## IBPS PO Prelims 2018 Quantitative Aptitude Test-Day 2

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- Question 1 of 10
##### 1. Question

Rahim covers 1/3rd of his total journey by Bus, 1/6th of his total journey by Car and rest of the journey by Train. If the speeds of the Train, Car and Bus are 60 km/hr, 80 km/hr and 50 km/hr respectively then what was the average speed of Rahim during entire journey?

CorrectLet the total distance be 6d km.

Hence the distance covered by bus = 2d km

Since speed of the bus = 50 km/hr

Hence time taken by bus, T1 = 2d/50 = d/25 hours

Similarly time taken by car, T2 = d/80 hours

Similarly time taken by train, T3 = 3d/60 = d/20 hours

Average speed = Total distance/Total time

6d

d/25 + d/80 + d/20

= 58.5 km/hrIncorrectLet the total distance be 6d km.

Hence the distance covered by bus = 2d km

Since speed of the bus = 50 km/hr

Hence time taken by bus, T1 = 2d/50 = d/25 hours

Similarly time taken by car, T2 = d/80 hours

Similarly time taken by train, T3 = 3d/60 = d/20 hours

Average speed = Total distance/Total time

6d

d/25 + d/80 + d/20

= 58.5 km/hr - Question 2 of 10
##### 2. Question

A sum of money is to be distributed among Anirban, Bimal, Kishore, Jhalak and are in the proportion of 5 : 2 : 4 : 3. If Kishore gets Rs. 4000 more than Jhalak, what is Bimal ‘s share?

CorrectLet the shares of Anirban, Bimal, Kishore and Jhalak be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

IncorrectLet the shares of Anirban, Bimal, Kishore and Jhalak be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

- Question 3 of 10
##### 3. Question

The salaries of Ajay, Vijay and Chandu are in the ratio 7:5:4. Their savings are in the ratio 5:3:2. If Vijay saves Rs 15000, which is 30 % of his income, find the expenditure of Chandu. (Salaries = Savings+ expenditure)

CorrectLet their savings be 5n, 3n ,2n.

Vijay saves 15000, so 3n =15000 => n =5000

So savings of Ajay and Chandu are 25000 and 10000 respectively.

IncorrectLet their savings be 5n, 3n ,2n.

Vijay saves 15000, so 3n =15000 => n =5000

So savings of Ajay and Chandu are 25000 and 10000 respectively.

- Question 4 of 10
##### 4. Question

A sum of money was lent at 10% per annum compounded annually for 2 years. If the interest was compounded half-yearly, he would have received Rs. 440.50 more find the sum.

CorrectLet the sum be x.

When compounded annually,

Amount = x(1+10/100)^2 =121x/100

When compounded half-yearly,

Amount = x(1+5/100)^4 =194481x/160000

Now,

194481x/160000 -121x/100 =440.5

=> 881x/160000 = 440.5

=> x = 80000

IncorrectLet the sum be x.

When compounded annually,

Amount = x(1+10/100)^2 =121x/100

When compounded half-yearly,

Amount = x(1+5/100)^4 =194481x/160000

Now,

194481x/160000 -121x/100 =440.5

=> 881x/160000 = 440.5

=> x = 80000

- Question 5 of 10
##### 5. Question

A father distributes his property of Rs 81000 among his three sons. The first son gets 2/9th of the property and the remaining property is divided among the two sons in the ratio of 4 : 3 (Second son to the third son). What is the total amount that second son will get?

CorrectAccording to the question 2/9th of the property is given to first Son.

Now, 7/9th part of the property is divided into the ratio of 4:3.

Hence the part of property that second son will get = 4/9th

Hence total amount which second son will get =81000 x 4/9 = Rs 36,000.

IncorrectAccording to the question 2/9th of the property is given to first Son.

Now, 7/9th part of the property is divided into the ratio of 4:3.

Hence the part of property that second son will get = 4/9th

Hence total amount which second son will get =81000 x 4/9 = Rs 36,000.

- Question 6 of 10
##### 6. Question

In the following question two equations are given in variables x and y. You have to solve these equations and determine relation between x and y.

- 38x
^{2}– 3x – 11 = 0 - 28y
^{2}+ 32y + 9 = 0

Correct38x

^{2}+ 19x – 22x – 11 = 0⇒ 19x(2x + 1) – 11(2x + 1) = 0

⇒ (2x + 1)(19x – 11) = 0

⇒ x = -1/2 or x = 11/19

Equation 2:

28y

^{2}+ 14y + 18y + 9 = 0⇒ 14y(2y + 1) + 9(2y + 1) = 0

⇒ (2y + 1)(14y + 9) = 0

⇒ y = -1/2 or y = -9/14

When x = -1/2, y = -1/2, then x = y

When x = -1/2, y = -9/14, then x > y

When x = 11/19, y = -1/2, then x > y

We will solve both the equations separately.

Equation I:

When x = 11/19, y = -9/14, then x > y

∴ x ≥ y

IncorrectWe will solve both the equations separately.

Equation I:

38x

^{2}+ 19x – 22x – 11 = 0⇒ 19x(2x + 1) – 11(2x + 1) = 0

⇒ (2x + 1)(19x – 11) = 0

⇒ x = -1/2 or x = 11/19

Equation 2:

28y

^{2}+ 14y + 18y + 9 = 0⇒ 14y(2y + 1) + 9(2y + 1) = 0

⇒ (2y + 1)(14y + 9) = 0

⇒ y = -1/2 or y = -9/14

When x = -1/2, y = -1/2, then x = y

When x = -1/2, y = -9/14, then x > y

When x = 11/19, y = -1/2, then x > y

When x = 11/19, y = -9/14, then x > y

∴ x ≥ y

- 38x
- Question 7 of 10
##### 7. Question

Find the relation between x and y from the two given quadratic equations.

6x

^{2}– x – 1 = 0

2y^{2}– 5y + 2 = 0CorrectFor eq 1:

6x

^{2}– x – 1 = 0⇒ 6x

^{2}– 3x + 2x – 1 = 0⇒ 3x(2x – 1) + 1(2x – 1) = 0

⇒ (2x – 1)(3x + 1) = 0

⇒ x = 1/2, -1/3

For eq 2:

2y

^{2}– 5y + 2 = 0⇒ 2y

^{2}– 4y – y + 2 = 0⇒ 2y(y – 2) – 1(y – 2) = 0

⇒(2y – 1)(y – 2) = 0

⇒ y = 1/2, 2

When x = -1/3, y = 1/2, then x < y

When x = -1/3, y = 2, then x < y

When x = 1/2, y = 1/2, then x = y

When x = 1/2, y = 2 then x < y

∴ x and y are related as x ≤ y

IncorrectFor eq 1:

6x

^{2}– x – 1 = 0⇒ 6x

^{2}– 3x + 2x – 1 = 0⇒ 3x(2x – 1) + 1(2x – 1) = 0

⇒ (2x – 1)(3x + 1) = 0

⇒ x = 1/2, -1/3

For eq 2:

2y

^{2}– 5y + 2 = 0⇒ 2y

^{2}– 4y – y + 2 = 0⇒ 2y(y – 2) – 1(y – 2) = 0

⇒(2y – 1)(y – 2) = 0

⇒ y = 1/2, 2

When x = -1/3, y = 1/2, then x < y

When x = -1/3, y = 2, then x < y

When x = 1/2, y = 1/2, then x = y

When x = 1/2, y = 2 then x < y

∴ x and y are related as x ≤ y

- Question 8 of 10
##### 8. Question

Find the relation between x and y from the given two quadratic equation?

X

^{2}– 2x – 3 = 0Y

^{2}+ 7y + 6 = 0CorrectFor eq 1:

X

^{2}– 2x – 3 = 0⇒ x

^{2}– 3x + x – 3 = 0⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x + 1)(x – 3) = 0

⇒ x = -1, 3

For eq 2:

Y

^{2}+ 7y + 6 = 0⇒ y

^{2}+ 6y + y + 6 = 0⇒ y(y + 6) + 1(y + 6) = 0

⇒ (y + 1)(y + 6) = 0

⇒ y = -1, -6

When x = -1, y = -1, then x = y

When x = -1, y = -6, then x > y

When x = 3, y = -1, then x > y

When x = 3, y = -6, then x > y

∴ x and y are related as x ≥ y

IncorrectFor eq 1:

X

^{2}– 2x – 3 = 0⇒ x

^{2}– 3x + x – 3 = 0⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x + 1)(x – 3) = 0

⇒ x = -1, 3

For eq 2:

Y

^{2}+ 7y + 6 = 0⇒ y

^{2}+ 6y + y + 6 = 0⇒ y(y + 6) + 1(y + 6) = 0

⇒ (y + 1)(y + 6) = 0

⇒ y = -1, -6

When x = -1, y = -1, then x = y

When x = -1, y = -6, then x > y

When x = 3, y = -1, then x > y

When x = 3, y = -6, then x > y

∴ x and y are related as x ≥ y

- Question 9 of 10
##### 9. Question

In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer.

- 7x
^{2}+ 26x – 45 = 0 - 3y
^{2}+ 7y – 48 = 0

CorrectWe will solve both the equations separately.

Equation I:

7x

^{2}+ 26x – 45 = 0⇒ 7x

^{2}+ 35x – 9x – 45 = 0⇒ 7x(x + 5) – 9(x + 5) = 0

⇒ (7x – 9) (x + 5) = 0

⇒ x = 9/7 = 1.28 or x = – 5

Equation II:

3y

^{2}+ 7y – 48 = 0⇒ 3y

^{2}+ 16y – 9y – 48 = 0⇒ 3y(y – 3) + 16(y – 3) = 0

⇒ (3y + 16) (y – 3) = 0

⇒ y = – 16/3 = – 5.33 or, y = 3

Comparing the values of x and y, we get,

Values of x are greater than one value of y and less than one value of y, therefore relation between x and y cannot be determined.

IncorrectWe will solve both the equations separately.

Equation I:

7x

^{2}+ 26x – 45 = 0⇒ 7x

^{2}+ 35x – 9x – 45 = 0⇒ 7x(x + 5) – 9(x + 5) = 0

⇒ (7x – 9) (x + 5) = 0

⇒ x = 9/7 = 1.28 or x = – 5

Equation II:

3y

^{2}+ 7y – 48 = 0⇒ 3y

^{2}+ 16y – 9y – 48 = 0⇒ 3y(y – 3) + 16(y – 3) = 0

⇒ (3y + 16) (y – 3) = 0

⇒ y = – 16/3 = – 5.33 or, y = 3

Comparing the values of x and y, we get,

Values of x are greater than one value of y and less than one value of y, therefore relation between x and y cannot be determined.

If x > y

- 7x
- Question 10 of 10
##### 10. Question

In the following question two equations are given in variables x and y. You have to solve these equations and determine relation between x and y.

X

^{2}+ 2x – 8 = 0Y

^{2}+ 8y + 16 = 0CorrectFor eq 1:

X

^{2}+ 2x – 8 = 0⇒ x

^{2}+ 4x – 2x – 8 = 0⇒ (x + 4)(x – 2) = 0

∴ x = −4 or x = 2

For eq 2:

Y

^{2 }+ 8y + 16 = 0⇒ y

^{2}+ 4y + 4y + 16= 0⇒ (y + 4)

^{2}= 0∴ y = −4

IncorrectFor eq 1:

X

^{2}+ 2x – 8 = 0⇒ x

^{2}+ 4x – 2x – 8 = 0⇒ (x + 4)(x – 2) = 0

∴ x = −4 or x = 2

For eq 2:

Y

^{2 }+ 8y + 16 = 0⇒ y

^{2}+ 4y + 4y + 16= 0⇒ (y + 4)

^{2}= 0∴ y = −4