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Ram and Mohan are friends. Each has some money. If Ram gives ₹ 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ₹ 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?- a)₹62,₹34
- b)₹6,₹2
- c)₹170,₹124
- d)₹43,₹26
Correct answer is option 'A'. Can you explain this answer?
Ram and Mohan are friends. Each has some money. If Ram gives ₹ 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ₹ 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?
a)
₹62,₹34
b)
₹6,₹2
c)
₹170,₹124
d)
₹43,₹26
|
Nilesh Banerjee answered |
Given Information:
Ram gives 30 to Mohan, and Mohan gives 10 to Ram. After these transactions, Mohan has twice the money left with Ram, and Ram has thrice as much as is left with Mohan.
Let's Assume:
Let the initial amount of money with Ram be R and Mohan be M.
Equations:
1. M + 30 = 2(R - 30)
2. R + 10 = 3(M - 10)
Solving the Equations:
1. M + 30 = 2R - 60
M = 2R - 90
2. R + 10 = 3M - 30
R = 3M - 40
Substitute R and M:
3. Substitute M from equation 1 into equation 2:
R = 3(2R - 90) - 40
R = 6R - 270 - 40
R = 6R - 310
5R = 310
R = 62
4. Substitute R = 62 into equation 1:
M = 2(62) - 90
M = 124 - 90
M = 34
Conclusion:
Ram has 62 units of money, and Mohan has 34 units of money. So, the correct answer is option A: 62, 34.
Ram gives 30 to Mohan, and Mohan gives 10 to Ram. After these transactions, Mohan has twice the money left with Ram, and Ram has thrice as much as is left with Mohan.
Let's Assume:
Let the initial amount of money with Ram be R and Mohan be M.
Equations:
1. M + 30 = 2(R - 30)
2. R + 10 = 3(M - 10)
Solving the Equations:
1. M + 30 = 2R - 60
M = 2R - 90
2. R + 10 = 3M - 30
R = 3M - 40
Substitute R and M:
3. Substitute M from equation 1 into equation 2:
R = 3(2R - 90) - 40
R = 6R - 270 - 40
R = 6R - 310
5R = 310
R = 62
4. Substitute R = 62 into equation 1:
M = 2(62) - 90
M = 124 - 90
M = 34
Conclusion:
Ram has 62 units of money, and Mohan has 34 units of money. So, the correct answer is option A: 62, 34.
In the certain party, there was a bowl of rice for every two guests , a bowl of juice for every three of them and a bowl of meat for every four of them. If there were all 65 bowls of food , then how many guests were there in the party ?- a)65
- b)24
- c)60
- d)48
Correct answer is option 'C'. Can you explain this answer?
In the certain party, there was a bowl of rice for every two guests , a bowl of juice for every three of them and a bowl of meat for every four of them. If there were all 65 bowls of food , then how many guests were there in the party ?
a)
65
b)
24
c)
60
d)
48
|
Bayshore Academy answered |
Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15
The Total number of Guests = 2a = 3b = 4c = 60
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15
The Total number of Guests = 2a = 3b = 4c = 60
On March 1st 2016 , sherry saved ₹ 1. Everyday starting from March 2nd 2016, he save ₹1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.- a)17th March 2016
- b)18th March 2016
- c)26th March 2016
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
On March 1st 2016 , sherry saved ₹ 1. Everyday starting from March 2nd 2016, he save ₹1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.
a)
17th March 2016
b)
18th March 2016
c)
26th March 2016
d)
None of these
|
Ssc Cgl answered |
According to the question,
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.
n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.
Chapter doubts & questions for Linear Equations - Mathematics for RRB NTPC / ASM 2025 is part of RRB NTPC/ASM/CA/TA exam preparation. The chapters have been prepared according to the RRB NTPC/ASM/CA/TA exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for RRB NTPC/ASM/CA/TA 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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