MCQ: Linear Equations - 1 - SSC CGL MCQ

# MCQ: Linear Equations - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Linear Equations - 1

MCQ: Linear Equations - 1 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Linear Equations - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Linear Equations - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Linear Equations - 1 below.
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MCQ: Linear Equations - 1 - Question 1

### How many 3-digits numbers are completely divisible by 6?

Detailed Solution for MCQ: Linear Equations - 1 - Question 1

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 = tn
Then tn = 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

MCQ: Linear Equations - 1 - Question 2

### A man arranges to pay off a debt of Rs.3600 in 40 annual installments which form an Arithmetic Progression (A.P). When 30 of the installments are paid, he dies leaving one-third of the debt unpaid . Find the value of the first installment.

Detailed Solution for MCQ: Linear Equations - 1 - Question 2

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 = Sn = 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

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MCQ: Linear Equations - 1 - Question 3

### A boy agrees to work at the rate of 1 rupees on the first day, 2 rupees on the second day. Four rupees on the third day and so on . how much will the boy get if he starts working on the 1st of February and finishes on the 20th of February ?

Detailed Solution for MCQ: Linear Equations - 1 - Question 3

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)
Sn = 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;
Sn = 1(220 ? 1) /(2 ?1)
Sn = (220 ? 1) /1
Sn = (220 ? 1)
Sn = 220 ? 1

MCQ: Linear Equations - 1 - Question 4

If the mth term of an Arithmetic Progression (AP) is 1/n and nth term is 1/m, then find the sum of mn terms.

Detailed Solution for MCQ: Linear Equations - 1 - Question 4

let us assume a be the first term and d the common difference.
According to Question,
mth term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for mth term,
⇒ a + ( m - 1 ) x d = tm
⇒ a + ( m - 1 ) x d = 1/n
⇒ a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for nth term,
⇒ a + ( n - 1 ) x d = tn
⇒ 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)
Smn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
Smn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
Smn = mn/2 [2/mn + 1 - 1/mn ]
Smn = mn/2 [ 1 + 1/mn ]
Smn = mn/2 [ ( mn + 1 )/mn ]
Smn = mn/2 (1 + mn)/mn
Smn = 1/2 (1 + mn)
Smn = (1 + mn)/2 = (mn + 1)/2

MCQ: Linear Equations - 1 - Question 5

If 24 is subtracted from a number, it reduces to its four-seventh. what is the sum of the digits of that number ?

Detailed Solution for MCQ: Linear Equations - 1 - Question 5

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

MCQ: Linear Equations - 1 - Question 6

A driver's income consists of his salary and tips. during one week his tips were 5/4 of his salary. what fraction of his income came from tips?

Detailed Solution for MCQ: Linear Equations - 1 - Question 6

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

MCQ: Linear Equations - 1 - Question 7

Ram and Mohan are friends. Each has some money. If Ram gives Rs. 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives Rs. 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?

Detailed Solution for MCQ: Linear Equations - 1 - Question 7

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

MCQ: Linear Equations - 1 - Question 8

The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are

Detailed Solution for MCQ: Linear Equations - 1 - Question 8

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.

MCQ: Linear Equations - 1 - Question 9

The numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers be in the ratio 3 : 4, the numbers are

Detailed Solution for MCQ: Linear Equations - 1 - Question 9

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

MCQ: Linear Equations - 1 - Question 10

The sum of three consecutive odd numbers is 20 more than the first of these numbers.What is the middle number ?

Detailed Solution for MCQ: Linear Equations - 1 - Question 10

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

MCQ: Linear Equations - 1 - Question 11

The present ages of Vikas and Vishal are in the ratio 15:8. After ten years , their ages will be in the ratio 5:3. Find their present ages

Detailed Solution for MCQ: Linear Equations - 1 - Question 11

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

MCQ: Linear Equations - 1 - Question 12

The sum of the ages of a father and his son is 4 times the age of the son. If the average age of the father and the son is 28 years, What is the son's age?

Detailed Solution for MCQ: Linear Equations - 1 - Question 12

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.

MCQ: Linear Equations - 1 - Question 13

The sum of three consecutive even numbers is 14 less than one-fourth of 176. What is middle number?

Detailed Solution for MCQ: Linear Equations - 1 - Question 13

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

MCQ: Linear Equations - 1 - Question 14

When 20 is subtracted from a number , it reduces to seven- twelve of the number. What is the sum of the digit of the number?

Detailed Solution for MCQ: Linear Equations - 1 - Question 14

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

MCQ: Linear Equations - 1 - Question 15

The cost of 5 pens and 8 pencils is ₹31.What would be the cost of 15 pens and 24 pencils ?

Detailed Solution for MCQ: Linear Equations - 1 - Question 15

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

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests