# Ratios and Proportions – Quiz 6

Home > > Tutorial
5 Steps - 3 Clicks

# Ratios and Proportions – Quiz 6

### Introduction

Ratios and Proportions is one of important topic in Quantitative Aptitude Section. In Ratios and Proportions – Quiz 6 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Ratios and Proportions – Quiz 6 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

In a house, there are dogs, cats and parrot in the ratio 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7, what is the minimum of pets in the house?
A. 945 B. 630 C. 250 D. 238

A
If three kinds of pets are taken be 3k,7k and 5k respectively, then 7k − 3k = 63p (where p is any positive integer).
As the number is a multiple of both 9 and 7, it has to be multiple of 63.
⇒ k = $\frac{63p}{4}$
Minimum value of p for which k is a natural number is 4.
Thus, k = 63.
Hence, the number of pets = 15k = 945

### Q2

A fort has provisions for 60 days. If after 15 days 500 men strengthen them and the food lasts 40 days longer, how many men are there in the fort?
A. 3500 B. 4000 C. 6000 D. None

B
Let there be ′x′ men in the beginning so that after 15 days the food for them is left for 45 days.
After adding 500 men the food lasts for only 40 days.
Now (x + 500) men can have the same food for 40 days.
Therefore by equating the amount of food we get,
45x = (x + 500) × 40
5x = 20,000
x = 4000
Therefore there are 4,000 men in the fort.

### Q3

If N>1000, what could be the least value of N?
A. 1249 B. 1023 C. 1202 D. 1246

D
Suppose N = 5x + 1
A took (x + 1) biscuit.
Now 4x is of the form 5y + 1 then x must be in the form 5z + 4
⇒ 4(5z + 4) = 5y + 1
⇒ y = 4z + 3 and x = 5z + 4
The ratio of number of biscuits that A and B took is
[(5z + 4) + 1]:[(4z + 3) + 1]= 5 : 4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5 : 4
Let the number of biscuits that A, B, C and D took be a, b, c and d respectively.
a : b = b : c = c : d = 5 : 4
a : b : c : d = 125 : 100 : 80 : 64
⇒a = 125k
⇒x = 125k − 1 and N = 5x + 1 = 625k − 4
As, N > 1000, the least value of N is when k = 2
⇒N = 1246

### Q4

What number has a 5:1 ratio to the number 10?
A. 42 B. 50 C. 55 D. 62

B
5:1 = x: 10
x = 50

### Q5

If a : b = 7 : 5, b : c = 9 : 11, find a : b : c?
A. 63 : 14 : 55 B. 63 : 45 : 55 C. 14 : 14 : 15 D. 7 : 14 : 15

B
a: b = 7: 5
b : c = 9: 11
a : b : c = 63 : 45 : 55