GRE Numeric Entry SAMPLE QUESTIONS-9
GRE Numeric Entry Sample Questions
GRE Numeric Entry Sample Questions 1 | GRE Numeric Entry Sample Questions 2 | GRE Numeric Entry Sample Questions 3 | GRE Numeric Entry Sample Questions 4 | GRE Numeric Entry Sample Questions 5 | GRE Numeric Entry Sample Questions 6 | GRE Numeric Entry Sample Questions 7 | GRE Numeric Entry Sample Questions 8 | GRE Numeric Entry Sample Questions 9 | GRE Numeric Entry Sample Questions 10
1. Question :Sam deposited $30000 in a bank at 10% per annum and $40000 in another bank at a certain rate of interest per annum. What was this rate of interest if the rate of interest on the whole amount was (78/7)%?
Correct Answer: 12
Explanation:Simple interest is given by
SI= P*R*T/100, where P, R and T are the principle, rate and time.
Let the time be 1 year
SI for 30000= 30000*10*1/100 = 3000
Let the interest on Rs.40000 be x%
SI for 40000= 40000*x*1/100 = 400x
Total SI = 3000 + 400x
Total principal = 30000 + 40000 = 70000
From the given conditions
3000 + 400x = 70000*78/7*1/100
3000 + 400x = 7800
400x = 7800-3000
x = 4800/400 = 12
The required rate is 12%
2. QuestionWhat is the sum of all integers between 83 and 718 which are multiples of 5?
Correct Answer: 50800
Explanation:The numbers form an AP with first term as 85 and last term 715.
Let the number of terms be n and the common difference will be 5.
The last term is given by a+(n-1)d, where a is the first term and d is the common difference
715 = 85 + (n-1)5
n=126+1=127
Sum of n terms of an AP with a as the first term and l as the last term is given by
Sn = (n/2)(a+l)
Sum of all multiples of 5 lying between 83 and 718 = 127(85+715)/2
= 127*400 = 50800
3. Question :A car moves 16 meters in the first minute, 48 meters in the second minute, 80 meters in the third, 112 meters in the fourth and so on. How far in meters from its starting point is it after 11 minutes?
Correct Answer: 1936m
Explanation:The distances moved in first, second, third and further minutes are 16, 48, 80, 112…
These form an AP with first term 16 and common difference 32.
We have to find the sum of 11 terms
The sum of n terms of an AP with n terms with a as its first term and d as its common difference is given by
Sum = (n/2)[2a+(n-1)d]
= (11/2)[2*16+(11-1)*32]
= (11/2)(32+320)
=1936m
GRE Numeric Entry Sample Questions
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