following information carefully and answer the questions below it.
Three small children Sonu, Monu and Tonu went on a picnic with their dog Jhony. They carried with them few chocolates, which none of them incidentally counted on their way. They rested under a tree and slept for a while. After some time Sonu woke up, gave one chocolate from the total to Jony and distributed the remaining into three equal parts, ate his share and slept. After Some time, Monu woke up, gave one chocolate to Jony and distributed the remaining into three equal parts, ate his share and slept. After some time Tonu woke up and repeated the same. A little later all of them woke up together, gave one chocolate from the total to Jhony and divided the remaining chocolates among them and each one ate his share. By chance we know that the total number of chocolates were less than 150 in the beginning and they didn’t break any chocolate
Monu and Tonu ate the chocolates in the ratio.
Correct
Ans-c
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
Incorrect
Ans-c
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
-
Question 5 of 5
5. Question
Direction : Read the following information carefully and answer the questions below it.
Three small children Sonu, Monu and Tonu went on a picnic with their dog Jhony. They carried with them few chocolates, which none of them incidentally counted on their way. They rested under a tree and slept for a while. After some time Sonu woke up, gave one chocolate from the total to Jony and distributed the remaining into three equal parts, ate his share and slept. After Some time, Monu woke up, gave one chocolate to Jony and distributed the remaining into three equal parts, ate his share and slept. After some time Tonu woke up and repeated the same. A little later all of them woke up together, gave one chocolate from the total to Jhony and divided the remaining chocolates among them and each one ate his share. By chance we know that the total number of chocolates were less than 150 in the beginning and they didn’t break any chocolate
How many more chocolates did monu eat, than Jhony?
Correct
Ans-a
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
Incorrect
Ans-a
Let the number of chocolates in the beginning be x.
So when Sonu woke up,
he left (2/3)(x-1) for the others.
When Monu woke up he left.
2/3[(2/3)(x-1)-1] = (4x-10)/9
while Tonu left = [(8x-38)/27]
Finally they divided[(8x-38)/27] -1-1 into three equal parts.
So [(8x-38-27)/27] or [(8x-65/27] was divided
into three equal parts. This means
[(8x-65)/27] is divisible by 3 or 8x – 65 is a
multiple of 81. Let 8x-65 = 81 n where n is an integer
So 81n+ 65 should be divisible by 8. It has to be an even number, so 81n should
be odd number. Trying with 1, 3, 5…….
we get 7 and 15 as the two initial possible
number but n = 15 will give us an initial
starting number x= 160, which is not
acceptable. So, n= 7 is the accepted
value. So in the beginning there were 79 chocolates with the given conditions. We can distribute them accordingly. The table below shows number of chocolates eaten by each.
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