Linear Equations - 1 - Free MCQ Practice Test with solutions, SSC CGL Quant

Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + (n - 1) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + (n - 1) x d = 12 + 37
a + (6- 1) x d = 49
a + 5d = 49................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 49 - 37
5d - 3d= 12
2d =12
d = 6
Put the value of d in equation (1), we will get
a + 3 x 6 = 37
a = 37 - 18
a = 19
Second term = a + (n - 1) x d = 19 + (2 - 1) x 6 = 19 + 6 = 25
Eight term = a + (n - 1) x d = 19 + (8 - 1) x 6 = 19 + 42 = 61
Sum of Second and Eight term = 25 + 61 = 86
Sum of Second and Eight term = 86
Answer is 86.

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.

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.

Let us assume the second number is a and the difference between consecutive numbers is d.
According to Arithmetic progression,
First number = a - d
Second number = a
Third number = a + d
According to question,
Sum of the all three numbers = 15
a - d + a + a + d = 15
3a = 15
a = 5
Again according to given question,
sum of square of the 3 numbers = 83
(a - d) ² + a ² + (a + d) ² = 83
apply the algebra formula
a ² + d ² - 2ad + a ² + a ² + d ² + 2ad = 83
3a ² + 2d ² = 83
Put the value of a in above equation.
3 x 5 ² + 2d² = 83
3 x 25 + 2d² = 83
75 + 2d² = 83
2d ² = 83 - 75
2d ² = 8
d ² = 8/2
d ² = 4
d = 2
Put the value of a and d in below equation.
First number = a - d = 5 - 2 = 3
Second number = a = 5
Third number = a + d = 5 + 2 = 7
The smallest number is 3. 

First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
where a = First Number , l = Last Number and d = difference of two consecutive numbers.
Let the number of terms be n. So Last term = t_n
Then t_n = 996
Use the formula for n terms of arithmetic progression.
∴ a + ( n - 1) x d = 996
⇒ 102 + (n - 1) x 6 = 996
⇒ 6(n - 1) = 894
⇒ (n - 1) = 149
⇒ n = 150
∴ Numbers of terms = 150 

Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
Apply the formula Sum of first n terms in an Arithmetic Progression = S_n = n/2[ 2a + (n?1)d ]
where a = the first term, d = common difference, n = number of terms.
n/2[ 2a + (n?1)d ] = 3600
Put the value of a, n and d from question,
40/2[ 2a + (40 ?1)d ] = 3600
20[ 2a + 39d ] = 3600
[ 2a + 39d ] = 3600/20 = 180
2a + 39d = 180...........................(1)
Again according to given question,
After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
After paying the 30 installments the unpaid amount = 1200
So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
Sum of 30 installments = 2400
30/2[ 2a + (30 ?1)d ] = 2400
[ 2a + (30 ?1)d ] = 2400 x 2/30
2a + 29d = 80 x 2 = 160
2a + 29d = 160..........................(2)
Subtract the Eq. (2) from Eq. (1), we will get
2a + 39d - 2a - 29d = 180 - 160
10d = 20
d = 2
Put the value of d in Equation (1), we will get
2a + 39 x 2 = 180
2a = 180 - 78
a = 102/2
a = 51
The value of first installment = a = 51 

According to given question,
First day 1 rs, second day 2 rs, third day 4 rs.....
1 , 2 , 4............ and so on
The given series is in geometric progression (GP).
1, 2¹, 2²,.............
Sum of first n terms in a geometric progression(GP)
S_n = a (rⁿ ? 1) /(r ?1)
if r > 1 and where a= the first term, r = common ratio,n = number of terms
As per the given question a = 1 , r = 2 and n = 20;
S_n = 1(2²⁰ ? 1) /(2 ?1)
S_n = (2²⁰ ? 1) /1
S_n = (2²⁰ ? 1)
S_n = 2²⁰ ? 1 

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
⇒ a + ( m - 1 ) x d = t_m
⇒ a + ( m - 1 ) x d = 1/n
⇒ a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for n^th term,
⇒ a + ( n - 1 ) x d = t_n
⇒ a + ( n - 1 ) x d = 1/m
⇒ a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - a - md + d = 1/m - 1/n
⇒ nd - md = 1/m - 1/n
⇒ d(n - m) = 1/m - 1/n
⇒ d(n - m) = (n - m)/mn
⇒ d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
⇒ a + md - d = 1/n
⇒ a + ( m x 1/mn ) - 1/mn = 1/n
⇒ a + 1/n - 1/mn = 1/n
⇒ a = 1/n - 1/n + 1/mn
⇒ a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
S_mn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
S_mn = mn/2 [2/mn + 1 - 1/mn ]
S_mn = mn/2 [ 1 + 1/mn ]
S_mn = mn/2 [ ( mn + 1 )/mn ]
S_mn = mn/2 (1 + mn)/mn
S_mn = 1/2 (1 + mn)
S_mn = (1 + mn)/2 = (mn + 1)/2 

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
⇒ n - 4n/7 = 24
⇒ (7n - 4n)/7 = 24
⇒ 3n/7 = 24
⇒ 3n = 24 x 7
⇒ n = 24 x 7/3
⇒ n = 8 x 7
⇒ n = 56
Sum of the digits of the number = 5 + 6 = 11

First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
⇒ a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
⇒ 10 + (n - 1) x 7 = 94
⇒ 10 + 7n - 7 = 94
⇒ 7n + 3 = 94
⇒ 7n = 94 - 3 = 91
⇒ n = 91/7
⇒ n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676 

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t₁ = a + ( n - 1 ) x d
t₁ = a + (1 - 1) x d = a + 0 x d = a
Fifth term t₅ = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10
⇒ a + a + 4d = 10
⇒ 2a + 4d = 10
⇒ a + 2d = 5.............................(1)
Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
⇒ S₁₀ = 10/2[ 2a + ( 10 - 1 ) x d ]
⇒ S₁₀ = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99
⇒ 10a + 45d - a = 99
⇒ 9a +45d = 99
⇒ a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
⇒ a + 5d - a - 2d = 11 - 5
⇒ 3d = 6
⇒ d = 2
Put the value of d in equation (1), we will get
⇒ a + 2 x 2 = 5
⇒ a + 4 = 5
⇒ a = 5 - 4
⇒ a = 1
Since t_n = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t₃ = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5. 

Let us assume the number is a.
Given that 7 ¹/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
⇒ a/2 + a/5 = a/3 + 22/3
⇒ a/2 + a/5 - a/3 = 22/3
⇒ (15a + 6a - 10a)/30 = 22/3
⇒ (15a + 6a - 10a) = 30 x 22/3
⇒ 11a = 10 x 22
⇒ a = 10 x 2
⇒ a = 20 

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 

Let us assume the numbers are a and b.
According to question,
sum of two numbers = 25
a + b = 25......................(1)
difference of two numbers = 13
a - b = 13........................(2)
add the Equation (1) and (2)
a + b+ a - b = 25 + 13
⇒ 2a = 38
⇒ a = 19
Put the value of a in equation in (1)
19 + b = 25
⇒ b = 25 - 19
⇒ b = 6
Product of the numbers = a x b
put the value of a and b,
⇒ Product of the numbers = 19 x 6
⇒ Product of the numbers = 114

Let us assume the salary of driver be ₹ R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
⇒ Total Income during one week = R + (5R/4)
⇒ Total Income during one week = (4R + 5R)/4 = 9R/4
Required Fraction = Tips in a week/Total Income in a week
⇒ Required Fraction = 5R/4 / 9R/4
⇒ Required Fraction = 5R/4 x 4/9R
⇒ Required Fraction = 5/9

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