All Exams > SSC CGL > Quantitative Aptitude for SSC CGL > All Questions
All questions of Linear Equations for SSC CGL Exam
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
 | Pranab Goyal answered |
Understanding Sherry's Savings
Sherry starts saving on March 1st, 2016, with an initial amount of 1 unit. Starting from March 2nd, he saves an additional 1 unit every day. This means his savings for each day can be calculated as follows:
- **March 1st**: 1 unit
- **March 2nd**: 2 units
- **March 3rd**: 3 units
- **March 4th**: 4 units
- ...
- **March n**: n units
Calculating Total Savings
To find the total savings by a specific date, we can use the formula for the sum of the first n natural numbers:
\[ \text{Total Savings} = 1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2} \]
Where n is the total number of days since March 1st.
Perfect Squares
We want to identify the first date after March 1st, 2016, where the total savings becomes a perfect square.
Checking Dates
1. **March 1**: \( \text{Total} = 1 \) (Perfect Square)
2. **March 2**: \( \text{Total} = 1 + 2 = 3 \) (Not a Perfect Square)
3. **March 3**: \( \text{Total} = 1 + 2 + 3 = 6 \) (Not a Perfect Square)
4. **March 4**: \( \text{Total} = 10 \) (Not a Perfect Square)
5. **March 5**: \( \text{Total} = 15 \) (Not a Perfect Square)
6. **March 6**: \( \text{Total} = 21 \) (Not a Perfect Square)
7. **March 7**: \( \text{Total} = 28 \) (Not a Perfect Square)
8. **March 8**: \( \text{Total} = 36 \) (Perfect Square: \(6^2\))
The first date after March 1st when the total savings is a perfect square is **March 8th, 2016**.
Conclusion
Since the options provided do not include March 8th, the correct answer is **None of these** (option 'D').
Sherry starts saving on March 1st, 2016, with an initial amount of 1 unit. Starting from March 2nd, he saves an additional 1 unit every day. This means his savings for each day can be calculated as follows:
- **March 1st**: 1 unit
- **March 2nd**: 2 units
- **March 3rd**: 3 units
- **March 4th**: 4 units
- ...
- **March n**: n units
Calculating Total Savings
To find the total savings by a specific date, we can use the formula for the sum of the first n natural numbers:
\[ \text{Total Savings} = 1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2} \]
Where n is the total number of days since March 1st.
Perfect Squares
We want to identify the first date after March 1st, 2016, where the total savings becomes a perfect square.
Checking Dates
1. **March 1**: \( \text{Total} = 1 \) (Perfect Square)
2. **March 2**: \( \text{Total} = 1 + 2 = 3 \) (Not a Perfect Square)
3. **March 3**: \( \text{Total} = 1 + 2 + 3 = 6 \) (Not a Perfect Square)
4. **March 4**: \( \text{Total} = 10 \) (Not a Perfect Square)
5. **March 5**: \( \text{Total} = 15 \) (Not a Perfect Square)
6. **March 6**: \( \text{Total} = 21 \) (Not a Perfect Square)
7. **March 7**: \( \text{Total} = 28 \) (Not a Perfect Square)
8. **March 8**: \( \text{Total} = 36 \) (Perfect Square: \(6^2\))
The first date after March 1st when the total savings is a perfect square is **March 8th, 2016**.
Conclusion
Since the options provided do not include March 8th, the correct answer is **None of these** (option 'D').
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
 | EduRev SSC CGL answered |
Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
⇒ R + 10 = 3M - 30
⇒ 3M - R = 10 + 30
⇒ 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
⇒ 6M - M = 170
⇒ 5M = 170
⇒ M = 170/5
⇒ M = 34
Put the value of M in equation (1), we will get
⇒ 2R - 34 = 90
⇒ 2R = 90 + 34
⇒ 2R = 124
⇒ R = 124/2
⇒ R = 62
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
⇒ R + 10 = 3M - 30
⇒ 3M - R = 10 + 30
⇒ 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
⇒ 6M - M = 170
⇒ 5M = 170
⇒ M = 170/5
⇒ M = 34
Put the value of M in equation (1), we will get
⇒ 2R - 34 = 90
⇒ 2R = 90 + 34
⇒ 2R = 124
⇒ R = 124/2
⇒ R = 62
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
A person on tour has total ₹360 for his daily expenses. He decides to extend his tour programme by 4 days which leads to cutting down daily expenses by ₹3 a day. The numbers of days of tour are :- a)15
- b)20
- c)18
- d)16
Correct answer is option 'B'. Can you explain this answer?
A person on tour has total ₹360 for his daily expenses. He decides to extend his tour programme by 4 days which leads to cutting down daily expenses by ₹3 a day. The numbers of days of tour are :
a)
15
b)
20
c)
18
d)
16
 | Ssc Cgl answered |
Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
⇒ ED = 360
⇒ E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
⇒ (D + 4) x ( (360 - 3D)/D ) = 360
⇒ (D + 4) x ( (360 - 3D) ) = 360 x D
⇒ (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
⇒ 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
⇒ 360 x D - 3D + 1440 - 12D = 360D
⇒ - 3D + 1440 - 12D = 360D - 360 x D
⇒ - 3D + 1440 - 12D = 0
⇒ 3D - 1440 + 12D = 0
⇒ D - 480 + 4D = 0
⇒ D + 4D - 480 = 0
⇒ D + 24D - 20D - 480 = 0
⇒ D(D + 24) - 20(D + 24) = 0
⇒ (D + 24) (D - 20) = 0
Either (D + 24) = 0 or (D - 20) = 0
So D = - 24 or D = 20
But days cannot be negative so D = 20 days.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
⇒ ED = 360
⇒ E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
⇒ (D + 4) x ( (360 - 3D)/D ) = 360
⇒ (D + 4) x ( (360 - 3D) ) = 360 x D
⇒ (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
⇒ 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
⇒ 360 x D - 3D + 1440 - 12D = 360D
⇒ - 3D + 1440 - 12D = 360D - 360 x D
⇒ - 3D + 1440 - 12D = 0
⇒ 3D - 1440 + 12D = 0
⇒ D - 480 + 4D = 0
⇒ D + 4D - 480 = 0
⇒ D + 24D - 20D - 480 = 0
⇒ D(D + 24) - 20(D + 24) = 0
⇒ (D + 24) (D - 20) = 0
Either (D + 24) = 0 or (D - 20) = 0
So D = - 24 or D = 20
But days cannot be negative so D = 20 days.
The sum of three consecutive multiples of 3 is 72. What is the largest number ?- a)21
- b)24
- c)27
- d)36
Correct answer is option 'C'. Can you explain this answer?
The sum of three consecutive multiples of 3 is 72. What is the largest number ?
a)
21
b)
24
c)
27
d)
36
| | EduRev SSC CGL answered |
Let us assume the number be 3p , 3(p+1) and 3(p+2)
According to question,
3p + 3(p+1) + 3(p+2) = 72
⇒ 3p + 3p + 3 +3p + 6 = 72
⇒ 9p +9 = 72
⇒ 9p = 72 - 9
⇒ 9p = 63
⇒ p = 63/9 = 7
∴ Largest number = 3(p + 2)
Put the value of p in above equation.
⇒ Largest number = 3 x (7 + 2)
⇒ Largest number = 3 x 9
⇒ Largest number = 27
According to question,
3p + 3(p+1) + 3(p+2) = 72
⇒ 3p + 3p + 3 +3p + 6 = 72
⇒ 9p +9 = 72
⇒ 9p = 72 - 9
⇒ 9p = 63
⇒ p = 63/9 = 7
∴ Largest number = 3(p + 2)
Put the value of p in above equation.
⇒ Largest number = 3 x (7 + 2)
⇒ Largest number = 3 x 9
⇒ Largest number = 27
Chapter doubts & questions for Linear Equations - Quantitative Aptitude for SSC CGL 2026 is part of SSC CGL exam preparation. The chapters have been prepared according to the SSC CGL exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for SSC CGL 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Linear Equations - Quantitative Aptitude for SSC CGL in English & Hindi are available as part of SSC CGL exam. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free.
Quantitative Aptitude for SSC CGL319 videos|366 docs|157 tests |
