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Quantitative Aptitude Question on Averages

There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is

Answer

Let the four numbers be a,b,c,da, b, c, d.

Step 1. The average of the first two numbers is a+b2\frac{a+b}{2}, and it is 1 more than aa, so:
a+b2=a+1    a+b=2a+2    b=a+2\frac{a+b}{2} = a+1 \implies a+b = 2a+2 \implies b = a+2.

Step 2. The average of the first three numbers is a+b+c3\frac{a+b+c}{3}, and it is 2 more than the average of the first two, so:
a+b+c3=a+b2+2    a+b+c3=a+1+2=a+3\frac{a+b+c}{3} = \frac{a+b}{2} + 2 \implies \frac{a+b+c}{3} = a+1+2 = a+3.
a+b+c=3(a+3)=3a+9    c=3a+9(a+b)=3a+9(a+a+2)=2a+7a+b+c = 3(a+3) = 3a+9 \implies c = 3a+9 - (a+b) = 3a+9 - (a+a+2) = 2a+7.

Step 3. The average of the first four numbers is a+b+c+d4\frac{a+b+c+d}{4}, and it is 3 more than the average of the first three numbers, so:
a+b+c+d4=a+b+c3+3    a+b+c+d4=a+3+3=a+6\frac{a+b+c+d}{4} = \frac{a+b+c}{3} + 3 \implies \frac{a+b+c+d}{4} = a+3+3 = a+6.
a+b+c+d=4(a+6)=4a+24a+b+c+d = 4(a+6) = 4a+24
    \implies d=4a+24(a+b+c)=4a+24(a+a+2+2a+7)=15.d = 4a+24 - (a+b+c) = 4a+24 - (a+a+2+2a+7) = 15.

The numbers are a,a+2,2a+7,15a, a+2, 2a+7, 15. The largest number is 15, and the smallest is aa.
Thus, the difference is:
15a=1515 - a = 15.

Explanation

Solution

Let the four numbers be a,b,c,da, b, c, d.

Step 1. The average of the first two numbers is a+b2\frac{a+b}{2}, and it is 1 more than aa, so:
a+b2=a+1    a+b=2a+2    b=a+2\frac{a+b}{2} = a+1 \implies a+b = 2a+2 \implies b = a+2.

Step 2. The average of the first three numbers is a+b+c3\frac{a+b+c}{3}, and it is 2 more than the average of the first two, so:
a+b+c3=a+b2+2    a+b+c3=a+1+2=a+3\frac{a+b+c}{3} = \frac{a+b}{2} + 2 \implies \frac{a+b+c}{3} = a+1+2 = a+3.
a+b+c=3(a+3)=3a+9    c=3a+9(a+b)=3a+9(a+a+2)=2a+7a+b+c = 3(a+3) = 3a+9 \implies c = 3a+9 - (a+b) = 3a+9 - (a+a+2) = 2a+7.

Step 3. The average of the first four numbers is a+b+c+d4\frac{a+b+c+d}{4}, and it is 3 more than the average of the first three numbers, so:
a+b+c+d4=a+b+c3+3    a+b+c+d4=a+3+3=a+6\frac{a+b+c+d}{4} = \frac{a+b+c}{3} + 3 \implies \frac{a+b+c+d}{4} = a+3+3 = a+6.
a+b+c+d=4(a+6)=4a+24a+b+c+d = 4(a+6) = 4a+24
    \implies d=4a+24(a+b+c)=4a+24(a+a+2+2a+7)=15.d = 4a+24 - (a+b+c) = 4a+24 - (a+a+2+2a+7) = 15.

The numbers are a,a+2,2a+7,15a, a+2, 2a+7, 15. The largest number is 15, and the smallest is aa.
Thus, the difference is:
15a=1515 - a = 15.