Averages: CAT previous year questions

Solution

Sum of days 1–7 $= 7 \times 60 = 420$; sum of days 1–8 $= 8 \times 63 = 504$, so day 8 $= 84$.

Day 9 $= 84 - 11 = 73$, so the sum of days 1–9 $= 577$.

Sum of days 2–9 $= 8 \times 66 = 528$.

Day 1 $= 577 - 528 = 49$.

Solution

Total salary $= 30 \times 60000 = 1800000$.

A $5\%$ rise in the average means the total rises by $90000$.

This equals $20\%$ of the managers' total, so managers' total $= 450000$.

Engineers' total $= 1350000$; average $= \dfrac{1350000}{25} = 54000$.

Solution

Let the numbers be $a, b, c, d$.

$\dfrac{a + b}{2} = a + 1 \Rightarrow b = a + 2$; the average of two $= a + 1$.

Average of three $= a + 3 \Rightarrow c = 3(a + 3) - (2a + 2) = a + 7$.

Average of four $= a + 6 \Rightarrow d = 4(a + 6) - (3a + 9) = a + 15$.

Numbers: $a, a + 2, a + 7, a + 15$. Largest − smallest $= 15$.

Solution

Let the bonus totals be $F$ (first 10), $M$ (middle 20) and $L$ (last 10).

$F + M = 30 \times 40000 = 1.2$ million; $M + L = 1.8$ million; $F + L = 20 \times 50000 = 1$ million.

Adding: $2(F + M + L) = 4$ million, so the total is 2 million: $M = 1$, $F = 0.2$, $L = 0.8$ (million).

2023: $2F + M + 3L = 0.4 + 1 + 2.4 = 3.8$ million.

Average $= \dfrac{3800000}{40} = 95000$.

Solution

Let the numbers be $a < b < c$ with $a + b + c = 84$.

New numbers $a + 7, b, c - 10$ have sum 81 and mean 27, which is $b + 2$, so $b = 25$.

$(c - 10) - (a + 7) = 64 \Rightarrow c - a = 81$, and $a + c = 84 - 25 = 59$.

$c = \dfrac{59 + 81}{2} = 70$.

Solution

Let each girl score $g$ and each boy $b$: $4g + 6b = 240 \Rightarrow g = \dfrac{120 - 3b}{2}$.

Condition $b \leq g \leq 2b$:

$g \geq b \Rightarrow b \leq 24$; $g \leq 2b \Rightarrow b \geq \dfrac{120}{7} \approx 17.14$.

Total of 2 girls and 6 boys $= 2g + 6b = 120 + 3b$, which ranges over $[171.43, 192]$.

Integers from 172 to 192: $21$ values.

Solution

Total money $= 352n$.

After two people take 506 each, the rest get $\dfrac{352n - 1012}{n - 2} \leq 330$.

$352n - 1012 \leq 330n - 660 \Rightarrow 22n \leq 352 \Rightarrow n \leq 16$.

Maximum $n = 16$.

Solution

Let the average of A, B, C be $a$ (total $3a$).

With D: $\dfrac{3a + D}{4} = a - x \Rightarrow D = a - 4x$.

With E: $\dfrac{3a + E}{4} = a + 2x \Rightarrow E = a + 8x$.

$E - D = 12x = 12 \Rightarrow x = 1$.

Solution

Let there be $n$ original students (average $w$) and $m$ new students (average $w + 3$).

$nw + m(w + 3) = (n + m)(w + 0.6) \Rightarrow 3m = 0.6(n + m) \Rightarrow 2.4m = 0.6n$.

$n : m = 4 : 1$.

Solution

Sum of the three integers $= 39$.

$\dfrac{39 + n}{4}$ must be an odd integer, so $39 + n = 4 \times$ odd.

$n = 1$ gives 40 ($\div 4 = 10$, even) ✗. The next multiple of 4 is 44 ($\div 4 = 11$, odd) ✓, giving $n = 5$.

Solution

Replacing $a_1$ by $6a_1$ adds $5a_1$ to the sum: $5a_1 = (400 - 300)N \Rightarrow a_1 = 20N$.

In a non-decreasing sequence the first term is at most the average: $20N \leq 300 \Rightarrow N \leq 15$.

$N = 1$ is impossible (then $a_1$ would be both 20 and 300).

So $N = 2, 3, \dots, 15$: $14$ possible values of $a_1$.

Solution

Yearly savings target $= 12 \times 550 = 6600$.

First 9 months: $9 \times (4000 - 3500) = 4500$.

Needed in the last 3 months $= 2100$, i.e. 700 per month.

Earnings $= 3700 + 700 = 4400$.

Solution

Total marks $= 5 \times 38 = 190$. Three students score above 32; Amit and one other ($X$) score at most 32, with Amit $< X \leq 32$.

The top three are distinct and at most 50, so their sum $\leq 50 + 49 + 48 = 147$; hence Amit $+ X \geq 43$.

Also Amit $+ X \leq 31 + 32 = 63$.

Highest Amit: $31$ (with $X = 32$, top three summing to 127, e.g. 40, 43, 44).

Lowest Amit: $43 - 32 = 11$.

Difference $= 31 - 11 = 20$.

Solution

Two smallest sum to 28 and two largest sum to 56. To make the middle two as large as possible:

Second smallest as large as possible: $13, 15$. Fifth number as large as possible: $27, 29$.

Middle two must lie between 15 and 27: take $25, 26$.

Sum $= 28 + 51 + 56 = 135$; average $= 22.5$.

Solution

Let section B have $n$ students and A have $n - 10$.

Combined average $= \dfrac{32(n - 10) + 60n}{2n - 10} = \dfrac{46n - 160}{n - 5} = 46 + \dfrac{70}{n - 5}$.

So $n - 5$ divides 70, and A $\geq 1$ means $n - 5 \geq 6$: $n - 5 \in \{7, 10, 14, 35, 70\}$.

A $= (n - 5) - 5 \in \{2, 5, 9, 30, 65\}$.

Difference $= 65 - 2 = 63$.

Solution

Let the family spend $X$ in each of the first three months and $\dfrac{X}{2}$ in each of the last two.

Onion bought: $\dfrac{X}{10} + \dfrac{X}{20} + \dfrac{X}{25} + \dfrac{X}{100} + \dfrac{X}{100} = 0.22X$ kg.

Total spend $= 4X$.

Average $= \dfrac{4X}{0.22X} \approx 18.2$ per kg, closest to 18.

Solution

Let hospital A admit $a$ patients, so B admits $a + 21$.

$\dfrac{200}{a} = \dfrac{152}{a + 21} + 3$.

$200(a + 21) = 152a + 3a(a + 21) \Rightarrow 3a^2 + 15a - 4200 = 0 \Rightarrow a^2 + 5a - 1400 = 0$.

$(a - 35)(a + 40) = 0 \Rightarrow a = 35$.

Solution

Let him have played $n$ matches with $g$ goals.

$\dfrac{g + 1}{n + 10} = 0.15$ and $\dfrac{g + 2}{n + 10} = 0.2$.

Subtracting: $\dfrac{1}{n + 10} = 0.05 \Rightarrow n + 10 = 20 \Rightarrow n = 10$.

Solution

Total $= 25 \times 50 = 1250$.

To maximise the toppers' score, keep the other 20 distinct scores as low as possible: $30, 31, \dots, 49$.

Their sum $= \dfrac{20(30 + 49)}{2} = 790$.

Toppers: $5t = 1250 - 790 = 460 \Rightarrow t = 92$ (above 49 ✓).

Solution

$A + \dfrac{B + C}{2} = 5 \Rightarrow 2A + B + C = 10$.

$B + \dfrac{A + C}{2} = 7 \Rightarrow A + 2B + C = 14$.

Subtracting: $B - A = 4$. Then $2A + (A + 4) + C = 10 \Rightarrow 3A + C = 6$.

Positive integers: $A = 1$, $C = 3$, $B = 5$.

$A + B = 6$.

Solution

Let the lowest score be $L$ and the highest $H$. Lowest nine sum $= 378$, highest nine sum $= 423$, so $H - L = 45$ and the total $= 423 + L$.

Maximum: $L$ cannot exceed the average of the middle eight: $L \leq \dfrac{378 - L}{8} \Rightarrow L \leq 42$. Max mean $= \dfrac{465}{10} = 46.5$.

Minimum: $H$ must be at least the average of the middle eight: $L + 45 \geq \dfrac{378 - L}{8} \Rightarrow L \geq 2$. Min mean $= \dfrac{425}{10} = 42.5$.

Difference $= 4$.

Solution

Total over $n + 2$ innings $= 29(n + 2)$; the last two add $38 + 15 = 53$.

First $n$ innings: $29n + 58 - 53 = 30n \Rightarrow n = 5$, total $= 150$.

Each of these is less than 38 (at most 37). To make the lowest score as small as possible, let the other four be 37:

$x = 150 - 4 \times 37 = 2$.