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

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

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

Let us assume the numbers are a and b.
According to question,
a² - b² = 256000 .......................(1)
and a + b = 1000 .........................(2)
On dividing the Equation (1) with Equation (2),
(a² - b²)/(a + b) = 256000/1000
(a² - b²)/(a + b) = 256
(a + b)(a - b)/(a + b) = 256
a - b = 256...............................(3)
Add the equation (2) and (3), we will get
a + b + a - b = 1000 + 256
2a = 1256
a = 628
Put the value of a in equation (2), we will get
658 + b = 1000
b = 1000 - 628
b = 372
So answer is 628 , 372.

Let us assume the ratio term is r.
then the number will be 3r and 4r.
According to question,
(4r)² + 224 = 8 x (3r)²
16r² + 224 = 8 x 9r²
16r² + 224 = 72r²
72r² - 16r² = 224
56r² = 224
r² = 224/56
r² = 4
r = 2

First number = 3r = 3 x 2 = 6
Second number = 4r = 4 x 2 = 8

Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
⇒ a + a + 2 + a + 4 - a = 20
⇒ 2a + 6 = 20
⇒ 2a + 6 = 20
⇒ 2a = 20 - 6
⇒ 2a = 14
⇒ a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9

Method 1
Let us assume the ratio factor is x.
Therefor the present ages of Vikas and Vishal be 15x years and 8x years.
After 10 years
vikas's age = 15x + 10 and Vishal' age = 8x + 10
According to question,
(15x+10)/(8x+10) = 5/3
⇒ 3(15x + 10) = 5(8x + 10)
⇒ 45x + 30 = 40x + 50
⇒ 5x =20
⇒ x = 20/5
⇒ x = 4
Therefore Present age of Vikas = 15x = 15 x 4 = 60 years
and Present age of Vishal = 8x = 8 x 4 = 32 years

Method 2
Let us assume the present age of Vikas = x years and Vishal's present year = y years
According to question,
Present age of Vikas/ Present age of Vishal = 15/8
⇒ x/y = 15/8
⇒ 8x = 15y
⇒ 8x - 15y = 0
⇒ x = 15y/8 .................................(1)
After 10 years Vikash age = x + 10 and vishal age = y + 10
Ratio of age after 10 years = 5/3
(x + 10)/(y + 10) = 5/3
⇒ 3(x + 10) = 5(y + 10)
⇒ 3x + 30 = 5y + 50
⇒ 3x - 5y = 50 - 30
⇒ 3x - 5y = 20 ................................(2)
Put the value of x from equation (1) in above equation (2).
⇒ 45y/8 - 5y = 20
⇒ 45y - 40y = 20 x 8
⇒ 5y = 20 x 8
⇒ y = 4 x 8 = 32
Put the value of Y in equation (1)
x = 15 x 32/8 = 15 x 4 = 60

Therefore Present age of Vikas = x = 60 years
and Present age of Vishal = y = 32 years

Let us assume father age is F and his son age is S.
According to question,
The sum of the ages of father and his son is 4 times the age of the son,
F + S = 4S
F = 3S.............. (1)
Average age of father and son is 28.
(F + S)/2 = 28
F + S = 56
Put the value of F from equation from (1),
3S + S = 56
4S = 56
S = 14 years
Age of Son = 14 years.

Let the middle number = x , then first number = x - 2 and last number = x + 2
(x - 2) + x + (x + 2) = 176/4 - 14
x - 2 + x + x + 2 = 44 - 14
3x = 30
x = 10
So middle number = x = 10

Let the number be x.
According to question,
x - 20 = 7x/12
⇒ x - 7x/12 = 20
⇒ (12x - 7x)/12 = 20
⇒ 5x/12 = 20
⇒ x = 20 x 12/5
⇒ x = 4 x 12
⇒ x = 48
Sum of the digits of 48 = 4 + 8 = 12

Cost of 5 pens + 8 pencils = Rs.31
On multiplying by 3
15 pens + 24 pencils = 3 x 31 =
Rs.15 pens + 24 pencils = 93