MAT PO CAT

# Quantitative Aptitude

## Sample Paper 6

Question 1: The ratio of two numbers is 3:7 and their HCF is 6. What is the sum of the numbers?

- 32
- 42
- 52
- 60

**Answer:** (d) 60

**Explanation:** Out of the given options, only options b and d are divisible by 6 (HCF of numbers). The sum of the ratio is `3+7=10`. Only option d is divisible by 10. Hence, option d is the correct answer.

Question 2: What is the least number which needs to be subtracted from 11, 15, 21 and 30 so that the resultant numbers will be in proportion?

- 1
- 2
- 3
- 3

**Answer:** (c) 3

**Explanation:** Let us assume the least number = x

Then, `(11-x)/(15-x)=(21-x)/(30-x)`

Or, `(11-x)(30-x)=(21-x)(15-x)`

Or, `330-11x-30x+x^2=315-21x-15x+x^2`

Or, `330-41x=315-36x`

Or, `330-41x+36x=315`

Or, `330-5x=315`

Or, `5x=330-315=15`

Or, `x=(15)/5=3`

Question 3: A shopkeeper claims to sell chickpeas at the cost price. But he weighs only 800 grams for every 1 kg. What is the profit percentage of the trader?

- 25%
- 15%
- 17.5%
- 20%

**Answer:** (a) 25%

**Explanation:** Let us assume that cost price of 1 kg chickpeas = Rs. 100

So, selling price of 800 g chickpeas = Rs. 100

So, selling price of 1 kg chickpeas `= (100)/(800)xx1000=125`

So, profit `=125-100=25%`

Question 4: The selling price of 20 bananas is equal to the cost price of 25 bananas. Find the percentage profit or loss.

- 20% loss
- 25% profit
- 15% loss
- 17.5% profit

**Answer:** (b) 25% profit

**Explanation:** Let us assume that cost price of 25 bananas = Rs. 100

So, selling price of 20 bananas = Rs. 100

So, selling price of 25 bananas `=(100)/(20)xx25=125`

So, profit = 25%`

Question 5: The cost of 100 pens and 60 pencils is Rs. 540 and the cost of 60 pens and 100 pencils is Rs. 420. What is the cost of one pen and one pencil?

- 7
- 6
- 8
- 10

**Answer:** (b) 6

**Explanation:** Let us assume, the cost of one pen is x and that of one pencil is y.

Then, `100x+60y=540` …………(1)

And, `60x+100y=420` ……………..(2)

Let us take equation (1)

`100x+60y=540`

Or, `10x+6y=54`

Or, `5x+3y=27`

Or, `5x=27-3y`

Or, `x=(27-3y)/5`

Substituting this value of x in equation (2) we get

`60xx(27-3y)/5+100y=420`

Or, `6xx(27-3y)/5+10y=42`

Or, `3xx(27-3y)/5+5y=21`

Or, `(81-9y)/5+5y=21`

Or, `(81-9y+25y)/5=21`

Or, `81+16y=105`

Or, `16y=105-81=24`

Or, `y=(24)/(16)=3/2`

Substituting the value of y in equation (1) we get:

`100x+60xx3/2=540`

Or, `100x+90=540`

Or, `100x=540-90=450`

Or, `x=4.5`

So, `x+y=4.5+3/2=6`